Vector Potential Multipole Expansion

[physwiz222]

TL;DR Summary

Why is the monopole term 0 in the multipole expansion for Magnetic Vector Potential A.

when you do a multipole expansion of the vector potential you get a monopole, dipole, quadrupole and so on terms. The monopole term for a current loop is μI/4πr*∫dl’ which goes to 0 as the integral is over a closed loop. I am kinda confused on that as evaulating the integral gives the arc length but it should go to 0. For example for a circular loop ∫dl’ just gives 2πr’ which is the circumference, not 0. dl’ in Spherical coordinates is <dr, rdθ, rsinθdφ> So if we take θ=π/2 and integrate dφ As the current is in the φ direction we end up with 2πr’. Could someone show how the integral goes to 0 like actually show rigorously by evaluating it and proving that it goes to 0 and explain any mistake I made in the integration.

 

[vanhees71]

I'm not sure, what you are really considering. I guess it's magnetostatics. Then the multipole expansion for the magnetic field, given a current density $\vec{j}(\vec{x})$ which is $0$ outside and at the boundary of a sphere of radius $a$ can be derived as follows:

It all starts, of course, with Maxwell's equations in differential form, and I'm assuming the fundamental vacuum form, i.e., $\vec{j}$ is the total current density. Then you have (in SI units)
$$\vec{\nabla} \times \vec{B}=\mu_0 \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0.$$
From the 2nd equation you know that there's a vector potential $\vec{A}$ such that
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Plugging this into the 1st Maxwell equation (Ampere's Law) you get
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\mu \vec{j}.$$
That can be rewritten as
$$\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=\mu_0 \vec{j}. \qquad (*)$$
Now the vector potential is determined from the physics only up to a "gauge transformation", i.e., you can add any gradient field to $\vec{A}$ without changing the physics at all. This means you can impose some convenient constraint on $\vec{A}$. In magnetostatics the most convenient "choice of gauge" is the Coulomb gauge, which demands
$$\vec{\nabla} \cdot \vec{A}=0.$$
Then (*) simplifies to
$$-\Delta \vec{A}=\mu_0 \vec{j}.$$
For the Maxwell equations of magnetostatics to be solvable at all you must have
$$\vec{\nabla} \cdot \vec{j}=\frac{1}{\mu_0} \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B})=0,$$
which just describes the conservation of charge for a static configuration.

The solution for the potential is now given by the Green's function of $-\Delta$, which is well-known from electrostatics:
$$\vec{A}(\vec{x})=\mu_0 \int_{B_a} \mathrm{d}^3 x' \frac{\vec{j}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
We also know from electrostatics that the Green's function can be expanded in terms of spherical harmonics. Using the notation $r_<=\text{min}(r,r')$ and $r_>=\text{max}(r,r')$ it reads
$$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|} = \sum_{\ell=0}^{\infty} \frac{1}{2 \ell +1} \frac{r_{<}^{\ell}}{r_{>}^{\ell+1}} \sum_{m=-\ell}^{\ell} \text{Y}_{\ell m}(\vartheta,\varphi) \text{Y}_{\ell m}^*(\vartheta',\varphi').$$
Now for $r=|\vec{x}|>a$ (i.e., for a point outside of the current distribution) you thus get
$$\vec{A}(\vec{x}) = \sum_{\ell=0}^{\infty} \frac{\mu_0}{(2 \ell +1) r^{\ell+1}} \sum_{m=-\ell}^{\ell} \text{Y}_{\ell m}(\vartheta,\varphi) \int_{B_a} \mathrm{d^3 x'} \mathrm{d} \varphi r^{\prime \ell} \text{Y}_{\ell m}^*(\vartheta',\varphi') \vec{j}(\vec{x}').$$
Defining the "vector multipole moments" (I'm not sure, where to find this in the literature in this way, so maybe my convention concerning coefficients is differen from the literature) such that
$$\vec{A}(\vec{x}) = \frac{\mu_0}{4 \pi} \sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{\ell} \frac{\vec{M}_{\ell m}}{r^{\ell+1}},$$
you have
$$\vec{M}_{\ell m} = \frac{4 \pi}{2 \ell+1} \int_{B_a} \mathrm{d}^3 x' r^{\prime \ell} \text{Y}_{\ell m}^*(\vartheta',\varphi') \vec{j}(\vec{x}').$$
Now let's look at the "monopole term", i.e., $\ell=m=0$. Since $Y_{00}=1/\sqrt{4 \pi}$ you simply need the integral
$$\vec{M}_{00} = \sqrt{4 \pi} \int_{B_a} \mathrm{d}^3 x' r^{\prime \ell} \vec{j}(\vec{x}').$$
To show that this vanishes, given that $\vec{\nabla} \cdot \vec{j}=0$. Just consider
$$\partial_k (x_j j_k)=\delta_{jk} j_k + x_j \partial_k j_k=j_j$$
and use Gauss's integral theorem
$$\int_{B_a} \mathrm{d}^3 x \partial_k (x_j j_k) = \int_{B_a} \mathrm{d}^3 x j_j = \int_{\partial B_a} \mathrm{d}^2 f_k (x_j j_k)=0,$$
because by assumption $\vec{j}(\vec{x})=0$ for $r=|\vec{x}| \geq a$. Thus you indeed have $M_{00}=0$, and the expansion starts with the dipole contribution.

