The physics of length contraction
Question 1
 Is it possible to establish an absolute rest frame ?

Background question 1
One of the concepts of the Special Relativity is a rest frame
A frame in uniform motion in which an observer is at rest is called a rest frame. At rest means that the observer has a speed 0 relative in its (rest) frame. The length of a rod at rest in its (rest) frame is called its proper length. The time of a clock at rest in its (rest) frame is called its proper time.
A rest frame is important
 Because all the laws of Physics are the same in every rest frame.
 For example: The length of any rod with has a speed v in its frame has a length of L = L0*SQR(1vē/cē). This is called Length Contraction.
 For example: The time of a clock which has a speed v in its frame shows a time of T = T0*SQR(1vē/cē). This is called Time Dilation.
Special Relativity does not use the concept of an absolute rest frame.
The concept absolute rest frame implies that not all the frames in uniform motion are identical i.e. not all the rest frames are identical. In fact it assumes that there is one frame which is in absolute rest. Only in that frame the above formulas apply. The speed v measured in the absolute rest frame is called absolute velocity.
Studying the Moving Train Experiment by Ray d'Inverno in the book "Introducing Einstein's Relativity" and the concept of "The Relativity of simultanity" at page 23 I come to the conclusion that part of this experiment is not as clear as the author seems to believe.
IMO this train (carriage) experiment is rather complex, much more is at stake.
IMO what the experiment shows is that it is possible to detect an absolute frame if you take signal strength into account.
The same problem was discussed in the newsgroup: news:sci.physics.relativity. See: Train Thought Experiment
Description of Experiment One with Two Trains
In order to understand why I think that this experiment demonstrates absolute rest frame and absolute velocity let us study the experiment in a more symmetrical fashion.
Starting point are two trains AB and A'B' and one track of indefinite length.
In the book only A'B' is a train or carriage.
Train AB is at rest. Train A'B' has a constant speed v to the right.
In the initial state at t=0 train A'B' is left from train AB.
The following sketch shows what happens at t=t when the two trains meet.

v>
A'M'B'
X1X2 Track
AMB
\ / v=0
\ /
\ /
\ /
\ /
\ /
\ /
M


Figure 1
At train AB there is one Observer M. At train AB there is an Observer M'. Each observer
stands at the center of the train.
When the two trains meet there is one spark at point X1 and one spark at point X2.
In the book at point X1 and X2 there are Electrical Devices at the Track and Lamp at point A' and point B' of the moving train.
When the moving train A'B' goes over these Electrical Devices they fire and activate the lamps. I use sparks, which is the same.
In the book what will happen next is that M sees the two sparks simultaneous and M' not.
The reason for M' is that M' moves to the right. M' sees the spark from B' before seeing
the spark from A'. This is called Relativity for Simulaneity. Here the experiment in the book ends.
In reality the experiment is not as easy as it is.
 Because train A'B' is a moving, Length Contraction is at stake. That means the length L0' of train A'B' at rest is longer than the length L0 of train AB. What this means
that for each specific length L0' there is only one specific speed v for which this experiment works.
 The length L0' of A'B' in rest frame of AB has to be L0/SQR(1vē/cē)
 For observer M' length contraction is not an issue. When You start the Experiment at t=0 with M' equidistant between A'and B' with v=0 than M' will stay centered (equidistant) between A' and B' because length contraction wil be equal for the distance A'M' and M'B'
 You can place a clock near A and B. In order to use those clocks they have to be synchronised. Clock synchronisation means you sent a pulse signal from M. When an observer at A sees the pulse he resets his clock. For an observer at B the same applies.
With those clocks (as part of train AB) you will be able to establish, that the time when A meets A' and B meets B', is the same.
 With clocks at A' and B' you can do the same. Again those clocks have to be synchronised. It is important to note that clock synchronisation is done with speed v.
With those clocks (as part of train A'B') you will be able to establish, that the time when A meets A' and B meets B', is not the same.
 Now we come to the real issue of this experiment: What is the position of M' in order to see the sparks from A' and B' simultaneous?
When M' is equidistant from A' and B' than M' will see the spark from B' first, because the train moves to the right. That means in order to see both sparks simultaneous M' has to move towards the left. To establish the correct position will take a lot of tests.
 In order to establish that M' is now not any more equidistant we use a rod with equal length of train A'B'. This rod moves with the train. We cut this rod at M' in two parts. When you place those 2 parts side by side you will see that they are not of equal length.
When you stop train A'B', with v=0, the same is true; both parts are not equal.
 You can do the same experiment with a third train A3B3, with obsever M3, with speed v3 and with a rod of the same length as the train. Again you establish the position where M3 sees the two sparks simultaneous and you cut the rod in two parts. The result is identical as with train A'B': the two parts are not equal.
 You can do the same experiment with any train AnBn. The result is always the same for each rod: the two parts are not equal.
 This raises an interesting observation: For all the trains used we establish the position where the observer sees the two sparks simultaneous and we see that the observer is not in the center. The only exception is train AB where the observer sees the sparks simultaneous. What makes this train so special?
 IMO there is nothing special about this train. We only used train AB as our reference frame and we specified that is speed is zero.
You can also use train A'B' (with speed v') as your reference frame. Nothing will change in your experiment. The observer M' stays at same position and will see the sparks simultaneous, but the length of the 2 parts of the rod is still not equal.
 IMO if you use a rod with train AB and you cut the rod in two parts, where M is, than you will see that both parts are not equal.
 IMO the above sketch is misleading. IMO there is only one frame where the above sketch is true. There is not any position on Earth where that is true i.e. where the rods are of equal length and where you see the sparks simultaneous.
Experiment One with Two Trains, M simultaneous, Frame M
De following, rather complicated sketch, tries to explain this.
The bottom half shows what happens for train AB. Later time goes down
The top half shows what happens for train A'B'. Later time goes up.

