# Question about photons traveling on same path

1. Jan 11, 2016

### stever

if photons are traveling on the same path, and in only one direction, are they sometimes passing each other? Or are they staying in sequence? Assume the sources for these photons are variously moving in relation to each other. .

2. Jan 11, 2016

### HallsofIvy

Staff Emeritus
Since all photons travel at the same speed, c, regardless of the speed of their sources, no, one photon cannot pass another.

3. Jan 11, 2016

### stever

does this mean that the speed of the photons in relation to their sources can vary if the source's speeds vary?

4. Jan 11, 2016

### Staff: Mentor

No. A flash of light travels at $c$ relative to everything - its source, its target, and anything else in its vicinity - no matter how these are moving relative to one another.

(You'll notice that I said "flash of light" instead of "photon". It's not a big deal in the relativity forum because everyone here automatically makes the mental substitution when we see the word "photon"... but photons aren't what you're thinking they are).

5. Jan 11, 2016

### Ibix

Depends what you mean by "speed of the photons in relation to their sources"..

If you mean "what speed will an observer riding on a flashlight measure for the light being emitted", then this is always c.

If you mean "what rate of change of separation will some other observer (moving relative to the flashlight) measure between the flashlight and the light it emits", then from this perspective the distance between the flashlight and the light it emits will grow at c-v, not c. This is becaue they, too, will measure the light to be travelling at c.

6. Jan 11, 2016

### stever

If a flashlight varies in speed, what becomes different about the flashlight so that the speed of light stays the same for the observer on the flashlight?

7. Jan 11, 2016

### Mister T

Nothing. There's no difference between a flashlight in uniform motion and a flashlight at rest. The Principle of Relativity asserts that all inertial reference frames are the same, so if there were something different about a moving flashlight that thing that's different would give us a way to distinguish between two inertial frames.

And there's nothing different about the light, either. It's the speed itself that's invariant, or in other words the same in all frames of reference.

8. Jan 11, 2016

### Staff: Mentor

Suppose that I am at rest and shining a flashlight in front of me. Meanwhile, you are flying past me at .5c and shining a flashlight in front of you. Your speed relative to me is .5c; the speed of the light from both your flashlight and mine relative to me is c; and the speed of the light from both your flashlight and mine relative to you is c.

However, we could just as well describe this situation from your point of view: I'm moving backwards at .5c while you are at rest. The speed of the light from both flashlights relative to both of us is still c. Thus, there's nothing different about either flashlight. They're both acting the same way and it makes no difference which one we consider to be the one that's moving.

This might be a good time to mention that if A is moving at speed $u$ relative to B, and B is moving at speed $v$ relative to C, A's speed relative to C is not the $(u+v)$ that you would expect from classical physics; it is $\frac{u+v}{1+uv/c^2}$. It's an interesting exercise to try setting $u$ or $v$ to c - that's the flashlight problem you posed.

9. Jan 12, 2016

### stever

I think it is being said that changes to a moving object are only apparent to observers in motion relative to that object; the moving object itself can see no difference in itself.

what does a moving observer think or do that would explain how objects, moving in relation to said observer, calculate light to be at c for themselves?

10. Jan 12, 2016

### HallsofIvy

Staff Emeritus
Surely that is not what you meant to say? The idea of objects being able to "calculate light to be at c for themselves" smacks of mysticism! A basic "axiom" of relativity, based on experimental evidence, is that the speed of light, relative to any object, is a constant, c.

11. Jan 12, 2016

### stever

light speed is always calculated. how is it calculated to be at c when for all moving observers it seems not to be at c (for that object).

12. Jan 12, 2016

### Staff: Mentor

I'm not sure I understand what you mean when you say "for all moving observers it seems not to be at c" - everyone in this thread has been saying that the speed of light is c for all observers.

You might want to try reading about the Michelson-Morley experiment, which was first done well before the discovery of relativity. The idea is simple enough: the earth is moving through space at many thousands of kilometers an hour, and we can send flashes of light both parallel to and perpendicular to the direction of that motion. Does one of the flashes cover more distance in a given time than the other?

