*Duration and Simultaneity*]

time, as measured by the speeding space traveler, slows down compared to time as measured by a friend left home on Earth. This is sometimes described as the "twin paradox": two identical twins, one of whom goes off on a voyage close to the speed of light, and the other one stays home. When the space-traveling twin returns home, he or she has aged only a little, while the twin who has remained at home has aged at the regular pace. So we have two identical twins who may be decades apart in age. Or maybe the traveling twin returns in the far future, if you go close enough to the speed of light, and everybody he knows, everybody he ever heard of has died, and it's a very different civilization. ("Sagan on Time Travel.")

*Duration and Simultaneity*for his account]. He strictly adopts special relativity's premise that there is no absolute frame of reference. As the space-bound twin flies-off (at a constant speed), he feels stationary, while it appears to him that the earth flies-away into the distance. So according to his calculations, he predicts that his earth-bound brother will be younger than him. On what grounds have we to say that one is really younger than the other?

*Inside Relativity*, Mook & Vargish explain the paradox this way:

Let the twins move apart for some time so that the difference in aging measured by each is substantial; then bring the twins back together and let them stand side by side and compare their appearances. Each will say that his sibling is younger than he, but both cannot be right. Which twin is, in fact, older, or is there any age difference seen at all? (Mook & Vargish 110c)

The resolution of the "paradox" depends upon a careful analysis of how the experiment actually can be carried out. When the twins separate on the tracks it is true that each will claim the other is aging more slowly than he, but notice that this is a conclusion based onmeasurementsmade of the other twin. Unless we change the direction of motion of at least one twin (that is, accelerate the twin), the two twins will continue to separate forever, and we cannotdirectlycompare their two ages (this is like bringing two events to the same point in space so that simultaneity can be determined directly). But since we do want to compare the twins' ages directly by comparing their simultaneous appearance, suppose that we accelerate twin #1 and bring him back to stand at rest with respect to twin #2. Notice that now the two twins no longer have identical histories. One twin has undergone an acceleration (and the twins can tell which has been accelerated because the accelerated twin moved in a non-inertial frame for a time and so experienced some sort of force, as described in section 2.4). As we will see in the following chapter, in accelerating, twin #1's clock will be slowed relative to twin #2. When the twins are standing at rest with respect to one another, the twin who suffered the acceleration will have aged less than the other. (Mook & Vargish 110b-111).

The standard textbook approach treats the twin paradox as a straightforward application of special relativity. Here the Earth and the ship are not in a symmetrical relationship: the ship has a "turnaround" in which it undergoes non-inertial motion, while the Earth has no such turnaround. Since there is no symmetry, it is not paradoxical if one twin is younger than the other. Nevertheless it is still useful to show that special relativity is self-consistent, and how the calculation is done from the standpoint of the traveling twin.Special relativity does not claim that all observers are equivalent, only that all observers at rest in inertial reference frames are equivalent. But the space ship jumps frames (accelerates) when it performs a U-turn. In contrast, the twin who stays home remains in the same inertial frame for the whole duration of his brother's flight. No accelerating or decelerating forces apply to the homebound twin.There are indeed not two but three relevant inertial frames: the one in which the stay-at-home twin remains at rest, the one in which the traveling twin is at rest on his outward trip, and the one in which he is at rest on his way home. It is during the acceleration at the U-turn that the traveling twin switches frames. That is when he must adjust his calculated age of the twin at rest. (Twin Paradox)

Some physicists claim that the situation would in practice not be symmetric as one observer has to turn around in order to compare the clocks (see for instance http://www.phys.vt.edu/~jhs/faq/twins.html), but it is clear that this argument does not hold water as the time dilation should already be apparent before one observer turns around. (Smid Time Dilation and Twin Paradox Debunked)

The clocks are started and stopped in each reference frame simultaneously (by definition, the contact starts each clock from zero) and both clocks will thus show identical times (for identical rods and clocks) after having been stopped (and after the clocks have been stopped, it is obviously irrelevant if A or B (or both) turn around to compare the clocks). (Smid Time Dilation and Twin Paradox Debunked)

It should also be pointed out that the signal propagation time from the trigger points to the clocks is actually irrelevant here (whatever way of transmitting the signal is used): if the corresponding distances are identical in both systems, then the delay times will also be identical and there won't be any difference in the clock readings afterwards (Time Dilation and Twin Paradox Debunked)

According to relativity, clock differences should arise solely due to a constant relative velocity, but as this leads to the twin paradox, a hand-waving argument is being made that one system has to turn around (i.e. accelerate) to compare the clock (which of course doesn't resolve the paradox, as a) you could in principle make the period of acceleration arbitrarily short, and b) it would violate causality as the clock difference would have to build up already before the acceleration occurs). (Smid Relativity Discussion Forum page 4)

*Inside Relativity*. Princeton: Princeton University Press, 1987.

Hey man! It's me Jeremy Cubas. I know this is an older post, but I just got a chance to read it. It was a great read.

ReplyDeleteThe error in the moving clocks "paradox" is the same as the error is most problems that claim to "debunk" relativity -- it ignores the relativity of simultaneity, acting as though distant events simultaneous in one reference frame are also simultaneous in another reference frame. Relativity says that they are not.

ReplyDeleteSince you're using distant switches to start and stop the clocks, you have to begin by asking: in what reference frame is the switch event simultaneous with the clock starting or stopping?

What each of the two clocks is going to see is:

a) the other clock has shorter arms,

b) the other clock ticks slower,

c) the other clock starts before it starts, and

d) the other clock ends after it ends.

I'd love to see you make an animation that illustates this.

Dear avalonxq,

DeleteThanks very much for taking a look at the material and explaining your concerns. I am getting in touch with Prof Smid to discuss the matter, and hopefully he can reply. It seems that now he is away until May 12, so maybe check back sometime after that.

All the best,

Corry

We know what Relativity says. I quote from Einstein's original 1905 paper (see §4)

Delete"From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2*t*v^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B."

We also know that (uniform) motion is only relative, so any of the two observers could claim to be the one at rest. This gives rise to the well known twin paradox, because both observers could not possibly agree about the clock rates on the basis of the above claim. Relativity 'resolves' this logical paradox by asserting that in practice the situation will be asymmetrical as one of the observers has to turn around to compare the clocks. The thought experiment suggested on the page Time Dilation and Twin Paradox Debunked (which has been discussed here) is however perfectly symmetric, so in this case logical paradox can not be resolved by means of some dubious asymmetry arguments, i.e. the claim of moving clocks running slower can be logically excluded.

Of course, this only illustrates the fact that what Relativity says is false. It does not address the causes of the falsehood. For this one has to examine in more detail how the Lorentz transformation is actually derived. This is done for instance on the page Mathematical Inconsistencies in

Einstein's Derivation of the Lorentz Transformation.

Thomas

Thomas Smid is making to separate claims here, and pretending that they are the same.

ReplyDeleteFirst he says:

"The thought experiment suggested on the page Time Dilation and Twin Paradox Debunked (which has been discussed here) is however perfectly symmetric, so in this case logical paradox can not be resolved by means of some dubious asymmetry arguments,"

This is accurate. If the situation is perfectly symmetrical, the clocks will display the same time at the end of the interaction. This is also consistent with special relativity.

But then he makes the second claim:

"i.e. the claim of moving clocks running slower can be logically excluded."

This is incorrect. Each reference frame DOES measure the other's clocks as running slower, and the fact that the symmetry in their final measurement is maintained does not discount that.

Thomas Smid presents this thought experiment as an example contrary to special relativity, but in fact special relativity says there's no contradiction or paradox. Does Thomas Smid disagree? If not, then why use an example that has a known outcome and that fails to contradict or disagree with anything?

Can Thomas Smid present any thought experiment where physics as described in special relativity actually does create a paradox or an inconsistency with known measurements? If this clock experiment is any indication, it appears that he cannot.

avalonxq says:

Delete"If the situation is perfectly symmetrical, the clocks will display the same time at the end of the interaction. This is also consistent with special relativity.".

He implies therefore that the clock rate only differs in the two reference frame if the situation is asymmetric.

a) this asymmetry requirement is not at all mentioned in Einstein's paper (I quoted the relevant paragraph above), so I wonder where he gets this from.

b) it should anyway be clear that such a requirement would violate causality as the final clock readings must be the result of the accumulated clock rates for the whole of the time the clocks are moving; they can not be determined just by the fact that at some stage towards the end (which in principle could be made arbitrarily short) some event occurs that breaks the symmetry.

Thomas

Thomas Smid says:

Delete"it should anyway be clear that such a requirement would violate causality as the final clock readings must be the result of the accumulated clock rates for the whole of the time the clocks are moving; they can not be determined just by the fact that at some stage towards the end (which in principle could be made arbitrarily short) some event occurs that breaks the symmetry."

I asked earlier if you were capable of generating such a scenario, which WOULD be a paradox for relativity. Can you? If not, why not?

I challenge you to actually create a scenario, with actual numbers, and demonstrate the paradox -- that is, demonstrate a problem for which relativity is inconsistent or does not match observed results. Heck, if you're so certain an asymmetry at the end has no difference on your hypothetical but demonstrates the paradox, use that.

Choosing a problem that gives the same answer whether relativity is true or not, doesn't demonstrate anything.

avalonxq says:

Delete"Choosing a problem that gives the same answer whether relativity is true or not, doesn't demonstrate anything"

You first should make clear what Relativity actually says with regard to this problem in comparison to a situation containing some asymmetry (note that I said "what Relativity says", not "what avalonxq says", so please cite any peer reviewed Relativity paper where your claim is theoretically substantiated).

Thomas

You first need to set out your scenario explicitly and fully (no hand-waving the superlight signals). Then we can discuss what special relativity says about it.

DeleteThe scenario has been set up sufficiently explicitly to formulate the paradox. As clearly pointed out above, it is due to symmetry alone, and independent of how fast the signals travel in the two timing devices (assuming they are physically identical).

DeleteIn your first post (May 3) you said then

"The error in the moving clocks "paradox" is the same as the error is most problems that claim to "debunk" relativity -- it ignores the relativity of simultaneity, acting as though distant events simultaneous in one reference frame are also simultaneous in another reference frame. Relativity says that they are not."

Whilst it is true that the Lorentz transformation formula contains a 'relativity of simultaneity' offset term, this term drops out if you consider the time difference of two clocks that accumulates over a certain time. Just go to Einsteins 1905 paper and have a look at Sect.4. The time difference is proportional to a factor depending only on the square of the relative velocity of the clocks, not the direction of this velocity nor the location of the clocks.

So again, on what basis are you saying that in this case Relativity would not predict a difference in the clock readings, but in other (asymmetric) scenarios it would?

Thomas

You say this:

Delete"The scenario has been set up sufficiently explicitly to formulate the paradox."

I disagree. Please lay out your scenario explicitly and then use the appropriate equations to demonstrate the paradox. If there is one, you should have no trouble showing it explicitly.

I expect that when you apply actual values to your scenario and then use the appropriate transformations that Einstein expresses for special relativity, you'll find no paradox at all. But I am happy to be shown to be wrong.

This comment has been removed by the author.

DeleteThe thought experiment has been laid out in detail, and there is even an animated illustration on this page that should practically make a real experiment redundant. Still, you have so far only come up with nothing more than hand-waving claims that in this case it would agree with Relativity, but in other (asymmetric) cases not.

DeleteI shall therefore make it easier for you by considering a more straightforward scenario:

we have a clock A at rest at location X of the coordinate system and another clock B located at the origin but moving with velocity v in the positive x-direction. This means they will collide at a certain time and both clocks will stop on impact. Assuming that both clocks are synchronized to t=o when B is at the origin, the question is what time will either clock show when you examine them (assume that the impact just stopped the clocks, but that they are still readable)?

This is a straightforward and unambiguous question to which Relativity should be able to give a straightforward and unambiguous answer.

Thomas

I just would like to make a further comment before avalonxq wastes his (and others') time on trying to come up with further false arguments here:

ReplyDeletein his reply of May 3, he says:

"The error in the moving clocks "paradox" is the same as the error in most problems that claim to "debunk" relativity -- it ignores the relativity of simultaneity, acting as though distant events simultaneous in one reference frame are also simultaneous in another reference frame. Relativity says that they are not."

This is actually a false claim, because even though the Lorentz transformation formula t'=γ(t-v*x/c^2) contains formally the 'simultaneity offset' term v*x/c^2, this term disappears for a uniform motion x=v*t, because then t'=t/γ (as is seen easily by inserting γ= 1/√[1-(v/c)^2]). In §4 of his paper On the Electrodynamics of Moving Bodies, Einstein actually derives the same expression for the alleged transformation of the clock readings, which evidently depends only on the absolute value of the speed and neither on its direction nor the x-coordinate (i.e. it does not depend on the 'relativity of simultaneity'). Even if the velocity reverses at some point, this has no impact on the clock reading as t' depends only on v^2 (by re-defining the origin x=0 as the reversal point for the 'return-leg' the same expression t'= t/γ does indeed apply to the outward and return legs). So even such an alleged 'asymmetrical' situation is in fact symmetric if the Lorentz transformation is applied correctly. In other words, the claim of time dilation shows in itself that that Relativity is an internally inconsistent theory.

I have added this argument now to my web page Time Dilation and Twin Paradox Debunked.

Thomas

Yet again, you have created a scenario that assumes universal simultaneity -- something that relativity specifically refutes.

ReplyDeleteA's reference frame and B's reference frame will not agree as to which events are "simultaneous", and therefore they will not agree as to what A reads "at the same instant that" B passes the origin reading t=0. In order for the scenario to be fully defined, you have to clarify whether they are "synchronized to t=0 when B is at the origin" relative to the reference frame of A or the reference frame of B. Whichever you choose, A WILL not read t=0 when B reads t=0 according to the other frame (the one you didn't choose).

This is a direct consequence of the Lorentz equation relating time to position.

avalonxq said:

Delete"In order for the scenario to be fully defined, you have to clarify whether they are "synchronized to t=0 when B is at the origin" relative to the reference frame of A or the reference frame of B"

Whatever. Assume for this purpose they are synchronized in the rest frame of clock A. You could assume for instance that the moving clock B makes contact with a further clock at the origin of frame A (lets call this clock A1 and the other one, resting at location X, A2) and that this contact transmits the time from A1 to B (assume both clocks and contacts infinitesimally small, so the transmission of the time stamp is instantaneous). Define this instant as t=0 and assume also that A2 has been synchronized with A1 beforehand.

So again my question: what will in this case the clocks A2 and B show when examined after their collision at location X?

Thomas

In reference frame A, B passes A1 at time t=0, at which time A2 reads 0. B reaches A2 at time t=x/v, so A2 reads x/v. B is running slow by factor sqrt(1-v^2) (which I will call y for convenience), and so B reads y*x/v.

ReplyDeleteIn reference frame B, A1 passes B at time t=0, at which time A2 reads v*x. At t=0, A2 is distance y*x away from B, moving at velocity -v. They collide at time t=y*x/v, so B reads y*x/v. A2 is running slow by factor y, and so A2 reads v*x + y^2*x/v.

To simplify the reading on A2 according to reference frame B, y^2 = (1-v^2), and so...

v*x + y^2*x/v = [v^2*x + (1-v^2)*x]/v = [v^2*x + x - v^2*x]/v = x/v.

Or, in other words, the clocks read the same regardless of which reference frame you make the calculations in -- provided you remember that A2 does not read 0 at t=0 according to reference frame B, which is the traditional "mistake" made in this problem (not a problem at all if one remembers there's an x-term in the Lorentz time equation).

Are you willing to carry out the calculations yourself, and then identify where you believe the contradiction lies?

avaonxq said:

ReplyDelete"in other words, the clocks read the same regardless of which reference frame you make the calculations in"

But you didn't change the reference frame in yor calculation, you merely used the backward-transformation

(2) x=γ(x' + vt') ; t = γ(t' + vx'/c²)

instead of the forward-transformation

(1) x'=γ(x - vt) ; t' = γ(t - vx/c²) .

