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Simultaneity (Special Relativity)

  1. Feb 22, 2010 #1
    Hi guys,

    I'm studying for a midterm and i decided to do some problems for practice and I'm stuck on a fairly simple one

    "The space and time coordinates of two events as measured in a frame S are as follows:
    Event 1 (x,0,0) @ t=x/c
    Event 2 (2x,0,0) @ t=x/2c

    There exists a frame in which these events occur at the same time. Fin d the velocity of this frame with respect to S."

    The solution is -c/2

    For my solution I set event 1 as S and event 2 as S'
    For them to be simultaneous I also set t=t'
    I equated the L.E transformations and my solution is -c/6

    Any help would be greatly appreciated

    P.S I also tried drawing a Minkowskian diagram and figuring out the slope of the time-constant line (linking the two-event points) however this gives me the x'-axis equation and not the t'-axis equation which is what i would need to figure out the velocity.
     
  2. jcsd
  3. Feb 23, 2010 #2
    Hi, I hope this isn't too late. By the way, you might get more of a response if you post a relativity question in the "special and general relativity" forum. Thinking in terms of a Lorentz transformation:

    [tex]t'=\gamma\left ( \frac{x}{c}-\frac{x}{c}\frac{v}{c}\right )=\gamma\left ( \frac{x}{2c}-\frac{2x}{c}\frac{v}{c} \right )[/tex]

    [tex]x-x\frac{v}{c}=\frac{x}{2}-2x\frac{v}{c}[/tex]

    [tex]\frac{x}{2}=-x\frac{v}{c}[/tex]

    [tex]v=-\frac{c}{2}.[/tex]

    We can also do it the other way you suggested, but have to be careful as the convention is to measure time on the vertical axis of a Minkowski diagram, which differs from the usual convention in elementary physics and calculus books. If you turn the diagram on it's side, with x increasing in the vertical direction, the slope of the line through the two events is

    [tex]\frac{2x-x}{\frac{x}{2c}-\frac{x}{c}}=-2c.[/tex]

    (Negative because the t axis is now increasing to the left.) The slope is more than c because it's a spacelike line; in other words, it isn't possible for an object to be present at both events. The slope representing the world line of an object in whose rest frame these events are simultaneous will be the inverse of this, except that we don't invert c because c is the scale factor expressing the difference in scale between the x and t axes, and we only want to invert the slope of the line, not interchange the scales of the x and t axes. So the velocity of the frame where the two events have the same time coordinate is

    [tex]-\frac{c}{2}.[/itex]
     
  4. Feb 23, 2010 #3
    here are two equations also maybe to memorize cause they are very simple:
    for the same position, use:

    $\frac{v}{c}=\frac{\Delta x}{\Delta ct}$

    for the same time:
    $\frac{v}{c}=\frac{\Delta ct}{\Delta x}$

    and insert your space time intervals, this gets you your answer I think. :smile:
     
  5. Feb 23, 2010 #4
    Together, your equations imply that [itex]v=c[/itex]. This can't be the case here because the question asked for the speed of a (from the context I assume inertial) reference frame, relative to another.
     
  6. Feb 23, 2010 #5
    no, when you use the "same time" formula, you get (-x/2)/x=v/c, which is the answer, v=-c/2

    im not sure how else to tell you this..., tell me how v=c?, they are for two different situations, you cant use them together...
     
  7. Feb 23, 2010 #6
    Oh, I see, sorry. Just me being dense ;-)

    I mistook your "for the same time" as meaning that these equations were simultaneous equations, both true:

    [tex]\frac{v}{c}=\frac{\Delta x}{\Delta ct}=\frac{\Delta ct}{\Delta x}.[/tex]

    Multiplying by [itex]\Delta x \cdot \Delta ct[/itex], the equality on the right becomes

    [tex]\left(\Delta x\right)^2=\left(\Delta ct\right)^2[/tex]

    [tex]\therefore \Delta x=\Delta ct.[/tex]

    So

    [tex]\frac{v}{c}=\frac{\Delta x}{\Delta ct}=\frac{\Delta x}{\Delta x}=1.[/tex]

    Therefore [itex]v=c[/itex]. But I see now that isn't what you had in mind. You meant that we can use the second of the equations you gave to work out the velocity of an inertial frame where the two events happen at the same time, in this case:

    [tex]\frac{\Delta ct}{\Delta x} = -\frac{1}{2} = \frac{v}{c},[/tex]

    where [itex]\Delta ct[/itex] is the difference in time between the two events, and [itex]\Delta x[/itex] the spatial separation between them.

    And your first equation was for solving a different problem: finding the velocity of an inertial frame where two events happen in the same place.
     
    Last edited: Feb 23, 2010
  8. Feb 23, 2010 #7
    yes, sorry for being unclear.
     
  9. Feb 23, 2010 #8
    That's okay. I'd just been working on an epic vector algebra problem that must have caused my brain to go on strike...
     
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