# Current ambiguity symmetry

1. Oct 19, 2009

### jostpuur

It is well known, that a point charge $q\delta^3(\boldsymbol{x}-\boldsymbol{x}')$ creates the same electric field $\boldsymbol{E}(\boldsymbol{x})$ as any spherically symmetric charge density $\rho(\boldsymbol{x})$ around the point $\boldsymbol{x}'$, with the right total charge, for the points $\boldsymbol{x}$ that lie outside the support of $\rho$.

Is there an equivalent property for the magnetic fields? How is it formulated?

I've heard that there exists several different current distributions which give rise to equal magnetic fields (for some points in space), but I've never seen what these equivalent current densities really would have to look like.

2. Oct 19, 2009

### Bob_for_short

Try integral forms of the field solutions.

3. Oct 19, 2009

### jostpuur

$$\boldsymbol{E}(\boldsymbol{x}) = \frac{1}{4\pi \epsilon_0} \int d^3y\; \rho(\boldsymbol{y}) \frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}$$

$$\boldsymbol{B}(\boldsymbol{x}) = \frac{\mu_0}{4\pi} \int d^3y\; \frac{\boldsymbol{j}(\boldsymbol{y}) \times (\boldsymbol{x}-\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}$$

I don't know how these help me. I don't even know what these help in the case of electric field. If I replace the current density $\rho$ with some convolution

$$\overline{\rho}(\boldsymbol{x}) = \int d^3z\; K(\boldsymbol{x}-\boldsymbol{z})\rho(\boldsymbol{z}),$$

how do you start proving some invariance properties of the electric field, using spherical symmetry of the kernel $K$?

4. Oct 19, 2009

### jostpuur

Actually the electric field problem can be dealt with. If one wants to prove that

$$\overline{\boldsymbol{E}}(\boldsymbol{x}) = \frac{1}{4\pi \epsilon_0} \int d^3y\; \overline{\rho}(\boldsymbol{y}) \frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^3} = \cdots = \boldsymbol{E}(\boldsymbol{x}),$$

one has to show that

$$\int d^3r\; K(\boldsymbol{r}) \frac{\boldsymbol{x} - \boldsymbol{z} - \boldsymbol{r}}{\|\boldsymbol{x} - \boldsymbol{z} - \boldsymbol{r}\|^3} = \frac{\boldsymbol{x}-\boldsymbol{z}}{\|\boldsymbol{x}-\boldsymbol{z}\|^3}.$$

I don't know a nice way to do that, but if one knows how to prove that a uniform spherical charge distribution creates an equal electric field as a point charge, then it can be used to carry out the integral over $\boldsymbol{r}$.

But the integral representations did not yet turn out to be helpful for my current charge problem.

5. Oct 19, 2009

### Bob_for_short

For electrical filed it should be reduced to the total charge multiplied by an electric filed of a point at the charge "center", if the integration region is larger than the charge cloud.

6. Oct 19, 2009

### jostpuur

Are you trying to say that electric fields are always like $\propto \boldsymbol{r}/r^3$, provided that the charge distribution is in some bounded domain and we are interested in the field outside the domain? That claim would certainly not be true.

7. Oct 19, 2009

### Bob_for_short

No, the charge distribution should be, of course, spherically simmetric: ρ(r) but not obligatorily r-independent.

If you have a non-symmetric charge distribution ρ(r), the lectric field can be represented as a superposition of fileds of a monopole, dipole, quadrupole, etc., sources.