Both transformations are exactly equivalent (you can derive one algebraically from the other). So in both sets of equations the unprimed variables (the rest frame) is associated with the A-clocks, and the primed variables (the moving frame) with the B-clock. However, in order to check whether the Lorentz transformation implies a twin paradox, one has to additionally apply the principle of relativity, which states that (uniform) motion is relative and each observer can consider himself to be at rest (not only this, but the physics can not possibly depend on the orientation of the coordinate system or its positioning or the units used either).

So in this sense, consider the scenario with clock B instead of A defining now the rest frame, assuming that B is at position X (as defined by B) in this reference frame, with A1 passing B at t=0 (moving with velocity v) and A2 at the origin also moving with velocity v (i.e. according to (1) we have t=t'=o for the initial readings). This means that the collision between A2 and B occurs after

t=X/v

for clock B and

t' = γ(X/v - vX/c²) = X/v/γ

for clock A (note again that the role of the prime is now reversed).

This means that B concludes now that for A less time has elapsed than for him, whereas with the rest frame reversed, A concluded that for B less time elapsed than for him, i.e. we have a logical contradiction.

Thomas

"But you didn't change the reference frame in yor calculation,"

DeleteI absolutely did change the reference frame in my calculation.

First, I calculated things in reference frame A. In reference frame A, B moves at velocity v from A1 to A2, which are x apart; B passes A1 at t=0, and B and A2 both read 0 at t=0. Those are the conditions of the problem as you stated them.

Second, I calculated things in reference frame B. in reference frame B, A1 and A2 move past B at velocity -v, and are y*x apart. A1 passes B at t=0, at which point B reads 0. We also know from the Lorentz transformations that A2 reads v*x at t=0. Again, this is simply a product of the way the problem is stated. The formulation and the answer are both completely unambiguous in special relativity.

"However, in order to check whether the Lorentz transformation implies a twin paradox, one has to additionally apply the principle of relativity, which states that (uniform) motion is relative and each observer can consider himself to be at rest (not only this, but the physics can not possibly depend on the orientation of the coordinate system or its positioning or the units used either)."

I used all of this. The physics is consistent in both reference frames A and B, and the answer is no way dependent on the coordinate system orientation or the unites. The answer is dependent on the starting conditions that we set, particularly what the clocks are set to read as.

And again, I'm forced to remind you that simultaneity is not absolute: particularly, any two distant events that are simultaneous in one inertial reference frame WILL NOT BE simultaneous in another inertial frame moving relative to the first frame. So, having set A2 and B to 0 at t=0 for frame A, we have dictated what A2 and B read at t=0 for frame B. The clock readings themselves are not coordinate-dependent, but what clock reading lines up with what indexed instant of time entirely depends on what reference frame you're in.

"This means that B concludes now that for A less time has elapsed than for him, whereas with the rest frame reversed, A concluded that for B less time elapsed than for him, i.e. we have a logical contradiction."

Except there is no contradiction, as I demonstrated. Yes, A2 experiences less time than B in reference frame B (specifically, B experiences y*x/v and A experiences y^2*x/v). But since A2 started at a reading above zero, it still reads more time than B when the two meet. Gratifyingly, it reads the same as the calculations from reference frame A show that it reads. There is no contradiction.

I notice that you haven't done what I did and actually explicitly give the numbers you believe apply here. Please do so. What do you assert that A2 reads at t=0 in reference frame B? What do you assert is the distance between A1 and A2 in reference frame B? If you get some other reading than what I have calculated, please show your work.

avalonxq said:

Delete"Second, I calculated things in reference frame B."

As I said above already, you did not consider B as the rest frame. You first transformed from the rest frame A to the moving frame B, then back from the moving frame B to the rest frame A. In both parts of your calculation the rest frame (i.e. the unprimed) variables refer to frame A, hence it is no surprise that you get the same result. To demonstrate a possible twin paradox you have to use B as your rest frame in your second part. Do you agree with this or not?

Thomas

"As I said above already, you did not consider B as the rest frame."

DeleteI certainly did consider B as a rest frame. Again, I consider A1 and A2 moving past B at velocity -v.

Now, our goal is to consider THE SAME SITUATION from reference frame B, correct? We're not just looking at creating a second, different situation? Because if so, then our values for things in reference frame B are constrained by what they were in reference frame A.

Is your intent to consider the situation where A2 and B both read 0 according to reference frame B? Because in that case, you're dealing with a different set of clocks. As Special Relativity makes clear, distant events simultaneous in one reference frame are not simultaneous in other frames.

If you disagree that, according to reference frame B, A2 reads x*v at time t=0, please explain what value you believe it reads. My value is the single, objective answer that relativity requires.

Again, having set the starting values for reference frame A (the distance between A1 and A2 is X, B is moving at v, A2 and B read 0 at t=0, B passes A1 at t=0), these values are set in reference frame B. We can't just define them as any values we want.

If you give a different answer, please stop alluding to it and actually explicitly write it out.

avalonxq said:

Delete"I certainly did consider B as a rest frame".

No, sorry, you did not. The transformation from a

restframe into the moving frame is(1) x'=γ(x - vt) ; t' = γ(t - vx/c²) .

You used

(2) x=γ(x' + vt') ; t = γ(t' + vx'/c²)

which is the back-transformation from the

movingframe (in this case frame B) into the rest frame.Even though you can make the latter formally identical by exchanging the primed and unprimed variables and reversing the value of v, this is not the same, as in (2) x' and t' are not independent variables but those determined by (1) (as you yourself strongly insisted in your last post).

Effectively, what you have been doing is something like the following:

you make a claim that a>b (your first part), then you find that (not surprisingly) this implies b<a (your second part), but obviously this holds regardless of whether the statement a>b is is true in the first place.

In this sense, from your measured values x,t in frame A you calculate a set of values x',t' that you

thinkthe observer in frame B would measure for the corresponding events according to your proposed (but unproven) transformation formula, then you transform in (2) these kind of hypothetical values back by the algebraically reverse formula. At no point in this process do you actually allow B to make his measurements and apply the same rules in reverse. Never mind the actual experimental verification, but by not even allowing B to theoretically fully consider the situation independently himself (i.e. with the roles fully reversed), the whole claim effectively becomes un-falsifiable, which therefore disqualifies it as a scientific theory.avalonxq said:

"As Special Relativity makes clear, distant events simultaneous in one reference frame are not simultaneous in other frames".

Again, this is something you claim on the basis of the belief that the fictitious values calculated through the Lorentz transformation (1) for reference frame B from measurements in frame A do actually correspond to actual values measured by B. And it is in fact easily disproved: just consider a setup with a fourth clock B2, where at a certain moment A1 passes B1 and A2 passes B2 (similar to the situation suggested by the animated graphic on this page, only with four clocks (at the ends) rather than 2 clocks). Since the events defined by the passing of the clocks are absolute facts independent of the reference frame (if e.g. A1 passes B1, then B1 passes A1) it is clear that both events will be simultaneous in frame A as well as B. That is the only single and objective answer possible in this case.

Thomas

You have yet to actually perform the calculations necessary to demonstrate the paradox.

DeleteI would appreciate it if you would please take the time to actually answer my question -- according to reference frame B, what do you claim that A2 reads at t=0? I claim that A2 reads x*v; what do you claim it reads?

Without having given any actual results yourself, you then alter the scenario again and YET AGAIN fail to define your problem precisely. You say this:

"And it is in fact easily disproved: just consider a setup with a fourth clock B2, where at a certain moment A1 passes B1 and A2 passes B2 (similar to the situation suggested by the animated graphic on this page, only with four clocks (at the ends) rather than 2 clocks)."

Do you claim that B2 passes A2 at t=0 in reference frame A or in reference frame B? It cannot do both, because the distance between A1 and A2 is only equal to the distance between B1 and B2 in one reference frame or the other (but, again, not in both).

"Since the events defined by the passing of the clocks are absolute facts independent of the reference frame (if e.g. A1 passes B1, then B1 passes A1) it is clear that both events will be simultaneous in frame A as well as B."

It is certainly not clear, and you haven't shown it to be true. Again, the event where A1 passes B1 will not be simultaneous to the event where A2 passes B2 IN BOTH FRAMES. You have to choose one, and then the events are not simultaneous to the other.

You have yet to demonstrate a paradox. Again, if you believe you have one, finish carrying out the calculations yourself, and I will happily show you where I believe you have neglected to make a transformation as special relativity actually requires.

I believe even an undergraduate student with a year or two of modern physics will be able to corroborate my calculations. I do not know if they would agree with yours, because you have not yet made any. Please do so.

avalonxq said:

ReplyDelete"You have yet to actually perform the calculations necessary to demonstrate the paradox".

I did the calculation already back in my post of June 15, but you replied to it (on June 16):

"Again, having set the starting values for reference frame A (the distance between A1 and A2 is X, B is moving at v, A2 and B read 0 at t=0, B passes A1 at t=0), these values are set in reference frame B. We can't just define them as any values we want"

In other words, you are having A dictate B how to set his clocks so that no contradiction arises, and naturally B would then 'agree' with the conclusions of A. But B can in this case not be considered an independent observer, so you are not examining a truly symmetric setup. I have shown in my calculation that a truly symmetric setup (where B is allowed to perform exactly the same steps as A) will lead to a paradox regarding the claim of different clock rates.

It should be clear from all this that the Lorentz transformation is in reality only a one-way transformation that, ironically, does not allow the principle of relativity to be applied in a way that the situation is considered independently in a different coordinate system (because then inevitably the paradoxes would become apparent).

Thomas said:

"Since the events defined by the passing of the clocks are absolute facts independent of the reference frame (if e.g. A1 passes B1, then B1 passes A1) it is clear that both events will be simultaneous in frame A as well as B."

avalonxq said:

"It is certainly not clear, and you haven't shown it to be true. Again, the event where A1 passes B1 will not be simultaneous to the event where A2 passes B2 IN BOTH FRAMES. You have to choose one, and then the events are not simultaneous to the other".

I am sure you will agree that a sufficient condition for both events being simultaneous in frame A is that the clocks A1 and B1 as well as A2 and B2 appear to be match up in snapshot taken at the corresponding time (to make things even clearer assume that the clocks are in physical contact at this moment).

You should explain to me then how at this instant the clocks should possibly not match up (i.e. showing to have contact) in a snapshot taken in frame B. It would be like giving another person the hand, but that person not giving you the hand (yes, I know the Lorentz transformation demands it, but the questions is whether it has anything to do with reality).

Thomas

"In other words, you are having A dictate B how to set his clocks so that no contradiction arises, and naturally B would then 'agree' with the conclusions of A."

DeleteIt is not A that dictates how B's clocks are set up; it's the problem itself that does so.

"But B can in this case not be considered an independent observer, so you are not examining a truly symmetric setup."

You're going to have to explain what you mean by "independent observer" and "truly symmetric setup".

We can't just choose arbitrary values; we have to choose the values that the problem requires. This isn't unique to relativity; this is true in any problem with defined variables. Otherwise you can cause a contradiction in any physical laws just by choosing different arbitary starting variables.

We have, at time t=0 in frame A, clock B passing point 0 at velocity v and reading 0. We have, at time t=0 in frame B, clock A2 stationary at point x and reading 0.

Based on that information alone, we also know that at time t=0 in frame B, clock A2 is moving at -v, is passing point y*x, and reads v*x. If you choose any other values for clock A2 at time t=0, you are not following Special Relativity.

Please, explicitly tell me what values you believe clock A2 has at time t=0, and explain how you got them using relativity. If your next post does not do so, I will respond by quoting this request again.

"I have shown in my calculation that a truly symmetric setup (where B is allowed to perform exactly the same steps as A) will lead to a paradox regarding the claim of different clock rates."

In other words, you run the experiment a first time, synchronizing the clocks in frame A (B sees that the clocks are not synchronized).

Then you go back and run the experiment a second time, synchronizing the clocks in frame B (A sees that the clocks are not synchronized).

You get different values for these different setups, and call it a contradiction.

Anyone can do that with any law of physics; it means nothing. You need to show how the two scenarios are the same according to relativity, or you haven't proven anything.

Typographical error, fixed below:

Delete"We have, at time t=0 in frame A, clock B passing point 0 at velocity v and reading 0. We have, at time t=0 [in frame A], clock A2 stationary at point x and reading 0."

You didn't answer my rebuttal of your claim that two events can not be simultaneous in two different reference frames.

DeleteI am sure you will agree that a sufficient condition for both events being simultaneous in frame A is that the clocks A1 and B1 as well as A2 and B2 appear to be match up in snapshot taken at the corresponding time (to make things even clearer assume that the clocks are in physical contact at this moment).

You should explain to me then how at this instant the clocks should possibly not match up (i.e. showing to have contact) in a snapshot taken in frame B. It would be like giving another person the hand, but that person not giving you the hand (yes, I know the Lorentz transformation demands it, but the questions is whether it has anything to do with reality)."

avalonxq said:

"It is not A that dictates how B's clocks are set up; it's the problem itself that does so."

What is the problem then? The task is simply how the clock rates of two moving clocks can be compared to each other. This as such is just an experimental task and for this it isn't even necessary to address the causes of any potential difference between two clock rates (of which there could be many, like the clocks actually not being identical, the battery running flat, or mechanical forces due to e.g. the clock moving through a magnetic field etc.).

And this task is indeed completely straightforward: if clock B1 moves from clock A1 to clock A2, and assuming it is synchronized with A1 at departure, and also assuming that at this time A2 and A1 are synchronized, then B1 can be directly compared with A2 on arrival and any difference between the clock rates is immediately obvious. The point is that in order to yield a correct reading, A2 and A1

mustbe synchronized in the rest frame (A-frame) as we want to know the difference between the clock rates of A1 and B1, and in order for a direct comparison of the readings of B1 and A2 being possible at all, A2 (the clock at the destination point) must effectively be a copy of A1 (the clock at the departure point).And this procedural principle must apply to any observer frame. So if you consider the situation in the rest frame of B1 (i.e. let instead A2 and A1 move), then the synchronization must be exactly equivalent: if you assume a further auxiliary clock B2 in the (now) rest frame B coinciding and being synchronized with A2 at t=0 and also B1 being synchronized with B2 at this moment, then again, A2 can directly compare its clock reading with B1 on arrival and thus make a judgement regarding the clock rates. And since both in this and the previous case the initial distance as well as the relative velocity between the stationary and moving clocks are identical, the situation is exactly equivalent to the previous case, with the same clocks being used, the clocks started and stopped at the same times, just considered in a different reference frame.

Again, this is an objective measuring procedure for the clock rates, and it is not dependent on any specific causes for a possible difference in the clock rates.

Thomas

"I am sure you will agree that a sufficient condition for both events being simultaneous in frame A is that the clocks A1 and B1 as well as A2 and B2 appear to be match up in snapshot taken at the corresponding time (to make things even clearer assume that the clocks are in physical contact at this moment)."

ReplyDeleteSpell this out, please. Exactly where is the "snapshot" taken from, and what exactly does it show?

You appear now to be saying that you refuse to agree that Special Relativity is correct because of factors OUTSIDE of Relativity itself -- that is, you're denying the theory rather than assuming it. Do you concede, then, that the theory is self-consistent? You were entirely unable to show inconsistent calculations that would lead to any contradiction or paradox, and I was able to give you a single, unambiguous answer to your problem. You were not able to give me a second one.

Which reminds me...

Please, explicitly tell me what values you believe clock A2 has at time t=0, and explain how you got them using relativity. If your next post does not do so, I will respond by quoting this request again.

You say:

"And this task is indeed completely straightforward: if clock B1 moves from clock A1 to clock A2, and assuming it is synchronized with A1 at departure, and also assuming that at this time A2 and A1 are synchronized, then B1 can be directly compared with A2 on arrival and any difference between the clock rates is immediately obvious. The point is that in order to yield a correct reading, A2 and A1 must be synchronized in the rest frame (A-frame) as we want to know the difference between the clock rates of A1 and B1, and in order for a direct comparison of the readings of B1 and A2 being possible at all, A2 (the clock at the destination point) must effectively be a copy of A1 (the clock at the departure point)."

I agree with this, and did exactly this.

