The Absurd Assertions of the SR experts
In the newsgroup news:sci.physics.relativity the following thought experiment was explained:
Consider two trains AF and F'A' with equal speeds v+ and v-
In the rest frame of the track the length of each train is L0.
+v -v
-------> <-------
A' F' F A
When they meet the situation is the situation is the following:
-v
F<------A* *=explosion
A'------>F'
+v
- Assertion one: When they meet the length L of FA and the length L' of F'A' are equal
- Assertion two: L = L' = L0*sqr(1-v^2/c^2)
- When they meet at t1 at the position of A and F' there is an explosion
- An Observer at A' sees the explosion at t2.
- An Observer at F sees the explosion at t3.
- Assertion three: t2=t3.
Around this last third assertion a whole discussion is going on. IMO the explanation is as follows:
- If you consider the whole experiment from the track frame, that means all the clocks involved are synchronised in that frame, than that assertion is not true. t2 is less than t3. A' sees the explosion before F.
- If you consider the experiment from the train frame, this means that the clocks in the train A'F' are synchronised in its own train frame and the clocks in the train AF are synchronised in its own train frame, than the assertion is true. For a more detailed explanation see The Absurd Assertions of the SR experts - comment 1.
However IMO the first assertion is much more challenging: How do we proof that length contraction takes place, how do we demonstrate that the speed of both trains is equal, and how to prove that the length of both trains are equal.
IMO, In order to discover how nature works you have to perform all experiments in one frame. That is the easiest way to measure in an uniform way and to discover the rules that link all those measurements together. In a second pass you can review those same experiments in a different frame.
A thought experiment with different trains
The purpose of this experiment is to challenge if the length of two trains with the same speed but in opposite directions is equal
v=0 +2v
G<----------B B'---->G'
........................ track of A'F'
A<--------->F A'------->F'
L0 v=0 +v
............................................. track of AF
The above picture shows four trains. All are discussed in the same frame
- The train AF. This is a train at rest. It length is L0.
- The train A'F'. This train has a speed v to ->. Its length is L'. L'=L0*SQR(1-v^2/c^2)
- The train B'G'. This train has a speed 2v=2*v to the right. Its lenghth is M'. M'=L0*SQR(1-4*v^2/c^2).
- The train GB. This train is also at rest. Its length is L0 i.e. the same as AF.
The trains GB and B'G' are special. Both trains are placed on a track on top of the train A'F'.
- The train B'G' moves to the right relative from the train A'F'.
- The train GB moved to the left relative from the train A'F'.
The whole story about this experiment is as follows:
- In the Initial state all the four trains are at rest with the train AF
- Then the train A'F' started to move to the right with speed v
- A certain distance away the train GB started to move to the left, which such a speed
that its distance does not increase to the train AF
- At that same moment the train B'G' started to move to the right, which a speed of 2*v
relative to the train AF
- In order to control the speed of each train there are clocks near the track of AF at equal distances, synchronised in the frame of AF
Now we are going to consider exactly the same situation in the frame of A'F'.
In the following picture the length drawn as the previous one, but this can be wrong !
v=w v=w'
G<----------B B'---->G'
......................... track of A'F'
A<----------F A'<------>F'
L0 v=v v=0
............................................. track of AF
The above picture again shows four trains. The rest frame is the train A'F'
- The train A'F'. This is the train at rest. Its length is L'.
- The train AF. This train has a speed v to <-. Its length is L0. L0=L'/SQR(1-v^2/c^2)
- The train B'G'. This train has a speed v=w' to the right. Its lenghth is M'. M'=L'*SQR(1-4*v^2/c^2)/SQR(1-v^2/c^2)
- The train GB. This train has a speed of v=w to the right. Its length is M i.e. the same as AF.
In the frame A'F' there are three issues: Length, time and Speed
The above text immediate requires two comments.
- First the mathematics may be is not right. Most probably the values for the speed v and 2*v are not correct. A similar problem arises with the value of w'. Is it for example correct to write M' (length of B'G') = L1*SQR(1-w'^2/c^2) ?