 

[kuruman]

Could someone show how the integral goes to 0 like actually show rigorously by evaluating it and proving that it goes to 0 and explain any mistake I made in the integration.

Sure. The monopole term of the magnetic vector is a vector which I would write as $\vec M_{00}.$ With $K=\frac{\mu_0 I}{4\pi r}$ we have $\vec M_{00}=K\int d\vec s$. Let's assume that the loop is in the xy-plane and has radius $R$. Then in standard polar coordinate notation, $d\vec s=R(-\sin\phi)~d\phi~\hat x+R\cos\phi~d\phi ~\hat y +0~\hat z.$ The components of the monopole are $$
\begin{align} & M_{00,x} =K\int_0^{2\pi} R(-\sin\phi)~d\phi =0 \nonumber \\
& M_{00,y} =K\int_0^{2\pi} R\cos\phi~d\phi =0 \nonumber \\
& M_{00,z} =K\int_0^{2\pi} 0 =0. \nonumber \\ \end{align}$$ A vector the components of which are all equal to zero is the zero vector. The mistake you made is that you didn't calculate the components separately. You did not interpret $\int d\vec s$ as three integrals in one. When you add vectors, in this case the contributions to the components of $d\vec s$, you add the contributions to the component in a given direction separately from the contributions in the other directions.

 

[physwiz222]

So basically you have to convert that φ unit vector into x and y components. Makes sense. Btw how do u do vector displacement integrals. I have looked online but can’t find anything I just find Line Integral of Vector and Scalar fields but never vector displacement.

 

[physwiz222]

Sure. The monopole term of the magnetic vector is a vector which I would write as $\vec M_{00}.$ With $K=\frac{\mu_0 I}{4\pi r}$ we have $\vec M_{00}=K\int d\vec s$. Let's assume that the loop is in the xy-plane and has radius $R$. Then in standard polar coordinate notation, $d\vec s=R(-\sin\phi)~d\phi~\hat x+R\cos\phi~d\phi ~\hat y +0~\hat z.$ The components of the monopole are $$
\begin{align} & M_{00,x} =K\int_0^{2\pi} R(-\sin\phi)~d\phi =0 \nonumber \\
& M_{00,y} =K\int_0^{2\pi} R\cos\phi~d\phi =0 \nonumber \\
& M_{00,z} =K\int_0^{2\pi} 0 =0. \nonumber \\ \end{align}$$ A vector the components of which are all equal to zero is the zero vector. The mistake you made is that you didn't calculate the components separately. You did not interpret $\int d\vec s$ as three integrals in one. When you add vectors, in this case the contributions to the components of $d\vec s$, you add the contributions to the component in a given direction separately from the contributions in the other directions.

So basically you have to convert that φ unit vector into x and y components. Makes sense. Btw how do u do vector displacement integrals. I have looked online but can’t find anything I just find Line Integral of Vector and Scalar fields but never vector displacement.

 

[kuruman]

Btw how do u do vector displacement integrals.

I just showed you. An integral is a sum. How do u add vectors that point in different directions? U add all the x-components together, all the y-components together and all the z-components together. Is there another way?

 

[physwiz222]

I just showed you. An integral is a sum. How do u add vectors that point in different directions? U add all the x-components together, all the y-components together and all the z-components together. Is there another way?

Makes sense

 

[pasmith]

So basically you have to convert that φ unit vector into x and y components. Makes sense. Btw how do u do vector displacement integrals. I have looked online but can’t find anything I just find Line Integral of Vector and Scalar fields but never vector displacement.

Perhaps consider $d\vec s = \hat t\,ds$ where $\hat t$ is the unit tangent vector in the direction of travel along the curve.

 

[kuruman]

Perhaps consider $d\vec s = \hat t\,ds$ where $\hat t$ is the unit tangent vector in the direction of travel along the curve.

And then what? It's a dead end because you cannot take $\hat t$ out of the integral. You can argue that the resultant of a bunch of vectors put tip-to-tail to form a closed loop adds to zero, however OP wants a rigorous proof, i.e. actually performing an integral which means a sum, which means adding components separately.