# Inequalities. Special relativity.

1. Apr 10, 2015

### LagrangeEuler

Lorentz transformation. Time separation between two events in system S and system S' is given by Lorentz transformations

$$\Delta t'=\frac{\Delta t-\frac{u \Delta x}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}$$
If event 2 is caused by event 1 in system S then $\Delta t=t_2-t_1$. But it is possible to choose $\Delta x$ in such way that $\Delta t'$ is negative. Which is very hard to understand. $\Delta t'<0$ only when $\frac{u \Delta x}{c^2}>\Delta t$ or

$$\frac{u}{c}>\frac{c\Delta t}{\Delta x}$$
Now I am confused. $\Delta x$ is space separation between event 1 and event 2 and $\Delta t$ is time separation between events. Is it possible to choose some reference system S' in such way that event 2 occurs before event 1, even in case when event 2 is consequence of event 1?

2. Apr 10, 2015

### phinds

Causality is not frame dependant, so no you can't pick a different frame that changes the order of cause/effect, you can only change the apparent time between cause and effect.

3. Apr 10, 2015

### Orodruin

Staff Emeritus
To add to what phinds said: It is possible to change the order only of spacelike separated events. Such events are not within each other's light cones and can therefore never have a causal relationship (either cannot affect the other). Such causes would propagate at speeds larger than the speed of light.

4. Apr 10, 2015

### DrGreg

If event 2 is caused by event 1 then$$\frac{|\Delta x|} {|\Delta t|} < c$$

5. Apr 10, 2015

### Matterwave

You can see it in your original inequality: $$\frac{u}{c}>\frac{c\Delta t}{\Delta x}$$

Rearrange this inequality and you will get $u\frac{\Delta x}{\Delta t} > c^2$. So you need either $u>c$ or $\frac{\Delta x}{\Delta t}>c$. In the former case, you are making an invalid Lorentz transformation since you are transforming to a frame moving faster than the speed of light, and in the latter case the two events can not be causally connected since they are space-like separated.

6. Apr 11, 2015

### LagrangeEuler

What is now problematic for me. For example observer sits on x-axis of reference frame S. He see red light at the distance of $1210m$ and for example $4.96 \mu s$ later blue light at the distance $510m$. If I do not have information that two events are in some corelation how could I know that maybe in some reference frame S' with velocity $u\vec{e}_x$ relative to S some observer can/cannot register blue light before red light and is it or not that forbiden to observer in S'?

7. Apr 11, 2015

### Staff: Mentor

If two events are spacelike separated in one frame, they will be spacelike-separated in all frames. You can get to this conclusion by doing some algebra with the Lorentz transformations while respecting the constraint that the relative speed between any two frames must be less than $c$.

Similar algebraic crunching will show that if two events are timelike-separated, the constraint that the relative speed between any two frames must be less that $c$ implies that the ordering of the events will be the same in all frames.

8. Apr 11, 2015

### Ibix

You need to be a little careful with "see" in this context. Do you mean that the observer receives a pulse of light from the red source at time zero, and a pulse of light from the blue source at time 4.96μs, or do you mean that the red source emits a pulse at time zero and the blue source emits a pulse at time 4.96μs? The former is what one would typically mean by "see", but the latter is what the Lorentz transforms will help you to relate[1].

I will assume that you mean that a red source at x=1210m emits a pulse that is received by an observer at x=0 at time t=0, and a blue source at x=510m emits a pulse that is received by the same observer at time t=4.96μs. We can calculate backwards and determine that the red pulse was emitted at time t=0-1210/c=-4.0μs and the blue pulse at time t=4.96μs-510/c=3.3μs. That is to say, the emission events were at (t1,x1)=(-4×10-6,1210) and (t2,x2)=(3.3×10-6,510), according to observers at rest in frame S. For observers at rest in any other frame, you would need to apply the Lorentz transforms to get the coordinates that they would record. However, whatever frame you are in, you will find that the interval, Δs2:
$$\Delta s^2=(c\Delta t)^2-(\Delta x^2+\Delta y^2+\Delta z^2)$$
is always the same ("invariant"). The sign of that quantity will tell you whether the events are time-like separated (in which case everyone will agree on their time ordering but not necessarily their space ordering) or space-like separated (in which case everyone will agree on their space ordering, but not necessarily their time ordering). WARNING: There is no agreement about the sign convention in Δs2. Some people define it as I did; some people define it with the opposite sign. In the convention I have used, positive would mean time-like separated, and the two events could be causally linked. Negative would mean space-like separated, and the two events could not be causally linked. Zero would mean null separated, and the events could only be causally connected by light rays.

[1] That is a simplification - the Lorentz transforms applied to the time you receive the light pulses will tell you the times and places that moving observers record you as receiving the pulses. Since you are probably interested in the emission events, however, this isn't terribly interesting.

9. Apr 13, 2015

### LagrangeEuler

I can not understand this without a numbers. Lets try to add some numbers. My questions is given by bold letters. Let system S' is moving along x-axis with velocity $u=0.8c$.
In system S we have two events, event 1 $(x_1,t_1)$ and event 2 $(x_2,t_2)$, such as $\Delta x=3\cdot 10^8$ and $\Delta t=5s$.

Then $\Delta x'=\frac{\Delta x-u\Delta t}{\sqrt{1-\frac{u^2}{c^2}}}=-15\cdot 10^{8}m$. Could you explain me this minus? Does that means that event 2 is closer to man which sits in S'?

And $\Delta t'=\frac{\Delta t-\frac{u \Delta x}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}=7s$

So now I want to calculate $\Delta s^2$ in both systems.
$(c\Delta t)^2-(\Delta x)^2=225\cdot 10^{16}-9\cdot 10^{16}=216\cdot 10^{16}m^2$
$(c\Delta t')^2-(\Delta x')^2=441\cdot 10^{16}-225 \cdot 10^{16}=216 \cdot 10^{16}m^2$
So it is really a scalar. If I understand you well if I have two events and I want to get space like separation $\Delta x$ needs to be huge. In that case even light can not get from one part of space to the other and because of that event 1 and 2 are not in any corelation. Right?

10. Apr 13, 2015

### Ibix

It's important to remember (and easy to lose sight of) the fact that observers at rest in one frame are moving in the other. In the frame S, your events are one light second apart in the x direction and 5s apart in the time direction. An observer at rest in S' (doing 0.8c in S) travels 4 light seconds in those 5s. If he is right beside the first event, he will be 3 light seconds past the second event when it happens. If you repeat the calculation with your frame S' moving at less than c/5 then you will find that Δx' is positive because the observer a rest in S' won't cross the light second gap in the five seconds he has.

You might find it easier to visualise with Δx=0. Imagine I am standing on the platform while you pass by in a train. At the instant you pass me, I let off a firework. Five seconds later I set off another. According to me (at rest in S) both fireworks went off in the same place (Δx=0). According to you (at rest in S') the first one went off right beside you ($x'_1=0$), but the second one went off somewhere behind you($x'_2<0$). That gives you a negative Δx'.

Strictly speaking you haven't proved that yet. Formally, you need to write $(c\Delta t')^2-(\Delta x')^2$ and use the Lorentz transforms to show that $(c\Delta t')^2-(\Delta x')^2=(c\Delta t)^2-(\Delta x)^2$ for any (x,t).

The condition is that $\Delta x>c\Delta t$. So either you need a very large x difference or a very small time difference. For example, two events one meter apart (in some frame) are space-like separated as long as the time difference (in the same frame) does not exceed 3.33ns.