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Consequences of space-/time-/light-like separations

  1. Jul 23, 2015 #1
    I'm trying to prove the following statements relating to space-like, time-like and light-like space-time intervals:

    1. There exists a reference frame in which two space-time events are simultaneous if and only if the two events are space-like separated.

    2. There exists a reference frame in which two space-time events are coincident at a single spatial point if and only if the two events are time-like separated.

    3. If two events are light-like separated, then there are no reference frames in which they are simultaneous or coincident at a single spatial point.

    I can't seem to find any notes that a can verify my attempt with so I'm hoping that people won't mind taking a look at my workings on here to see if they're correct.


    Consider the space-time interval [tex]\Delta S^{2}= (\Delta x^{0})^{2}-(\Delta\mathbf{x})^{2}[/tex] where we use the metric signature ##(+,-,-,-)##.
    We then perform a spatial rotation of our coordinate system such that one of the space-time events ##x^{\mu}## is located at the origin of our reference, [tex]x^{\mu}=(0,0,0,0)[/tex] The other space-time event that we consider ##y^{\mu}## is then spatially aligned along the ##z##-axis in our coordinate system, [tex]y^{\mu}=(y^{0},0,0,y^{3})[/tex]

    Given this, we shall now consider each of the (numbered) cases above.

    1. Space-like interval :

    First, let the two events ##x^{\mu}## and ##y^{\mu}## be simultaneous in our inertial frame of reference ##S##, i.e. ##x^{0}=0=y^{0}##. It then follows that the space-time interval between them is given by [tex]\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(0-0)^{2}-(0-y^{3})^{2}=-(y^{3})^{2}[/tex] Now, ##(y^{3})^{2}>0## and so clearly ##\Delta S^{2}<0##. Therefore, if the two events are simultaneous in $S$, then they are space-like separated.

    Next, let the two events ##x^{\mu}## and ##y^{\mu}## be space-like separated. We have then that [tex]\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}<0\quad\Rightarrow\quad\vert y^{0}\vert < \vert y^{3}\vert[/tex] Thus, we can choose ##\beta =v=\frac{y^{0}}{(y^{3})}## (in units where ##c=1##). From this, we see that ##\beta <1## as required. Performing a Lorentz boost along the ##z##-axis we can relate the coordinates ##x^{\mu}## and ##y^{\mu}## in ##S## to their expressions in another inertial frame ##S'## [tex]x'^{0}=\gamma\left(0-\beta 0\right)=0\; , \qquad x'^{3}=\gamma\left(0-\beta 0\right)=0[/tex] and [tex]y'^{0}=\gamma\left(y^{0}-\beta y^{3}\right)=0 \; , \qquad y'^{3}=\gamma\left(y^{3}-\beta y^{0}\right)[/tex] where ##\gamma =\frac{1}{\sqrt{1- \beta^{2}}}##.

    Hence, in ##S'## we see that the two space-time events have the following coordinates [tex]x'^{\mu}=(0,0,0,0) , \qquad y'^{\mu}=(0,0,0,y'^{3})[/tex] and thus are simultaneous in this frame.




    2. Time-like interval :

    This follows a very similar approach to the space-like case.
    First, let the two events ##x^{\mu}## and ##y^{\mu}## be coincident at a single spatial point in our inertial frame of reference ##S##, i.e. ##\mathbf{x}=\mathbf{0}=\mathbf{y}##. It then follows that the space-time interval between them is given by [tex]\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(0-y^{0})^{2}-(0-0)^{2}=(y^{0})^{2}[/tex] Now, ##(y^{0})^{2}>0## and so clearly ##\Delta S^{2}>0##. Therefore, if the two events are spatially coincident in $S$, then they are time-like separated.

    Next, let the two events ##x^{\mu}## and ##y^{\mu}## be time-like separated. We have then that [tex]\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}>0\quad\Rightarrow\quad\vert y^{0}\vert > \vert y^{3}\vert[/tex] Thus, we can choose ##\beta =v=\frac{y^{3}}{(y^{0})}## (in units where ##c=1##). From this, we see that ##\beta <1## as required. Performing a Lorentz boost along the ##z##-axis we can relate the coordinates ##x^{\mu}## and ##y^{\mu}## in ##S## to their expressions in another inertial frame ##S'## [tex]x'^{0}=\gamma\left(0-\beta 0\right)=0\; , \qquad x'^{3}=\gamma\left(0-\beta 0\right)=0[/tex] and [tex]y'^{0}=\gamma\left(y^{0}-\beta y^{3}\right) \; , \qquad y'^{3}=\gamma\left(y^{3}-\beta y^{0}\right)=0[/tex] where ##\gamma =\frac{1}{\sqrt{1-\beta^{2}}}##.

    Hence, in ##S'## we see that the two space-time events have the following coordinates [tex]x'^{\mu}=(0,0,0,0)\; , \qquad y'^{\mu}=(y'^{0},0,0,0)[/tex] and thus are spatially coincident in this frame.



    3. Light-like interval :

    In this last case it is trivial, as given two events ##x^{\mu}## and ##y^{\mu}##, if they are light-like separated, then [tex]\Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}=0[/tex] and thus it is impossible to find a frame in which they are either simultaneous, or spatially coincident, as either one would change the interval into a space-like or a light-like interval. The interval is Lorentz invariant, so this clearly cannot be the case.
     
    Last edited: Jul 23, 2015
  2. jcsd
  3. Jul 23, 2015 #2

    DEvens

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    What part are you having trouble with? It looks ok.
     
  4. Jul 23, 2015 #3
    Nothing in particular, I just wasn't sure whether I'd constructed my argument correctly - particularly whether I've done the Lorentz transformation parents correctly?! Was seeking clarification more than anything really.
     
  5. Jul 23, 2015 #4

    Mentz114

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    If we have two points ##P(t_1,x_1)## and ##Q(t_2,x_2)## with ##\Delta t = t_2-t_1,\ \Delta x=x_2-x_1## the transformed interval ##\Delta t' - \Delta x'## is
    ##\gamma \Delta t +\beta\gamma \Delta x - (\gamma \Delta x +\beta\gamma \Delta t )= \gamma(1-\beta)(\Delta t - \Delta x)##

    From this you can work out quickly the answers to your question. I can't see anything wrong with what you've done, I'm just proposing slightly more compact way.
     
  6. Jul 23, 2015 #5
    Ok cool, thanks for the tip. Glad I've understand it correctly.
     
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