# Conceptually understanding light beam emissions +light clock

1. May 27, 2015

### Micheth

No matter how many explanations of SR I try to read this honestly really bugs me :-)
Perhaps someone here can provide a different perspective that will make it click for me, i don't know...

The way I model the problem is thus:

A
C
B

there being 2 inertial frames, one for A&B, and another for C.
A & B, at complete rest with respect to each other, and separated by an arbitrary long distance (let it be many light years), are sending light beams to each other, effectively constituting a light clock.

We may let A/B be the reference frame, with C approaching them at a fixed speed (no acceleration).
At the moment when C transects the light beams being transmitted between A and B, it must necessarily lie on the virtual line between them, since the light beams are never anywhere outside of that line.
Indeed, if A/B wished to send a signal (using the same light beam) that would reach C, then they would need to calculate the appropriate time prior to C's transection of that line, and emit the beam accordingly in advance.
(B would not even be necessary, but just for reference.)

Now, this may be where I'm mixed up but it certainly seems as if the following must be true if we now let C be the reference frame (considered to be at rest).
Now, in this case, for the signal to reach C, A would not emit the signal beforehand, but at the point where C lies on the virtual line between A and B (as calculated by A&B).
That beam would travel along that virtual line that existed between A and B at the moment A emitted the signal, eventually reaching C at some later time after A and B have moved onward (to the right).
Therefore, it seems C transected a line that is no longer between A and B, which contradicts what seemed to happen if we consider A/B at rest...

(Obviously C would "see" A and B along the light beam's line, merely because of the time it took their light to get to him, but knowing the speed of light, he would know A and B are actually beyond that point by now).

Stated slightly differently, to me it seems that in A/B rest frame, A/B could in fact receive a reply signal from C (C reflects the light beam (replies) at the instant it transects line A-B), whereas in the C rest frame, A/B could never intercept a reply (reflection) from C.

This of course renders the light clock analogy (ala SR) completely unintelligible anymore :-(

The only thing I can try to muster is that if you consider yourself at rest and the other frame moving, then somehow that frame imparts lateral momentum to emitted light beams but that can't be right I think.
(I think, I dunno.)

2. May 27, 2015

### Micheth

My A-B-C positions didn't work with this formatting.
I meant it like this:

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

3. May 27, 2015

### A.T.

If the signal is supposed to pass C and reach B, then it has to be emitted beforehand. You are confusing differently aimed signals.

Last edited: May 27, 2015
4. May 27, 2015

### Micheth

Hi,
I mean that in C's rest frame, I think that any signal that A/B sent and received by C would have to have been sent when C was between A&B, not beforehand.
(Once C receives that light, A&B are already far past that point)

Unless, that is, the beam sent out perpendicular to A/B's direction of travel continues to move in the direction of their travel even after leaving A (or B).
(But I thought it doesn't work that way.)

5. May 27, 2015

### harrylin

I don't fully follow what you wrote. However the above last part may be pertinent: Light does have momentum, and its momentum is conserved just as its velocity in free space.

For me the simplest way to look at it is as follows: consider that laser pointer light is initially emitted by A. Relative to a system in which A is in rest, the light propagates straight down. Consequently, in a system relative to which A is laterally moving, the light must propagate under an angle because the lateral position of the laser pointer is changing while the light propagates downward inside the laser pointer. Else the light would propagate parallel to the wall of the laser pointer according to the first system, but it would traverse the wall of the laser pointer according to the second system!

By the way, if you study "the" Michelson-Morley experiment then you also see the difference in angle. See the sketch here:
https://en.wikipedia.org/wiki/Michelsonâ€“Morley_experiment#Light_path_analysis_and_consequences

6. May 27, 2015

### Micheth

If C weren't there the signal would reach B, but I'm not really thinking of B so much as how the signal relates to A and C (being reflected by C). But since from C's frame A has moved on since then, C's reflection never reaches A, though it ought to in the other frame...

7. May 27, 2015

### A.T.

That is obviously not the case. In particular the signal that can also reach B after passing C cannot be the one sent when C is between A&B.

Why would it be sent out perpendicular to A/B's direction of travel?

But you should, as B defines where the signal from A goes to. Otherwise you arenâ€™t comparing the same signal.

8. May 27, 2015

### Micheth

Yes, that's exactly what I want to pin down!
So, it is true that for a moving body that is "laterally" moving, and emits a light beam, that light beam propagates at an angle, the beam being imparted with the lateral movement of whatever emitted it?
This is then true for any moving body, right? For example, if I accelerate for a certain period of time and then, continuing on a constant velocity, and turn on a laser beam perpendicular to the direction of my movement, that all the photons emitted for that laser will always be parallel to the laser pointer, and never lag behind me?

That is, if I'm passing a planet and want to send a signal, I have to send it before I'm in line with the planet because any photons I emit keep traveling in my direction in addition to the perpendicular direction? Is that correct?
(I was thinking, perhaps erroneously, that after you emitted a photon, it just continues in that direction and doesn't care what direction or speed I was moving when I emitted it.)

9. May 27, 2015

### A.T.

Yes

Yes

It does continue in the direction it was emitted, but that emission direction is frame dependent.

10. May 27, 2015

### Micheth

Although when I say "continues in the direction it was emitted", I mean like this:

A ---> A
.
.
.
o

and not
A--->A
--.
---.
----.
-----o

But obviously the 2nd case is true, if an emitted photon doesn't lag behind A.

11. May 27, 2015

### A.T.

"Continues in the direction it was emitted" just means that it moves in a straight line. But that line can have different orientations in different frames.