# Faster than light communication and signals moving backwards in time

1. Feb 4, 2006

### robousy

Hi,

Can someone help explain why something that has the ability to move faster than c can potentially send a signal backwards in time - thus ruling out superluminal travel.

If a 'superluminal' ray starts out at t=0 then even if it does travel faster than light, any information it transmits anywhere in space will reach that point at t>0.

As I understand things the 'backwards in time' signalling occurs when you consider Lorentz Invariance but I would like this to be explained a little more.

Thanks

2. Feb 4, 2006

Staff Emeritus
Here's an old example that I've posted before. I'm going to assume the signal is instantaneous, rather than just FTL to simplify the discussion, but you can work through the example with FTL and show that it works the same.

Suppose you send an instantaneous signal to your friend who is a light year away and moving away from you at high sublight speed, and he reflects it back to you, also instantaneously.

Now in a spacetime diagram with x-axis the space direction between you and your friend and y-axis for time, the signal travels along your x-axis from you to your friend without making any movment in the y-direction, because it's instantaneous, and no time passes for it. So two points on the x-axis, for you and your friend and a straight line between them for the signal.

Now your friend, because she is traveling fast relative to you, has her own x-axis, which is tilted relative to yours. The angle of tilt between your x-axis and hers depends on the speed she is going relative to you (or you to her; it's symmetric). So her x-axis, passing through her point on your x-axis and making an angle with it, intersects your y-axis in your past. When she sends the instantaneous signal back to you it travels along her x-axis arrives in your past, before you sent the signal!

Do draw the picture and convince yourself that what I've said here is kosher relativity.

3. Feb 4, 2006

### JesseM

Because different inertial frames in relativity define simultaneity differently, anything going faster than light but forwards in time in one frame will be going backwards in time in another (ie the order of the events of the signal being emitted and received will be reversed in this frame). And so if tachyons obeyed the requirement that the laws of physics should look the same in all inertial frames (lorentz-invariance), this means it should be possible for them to go backwards in time in any frame.

Imagine we are moving apart at 0.9c, and I send you a tachyon signal which travels faster-than-light in my frame, but backwards-in-time in yours. Then you send me a reply which travels faster-than-light in your own frame, but backwards-in-time in mine. In this case it will be possible for your reply to reach me before I sent the original signal. Try drawing the lines of simultaneity for two observers moving apart at relativistic velocities, and you may get a better idea of how this works. Or you could look at the diagrams on this page:

http://www.theculture.org/rich/sharpblue/archives/000089.html

4. Feb 5, 2006

### Longstreet

Here is the lorentz transform for time between two observers:

$$t' = \frac{t - x(u/c^2)}{\sqrt{1-u^2/c^2}}$$

Say we have one reference where two people are exchanging signals in the solar system. All the parties here are at rest relative to each other. t represents at what time the signal is either transmitted or received in their frame. x is the distance seperating the people signalling each other.

Now suppose there is a space ship passing by parallel to the line of communication going at a significant fraction of c, which is represented by u. t' represents the time of transmission or reception in their frame.

Lets plug in some quick numbers just to test the equation:

$$t_1 = 0s$$ (time of transmission)
$$t_2 = 1s$$ (time of reception)
$$x_1 = 0m$$ (place of transmission)
$$x_2 = 3*10^8m$$ (place signal was received)
$$u = 0.6c$$ (relative velocity of spaceship)

We have two t' to calculate, one for transmission time and one for reception time. When I punch this into my calculator I get this result:

$$t'_1 = 0s$$ (time of transmission for spaceship)
$$t'_2 = 0.5s$$ (time of reception for spaceship)

This is assuming the signal traveled at c exactly. What if we said it traveled at 2*c? In other words change the reception time:

$$t_1 = 0s$$ (time of transmission)
$$t_2 = 0.5s$$ (time of reception)

Now lets see what the spaceship would see. Here is what I get from my calculator:

$$t'_1 = 0s$$ (time of transmission for spaceship)
$$t'_2 = -0.125s$$ (time of reception for spaceship)

It apears that the signal was received 0.125s before it was transmitted, at least from the spaceship.

First you can notice that if $$x(u/c^2)$$ is non-zero and positive, and t was less than that value, then t' will be negative. You might say, hey, lets just change the sign of the coordinate system? Well, that's a problem because if you flip the direction of x, then you flip the sign of the velocity for the ship too. The two signs would cancel out if you did this resulting in the same number.

Last edited: Feb 5, 2006
5. Feb 6, 2006

### robousy

Thanks for the replies everyone.

I have another question and it is really with regards to everyones answer so far which essentially boils down to the Lorentz transformations which can be neatly explained using space-time diagrams.

Is it possible that the LT's break down at superluminal velocity?

Perhaps I'm just not thinking about this like a physicist but the idea of a signal propogating backwards in time seems absurd.

I know relativity has been tested and verified so do not stand to contest this wonderful theory - but does anyone think that it is more likely that the LT's break down in this regime.

