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Similar to the Einstein train experiment

  1. Nov 16, 2013 #1
    Please help with this. This is similar to the Einstein train experiment.

    When C and C' are at the same place, lightning strikes at their location.

    Both survive though. Assume another prime B' behind C' When B' is at the same location as C, the lightning is at (A,0,0) and (-A,0,0) in the C frame. These are simultaneous events. But, they are not simultaneous to B'. So, C says the lightning is at A and -A when B' and C are together but B' says the lightning cannot be at both A' and -A' (these are located at A and -A respectively when B' and C are together).

    So, here is my problem. How can B' and C disagree, when they are at the same place, on the distance the lightning traveled along the positive x-axis and negative x-axis?

    Thanks in advance.
     
  2. jcsd
  3. Nov 16, 2013 #2

    mfb

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    Where is the problem? Different observers with different velocity disagree on the simultaneity of events and on their coordinate systems. Their axes are different in spacetime.
     
  4. Nov 16, 2013 #3

    I am not seeing the answer here.
    B' and C are at the same place when C determines the lightning strike is a distance A from both B' and C. But, B' cannot agree with this because B' does not agree the A and -A events are simultaneous.

    However, that mean B' and C must disagree on the distance the lightning strike has traveled when they are together. That means the light must be at 2 different locations along the common x-axis when B' and C are together. How can that be true?

    Or, B' must say when they are together, the lightning is not at A and C must say it is at A. How does this work?
     
  5. Nov 16, 2013 #4

    mfb

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    There is no common x-axis for B' and C, as they are moving relative to each other.
     
  6. Nov 16, 2013 #5
    I am a little confused. They do not share the same x coordinates for events, but they share the same x-axis.

    Is this false?
     
  7. Nov 16, 2013 #6

    mfb

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    Not in spacetime. The direction might be the same, but the scale and the simultaneity is not. Just look at a Minkowski diagram.
     
  8. Nov 17, 2013 #7

    ghwellsjr

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    Yes, diagrams help a lot in situations like this. Here's a diagram showing the rest frame of the ground observer C in blue. The black train locomotive is C' and the red caboose is B'. The train is traveling at 0.6c. As you said, when the black C' locomotive reaches the blue C ground observer (at the beginning of the scenario which is the origin of the frame), lightning strikes their location and emits two green light flashes in both directions:

    attachment.php?attachmentid=64013&stc=1&d=1384674689.png

    When the red caboose, B', reaches the ground observer, C, we can note the Coordinate Positions of the progress of the two green light flashes in the ground frame and they have reached plus and minus 10 thousand feet. All three events are at a Coordinate Time of 10 microseconds.

    Now let's transform the coordinates of all the events into the rest frame for the train:

    attachment.php?attachmentid=64014&stc=1&d=1384674689.png

    We can see that the event of the red caboose coinciding with the blue ground observer occurs at the Coordinate Time of 12.5 microseconds but the events for the progress of the flashes occur at Coordinate Times of 20 and 5 microseconds making none of these events simultaneous.

    Does this clear up all your confusion?
     

    Attached Files:

  9. Nov 17, 2013 #8
    You have a very nice way of putting this problem. So far, everything above is good.

    When the red caboose, B', reaches the ground observer, C, how can they disagree on the distance the lightning is from their common location?

    Code (Text):

    Initial conditions              | Lightning strike
                                    C
                              B'    C'
     
    Code (Text):

    After motion                              
     | Lightning location                                  | Lightning location
    -A                        C                            A
                              B'    C'                                      
     
    So, when B' and C are at the same place, C claims they are equidistant to the lightning locations of A and -A. B', however, claims A calculated in its space-time occurs before -A in its space-time.

    Therefore, for example, it is most likely that B' thinks in its time that the lightning strikes are further down the positive x-axis than event A. So, if we were to take the C' frame's A' and translate it to C space-time, say A2, we would find A2 > A. But, that means since A2 and A are in the same space-time, then when C and B' are together, the lightning is at 2 different C space-time positive x-axis locations.

    But, this makes no sense, however I can't see what is false above.
     
    Last edited: Nov 17, 2013
  10. Nov 17, 2013 #9

    PAllen

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    Every event exists in every frame. A2 and A, as you've defined them, are simply different events. They are at different space time coordinates per B'/C', and they remain at different spacetime coordinates for C. Absurd would be if they were different for B'/C' but the same for C.
     
  11. Nov 18, 2013 #10

    ghwellsjr

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    Correct.

    Correct. Here's a spacetime diagram to show the new event A' that is simultaneous to when B' and C are at the same place in the rest frame for C':

    attachment.php?attachmentid=64046&stc=1&d=1384758841.png

    Correct. Here's the spacetime diagram we get when we transform the previous frame to the rest frame of C and we change the name of the A' event to A2:

    attachment.php?attachmentid=64047&stc=1&d=1384758841.png

    I think PAllen's explanation should cover it.
     

    Attached Files:

  12. Nov 18, 2013 #11

    I am not following you. I will say I think and then you come back and correct it so that it all makes sense.

    - The C frame says the cos(0) lightning beam is at A when B' and C are colocated.
    - The B' observer says the cos(0) lightning beam is at some A' when B' and C are colocated.
    - All space-time coordinates map from one frame to the other.
    - So, either Map(A',0,0,A'/c) = (A,0,0,A/c), Map(A',0,0,A'/c) > (A,0,0,A/c) or Map(A',0,0,A'/c) < (A,0,0,A/c), trichotomy property for light beams.
    - B' believes there are cos'(\pie) and cos'(0) events that are simultaneous but different from C's simultaneous events.

