# Help with a little Gedanken experiment please?

1. Feb 15, 2005

### cepheid

Staff Emeritus
Help with a little "Gedanken" experiment please?

Hello.

I was reading the chapter on special relativity from Brian Greene's The Elegant Universe, just to aid my understanding of the basics. SR isn't coming easily to me. During the discussion of the speed of light, and the build up to the postulate that the speed of light is constant in all inertial frames of reference, Green gives an example of two people playing catch. He does so to illustrate the pitfalls of the Galilean addition of velocities when applied to situations involving light. Basically the scenario is a guy's playing catch with his friend when a thunderstorm arrives and they duck for cover. When it passes, they resume the game but the friend has gone wild for some reason and tosses the guy a hand grenade instead of the ball. They are initially separated by distance d.

The guy turns to run away, and Green notes that our common sense would suggest that the velocity of approach of the grenade is the velocity with which it is thrown minus the velocity at which the guy flees (call this v). Then Green asks us to reconsider the situation replacing the grenade with a "laser gun". He points out that, absurd as it may seem, the shooter sees the photons chasing the guy at c, and the guy sees them approaching at c, NOT at c-v (for this would violate the principle of relativity: Maxwell's electrodynamics is supposed to be frame invariant). Nevertheless, it seemed paradoxical to me, even though I have studied SR in school before, when I swallowed it without question.

It occured to me that if I could nail down exactly why I thought it was crazy, my understanding of how SR requires us to alter our notions of space and time would deepen. Upon a second reading, something *clicked*. I thought of it this way: in the frame of the fleeing guy, we regard him as stationary, jogging on the spot, with the world whizzing by him at v. His crazy friend is receding behind him at v, but this relative motion between observer and source doesn't affect the velocity of the oncoming photons. So assuming the guy turned to flee just as his friend fired, then the very first photon to emerge would have only distance d to cover (remember the fleeing dude is stationary in this frame). In the other frame (the rest frame of the shooter), the photon is still chasing him at c, but has a longer distance to cover! So given that the laser beam travels at the same velocity in both frames, but has to go farther in the latter, it would seem that the shooter would predict that more time would elapse before the guy got hit than the guy himself would predict. In other words, the fleeing guy predicts he will get hit BEFORE the shooter's predicted time! It's nuts. From the point of view of both observers, the guy has to either have been hit already, or not yet! THEN I realised what the resolution was. My statement above:

"In other words, the fleeing guy predicts he will get hit BEFORE the shooter's predicted time!"

is valid only if you consider time to be absolute and immutable, ticking away the seconds dispassionately, to use the colourful language of Brian Green. If you don't make that assumption about time, then you can instead conlude that the shooter sees the guy get hit "when" the guy feels himself get hit, but the elapsed time up to that event is different as measured by the two friends, ie they don't agree on how long it took. The shooter thinks it took longer. The fleeing guy's watch appears to have been running slow (to the shooter), for he has registered less elapsed time. It may seem mundane to you guys, but this was a profound realization for me, because I had gone from the postulate that c is constant in all inertial frames STRAIGHT to the conclusion that the elapsed times for an event to occur would be different as perceived by observers in relative motion at constant velocity. There was no hoopla in arriving at the conclusion. Much to my satisfaction, it arose as a necessary consequence of the postulate. The relative nature of space and time no longer seemed so hard to believe.

That having been said, I wanted to make sure there were no flaws in my reasoning. The only way to do that, I thought, was to set up the situation, and try to derive the exact factor by which time had slowed down for the fleeing guy as perceived by the shooter. I hoped that the answer I would get would be the familar relativistic gamma, but I haven't had any success. Sorry for the huge preamble, here's my work:

Fleeing guy's frame S' (moves at speed v wrt shooter's frame: S)

The photon that is emitted just as the guy turns to run has only to cover distance d (because the guy is stationary in this frame). So the time to impact will be:

$$t' = d/c$$

S-frame:

The shooter sees the photon close the gap between himself and the target, but since the target is in motion, that distance is larger. If it takes time t to impact, it has to cover a distance:

$$ct = d + vt$$

(note, My entire problem may be that the RHS of this eqn is wrong, but I don't know how to correct it. My reasoning leads me to arrive at that expression.)

$$ct - vt = d = ct'$$

$$t(c - v) = ct'$$

$$t = \frac{ct'}{c - v}$$

$$t = \frac{t'}{1 - v/c}$$

Umm...that's not the right answer. It should be t as measured by the shooter is GAMMA times the fleeing guy's proper time, right? Can somebody please help me fix this up?