/ / /
/ . .
/ ./ ./
/ . / . /
. . / . /
. / . . / . /
. / .x. / . /
. . . . /
/ .. / .. / v>
/. . /. . /
A'M'B'
X1X2 Track
AMB
 .  .  v=0
 .  . 
 .  . 
 .. 
 X 


Figure 2: M simultaneous, Frame M (train AB)
 The vertical lines in the bottom half, shows the position of the points A, Observer M and B.
 The line AX shows the position of the spark, originating from A.
 The line BX shows the position of the spark, originating from B.
 The Observer M sees the sparks simultaneous at point X.
 The dashed lines in the top half, shows the position of the points A', Observer M' and B'.
 The line A'x shows the position of the spark, originating from A'.
 The line B'x shows the position of the spark, originating from B'.
 The Observer M' should have been at position x inorder to see the sparks from A' and B' simultaneous.
 The observer M' sees the sparks from B' at the moment tM'B' when the line B'x crosses the dashed line through M'
 The observer M' sees the sparks from A' at the moment tM'A' when the line A'x crosses the dashed line through M'. tM'A'>tM'B'
 The two dotted lines originating from M' demonstrate Clock Synchronisation for both A' and B'. Clock Synchronisation starts by M' sending a Synchronisation Pulse. Clock synchronisation of those clocks is done before both train meets. The train A'B' should have its final speed v.
 A' has to reset his clock when he receives the Synchronisation Pulse. This is the moment when the left dotted line (originated from M') meets the dashed line through A'
 B' has to reset his clock when he receives the Synchronisation Pulse. This is the moment when the left dotted line (originated from M') meets the dashed line through B'
 The moment that A' meets A is tA' (with the clock of A', after clock synchronisation). The moment that B' meets B is tB'. (with the clock of B', after clock synchronisation).
 What the sketch shows is that tA'>tB'. The difference between tA'tB' = tM'A'tM'B'
Experiment Two with Two Trains, M' simultaneous, Rest frame M
The following sketch shows that it is possible that observer M' sees the sparks simultaneous.

/ / /
/ / /
/ / /
/ x .
/ ./. ./
. / . / .. /
. / . / . B2
. / . / . /
/.. / . /  v>
/. . /. / 
A'M'B' 
X1X2 Track
AMB
 .   v=0
 .  
 .  
 . B2
 . . 
  . . 
  .X 
 . 
 . 


