# Relative speed of a photon - mk2

1. Jul 3, 2007

### michael879

This thread was snipped and moved from the thread "conflict between electrodynamics and classical relativity" orignally posted by michael879.

Last edited by a moderator: Jul 12, 2007
2. Jul 12, 2007

### alvaros

In Maxwell laws, what does x mean ?

Supose a photon emitted by an atom.

The atom is moving in relation to other atom buts it is at rest respect to
another one.

The photon is an entity different from the atoms ( It has energy, wrote
Einstein ).

The photon ( or the wave ) is moving in relation to what ? To itself ?

May be the atoms dissapear after the photon is created.

I think this is basic for all the argument of ( special ) relativity
and I wish to find a link where this is explained ( whithout complex
numbers ).

If somebody could help me, thanks.

3. Jul 12, 2007

### pervect

Staff Emeritus
All motion is relative. But in this special case, it turns out that an observer located at rest relative to either of the two atoms you mention will measure the same speed for light (i.e. the same speed for the photon).

I get the impression you may be using a philosophy that assumes absolute space, that's the only way your question makes any sense to me. Unfortunately for you, if my impression is correct, there isn't any absolute space or time in relativity.

4. Jul 12, 2007

### rbj

dunno specifically what you mean by the "first principle of SR". if you mean that the laws of physics are exactly the same for any inertial observer, i would agree and say that this is a fundamental axiom. but if it is that the speed of electromagnetic waves(a.k.a. "light") are the same for any unaccelerated observer, i would say that given the axiom that the laws of physics are the same, then it is that Maxwell's equations are those laws of physics that, if identical in every respect for any and every inertial observer, is the principle from where you deduce that the speed of light in vacuo is the same for every observer.

so you have this changing E field that is causing this changing B field that is causing this changing E field that is causing this changing B field that is causing this changing E field that is causing this changing B field and all that is, by solving Maxwell's equations, propagating at this speed $c = 1/\sqrt{\epsilon_0 \mu_0}$. would two inertial observers, both moving relative to the other and both observing the same little beam of light (doesn't matter which of them might be holding the flashlight), would that not be the same for each observer? why should one observer be preferred over the other (which is what would have to be the case if one measured c to be different than the other) so that this one observer has the more "correct" value for c?

5. Jul 12, 2007

### Mentz114

No. We're not talking about moving objects or collisions but light being emitted. In the frame of the guy lighting the candle Maxwells laws work and the speed of the light is c in that frame.

I want to read RBJ's post before I say anymore.

[later]
r b-j, thanks for the input. But you guys are using post-Einstein ideas, mainly 'no aether'.

Anyhow, it's not that important, this is all history now.

6. Jul 12, 2007

### rbj

but that concept of "aether" was an unnatural workaround or "kludge" concocted so that the pre-Einstein guys could get a grip on how E&M was propagating. they fully thought that the velocities added in such a way that the two observers would have to observe the speed of that beam of light to be different, just as you and i would if you, traveling in a 100 km/hr vehicle tossed out, in front of you, a ball at a speed (observed by you and others in your vehicle) at 40 km/hr. i, standing at the side of the road, would observe that ball to move at 140 km/hr. they thought that logically, it would be the same for light, but the only way for that to happen is for there to be a stationary frame of reference for light to travel in (and the name for that was the "aether") and that was a concoction, a kludge. all's Einstein did was toss that kludge out and then ask, "if both observers are absolutely equal qualitatively, and observe the same speed for that propagating EM wave, what parameter (that was thought to be invariant) must give for that to happen", and the answer he came up with is time (or rate of clock ticking) and length colinear with the direction of motion, and inertial mass (defined to be momentum divided by velocity, and not to be confused with invariant mass), and energy, etc. all of those things have to be observed differently for the two different observers in order for them to perceive the speed of propagation of the EM wave to be the same.

it's not just the speed of EM. it's the speed of all ostensibly instantaneous actions (like gravity from a classical perspective) that is this c including the physical things we do to communicate information from one place to another. all of that is limited in speed by c.

7. Jul 12, 2007

### JesseM

We're talking about moving charges. Do you think Maxwell's laws are incapable of predicting the speed of electromagnetic waves emitted by an oscillating charge whose average position (i.e. not worrying about the small motion caused by the oscillation) is moving in the frame you're using (the frame where you assume Maxwell's laws are correct), without transforming into the frame where the charge's average position is at rest? Do you think that in order to calculate the magnetic field of a moving charge, you need to transform into a frame where it's at rest? Maxwell's laws work just like classical dynamics or every other set of laws in physics--you never need to switch reference frames to predict what will happen, although in some cases it might make the math simpler.
In relativity, yes. But before relativity, physicists imagined that Maxwell's laws wouldn't work in that guy's frame unless he was at rest relative to the aether.

8. Jul 12, 2007

### rbj

cross product of vectors (specifically in Maxwell's Eqs. the cross product with the "del" or "nabla" operator which defines the "curl" operation).

probably there are posts in the past here that would help. i dunno.

9. Jul 12, 2007

### Mentz114

Guys, please lay off me ! I'm not disagreeing with anything you say. We were talking about how these ideas evolved - I hope there's no argument about relativity going on here.

Thanks for the robust discussion, I'm not clear yet what you're trying to tell me but but I'll be thinking about it.

Last edited: Jul 12, 2007
10. Jul 12, 2007

### JesseM

Aren't you disagreeing about Maxwell's laws, though? You seemed unconvinced that Maxwell's laws could give you the speed of an electromagnetic wave from a moving emitter without the need to transform into the emitter's frame, and you also seemed unconvinced that Maxwell's laws demand the speed of the wave from a moving emitter would be c in any frame where these laws hold.

11. Jul 12, 2007

### Mentz114

It would greatly enhance my education to see this properly demonstrated - with a luminiferous aether and pre-SR concepts.

12. Jul 12, 2007

### JesseM

Maxwell's laws are just defined by a set of equations (relating the electric field, the magnetic field, and the distribution and motion of charge), they have nothing to do with whether you believe in aether or not. The aether concept is just the idea that these equations are only obeyed in the rest frame of the aether, and in other frames electromagnetism would obey different equations (Maxwell's laws modified by a Galilei transformation). Relativity says that these equations are obeyed in every inertial frame.

If you want to see a basic derivation of the fact that, in any frame where these equations are obeyed, the only vacuum wave solutions to the equations describe electromagnetic waves which move at exactly c, you could look http://www.phy.syr.edu/courses/PHY312.00Spring/notes/hand/2EMwaves.pdf [Broken], for example. Showing the exact form of electromagnetic waves emitted by an oscillating charge (or a charge undergoing any sort of acceleration) is trickier, but the above is enough to establish that these waves must have a velocity of c in a vacuum, if we're in a frame where Maxwell's equations hold.

Last edited by a moderator: May 3, 2017