# How do maxwell's equations show that speed of light is constant

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1. Jun 23, 2014

### Sreenath Skr

Constant for all observers?

I have heard that maxwell showed that the speed of light is constant for all observers even before Einstein did. Is that true?

If not, then how can we say maxwells equation shows the speed of light is constant?

2. Jun 23, 2014

### Simon Bridge

If you look up "Maxwell's equations" you will see that none of them explicitly include the speed of light in a vacuum.

However - Maxwels equations can be combined to produce a wave equation for electromagnetic waves.

That equation can be used, like any wave equation, to predict what speed EM waves should travel at in a vacuum. Do the maths, and that speed turns out to be given by: $$v=\frac{1}{\sqrt{\epsilon_0\mu_0}}$$

Since the permittivity and permiability of free space are (were presumed in Maxwell's day to be) constants, this is a big hint that the EM wave speed is also a constant.

When you crunch the numbers you get $v=c$ the speed of light.
... of course it's exact these days, but in Maxwell's day these were things that you measured separately.

For details see:

3. Jun 23, 2014

### Bill_K

If you look them up in Gaussian units, you will find that they do.
https://en.wikipedia.org/wiki/Maxwell's_equations#Equations_in_Gaussian_units

The so-called "permittivity and permeability of free space" are merely artifacts of the SI system of units. Especially the latter, which has the exact, preordained value of 4π x 10-7.

4. Jun 23, 2014

### Simon Bridge

Hmmm... did Maxwell use gaussian units?

5. Jun 23, 2014

### Staff: Mentor

No. I think BillK's point is that now that we know what to look for, it's a lot easier to consider using units that don't obscure the physics... And obscure it they do, or OP wouldn't have needed to ask.

6. Jun 23, 2014

### Staff: Mentor

SimonBridge's answer pretty much says it all. The only that I can add is that Maxwell's equations and the predicted speed of light were published in 1861. For almost a half-century after that, the great unsolved problem of physics was how to reconcile Maxwell's equations with Newtonian/Galilean classical physics; and it was this problem that Einstein solved with special relativity in 1905.

7. Jun 23, 2014

### Integral

Staff Emeritus
Einstein did not show that the speed of light is constant. He ASSUMED is was constant, based on Maxwell's prediction, which was common knowledge at that time.

8. Jun 24, 2014

### bcrowell

Staff Emeritus
Maxwell's equations do not predict that the speed of light is the same for all observers. They are stated in a certain frame of reference, and they predict that light has a certain speed in that frame. They are silent on what happens in other frames. To decide what happens in other frames, you have to decide how measurements of space and time change when you change frames. In Maxwell's era, everyone believed that changing the frame of reference was defined in terms of the Galilean transformation, and that meant that Maxwell's equations would have their simplest form in a certain special frame, and change their form when you changed to some other frame. The special frame was interpreted as the frame of the ether. According to that interpretation, the speed of light would *not* be the same in all frames of reference.

9. Jun 25, 2014

### PhilDSP

Another way to put this, which sort of echoes the last post and others, is that the Maxwell equations as we know them apply only for charges or bodies that are not moving with respect to the bodies that are emitting radiation.

Maxwell initially attempted to expand the equations to support moving bodies but he used rather complicated Eulerian formulations which were not successful. He didn't publish anything further on the subject. This is one reason why Einstein's 1905 article was titled "On the Electrodynamics of Moving Bodies".

10. Jun 25, 2014

### PAllen

No, this isn't right. Maxwell's equations apply to charges in all states of motion (else there would be no magnetism, no wave equation!), as long as the measurements are made in the special type of frame in which they are valid. Circa 1880, that special frame was assumed to be the aether frame. SR established that it was any inertial frame.

Last edited: Jun 25, 2014
11. Jun 25, 2014

### PhilDSP

I meant, of course, the use of the Maxwell equations without the application of the Lorentz transform. The LT effectively performs the moving body calculations. The Maxwell equations do handle non-relativistic motion of non-radiating charges (near field effects) also, but not motion involved with radiation between source and sink.

