Questions about Einstein SR interpretation

[jtbell]

SR theory and æther theory are indistinguishable from each other.

Better: "SR theory and a particular version of aether theory are indistinguishable from each other."

 

[Simplyh]

What specific section of the 1905 paper (viewable here)


There is no "conversion factor" of the type you suggest: ... The correct equation is:

t' = gamma*(t - vx/c^2)

where v is the velocity of the train in the ground frame, and gamma = $$\frac{1}{\sqrt{1 - v^2/c^2}}$$

So for example say the train is moving with v=0.6c, meaning gamma=1.25. Then say lightning strike A happens at position x=0 light-seconds, t=0 seconds in the ground frame, while lightning strike B happens further along the x-axis at the same moment, at position x=8 light-seconds, t=0 seconds. In that case the time coordinate t' of strike A in the train frame would be:

t' = 1.25 * (0 - 0.6*0) = 0 seconds

While the time coordinate t' of strike B in the train frame would be:

t' = 1.25 * (0 - 0.6*8) = 1.25*(-4.8) = -6 seconds

So you can see that according to the SR equation above (which is just part of the general Lorentz transformation for relating coordinates in one frame to coordinates in another), strike A and B happened 6 seconds apart in the train frame.

Any two events with coordinates (x,t) equal to (0,0) and (x,0), with x <>0 are not simultaneous in the same reference frame.
If we have two events at distance, then “... it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an ``A time'' and a ``B time.'' We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A. Let a ray of light start at the ``A time'' from A towards B, let it at the ``B time'' be reflected at B in the direction of A, and arrive again at A at the ``A time'' .
In accordance with definition the two clocks synchronize if TB – TA = T'A – TB “(Einstein – 1905 paper).
So, according to this definition from Einstein any events (x,t) are only simultaneous in the same reference frame if tx = x/c. This means that your events would only be simultaneous if one had coordinates (0,0) and the other (8,8/c). Applying now the Lorentz transformation for time we get:
t' = 1,25 * (8/c – 0,6*8/c) = 1,25*8/c(1-,6) = 1,25*8/c*,4 = 4/c
Let's remember that a lightning took place at coordinates (8,8/c) in the platform frame. The same lightning has as coordinates in the train frame (x';t') = (x';4/c). But to be simultaneous with event (0,0) in the train frame, the time at that point must be t' = x'/c. So:
x'/c = gamma* (x-vt)/c = gama*(x-vx/c)/c = 1,25*(8-0,6c8/c)/c = 1,25*(8/c*(1-,6)) = 1,25*8/c*,4 = 4/c
And so both lightenings are simultaneous in both reference frames.
Indeed, this is equivalent to say that if S and S' are two inertial systems and we place the origin of S at any event in space-time, then, the corresponding event on S' will be given by the contraction factor. Or, in other words: for any event in space-time S, with coordinates (x,y,z,t), we'll have a corresponding event (x',y',z',t'), on space time S', whose coordinates are given by the contraction factor.
Let's please go back to the 1905 experiment which is easier to understand because everything happens in the same reference frame.
We have a solid arm with two clocks and two observers at extremities A and B. At A there is a light source; at B a mirror. Let a light beam come from A, be reflected at B and go back to A. Time for light to travel from A to B should always be equal to time for light to travel from B to A because all happens in the same reference frame. But it is not so: if the frame is stationary relatively to Earth this is correct; if the frame displaces relatively to Earth than this is not correct. This obviously means that light does not travel at the same speed in all directions in that inertial system whenever that system travels relatively to Earth. Why?

 

[JesseM]

Any two events with coordinates (x,t) equal to (0,0) and (x,0), with x <>0 are not simultaneous in the same reference frame.

What do you mean "in the same reference frame"? They are certainly simultaneous in the frame where the first event had coordinates (0,0) and the second had coordinates (x,0), because all "simultaneous" means is "same time coordinate", and both have a time coordinate of 0 in this frame. In a second inertial frame in motion relative to this one, the events would be assigned different coordinates and their time coordinates might no longer be equal, in which case they'd be non-simultaneous in the second frame.

