Is the Vector Potential of a Toroid Nonzero Outside the Toroid?

[Somali_Physicist]

I am sure that the vector potential of a toroid isn't 0 even though its magnetic field is , does anyone have a derivation for the its the vector potential at a point P(x,y,z) outside the toroid ? i expect that since its curl is 0 we have a general form of :

A = f(!x)e_x + g(!y)e_y+m(!z)e_z
Such that a general function L(!k) implies that L is not the parameter k dependent.

intuitive feel:
If we rotate the torous on its back facing a zy plane we get an infinite stack of circles with a hole.Integrating over these stacks from 0 to the thickness r allows us to add up all the vector potentials.This sum is the total vector potential.An infinite stack of circles must be a cylinder therefore if we found the vector potential of a cylinder 1 - vector potential of an emptying cylinder.

A(c1) - A(c2) = A(c)

uNI/4π ( ∫1/r₁ dl - ∫1/r ₂dl) =A(c)
if we imagine a line to the peak of each cylinder we have two triangles which we can use to relate dl tp x,y,z from there i would assume we use cylindrical coordinates to get solvable integral.However this seems long winded and prone to error.

Does anyone know how to do this?

 

[Dale]

I am sure that the vector potential of a toroid isn't 0 even though its magnetic field is

You can always just do a gauge transformation to zero in that case.

 

[Charles Link]

The current going around the toroid has to necessarily proceed in one direction, either clockwise or counterclockwise. Thereby in a non-deal toroid you will never have precisely zero magnetic field outside the toroid. $\\$ For the ideal case, (where it will not have this residual clockwise or counterclockwise current), of a surface current per unit length (the surface current per unit length will be slightly less in the outer perimeter than it is in the inner perimeter), I think you may be able to show with some effort that $\vec{A}(x)=C \int \frac{\vec{J}(x')}{|x-x'| } \, d^3x'$ is approximately zero outside of the toroid. How close it is to zero is something that you might find in a google. I do think if $\vec{B} =0$, you would find that this integral for $\vec{A}$ is also zero. Applying Ampere's law in a circle at radius $r$ anywhere in the x-y plane outside the toroid does give the result of zero magnetic field for $B_{\phi}$. Perhaps the complete $\vec{B}$ is also precisely zero anywhere outside of the toroid. It's an interesting question.

 

[Somali_Physicist]

You can always just do a gauge transformation to zero in that case.

I am an undergraduate, any advice on where I can begin to study on these transformations?

 

[Dale]

I am an undergraduate, any advice on where I can begin to study on these transformations?

Here is one resource that I use frequently for this type of material:

http://farside.ph.utexas.edu/teaching/em/lectures/node45.html

 

[vsv86]

There is in fact a reasonably large body of literature on the vector potential and radiation of toroids. This is motivated by Aharonov-Bohm effect and anapole nonradiating configurations. You can find exact solutions if you use Legendre functions of the second kind (e.g. G. N. Afanasiev, "Static and non-static electrical solenoids", J. Phys. A Math. Gen. 26 (1993) 731-742).

If you do not want exact solutions, the maths becomes much easier. I could produce it here for a point-like torus for example.

You can always just do a gauge transformation to zero in that case.

It is not that simple. You can remove vector potential with a gauge transformation at any point you choose, but there is no gauge transformation that will remove the vector potential everywhere at the same time. Aharonov-Bohm effect is an experimentally testable consequence of this. Also, bear in mind that non-static toroidal solenoids will in fact radiate.

 

[vsv86]

It is not that simple.

Perhaps I should clarify. Imagine a toroidal solenoid with static current flowing along its minor loops. The magnetic field inside the solenoid is non-zero. The magnetic field outside the solenoid is zero. The electric field is zero everywhere. Now take a loop integral of the vector potential $\oint_C dl \vec{\hat{l}}.\vec{A}$, and let loop go through the hole in the torus, and then around it. As a result the electromagnetic field will be zero everywhere along the loop, yet the value of the loop integral will be equal to the magnetic flux through the toroidal solenoid (by Stokes theorem), and will not be zero. No gauge transformation is going to change this, i.e. there will always be a region with non-zero vector potential somewhere along this loop.

