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Consider the following two points, or events as they are more commonly called, in SpaceTime:

Event 1: $(x,t) = (0,0)$

Event 2: $(x,t) = (a,0)$

As you can see they are merely two separate locations, Event 2 could be any point in space. The word 'event' here is a misnomer, because it carries with it the connotation of time, and the two events are simultaneous in the unprimed frame. In other words the events are simply separate points in space.

For two coordinate systems in standard configuration, the general expression from the Lorentz transformations relating the t' coordinate to the t coordinate with $c=1$ is:

$t' = \gamma(t - vx)$

So

$t'_0 = \gamma(t_0 - vx_0)$

$t'_0 + ∆t' = \gamma(t_0 + ∆t -v(x_0+∆x) )$

Taking the difference yields

$∆t' = \gamma(∆t - v∆x )$

So for the two events at the start of my question, that are simultaneous in the unprimed frame, we have

$∆t' = \gamma(0- va )$

According to the relativity of simultaneity, what is one moment in time in the unprimed frame, corresponds to two moments in time in the primed frame.

This is all because $∆t'$ isn't zero, and that is the case because $\Delta t'$ is a function of spatial coordinates.

Now suppose there is a particle at location $(a,0)$ whose $instantaneous$ speed is greater than zero. Let its instantaneous speed be the constant $v_0$. Define $a=v_0 dt >0$. Therefore $∆t'=\gamma(-vv_0dt)$. Therefore $∆t' <0$, but amounts of time are strictly non-negative.

Suppose $\Delta t′<0$ simply means that event 2 occurred before event 1 in the primed frame, and $\Delta t′>0$ means that in the primed frame event 2 occurred after event 1.

Let the particle start out at $a=-dx$, amount of time $2dt$ before event 1, and be moving to the right in the unprimed frame with speed $v_0$. Call that moment Event 0. Clearly $dx>0$ and $dt>0$.

By that supposition Event 0 occurred after Event 1, and Event 1 occurred after Event 2 in the primed frame. But the particle is moving rightwards faster than v. Therefore in the primed frame event 0 is before event 1, and event 1 is before event 2. Therefore event 1 is before and after event 2 in the primed frame.

To rigorously prove this requires the use of the Einstein velocity addition rule.

$u' =\frac {u-v}{1-uv}$

The Lorentz transformations give $t=\gamma (t'+vx')$

So $dt=\gamma(dt'+vdx')$

So $dt=\gamma dt'(1+v\frac {dx'}{dt'})$

Therefore Event 0 is $ (-v_0\gamma dt'(1+v\frac{dx'}{dt'}),-\gamma 2dt'(1+v\frac{dx'}{dt'})$

Now $\frac {dx'}{dt'}= \frac {u-v}{1-uv}$

And $u=\frac {dx}{dt}=v_0$.

So my question is,"do the Lorentz transformations lead to negative amounts of time?"

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    $\begingroup$ Therefore $∆t' <0$, but amounts of time are strictly non-negative. Why do you think a time difference can’t be negative? $\endgroup$
    – Ghoster
    Feb 10 at 22:30
  • $\begingroup$ I am thinking of $\Delta t'$ as an amount of time, not a time difference with arbitrary start moment and end moments. I am thinking time flows unidirectionally, so the end moment must come after the start moment. $\endgroup$
    – lee pappas
    Feb 10 at 22:37
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    $\begingroup$ You will need to revise your thinking. Time flowing in one direction is unrelated to the temporal ordering of spacelike-separated events in different Lorentz frames. $\endgroup$
    – Ghoster
    Feb 11 at 3:57
  • $\begingroup$ But if I do that we get the contradiction I stated below. I will restate it here. Let the particle start out at a=-dx, dt before event 1, and be moving to the right with speed v0. Call that moment Event 0. By your logic Event 0 occurred after Event 1, and Event 1 occurred after Event 2. But the particle is moving rightwards faster than v. Therefore in the primed frame event 0 is before event 1, and event 1 is before event 2. $\endgroup$
    – lee pappas
    Feb 11 at 10:03
  • $\begingroup$ That is not a contradiction. You now have three different answers telling you the same thing. $\endgroup$
    – Ghoster
    Feb 11 at 17:55

4 Answers 4

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The word 'event' here is a misnomer, because it carries with it the connotation of time, and the two events are simultaneous in the unprimed frame

The word event is correct, not a misnomer. The fact that these two events are simultaneous doesn’t mean that they are not events. For example, one event could be me getting stung by a bee and the other could be you clicking the button to submit this question. Even if those two events happened at the same time they are still events and not locations. There were things that happened at each location before and after these events.

