^{3}?

I don't know, but here are some observations:

**Spoiler:**

**Moderators:** gmalivuk, Moderators General, Prelates

What E-field is produced by a charge distribution which is uniform in all of R^{3}?

I don't know, but here are some observations:

**Spoiler:**

I don't know, but here are some observations:

Blatm wrote:What E-field is produced by a charge distribution which is uniform in all of R^{3}?

I don't know, but here are some observations:Spoiler:

Strictly speaking, that's all you really need. I would go with something like:

[math]E = \frac{\rho}{\varepsilon_0} \cdot \left(x, y, z\right) + \left(F(y, z), G(x, z), H(x, y) \right)[/math]

The constant you can add can be any funny function you want, but the trivially easy solution is the one

where the field strength is proportional to the distance from the origin.

http://en.wikipedia.org/wiki/DSV_Alvin#Sinking wrote:Researchers found a cheese sandwich which exhibited no visible signs of decomposition, and was in fact eaten.

I like this question. Somehow I ended up nerd-sniping myself because of it for half the day and I'm still not sure about the answer:

Let me point out a few things:

This is not entirely correct. The ansatz for the E field given here will have in general a non-vanishing curl, which is bad in the absence of a magnetic field. You'll need to impose something like

[math]\partial_y F(y,z) = \partial_x G(x,z)[/math]

and similar for the other derivatives. I would guess that once you go through the motions, you'll find that all the functions F, G, H are constant and not arbitrary constant functions.

However, I think there's something wrong with Gauss's law in this context. The reason is simple: The space is still homogeneous and isotropic, so it is clear that there can be no force on any test charge placed into it. Hence, the electric field has to be zero. So something must be wrong with Gauss's law.

I have only an idea what that might be. Relating the differential form of Gauss's law with the integral form makes use of Stokes' theorem. Stokes theorem on non-compact manifolds assumes (to my knowledge) that the function you are integrating over has compact support. In our case, the charge is clearly not with compact support. But this is clearly not the complete picture.

Anybody got a better idea?

Let me point out a few things:

Sagekilla wrote:Strictly speaking, that's all you really need. I would go with something like:

[math]E = \frac{\rho}{\varepsilon_0} \cdot \left(x, y, z\right) + \left(F(y, z), G(x, z), H(x, y) \right)[/math]

The constant you can add can be any funny function you want, but the trivially easy solution is the one

where the field strength is proportional to the distance from the origin.

This is not entirely correct. The ansatz for the E field given here will have in general a non-vanishing curl, which is bad in the absence of a magnetic field. You'll need to impose something like

[math]\partial_y F(y,z) = \partial_x G(x,z)[/math]

and similar for the other derivatives. I would guess that once you go through the motions, you'll find that all the functions F, G, H are constant and not arbitrary constant functions.

However, I think there's something wrong with Gauss's law in this context. The reason is simple: The space is still homogeneous and isotropic, so it is clear that there can be no force on any test charge placed into it. Hence, the electric field has to be zero. So something must be wrong with Gauss's law.

I have only an idea what that might be. Relating the differential form of Gauss's law with the integral form makes use of Stokes' theorem. Stokes theorem on non-compact manifolds assumes (to my knowledge) that the function you are integrating over has compact support. In our case, the charge is clearly not with compact support. But this is clearly not the complete picture.

Anybody got a better idea?

Another problem with the charge distribution: As we know, fields from a charge can only expand with c. To "see" the uniform charge density in an infinite space, you have to wait an infinite time.

Actually, this might be a way out of the problem: Assume this charge density just appeared at some time t=0 in our lab system. In that case, every point in space sees a ball with a radius of ct around it, so every point is free of electric fields. This does not violate Gauss' law, as this requires charge conservation (like the Maxwell equations, which can be used to derive it).

Actually, this might be a way out of the problem: Assume this charge density just appeared at some time t=0 in our lab system. In that case, every point in space sees a ball with a radius of ct around it, so every point is free of electric fields. This does not violate Gauss' law, as this requires charge conservation (like the Maxwell equations, which can be used to derive it).

mfb wrote:To "see" the uniform charge density in an infinite space, you have to wait an infinite time.

Well, yes. But if you're considering it as a problem in electrodynamics, you're kind of cheating. I mean, it's a non-sensical problem anyways, so if you assume that there's already an infinite amount of charge in your space, you can as well assume that you have waited an infinite time for the field to establish itself - otherwise you'll run into trouble with more physical problems such as infinite charged rod and infinite charged plane.

On a more mathsy note, you can actually see what goes wrong when deriving Gauss's law from the Lagrangian. Using four-component notation and differential forms, you can write this down quite quickly as (F = dC, greek "rho" is the charge density)

[math]L = \int_{R^{3,1}} \frac{1}{2} F \wedge \star F + \rho C \wedge dx \wedge dy \wedge dz[/math]

If you then do the variation (varying the potential C), you get

[math]0 = \int d (\delta C \wedge \star F) - \delta C \wedge (d \star F - \rho d x \wedge d y \wedge d z)[/math]

So usually you would discard the total derivative term and pick up your Maxwell equation from the remainder (If you're not used to four-component notation: This is the curl of the E field). However, you can only discard the total derivative term if it vanishes at spatial infinity, which it clearly doesn't.

Users browsing this forum: No registered users and 3 guests