 

[Somali_Physicist]

There is in fact a reasonably large body of literature on the vector potential and radiation of toroids. This is motivated by Aharonov-Bohm effect and anapole nonradiating configurations. You can find exact solutions if you use Legendre functions of the second kind (e.g. G. N. Afanasiev, "Static and non-static electrical solenoids", J. Phys. A Math. Gen. 26 (1993) 731-742).

If you do not want exact solutions, the maths becomes much easier. I could produce it here for a point-like torus for example.
It is not that simple. You can remove vector potential with a gauge transformation at any point you choose, but there is no gauge transformation that will remove the vector potential everywhere at the same time. Aharonov-Bohm effect is an experimentally testable consequence of this. Also, bear in mind that non-static toroidal solenoids will in fact radiate.

Ye , i wasnt getting anywhere with gauge transformations. As for the effects you mentioned i will make sure to search them up when I am not drowning in study.
A non exact solution would suffice i really like to feel the solutions instead of simply seeing it, once i can see its general trend it should be trivial to acquire actual solution or at least understand it.

 

[vsv86]

Hi Somali_Physicist

Sorry for delay. So let's look at the vector potential etc from a torus. This is best done in several steps. First you get a good expression for the current density, then the vector potential.

Current density
Current density itself is best broken into two steps. First one finds the current density of a magnetic dipole, then the torus.

Current density: magnetic dipole

Let there be a loop of current located at point $\vec{r}_m$ in space. The current in the loop is circularing around the unit vector $\hat{\vec{\zeta}}$, and there are two more unit vectors $\hat{\vec{\xi}}$ and $\hat{\vec{\chi}}$ such that $\hat{\vec{\xi}}.\hat{\vec{\chi}}=0$ and $\hat{\vec{\zeta}}=\hat{\vec{\xi}}\times\hat{\vec{\chi}}$. You can think of loop of current as consisting of many separate current density elements, all arranged in a loop. What is the expression for the single current density element that extends from point $\vec{r}_{c1}$ to $\vec{r}_{c2}$

$\vec{J}_c\left(\vec{r}\right)=I \int_{\vec{r}_{c1}}^{\vec{r}_{c2}} dl \: \hat{\vec{l}} \: \delta^{\left(3\right)} \left( \vec{r}-\vec{r}\left(l\right) \right)$

i.e. you express the current density as the integral of the 3d delta function along the line segment from point $\vec{r}_{c1}$ to $\vec{r}_{c2}$. Here $\hat{\vec{l}}$ points in the direction of the segment, vector $\vec{r}\left(l\right)$ denotes the position on the segment and and $I$ is a suitable constant with units of current (i.e. amps). It is easy to see that this expression does behave as what we need. It has units of Current per area, it is point-like and it points along the segment we want.

Next we want to cover out loop with these current elements. Let's say the loop radius is $R_1$ then the integral along the length of the loop is $\int_{\vec{r}_{c1}}^{\vec{r}_{c2}} dl \dots \to \int_{0}^{2\pi} (R_1d\phi)\dots$.

What is the correct expression for $\vec{r}(l)$, i.e. position on the loop? Well, every point on the loop can be accessed by moving from point $\vec{r}_m$ (the centre of the loop) along unit vectors $\hat{\vec{\xi}}$ and $\hat{\vec{\chi}}$ is such a way as to keep the overall distance from $\vec{r}_m$ equal to $R_1$. So every point on the loop is given by $\vec{r}(l) \to \vec{r}_m+R_1 \hat{\vec{\rho}}$, where $\hat{\vec{\rho}}=\hat{\vec{\xi}}\cos\phi+\hat{\vec{\chi}}\sin\phi$ (this expression defines angle $\phi$).

