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

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Discussion Overview

The discussion revolves around the properties of space-like, time-like, and light-like separations in space-time intervals, focusing on their implications in different reference frames. Participants explore theoretical aspects and mathematical reasoning related to these separations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to prove that two space-time events are simultaneous if and only if they are space-like separated, using the space-time interval and Lorentz transformations.
  • Another participant agrees with the initial claims but seeks clarification on the correctness of the Lorentz transformation steps.
  • A third participant suggests a more compact method for deriving the results related to the space-time intervals, indicating that the original approach is valid.
  • Concerns are raised about whether the arguments constructed are correct, particularly regarding the application of Lorentz transformations.
  • It is noted that if two events are light-like separated, then there are no reference frames in which they can be simultaneous or coincident at a single spatial point, as this would alter the interval type.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the claims regarding space-like and time-like separations, but there is some uncertainty about the correctness of the Lorentz transformation steps. The discussion remains unresolved regarding the best approach to proving the statements.

Contextual Notes

Some participants express uncertainty about specific mathematical steps and the application of transformations, indicating potential limitations in their understanding or execution of the concepts discussed.

"Don't panic!"
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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 \Delta S^{2}= (\Delta x^{0})^{2}-(\Delta\mathbf{x})^{2} 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, x^{\mu}=(0,0,0,0) The other space-time event that we consider ##y^{\mu}## is then spatially aligned along the ##z##-axis in our coordinate system, y^{\mu}=(y^{0},0,0,y^{3})

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 \Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(0-0)^{2}-(0-y^{3})^{2}=-(y^{3})^{2} 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 \Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}&lt;0\quad\Rightarrow\quad\vert y^{0}\vert &lt; \vert y^{3}\vert 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'## x&#039;^{0}=\gamma\left(0-\beta 0\right)=0\; , \qquad x&#039;^{3}=\gamma\left(0-\beta 0\right)=0 and y&#039;^{0}=\gamma\left(y^{0}-\beta y^{3}\right)=0 \; , \qquad y&#039;^{3}=\gamma\left(y^{3}-\beta y^{0}\right) where ##\gamma =\frac{1}{\sqrt{1- \beta^{2}}}##.

Hence, in ##S'## we see that the two space-time events have the following coordinates x&#039;^{\mu}=(0,0,0,0) , \qquad y&#039;^{\mu}=(0,0,0,y&#039;^{3}) 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 \Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(0-y^{0})^{2}-(0-0)^{2}=(y^{0})^{2} 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 \Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}&gt;0\quad\Rightarrow\quad\vert y^{0}\vert &gt; \vert y^{3}\vert 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'## x&#039;^{0}=\gamma\left(0-\beta 0\right)=0\; , \qquad x&#039;^{3}=\gamma\left(0-\beta 0\right)=0 and y&#039;^{0}=\gamma\left(y^{0}-\beta y^{3}\right) \; , \qquad y&#039;^{3}=\gamma\left(y^{3}-\beta y^{0}\right)=0 where ##\gamma =\frac{1}{\sqrt{1-\beta^{2}}}##.

Hence, in ##S'## we see that the two space-time events have the following coordinates x&#039;^{\mu}=(0,0,0,0)\; , \qquad y&#039;^{\mu}=(y&#039;^{0},0,0,0) 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 \Delta S^{2}=(x^{\mu}-y^{\mu})^{2}=(y^{0})^{2}-(y^{3})^{2}=0 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.
 
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What part are you having trouble with? It looks ok.
 
DEvens said:
What part are you having trouble with? It looks ok.

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.
 
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"Don't panic!" said:
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.

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.
 
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Mentz114 said:
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.

Ok cool, thanks for the tip. Glad I've understand it correctly.
 
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