If we have two events at distance, then “... it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an ``A time'' and a ``B time.'' We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A. Let a ray of light start at the ``A time'' from A towards B, let it at the ``B time'' be reflected at B in the direction of A, and arrive again at A at the ``A time'' .
In accordance with definition the two clocks synchronize if TB – TA = T'A – TB “(Einstein – 1905 paper).
So, according to this definition from Einstein any events (x,t) are only simultaneous in the same reference frame if tx = x/c.

How do you get that from Einstein's definitions? The units in your equation tx = x/c don't make sense since the left side has units of time*distance (unless you mean to write t_x, the x just being a subscript) while the right has units of time. And your equation doesn't seem to make sense as a definition of simultaneity, because it seems to deal only with a single set of x and t coordinates, while simultaneity always deals with a pair of events with different coordinates, we might call them (x1, t1) and (x2, t2). And Einsteins' definition above was really dealing with three different events which could be labeled (XA, TA), (XB, TB) and (X'A, T'A). The first event is light being sent from clock A, the second event is the light beam being reflected from clock B, and the third event is the reflected light being received by clock A. Since these coordinates are supposed to in the rest frame of the clocks, XA = X'A, the distance between clocks A and B is (XB - XA), and if the clocks are synchronized in this frame then TB - TA = (XB - XA)/c. Of course, this assumes the coordinate system we are using to describe the events is the same as the one where the clocks are at rest and we want to synchronize them, if we want to apply Einstein's definition using coordinates of a frame that is not the rest frame of the two clocks, translating the definition into equations involving the frame's coordinates would be more complicated (see below for an example).

This means that your events would only be simultaneous if one had coordinates (0,0) and the other (8,8/c).

You mean, they would be simultaneous in the train frame if those were the coordinates in the platform frame? Again you can only talk about events being "simultaneous" relative to a particular choice of frame, never in an absolute sense!

But even if that is what you meant, your math is wrong. If the train is moving at 0.6c relative to the platform frame, and its length is 8 light-seconds in the platform frame, then the event with coordinates in the platform frame x=12.5, t=7.5 will occur at the front end of the train and will be simultaneous in the train frame with the event at the back end of the train at coordinates x=0, t=0 in the platform frame. In other words, if a clock A at the back of the train reads 0 at x=0, t=0, then a clock B at the front of the train must read 0 at x=12.5, t=7.5 in order for it to be "synchronized" with clock A in the train's own rest frame.

With a little analysis we can show that this does make sense in terms of Einstein's definition. Suppose we have a clock A at the back of the train and a clock B at the front, and their worldlines in the coordinates of the platform frame are giving by x=0.6c*t for clock A, and x=0.6c*t + 8 for clock B (these equations ensure that they are both moving at 0.6c, and that at t=0 clock A is at x=0 while clock B is at x=8). Suppose at t=-12.5 seconds in the platform frame, a light signal is sent from clock A; by the equation x=0.6c*t, clock A must be at position x=-7.5 light-seconds when the signal is sent. We assume that at x=0, t=0 clock A read a time of zero, but clock A is running slow by a factor of $$\sqrt{1 - 0.6^2}$$ = 0.8 in the platform frame, so 12.5 seconds earlier at t=-12.5 seconds, clock A read a time of 0.8*-12.5 = -10 seconds, and we call this reading TA in Einstein's notation.

If a light beam is emitted at x=-7.5, t=-12.5 in the platform frame, the light beam's position as a function of time in this frame must be x = c*t + 5 (this ensures that the light moves at a coordinate speed of c, and that at t=-12.5 it is at position x=-12.5 + 5 = -7.5). So if the front of the train has position as a function of time given by x=0.6c*t + 8, then to figure out when the light beam will catch up with the front, we set these equal, giving c*t + 5 = 0.6c*t + 8, and solving for t gives t = (8 - 5)/(1c - 0.6c) = 3/0.4c = 7.5 seconds. If you plug this time into either the light beam equation or the equation for the front of the train, you find that the x-coordinate of the light catching up to the front of the train must be x=12.5 light-seconds. And remember, I said above that the clock B should read 0 at exactly these coordinates in the platform frame in order for it to be "synchronized" with clock A in the train frame. So, in Einstein's notation we can say that TB = 0, since TB is supposed to be the time on clock B when the light hits it and is reflected back to A.