"And this procedural principle must apply to any observer frame. So if you consider the situation in the rest frame of B1 (i.e. let instead A2 and A1 move), then the synchronization must be exactly equivalent:"

This is entirely wrong.

Again, relativity says that if two distant events are simultaneous in one frame, they are not simultaneous in another.

You can't synchronize A1 and A2 in both reference frame A and reference frame B. If you synchronize them in reference frame A, an observer in reference frame B will not see them as synchronized.

The problem drills down to this:

At the momement that B passes A1 heading toward A2 at velocity v, B and A1 both read 0. A1 observes that A2 also reads 0. What does B observe that A2 reads?

You want to say the answer is 0, but the correct answer is x*v. It is the ONLY answer that Relativity gives.

You can continue to deny that relativity of simultaneity is TRUE, but you have provided no evidence whatsoever that it's INCONSISTENT.

avalonxq said:

Delete"Spell this out, please. Exactly where is the "snapshot" taken from, and what exactly does it show"

You don't necessarily have to understand it as a photographic snapshot, just take it as a statement that a certain moment in frame A two clocks A1 and A2 resting in that frame coincide in their location (i.e. are in contact with) two clocks B1 and B2 respectively that are at rest in frame B. This means the contact events A1/B1 and A2/B2 are (by definition) simultaneous in frame A. The question is how do these two events represent themselves in frame B?

avalonxq said:

"You appear now to be saying that you refuse to agree that Special Relativity is correct because of factors OUTSIDE of Relativity itself -- that is, you're denying the theory rather than assuming it. Do you concede, then, that the theory is self-consistent?"

Have a look these two equations:

1=2

2=1

They are consistent with each other, but nonetheless they are false as they are not consistent with the axioms of mathematics. You always have to look at the consistency in a wider context. In this case it means that the fundamental axioms of physics must hold, and one of these fundamental axioms is the 'principle of relativity' which says that (uniform) motion is a subjective concept and can as such not possibly result in any objective physical effects. Yet Relativity claims exactly that.

avalonxq said:

"Which reminds me...

Please, explicitly tell me what values you believe clock A2 has at time t=0, and explain how you got them using relativity."

As I have indicated above already, in order for B to be able to compare his clock with those in frame A, it is absolutely necessary that both A1 (the clock at B's departure point) and A2 (the clock at B's arrival point) are synchronized with each other in frame A. Otherwise he would not be able to simply compare his clock with A2 and relate this to the clock rate of A1. So naturally, it only makes sense if the rest frame clocks are synchronized in the rest frame. Otherwise B would effectively be comparing apples with oranges.

The point is that if instead A2 considers itself the moving clock (which it can do as any (uniform) motion is just a question of reference frame), one would have to demand correspondingly that the departure and arrival clocks are synchronized in B's frame, hence giving rise to a paradox if it is claimed that the relative motion leads to a difference in the clock rates.

Thomas said:

"And this procedural principle must apply to any observer frame. So if you consider the situation in the rest frame of B1 (i.e. let instead A2 and A1 move), then the synchronization must be exactly equivalent:"

avalonxq said:

"This is entirely wrong.

Again, relativity says that if two distant events are simultaneous in one frame, they are not simultaneous in another."

a) being in contradiction to Relativity does not necessarily mean it is wrong

b) as I have pointed out above already, the Lorentz transformation only knows one rest frame, the transformed coordinates hold for the moving frame. Algebraically back-transforming these hypothetical coordinates into the rest frame will always be consistent, whether or not the original transformation leads to any logical paradoxes in reality (namely if applied correspondingly by any other observer).

Thomas

"As I have indicated above already, in order for B to be able to compare his clock with those in frame A, it is absolutely necessary that both A1 (the clock at B's departure point) and A2 (the clock at B's arrival point) are synchronized with each other in frame A. Otherwise he would not be able to simply compare his clock with A2 and relate this to the clock rate of A1. So naturally, it only makes sense if the rest frame clocks are synchronized in the rest frame. Otherwise B would effectively be comparing apples with oranges.

DeleteThe point is that if instead A2 considers itself the moving clock (which it can do as any (uniform) motion is just a question of reference frame), one would have to demand correspondingly that the departure and arrival clocks are synchronized in B's frame, hence giving rise to a paradox if it is claimed that the relative motion leads to a difference in the clock rates."

This is a dodge to my question which is: A1, B, and A2 all read 0 at time t=0 in reference frame A. What time does A2 read at t=0 in reference frame B.

You claim you're identifying a contradiction in relativity, but your answer is 0, and relativity's answer is a very specific answer that is not 0.

You have yet to perform the calculations to demonstrate a contradiction; I'm still waiting for you to do so. As has been pointed out repeatedly, the reason that there's no contradiction even though each reference frame measures the other as moving slow, is becomes the reference frames don't agree on initial synchronization (and according to relativity they never will).

As for your snapshot, I will remind you that if a light signal is emitted from A1 and B1 at the moment they intersect, A2 and B2 won't even receive this signal at the same time in either frame, and due to their relative movement will not interpret the signal as representing the same length of time ago.

So you have yet to establish any physical mechanism in which to somehow synchronize events in two different reference frames simultaneously -- something relativity says will not happen. I invite you to make another attempt if you so choose.

avalonxq said:

Delete"This is a dodge to my question which is: A1, B, and A2 all read 0 at time t=0 in reference frame A. What time does A2 read at t=0 in reference frame B."

It is not a dodge. As should be clear from the very section you quoted, your question is irrelevant. I'll try make it even clearer:

assume you make a train journey from station A1 to A2. If you want to know the time your journey takes according to the station clocks, you read the clock of A1 on your departure and you read the clock of A2 on your arrival. Assuming that the railway company have properly synchronized all their station clocks, you know then the time the journey took according to the station clocks. So again, you read A1 on your departure and you read A2 on your arrival. The time of A2 on your departure doesn't come into it at all. Now if you have your own watch with you on the train and wonder whether it will indicate a different journey time, then you read this as well on your departure and again on your arrival. A comparison with the other readings will then show whether your watch has a different clock rate to the station clocks or not. So there is no thing like 'the time of A2 in reference frame B' entering into the problem; you take one set of readings of clocks in the A-frame, you take another, unrelated set of readings of your clock in the B-frame. And with reversed roles (by e.g. comparing one of the station clocks to two clocks at the front and back of the train) the same would hold. As I said before, this all is simply a general measuring procedure and independent on the assumptions for the cause of any difference in the clock rates. The point is that

anytheory suggesting the thus measured difference in the clock rates would depend on the relative velocity between the A and B clocks leads to a logical paradox.avalonxq said:

"As for your snapshot, I will remind you that if a light signal is emitted from A1 and B1 at the moment they intersect, A2 and B2 won't even receive this signal at the same time in either frame, and due to their relative movement will not interpret the signal as representing the same length of time ago."

Just assume that we take a photograph in frame A from halfway between A1 and A2 perpendicular to the direction of motion, and that this photo shows A1/B1 as well as A2/B2 matching up (touching). The question is, will you able to get any photo taken in frame B showing all the 4 contact points matching up as well? If not, why not?

Thomas

"assume you make a train journey from station A1 to A2. If you want to know the time your journey takes according to the station clocks, you read the clock of A1 on your departure and you read the clock of A2 on your arrival. Assuming that the railway company have properly synchronized all their station clocks, you know then the time the journey took according to the station clocks. So again, you read A1 on your departure and you read A2 on your arrival. The time of A2 on your departure doesn't come into it at all. "

ReplyDeleteIt does if you want to know whether the station clocks are running fast, or slow, or what they're doing in frame B.

You're dodging the question because you know the answer resolves the paradox. A2 isn't synchronized with A1 in frame B. That's what relativity predicts, that's what the evidence shows, and no matter how many times you change the thought problem, you've yet to come up with one that actually shows any contradictory answers. The descriptions remain logically consistent the whole time.

"Just assume that we take a photograph in frame A from halfway between A1 and A2 perpendicular to the direction of motion, and that this photo shows A1/B1 as well as A2/B2 matching up (touching)."

What you're describing is that we're going to have a point A3, at rest in frame A and halfway between A1 and A2 (we can move A3 perpendicularly off the line between A1 and A2 or not; all it does is complicate the math).

Then, at time t=0 in frame A, A1/B1 and A2/B2 each send light signals. A3 gets both of the light signals at the same time, and knowing the distances to A1 and A2 are the same, concludes that the clocks all met simultaneously -- in frame A.

The problem is, what does point B3, which is directly halfway between B1 and B2 and at rest in frame B, see?

Well, according to reference frame A, the signals are emitted at t=0 when B3 coincides with A3, and then B3 keeps moving forward. So B3 will see the A2 signal first, and then the A1 signal.

In B3's own frame, then, we have it receiving the B2 signal first, and it knows that B2 and B1 are the same distance away. Therefore, in frame B, the point B3 concludes that A2 passed B2 BEFORE A1 passed B1 -- just as the calculations would show.

There is, of course, a point along B that matches up with A3 at the point in time when A3 receives both signals; lets call this point B3'. This is the point sees the A1/B1 and the A2/B2 signals at the same time. But note that B3' IS NOT HALFWAY BETWEEN B1 and B2. Instead, B3' is much closer to B1 than B2, because of the movement that occurs between the time the match-up signals are sent and the time they arrive.

So, looking at this event in reference frame B, we have B3' receiving the B1 signal at the same time as the B2 signal, but B2 is father way than B1. Therefore, according to B3' as measured in reference frame B, the B2/A2 match-up occurs first.

I want to make this point separately so you can see it clearly: There is no point at rest in reference frame B that is both equidistant from B1 and B2, and will see both match-up signals at the same time. Every single point moving with B that is equidistant will see the A2/A1 match-up signal first. Every single point moving with B that sees the signals at the same time is closer to B1 than to B2. AND BOTH REFERENCE FRAMES A AND B AGREE ON THESE FACTS. Even in reference frame A, we can see that equidistant B points don't see the signals at the same time and B points that see the signals at the same time are closer to B1.

It is therefore a consequence of the speed of light being used as a constant standard for measuring time and distance, that different reference frames don't measure the same events as simultaneous.

Thomas said:

Delete"The time of A2 on your departure doesn't come into it at all. "

avalonxq said:

"It does if you want to know whether the station clocks are running fast, or slow, or what they're doing in frame B."

The clocks in one frame can not possibly know anything about those in the other frame. They are completely independent. You just compare the reading of your B clock with the local A clock (A1 at the start of your journey and A2 at the end) to detect any possible difference in the clock rates. This is in principle how the Haefele-Keating experiment was indeed done: the B clock was synchronized with the A clock, then put on an airplane, flown a couple of times around the earth, brought back to the A clock (which stayed at the ground) and then compared to the latter. Note again that this is just a general experimental procedure; the same time difference would have been discovered even if the theory of Relativity wouldn't have been invented yet; so your arguments based on your specific application of the Lorentz transformation completely miss the point that the experimental procedure as such is liable to the twin paradox if its results are explained in terms of the relative velocity alone.

avalonxq said:

"You're dodging the question because you know the answer resolves the paradox. A2 isn't synchronized with A1 in frame B. That's what relativity predicts, that's what the evidence shows, and no matter how many times you change the thought problem, you've yet to come up with one that actually shows any contradictory answers. The descriptions remain logically consistent the whole time."

As I said already, just algebraically reversing the transformation equations back from the moving frame into the rest frame does not even address (let alone resolve) the paradox. The paradox can only arise if you allow both frames to

independentlyconsider themselves as the rest frame (i.e. if you assume the validity of the principle of relativity). Of course, if you don't allow this but dictate the other observer how to set his clocks so that it fits your purpose, then obviously the result appears consistent, but this then clearly violates the principle of relativity (and thus creates another inconsistency with the initial assumptions).If experimental evidence does seem to support the time dilation theory, then this could only be explained in terms of actual physical forces (e.g. due to motion in a magnetic field) affecting the clocks differently.

avalonxq said:

"Well, according to reference frame A, the signals are emitted at t=0 when B3 coincides with A3, and then B3 keeps moving forward. So B3 will see the A2 signal first, and then the A1 signal."

That would actually contradict the invariance of c, according to which the speed of light should be independent of the velocity of the source, i.e. the signals from B1 and B2 should reach B3 at the same time if they are emitted from the same distance.

As for the rest of your treatment of this thought experiment, I wanted at first to reply to certain other points, but I think that this discussion would get way too complicated for posting here, considering that this is a philosophy blog and non-physicists should be able to understand the arguments as well. As you will know, signal propagation effects can muddle up things already in non-relativistic physics, so for the discussion of the logical consistency of relativistic effects alone, it is advisable to exclude them altogether. So I would like to ask you, as already suggested initially, to understand 'snapshot' not as a photographic one but merely as a 'freeze' of the scenario at a certain time in either reference frame. In this sense, please consider the following situation:

In frame A, at time t=0, A1 and B1 match up at the origin (x=0), whereas A2 and B2 match up at location x=X. What does this situation look like in frame B at t'=0?

Thomas

"In frame A, at time t=0, A1 and B1 match up at the origin (x=0), whereas A2 and B2 match up at location x=X. What does this situation look like in frame B at t'=0?"

ReplyDeleteAt t'=0 in reference frame B, A1 and B1 match up at the origin (x=0). A2 has already passed B2; they met up earlier when A2 and B2 both read 0.

This result is fully consistent. In reference frame B, the distance between A1 and A2 is shorter than the distance between B1 and B2. And, as explained previously, A2 has already ticked to x*v at t'=0. Similarly B2 reads x*v/y, and hence they met at t'=-x*v/y.

Again, there is no inconsistency here, and all the numbers hang together. It also works if you start sending light signals from any event to any other event; BOTH REFERENCE FRAMES will agree on the order that those light signals are received, which will allow any reference point in A or B to correctly measure the events exactly as described here.

You still haven't demonstrated any inconsistency, because there isn't one. Everything works together; each of the transformations corroborates the others.

You said:

"As I said already, just algebraically reversing the transformation equations back from the moving frame into the rest frame does not even address (let alone resolve) the paradox. The paradox can only arise if you allow both frames to independently consider themselves as the rest frame (i.e. if you assume the validity of the principle of relativity)."

But you're not just wanting to allow both frames to independently consider themselves as the rest frame. You're also wanting to ignore the fact that the starting conditions have to match in both frames. I don't know how many times I can say this before you at least acknowledge it -- choosing the starting values in frame A also dictates what those starting values are in frame B if you want to consider A SINGLE SCENARIO from two perspectives; you can't just choose the starting scenarios fresh.

Considering the following scenario (reductio ad absurdum time). I claim that "distance" is a ridiculous concept in physics, and seek to show a "contradiction". Here's my scenario:

Consider observer A and B, who are 100 meters apart along the x axis.

Each of them seeks to measure the distance between points C and D.

A measures that C is +20 meters from him along the x axis, and D is +80 meters from him along the x axis. He therefore calculates the distance CD to be 60 meters.

B measures that C is +20 meters from him along the x axis, and D is -20 meters from him along the x axis. He therefore calculates the distance CD to be 40 meters.

Contradiction! Distance doesn't exist.

The flaw is left to the reader.

avalonxq said:

Delete"choosing the starting values in frame A also dictates what those starting values are in frame B if you want to consider A SINGLE SCENARIO from two perspectives; you can't just choose the starting scenarios fresh."

But I told you already several times that you are not actually considering the situation from two perspectives: you have coordinates

measured in frame A, these thenprojectedby your suggested transformation formula into frame B, and these projected A-coordinates projected back by the reverse formula into frame A. At no point has B any knowledge about this let alone are any coordinatesmeasured in frame Binvolved here. And the latter is what you have to do if you want to address the twin-paradox.avalonxq said:

"This result is fully consistent. In reference frame B, the distance between A1 and A2 is shorter than the distance between B1 and B2. And, as explained previously, A2 has already ticked to x*v at t'=0. Similarly B2 reads x*v/y, and hence they met at t'=-x*v/y."

I was asking you to consider a 'snapshot' with t=t'=0 throughout either reference frame.

