# 'Speed of Light' Thought Experiment

1. Oct 26, 2004

### Thomas2

Consider the following situation: a car is moving away from a stationary observer with velocity V. At distance S it triggers a contact in the road which sends a light signal back to the observer from the moving car. The question is how long will it take for the signal to reach the observer ? (The time could be measured by having synchronized clocks at distance S (the trigger point) and at the observer, the former being stopped when the car passes the contact (and emits the light signal), the latter when the light signal arives at the observer; the difference of the clock readings is then the light travel time).

Last edited: Oct 26, 2004
2. Oct 26, 2004

### James R

The light signal has to travel distance S at speed c, so the time of arrival of the signal is S/c, right?

Of course, the question of how the trigger manages to start the observer's clock instantaneously is a potential problem, but I assume that's not important.

3. Oct 26, 2004

### Thomas2

Just to avoid confusion: the light signal is sent out from the moving car (I have clarified the opening post in this respect).

4. Oct 26, 2004

### James R

I already assumed the light signal was sent from the moving car, so my previous answer stands.

Another problem is that you haven't specified the reference frame in which the observer's clock is started simultaneously with the emission of the light signal. My answer assumes the emission and the clock start are simultaneous in the observer's frame.

What are you having trouble with?

5. Oct 26, 2004

### Thomas2

I have modified the example in my opening post somewhat so that there are no 'simultaneity' issues with the timing procedure.

OK, if you don't have any problems with it and your suggestion is the time is S/c then we would agree (the light travel time depends only on S but not on the velocity V).
But one can consider the whole thing from the viewpoint of the car as well : when passing the contact at distance S from the observer, the car emits the light signal. The observer is receding now with velocity V, but still the signal should reach him within a time S/c despite the fact that the distance has increased in the meanwhile from S to S*(1+V/c).

Last edited: Oct 26, 2004
6. Oct 26, 2004

### Janus

Staff Emeritus
No, from the car's viewpoint, the light will take

$$\frac{S\sqrt{1-\frac{V^2}{c^2}}}{c-V}$$

to reach the observer, assuming that S is the proper distance between the contact and observer as measured by the observer.

7. Oct 26, 2004

### RandallB

Wow when & how did S to Observer get longer from the car view??
I thought it got shorter from the moving view.
Plus time runs slower in the moving car.
The little shorter length divided by the shorter time still = c

What is the speed (c-V)
Is there something going near the speed of light here?

RB

Last edited: Oct 26, 2004
8. Oct 26, 2004

### Janus

Staff Emeritus
Where did anyone say differently?
Time runs slower for the car as measured from the frame of the contact and observer, but at this point we are talking about the frame of the Car, in which the time oif the car runs normally.
(c-V) is the difference between the speed of light and the relative speed between the car and observer. The car sees the light travel at c wrt itself at c and the observer recede at V. Thus the "apparent" relative speed between the light and observer according to the car is (c-V), Divide this into the distance that the car measures between contact and observer and you get how long it takes from the car's frame for the light to "catch up" to the "receding" observer.

9. Oct 27, 2004

### James R

The travel time for the light will be different according to the ground-based observer and a car-based observer.

The time for the ground observer is just S/c, as I said before.

In the car's reference frame, the distance S is contracted to a distance of

$$S\sqrt{1-(v/c)^2}$$

where v is the car's speed, and c is the speed of light. Therefore, the time taken, according to the car, is the above distance divided by c.

It is NOT divided by (c-v), as Janus said, since the car observer still sees the light travelling at c, despite the car's motion.

10. Oct 27, 2004

### Thomas2

I should have made the reversal of the situation differently:
assume that the light signal is sent from the observer to the car rather than the other way around (the signal is sent out (and the observer clock stopped) when the car is at distance S and the car stops one of a number of synchronized clocks alongside the road when the signal is received).
Since it cleary can not make any difference for the light travel time whether you send a signal from A to B or B to A (even if both are moving relatively to each other) the travel time should then still be S/c here.

11. Oct 27, 2004

### Staff: Mentor

Is this your setup: A ground observer at location x = 0. When a car traveling at speed v passes position x = S, the ground observer initiates a light signal from x = 0 to the car. The signal reaches the car when the car is at a position x = S' (where S' > S). (All distances measured in the ground frame.)

Huh? How could it not make a difference? If the signal is sent from ground observer to car, it must travel a distance S'. But if sent from car to ground observer, it need only travel S. (All distances measured in the ground frame.)

12. Oct 27, 2004

### Thomas2

The question was not at what distance the car is when the light signal reaches it, but what the clock shows at this moment.
Your conclusion that the larger distance S'>S is also associated with a larger time T'>T is based on the usual velocity addition rule, which however (as I am sure you are aware) can not be applied to light signals (any such expression as "c-v" contradicts the invariance property for the propagation of light). Only a velocity independent travel time T'=T= S(t=0)/c would be truly independent of any motion of emitter or observer (as the invariance principle for light requires).