If the light beam is reflected at these coordinates, then its speed remains constant but its direction changes, so its new position as a function of time is given by x = -c*t + 20 (this ensures that the light is now moving at -c along the x-axis, and that at t=7.5 it has position x=-7.5 + 20 = 12.5 light-seconds). And the position as a function of time for clock A was x=0.6c*t, so to find when the reflected light gets back to clock A we set them equal, giving -c*t + 20 = 0.6c*t, solving for t gives t = 20/(0.6c + 1c) = 12.5 seconds. Since clock A read 0 at t=0, and it's running slow by a factor of 0.8, at t=12.5 seconds clock A reads a time of 12.5*0.8 = 10 seconds. So in Einstein's notation, we have T'A = 10 seconds.

Putting it all together, with TA = -10 seconds, TB = 0 seconds, and T'A = 10 seconds, you can see that Einstein's equation TB – TA = T'A – TB is indeed satisfied here. Thus, in order for clocks A and B to be synchronized in their own frame, if clock A reads 0 at x=0, t=0 in the platform frame, clock B must read 0 at x=12.5 light-seconds, t=7.5 seconds.

Now to check that this matches the Lorentz transformation. Transforming from the platform frame to the train frame, x=0, t=0 in the platform frame corresponds to x'=0, t'=0 in the train frame. As for x=12.5, t=7.5, plugging into the Lorentz transformation gives:

x' = gamma*(x - vt) = 1.25 * (12.5 - 0.6*7.5) = 10
t' = gamma*(t - vx/c^2) = 1.25 * (7.5 - 0.6*12.5) = 0

So, both events have a time coordinate of t'=0 in the train frame, so they are indeed simultaneous in this frame according to the Lorentz transformation.

Let's please go back to the 1905 experiment which is easier to understand because everything happens in the same reference frame.
We have a solid arm with two clocks and two observers at extremities A and B. At A there is a light source; at B a mirror. Let a light beam come from A, be reflected at B and go back to A. Time for light to travel from A to B should always be equal to time for light to travel from B to A because all happens in the same reference frame.

"time" in what frame? If the device is moving in the direction of B in your frame, then in your frame the time from A to B is not equal to the time from B to A, because B is moving away from the light and A is moving towards the light, therefore the light has a longer distance to travel to go from A to B then it does to go from B to A, and since it travels at the same speed in both directions in your frame, the time to go from A to B must be larger. But if there are clocks attached to A and B, and these clocks are "synchronized" in the device's own frame, then if TA is the reading of the clock at A when the light leaves it, TB is the reading of the clock at B when the light hits it, and T'A is the reading of the clock at A when the light returns to it, the equation (TB - TA) = (T'A - TB) will be satisfied. In your frame this is because the clocks are out-of-sync, so even though the time in your frame between the first pair of events is different than the time in your frame between the second pair, the actual readings on the clocks at each event will satisfy (TB - TA) = (T'A - TB). In the device's own rest frame the clocks are "really" synchronized", and the reason (TB - TA) = (T'A - TB) is that the coordinate time between each pair of events really is the same in the device's frame, since the device is at rest and the light travels the same distance in both directions in this frame.

But it is not so: if the frame is stationary relatively to Earth this is correct; if the frame displaces relatively to Earth than this is not correct. This obviously means that light does not travel at the same speed in all directions in that inertial system whenever that system travels relatively to Earth.

Wrong, see above. The light does travel at the same speed in all directions in the device's rest frame, since that frame defines distance/time in terms of distance as measured by something at rest in the frame (like the device itself), and assigns time coordinates to events using clocks which are considered to be "synchronized" in that frame (like the pair of clocks at either end of the device). So the distance between the two pairs of events above will be equal, and the time between each pair will also be equal since it is true of the clock readings TA, TB and T'A that (TB - TA) = (T'A - TB).

 

[ghwellsjr]

Better: "SR theory and a particular version of aether theory are indistinguishable from each other."