13. Jan 12, 2016

### stever

Ibix said above: if you mean "what rate of change of separation will some other observer (moving relative to the flashlight) measure between the flashlight and the light it emits", then from this perspective the distance between the flashlight and the light it emits will grow at c-v, not c.

14. Jan 12, 2016

### Ibix

Yes. Light always travels at c.

15. Jan 12, 2016

### stever

Ibix said above: "if you mean "what rate of change of separation will some other observer (moving relative to the flashlight) measure between the flashlight and the light it emits", then from this perspective the distance between the flashlight and the light it emits will grow at c-v, not c."

16. Jan 12, 2016

### Staff: Mentor

OK, I think I see what you mean. The important thing here is that c-v is the difference between the speed of the flashlight relative to the observer (that's v) and the speed of the light relative to the observer (that's c). It's natural, based on intuition from a lifetime of living around slow-moving objects, to assume that it must also be the speed of the light relative to the flashlight - but that's an assumption, and one that turns out not to be correct.

I mentioned the relativistic rule for adding speeds above. We can try it here: We see a flash of light moving at speed c. What is the speed of that light flash relative to a flashlight moving at speed v relative to us? Using the formula above, it is:$$\frac{v+c}{1+vc/c^2}=\frac{v+c}{1+v/c}=\frac{c(v+c)}{c(1+v/c)}=\frac{c(v+c)}{c+v}=c$$

17. Jan 12, 2016

### Staff: Mentor

Hit post too soon... I was going to add:

That the speed of light is the same for all observers has some very surprising consequences, and the relativistic velocity addition formula is far from the most surprising. When we start with the assumption (solidly supported by more than a century and a half of experiments) that the speed of light is the same for all inertial observers, we find ourselves drawn step by logical step to:
1) Relativity of simultaneity, essential for making sense of everything else (google for "Einstein train simultaneity").
2) The Lorentz transformations which relate the times and positions measured by observers in motion relative to one another.
3) The velocity addition formula I've been using, which is derived from the Lorentz transforms.
4) Length contraction and time dilation, also derived from the Lorentz transformations.
5) The impossibility of travelling faster than light.
6) That gravitational waves must also travel at the speed of light.
......
And it just gets more interesting from there. Part of the charm of special relativity is the mathematical price of admission is quite reasonable - everything I describe above can be competently covered with nothing more than high school math and an optional dash of first-year calculus.

18. Jan 12, 2016

### stever

there is a difference. i am not asking about an observers own c. i mean how can observers in motion explain the c speed for relative moving objects. On the face of it, the light speed would seem to be more or less than c for others, but those others measure there own at c. the question is how to get from the appearance or newtonian expectation to the reality. not only must there be some reason why this happens, but also a way to do the math.

19. Jan 12, 2016

### Staff: Mentor

You'd think so, wouldn't you? That certainly squares with our intuition and it is EVERYONE's first reaction to relativity. It just happens not to be correct, although it is such a good approximation for speeds that are small compared with that of light that we seldom notice.

It's a worthwhile exercise to apply that velocity addition formula to a bullet fired at 300 meters per second (.000001c) from a car moving at 30 meters/second (.0000001c) - you'll see why it's reasonable to just add the speeds and not mess with that $uv/c^2$ correction in the denominator.

There is. I sort of hinted at it above: Start with the assumption (confirmed by enough experiments that we can reasonably choose to run with it and see where it takes us) that the speed of light is constant for all observers; then derive the Lorentz transformations from that assumption; and then watch the velocity addition rule fall out of that.

It's historically quite interesting to follow in Einstein's footsteps (the seminal paper was published in 1905 - "On the electrodynamics of moving bodies", and Google will quickly find it for you). However, the much more modern "Spacetime Physics" by Taylor and Wheeler comes with the advantage of a century of hindsight and refinement and is a much easier way of learning the key concepts and the math.

20. Jan 12, 2016

### stever

is there a way to start with c+v or c-v that one would have expected and then produce the c by means of a verbal description or simple math as to why or how it happens?

Last edited: Jan 12, 2016