13. Oct 27, 2004

### Staff: Mentor

Nonsense. The car moves, so of course it takes longer (as measured by the ground observer) for a signal from the ground observer to the reach the car than it would for a signal from the car to reach the ground observer. (I assume that in both scenarios the car is the same distance from the ground observer when the signal is initiated.)

And what makes you think that figuring out the distance involves some kind of "velocity addition rule" or that having an expression of "c-v" somehow contradicts the invariance of the speed of light?? Get a grip! In figuring out the time of travel of the signal, both frames would make use of the invariance of light speed. Show me where someone is claiming otherwise.

It's trivial. If the ground observer measures the time for a light signal from him to the car he must start with the distance the light must travel in that time, thus: ct = S + vt, so t = S/(c-v).

Of course, if the signal went from car to ground observer, it would only take t = S/c (according to the ground observer), since the car's motion would not affect the distance the light has to travel.

14. Oct 28, 2004

### Thomas2

This is the formula you would use to take into account the relative velocity of normal objects: if you have one object moving with velocity v1 and a second moving with v2 (starting when the other object is at S), you would indeed have v2*t = S + v1*t if you want to find the time when both have the same location, but this is the usual vectorial velocity addition (Galilei transformation) which can not be applied to light signals, i.e. you can not set v2=c here.

As I said already, the time for the signal to get from A to B should always be equal to the time to get from B to A, whether it is for normal objects or light (this should be a straightforward consequence of the principle of relativity and the isotropy of space and time).
For light signals, a velocity dependence appears only in form of the Doppler shift of the light frequency but it should not appear in form of different light travel times.

15. Oct 28, 2004

### pervect

Staff Emeritus
But if I'm understanding this problem correctly, there are three points involved, A,B, and C - not two points. So I don't see how this comment is relevant.

Let A be the fixed observer, and B be the car, and S be a bump in the road.

In one case we have

A--------------B-------C
---------------S

When the car B reaches point S, a signal is sent from point A towards B. This has to be pre-arranged based on knowledge of how fast the car is moving, the car reaching point B cannot cause a signal to be sent from point A instantaneously.

The signal reaches the car at some later point C after the car has moved.

In the second case we have

A----------------B
-----------------S

When car B reaches point S, it causes the car to emit a signal. This signal travels from point B back to point A. Here, there is an actual cause and effect relatinoship between the car being at point S and the signal being emitted. And there is no point C, there is only point A and point B.

The time it takes for light to travel from point B to point A is not the same as it takes light to travel from point A to point C.

16. Oct 28, 2004

### Staff: Mentor

Sorry, but this has nothing whatsoever to do with addition of velocities. Both the speed of light (v2 = c) and the speed of the car (v1 = v) are measured in the same frame. (It is just an application of "distance = speed x time"; as long as all measurements are made from the same frame, no problem.)
Just because you keep saying it, doesn't make it true. See pervect's post illustrating that the distance traveled is different in each case.
The speed of light is invariant, but the time it takes for light to travel depends on the distance involved.

Last edited: Oct 28, 2004
17. Oct 28, 2004

### Thomas2

No, there are in fact only two points involved here: the observer A and the car B (betweeen which the light signal is exchanged). Point C only enters into the picture if you assume that the light signal is an independent entity that can be classically described as having a velocity c with regard to A and c-v with regard to B (see also my reply to Doc Al below).

18. Oct 28, 2004

### Thomas2

This is exactly the point: by plotting the velocity of the car v and the velocity of the light signal c in the same frame, you are implying that the light signal moves relative to the car with velocity c-v, which contradicts the invariance of c.
As I said already, it is not the question here when a light signal in the observers frame reaches a point at which the car happens to be as well at that moment, but when the light signal reaches the car in its frame. These two things would only be identical if you have the usual vectorial velocity addition (which however does not hold for light signals). The usual definition of 'speed'=distance/time does simply not apply here.

19. Oct 28, 2004

### Staff: Mentor

The car moves!
Nonsense! Both ground observers and car observers measure the speed of light to be c with respect to themselves. (That's what invariance of light speed means!) That "c-v" is not the speed of light as measured in the car frame. That "c-v" is not the speed of any "thing", it's just the rate at which the ground observer sees the light gain on the car. You do realize that no matter how fast the car moves, the light will catch up... don't you?

20. Oct 28, 2004

### Staff: Mentor

Nonsense. See my post above to learn what "c-v" actually means.
In the car's frame, the light travels a distance of $S/\gamma$, so the travel time is just $S/(\gamma c)$ according to the car frame.
Nonsense. If the signal is now sent from car to ground observer, then according to the car frame the distance the light must travel is: $ct = S/\gamma + vt$. So, in that case, the travel time according to the car frame is $S/[\gamma (c-v)]$.
Nonsense. You just don't know what you're doing, I'm afraid.