I was paraphrasing the quote from JesseM's link (see post #35) which says "all are experimentally indistinguishable from SR", not just one particular theory.

 

[JesseM]

I was paraphrasing the quote from JesseM's link (see post #35) which says "all are experimentally indistinguishable from SR", not just one particular theory.

But they were referring to a particular class of aether theories (the ones I would call 'interpretations' rather than 'theories') which can't be ruled out by the experiments listed, they weren't saying that all aether theories in general are indistinguishable from SR. From the link:

Note that while these experiments clearly use a one-way light path and find isotropy, they are inherently unable to rule out a large class of theories in which the one-way speed of light is anisotropic. These theories share the property that the round-trip speed of light is isotropic in any inertial frame, but the one-way speed is isotropic only in an æther frame. In all of these theories the effects of slow clock transport exactly offset the effects of the anisotropic one-way speed of light (in any inertial frame), and all are experimentally indistinguishable from SR.

In particular, imagine an aether theory where Lorentz contraction of objects moving relative to the aether frame still happens, but time dilation of clocks moving relative to the aether frame doesn't happen. In this case all experiments which look at the two-way speed of light in different directions, like the Michelson-Morley experiment, would still show it has the same two-way speed in all directions regardless of the motion of the device (without time dilation the precise value of that two way-speed would be different depending on the device's motion but then the Michelson-Morley experiment didn't try to measure the value, just whether it was the same or different along different arms of the device). But experiments like the ones listed, where the clocks were synchronized at a single location and then moved apart, would give different results than SR predicts when the one-way speed was measured.

 

[JesseM]

Did you mean predictions still in the future or predictions in Einstein's future that have now been fulfilled?

Predictions about all possible fundamental laws discoverable by experiment, predictions which have so far proven correct when we look at all the most fundamental laws of physics found since (the standard model of particle physics, for example). In modern terms I would say he was predicting that all the fundamental laws of physics would turn out to be Lorentz-invariant, that is, if you write the equations of the laws of physics in the coordinates of one inertial frame and then apply the Lorentz transformation, the equations in the new frame will be unchanged (and of course the Lorentz transformation can itself be derived from the two postulates of SR). This is a physical prediction which could quite easily be falsified, since it's possible to write down all sorts of equations that are not Lorentz-invariant (Newtonian gravity for example), and thus if you used the Lorentz transformation

Either way, this is the first I have ever heard of this new definition of "postulate" and I totally disagree. Einstein was not making a prediction, he was stating an assumption that could not be proven or measured either then or now.

It's not an "assumption" that the laws of physics are Lorentz-invariant. And if they weren't (if Newton's gravity was correct, say), then it would be impossible to come up with a coordinate transformation giving a family of coordinate systems that satisfied both of the two postulates. Do you disagree?

Didn't you read your own link? It says: "In all of these theories the effects of slow clock transport exactly offset the effects of the anisotropic one-way speed of light (in any inertial frame), and all are experimentally indistinguishable from SR."

See my previous post #40, they weren't talking about all possible aether theories.

I wasn't talking about an "interpretation" of SR, I was talking about what scientists believed before SR. You could say SR was the new alternative physical theory to the æther theory that everyone believed in prior to SR. But, as your link pointed out, SR theory and æther theory are indistinguishable from each other.

No they aren't, since before SR scientists had no reason to suppose that all physical rulers would shrink by gamma when moving relative to the ether (even ones not based on electromagnetic forces), or that all physical clocks would slow down by gamma when moving relative to the ether. If you add these assumptions then aether theories do become experimentally indistinguishable from SR, but they would be rather strange assumptions prior to SR, there's no a priori reason to think they'd be true in an aether theory.

 

[ghwellsjr]

Prior to SR in 1905, time dilation and length contraction were the explanations for the null result of MMX and some follow-on experiments. It was the presumed physical contraction of the apparatus that was believed to have caused the null result.

It is true that Maxwell did not correctly understand the implications of his equations and believed in an absolute æther and that an experiment could have detected an æther wind, but that shouldn't detract from the fact that the prediction could have been made from his equations that there wouldn't have been a detectible æther. Who knows how he would have responded to MMX since he died even before the earlier experiment that Michelson performed. Maybe he would have tenaciously hung on to the idea of an æther or maybe he would have been the one to have invented Special Relativity many years before Einstein. Who knows?

