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Very very simple question in SR

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Referring to the drawing attached:
    Two rods of the same length ( say 3 m ).
    Rod A has a velocity of v = 0.866 c in x.
    Rods are very close in y.
    The drawing is taken from the point of view of rod B.
    1 - Is the drawing correct ?
    2 - Could we say that the right ends of the rods are at the same point and at the same time ?
    3 - Could we say that the left end of rod A is at the same point and at the same time that the middle ( 1.5 m ) point of rod B ?


    2. Relevant equations

    l' = l . [tex]\gamma[/tex]



    3. The attempt at a solution

    The attempt at a solution is the drawing.
     

    Attached Files:

  2. jcsd
  3. Nov 13, 2007 #2
    No answer ?

    The next question would be: what happens from the point of view of rod A ? In this case:
    - rod B would be shorter than rod A
    - the right end of rod B would be at the same point and at the same time than the middle point of rod A.
     
  4. Nov 13, 2007 #3

    nrqed

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    Yes to al your questions at the condition that by "at the same time" you mean at the same time as measured in the frame of B.
     
  5. Nov 13, 2007 #4

    nrqed

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    From the point of view of A, rod B is shorter, yes. I can't quite make sense of your second sentence, it's incomplete.
     
  6. Nov 14, 2007 #5
    Ive made another drawing from the point of view of rod A.

    Then:

    At the same readings on clocks A and B the right ends of the rods are at the same point.
    ( suppose both clocks give "0" )
    But from the point of view of B its middle point is at the same point than the left point of rod A. ( first drawing, "dibujo1.bmp" ) and
    From the point of view of A its middle point is at the same point than the left point of rod B. ( second drawing, "dibujo2.bmp" )
     

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  7. Nov 14, 2007 #6

    nrqed

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    Again your sentences are not complete (sorry, I am not trying to be difficult, it's just that most of the apparent paradoxes in SR arise from a careless use of words which lead to misunderstandings so I am just trying to be careful with the wording).

    here would be two correct statements:

    But from the point of view of B and at t=0 as measured by B its middle point is at the same point than the left point of rod A.

    From the point of view of A abd at t=0 as measured by A its middle point is at the same point than the left point of rod B.


    The reason those two statements don't contradict each other is that two events which are simultaneous in one frame are not simultaneous in another frame.
     
  8. Nov 14, 2007 #7

    Doc Al

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    OK. So when the right ends of A and B overlap, the clocks at those locations read the same time t = 0. No problem.

    Just to be clear, let's imagine that each rod has three synchronized clocks located at the left end, middle, and right end. Let's call the clocks AL, AM, AR and BL, BM, BR.

    So, to restate what we already stated: When clocks AR and BR overlap, they read the same reading, t = 0.

    No problem. The first drawing is from B's point of view: According to B, at the instant shown all three of B's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod A is aligned with the middle of B. Note that clock AL does not read t = 0. (According to B, clock AL is behind clock AR.)

    No problem. This drawing is from A's point of view: According to A, at the instant shown all three of A's clocks read t = 0, the right sides of both rods are aligned, and the left side of rod B is aligned with the middle of A. Note that clock BL does not read t = 0. (According to A, clock BL is ahead of clock BR.)

    So both diagrams are perfectly consistent.

    (nrqed beat me to it!)
     
  9. Nov 15, 2007 #8
    nrqed:
    Yes, this is what I meant.

    Doc Al:
    I suppose you are using t' = γ(t-vx/c²)
    "According to B, clock AL is behind clock AR"
    "According to A, clock BL is ahead of clock BR"

    1 - The change in sign comes from the change in the direction of velocity "v" ?
    2 - Anyway, its very difficult to understand that movements from left to right have different results than right to left.
    Why the left clock ( A ) that moves to the left is behind and
    the left clock ( B ) that moves to the right is ahead ?
     
  10. Nov 15, 2007 #9

    Doc Al

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    clock desynchronization

    This is the relativity of simultaneity. The rule of thumb for moving clocks is: Moving clocks, synchronized in their rest frame but separated by a distance D along their direction of motion, are not synchronized in the stationary frame; The front clock lags the rear clock by an amount:
    [tex]T = Dv/c^2[/tex].

    I think this is easier to see using:
    [tex]t = \gamma(t' + vx'/c^2)[/tex]

    From A's point of view: What do B's clocks read when the time is t = 0 on A's clocks? The above transformation gives us:
    [tex]0 = \gamma(t' + vx'/c^2)[/tex]
    Or:
    [tex]t' = -vx'/c^2[/tex]

    Clock BR is at x'=0, so clock BL is at some negative value of x' making it have a time t' > 0.

    Sure.
    They don't! The physics works the exact same way. Of course the velocity (and coordinate system) is different depending upon which direction you are moving, but the effect is completely symmetric. To see this, just rotate both of your diagrams by 180 degrees. Do you think the results would be different?
    See my explanation above.
     
  11. Nov 16, 2007 #10
    This is a very simple and clear explanation ( at least to me )

    Ive rotated the diagrams 180 degrees and the results are OK.

    Thanks.