What is the correct expression for $\hat{\vec{l}}$? Vector $\hat{\vec{l}}$ is along the loop, i.e. if you move along it (for a bit), you don't get closer to the loop's centre, so it must be perpendicular to the radial vector $\hat{\vec{\rho}}$. We can call this perpendicular vector $\hat{\vec{\phi}}=-\hat{\vec{\xi}}\sin\phi+\hat{\vec{\chi}}\cos\phi$, so $\hat{\vec{l}} \to \hat{\vec{\phi}}$. Thus the expression for the current density of the loop of current becomes:

$\vec{J}_m\left(\vec{r}\right)= I R_1 \int_{0}^{2\pi} d\phi \: \hat{\vec{\phi}} \: \delta^{\left(3\right)} \left( \vec{r}-\left( \vec{r}_m+R_1 \hat{\vec{\rho}} \right) \right)$

Now comes the technical bit.Lets treat the delta function as a conventional function and write it as 'Taylor' series for small R_1. (I know delta function is a generalized function etc etc. Here we are differentiating it in the generalized sense, i.e. the derivative will be applied to the well-behaved test-function at some later time in evaluation).

$\vec{J}_m\left(\vec{r}\right)= I R_1 \int_{0}^{2\pi} d\phi \: \hat{\vec{\phi}} \: \left[ \delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right) - R_1 \hat{\vec{\rho}}.\vec{\nabla}\delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right) \right]$.
If you evaluate the first term you will find that it vanishes since $\int_{0}^{2\pi} d\phi \: \hat{\vec{\phi}} =0$, so let's only write the second term and expand the radial and azimuthal vectors:

$\vec{J}_m\left(\vec{r}\right)= -I {R_1}^2 \int_{0}^{2\pi} d\phi \: \hat{\vec{\phi}} \: \hat{\vec{\rho}}.\vec{\nabla}\delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right)$
$\vec{J}_m\left(\vec{r}\right)=- I {R_1}^2 \int_{0}^{2\pi} d\phi \: \left( -\hat{\vec{\xi}}\sin\phi+\hat{\vec{\chi}}\cos\phi \right) \left[ \left( \hat{\vec{\xi}}\cos\phi+\hat{\vec{\chi}}\sin\phi \right).\vec{\nabla}\delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right)\right]$

Now integrate with respect to $\phi$ noting that only $\int_{0}^{2\pi} \sin^2 \phi=\int_{0}^{2\pi} \cos^2 \phi =\pi$ whilst other mixed terms will be zero.

$\vec{J}_m\left(\vec{r}\right)=- \pi I {R_1}^2 \left[ \hat{\vec{\chi}}\left(\hat{\vec{\xi}}.\vec{\nabla}\delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right) \right) - \hat{\vec{\xi}}\left(\hat{\vec{\chi}}.\vec{\nabla}\delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right) \right) \right]$

$\vec{J}_m\left(\vec{r}\right)=\vec{\nabla} \times m \hat{\vec{\zeta}} \delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right)$

Where we used the standard convention for defining the magnetic moment $m=Area \times Current=\pi {R_1}^2 I$. Let's make it a vector $\vec{m}=m \hat{\vec{\zeta}}$, so where is the final current density of a point-like magnetic dipole (current loop) at $\vec{r}_m$.

$\vec{J}_m\left(\vec{r}\right)=\vec{\nabla} \times \vec{m} \delta^{\left(3\right)} \left( \vec{r}-\vec{r}_m \right)$

At this point I will stop. And return to derrivation at some later point. Are you happy with this bit?

 

[vsv86]

OK, the TC seems to be gone. I will finish what I have started just in case someone else needs it.

In the last topic I have shown that the current density due to a single current loops is the curl of the delta function. Now, a torus is essentially a loop of magnetization, and magnetization is the density of magnetic dipoles per volume. One can therefore simply repeat the whole derivation I did before, but this time on the magnetization itself. Since the process is exactly the same as before I will simply quote the end result.

