The Absurd Assertions of the SR experts

           +v                   -v 
       ------->              <-------
       A'     F'             F      A

                  -v  
              F<------A*       *=explosion
             A'------>F'            
                  +v

1. Assertion one: When they meet the length L of FA and the length L' of F'A' are equal
2. Assertion two: L = L' = L0*sqr(1-v^2/c^2)
3. When they meet at t1 at the position of A and F' there is an explosion
4. An Observer at A' sees the explosion at t2.
5. An Observer at F sees the explosion at t3.
6. Assertion three: t2=t3.

• 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

   
                          v=0          +2v
                    G<----------B   B'---->G'                        
                    ........................ track of A'F'
   A<--------->F           A'------->F'
      L0  v=0                  +v
............................................. track of AF

1. The train AF. This is a train at rest. It length is L0.
2. The train A'F'. This train has a speed v to ->. Its length is L'. L'=L0*SQR(1-v^2/c^2)
3. 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).
4. The train GB. This train is also at rest. Its length is L0 i.e. the same as AF.

• 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'.

• 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

   
                         v=w          v=w'
                    G<----------B   B'---->G'                        
                    ......................... track of A'F' 
   A<----------F           A'<------>F'
      L0  v=v                   v=0
............................................. track of AF

1. The train A'F'. This is the train at rest. Its length is L'.
2. The train AF. This train has a speed v to <-. Its length is L0. L0=L'/SQR(1-v^2/c^2)
3. 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)
4. The train GB. This train has a speed of v=w to the right. Its length is M i.e. the same as AF.

1. 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) ?
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.

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

 
                     /           /  x t3        \
                    /           /    \.          \ 
                   /           /      \ .         \ 
                  /           /        \  .        \ 
              t3 x           /          \   .       \            
                /  .        /            \    .      \ 
               /     .     /              \     .     \    
              /        .  /                \      .    \ 
             /           x t2            t1 x       .   \ 
            /          ./                    \.       .  \
           /         . /                      \ .       . \ 
          /        .  /                        \  .       .\
      t1 x       .   /                          \   .       x t2
        /  .   .    /                            \    .   .  \
       /     .     /                              \     .     \
      A'    t0    F'                               F    t0     A 
         ------>                                     <-------                            

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

 
                    <--               --->          
                   B    G           B'  G' 
               t6  |    |           /   /
                   |    |        /   /
             t5    |    |     /   / 
                  |     |  /      /
                 |      X       /
                 |    / |    / 
              t4 | /    | /             
                /      / 
               /      x         
           t3 /     ./ 
             /    . /           
            x   .  /                
           /  .   /                   
       t2 /      /   
         /        /
        /          /
       /           /
    t1|           |
      x           x 
      | .       . |
      |   .   .   |
      |     .     |
      A'    t0    F'                                                      
         -----> 
           v