    But one question that I see now is: the same ( real ) clock, at the same point, is seen different if you are moving or not. So, at the same point, you see different events, depending on your velocity.
     
  12. Nov 16, 2007 #11

    Doc Al

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    I don't quite know what you mean by seeing different events at the same point. An "event" is something that happens at a particular place and time, but the measurement of the position and time of that event depends on who's doing the observing.

    In this example, what the the two observers disagree about is whether the "events" (say the left-end and right-end clocks showing 2 pm) happen at the same time or not. A says B's clocks strike 2pm at different times (according to A's clocks). Of course B says the same thing about A's clocks.
     
  13. Nov 17, 2007 #12
    I call one event "A real clock whit its hand pointing to 10 s" ( Observer 1 )
    Another event "The same real clock whit its hand pointing to 0 s" ( Observer 2 )

    If the clock is programmed to explode at +10s Observer 1 is seeing the explosion and Observer2 will see the explosion ( at another point ) ...

    unless Observer 2 destroys the clock ( they are at the same point, can he destroy the clock ? )
     
  14. Nov 17, 2007 #13

    nrqed

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    The point is that an event is not only something that oocurs at one specific time but also at one specific location. two observers in different frames will assign different coordinates (values of x and t) to a single event but they both agree that the event occured at a specific time and location.


    Now, a clock indicating 10 seconds is a valid event. Everybody (in all frames!) will agree that this event occurred and that they saw that clock indicating 10seconds when it was at some location in space. Nobody can disagree on the event itself . when all observers assign a time and a value of x to this particular event, they will all have different results (and values of time which may have nothing to do with the value indicated by the clock).
     
  15. Nov 18, 2007 #14
    nrqed:
    I dont understand what you mean, but in the given example the two events occurs at the same point ( I dont know why you talk about x coordinates ),
    and the two events are: a real clock with its hand at one position ( 0 s ) and the same clock with its hand at another position ( 10 s ).

    Are you saying that there is something called "time" that has nothing to do with clocks ?
    How do you measure this "time" ?
     
  16. Nov 18, 2007 #15

    nrqed

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    whn you say that the two events occur at the same "point" , you are making a statement which depends on the frame. There is only one frame of reference for which the two events occur at the same point (i.e. at the same value of x...I am considering only frames moving in one dimension, along the x axis) and that's the frame attached to the clock. Look, if I am moving at some speed relative to the clock, the two events will NOT occur at the same value of x in my frame! Between the instant the clock will indicate 0 and 10 seconds, the clock will have moved with respect to me!!! So the two events will occur at different locations fo rme. You were restricting yourself to the frame of the clock!

    Yes, you measure time using clocks, but you must use clocks in yoru frame of reference, that is at rest with respect to you! . If I see a flying clock zooming by me, I cannot use this to measure time in my frame
     
  17. Nov 18, 2007 #16
    I dont know if you dont understand or you dont want to understand.

    Doc Al:
    What do you think "are aligned" means ?
     
  18. Nov 18, 2007 #17

    nrqed

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    I guess I don't understand special relativity.

    Goodbye
     
  19. Nov 18, 2007 #18

    Doc Al

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    I beg your pardon? You might wish to rephrase that. You are the one looking for help, not us.

    Two points on the the moving rods "are aligned" when they are momentarily passing each other.

    For example: When the right ends of each rod pass each other ("are aligned"), the clocks at those points (AR & BR) both read t = 0. Everyone agrees that that is true: that those clocks really read those times as they passed each other. What they disagree about is the time that the other clocks read at that moment.
     
  20. Nov 19, 2007 #19
    You seem to be a lawyer, not a physicist.
    Yes, Im looking for help, so, please, help me, and tell me if the following sentences are true or false.

    What do you think "are aligned" means ?
    1 - It means that they are at the same point. ( True / False )
    2 - Time is measured by clocks, so time = reading of a clock ( T / F )
    3 - One clock, at rest, seeing by an observer at rest, gives a time, but the same clock seeing by a moving observer, gives a different time ( both observers at the same point ) ( T / F )

    Last, Doc Al:
    .
    You introduce a new concept ( to me ) in SR: the moment
     
  21. Nov 19, 2007 #20

    Doc Al

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    Since, as is evident from this thread, SR is a subtle and tricky subject one must be precise. If you are serious about learning relativity (or physics in general), you'd better get used to being precise.
    I already told you how I used the term. Go back and reread what I wrote earlier.
    The way I used the term, the ends of the rod are momentarily "aligned" when they pass each other. When the ends are so "aligned" they are at the same point.
    Not sure I understand the second part of that statement. Every frame uses their own clocks to measure time.
    I don't understand your question; If I misinterpret your question, please restate with a clear, specific example. I don't know what you mean by "both observers at the same point". Do you mean that the observations take place as the two observers--each with their own clock--pass each other and look at each other's clock? If that's what you mean, then the answer is clear: If observer A's clock reads a certain time--as observer B passes it--then moving observer B will also see A's clock read that same time. How could it be otherwise?
    SR deals with time and simultaneity, so the idea of "now" should not be new.
     
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