You are right that according to the Lorentz transformation the distance between A1 and A2 is shorter than the distance between B1 and B2 (namely by a factor 1/γ), so with A1 and B1 matching up at the origin in both frames at t=t'=0, A2 and B2 would match up at (say) x_A in reference frame A but at x_B = x_A*γ in reference frame B. But now let B consider the situation with his reference frame taken as the rest frame: with B2 located at x_B = X_A*γ, he would then conclude that at t'=0, x_A = x_B*γ = x_A*γ², in contradiction to the original assumption that A2 is located at x_A.

Thomas

"A2 and B2 would match up at (say) x_A in reference frame A but at x_B = x_A*γ in reference frame B."

ReplyDeleteYes, and don't forget to calculate WHEN this event happens. The Lorentz equations include a distance component, so if t=t'=0 at the origin, t=0 and t'=0 do not match up at any point other than the origin.

And, when you calculate the point in time in reference frame B at which A2 and B2 meet, you see that there's no contradiction.

"But now let B consider the situation with his reference frame taken as the rest frame: with B2 located at x_B = X_A*γ, he would then conclude that at t'=0, x_A = x_B*γ = x_A*γ², in contradiction to the original assumption that A2 is located at x_A."

Sorry, but you've yet again failed to convert the time coordinates. There is no inconsistency in the distances, because A2 and B2 do not pass each other at t'=0. They pass each other at t'=-x*v/y.

Your contradiction relies entirely on your refusal to actually accept that simultaneity is relative and convert the times as necessary. When ysou do so, there is no contradiction. Special Relativity gives us exactly one solution to the problem you set regardless of which way we "look" at it. Only by changing the problem to a different problem (such as by syncing the clocks in a different reference frame, something that relativity tells you will unsync the clocks in the first reference frame) do you ever arrive at a contradiction.

avalonxq said:

Delete"Sorry, but you've yet again failed to convert the time coordinates. There is no inconsistency in the distances, because A2 and B2 do not pass each other at t'=0. They pass each other at t'=-x*v/y."

Forget about the timings for a moment.

Do you agree that the distance of the markers B1 and B2 in reference frame B (i.e. their 'proper' distance) is X_A*γ if it is measured in reference frame A as X_A (as indicated by matching up with the markers A1 and A2 at positions 0 and x_A respectively)?

Thomas

Thomas said:

Delete"Forget about the timings for a moment.

Do you agree that the distance of the markers B1 and B2 in reference frame B (i.e. their 'proper' distance) is X_A*γ if it is measured in reference frame A as X_A (as indicated by matching up with the markers A1 and A2 at positions 0 and x_A respectively)?"

Yes. The distance between B1 and B2 in reference frame B is y*x. Furthermore, the distance between A1 and A2 in reference frame B is x/y.

Wait, I got that backwards.

DeleteThe distance between B1 and B2 is reference frame B is x/y. The distance between A1 and A2 is y*x.

avalonxq said:

Delete"Yes. The distance between B1 and B2 in reference frame B is y*x. Furthermore, the distance between A1 and A2 in reference frame B is x/y.

Wait, I got that backwards.

The distance between B1 and B2 is reference frame B is x/y. The distance between A1 and A2 is y*x."

I am not sure how you got to either statement. The

assumptionwas that in reference frame A (the rest frame) the points A1 and A2 have the proper distance x_A = x. If A1 and A2 match up with B1 and B2 respectively, then,according to the Lorentz transformation, B1 and B2 have the proper distance x_B = γ*x in the moving frame B (in reference frame A this distance is thus measured contracted to x).Now note that these proper distances x_A and x_B are (assumed as) constant i.e independent of time. So if in frame A, at time t=0, A1/B1 match up at the origin and A2/B2 at x, then in frame B, at t'=0, they match up at the origin and γ*x (note that I have chosen the origin as the location for one pair only for convenience; as it is only the proper distances that are of relevance here, one could shift the whole scenario also to any other location).

The point here is that the proper distance γ*x in reference frame B is not a quantity actually measured in reference frame B. It is the

proper distance x in frame A projected into frame Bvia the Lorentz transformation. Yet this proper distance does not match up with proper distance x in frame A anymore if we consider B as the rest frame, because if B measures a distance γ*x between the matching pairs, then according to the Lorentz transformation he would conclude that the corresponding proper distance in frame A is γ²*x instead of the true value x.And before you try to bring in the timing offsets due to the relativity of simultaneity again, note that you simply can not assume the two 'events' as independent here. The match-ups of the two pairs are strictly correlated due to the constancy of their distance. So if you would mismatch the pair at x whilst keeping the pair at the origin matched, you would have changed the proper distance.

Thomas

"I am not sure how you got to either statement. The assumption was that in reference frame A (the rest frame) the points A1 and A2 have the proper distance x_A = x."

DeleteThat's correct.

"If A1 and A2 match up with B1 and B2 respectively,"

And that's incorrect; there is no point in time in reference frame B where A1 and A2 match up with B1 and B2 respectively.

In reference frame B, the distance between A1 and A2 is different than the distance between B1 and B2: the distance between A1 and A2 is y*x, but the distance between B1 and B2 is x/y.

You keep telling me to ignore the time offsets, but I'm telling you that ignoring the time offsets is what is causing your error!

The time offset due to the relativity of simultaneity is irrelevant if you just consider the distances between the markers. As you agreed yourself, if A measures the distance of the B-markers as X, this corresponds in B to the 'proper distance' γ*X. But from B's point of view, this distance does not match up again with the proper distance X in frame A but rather with γ*(γ*X). If you would continue this measurement process indefinitely, you would thus end up with an infinite proper distance. This shows thus that, as I mentioned before, the Lorentz transformation is essentially a one-way transformation from the rest frame into the moving frame, and does actually not allow a change of the rest frame at all. Only by violating in this sense the principle of relativity (according to which each observer can consider himself to be at rest), is the twin paradox 'resolved'.

DeleteThomas

Sorry, Thomas, but relativity doesn't work that way.

DeleteWhen you move from a reference frame where the B clocks are moving at v to one in which they are at rest, the distance between them increases by a factor of 1/y. When you move from a reference frame where the B clocks are at rest to one in which they are moving at v, the distance between them increases by a factor of y.

Similarly, when you move from a reference frame where the A clocks are moving at v to one in which they are at rest, the distance between them increases by a factor of 1/y. When you move from a reference frame where the A clocks are at rest to one in which they are moving at v, the distance between them increases by a factor of y.

This means that in the A frame, the distances are both x, while in the B frame, the B distance is x/y while the A distance is y*x. Move back to the B frame to the A frame, and you move back to y*x/y and y*x/y respectively -- that is, x and x again. There is no situation in which this y^2 thing happens.

If you disagree, please spell out the derivation in more detail -- that is, define the starting conditions, give the equation, and the outcome in each step of the process. Because as is, you have not illustrated any contradiction in either direction.

Ick, I did it again. I really wish I could edit these posts.

DeleteWhen you move from a reference frame where the B clocks are moving at v to one in which they are at rest, the distance between them increases by a factor of 1/y. When you move from a reference frame where the B clocks are at rest to one in which they are moving at v, the distance between them DEcreases by a factor of y.

Similarly, when you move from a reference frame where the A clocks are moving at v to one in which they are at rest, the distance between them increases by a factor of 1/y. When you move from a reference frame where the A clocks are at rest to one in which they are moving at v, the distance between them DEcreases by a factor of y.

avalonxq says:

Delete"Sorry, Thomas, but relativity doesn't work that way.

When you move from a reference frame where the B clocks are moving at v to one in which they are at rest, the distance between them increases by a factor of 1/y. When you move from a reference frame where the B clocks are at rest to one in which they are moving at v, the distance between them DEcreases by a factor of y."

First of all, you have got the figures the wrong way around: since γ>=1, a factor γ increases the length, whereas 1/γ decreases it.

Secondly, you are missing the point here by just transforming the distance between the B-clocks back and forth; this obviously will always work as it amounts to an algebraic tautology, but the whole consideration here was about A and B clocks mutually matching up in their locations, and in this respect I wouldn't say that Relativity 'works' if it predicts that

at the same time instant t=t'=0, in frame A, A1 and A2 match up with B1 and B2 respectively, but in frame B, B1 matches up with A1, yet B2 with some other marker at a distance different from that of A2.Also, consider the following: if we have two observers in each reference frame, one at A1 and one at A2, both of which define the origin of their reference frame to be at their location, then in frame A both observers would still agree that at time t=0 we would have A1/B1 and A2/B2 matching up (assuming that all clocks are synchronized in A). Transformed into frame B however, this would mean that the observer who defines the origin at A1 would have to assume that at t'=0 in reference frame B, B1 and A1 match up (whereas B2 and A2 don't), yet the observer who defines the origin at A2 would have to assume that B2 and A2 match up in frame B (whereas B1 and A1 don't) (note that t'=0 denotes the same time instant in either case, as all the clocks in B are synchronized as well).

So it is not only by changing the inertial frame, but already by changing the position

within one inertial framethat contradictions arise here.Thomas

"First of all, you have got the figures the wrong way around: since γ>=1, a factor γ increases the length, whereas 1/γ decreases it."

DeleteNo, I have it right. The problem is that I think you're using γ = 1/y. If you remember, y = (1 - v^2), so y <= 1.

"Secondly, you are missing the point here by just transforming the distance between the B-clocks back and forth; this obviously will always work as it amounts to an algebraic tautology, but the whole consideration here was about A and B clocks mutually matching up in their locations, and in this respect I wouldn't say that Relativity 'works' if it predicts that at the same time instant t=t'=0, in frame A, A1 and A2 match up with B1 and B2 respectively, but in frame B, B1 matches up with A1, yet B2 with some other marker at a distance different from that of A2."

But that's exactly what relativity DOES say, what it predicts, and what all data appear to show to be true. And since you've not identified any paradox that comes from this prediction, I don't see why you say it doesn't 'work'.

"Also, consider the following: if we have two observers in each reference frame, one at A1 and one at A2, both of which define the origin of their reference frame to be at their location, then in frame A both observers would still agree that at time t=0 we would have A1/B1 and A2/B2 matching up (assuming that all clocks are synchronized in A)."

True so far.

"Transformed into frame B however, this would mean that the observer who defines the origin at A1 would have to assume that at t'=0 in reference frame B, B1 and A1 match up (whereas B2 and A2 don't), yet the observer who defines the origin at A2 would have to assume that B2 and A2 match up in frame B (whereas B1 and A1 don't) (note that t'=0 denotes the same time instant in either case, as all the clocks in B are synchronized as well)."

Incorrect.

This is where you, again, insist on something that relativity says isn't true -- in reference frame B, the clocks B1 and B2 are not synchronized! So, if we define t'=0 at reference frame B as being the instant when A1 passes B1, then A2 passes B2 at some instant prior to that and both the A2 and B2 clocks read greater than 0.

You keep insisting that B1 and B2 are synchronized, and that's your mistake (not Relativity's). Again, two distant clocks that are synchronized in one reference frame are not synchronized in another frame, and you already synchronized them in A so we KNOW they're not synchronized in B.

We can, instead, talk about t''=0 at the point when A2 passes B2, but again this is a different point in time than when t'=0. In fact, the equation t''= t' - x*v/y will hold for all t', as it's just a straight offset (basing your timing on the B2 clock rather than the B1 clock).

You seem to be claiming there's some symmetry here, but of course there's not -- if you try to swap the rolls of A1 and A2 in the problem, you also have to reverse the sign of your velocity from -v to v. This won't change your y factor which uses v^2, but it will change your time offsets when tranforming to the B frame, in the obvious way. Now A2 passes B2 at time t''=0, but A1 doesn't pass B1 until t'' = x*v/y.

"So it is not only by changing the inertial frame, but already by changing the position within one inertial frame that contradictions arise here."

You still haven't identified a single contradiction; every outcome I've presented to you is consistent with every other, and the only way you've been able to even suggest an inconsistency is by insisting something that Relativity specifically denies like refusing to convert the time from one frame to another or insisting a synchronization that's not there.

Can you demonstrate where your double-y (or double-gamma) is derived from by first carefully laying out your problem, then presenting the equations you're going to use, and then showing your results? Up to this point, you seem unwilling to actually go through the full process of working out the problem yourself. I'd be happy to point out any further mistaken assumptions or erroneous calculations along your way.

Deleteavalonxq said:

Delete"You keep insisting that B1 and B2 are synchronized, and that's your mistake (not Relativity's)."

Each reference frame must have all their clocks synchronized by default, otherwise there would be no point in keeping clocks at all.

So one has to assume that all clocks in A are synchronized with each other and all clocks in B are synchronized with each other. If then A1 and B1 synchronize at the origin, clocks A2 and B2 matching up at any other point must synchronize as well. This is the logical consequence of the transitivity of clock synchronization (note that I don't even have to imply that the clocks

staysynchronized; it is just about the clock readings at a specific moment here).And if on the basis of the Lorentz transformation A concludes that the B clocks do

notshow the same time, then this proves that this conclusion is simply false (as it contradicts the original axioms).Thomas

"So one has to assume that all clocks in A are synchronized with each other and all clocks in B are synchronized with each other."

DeleteWe can certainly modify our initial problem to require this, as I did in my Spaceship A and B description linked to below.

But if A1 and A2 are synchronized in frame A and B1 and B2 are synchronized in frame B, then A1 and A2 are not synchronized in frame B and B1 and B2 are not synchronized in frame A. No matter what positions you start them with, there is simply no instant where they all four match up. If you set A1, A2, and B1 all to zero, then B2 will be nonzero.

"If then A1 and B1 synchronize at the origin, clocks A2 and B2 matching up at any other point must synchronize as well."

Certainly not, no. If you're requiring B1 and B2 to be synchronized in frame B, then they are not synchronized in frame A. A2 and B2 will not synchronize.

"This is the logical consequence of the transitivity of clock synchronization (note that I don't even have to imply that the clocks stay synchronized; it is just about the clock readings at a specific moment here)."

This doesn't make sense; what do you mean by "transitivity of clock synchronization"? You think you can magically ignore the Lorentz time factor and just pretend B2 is simultaneously synchronized with B1 in BOTH reference frame A and reference frame B, in condradiction to Relativity? What is your basis for this claim?

avalonxq said:

Delete"This doesn't make sense; what do you mean by "transitivity of clock synchronization"? You think you can magically ignore the Lorentz time factor and just pretend B2 is simultaneously synchronized with B1 in BOTH reference frame A and reference frame B, in condradiction to Relativity? What is your basis for this claim?"

I know that it contradicts Relativity. That's the reason why I mentioned it. You seem to assume that anything which contradicts Relativity must a priori be wrong. So then tell me why you think the following argumentation is wrong (and please don't say 'because it contradicts Relativity'; I want a fundamental logical argument).

Axiom 1: All clocks in reference frame A can be assumed to be

unconditionallytied in which each other (meaning that they are kept synchronized at all times by appropriate means in that particular reference frame alone).Axiom 2: All clocks in reference frame B can be assumed to be

unconditionallytied in which each other.Theorem: If any clock in A is tied in with a clock in B (for instance by showing the same time when the two are in the same location),

allthe clocks in both frame A and B will beunconditionallytied in with each other.Thomas

"I know that it contradicts Relativity. That's the reason why I mentioned it. You seem to assume that anything which contradicts Relativity must a priori be wrong."

DeleteNot at all. Remember that the purpose here is to show the assumptions of relativity to be logically inconsistent. To do so, you must start with those assumptions, and not your own assumptions that are contrary to those assumptions.

"So then tell me why you think the following argumentation is wrong (and please don't say 'because it contradicts Relativity'; I want a fundamental logical argument)."

Let me re-write your Axioms 1 and 2 and confirm that you agree with the re-write.

Axiom 1: All clocks at rest in reference frame A can be assumed to be unconditionally tied in which each other (meaning that they are kept synchronized at all times by appropriate means in reference frame A alone).

Axiom 2: All clocks at rest in reference frame B can be assumed to be unconditionally tied in which each other (meaning that they are kept synchronized at all times by appropriate means in reference frame B alone).