 

[Dale]

JesseM and ghwellsjr, I don't really follow your present argument, but a postulate is an assumption by definition.

 

[JesseM]

Prior to SR in 1905, time dilation and length contraction were the explanations for the null result of MMX and some follow-on experiments. It was the presumed physical contraction of the apparatus that was believed to have caused the null result.

Only length contraction was needed to explain the null result of the MMX--are you saying that anyone had actually proposed that all clocks also experience time dilation when moving relative to the aether? I know Lorentz gave the Lorentz transformation prior to Einstein, but my impression was that he just saw it as a sort of interesting mathematical fact that Maxwell's laws would be invariant under such a transformation, I don't think anyone prior to Einstein proposed that all physical laws would be Lorentz-invariant and thus that it would be absolutely impossible for any experiment to give a different result depending on one's velocity relative to the aether.

It is true that Maxwell did not correctly understand the implications of his equations and believed in an absolute æther and that an experiment could have detected an æther wind, but that shouldn't detract from the fact that the prediction could have been made from his equations that there wouldn't have been a detectible æther.

No, it couldn't. It is perfectly conceivable that purely electromagnetic phenomena will be the same under the Lorentz transformation, but that the forces governing the matter that makes up physical clocks and rulers are not purely electromagnetic, and that the equations for at least some of these non-electromagnetic forces would not be Lorentz-invariant. In this case one might have rulers that didn't shrink when moving relative to the aether, and clocks that didn't slow down. Einstein's insight was in predicting that all fundamental physical laws would turn out to be Lorentz-invariant too, not just electromagnetic ones.

Who knows how he would have responded to MMX since he died even before the earlier experiment that Michelson performed. Maybe he would have tenaciously hung on to the idea of an æther or maybe he would have been the one to have invented Special Relativity many years before Einstein. Who knows?

True, we don't know, but my point is not to denigrate Maxwell, just to point out that in actual history Einstein was the first to publish this idea, and that taken together the two postulates of SR do constitute an actual physical prediction (that all fundamental laws are Lorentz-invariant) and cannot simply be considered "true by definition".

 

[JesseM]

JesseM and ghwellsjr, I don't really follow your present argument, but a postulate is an assumption by definition.

What do you mean by "assumption", are you suggesting that Einstein's postulates don't constitute an actual physical prediction about the way the laws of physics work? As I just said to ghwellsjr, my point is that taken together the two postulates are equivalent to the modern statement that all the fundamental laws of physics are Lorentz-invariant, and I'd say that this is a physical prediction which we could easily imagine being falsified by experiment (in which case it would be impossible to define a set of 'inertial frames' in which both postulates of SR would hold true). Do you disagree?

 

[Dale]

As I just said to ghwellsjr, my point is that taken together the two postulates are equivalent to the modern statement that all the fundamental laws of physics are Lorentz-invariant, and I'd say that this is a physical prediction which we could easily imagine being falsified by experiment (in which case it would be impossible to define a set of 'inertial frames' in which both postulates of SR would hold true). Do you disagree?

I agree that the two postulates are equivalent to the modern statement that all the fundamental laws of physics are Lorentz-invariant. I just don't know why you wouldn't consider that a (testable) assumption.

 

[JesseM]

I agree that the two postulates are equivalent to the modern statement that all the fundamental laws of physics are Lorentz-invariant. I just don't know why you wouldn't consider that a (testable) assumption.

I would definitely consider it testable! If for you an "assumption" can be tested then I have no problem with your use of the word. But ghwellsjr seemed to have been using the words "assumption" and "postulate" differently in post #35, to mean something that is true as a matter of the definitions used, and is not testable:

Either way, this is the first I have ever heard of this new definition of "postulate" and I totally disagree. Einstein was not making a prediction, he was stating an assumption that could not be proven or measured either then or now. His postulate was the same as saying, "in any inertial frame, you can assume that you are at rest with the æther", which is a ridiculous thing to say, but it works just as well as the previously assumed postulate which was "you can assume that there exists just one inertial frame in which you can be at rest with the æther".

