# In an electrodynamic system, does the entropy of the field increase?

1. Feb 17, 2005

### Crosson

Consider an electric field throughout the region of an electrodynamic system:

E [x,y,z,t]

The function, E [x,y,z,t], is a macrostate. Corrresponding to it are all the ways the charge in the system could be arranged ( p [x,y,z,t])so as to produce this macrostate. The set of p [x,y,z,t]s corresponding to a given E [x,y,z,t] are called microstates.

QM places restrictions on the number of distinct pairs of (momentum, position) that a particle can have. Therefore, in the EM system described above, there are a finite number of microstates corresponding to each macrostate.

Since the entropy is essentially the number of microstates corresponding to the current macrostate, is it true that the entropy of an electrodynamic system tends to increase?

This would be a most fabulous result.

2. Feb 17, 2005

### Physicsguru

Out of curiosity, why do you say that this would be a most fabulous result?

Regards,

Guru

3. Feb 18, 2005

### Gokul43201

Staff Emeritus
I think you may have things a little mixed up. The charge distribution $\rho(\mathbf{r})$ fixes the field $\mathbf{E}(\mathbf{r})$.

And I don't understand how the rest of your argument works either...

4. Feb 18, 2005

### dextercioby

Nope,i think you should review the microstate-macrostate part for a STATISTICAL SYSTEM...

Daniel.

5. Feb 18, 2005

### zefram_c

If the electric field in a system is fully specified, the charge distribution is uniquely determined by Gauss' Law: $\nabla \cdot E = \rho$. So I don't see how you can have different microstates for a macrostate.

6. Feb 19, 2005

### Crosson

Sorry guys, this only a thought for a thought at this stage.

You are right Zefram, the electric field does uniquely determine the charge density locally. So to pose my question properly, I have to say:

The E field out side of a spherical region is givent by E[x,y,z,t]. This is the macrostate. Corresponding to it are all of the possible charge distributions inside the spherical region, the microstates. So now the question becomes, does the charge inside of the spherical region evolve so that the electrodynamic entropy of the field outside is a maximum?

The answer to the question is an immediate yes if we (like statistical mechanics) assume that all microstates are equally probable (the assumption works there but not here). So clearly, the result should be proved using the current laws of E and M.

The reason I think this would be a fabulous result is that I would like to develop electrodynamics in the following way:

1. Define the electric field by Gauss's law $\nabla \cdot E = \rho$

2. Assume that electrodynamic entropy always increases.

Then, because the charges have to move in order to increase the electrodynamic entropy, we show that (surprise) they move in such a way that the local charge density responds to the local electric field as:

F = qE

If all this could be done, I think the question "why does electrodynamic entropy tend to increase?" would then lead to new physics.

If this highly abstract (admittedly, I have no clue where to begin) approach to a non-problem bothers you guys, I am sorry.

7. Feb 19, 2005

### dextercioby

What is "electrodynamic entropy"...?Something you invented...?Give a reference to a book/peer reviewed article where this notion is explained...

Daniel.

8. Feb 20, 2005

### Crosson

I think if you read my previous post, I make it clear that this is my own half-assed idea.

When it first struck me, I thought the electrodynamic entropy of the field at a particular point in space might be increasing, but this problem seems more intractable.

I tell you guys what, respond if you have any thoughts about my idea. If you think it totally doesn't make sense/ doesn't work than I am not going to argue with you untill I do some more work and come up with an example.

9. Feb 20, 2005

### Gokul43201

Staff Emeritus
Isn't this violating the uniqueness of the solution to the Poisson Equation with Neumann boundary conditions ?

10. Feb 20, 2005

### dextercioby

Why Neumann boundary conditions...?I see no connection to that...It may be Dirichlet...

Daniel.

11. Feb 21, 2005

### Gokul43201

Staff Emeritus
It may be...I may have gotten those two mixed up. I mean the one where $\partial \Phi / \partial n$ is specified on a surface. Isn't that what the OP is doing anyway. By specifying E(x,y,z) everywhere outside some sphere, he is at least specifying on the sphere. But it's been a long time and I'm too tired now, so I may just be babbling.

12. Feb 21, 2005

### dextercioby

Yes,the normal derivative is part of the Neumann BC.Anyways,i think the OP has little knowledge of what SM really means...

Daniel.