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Negative squares using the space time interval invariance

  1. Sep 17, 2010 #1
    I'm new to this (wonderful) forum, and to SR too...
    I've a general question about the space time interval invariance.
    Say we have two points A and B, at rest each other, at distance AB.
    Now A and B simultaneously in their reference frame emit a flash of light.
    The space time interval between these two events is spacelike and equals -(AB)2.
    Now a fast spaceship C travels from A toward B and is reached by the two flashes at the middle point between A and B. For C the two flashes are not emitted simultaneously, and I want to use the invariance of the space time interval to compute T, that is the difference in time (measured by C) between the two flashes. The distance between A and B measured by C is shortened by the Lorentz contraction: ABc < AB. The space time interval measured by C should be:
    (CT)2 - (ABc)2
    and this number should equal -(AB)2. It results
    (CT)2 = (ABc)2 - (AB)2 < 0.

    So, how can I find T, as a square root of a negative number?

    Sorry if the question is stupid....
  2. jcsd
  3. Sep 17, 2010 #2


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    This is a space-time diagram of your scenario as I understand it. The first is from A and B's rest frame, the second is from the ship frame. I think you can read off the time between flashes on the diagram. The ship is travelling at 0.42c in the A,B frame.

    I think you must have made a mistake, if you're trying to take the sqrt of a negative number. With this signature (+,-,-,-) a timelike interval is negative, so just ignore the sign, take the square root, the put the sign back.

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  4. Sep 17, 2010 #3
    Thank you

    With this signature (+,-,-,-) a SPACElike interval is negative, and this is our case.

    And for sure (CT)2 = (ABC)2 - (AB)2 IS negative.

    What I find very interesting in your answer is when you say: "just ignore the sign, take the square root, the put the sign back"... Really so simple?

    Thank you very much!
  5. Sep 17, 2010 #4


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    I don't know if space-like intervals are invariant ( they are, see below) so ignore my advice.

    The flashes are at the same time in the A,B frame (tA-tB=0) but clearly not when the diagram is boosted by -0.412c. We're talking about a change in simultaneity here, and I'm not sure how proper length comes into it. I'll try and work it out later.


    Spacelike intervals are invariant, so it seems there's not much point in using one to explain a change in simultaneity. In my diagrams, the interval is sqrt(-100) in both frames. The quantity you want is one side of the triangle that defines the interval.

    If two events ( tA,xA), (tB,xB) are boosted by [itex]\gamma[/itex] the transformed time coords are (Y is gamma, b is beta)

    t'A = YtA+YbxA
    t'B = YtB+YbxB

    so in your scenario
    t'A - t'B = Yb(xA - xB)

    which shows that the change in simultaneity depends on the spatial separation, and the relative velocity.
    Last edited: Sep 17, 2010
  6. Sep 17, 2010 #5
    Thank you.

    My point (now) is not to *find* a number, but only to understand: By using the invariance of the spacetime interval, all quantities in SR kinematics can be found, as I've understood. So my question, that perhaps you have already answered, is: Why have I a negative square in this case?
    You said: "just ignore the sign, take the square root, the put the sign back" and perhaps this is the simple answer (but I don't really understand WHY...)

    Thanks again
  7. Sep 17, 2010 #6


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    The sign of a spacelike interval depends on the signature convention, so there's no harm in changing the sign and then taking the square root and putting sign back.

    I still can't see how to get the t'A - t'B in my last post without doing a Lorentz transformation. I'll have another try later.
  8. Sep 17, 2010 #7
    Very good! Now I can do my simple calculation without harm!
    Thank you!

    Here perhaps I can help you: From the invariance of the spacetime interval

    (CT)2 - (ABc)2 = (AB)2

    (and without problems arising from the 'negative square'...) one can immediately find T... or not?

    Last edited: Sep 17, 2010
  9. Sep 17, 2010 #8


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    I think that should be

    (CT)2 - (ABc)2 = -(AB)2


    (CT)2= (AB)2( 1/(1-b2) - 1 ) = b2Y2(AB)2

    as above. But by using a Lorentz contraction, have you not assumed that an LT connects the frames ?

    (it's late so don't assume I've done the math correctly).
  10. Sep 17, 2010 #9
    You're right, I forgot a minus sign...

    (CT)2 - (ABc)2 = - (AB)2

    that is

    (CT)2 = (AB)2 ([tex]\frac{1}{\gamma^{2}}[/tex] - 1) (<0 but no more a problem...)

    Yes, I must admit!
  11. Sep 17, 2010 #10


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    Now I am confused ( nothing new there ). This is wrong

    [STRIKE](CT)2= (AB)2( 1/(1-b2) - 1 ) = b2Y2(AB)2[/STRIKE]

    but is this ?

    Now I have doubts the validity of (ABC)=1/Y(AB). This sort of length contraction is because the moving frames perception of simultaneity when measuring the ends of a timelike interval. These events are not causally connected so can length contraction be applied ?
    Last edited: Sep 17, 2010
  12. Sep 18, 2010 #11
    I guess YES, but is perhaps better to wait for an explanation....

    Now I've less doubts, thank you...
  13. Sep 19, 2010 #12


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