The current density due to a point-like torus is:

$\vec{J}_T \left( \vec{r} \right)=\vec{\nabla}\times\vec{\nabla}\times\vec{T}\delta^{\left(3\right)}\left(\vec{r}-\vec{r}_T\right)$

Here $\vec{T}$ is what is known as toroidal dipole, a 3d vector around which the loop of magnetization circulates. Given the torus with major radius $\mathcal{R}$, minor radius $R$ and current $I$, its toroidal dipole magnitude is $T=I \pi R\mathcal{R}/2$

The question was about the static case, so let's stick to it. Applying the Green's function to solve $\nabla^2 \vec{A}=-\mu_0 \vec{J}$, one finds the vector potential due to torus:

$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \int d^{3}r' \frac{ \vec{J}_T \left( \vec{r'} \right) }{\left| \vec{r}-\vec{r'}\right|}$
$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \int d^{3}r' \frac{ \vec{\nabla}'\times\vec{\nabla}'\times\vec{T}\delta^{\left(3\right)}\left(\vec{r'}\right) }{\left| \vec{r}-\vec{r'}\right|}$

Next, one uses the fact that delta function vanishes everywhere apart from the origin, to integrate by parts:

$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \int d^{3}r' \delta^{\left(3\right)}\left(\vec{r'}\right) \vec{\nabla}'\times\vec{\nabla}'\times\frac{ \vec{T} }{\left| \vec{r}-\vec{r'}\right|}$

Since the Green function only depends on $\left| \vec{r}-\vec{r'}\right|$ one can change the differentiation variable $\vec{\nabla}'\to\vec{\nabla}$, take the derivatives out of the integral, and then evaluate the integral:

$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \vec{\nabla}\times\vec{\nabla}\times\frac{ \vec{T} }{r}$

Evaluating the above expression is a pain, so I will leave it up to the reader. One thing to point out here is that this expression is only valid for $r\neq0$. The origin needs more careful treatment. I will do this in the next post.

 

[Somali_Physicist]

OK, the TC seems to be gone. I will finish what I have started just in case someone else needs it.

In the last topic I have shown that the current density due to a single current loops is the curl of the delta function. Now, a torus is essentially a loop of magnetization, and magnetization is the density of magnetic dipoles per volume. One can therefore simply repeat the whole derivation I did before, but this time on the magnetization itself. Since the process is exactly the same as before I will simply quote the end result.

The current density due to a point-like torus is:

$\vec{J}_T \left( \vec{r} \right)=\vec{\nabla}\times\vec{\nabla}\times\vec{T}\delta^{\left(3\right)}\left(\vec{r}-\vec{r}_T\right)$

Here $\vec{T}$ is what is known as toroidal dipole, a 3d vector around which the loop of magnetization circulates. Given the torus with major radius $\mathcal{R}$, minor radius $R$ and current $I$, its toroidal dipole magnitude is $T=I \pi R\mathcal{R}/2$

The question was about the static case, so let's stick to it. Applying the Green's function to solve $\nabla^2 \vec{A}=-\mu_0 \vec{J}$, one finds the vector potential due to torus:

$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \int d^{3}r' \frac{ \vec{J}_T \left( \vec{r'} \right) }{\left| \vec{r}-\vec{r'}\right|}$
$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \int d^{3}r' \frac{ \vec{\nabla}'\times\vec{\nabla}'\times\vec{T}\delta^{\left(3\right)}\left(\vec{r'}\right) }{\left| \vec{r}-\vec{r'}\right|}$

Next, one uses the fact that delta function vanishes everywhere apart from the origin, to integrate by parts:

$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \int d^{3}r' \delta^{\left(3\right)}\left(\vec{r'}\right) \vec{\nabla}'\times\vec{\nabla}'\times\frac{ \vec{T} }{\left| \vec{r}-\vec{r'}\right|}$

Since the Green function only depends on $\left| \vec{r}-\vec{r'}\right|$ one can change the differentiation variable $\vec{\nabla}'\to\vec{\nabla}$, take the derivatives out of the integral, and then evaluate the integral:

$\vec{A}_T \left(\vec{r}\right)=\frac{\mu_0}{4\pi} \vec{\nabla}\times\vec{\nabla}\times\frac{ \vec{T} }{r}$

Evaluating the above expression is a pain, so I will leave it up to the reader. One thing to point out here is that this expression is only valid for $r\neq0$. The origin needs more careful treatment. I will do this in the next post.

Sorry i was busy with a project, ill make sure to see if i can try to understand your solution , thank you a lot.