If you agree that this re-write is acceptable, then I agree that Relativity would accept these axioms.

However, your theorem doesn't hold. You need to explain why Axioms 1 and 2 implies your theorem as stated.

avalonxq said:

Delete"Axiom 1: All clocks at rest in reference frame A can be assumed to be unconditionally tied in which each other (meaning that they are kept synchronized at all times by appropriate means in reference frame A alone).

Axiom 2: All clocks at rest in reference frame B can be assumed to be unconditionally tied in which each other (meaning that they are kept synchronized at all times by appropriate means in reference frame B alone).

If you agree that this re-write is acceptable, then I agree that Relativity would accept these axioms.

However, your theorem doesn't hold. You need to explain why Axioms 1 and 2 implies your theorem as stated."

Well, then add a third axiom, which should make it clearer:

Axiom 3: It is a sufficient condition for two clocks being synchronized if a (hypothetically) infinitely fast signal travelling from one to the other encounters them both showing the same time (note that I use the assumption of an infinitely fast signal here only for convenience; I could as well use the speed of light (or whatever) and then require a corresponding time offset).

So consider the following round-trip: the signal is emitted from A2 as it reads 0, then it reaches A1 reading 0 (because all the clocks in A are assumed synchronized); it reaches B1 reading 0 (by assumption of both synchronizing at the origin); it reaches B2 reading 0 (because all the clocks in B are assumed synchronized).

This means that all clocks here (including B2) must be synchronized with each other on the basis of these 3 axioms.

Thomas

Three things about your Axiom 3...

Delete1) Relativity doesn't allow infinite-speed signals. If your goal is to start with axioms consistent with relativity and then show a contradiction, please stick to things relativity agrees with. Use light speed signals with your proposed offset.

2) You're not being careful about your reference frames. Synchronized in what frame? Your example uses the bait-and-switch of sending a signal that shows synchronization in frame A, and then switching to frame B when you bounce it back. This is equivalent to assuming, yet again, that synchronized clocks in one frame are synchronized in all frames. Yet again, they are not. Simultaneity is relative. You continue to try to get around that fact but fail to do so.

Feel free to revise your third axiom, being careful to stay consistent about your frames. Then PLEASE PLEASE PLEASE set out your scenario IN FULL, not taking shortcuts with failing to define starting conditions and variables, and explain the result.

I have written out an extended version of our hypothetical using actual values. You can see it here:

ReplyDeleteyes2einstein.blogspot.com

I'd encourage you to take a look and comment. If I had the graphics skills I'd add diagrams showing the ships moving past each other, but for the moment you're stuck with text.

Let me know where you think a paradox sits. I think it all hangs incredibly well togather.

I had a look at it, but I am not quite sure what you are trying to say with it.

DeleteIn any case, I don't think it makes much sense to resort to numerical examples when even the fundamental principles have not been clarified yet.

Thomas

It demonstrates an example of how relativity works and forms no contradictions. Even though the two reference frames do not agree on whose clocks are running slower or which events are simultaneous, they nonetheless agree on the readings of all clocks at all events, and can even provide a measurement for why those clocks will read that way according to their own reference frame.

DeleteHave I made a mistake somewhere? If not, where is the contradiction? Where is the measurement that will demonstrate the Relativity gets it wrong?

As I said already earlier, proposing some function y=f(x) and then observing that this is consistent with the inverse function x=f^-1(y) does not in any way address the question whether y=f(x) is an equation which is consistent with any constraints that could be imposed on such an equation in the first place.

DeleteAs shown on my page(s) regarding the Derivation of the Lorentz transformation, assuming a velocity dependent transformation for the coordinates if light signals is indeed simply inconsistent with the constancy of the speed of light c. Suitably re-scaling and off-setting the original definitions for the space and time units in a velocity dependent way (as Einstein suggested) may formally recover the constancy of c, but only leads to all the other inconsistencies and paradoxes Relativity is known for.

Thomas

"Inconsistencies and paradoxes" that you have, so far, been completely unable to actually calculate or demonstrate, correct?

Delete

Deleteavalonxq said:"Relativity doesn't allow infinite-speed signals. If your goal is to start with axioms consistent with relativity and then show a contradiction, please stick to things relativity agrees with. Use light speed signals with your proposed offset."

Clocks can not only be synchronized by light signals (after all, what would you do in an environment without light?). You can use any kind of signals, even hypothetically infinitely fast ones (which indeed bring the synchronization principle to the point). You can even synchronize them without using any signal propagation at all (e.g. when you hit the start buttons of two stop-watches simultaneously with the flat of your hand). All synchronizations of this kind would have to be consistent with each other. If one is not, then there is something wrong with it.

avalonxq said:"You're not being careful about your reference frames. Synchronized in what frame? Your example uses the bait-and-switch of sending a signal that shows synchronization in frame A, and then switching to frame B when you bounce it back. This is equivalent to assuming, yet again, that synchronized clocks in one frame are synchronized in all frames. Yet again, they are not. Simultaneity is relative. You continue to try to get around that fact but fail to do so."

So then tell me please which part of this chain you don't agree with

1) clock A1 shows time t=0 when clock A2 shows time t=0

2) clock B1 shows time t'=0 when clock A1 shows time t=0

3) clock B2 shows time t'=0 when clock B1 shows time t'=0

or why this should not result in

4) clock B2 shows time t'=0 when clock A2 shows time t=0.

avalonxq said:"Inconsistencies and paradoxes" that you have, so far, been completely unable to actually calculate or demonstrate, correct?"

I have addressed inconsistencies repeatedly, but you have failed acknowledge them properly.

Besides, your own example at http://yes2einstein.blogspot.co.uk/ is inconsistent: according to your timelines of A and B, captain B would have aged by 2:45 in his own frame, but only 2:30 in frame A. This makes a factor 1.1 difference, but according to the time dilation formula it should be a factor 1.25.

P.S.: I shall be occupied with other matters for the next few days, so I might not be able to post any more replies until the latter part of next week.Thomas

"So then tell me please which part of this chain you don't agree with"

DeleteCertainly -- I disagree with you continuing to ignore the relativity of simultaneity. The problem is your ambiguous use of the word "when," which you use to mean "at the same time as."

In Relativity, this term has no universal meaning. You need to specify "at the same time in frame X," or you can end up equating two different things.

Let me include it in your chain, and you'll see the problem.

1) Clock A shows time t=0 at the same time in frame A as clock A2 shows time t=0

2a) clock B1 shows time t'=0 as the same time in frame A as clock A1 shows time t=0

2b) clock B1 shows time t'=0 as the same time in frame B as clock A1 shows time t=0

3) clock B2 shows time t'=0 at the same time in frame B as clock B1 shows time t'=0

With this in mind, hopefully you can see why your conclusion is not reached. Simultaneity is not transitive across frames; that's why it's called "relativity of simultaneity." And you have to cross frames in order to derive the conclusion you want, which is not allowed.

"I have addressed inconsistencies repeatedly, but you have failed acknowledge them properly."

I don't see how that's possible. Every alleged inconsistency you have addressed, I have demonstrated an error in your assumptions that accounts for it. You have yet to actually set up a problem that matches the assumptions of relativity but yet creates an inconsistent result.

"Besides, your own example at http://yes2einstein.blogspot.co.uk/ is inconsistent: according to your timelines of A and B, captain B would have aged by 2:45 in his own frame, but only 2:30 in frame A. This makes a factor 1.1 difference, but according to the time dilation formula it should be a factor 1.25."

If there is an error, do let me know and I will correct it. But I'm not seeing any error here.

The first and last recorded events for Captain B are his meeting Captain A and his receiving the B2 signal. In frame A, these events happen at 01:00 and 03:30, for a time difference of 02:30. In frame B, these events happen at 07:45 and 09:45, for a difference of 02:00. 2.5/2 = 1.25, the exact factor required by the formula.

Your mistake likely comes from attempting to use the 07:00 reading of the B1/A2 intersection as Captain B's "first" event, but Captain B isn't listed in that event and no position is given in ship A's timeline for Captain B's position when his clock reads 07:00.

Any other mistakes in my timeline? Can you ask a question or demonstrate an equation which will show any error? Again, I think this example shows pretty clearly that, even though individual time and distance measurements may differ, relativity stays entirely consistent throughout.

ReplyDeleteavalonxq said:

"Certainly -- I disagree with you continuing to ignore the relativity of simultaneity. The problem is your ambiguous use of the word "when," which you use to mean "at the same time as."

There is nothing ambiguous about what I said. It is indeed a direct consequence of the Lorentz transformation (including the relativity of simultaneity term). I'll give the explicit transformations again in this sense

t | t'=γ(t-vx/c²)-----------------------------------------

t_A2=0 | t_A1=0 (as v=0)

t_A1=0 | t_B1=0 (as x=0)

t_B1=0 | t_B2=0 (as v=0)

If you question this, then you question the Lorentz transformation (or even mathematical logic).

Having said this, if in the second step you choose a different location x for the transition point between the two systems, then you get obviously different values for t_B1 and t_B2, but this ambiguity only goes to show the inconsistency of the Lorentz transformation.

avalonxq said:

"Again, I think this example shows pretty clearly that, even though individual time and distance measurements may differ, relativity stays entirely consistent throughout."

It depends on what you mean by 'entirely consistent'. Note that in this case you have arbitarily chosen one space ship to be longer than the other. If you would choose ship B to be shorter than A (rather than longer) your time dilation effect would be the other way around i.e. the timeline of A would be longer.

And if indeed both ships have the same length (as originally suggested through the animated graphics near the top) then you run into the following contradiction: if you assume both ships to coincide at one end, then, due to the relative length contraction assigned to the other ship, one ship will conclude that at this moment the other ends have passed each other already, while the other will conclude that the other ends have not passed yet, that is we have a logical paradox.

Thomas

"There is nothing ambiguous about what I said. It is indeed a direct consequence of the Lorentz transformation (including the relativity of simultaneity term). I'll give the explicit transformations again in this sense

Deletet | t'=γ(t-vx/c²)

-----------------------------------------

t_A2=0 | t_A1=0 (as v=0)

t_A1=0 | t_B1=0 (as x=0)

t_B1=0 | t_B2=0 (as v=0)"

Yet again, you insist on leaving your terms ambiguous. How are you defining t_B1 and t_B2? If they are defined in frame A, then it's not true that v=0, and your third transform is wrong. If they are defined in frame B, then your third equation is correct but that doesn't matter, as t_B2 does not represent the time of B2 at t=0 in frame A (which is what you're trying to calculate).

You will apparently do anything to avoid using the actual equation, where neither x or v is zero for B2, which shows that t_B2 when t=0 in frame A is actually not 0.

"It depends on what you mean by 'entirely consistent'. Note that in this case you have arbitarily chosen one space ship to be longer than the other. If you would choose ship B to be shorter than A (rather than longer) your time dilation effect would be the other way around i.e. the timeline of A would be longer."

Yes. And?

"And if indeed both ships have the same length (as originally suggested through the animated graphics near the top) then you run into the following contradiction: if you assume both ships to coincide at one end, then, due to the relative length contraction assigned to the other ship, one ship will conclude that at this moment the other ends have passed each other already, while the other will conclude that the other ends have not passed yet, that is we have a logical paradox."

You're going to have to specify your new hypothetical better. We're now going to have A and B, and each ship measures itself as having the same length. Ship A has A1 at the front and A2 at the back; Ship B has B1 at the front and B2 at the back.

Now, are you talking about the moment when A1 lines up with B1, when A1 lines up with B2, when A2 lines up with B1, or when A2 lines up with B2?

I think what you're saying is that the ship will disagree about which occurs first - A1 meeting B2 or A2 meeting B1. They will measure these events as having occurred in a different order. But what's the paradox?

I wanted to emphasize a little more what's wrong with your application of the Lorentz equations.

DeleteThe main problem is that if you want a time coordinate in reference frame A, then your velocity value for the Lorentz equation must reflect the velocity relative to reference frame A.

So if you're trying to determine the frame A time coordinate for an object moving at velocity v, you need to use velocity v in your Lorentz equation.

So, again, trying to convert from B1 and B2 and using v=0 is going to result in a time coordinate that's still relative to frame B. You end up with t'_B2=0. If you actually want t_B2, you're going to have to use a non-zero v.

Deleteavalonxq said:

"Yet again, you insist on leaving your terms ambiguous. How are you defining t_B1 and t_B2? If they are defined in frame A, then it's not true that v=0, and your third transform is wrong. If they are defined in frame B, then your third equation is correct but that doesn't matter, as t_B2 does not represent the time of B2 at t=0 in frame A (which is what you're trying to calculate)."

No, you don't seem to understand:

t | t'=γ(t-vx/c²)-----------------------------------------

t_A2=0 | t_A1=0 (as v=0)

t_A1=0 | t_B1=0 (as x=0)

t_B1=0 | t_B2=0 (as v=0)

are 3 different Lorentz transformation, connecting 4 different reference frames: the first connects the rest frame of A2 with the rest frame of A1, the second the rest frame of A1 with the rest frame of B1, the third the rest frame of B1 with the rest frame of B2. Altogether this connects therefore the rest frames of A2 and B2.

Thomas said:

"It depends on what you mean by 'entirely consistent'. Note that in this case you have arbitrarily chosen one space ship to be longer than the other. If you would choose ship B to be shorter than A (rather than longer) your time dilation effect would be the other way around i.e. the timeline of A would be longer."

avalonxq said:

Yes. And?

Well, it goes to show that the timelines are actually not different because of the different clock rates as such (which as you admitted yourself, A and B would not agree about anyway) but because of the propagation times of the light signal you used here to define the events, in combination with the contracted lengths of the space ships.

I at least have not heard yet before that the reduced ageing of the travelling twin is in any way related to the length of his space ship.

Thomas

Deleteavalonxq said:

You're going to have to specify your new hypothetical better. We're now going to have A and B, and each ship measures itself as having the same length. Ship A has A1 at the front and A2 at the back; Ship B has B1 at the front and B2 at the back.

Now, are you talking about the moment when A1 lines up with B1, when A1 lines up with B2, when A2 lines up with B1, or when A2 lines up with B2?

I think what you're saying is that the ship will disagree about which occurs first - A1 meeting B2 or A2 meeting B1. They will measure these events as having occurred in a different order. But what's the paradox?"

Have a look at the this diagram

I have depicted the situation where the captains A1 and B1 are alongside each other, both as seen in frame A and B.

Although they travel at a relativistic velocity, they have a quick chat (being able to speak very fast):

A1: We have picked up your earlier distress call, and have already sent our engineer A2 into your space ship a while ago when the entrance at the other end lined up with your B2 entrance. He should be able to assist you with your problem.

B1: That's not possible. Our B2 entrance hasn't even reached your A2 entrance. It will still take a while until this happens.

A1: I can assure you that it has already happened. The opening of the airlock and the transfer of B2 was automatically programmed to take place a time L/v*(1-1/γ) ago when A2 and B2 matched up. So as we speak A2 is on your ship.

B1: I can assure you that as we speak A2 is not on our ship, as A2 and B2 haven't matched up yet. It will only happen in a time L/v*(1-1/γ) from now.

A1: It won't happen, because as I said, it has happened already, and also the airlock has been automatically permanently locked after this. Unfortunately, we haven't received the confirmation about this yet, as the regulation do not allow us to send faster-than-light signals, but when we have returned to our base, we will be able to check on which ship A2 actually is right now.

B1: OK, we will check this after return, but whatever the outcome, there seems to be a problem of explanation here.

Thomas

That's a great scenario! But you still haven't described the paradox.

DeleteAfter the A1/B1 conversation takes place, A1 will receive the A2/B2 signal, confirming that the A2/B2 transfer occurred at the expected time according to the ship A clock.

After (as defined in either frame) the A2/B2 signal has passed A1, B1 will receive the A2/B2 signal, confirming that the A2/B2 transfer occurred at the expected time according to the ship B clock.

There is a discrepency in the relative order of the A1/B1 and A2/B2 events, but not one that causes any paradox. Special relativity indicates that spacelike events have no absolute time order. But where's the paradox? There's no mistake in what either frame calculates, nor is there any discrepency or ambiguity in what either party should measure.

avalonxq said:

Delete"That's a great scenario! But you still haven't described the paradox."