    1) If Map(A',0,0,A'/c) = (A,0,0,A/c), then apply the same argument to -A and then B' thinks -A and A are simultaneous, which is wrong.
    2) If Map(A',0,0,A'/c) > (A,0,0,A/c), relativity concludes when B' and C are colocated, the cos(0) lightning beam is at two different C frame space-time locations since (A,0,0,A/c) is in the space-time of C and so is Map(A',0,0,A'/c), which can't be true for one light beam.
    3) If Map(A',0,0,A'/c) < (A,0,0,A/c), relativity concludes when B' and C are colocated, the cos(0) lightning beam is at two different C frame space-time locations since (A,0,0,A/c) is in the space-time of C and so is Map(A',0,0,A'/c), which can't be true for one light beam.
     
    Last edited: Nov 18, 2013
  13. Nov 18, 2013 #12

    ghwellsjr

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    I think your problem is that when you are going from A to -A, you are changing the sign of both the spatial term and the temporal term but the temporal term should be positive in both cases.

    Also, I'm not sure what you mean by cos(0) and cos'(\pie). Do you just mean the light beam that is going to the right for cos(0) and the light beam going to the left for cos'(\pie)? Did you really mean pi instead of pie? Since cos(0) = 1 and cos(pi) = -1, why don't you just say that you are referring to the beam going to the right which has positive x values and the beam going to the left has negative x values? And remember, they both of positive t values in all cases.

    If you just use the Lorentz Transformation process either with my specific numerical values or with your general symbolic variables and are careful to apply v as positive or negative depending on which way you are transforming (positive when going from the C frame to the B' frame and negative when going from the B' frame to the C frame) and make sure t is always positive then I think your problems will go away.
     
  14. Nov 18, 2013 #13
    Wow, misspelling I meant cos'(\apple pie), joke. Yes, cos(\pi). And, the direction of v is based on the motion of the frames not the light beams.

    We are not communicating.

    So, let's ignore -A for now which is a shorthand of writing (-A,0,0).

    We know frame C says when B' and C are colocated, the cos(0) beam is located at (A,0,0) with t=A/c.

    Now B' has an opinion where its cos(0) beam is located in its space-time. Let that be (A',0,0) with t'=A'/c. Then use the Lorentz Transformations on (A',0,0) with t'=A'/c, such that [itex]x=(x'+vt')\gamma[/itex]. So, [itex]x=(A' + vA'/c)\gamma[/itex].

    Either x=A, x<A or x>A.

    If x ≠ A, then when B' and C are colocated, the lightning strike is located at A and also x which are both C space coordinates. One light beam, can't be at 2 different places in the coordinates of the C frame when B' and C are colocated.

    Say x=A. Then apply the same argument to -A with -A' eliminating x2 ≠ -A leaving x2 = -A for the [itex]cos(\pi)[/itex] beam. But, A' and -A' are not simultaneous events in the C' frame, so that can't be true either.

    Therefore, to solve this problem, we need some way such that B' thinks it sees simultaneous events in its frame just like C does, but we don't trigger an error in which one light beam is at 2 different places in the space-time of C when B' and C are colocated.

    I can't think of a way.
     
  15. Nov 18, 2013 #14

    PAllen

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    Again these are two completely different events. Both of them exist for B'. Both of them exist for C. B' thinks the A' event is now and the A event is earlier. C thinks the A event is now and the A' event is later. Note, further, they both agree that A' is later than A (as required by causality since they are two different events in the history of the same light beam). They only differ on which is 'now'.

    This all goes back to your seeming refusal to accept that two coincident observers (in different states of motion) can disagree on what 'now' is. Not only can they, but they must if they both use the same procedure to define it.
     
  16. Nov 18, 2013 #15
    Can you please instead of asserting I refuse to accept different now's, which I have not done, instead answer the question?

    C claims A is the lightning location when B' and C are colocated.

    What does B' claim, when B' and C are colocated, the location of the lightning is in the coordinates of the C frame?

    Please answer this specific question since it is physical reality.
     
  17. Nov 18, 2013 #16

    PAllen

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    I've answered the question twice already. You refuse the answer and repeat the same mistake.

    The answer is that B' thinks the signal is at A' when they are colocated. This is a different event than what C calls event A precisely because what B' and C mean by 'now' are two different things. Both A' and A have separate coordinates for both B' and C. You have even been given a beautiful picture of this by
    ghwellsjr.

    Despite coinciding, and agreeing on 'now' for the single event of coincidence, B' and C disagree on what 'now' is everywhere else.
     
  18. Nov 18, 2013 #17
    You are still not answering the question.

    Try to translate A' to C coordinates.

    Use the lorentz transformations.

    Where does B' think the lightning strike is in the coordinates of C when C and B' are colocated.

    We already know that C thinks it is A. What does B' think it is in C space-time?

    That is the issue.
     
    Last edited: Nov 18, 2013
  19. Nov 18, 2013 #18

    PAllen

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    See Gwellsjr's picture. He already did it. The answer is the same whoever does it.
     
  20. Nov 18, 2013 #19
    I am not getting it.

    We all agree C says A when B' and C are colocated.

    Now, you have not given the view of B' for where the lightning is located in C coordinates when B' and C are colocated. That is a logical known for B' using lorentz transformations.

    Let's put it down in writing. I want to see this using math and writing.

    Surely this should be simple.
     
  21. Nov 18, 2013 #20

    PAllen

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    Gwellsjr put it in writing and a picture. He shows both A and A' per B', and A and A' (relabeled A2) per C. What part of this don't you get?
     
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