Magnetic vector potential of infinite wire

[siddharth]

I don't know if this thread belongs in Introductory Physics or here, so please feel free to move it if you wish.

The question is
Find the vector potential a distance r from an infinite straight wire carrying a current I

I know that the vector potential can be given by
\frac{\mu_0}{4 \pi} \int \frac{I \vec{dl}}{r}

The problem is that in this case, the wire goes to infinity. So that that doesn't work.

I tried it in another way. Since B = \nabla \times A and the direction of A is generally the direction of the current, it reduces to

\left(\frac{d A_r}{dz} - \frac{ d A_z}{dr}\right) \hat{e_{\phi}} = \frac{\mu_0 I}{2 \pi r} \hat{e_{\phi}}

ie,

\frac{d A_z}{dr} = -\frac{\mu_0 I}{2 \pi r}
and
A = -\frac{\mu_0 I}{2 \pi} ln\left(\frac{r}{r_0}\right) \hat{e_z}

Is that right? Also, how can we generally say that A is along the direction of the current?

 

[quantumdude]

I don't know if this thread belongs in Introductory Physics or here, so please feel free to move it if you wish.

No, it's fine here.

The question is
Find the vector potential a distance r from an infinite straight wire carrying a current I

I know that the vector potential can be given by
\frac{\mu_0}{4 \pi} \int \frac{I \vec{dl}}{r}

The problem is that in this case, the wire goes to infinity. So that that doesn't work.

Yes, it does. The improper integral you would obtain does in fact converge.

I tried it in another way. Since B = \nabla \times A and the direction of A is generally the direction of the current, it reduces to

\left(\frac{d A_r}{dz} - \frac{ d A_z}{dr}\right) \hat{e_{\phi}} = \frac{\mu_0 I}{2 \pi r} \hat{e_{\phi}}

ie,

\frac{d A_z}{dr} = -\frac{\mu_0 I}{2 \pi r}
and
A = -\frac{\mu_0 I}{2 \pi} ln\left(\frac{r}{r_0}\right) \hat{e_z}

Is that right?

I don't think so, but it looks like it's close. What's r_0?

Also, how can we generally say that A is along the direction of the current?

You needn't say that at all. Remember that the potentials are related to the physical fields via differential operators. That relationship affords you some gauge freedom. You can replace \vec{A} with \vec{A}+\nabla\chi provided that \chi is continuously differentiable. You will get the same magnetic field, even if \nabla\chi isn't in the direction of the current.

 

[siddharth]

Yes, it does. The improper integral you would obtain does in fact converge.

Does it?
Doing it that way, I get
A = \left(\frac{\mu_0 I}{4 \pi}\right) ln(\sec \theta + \tan \theta)
where \theta is the angle made between the lines joining the point where I'm finding the potential to the center of the wire and the ends.

So as \theta goes to pi/2 or -pi/2, doesn't it diverge?

I don't think so, but it looks like it's close. What's r_0?

r_0 is some point where I have defined A to be 0.

You needn't say that at all. Remember that the potentials are related to the physical fields via differential operators. That relationship affords you some gauge freedom. You can replace \vec{A} with \vec{A}+\nabla\chi provided that \chi is continuously differentiable. You will get the same magnetic field, even if \nabla\chi isn't in the direction of the current.

Yeah, I understand. But the thing is, in most cases A in fact does point along the direction of the current.

 

[Physics Monkey]

Hi siddharth,

The integral defining A does indeed diverge, however it is possible to extract a finite part which produces the correct physics. Extracting the finite parts from such integrals is a fairly common technique for dealing with these infinite geometries.

In your case the integral you want to do (after a change of variables) is \frac{\mu_0 I}{4 \pi} \int^{\Lambda}_{-\Lambda} dz \frac{1}{\sqrt{\rho^2 + z^2}}, which you can write as \frac{\mu_0 I}{2 \pi} \int^\Lambda_0 dz \frac{1}{\sqrt{\rho^2 + z^2}} = \frac{\mu_0 I}{2 \pi} \log{\left(\frac{\Lambda}{\rho} + \sqrt{1 + \frac{\Lambda^2}{\rho^2}} \right)}. For large cutoff you can approximate the argument of the log as \frac{\mu_0 I}{2 \pi} \log{\left(\frac{2 \Lambda}{\rho} \right)} = - \frac{\mu_0 I}{2 \pi} \log{\left(\frac{\rho}{\rho_0}\right)} + \frac{\mu_0 I}{2 \pi} \log{\left(\frac{2 \Lambda}{\rho_0}\right)}. Now look what has happened! Yes, the vector potential is formally infinite if you let the cutoff go to infinity, but all the physical results (magnetic field) are independent of the cutoff. You are able to extract the relevant physics from a formally divergent integral. Needless to say, this is a very powerful methodology that you will meet again and again as you progress through physics.

Also, the vector potential often points along the current because in the Coulomb gauge one has the formula \vec{A} = \frac{\mu_0}{4 \pi} \int d^3 x' \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}. It's clear then that as long as the current points in one direction, A will point in that direction in the Coulomb gauge.

Hope this helps.

 

[siddharth]

That's fantastic!
I'm curious, why does it work? How come the relevant physics comes out so nicely?

Thanks a ton for your help.

 

[quantumdude]

Does it?

I was wrong about that, as PM pointed out. You have to do some mathematical trickery that I had forgotten about (it's been a while).

Yeah, I understand. But the thing is, in most cases A in fact does point along the direction of the current.

As I said, that's a choice. You can choose the gauge that's convenient for you.

 

[Physics Monkey]

That's fantastic!
I'm curious, why does it work? How come the relevant physics comes out so nicely?

Isn't it cool! Physics is so awesome.

What's going on is essentially a separation of different scales. As far as charged particles close to a long wire are concerned, it doesn't matter how long the wire is as long as it's very long. You could never figure out what the length of the wire is by performing local experiments. Since the length of the wire is unobservable, you might suspect that it enters the vector potential only as a pure gauge term. This suspicion is confirmed when you carry out the calculation, and you can easily check for yourself that is indeed possible to eliminate \Lambda by choosing a different gauge. It is quite instructive to carry out calculations of this type for a variety of infinite geometries. You will find that the infinite geometry gives you some divergence, but the divergence is always unobservable. Such calculations give you glimpse of the very subtle and deep nature of field theory. It's all very beautiful stuff really.