Have I not? Then please tell us in which ship A2 is when the A1/B1 conversation takes place (i.e. at time t=t'=0).

Thomas

"Have I not? Then please tell us in which ship A2 is when the A1/B1 conversation takes place (i.e. at time t=t'=0)."

DeleteSimulataneity is relative. That means distant events don't happen at the same time as measured in different reference frames.

In reference frame A, the A2/B2 event takes place before the A1/B1 event, and so A2 is already on ship B at time t=0.

In reference frame B, the A2/B2 event takes place after the A1/B1 event, and so A2 is still on ship A at time t'=0.

Since time t=0 does not coincide with time t'=0 except at the A1/B1 intersection, there's no paradox -- and you still haven't described one.

Simultaneity is relative, and spacelike events don't have an "objective" chronoligical order that is the same in all frames. You still haven't explained what paradox this creates -- what causal inconsistency, what contradiction in calculation, or what irreconciliation in observation. Because there isn't one. Frames A and B have different coordinate axes with events occuring at different relative times, and it causes no actual paradoxes, inconsistencies, or other problems.

If you can find one, please share it.

Deleteavalonxq said:"Simultaneity is relative. That means distant events don't happen at the same time as measured in different reference frames."

But this is all taken into account in the diagram that I produced to illustrate the situation. You can consider this as a full representation of the Lorentz transformation (viewed from either reference frame taken as the rest frame). And it is exactly this which leads to the logical contradiction here.

Let me go through the argument again:

Spaceship A was programmed to transfer A2 over to spaceship B at A's ship-time t=-L/v*(1-1/γ) (when A2 and B2 did match up). So since this transfer was programmed to take place at this time, A1 knows

with absolute certaintythat A2 is not on board of his ship anymore (but instead on board of ship B) when he matches up with B1 at the later ship time t=0 (so he does not even need subsequent confirmation of the transfer). He communicates this fact to B1 during their conversation, so B1 knows now alsowith absolute certaintythat at this moment (that is B's ship-time t'=0) A2 is not on ship A anymore but on his own ship. However, from his point of view, this is impossible as B2 hasn't even reached the initial position of A2 yet (it will reach it at his ship time t'=L/v*(1-1/γ)), so A2 can not possibly have been transferred yet.This proves that the assumption of a frame dependent length contraction leads to a logical contradiction once you consider appropriate causal interactions between the two frames.

Thomas

Assume that A1 programmed the A2 transfer at 1:00 to occur at 02:00, knowing that A1 would meet B1 at 03:00.

DeleteYou claim that A1 knows at 03:00 that the A2 tranfer has occurred "with absolute certainty," even though A2 is too far away for A1 to have yet received ANY signal or confirmation from A2. That means that A1's "absolute certainty" must have existed from the moment he programmed A2. That means that, for instance, at 01:30, A2 also had "absolute certainty."

I want to make sure this point is clear -- A1 HAS NO MORE INFORMATION ABOUT WHETHER THE A2 TRANSFER OCCURS AT 03:00 THAN HE DID AT 01:30, so any "absolute certainty" he has about the transfer cannot be based on whether the event has actually happened yet.

Now, when A1 conveys to B1 his "absolute certainty" in an event that he can in no way confirm to have actually happened, you claim that B1 gets the same "absolute certainty" that A1 has. That's fine, because it's the same absolute certainty that A1 had at 01:30.

If A1's "absolute certainty" at 01:30, before the A2/B2 tranfer, does not "violate causality," then in what way does B's "absolute certainty" violate causality?

You said:

"This proves that the assumption of a frame dependent length contraction leads to a logical contradiction once you consider appropriate causal interactions between the two frames."

I'm still not seeing the logical contradiction. Can you explain what possible causal sequence of events will lead to a contradiction?

Specifically, can you explain something that B1 can learn about a "future event" that A1 didn't already know before that event occurred? I claim that the only actual information of events that A1 will have is of events that are in B1's past. If you have a mechanism to transfer future knowledge (that is, knowledge of events that have not yet occurred in B1's timeline and could not have been known before they actually happen), please share.

Deleteavalonxq said

"You claim that A1 knows at 03:00 that the A2 transfer has occurred "with absolute certainty," even though A2 is too far away for A1 to have yet received ANY signal or confirmation from A2. That means that A1's "absolute certainty" must have existed from the moment he programmed A2. That means that, for instance, at 01:30, A2 also had "absolute certainty."

I want to make sure this point is clear -- A1 HAS NO MORE INFORMATION ABOUT WHETHER THE A2 TRANSFER OCCURS AT 03:00 THAN HE DID AT 01:30, so any "absolute certainty" he has about the transfer cannot be based on whether the event has actually happened yet.

Any physical theory (apart from quantum physics) is by design deterministic, that is the state of the system at any later time is already implied at the start time. Of course, that may not always work out in reality as the model is only an approximation to reality, but as far the model itself is concerned, it is indeed a logical tautology i.e. everything is absolutely certain by design from the outset. So in this sense the absolute certainty about the transfer already exists before it happens. The problem here is that you have

twodeterministic models, each with a different sequence of events. And these two models can come into conflict with each other. Let me give you an example: you have an equation that, by itself, describes (with absolute certainty) the rectilinear motion of an object from point P1 to P2, and you have another equation that, by itself, describes (also with absolute certainty) the rectilinear motion of another object from point P3 to P4. So whilst each equation by itself would consider it absolutely certain that the object arrive at P2 and P4 respectively after a certain time, this neglects however the possibility that both paths intersect in a way that both objects collide with each other and thus would never get to their destination point. So the interaction of the two systems can render the individual models invalid. It is s similar situation here: whilst as such there may not be an obvious contradiction between the timelines in frames A and B, the whole scenario becomes inconsistent if the systems have points of actual physical interaction. So whilst nobody would expect to discover any kind of logical conflicts in practice if spaceship A is in one galaxy and spaceship B in another, a conflict arises here if there is the interaction and communication between the two ships as described above. And to demonstrate the 'twin paradox' the latter is obviously required.I realize that I may have implied this interaction between the two systems with my original argument, but have not explicitly clarified it or worked it into the example. In this sense, I intend to do a major rewrite of my page regarding the Time Dilation and Twin Paradox. I am not quite sure when I get around to do this, but hopefully within the next couple of weeks.

Thomas

"Any physical theory (apart from quantum physics) is by design deterministic, that is the state of the system at any later time is already implied at the start time. Of course, that may not always work out in reality as the model is only an approximation to reality, but as far the model itself is concerned, it is indeed a logical tautology i.e. everything is absolutely certain by design from the outset. So in this sense the absolute certainty about the transfer already exists before it happens."

DeleteAnd yet your paradox was based on the idea that B having "absolute certainty" at time t'=0 of the A2/B2 event, which happens later in reference frame B, was impossible.

Now you've conceded that this is not the case -- that there's no contradiction in B1, at time t'=0, learning about the arrangements that A1 had already made at a time prior to t'=0 and deciding that A2/B2 is an "absolute certainty" because of those preparations. Neither is B1 requires to believe that the A2/B2 event "already happened" just because A1 claims it to be so; after all, A1 can provide no proof that the A2/B2 proof "already happened" that is not also consistent with B1's assertion that the event "has not yet happened" (although A1 may have evidence that proves the event "will happen" with certainty, that is not a contradiction).

Do you concede, then, that your scenario again demonstrates no logical contradiction or paradox?

"It is s similar situation here: whilst as such there may not be an obvious contradiction between the timelines in frames A and B, the whole scenario becomes inconsistent if the systems have points of actual physical interaction."

You say that, and yet you again fail to demonstrate any.

I'm losing count - how many scenarios have we in fact covered here? Four? Five? And in each case, as soon as we look closely at what you claim is the paradox, we see that there's no paradox at all -- that relativity in fact has a single, unambiguous description of the situation that creates no problems in causality and no internal contradictions -- and then you move us on to another experiment where the same thing happens.

So instead of moving on to a new scenario, let's stick with our spaceship scenario. Where's the contradiction?

DeleteThomas said:

"Any physical theory (apart from quantum physics) is by design deterministic, that is the state of the system at any later time is already implied at the start time. Of course, that may not always work out in reality as the model is only an approximation to reality, but as far the model itself is concerned, it is indeed a logical tautology i.e. everything is absolutely certain by design from the outset. So in this sense the absolute certainty about the transfer already exists before it happens."

avalonxq said:

"And yet your paradox was based on the idea that B having "absolute certainty" at time t'=0 of the A2/B2 event, which happens later in reference frame B, was impossible."

You ignored the crucial second part of my argument. The absolute certainty as such is not an issue here at all (as I said, it is an implied feature of any physical theory). The point here is that you have

twoabsolute certainties: one that the A2/B2 eventhasalready happened, another that ithas nothappened. Both views can not be true at the same time. In order to make things a bit more drastic and thus clearer, just assume for instance that (referring again to this illustration) the first A2/B2 encounter (i.e. as seen from frame A) is programmed to trigger explosive devices that destroy both ships. So when A1 speaks to B1, the destruction must have already been initiated (taking the timeline of frame A). Of course, B1 can not believe this, as according to his timeline A2 and B2 haven't even met yet. He decides to reverse his ship on the spot so that A2 and B2 don't meet in his frame. So according to frame B, the ships will not be destroyed. If you don't think this is a paradox, then you have to explain how it is resolved.Thomas

Thomas said:

Delete"The point here is that you have two absolute certainties: one that the A2/B2 event has already happened, another that it has not happened. Both views can not be true at the same time."

Actually, they can. That's what the principle of relativity is all about -- different observers make different measurements, and both are correct.

A1 also has absolute certainty that A1 and A2 are at rest, while B1 and B2 are moving. B1 has

the oppositeabsolute certainty: B1 and B2 are at rest, while A1 and A2 are moving. Two contradictory certainties. Who's right? They both are.Thomas said:

"In order to make things a bit more drastic and thus clearer, just assume for instance that (referring again to this illustration) the first A2/B2 encounter (i.e. as seen from frame A) is programmed to trigger explosive devices that destroy both ships. So when A1 speaks to B1, the destruction must have already been initiated (taking the timeline of frame A)."

That's fine. Note, however, that the force of the explosion will only travel at sublight speeds. Still, obviously, the light of the explosion will reach them at the speed of light (some after the A1/B1 intersection).

Thomas said:

"Of course, B1 can not believe this, as according to his timeline A2 and B2 haven't even met yet. He decides to reverse his ship on the spot so that A2 and B2 don't meet in his frame."

I concede that if B1 has a magical, superluminal force that can

instantlychange the speed of B2, then there is an inconsistency in relativity. Of course, we don't have to bother with your thought experiment at all, since this forceitselfcontradicts relativity.I claim, however, that B1 must be limited to affecting B2 using signals and forces that travel at or below the speed of light.

If you limit yourself to such signals and forces, can you still create a paradox?

DeleteThomas said:

"The point here is that you have two absolute certainties: one that the A2/B2 event has already happened, another that it has not happened. Both views can not be true at the same time."

avalonxq said:

"Actually, they can. That's what the principle of relativity is all about -- different observers make different measurements, and both are correct."

You are again making an abstract statement that doesn't answer the question how the event should have happened as well as not happened at the considered moment in time (t=t'=0). If A1 and B1 disagree about the event in this sense, this proves that their assumptions are somewhere

notcorrect.avalonxq said:

A1 also has absolute certainty that A1 and A2 are at rest, while B1 and B2 are moving. B1 has the opposite absolute certainty: B1 and B2 are at rest, while A1 and A2 are moving. Two contradictory certainties. Who's right? They both are

They are both correct because motion as such is a symmetrical concept: if A moves relatively to B, then B moves relatively to A. So no contradiction here. However, if the length of A is greater than the length of B, the length of B can not be greater than the length of A. And the latter contradiction is what we are dealing with in case of the length contracted ships.

avalonxq said:

"That's fine. Note, however, that the force of the explosion will only travel at sublight speeds. Still, obviously, the light of the explosion will reach them at the speed of light (some after the A1/B1 intersection)."

That all doesn't matter for the question whether in the end (after the whole encounter is completed) the ships are destroyed or not.

avalonxq said:

"I concede that if B1 has a magical, superluminal force that can instantly change the speed of B2, then there is an inconsistency in relativity. Of course, we don't have to bother with your thought experiment at all, since this force itself contradicts relativity."

Forces are completely irrelevant in Special Relativity. 'Boosts' of this sort are assumed frequently in relativistic scenarios. Anyway, I was not necessarily thinking of an instant change of velocity but just a reversal before B2 reaches A2 (in B's frame). But it occurred to me now that this wouldn't actually work as during the reversal the length contraction would change and B2 match up with A2 anyway. So let's just assume a scenario where the clever (or not so clever) captain B1 has programmed some electronic defence system to be activated at t=t'=o (i.e. when A1 and B1 match up) that would prevent the explosion to be triggered when A2 and B2 match up (in his frame). However at this moment the explosion would have already happened a while ago in A's frame. So there is an obvious contradiction as to the question whether the ships are destroyed in the end or not.

Thomas

Thomas said:

Delete"Forces are completely irrelevant in Special Relativity. 'Boosts' of this sort are assumed frequently in relativistic scenarios."

Sure, but the assumption is that they're figured out ahead of time -- not that B1 can decide at t=0 to instantaneously affect the motion of the ship elsewhere.

Thomas said:

"Anyway, I was not necessarily thinking of an instant change of velocity but just a reversal before B2 reaches A2 (in B's frame)."

There is nothing B1 can choose to do at t'=0 that will affect B2

in any wayprior to the A2/B2 intersection.Thomas said:

"So let's just assume a scenario where the clever (or not so clever) captain B1 has programmed some electronic defence system to be activated at t=t'=o (i.e. when A1 and B1 match up) that would prevent the explosion to be triggered when A2 and B2 match up (in his frame)."

So B1 has set up, earlier, a defense system at B2 that will protect the ship from the A2/B2 explosion? That's fine; what's the paradox?

"However at this moment the explosion would have already happened a while ago in A's frame. So there is an obvious contradiction as to the question whether the ships are destroyed in the end or not."

There's still no contradiction. An observer in frame A or in frame B will see A1's and B1's preparations at A2 and B2 both as having happened prior to the A2/B2 event, and so both the bomb and the defense system will be taken into account according to observers in either frame.

Nothing that happens after the A2/B2 intersection, as observed in frame A, can possibly affect the A2/B2 intersection at t=0. An observer in frame B will agree with this, as every event that happens after the A2/B2 event according to frame A but before the A2/B2 event according to frame B

is also too far away to reach the A2/B2 event. This prevents any paradox.The scenario you orchestrated doesn't depend on relative motion at all. We could have you and I sitting in Chicago, when earlier A programmed a bomb in London and B programmed a shield to counteract the bomb. A would

believethe bomb had destroyed the city, but wouldjust be wrong, as he did not know about B's defenses.And since the A2/B2 event is too far away for A1 to have seen it at t=0, A1 just doesn't know yet that the bomb was counteracted. Again, no paradox.

But if, on the other hand, you intended the B1 defense system to be activated from B1 at t'=0, then whatever it is, it can't get to B2 in time to protect it from the bomb (unless you allow a magical, superluminal speed for the object or signal activating the defense, which I do not).

There is no paradox, because all reference frames agree on the order of all events in any causal relationship. It is not possible, under Special Relativity, to have a chain of events where any reference frame reverses the order of cause and effect, unless you allow superluminal communication or travel.

ReplyDeleteavalonxq said:

"So B1 has set up, earlier, a defense system at B2 that will protect the ship from the A2/B2 explosion? That's fine; what's the paradox?"