Therefore $Δt′<0$, but amounts of time are strictly non-negative.

$\Delta t$ is a time difference and it definitely can be negative. For example, when launching a rocket engineers at NASA will measure $\Delta t$ relative to liftoff. So they will say “t minus ten seconds” meaning that it is ten seconds before liftoff or equivalently negative ten seconds after liftoff.

So here $\Delta t’<0$ simply means that event 2 occurred before event 1 in the primed frame. Similarly, $\Delta t’>0$ would mean that in the primed frame event 2 occurred after event 1. Both positive and negative values are well defined and meaningful.

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  • $\begingroup$ Let the particle start out at a=-dx, dt before event 1, and be moving to the right with speed $v_0$. Call that moment Event 0. By your logic Event 0 occurred after Event 1, and Event 1 occurred after Event 2. But the particle is moving rightwards faster than v. Therefore in the primed frame event 0 is before event 1, and event 1 is before event 2. $\endgroup$
    – lee pappas
    Feb 10 at 22:24
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    $\begingroup$ @leepappas I am not sure about your math, but certainly there are some sets of events such that $t_0<t_1<t_2$ and $t’_0>t’_1>t’_2$ for any pair of distinct inertial frames $\endgroup$
    – Dale
    Feb 10 at 23:06
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The order of some events (those that are space like separated i.e. for which $c^2 \delta t^2 - \delta x^2 - \delta y^2 - \delta z^2 < 0$) is undetermined. In some frames event A comes before event B, in others they are simultaneous, and in others B comes before A. This is a well known consequence of the Lorentz transformation, which you've derived correctly.

Note that such events cannot affect one another by any influence that travels at or below light speed. So which one came first, if either, makes no difference at all: the events are necessarily independent. No paradox can arise from the temporal order changing, because information cannot travel faster than light.

Conversely, for events that are time like or light like separated, all observers agree on the temporal order of the events.

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In SR the time order of events can be reversed by a change of coordinates, provided the events are sufficiently far apart that light from one can't reach another. In that case, none of the events can have any effect on the others, so the order in which they happen is irrelevant. If the events are not that far apart, then they occur in the same order even if you switch reference frames- that ensures that causality is not effected by a switch in coordinates, as you might hope. The fact that negative amounts of time crop up is not a problem. If the time difference between two events is negative, that's just a convention which one happened before the other.

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Yes, according to the interval metric:

1. From the origin, to get to a timelike distance event (inside the null cone), you have to move at less than c (or sit in the same location at zero speed) and experience positive elapsed time. Sitting in the same place and not moving, you experience the maximum elapsed time. Any motion causes time dilation.

2. From the origin, to get to an event on the null cone, you'd have to move at c and experience zero elapsed time.

3. From the origin, to get to an event at a spacelike distance, you have to cross the null surface, exceed c, and experience negative elapsed time.

In reality, you can't do this because the speed of mass objects must always be less than c. You can never cross the null surface. That's why nothing in the actual universe experiences negative elapsed time.

It's also consistent with the idea that if you could travel faster than light, you'd be traveling subluminally, but in the reverse time direction, relative to the people and objects around you. You would perceive them as moving backwards in time, and they would perceive you as moving backwards in time.

Feynman says that normal matter moving in reverse time is antimatter.

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    $\begingroup$ Thank you Miss understands for your answer. My question to you is does that imply that a particle can be moving in two different directions simultaneously, if the Lorentz transformations are true? $\endgroup$
    – lee pappas
    Feb 12 at 14:54
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    $\begingroup$ Miss understands what do you mean by From the origin, "to get" to an event at a spacelike distance,? $\endgroup$
    – lee pappas
    Feb 12 at 18:43
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    $\begingroup$ @leepappas ? A path across an interval, to get from here to there. Mass can't do it. $\endgroup$
    – L Turner
    Feb 17 at 5:16
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    $\begingroup$ @leepappas "does that imply that a particle can be moving in two different directions simultaneously, " Yes. X and Y directions, for example = diagonal line. Time is a direction, like any other. You add it to the distance, just like other directions. Only you multiply it by c, unnecessary in natural units. And , oh yeah, it's negative. Hey, don't blame me, blame God. I'm just his humble interpreter. $\endgroup$
    – L Turner
    Feb 17 at 5:23

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