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Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Ostlie

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Because there is no such thing as absolute simultaneity, two observers in relative motion may disagree on which of two events A and B occurred first. Suppose, however, that an observer in reference frame S measures that event A occurred first and caused event B. For example, event A might be pushing a light switch, and event B might be a light bulb turning on. Prove that an observer in another frame S' cannot measure event B (the effect) occurring before event A (the cause). The temporal order of cause and effect is preserved by the Lorentz transformation equations. Hint: For event A to cause event B, information must have traveled from A to B, and the fastest that anything can travel is the speed of light.

    2. Relevant equations

    Suppose [itex](x_{A},y_{A},z_{A},t_{A}),(x_{B},y_{B},z_{B},t_{B})[/itex] are the coordinates in the coordinate system S, and [itex](x'_{A},y'_{A},z'_{A},t'_{A}),(x'_{B},y'_{B},z'_{B},t'_{B})[/itex] are the coordinates in the coordinate system S', which is traveling at speed [itex]u<c[/itex] in the positive [itex]x[/itex] direction. Then from the Lorentz transformation equations,
    [tex]t'_{A}=\frac{t_{A}-(u x_{A}/c^{2})}{\sqrt{1-(u^{2}/c^{2})}}[/tex]
    [tex]t'_{B}=\frac{t_{B}-(u x_{B}/c^{2})}{\sqrt{1-(u^{2}/c^{2})}}[/tex]
    [tex]x'_{A}=\frac{x_{A}-u t_{A}}{\sqrt{1-(u^{2}/c^{2})}}[/tex]
    [tex]x'_{B}=\frac{x_{B}-u t_{B}}{\sqrt{1-(u^{2}/c^{2})}}[/tex]
    [tex]y'_{A}=y_{A}[/tex]
    [tex]y'_{B}=y_{B}[/tex]
    [tex]z'_{A}=z_{A}[/tex]
    [tex]z'_{B}=z_{B}[/tex]

    Because B happened after A,
    [tex]t_{B} > t_{A}[/tex]

    Because event A caused event B,
    [tex]\frac{\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} + (z_{B}-z_{A})^{2}}}{t_{B}-t_{A}} \leq c[/tex]

    3. The attempt at a solution

    I need to prove that [itex]t'_{B} - t'_{A} > 0[/itex]

    [tex]t'_{B}-t'_{A} = \frac{(t_{B} - t_{A})-(u/c^{2})(x_{B}-x_{A})}{\sqrt{1-(u^{2}/c^{2})}}[/tex]

    What do I do now?
     
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  3. Jun 23, 2011 #2

    vela

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    Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

    Since the denominator is positive, you just need to show the numerator is positive. That is, prove that
    [tex]t_B-t_A > (u/c^2)(x_B-x_A)[/tex]
     
  4. Jun 23, 2011 #3
    Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

    I can't think of any way of proving that. Which equations should I use?
     
  5. Jun 23, 2011 #4

    vela

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    Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

    The one you already wrote:
    [tex]\frac{\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} + (z_{B}-z_{A})^{2}}}{t_{B}-t_{A}} \leq c[/tex]
     
  6. Jun 23, 2011 #5
    Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

    But those are squared terms.

    It implies that,

    [tex](x_{B} - x_{A})^{2} \leq c^{2} (t_{B} - t_{A})^{2}[/tex]

    So,

    [tex]\lvert x_{B} - x_{A} \rvert \leq c (t_{B} - t_{A})[/tex]
    [tex]\frac{u}{c^{2}} \lvert x_{B} - x_{A} \rvert < (t_{B} - t_{A})[/tex]

    This could be either [itex]\frac{u}{c^{2}} ( x_{B} - x_{A} ) < (t_{B} - t_{A})[/itex] or [itex]\frac{u}{c^{2}} ( x_{A} - x_{B} ) < (t_{B} - t_{A})[/itex] depending on whether [itex]x_{B} > x_{A}[/itex] or not.
     
  7. Jun 23, 2011 #6
    Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

    Oh, OK. If [itex]x_{A} > x_{B}[/itex], then,
    [tex]t_{B} - t_{A} > \frac{u}{c^{2}} (x_{A} - x_{B}) > \frac{u}{c^{2}} (x_{B} - x_{A})[/tex], since then[itex](x_{B} - x_{A})[/itex] is negative and [itex](x_{A} - x_{B})[/itex] is positive.

    Thanks!
     
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