This is what the disagreement between myself and ghwellsjr here was about.

 

[Simplyh]

You can discuss any frame you like. That is the whole point of the first postulate.

In the lab frame a proton from the left has a four-momentum of (3.5, 3.49999987, 0, 0) TeV/c and the proton from the right has a four-momentum of (3.5, -3.49999987, 0, 0) TeV/c so by conservation of four-momentum the product has a four-momentum of (7.0, 0, 0, 0) TeV/c corresponding to a mass of 7.0 TeV/c² and a kinetic energy of 0 TeV.

In the rest frame of the proton from the right a proton from the left has a four-momentum of
(25682.97778423260, 25682.97778423259, 0, 0) TeV/c and the proton from the right has a four-momentum of (0.00095393919, 0, 0, 0) TeV/c so by conservation of four-momentum the product has a four-momentum of (25682.97873817180, 25682.97778423259, 0, 0) TeV/c corresponding to a mass of 7.0 TeV/c² and a kinetic energy of 25675.97873817180 TeV.

You are right! I was wrongly assuming that Px would be zero also in the Proton's frame which of course is absurd. I stepped on this and decided to include it with only a brief superficial analyses. This is the result of rushing. Sorry: nothing can be concluded after this experiment about the relative speed of the two particles: according to SR they will continue to have a relative speed minor than c.

 

[Simplyh]

There have been experiments to measure the one-way speed of light and check that it's still c, see here.

I've tried to see where you mention but could not find any experiment measuring speed of light on a frame moving relatively to Earth. Could you please spare me time and identify the experiment? Thanks

 

[Simplyh]

What do you mean "in the same reference frame"? ... will also be equal since it is true of the clock readings TA, TB and T'A that (TB - TA) = (T'A - TB).

You understood me wrong a lot o times. Perhaps it's my fault. Of course in tx the x is a subscript. I don't know how to do subscripts here. Tx was meant to be all of them: t1, t2, …, tn.

You say: “Since these coordinates are supposed to in the rest frame of the clocks, XA = X'A, the distance between clocks A and B is (XB - XA), and if the clocks are synchronized in this frame then TB - TA = (XB – XA)/c.” This is quite correct and means coordinate time of B minus coordinate time of A equals coordinate x of B minus coordinate x of A divided by c. If we are comparing all events synchronized with event (0,0), than A coordinates will be (0,0) and B coordinates (x,(x-0)/c) = (x,x/c).
So, do we agree that, in the same reference frame, all events simultaneous to the origin (0,0) have coordinates (x, x/c)?

Of course I was applying that to the rest frame (the station, not the train). So I will ignore all your development assuming that I was applying it to the train frame which, we both agree, is wrong.

So you must apply the Lorentz transformation to the coordinates of B in S; not to the coordinate x of B and the coordinate t of A as you have done. If you do so and then apply Einstein's concept of simultaneity at distance you will see that events B' and A' are as simultaneous as B and A. But, the time coordinates of A and B are not the same; the same applies to the time coordinates A' and B'.

Let's go to the 1905 experiment. You say: “"time" in what frame? If the device is moving in the direction of B in your frame.” This is not correct: my frame is the devices' frame where everything is at rest even when it displaces relatively to Earth. All the objects in the experiment are at rest relatively to the devices frame! Please don't make any calculations; just tell me what is moving considering the device's frame? The observer in A, clock in A, light source in A? Observer in B, clock in B, mirror in B? My frame can be moving relatively to whatever I want but everything in the experiment will still be at rest relatively to that frame. If everything is at rest how can light speed not be the same on both directions?

 

[ghwellsjr]

I would definitely consider it testable! If for you an "assumption" can be tested then I have no problem with your use of the word. But ghwellsjr seemed to have been using the words "assumption" and "postulate" differently in post #35, to mean something that is true as a matter of the definitions used, and is not testable:

This is what the disagreement between myself and ghwellsjr here was about.

Please followup on any more discussion with me on this new thread:

https://www.physicsforums.com/showthread.php?p=2938903#post2938903