6. Feb 6, 2006

### JesseM

The Lorentz transforms are just equations for transforming between different coordinate systems, but in theory what could break down is the principle that the laws of physics work the same way in each of the coordinate systems given by the Lorentz transform (Lorentz-symmetry). For example, there might be a single preferred frame where tachyons could move arbitrarily fast in any direction but not backwards in time, which in other frames would look like a rule that tachyons have a limited FTL speed in one direction but can move backwards in time in the other, in such a way that actual time travel (visiting the past of your own worldline) is impossible. So in my frame I could visit a star 6 light-years away and get there 3 years in the past according to my frame's time-coordinate, but when returning in the other direction I could travel at a maximum of 2c in my frame so I couldn't return before I left. Most physicists would probably consider it unlikely that a basic symmetry like Lorentz-symmetry would turn out to be wrong at a fundamental level, though. Here's a usenet post on reasons physicists are skeptical about aether theories, most of the arguments would apply to any theory that introduces a preferred reference frame:

7. Feb 6, 2006

### Longstreet

Does anyone know the equation to see the return time given an arbitrary speed? I've been trying to make one but it's giving me non-sensical answers.

$$t_b = t_t+ \frac{1-k'(u/c)}{k'c - u}(x_r-x_b) + \frac{x_r-x_t}{kc}$$

t_t = transmit time
t_b = return time
x_t = transmit location
x_r = receiver location
x_b = return location
k = ftl constant for ftl test observer
k' = ftl constant for the other inertial observer
u = relative velocity between test and other observer

Not only is this giving me positive return times no matter how fast the k, but k lower than the speed of light result in time travel, or is completely undefined. I'll post by derivation if anyone is seriously interested in finding the problem. But I figure it's already been done somewhere.

8. Feb 6, 2006

### JesseM

Just to be clear, is the first observer at rest relative to the emitter? And is the reply signal moving at k' in the second observer's frame, while the original signal moved at k in the first observer's frame? If not, can you specify what k and k' are the velocities of?

9. Feb 6, 2006

### Longstreet

Observer A has a transmitter at x_t and a receiver at x_r set to send a signal in the +x direction. And observer B moves at u in the +x direction relative to A which has a transmitter and receiver set the send a signal in the -x direction.

B's transmitter is set to pass A's receiver located at x_r at time (x_r - x_t)/(k*c). So k is the multiple of c that the signal is traveling. The signal is then resent by B to it's receiver in it's frame. k' is the same thing but in B's frame. They can be the same but I made them seperate just for generality. B's receiver will be at x_b at time t_b when the signal is received. I'm not concerned about B's reference of the events. Essentially when B's receiver gets the signal at t_b it broadcasts it at c in A's frame, so x_b could be anywhere, although we're mostly interested when x_b = x_t (same starting and stoping point).

Last edited: Feb 6, 2006
10. Feb 6, 2006

### JesseM

Probably the simplest approach here is just to figure out how fast a signal will be moving in your frame if it is moving at k'c in the -x direction of a frame that is moving at u in a +x direction relative to you.

Say that in this other frame, a signal is emitted at x'=0 ,t'=0 and after a time interval of t0 in that frame it is at x'=-k'*c*t0, t'=t0. We can use the Lorentz transform to find the coordinates of these events in our frame:

$$x = \gamma (x' + ut')$$
$$t = \gamma (t' + ux'/c^2)$$
with $$\gamma = 1/\sqrt{1 - u^2/c^2}$$

So the coordinates of the first event are x=0, t=0 while the coordinates of the second are $$x = \gamma (-k'ct_0 + ut_0) = \gamma (u - k'c) t_0$$ and $$t = \gamma (t_0 - uk't_0 /c) = \gamma (1 - uk'/c) t_0$$. So, the speed is distance/time, or (u - k'c)/(1 - uk'/c). You could also find this just by using the formula for velocity addition in relativity with v = k'c.

Since k'c is faster than light and u is slower, no matter what k and u are we know the numerator u - k'c will be negative. The denominator will be positive when the signal is moving forward in time, negative when it's moving backwards in time. So if we know the signal was emitted by B at position $$x_r$$ and time $$t_r$$, and it had to go a distance of $$x_r - x_b$$ to B's receiver, then it should reach B's receiver at time $$t_r + (x_r - x_b)/(-v_b)$$, where $$v_b = (u - k'c)/(1 - uk'/c)$$ which again should be negative when the signal is moving forwards in time, which you can see from the above formula means the signal reaches B's receiver at a time greater than $$t_r$$, and positive when the signal is moving backwards in time, which means the signal reaches B's receiver at a time less than $$t_r$$.

So, all that's left is to note that $$t_r = (x_r - x_t)/kc$$ as you said earlier, and I would conclude that $$t_b = (x_r - x_t)/kc + (x_r - x_b)(1 - uk'/c)/(k'c - u)$$. This is the same thing you got, except for that extra $$t_t$$ in yours which doesn't fit with the statement that the first signal will reach A's receiver at $$(x_r - x_t)/kc$$...if you wanted the original signal to be sent at an arbitrary time $$t_t$$, then that should be $$(x_r - x_t)/kc + t_t$$. Apart from that minor issue I don't see a problem.

Last edited: Feb 6, 2006
11. Feb 7, 2006

### Longstreet

Oh ya, I meant $$t_r = t_t + (x_r - x_t)/kc$$. I see now that the graph actually does make sense. I got scared by the undefined part when k*c <= u. But I guess that makes sense since if B tried to send a message back to $$x'_t$$ slower than the relative velocity it wouldn't make it.

Graph:
$$t_b$$ on vertical and $$k=k'$$ on the horizontal where $$x_r - x_t = x_r - x_b = 1cs$$ (1 light second), $$t_t = 0$$ and u = 0.6c. It does (edit: it already violated causality in B's frame. I'm concerned with A's perspective) start time traveling around k > 3.

Thanks,

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Last edited: Feb 7, 2006