# Maxwell eqs invariant under other transforms

1. Mar 23, 2010

### zwoodrow

Has anyone ever seen a proof that lorentz transforms are the only transforms for maxwells equations to remain invariant between two reference frames moving at a uniform velocity with respect to each other?

2. Mar 23, 2010

### Meir Achuz

The proof is simple.
Max's equations contain the constant c.
This leads to the speed of light being c in any coordinate system.
The LT is the only transformation between two frames moving at a uniform velocity with respect to each other that preserves the speed c.

3. Mar 23, 2010

### zwoodrow

ive done that derivation before, i was wondering if anyone has proved that the lorentz transform is the unique solution.

4. Mar 23, 2010

### Naty1

5. Mar 23, 2010

### bcrowell

Staff Emeritus
This statement isn't technically true as written. The full set of transformations that leave Maxwell's equations form-invariant is the Poincare group, which is larger than the group of Lorentz boosts.

If you want to change the statement to a statement that the Poincare group is the only possible group, then I think you might still need to clarify your assumptions in order to rule out doubly special relativity: http://en.wikipedia.org/wiki/Doubly_special_relativity

6. Mar 24, 2010

### PhilDSP

That's an interesting proof, but it only applies to a pure vacuum. The addition of any charged object changes the permittivity (from a constant that produces the speed of light to a tensor producing dispersion and a different group velocity)

Last edited: Mar 24, 2010
7. Mar 24, 2010

### bcrowell

Staff Emeritus
I think there's a big gap in your argument between this step:
and this one:
The final statement asserts uniqueness, but you haven't proved uniqueness.

8. Mar 24, 2010

### Naty1

Is this generally accepted?? Never knew about it if accurate...

9. Mar 24, 2010

### Naty1

Exactly as discussed in Wikipedia, post #4...

10. Mar 24, 2010

### Mentz114

I am reliably informed ...

11. Mar 25, 2010

### PhilDSP

Yes, look at the equation for Gauss' law. It is almost always given in 2 variations, one for use in a vacuum using the symbol E (electric displacement) and the second with D (electric displacement field) for general use.

If the medium, ie. vacuum, contains no dielectric materials and is isotropic, non-dispersive and uniform then

$$D = \varepsilon_0E$$

where $$\varepsilon_0$$ is the vacuum permittivity.

Since the presence of any charged particle effectively makes the medium dispersive, one must then determine the actual value of the permittivity which will depend on many factors such as the positions of the charges and presence of EM fields. Macroscopically, within a uniform material, one can experimentally determine an approximate value.

Last edited: Mar 25, 2010
12. Mar 25, 2010

### clem

The presence of a medium provides a preferred system, the rest system of the medium, so the Lorentz transformation does not have the same meaning as it does in vacuum.

13. Mar 25, 2010

### PhilDSP

We may not be entirely synchronized on the meaning of "medium". I'm thinking of it just as Maxwell, Heavyside, Thomson and others did a century ago as the *thing* that fluccuates when an EM wave is in motion. We know from Poincare's analysis that such a medium must be stationary in all frames with regard to the object that either emits or receives radiation.

In that regard there is no difference between the medium in a vacuum or within a steel bar for example. Even the steel bar consists volume-wise almost entirely of vacuum.