# Is $\vec A$ (with 0 div.) for uniform $\vec B$ unique?

Tags:
1. Aug 19, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I have a difficulty in the last part of the problem.
Let's assume that gradient of a scalar function $\lambda$ could be added to the vector potential $\vec A$.
Then, acc. to the condition $\nabla . \vec A = 0$,
we get,
$\nabla ^2 \lambda = 0$
The above is true for the whole space. So, this means that there is no source for $\lambda$ in the whole space. This leads to , $\nabla \lambda = 0$.
Hence, the result is unique.

But, the uniqueness theorem says that for a vector quantity $\vec A$, if its divergence and curl is given and its normal component on a boundary of a region is specified , then within this region, $\vec A$ is uniquely defined.

Here, if I assume that $\vec A$ goes to zero at infinity, then using uniqueness theorem, I can say that $\vec A$ is specified uniquely.
But, without this assumption, I can't apply uniqueness theorem.Right?
Does the information that "$\vec B$ is uniform" help in anyway?

2. Aug 19, 2017

### Staff: Mentor

Why?
Correct, because A is not uniquely defined before that.
It is not relevant for the question if A is unique.

3. Aug 19, 2017

### Pushoam

I consider $\lambda$ as an electrostatic potential. Now, if $\nabla ^2 \lambda = 0$ is valid for all space →then there is no charge in the whole space i.e charge density is 0 in the whole space, then electric field i.e. $-\nabla \lambda$ in the whole space has to be zero. Isn't this argument correct?

4. Aug 19, 2017

### Staff: Mentor

The question is purely about magnetic fields, electric fields do not matter.

There are solutions to $\nabla ^2 \lambda = 0$ that have $-\nabla \lambda \neq 0$. Fixing boundary conditions removes them.

5. Aug 19, 2017

### Pushoam

Sorry for being not clear. I took $\lambda$ as electric potential analogically.
But, if there is no source for $\lambda$ in all space, how can there be non-zero $\nabla \lambda$?
Consider the following eqn.:
$\nabla^2 v =0$ ,if there is no source in all space i.e. charge density is 0 everywhere, then $- \nabla V$ has to be 0.
In a similar way, $\nabla^2 \lambda =0$ ,if there is no source for $\lambda$ in all space, $\nabla \lambda = 0$.

6. Aug 19, 2017

### Pushoam

What does"" Fixing boundary conditions"" remove?

7. Aug 19, 2017

### Staff: Mentor

In the analogy to an electric potential, it would be a potential that changes linearly in some direction. $\lambda = \vec r \vec d$ with an arbitrary vector d.
Requiring that $A(r) \to 0$ for $r \to \infty$, for example.

8. Aug 21, 2017

### Pushoam

I can't understand that if there is no charge in the whole universe, how there can be a changing potential. Potential difference is due to the presence of charge, right?
But your argument says that even if there exist no charge, there exist potential difference?
I guess now I understood, in case of potential difference, it is assumed that potential difference at infinity is 0( which is the general case). Under this assumption, $\nabla ^2 \lambda = 0$ gives $\nabla \lambda = 0$.
Here, in case of magneto-statics, I can take $\lambda$ such that $\nabla \lambda$ is not 0 at infinity, so $\vec A$ is not unique .

9. Aug 21, 2017

### Staff: Mentor

Don't take the electric analogy too far. The $\lambda$ I suggested changes A, but it does not change B. It does not have any physical relevance, it is purely a gauge freedom.
In the electric analogy this would be a uniform electric field everywhere in space. This satisfies the Maxwell equations, and there is no charge needed, but this would have an observable consequence of course.

10. Aug 21, 2017

### Pushoam

Does this mean that there can be electrostatic field without any charge?

Don't the above eqns. say that if there is no charge, there is no potential difference?
what kind of consequence, for example?

11. Aug 21, 2017

### Staff: Mentor

If the field is the same everywhere: Yes.
It uses the implicit assumption that the electric field vanishes for large distances. That is fine for electric fields because we know there is no uniform electric field all over space. It does not work any more if you study the magnetic vector potential, because we cannot measure that directly.