Last edited: Jun 25, 2014
12. Jun 25, 2014

### WannabeNewton

What? The Larmor formula is exactly for non-relativistic radiating charges and it is obtained directly from Maxwell's equations.

13. Jun 25, 2014

### PhilDSP

Interesting! On a first look it seems to invoke near field effects and acceleration of the charge in the near field. This is worth looking at in much deeper detail. But it also implicitly involves more than one application of the Maxwell equations at one time to calculate retarded potentials. So some assumptions are being made in how the Maxwell equations apply to different times rather than a single instance of time.

14. Jun 25, 2014

### WannabeNewton

Sorry I don't understand; what do you mean by this?

15. Jun 25, 2014

### PAllen

I still don't get what there is to debate. Maxwell's equations are (classically) exactly correct for any motion of any charges in any given inertial frame. You can have 10 charges oscillating in different ways with speeds near c, and (in principle) describe all radiation an field strength anywhere using Maxwell's equations. The only change from 1880 understanding is that it was, at that time, assumed they only applied (exactly, all cases) in the aether frame, and to apply them in some other frame you would have to figure out how they transform under a Galilean frame change. SR changed this understanding that they apply exactly in all inertial frames, and the the LT was the right transform of the description in one frame to the description in another.

[edit: of course, in 1880s understanding, when it came to describing the response of matter to field, you would have an issue of what classical dynamics to use - Newtonian or relativistic. But I see this as a separable issue from the description of field evolution given some specified motion of charge distribution.]

16. Jun 26, 2014

### PhilDSP

Maybe the best way of putting it is that the Maxwell equations accurately describe field values based on the position(s) and motion(s) of charged particles. But they don't describe the position(s) and motion(s) of the charged particles given the field values.

That means that other equations such as the Lorentz force law must be used to determine the effect of changing field values on bodies. So there is a missing description (in the Maxwell equations) of how radiation from a moving particle affects other charged particles. A naive application of the Lorentz force law isn't sufficient because the charged particle motion is affected by the fields which are affected by the particle and so on recursively...

That description in the form of extensions or modifications to the Maxwell equations had been given by a number of early physicists: Maxwell (who aborted the attempt), Heaviside, Hertz, Cohn, Bucherer and Ritz. Each of those solutions are somewhat unique but all, I believe, employ material derivatives rather than partial derivatives for changing field values.

Last edited: Jun 26, 2014
17. Jun 26, 2014

### PhilDSP

I mean that, to solve the problem that the Larmor formula addresses, the Maxwell equations need to be invoked iteratively. In most problems where iterative techniques are used to solve differential equations, such as the finite element method for example, the overlap of time and position initial values for each invocation is pretty straightforward. But in this case, since it involves retarded times as initial values analytically, it's not quite as straightforward.

We can also notice that the Larmor formula is single-ended. It gives the power of the radiation of an emitting particle. Or it gives the power of radiation absorbed by another particle. But it doesn't give us a description of the coupling between those 2 events.

Last edited: Jun 26, 2014
18. Jun 27, 2014

### homeomorphic

Volume II of Feynman's lectures in physics has a nice intuitive explanation of the speed of light for a certain plane wave, generated by a sheet of current that is suddenly turned on.

Here's a summary, which will be sort of vague because I don't want to take the time to describe the set-up in detail. If you just had that current from the start, all you'd get is a constant magnetic field parallel to the sheet. But when you suddenly turn the current on, it will take a while to get that magnetic field set up. The field is zero initially, but then attains its final value in a region spreading out from the current sheet. To calculate the speed of the resulting wave, you can just do two line integrals near the edge of the wave front to see what is going on there. Because the magnetic field is being switched on along the front, it is producing a changing magnetic flux through a loop perpendicular to it, near the wave front. According to Faraday's law that means there is an electric field. You can do another line integral for the electric field and the displacement current kicks in to tell you how the magnetic field is changing. Both the line integrals are equal to the changing flux, which depends on the velocity of the wave front. So, E = vB and B is proportional to vE, and from there, it's easy to calculate the speed of light, v = c.