Please note that I suggested the defence system is activated by B

at time t=t'=0, soafterthe explosion was already triggered in the A frame (it wouldn't actually be necessary to assume such a defence system as there still would be the contradiction that the ships have already exploded before the A2/B2 encounter in frame B, so there couldn't really be a second explosion anyway; I have just assumed the defence mechanism in order to keep things slightly clearer).In this sense, I have now created a new illustration that illustrates the situation at 3 different ship times (both from the viewpoint of frames A and B):

1) at time t=t'= -L/v*(1-1/γ) (when the explosion is automatically triggered by the A2/B2 encounter in frame A),

2) at time t=t'=0 (when A1 matches up with B1 in either frame and the defence system is automatically activated in frame B)

3) at time t=t'= +L/v*(1-1/γ) (when A2 matches up with B2 in frame B)

Note again that all the actions here are programmed (from the outset) to occur at the given times, so no communications of any sort are involved here.

And it is obvious that the outcome is very much different depending on whether you view the situation from the A frame or the B frame.

Thomas

So you acknolwedge, then that your paradox relies on B1 having the ability to instantaneously send a signal or force to B2, well in excess of the speed of light?

DeleteFor the second time, I concede that relativity creates paradoxes

if signals and forces are allowed to move at faster than the speed of light. Describing such paradoxes doesn't help anything, since we all know they're there.And now I'll ask the question again -- can you actually describe a paradox that doesn't rely on a magical superluminal travel?

Deleteavalonxq said:

"So you acknowledge then that your paradox relies on B1 having the ability to instantaneously send a signal or force to B2, well in excess of the speed of light?"

I don't know what gives you the impression that the scenario described and pictured in my previous post would involve sending any signals at all.

But anyway, this seems to be a suitable point to bring this discussion to some sort of conclusion, as I think the issue has been covered in more than sufficient depth and detail to allow the reader to make their own judgement here.

As I said already, I intend to substantially rewrite my page regarding the Twin Paradox, taking the arguments mentioned here into account. This should bring it in a form that is less objectionable for a Relativist, whilst being a lot more compact than this discussion.

I'll drop a note here once I have done this.

Thomas

Thomas said:

ReplyDelete"I don't know what gives you the impression that the scenario described and pictured in my previous post would involve sending any signals at all."

But, like each of the paradoxes he has attempted to formulate previously, this one either ignores the relativity of simultaneity or requires something to propogate over a distance instantaneously.

Notice in Thomas's most recent diagram that the A2/B2 explosion is shown spreading slowly, as is appropriate, while the "defense shield" somehow deploys instantly at all points along the ship B, and always simultaneous to the A1/B1 intersection in all frames. This is Thomas's "magical force" in this example -- some sort of magical defense shield that can be deployed at B1 and instanteously protect B2 without apparently having to travel or signal over any distance.

If, instead, Thomas had properly drawn the "defense shield" as spreading out from B1 at the speed of light (or slower) at the instant of the A1/B1 intersection, it would be clear that no paradox resulted.

Similarly, if Thomas had shown some earlier instant where B1 programmed the "defense shield" to activate at point B2 (by sending a light speed signal or some sublight message), it would be trivial to show that this event occurred before the A2/B2 intersection in both reference frames (and, in fact, in any arbitrary reference frame as well).

The only way that Thomas can create a paradox is by, just as previously, refusing to limit himself to sublight signals and communications.

And as was explained previously, the paradoxes entailed in relativity when allowing superlight travel or instant signaling are already well-known.

Since Thomas refuses to limit himself to scenarios that actually follow the assumptions of Special Relativity, we have to conclude that the problem is in his understanding of it.

Deleteavalonxq said:

"This is Thomas's 'magical force' in this example -- some sort of magical defense shield that can be deployed at B1 and instanteously protect B2 without apparently having to travel or signal over any distance.

This is not the scenario I suggested, where as I said all events are programmed in advance to be activated

locallyat the given times. So no signals are involved at all here.Please don't make up your own scenarios here to apply your arguments.

Thomas

Thomas says:

Delete"This is not the scenario I suggested, where as I said all events are programmed in advance to be activated locally at the given times."

Then your diagrams are wrong. If B2 had been programmed at some time prior to the A2/B2 intersection to put up a defense shield, that event would have proceeded the A2/B2 intersection

in both frames, including in A's reference frame.You can see it for yourself if you calculate the t value for the programming signal (or physical programmer, or whatever mechanism you use to program B2) reaching B2 in both reference frames. If the signal reaches B2 in time to stop the explosion in either reference frame, it is in time in both reference frames. There simply isn't a scenario where you switch reference frames and suddently the shield-programming event at B2 and the bombing event at B2 happen in reverse order.

Thomas said:

"Please don't make up your own scenarios here to apply your arguments."

It's not my scenarios here under discussion; it's yours. But so far your initial assumptions violate relativity or the outcome of your scenarios is entirely consistent. I've yet to see a paradox.

Deleteavalonxq said:

You can see it for yourself if you calculate the t value for the programming signal (or physical programmer, or whatever mechanism you use to program B2) reaching B2 in both reference frames. If the signal reaches B2 in time to stop the explosion in either reference frame, it is in time in both reference frames.

Please stop reinterpreting my scenario at your convenience. As should be evident from the illustration and what I said now repeatedly,

no signals of any form are used here at all. Assume that the explosion at the A2/B2 encounter at ship time t=t'= -L/v*(1-1/γ) is triggered locally by mechanical contact pins protruding from both ships, and that the contact pin of ship B has been programmed in advance to be withdrawn at ship time t'=0. So in frame B the explosion will never happen as there will be no contact between the two pins possible when A2 and B2 match up in this frame. Yet in frame A, the explosion (and thus the destruction of the ships) has already occurred.Thomas

Thomas said:

Delete"Assume that the explosion at the A2/B2 encounter at ship time t=t'= -L/v*(1-1/γ)..."

Please explain how you calculated that the A2/B2 encounter occurs at t=-L/v*(1-1/γ) in reference frame A and also at t'=-L/v*(1-1/γ) in reference frame B.

Specifically, if the A1/B1 encounter occurs at t=0 in reference frame A and at t'=0 at reference frame B, then the above cannot possibly be true.

This comment has been removed by the author.

Delete

Deleteavalonxq said:

Please explain how you calculated that the A2/B2 encounter occurs at t=-L/v*(1-1/γ) in reference frame A and also at t'=-L/v*(1-1/γ) in reference frame B.

This is not what the illustration says: the A2/B2 encounter occurs at t=

-L/v*(1-1/γ) in reference frame A, and at t'=+L/v*(1-1/γ) in reference frame B. This is simply because it is assumed that the A1/B1 encounter occurs at t=t'=0 and because each ship sees the other ship shortened by an amountL*(1-1/γ) and moving with velocity v.

Thomas

All right. If that is what you intend, I shall alter your above statement accordingly:

Delete"Assume that the explosion at the A2/B2 encounter

at ship A time t = -L/v*(1-1/γ) and at ship B time t' = L/v*(1-1/γ)is triggered locally by mechanical contact pins protruding from both ships, and that the contact pin of ship B has been programmed in advance to be withdrawn atship B time t'=0."If you agree with my edits, then the scenario at least is clear to this point.

Assuming that each ship's clocks can be read by the other ship, what time t will it be in reference frame A when the clock at B2 reads t'=0?

Deleteavalonxq said:

Assume that the explosion at the A2/B2 encounter at ship A time t = -L/v*(1-1/γ) and at ship B time t' = L/v*(1-1/γ) is triggered locally by mechanical contact pins protruding from both ships"

Look at the illustration again: these times correspond to two different instances in time (in either reference frame). So you would have two explosions here (when in fact the first one should have destroyed both ships already).

avalonxq said:

Assuming that each ship's clocks can be read by the other ship, what time t will it be in reference frame A when the clock at B2 reads t'=0?

The middle frame of the illustration should give you the answer: it is t=0 (the time when both ships are located such that A1 matches up with B1. You can assume here that both parties have agreed about this, so when A1 and B1 match up, A knows that at this moment the contact mechanism is disabled in B's ship; but of course this is of little interest to A as the explosion has already happened in his frame.

You probably wanted to hear something else, namely the result given by the Lorentz transformation formula for t'=0 and x'=L. This would indeed give a time which would precede the explosion in A (in the first frame), but again, this would be a different time instant, which even would violate causality, as at the moment where B disables the contact mechanism in his frame, the explosion has already happened in frame A.

It should be clear from this that the Lorentz transformation merely suitably offsets the events in time so that the effect of the Lorentz contraction of the moving reference frame is being compensates for. But this is done without any regard for any possible interaction between the two systems that would result in causality issues (like in this case).

Thomas

"Look at the illustration again: these times correspond to two different instances in time (in either reference frame). So you would have two explosions here (when in fact the first one should have destroyed both ships already)."

DeleteNo, they really don't. Your continued insistence that t=t' is contrary to Special relativity, specifically to the principle of relativity of simultaneity that you continue to ignore.

When an observer of reference frame A looks down the length of B, he doesn't observe a single time for all the clocks. The clocks all have different times, which correspond to the fact that events simultaneous in one reference frame are not simultaneous in another.

And what you seem to be unwilling to acknowledge is that if t=t'=0 at the A1/B1 interaction,

this is the only place on board the two ships where the clocks agree at time t=0 and at time t'=0.Based on your most recent mechanical pin scenario, an observer on ship A will see B2 retracting its pin when its clock reads t'=0,

which will be prior to the A2/B2 intersection. In other words, both frames agree that A2 retracted its pin before the explosion would have occurred, and both frames agree that the explosion didn't happen.This moment doesn't show up on your illustration because it happens earlier than your first illustration in reference frame A, but it happens nonetheless. We can draw it from ship A's perspective and it demonstrates unequivocally that the explosion never happens.

So, if we ignore faster-than-light signals (not allowed in relativity) and stop trying to make distant clocks synchronize in multiple reference frames (they won't according to relativity), we again end up with a consistent chain of events with no contradictions or paradoxes.

What I would really like to see is for you to draw your diagrams again, and this time include actual clock times for A1, B1, A2, and B2 in each diagram.

DeleteLooking at it from ship A's perspective, where all the clocks on ship A are synchronized and the clocks on ship B are different, we can even include the earlier instant where B2 reads t'=0 and retracts its pin, showing that it's earlier than all other events.

Then from ship A's perspective, we can show the A2/B2 intersection, showing the different clock times as they meet, and the lack of extended B2 pins which mean a lack of explosion.

Then from ship A's perspective, we can show the A1/B1 intersection, showing both clocks at 0 and a debate about whether the A2/B2 intersection has happened yet.

Then from ship B's perspective, we can show all the clocks on ship B synchronized and the A1/B1 intersection, showing all B clocks at 0 and the B2 pins retracting, while the A1 clock is at 0 and all the other A clocks are further along at different points.

Then we can show from ship B's perspective the A2/B2 intersection, where the B2 pin is retracting and there's no explosion and the clocks show different times (with the A2 clock showing exactly the time expected from its advancement from the previous illustration).

We can show it all, and by actually showing the different clocks rather than hiding that information, we can clearly see how there's no contradiction, unlike the equivocation your diagrams attempt right now.

So, consider it a dare: redraw your scenario,

including the clock times for both ends at each point. Make sure to draw the instant, from ship A's perspective, where B2 reads t'=0. And, with that illustration actually complete, make your paradox argument.

DeleteThomas said:

"Look at the illustration again: these times correspond to two different instances in time (in either reference frame). So you would have two explosions here (when in fact the first one should have destroyed both ships already)."

avalonxq said:

No, they really don't. Your continued insistence that t=t' is contrary to Special relativity, specifically to the principle of relativity of simultaneity that you continue to ignore.

When an observer of reference frame A looks down the length of B, he doesn't observe a single time for all the clocks. The clocks all have different times, which correspond to the fact that events simultaneous in one reference frame are not simultaneous in another."

Are you saying that the illustration is incorrect as far as the positions and lengths of the ships at the indicated ship-times is concerned? If so, please say where and why.

avalonxq said:

Based on your most recent mechanical pin scenario, an observer on ship A will see B2 retracting its pin when its clock reads t'=0, which will be prior to the A2/B2 intersection.

So you are proposing time travel into the past to prevent the explosion from happening in A's frame? Because that's effectively what it would constitute, considering that the time for A had already progressed to t=0 and the A2/B2 intersection (the explosion) happened earlier at t=-L/v*(1-1/γ). And even if you accept this time travel scenario, you wouldn't actually get to the required time (corresponding to B's pin withdrawal) if you define the beginning of the scenario indeed at A's A2/B2 intersection. It should be clear from this that for

anyevent of a given finite duration in a certain reference frame, the Lorentz transformation will never be able to give consistency across all reference frames without suitably extending/shifting the time interval (in contradiction to the original definitions). It is just a mathematical construct that doesn't know anything about causality and the constraints the latter imposes.Thomas

"So you are proposing time travel into the past to prevent the explosion from happening in A's frame?"

ReplyDeleteNot at all. Yet again, if you do the math, you'll see that ship A will observe B2 reaching t'=0 and retracting its pin prior to t=-L/v*(1-1/γ). No time travel required at all; just a simple application of the Lorentz equations combined with the scenario that you already spelled out (having B2 retract its pin at t'=0, which occured

beforethe A2/B2 interaction in all reference frames).True or false: According to the Lorentz equations, the event where the clock at B2 reads t'=0 occurs before the A2/B2 interaction? (The answer is "true," which is why your scenario never creates a paradox.)

"Because that's effectively what it would constitute, considering that the time for A had already progressed to t=0 and the A2/B2 intersection (the explosion) happened earlier at t=-L/v*(1-1/γ)."

B2 doesn't retract its pin at t=0 according to frame A,and there is no mathematical or logical way for you to arrive at that within the constraints and assumptions of relativity.I'll ask you the question again, because I have to believe that you have the math to answer it: when you transform t'=0 at B2 into a time coordinate in reference frame A, what time t do you get? Why are you unwilling to answer this question? Is it because you know the answer is a time prior to t=-L/v*(1-1/γ)?

Why do you refuse to admit that every scenario you come up with has a perfectly ordered, non-paradoxical result in special relativity?

And why won't you draw your illustration with clock readings on both ends? Is it because it makes your attempt at equivocation too obvious? I'll do it myself, if you'll be so kind as to supply your preferred values for the velocities and lengths of the ship.

You haven't answered my question where the errors in my illustration are. Maybe because there are none?

DeleteWhat you should realize is that the clocks are in principle completely irrelevant for the timings of the events here, as these are defined by the relative positions of the space ships and their relative velocity (even the B2-withdrawal of the pin at time t'=0 could be defined without a clock, for instance by B2 having a measuring tape fixed to A1 and withdrawing the pin if this has reeled off to a length L). The point is that for each time frame (as indicated by the times on the left hand side) the times in both frames are by definition identical (as the clocks on each ship are synchronized with each other). Just for the sake of this argument though, I have produced a new illustration which shows all the clocks in each time frame displaying the same time in this sense. So if the space ships had windows, the observers would see the

exactly identical timein the other ship. The Lorentz-transformed times (to which you have always been referring to) have nothing to do with this. I have indicated here the Lorentz-transformed times for a couple of instances by the dotted green lines. As is evident, apart from the A1/B1 encounter at t=t'=0, all events require a time jump either into the future or the past in order to compensate for the frame dependent Lorentz contraction. This not only would violate causality, but also would completely compromise the physical integrity of the ships as you would need different time jumps for each different section of the ship.Thomas

Thank you for drawing these clocks, and for finally making it abundently clear that you just simply don't understand, and refuse to assume, the relativity of simultaneity required by special relativity.

DeleteJust a glance at your diagram in which you make

noLorenz transformations of timeat allmakes your error obvious to everyone, and I would urgently request that you include these clocks on every one of your diagrams in the future, so that your error will be more obvious to point out to future observers.To put it simply, the relativity of simultaneity and the Lorentz tranformation mean that your A diagrams are not consistent with your B diagrams -- if one of them is assumed to be true, the other one cannot possibly be true as well. You can, of course, assert any contradiction and paradox you want when you start with contradictory premises.

"So if the space ships had windows, the observers would see the exactly identical time in the other ship."

Sorry, no. That's absolutely impossible according to relativity.

And once you assert that it's true, you're no longer "playing by the rules" of relativity, and any paradox you create is due to your own errors.

