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grav
2007-Sep-07, 09:28 PM
What is it about Maxwell's equations that determines that light always travels at 'c' to all observers? I can see how electric and magnetic fields set up in free space might always cause an observer that is stationary to the fields to receive electromagnetic radiation at 'c', but what is it that says that the electric and magnetic constants as well as 'c' will remain constant when an observer is travelling at a relative speed in respect to the field to begin with? (In layman's terms, please.)

Bearded One
2007-Sep-08, 12:46 AM
A basic rule in science is that the rules are the same for all observers in all inertial (non-accellerating) frames. For Maxwell's equations to be valid for all observers then c must be a constant for all observers. That's the basic gist of it.

korjik
2007-Sep-08, 02:29 AM
What is it about Maxwell's equations that determines that light always travels at 'c' to all observers? I can see how electric and magnetic fields set up in free space might always cause an observer that is stationary to the fields to receive electromagnetic radiation at 'c', but what is it that says that the electric and magnetic constants as well as 'c' will remain constant when an observer is travelling at a relative speed in respect to the field to begin with? (In layman's terms, please.)

If you do some algebra on the equations, you end up with a wave equation.

If you look up wave equation on wiki, you will see that there is a c^2 term. This is the speed of the wave, squared. When you do the algebra on Maxwells equations, you end up with

1/c^2=mu(0)epsilon(0)

where mu(0) is the magnetic permeability of free space, and epsilon(0) is the electric permittivity of free space.

This gives you that the speed of an EM wave in free space is dependent on the magnetic and electric constants

grav
2007-Sep-08, 02:53 AM
Thanks for the replies, but how do we know that mu(0) and epsilon(0) remain precisely the same even with a relative speed to the field?

hhEb09'1
2007-Sep-08, 03:23 AM
Thanks for the replies, but how do we know that mu(0) and epsilon(0) remain precisely the same even with a relative speed to the field?So far, that's the results of experiments, I suppose

publius
2007-Sep-08, 03:25 AM
There is some misconception about this in modern terms. Some will say that Maxwell implies SR because 'c' must be constant.

That is incorrect. Pre-MM, E and B fields were thought to be "stresses in a medium", generally called the '(a)ether', and those stresses can propagate as waves. This is similiar to sound, or any general disturbance that propagates through a medium.

Thus 'c' was understood to be relative to the medium. Maxwell in it's current form was then the (simplest) ether frame form. Do a Galilean transform, and you'd have what Maxwell supposedly looked like in a frame moving relative to the medium. This is exactly how sound waves behave. I've forgotten exactly but I think that makes Maxwell take the rather simple form of replacing the partial time derivatives with the "total" or, directional derivative, d/dt = @/@t + v dot del, where v is the velocity of the moving frame relative to the ether frame.

Don't hold me to that because I can't remember for certain, but I think that's the way it would play out.

Anyway, that's the way it was understood, and that was the "obvious", logical way to see it. But nature had a little trick up her sleeve, and that was discovered with the MM experiment. One way to put is all inertial observers think they are stationary with respect to the medium (which makes the notion of a "medium" as traditionally conceived be a pretty meaningless thing). Now, the ether frame c and Maxwell become something that must be invariant across all frames. The Lorentz transform of space and time is the one that makes Maxwell invariant.

So, I stress that Maxwell itself does not imply that c is constant. That was just something we stumbled onto, and said whoooa, space and time have to work differently than we thought in order to make this so.

-Richard

grav
2007-Sep-08, 04:00 AM
Thanks, Richard. That's interesting. So the MM experiment was truly the defining moment? It almost seems to me, then, that Einstein sort of lucked up upon relativity in a way, with the idea of the Lorentz contraction and all, since that particular experiment could also be explained by the speed of light including the speed of the source, but that would not agree with Relativistic Doppler and precession and so forth that has been observed since then. It seems to me that his main purpose was to connect it to what he describes in the first paragraph of his paper, that of electric and magnetic fields with a relative speed between them. What does he mean, though, when he says that when the conductor is moved toward the magnet, a current arises in the conductor, but when the magnet is moved toward the conductor, a current arises in the vicinity of the magnet? That would tell me something as well, if an observer were to measure the current to be in different points in space, since both cases could be considered from the standpoint of the conductor and magnet moving toward each other in the same way, but that the observer is stationary to one or the other instead.

[EDIT- Here's the paper, by the way. (http://www.fourmilab.ch/etexts/einstein/specrel/www/)]

grav
2007-Sep-08, 04:28 AM
In his paper, he says that "the observable phenomenom here depends only upon the relative motion of the conductor and magnet, whereas the customary view draws a sharp distinction between the two cases in which one or the other of these bodies is in motion." Now, does this mean that the customary view is that an electric field arises in the neighborhood of the magnet in one case and in the conductor in the other, but that this is not what is actually observed? He seems to describe it as if that that is what is observed.

publius
2007-Sep-08, 04:52 AM
Grav,

This is a pretty good question. The modern, post-Einstein/SR view is so ingrained nowadays that what Einstein was saying there seems rather odd. :) To appreciate it, we've got to get ourselves in the pre-MM "frame" of mind. That was the way Brother Heaviside sternly staring at you in my avatar viewed things, as well as Maxwell his own self.

Just like with sound, we have this medium, the ether, and these E and B field things are "relative to that". So when we're doing EM, and things are moving, what is naturally important is the velocity relative to that medium. The ether frame makes Maxwell take it's simplest form.

So, when we have a "moving magnet", what is naturally assumed to be meant is the magnet is moving with respect to the ether. We have a dB/dt, and therefore we have an electric field E in our medium. That electric field is what drives a current in a nearby conductor that is stationary with respect to the ether.

But, now hold the magnet stationary, and let the conductor move. We have no electric field in the ether. But we do have a Lorentz force, v x B acting on the moving charges in that conductor. That drives the current. And while he doesn't mention it, we could imagine both of them moving with respect to the medium, and having both E and v x B, which could cancel each other out.

What Einstein notes is that while these two cases are quite distinct, the result is the same. The "real" E in the first case is exactly equal to v x B in the second case. So he is hinting that what matters is *relative* velocity between the magnet and the conductor. But the current ether view makes a sharp distinction there.

Einstein did not simply "luck out". What he did here (as part of the "year of miracles") was to do away with the ether by saying that space and time themselves must act differently than "common sense", Galileo's transform, thinks they act. And this is why MM came up the way it did. All inertial observers see the speed of light as 'c', space and time transform to make that so, and so what matters in the case of EM induction is indeed the relative velocity between the magnet and the conductor. No ether needs to come into it. In the frame of the moving conductor, v x B is indeed an E for that observer. And an E is a v x B for others.

Any observer simply takes Maxwell in the simple form and solves them for what he sees, using the v's and r's, etc as he sees them. He need not worry about any medium.

-Richard

grav
2007-Sep-08, 05:40 AM
Thanks again. :) I think I'm starting to see what you're saying. I'll let that sink in for a bit. I expect I'll have more questions soon enough. ;)