19. Jul 27, 2014

### carrz

Weren't permittivity and permeability experimentally measured first, and how else could they have gotten into any of those equations to begin with?

20. Jul 27, 2014

### Simon Bridge

The constants get into those equations because the dimensions do not match otherwise ... something has to carry the extra units. So you measure, say, and electric force and range for a given pair of charges and you get a particular relationship.

You can choose units to make the constants any value you like.

I'm not sure that "merely" is a good description though.
Is the speed of light in a vacuum "merely" an artifact of the units chosen?
The speed of light can be any number we like after all.

21. Jul 27, 2014

### carrz

I'm pretty sure permittivity and permeability were first experimentally measured before they became a part of any equation, long before Maxwell. If they were arbitrary, how do you explain they just happen to hold the value of c?

22. Jul 27, 2014

### Staff: Mentor

Because all three depend on the same thing: the system of units. Once you have chosen your system of units then you have fixed all of the conversion factors. Any combination of conversion factors with the same dimensions is necessarily the same.

23. Jul 27, 2014

### carrz

Dependance on "system of units" does not explain how did permittivity and permeability become a part of any equation in the first place, nor it explains why would this equation be true:

By the way, it was not Maxwell who first realized connection between permittivity, permeability, and the speed of light. It was Kohlrausch and Weber in 1854 with their Leyden jar experiment who demonstrated that the ratio of electrostatic to electromagnetic units produced a number that matched the value of the then known speed of light.

http://en.wikipedia.org/wiki/Rudolf_Kohlrausch
http://en.wikipedia.org/wiki/Wilhelm_Eduard_Weber
http://en.wikipedia.org/wiki/On_Physical_Lines_of_Force

24. Jul 27, 2014

### Staff: Mentor

That equation has to be true because it's derived by solving Maxwell's equations for a particular set of boundary conditions. Permeability and permittivity appear in that solution if and only if you chose to write Maxwell's equations in a form in which they appear - and to make that choice is to choose a system of units.

We got into this mess in the first place because many systems of units were developed before Maxwell's equations were discovered. Using these systems in Maxwell's equations is perfectly legitimate, but it complicates the formulas without adding any new insight - just as doing relativity problems in the mks system so that the speed of light is $2.998\times{10}^8$ m/sec is harder but no more informative than using light-seconds for distance and seconds for time.

25. Jul 27, 2014

### Staff: Mentor

Yes, it does, although many people go through many courses in physics before they understand units and systems of units sufficiently to understand why. I will try to help as best as I can, but in the end there is no substitute for actually working a number of problems with different sets of units, like Gaussian, English, SI, Planck, and Geometrized units.

Suppose that you have some arbitrary (correct) physics equation a=b. Now, it is possible, in general, to use a system of units such that a and b have the same units. Such a system of units is called "consistent" with that equation. For example, SI units are consistent with Newton's second law: ∑f=ma where f is in newtons, m is in kilograms, and a is in m/s^2.

However, it is also possible to use other systems of units which are not dimensionally consistent with a given equation. In those systems of units you need to change the equation to a=kb, where k is a constant which changes the units on the right to match the units on the left. For example, you could express Newton's second law in US customary units as: ∑f=kma where f is in pounds-force, m is in avoirdupois pounds, a is in ft/s^2 and k is the constant 32.17 lbf s^2/(ft lbm).

The constant k is present only because of the system of units and is the factor that is required to convert the units on the left to the units on the right. Now, a given system of units may have several such conversion factors. Any combination of those conversion factors with the same base units is also itself a conversion factor and will therefore necessarily have the same units and the same value.

So, in SI units, c is the conversion factor between m and s (SI units are inconsistent with E=mc^2), μ0 is the conversion factor between kg m and s^2 A^2 (SI units are inconsistent with Ampere's law), and ε0 is the conversion between s^4 A^2 and kg m^3 (SI units are inconsistent with Coulomb's law). So 1/√(μ0 ε0) is an SI conversion factor between m and s, and must therefore match all other SI conversion factors between m and s, therefore it must equal c.

Last edited: Jul 27, 2014