Because it is grating to me to see something SO erroneous uncorrected, I would like to reiterate my offer. Please give me a value for the length of the ships and for their relative velocity, and I will produce a diagram that actually has the clocks showing the correct times. And I challenge you to find any contradiction or paradox in that diagram -- because, when calculated correctly (rather than just ignoring the rules as you do), there is none.

Deleteavalonxq said:

"Sorry, no. That's absolutely impossible according to relativity".

What you mean is that it is not consistent with relativity, and this is what you obviously would expect if relativity violates logic and your logic is right.

avalonxq said:

And once you assert that it's true, you're no longer "playing by the rules" of relativity, and any paradox you create is due to your own errors.

I asked you repeatedly to show me any errors in the diagram, but you haven't done so.

You also ignored what I said before and stated in the diagram, namely that the location of the ships in each frame are displayed for

identicalship times (whichby definitionmeans that all clocks for each of the three frames show identical times (with the Lorentz transformed times indicated by the green dotted line)).You also ignored my reminder that the clock times are anyway not relevant for defining the events here.

What you still don't seem to realize is that the Lorentz transformation willy-nilly transforms different points in space to different times. You may (wrongly) get away with this if you transform single uncorrelated events, but not if you transform physical structures like a space ship (as this will simply rip the ship apart in space and time). In my illustration I have done nothing that contradicts relativity, but just suitably displayed the resulting scenario to demonstrate this.

Thomas

"I asked you repeatedly to show me any errors in the diagram, but you haven't done so."

ReplyDeleteI certainly have done so. Your A and B diagrams are inconsistent with each other. You can see this immediately just from looking at the fact that, according to your first set of diagrams, A2 and B2 are next to each other and not next to each other both at the same time.

Yet again, give me actual values for the velocity and length and I will draw you diagrams where the times are actually correct.

"You also ignored what I said before and stated in the diagram, namely that the location of the ships in each frame are displayed for identical ship times (which by definition means that all clocks for each of the three frames show identical times (with the Lorentz transformed times indicated by the green dotted line))."

... which means that your diagrams are inconsistent with special relativity. Again, if you're not even going to make the effort to actually follow the rules of the theory you are trying to refute, you can't be taken seriously in your attempt to refute it.

You cannot have distant clocks synchronized in multiple different reference frames, because simultaneity in those reference frames is different. I have said this several times, just as special relativity requires, and yet you ignore it.

If you ignore an important principle of a theory and pretend like it doesn't exist, you won't be able to successfully describe the results of applying that theory.

"What you still don't seem to realize is that the Lorentz transformation willy-nilly transforms different points in space to different times."

It does nothing of the sort. It transforms events in spacetime in an entirely consistent way. You have yet to show any inconsistency when you actually follow the Lorentz equations; your problem is your complete unwillingness to do so.

I reiterate my offer yet again -- choose actual values for the velocity and length and I will give you the diagrams that show the single, consistent descriptions given by special relativity and the Lorentz equations. Then you can tell me where the paradox is.

Although Thomas has not accepted my offer, I decided to make an accurate diagram anyway. Here it is:

ReplyDeletehttp://yes2einstein.blogspot.com/#!/2012/09/ships.html

I assumed that:

a) A1 and B1 meet at 00:00,

b) each ship's clocks are synchronized in that ship's reference frame,

c) the ships' relative velocity is 0.6c, and

d) each ship is 150 light minutes long in its own frame.

From these four assumptions, the clock readings that I display are necessary in each diagram; no other clock reading is allowed or possible (and no other clock reading can be derived from relativity).

Feel free to look for a contradiction or paradox; there isn't one. That's the result you get when you follow relativity correctly without ignoring rules or making mistakes.

DeleteThomas said

"I asked you repeatedly to show me any errors in the diagram , but you haven't done so."

avalonxq said:

"I certainly have done so. Your A and B diagrams are inconsistent with each other. You can see this immediately just from looking at the fact that, according to your first set of diagrams, A2 and B2 are next to each other and not next to each other both at the same time."

They are inconsistent in this sense because the assumption of a velocity dependent length contraction is inconsistent. I have only depicted what the Lorentz transformation contains.

avalonxq said:

I decided to make an accurate diagram anyway. Here it is:

http://yes2einstein.blogspot.com/#!/2012/09/ships.html

........

Feel free to look for a contradiction or paradox; there isn't one.

There are many contradictions. I don't even want to address the more secondary (but still confusing) ones like the ship lengths not being drawn to scale or the distance travelled by the ships being irregular.

The crucial flaw is that your indicated times violate the constraint that on each ship all clocks should be synchronized i.e. show the same (ship)-time. I addressed this already a while ago in this thread: clock synchronization on ship A means t(A1)=t(A2) and clock synchronization on ship B means t'(B1)=t'(B2) at all times. If additionally you have the convention that t(A1)=0 <=> t'(B1)=0, then it is a trivial mathematical requirement that all clocks show zero at this moment. Otherwise some clocks would either not be synchronized (i.e. show the wrong time), or you would be associating different parts of the ships with different times. And the latter case is clearly paradoxical as well: if, in your top frame, you consider the point on ship A that matches up with the right side of ship B, then, if you assume the ships to be infinitesimally close, you should get a response to a signal sent out from A to B back in an infinitesimally short time. Yet, according to your drawing, the corresponding parts of ships A and B are separated by 1.5 hours in time (in the past at that as seen from A), so this could never happen.

Thomas

Thomas said:

Delete"There are many contradictions. I don't even want to address the more secondary (but still confusing) ones like the ship lengths not being drawn to scale or the distance travelled by the ships being irregular."

I'm sorry if my drawings are a little rough, but I do believe they are roughly correct. More importantly, the times are all correct based on the points that line up -- which is something that is simply not true in your diagram.

"The crucial flaw is that your indicated times violate the constraint that on each ship all clocks should be synchronized i.e. show the same (ship)-time."

And I've told you, over and over again, that this is exactly what relativity says will ABSOLUTELY NOT be true.

This is, in fact, that crucial principle that you continue to ignore and that fuels your so-called "paradoxes" for relativity -- you insist, contrary to the theory you claim to be representing, that distant clocks can be synchronized in multiple frames. They cannot. And your refusal to apply the Lorentz time transformation drives your continued error.

Which is why I said, please do us all a favor and continue to put clocks on the end points of all your "relativity" diagrams, as this will immediately show even beginners to Special Relativity exactly where your error is. When you claim to show clocks that are synchronized aboard a ship as also synchronized in a frame where the ship is moving, every first-year physics student immediately sees the mistake.

"I addressed this already a while ago in this thread: clock synchronization on ship A means t(A1)=t(A2) and clock synchronization on ship B means t'(B1)=t'(B2) at all times. If additionally you have the convention that t(A1)=0 <=> t'(B1)=0, then it is a trivial mathematical requirement that all clocks show zero at this moment."

Except that it's not mathematically trivial at all.

Let's spell this out formally to make it clear.

A1(0) is a specific event in space time, that is, the event when the A1 clock reads 0. Similarly, A2(0) is the event when the A2 clocks reads 0, and the same for B1() and B2().

We can start with the following premises:

1) t(A1(x)) = t(A2(x)) for any x

2) t'(B1(x)) = t'(B2(x)) for any x

3) t(A1(0)) = 0

4) t'(B1(0)) = 0

Those are the only actual premises given by your imposed starting conditions. If you can prove t(A2(0)) = 0 from the above, feel free. Of course, you can't, because it doesn't actually follow from your premises.

Simultaneity is relative, and distant events simultaneous in one reference frame are not simultaneous in other frames. This is what relativity says. It causes no paradoxes and contradicts no other premises.

"And the latter case is clearly paradoxical as well: if, in your top frame, you consider the point on ship A that matches up with the right side of ship B, then, if you assume the ships to be infinitesimally close, you should get a response to a signal sent out from A to B back in an infinitesimally short time."

DeleteThat's correct.

"Yet, according to your drawing, the corresponding parts of ships A and B are separated by 1.5 hours in time (in the past at that as seen from A), so this could never happen."

No, that's not what the diagram shows. When A2 and B2 meet, their clocks show different times; that doesn't mean that one of the ships is "in the past" relative to the other any more than I'm "in the past" when I'm on the phone with someone in Japan. The ship clocks just don't match.

And, as I've attempted to explain, this is because there is no way to synchronize all of a ship's clocks AND have its clocks synchronized with a moving ship. It's simply not possible; under relativity, there will always be a discrepancy.

Now that I've diagrammed the scenario correctly, I'm quite happy to conclude our dialog, if you will please promise to use clocks in your diagrams from now on. Your clocks, which clearly contradict relativity of simultaneity, make your errors obvious and make it clear to readers what part of relativity theory you have failed to apply.

I made some mistakes in my equations above for "proving" simultaneity. Allow me to rewrite.

DeleteExcept that it's not mathematically trivial at all.

Let's spell this out formally to make it clear.

A1(0) is a specific event in space time, that is, the event when the A1 clock reads 0. Similarly, A2(0) is the event when the A2 clocks reads 0, and the same for B1() and B2().

We can start with the following premises:

1) t(A1(x)) = t(A2(x)) = x for any x

2) t'(B1(x)) = t'(B2(x)) = x for any x

3) t'(A1(0)) = 0

4) t(B1(0)) = 0

Those are the only actual premises given by your imposed starting conditions. If you can prove t'(A2(0)) = 0 or t(B2(0)) = 0 from the above, feel free. Of course, you can't, because it doesn't actually follow from your premises.

Deleteavalonxq said:

A1(0) is a specific event in space time, that is, the event when the A1 clock reads 0. Similarly, A2(0) is the event when the A2 clocks reads 0, and the same for B1() and B2().

We can start with the following premises:

1) t(A1(x)) = t(A2(x)) = x for any x

2) t'(B1(x)) = t'(B2(x)) = x for any x

3) t'(A1(0)) = 0

4) t(B1(0)) = 0

Those are the only actual premises given by your imposed starting conditions. If you can prove t'(A2(0)) = 0 or t(B2(0)) = 0 from the above, feel free. Of course, you can't, because it doesn't actually follow from your premises.

It appears just you can't, because your equations 3) and 4) not only have nothing to do with 1) and 2) but are just plain wrong: the primed times only belong to the B-frame; in principle you could leave the primes off altogether, but I just wrote them because in general you may not have identical clock rates in the two frames, or the clocks may have started at different times. In any case, there are only four clocks here, so only 4 times to consider for the synchronization, namely the times mentioned in 1) and 2). And if at a specific location (in this case x-0), the left hand side of both 1) and 2) is zero, then so must be the right hand sides.

Thomas said:

"Yet, according to your drawing, the corresponding parts of ships A and B are separated by 1.5 hours in time (in the past at that as seen from A), so this could never happen."

avalonxq said:

No, that's not what the diagram shows. When A2 and B2 meet, their clocks show different times; that doesn't mean that one of the ships is "in the past" relative to the other any more than I'm "in the past" when I'm on the phone with someone in Japan.

So if you can confirm that the right side of ship B has as much physically existence as the left side at A's ship time 00:00, then please explain what the meaning and physical relevance of your annotation 22:30 is here (in your top frame). And explain how you manage to produce a diagram where the x-axis is apparently also a t-axis.

Thomas

This comment has been removed by the author.

Delete"It appears just you can't, because your equations 3) and 4) not only have nothing to do with 1) and 2) but are just plain wrong: the primed times only belong to the B-frame;"

DeleteI don't think it's clear what I meant by my terms, so I'll clarify.

t() represents times in the A frame, and t'() represents times in the B frame.

So equation 3 means "the time according to frame B when A1 reads 0 is t'=0," and equation 4 means "the time according to frame A when B1 reads 0 is t=0." This is to say, the clocks at both A1 and B1 read 0 at time 0 in both reference frames.

But what you're claiming, and what is clearly untrue according to special relativity, is that B2 will also read 0 at time t=0 according to frame A, and A2 will also read 0 at time t'=0 according to frame B. This simply isn't true, as the Lorentz tranformation equations give different times for these events.

And there is no "mathematically trivial" way for you to show that t'(A2(0))=0 or t(B2(0))=0, because both of those statements are false.

"In any case, there are only four clocks here, so only 4 times to consider for the synchronization, namely the times mentioned in 1) and 2)."

There are four locations. Each event that occurs in one of those locations has a time t in frame A and a time t' in frame B. If we care about properly describing all four locations in both reference frames, we have at least eight times to worry about -- for example, t(A1(0)), t(A2(0)), t(B1(0)), t(B2(0)), t'(A1(0)), t'(A2(0)), t'(B1(0)), and t'(B2(0)).

Now, by assumption, t(A1(0)) = t(A2(0)) = t(B1(0)) = t'(A1(0)) = t'(B1(0)) = t'(B2(0)) = 0. But this in no way proves that t(B2(0)) = 0 or t'(A1(0)) = 0, and in fact these two equations do not and cannot hold true under Special Relativity.

Feel free to prove them with "trivial mathematics" if you can.

"And if at a specific location (in this case x-0), the left hand side of both 1) and 2) is zero, then so must be the right hand sides."

Not in all reference frames. Clearly it is not generally the case that t=t', and your assertions contrary to Special Relativity simply don't hold water.

"So if you can confirm that the right side of ship B has as much physically existence as the left side at A's ship time 00:00, then please explain what the meaning and physical relevance of your annotation 22:30 is here (in your top frame)."

I don't know how much more clearly I can explain this. In A's reference frame, B is moving at 0.6c to the right. At the instant when all of A's clocks read 00:00, B1 reads 00:00 and B2 reads 22:30. This is a consequence of our premises, namely that B1 reads 00:00 by assumption and B2 is synchronized with B1 according to ship B's frame of reference.

This is exactly what the Lorentz equations and Special Relativity describe and require, and they lead to no paradoxes.

Deleteavalonxq said:

t() represents times in the A frame, and t'() represents times in the B frame.

This is not quite correct. As far as the synchronization equations are concerned, t() represents the time of A-clocks, t'() the times of B-clocks. So in principle the prime is redundant and you could just write for the synchronization condition between the 4 clocks here

1) t(A1)=t(A2)

2) t(B1)=t(B2)

If we now have the additional constraint that t(A1)=t(B1), then one must have t(A2)=t(B2).

Thomas said:

"In any case, there are only four clocks here, so only 4 times to consider for the synchronization, namely the times mentioned in 1) and 2)."

avalonxq said:

There are four locations. Each event that occurs in one of those locations has a time t in frame A and a time t' in frame B. If we care about properly describing all four locations in both reference frames, we have at least eight times to worry about.

Again, there are four clocks, and four clocks can only show four times, namely those mentioned in the above equations 1) and 2)

avalonxq said:

But what you're claiming, and what is clearly untrue according to special relativity, is that B2 will also read 0 at time t=0 according to frame A, and A2 will also read 0 at time t'=0 according to frame B. This simply isn't true, as the Lorentz transformation equations give different times for these events.

I am not claiming something. I am just observing that the Lorentz transformation is inconsistent with the constraint that the clocks in each reference frame should be synchronized. They can only be synchronized in one frame but not the other (unless the frames are not moving relatively to each other).

avalonxq said:

I don't know how much more clearly I can explain this. In A's reference frame, B is moving at 0.6c to the right. At the instant when all of A's clocks read 00:00, B1 reads 00:00 and B2 reads 22:30. This is a consequence of our premises, namely that B1 reads 00:00 by assumption and B2 is synchronized with B1 according to ship B's frame of reference.

How do you explain the two different time readings for B2 then? Assume that B2 broadcasts continuously its time by sending out time-stamp signals (similar like GPS satellites do). Assume two observers located infinitesimally close to B2 at the same instant, only one resting relatively to B2 and the other not. How can it be that the former reads 00:00 for the time stamp 00:00 and the latter 22:30 for what only could be the same time stamp?

Thomas

"Again, there are four clocks, and four clocks can only show four times, namely those mentioned in the above equations 1) and 2)"

I have one clock that can show any one of thousands of times, so this idea that "four clocks equals four times" is, in fact, your clever way of begging the quest