- A more difficult physical problem exists about the length of both B'G'and most important of BG.
- B'G' is a train with moves to the right relative from A'F'. Its length should be smaller accordingly to Special Relativity Theory. That is not so much of a problem.
- BG is a train which moves to the left relative from A'F'. Its length should also be smaller accordingly to Special Relativity Theory. However as we have seen if we consider this train in the reference frame AF than its length should be longer.
IMO in reference frame of train A'F' at a certain speed the length of train BG is larger than A'F', (while it should be smaller) which is in conflict with SR.
At that same speed the train the train B'G' is smaller
My first conclusion is that in the reference frame A'F', at the same speed, the trains BG and B'G' are not equal
My second conclusion is that the same is true in the reference frame AF; equal speeds are no quarantee for equal lengths.
There is also a program (demonstration) of this thought experiment.
To get a copy select: SR.BAS
To see the listing of the program select: SR.HTM
The Absurd Assertions of the SR experts - comment 1.
/ / x t3 \
/ / \. \
/ / \ . \
/ / \ . \
t3 x / \ . \
/ . / \ . \
/ . / \ . \
/ . / \ . \
/ x t2 t1 x . \
/ ./ \. . \
/ . / \ . . \
/ . / \ . .\
t1 x . / \ . x t2
/ . . / \ . . \
/ . / \ . \
A' t0 F' F t0 A
------> <-------
The above sketch shows two trains:
- At the left side the train A'F' moving to the right.
- At the right side the train FA moving to the left.
The sketch for the train A'F' consists of two parts:
- What happened during synchronisation
- What happened after the explosion
In order to synchronise the two clocks in the train A'F' we use a light flash at t0. The position of the light flash is in the middle of A'F'.
The light flash reaches the front F' of the train at t2.
The light flash reaches the back A' of the train at t1.
Synchronisation implies that at those instances the clock at A' and the clock at F' are both reset to zero.
In order to synchronise the two clocks in the train AF the same method is used. The position of the flash is in the middle of AF.
The explosion is at t2. At that moment, the position of the front F' of the train A'F' and the back A of the train FA, is the same.
The explosition reaches the back A' of A'F' at t3
The explosition reaches the front F of FA at t3
What the sketch shows is t3(A') in the reference frame of the train A'F' = t3(F) in the reference frame of train FA.
For a mathematical discussion select MATH.HTM
The Absurd Assertions of the SR experts - comment 2.
<-- --->
B G B' G'
t6 | | / /
| | / /
t5 | | / /
| | / /
| X /
| / | /
t4 | / | /
/ /
/ x
t3 / ./
/ . /
x . /
/ . /
t2 / /
/ /
/ /
/ /
t1| |
x x
| . . |
| . . |
| . |
A' t0 F'
----->
v
The above sketch follows SR and shows the following:
- At t0 the train A'F' has a speed of zero
At t0 I perform synchronisation for the left and right klock. The clocks show the track time. (old frame)
- At t1 the train starts towards the right. The length of the train slowly decreases.
- At t2 the train reaches it final speed v.
- At t3 I perform a second synchronisation. The clocks show now train time.
- At t4 I divide the train into parts. One part B'G' starts moving to the right
One part BG starts moving towards the left.
- At t5 the right train reaches it final speed +v in new frame
The left train has a speed -v in the new frame.
- At t6 we have the final situation.
The above sketch follows the SR concept.
That length of the train at t2 (With v=+v) is smaller than the length of then train at t0(With v=0).
That both trains at t6 (With v=+v and v=-v) are smaller than the length of the train at t2 (with v=0) in the frame of that frame.
IMO something is wrong with this sketch.
When you observe the train B'G' its speed in the track frame is 2*v and its length is the smallest. That is correct
When you observe the train BG its speed in the track frame is zero.
The length of BG should be the same as A'F' before it started to move
Around t5 the length of B'G' should not decrease (as shown) but the length of BG should increase. However this is inconflict with SR.
For a mathematical discussion select MATH.HTM
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Created: 15 January 2001
Last modified: 18 February 2001
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