Figure 3: M' simultaneous, Frame M (Train AB)
 The above figure shows that for observer M', in the moving train, to see the two sparks simultaneous train A'B' and train AB are not of equal length.
 The length L0' of A'B' in rest frame of AB has to be L0 * SQR(1vē/cē). For the derivation of this formula see below.
 The first spark in the frame AB is generated when A meets A'
 The second spark is generated a little later when the front part B' of train A'B'
makes contact with the front part B of train AB. This is point B2.
 The above figure shows that M does not see the two sparks simultaneous.
 In order for M to see the sparks simultaneous M has to be at position X with is not equidistant between AB
Derivation of L0 * SQR(1vē/cē).
 length of AB = 2*l0. length of A'B' is 2*l
 t1 = time of spark going from A'to x.
t1 = l / (cv)
 t2 = time of rod moving from B' to B (B2)
t2 = (2 *l0  2*l)/v
 t3 = time of spark moving from B (B2) to x.
t3 = l /(c+v)
 Because of equal arriving times: t1 = t2 + t3
l / (cv) = (2 *l0  2*l)/v + l /(c+v)
l / (cv)  l /(c+v) + 2*l /v = 2 *l0 / v
l * ( v / (cēvē) + 1/v) = l0/v
l * cē/(cēvē) = l0
l = l0 * ( 1  vē/cē)
 length of A'B' at rest is l = l0 * ( 1  vē/cē) / SQR(1vē/cē)
l = l0 * SQR(1vē/cē)
Experiment Two with Two Trains, M' simultaneous, Rest frame M'
The following figure shows the same situation as the above Figure 3, but now considering that the train A'B' is the rest frame.

  
  
 x 
 .. 
 .  . 
 .  . 
 .  .  v=0
A'M'B'
X1X2 Track
AMB
/ . ./ . ./
/ .. / .. / <v
/ . . . .
/ . / . . / .
/ . / .X. /
/ . /. /
/ . / /
/ / /


Figure 4: M' simultaneous, Frame M' (Train A'B')
 Figure 4 shows the same situation as Figure 3 (i.e. Observer M' sees the two sparks simultaneous), but now from reference frame A'B' or observer M'
 In fact figure 4 is identical as Figure 2 (i.e. Observer M sees the two sparks simultaneous, from reference frame AB.)
 However I do not agree that Figure 3 and Figure 4 are both correct. The issue is signal strength.
 In Figure 2 M receives both sparks simultaneous, but also with equal strength. If M' does not stand in the middle but at position x then M' will also receive both sparks simultaneous and with equal strength.
 In Figure 3 M' receives both sparks simultaneous, but not with equal strength. If M does not stand in the middle but at position X then M will also receive both sparks simultaneous, however not with equal strength.
 In Figure 4 M' receives both sparks simultaneous, but with equal strength. If M does not stand in the middle but at position X then M will also receive both sparks simultaneous, however with equal strength.
IMO Figure 4 is not a correct representation of the second experiment.
Answer question 1
IMO Figure 2 and Figure 3 are only valid for one particular case i.e. in one particular frame.
IMO they are only valid for a frame in absolute rest i.e. where v absolute is zero.
Reflection
My position is even stronger: The sketches do not show any experiment performed here on Earth. You have to take the speed of our Earth into account for both the trains AB and A'B'.
IMO, in the original description of the Moving Train Experiment by Ray d'Inverno, if you want see two simulataneous signals with equal strength you have to stay on the moving train and not on the track.
The effect of equal strength can also be demonstated as follows:
 Consider a light source, at position x, equidistant between two mirrors.
 Turn the light source ON and OFF. Observe what happens at position x: Two reflected ON signals will be received simultaneous, of equal strength.
 This effect is independent of the speed v of a rod where the mirrors are positioned at each end of the rod and x is equidistant between the two mirrors.
 In stead of two mirrors use two signal lamps. Signal reflection at each mirror is now simulated by using a signal lamp. When an ON (or OFF) light source signal is received, the signal lamp turned ON (or OFF).
 Observe what happens at position x: Two ON signals from the signal lamps will be received simultaneous, but (most probably) not of equal strength. The reason is the distance between the position of the signal lamps and position x. This distance is also a function of the absolute speed v. Only when v absolute is zero the strength of the ON signals from the two signal lamps will be identical.
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Created: 16 March 2001
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