2. Feb 15, 2005

### JesseM

The problem here is that if the distance is d in the S' frame, it's not d in the S frame. For one thing, there's Lorentz contraction, which says that if I see you running by me at velocity v carrying a ruler which has length L in your frame, then in my frame the ruler's length is shrunk down by $$L / \gamma = L \sqrt{1 - v^2/c^2}$$. Further complicating things is the "relativity of simultaneity", which says that if two events at different spatial locations happen at the same time-coordinate in one frame, then in another frame they have a different time-coordinate. So, if in the fleeing guy's frame, the laser beam was emitted at the same time that he was passing a certain rock on the ground, in the shooter's frame he was at a different position when the laser was emitted. The relativity of simultaneity is a consequence of the fact that each observer synchronizes his own set of clocks using the assumption that light travels at the same speed in all directions in his own frame--so if I am on a rocket in motion relative to you, and I turn on a light at the midpoint of the rocket and say clocks on either end are "synchronized" if each one reads the same time at the moment the light hits them, then in your frame these two clocks will be out-of-sync, since one clock was moving towards the point where the light was emitted while the other is moving away from it, so from your point of view the light will hit one clock before the other.

In relativity there is something called the "Lorentz transformation" which tells you how to transform from the coordinates of an event in one frame to the coordinates of the same event in another. If you have frame S which uses coordinates x,y,z,t and frame S' which uses coordinates x',y',z',t', and their spatial axes are all parallel and the origin of S' is moving along the x-axis of S with velocity v, and the origins of each frame coincided at time t=t'=0, then the transformation looks like:

$$x' = \gamma (x - vt)$$
$$y' = y$$
$$z' = z$$
$$t' = \gamma (t - vx/c^2)$$

And

$$x = \gamma (x' + vt')$$
$$y = y'$$
$$z = z'$$
$$t = \gamma (t' + vx'/c^2)$$

So suppose the laser is emitted at the origin at t=t'=0, and in the S' frame, the fleeing guy is passing a certain rock at x'=d at t'=0. Then in frame S, the event of the fleeing guy passing that rock would happen at:
$$x = \gamma d$$
$$t = \gamma (vd/c^2)$$

So at t=0 in frame S, the fleeing guy must have been at $$x = \gamma d - v * (\gamma (vd/c^2))$$ since he is moving at velocity v in this frame, so this insures that after a time of $$\gamma (vd/c^2)$$ he will have moved to the position $$x = \gamma d$$. Now you should have enough information to find what each frame predicts will be the time on the fleeing guy's watch when the laser catches up to him (remember that if frame S says the laser will catch up to him at time $$t_1$$, then it should predict the fleeing guy's watch will read $$t_1 / \gamma + \tau_0$$, where $$\tau_0$$ was the time on the fleeing guy's watch at t=0 in frame S.)

Last edited: Feb 15, 2005
3. Feb 16, 2005

### cepheid

Staff Emeritus
Yeah, I'm aware of the Lorentz transform. I should have mentioned that we have done SR in school in one course in modern physics, and are currently revisting it in electrodynamics. We delved into four vectors and everything, and naturally I had no idea what the !@#\$ was going on. Neither course offers sufficient time for it to really sink it for me. Mathematically I could just accept the Lorentz transform and be done with it, hoping that some deeper understanding would come later. But it bothers me that I can't arrive at the correct answer just using the postulates, and setting up this thought experiment. The Lorentz transform is fine and dandy. How do we arrive at it systematically? I'm aware that the derivation is in any text, so I won't ask anyone to answer that. I'll just research it further. Thanks for your help.

4. Feb 16, 2005

### Janus

Staff Emeritus
The problem with your approach is that it involves three effects of Relativity at once. A better way is to isolate on effect and work from there.

Example: instead of dealing with light traveling between two frames that are moving relative to each other, start with the light travleing between two points in the same frame, and compare how the other observer measures it. Also, in order to eliminate length contraction as a factor, have the light traveling at a right angle to the relative motion between the frames. The classical example would be to have the light travel between a source and a mirror in the same frame and back again. Deterine how long it takes for the light to make the round trip according the the frame of the source and mirror and compare it to how long it takes to make the same trip according to a frame with a relative motion with respect to the first.

This will give you the time dilation factor.

Then you can add a second mirror the same distance from the source(as measured by the source) as the first but aligned with the relative motion. Now, remembering that both light pulses must return to the source at the same time according to both frames, you can work out how far the second mirror must be from the source according to the relatively moving frame.

This will give you the length contraction factor.

5. Feb 16, 2005

### cepheid

Staff Emeritus
I think that is exactly what I was looking for. It's much appreciated. I will try it out. Thanks!

6. Feb 16, 2005

### JesseM

Another thing you can derive from scratch is the factor that an observer in one frame will see clocks in another frame being out-of-sync. Suppose I see two clocks flying by me at the same velocity v, and they are a distance x apart in my frame. If a light is turned on at their midpoint (at a distance x/2 from where each one is at that moment), then what time will I see it hit each one in my frame? Knowing that an observer at rest with respect to these clocks will see them as "synchronized" in their own frame if they both read the same time at the moment the light hits each one, I can figure out how much the clocks will be out-of-sync in my frame. The answer ends up being that I see the back clock ahead by $$\gamma (vx/c^2)$$, which gives a slightly more intuitive way of understanding the fact that if the fleeing guy sees himself a distance d from the laser at the moment it is fired, then in the shooter's frame he will be a distance of $$\gamma d - v * (\gamma (vd/c^2))$$ at the moment the laser is fired.