# Is the vector potential component $A_\phi$ for a dipole necessarily 0 due to symmetry about the $z$-axis?

Consider an electric dipole (+$$q(t)$$, -$$q(t)$$) [where say $$q(t)=q_0\cos\omega t$$] is placed along the $$\hat z$$ axis. In the spherical polar coordinates, its vector potential $$\vec{A} = A_r \hat r + A_\theta\hat\theta + A_\phi\hat\phi$$ at any point M said to only be a function of $$r$$ (radial distance) and $$\theta$$. However, from here my professor directly concluded that the component along $$\hat \phi$$ i.e., $$A_\phi$$ should be $$0$$ due to symmetry (while simplifying the expression for $$\nabla \times \vec A$$), but I didn't quite understand his reasoning.

Couldn't $$A_\phi$$ be a non-zero constant too? The only constraint I see on $$A_\phi$$ is that it should be independent of $$r$$ and $$\theta$$. Or is $$A_\phi = 0$$ simply a choice taking advantage of gauge invariance; that is, no matter what constant $$A_\phi$$ is it wouldn't change the curl. If so, how to prove it?

• something is not right here "Consider an electric dipole (+q, -q) placed along the $\hat z$ axis. In the spherical polar coordinates, its vector potential, etc." An electric dipole and its vector potential? – hyportnex Sep 12 '19 at 12:21
• @hyportnex Thanks for pointing that out. The charges are actually time-varying in this case. – S.D. Sep 12 '19 at 14:28
• OK, in that case recall that $\mathbf{A}=\int \frac{\mathbf{J}}{r}dV$ hence for current density parallel with $\hat z$ the vector potential is also parallel with the same, no $\phi$ component. Furthermore even if it were not so when $\mathbf{J}$ is confined within a bounded region $\mathbf{A}$ must go to zero as $r \rightarrow \infty$, no constant non-zero component allowed. – hyportnex Sep 12 '19 at 17:13
• @hyportnex Thanks! That should be an answer. :) – S.D. Sep 12 '19 at 17:14

Recall that $$\mathbf{A}(\mathbf{x})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}dV'$$ hence for current density that is parallel with $$\hat z$$ the vector potential can have only one component and be parallel with the same, there cannot be a $$\phi$$ component. Furthermore, even if that were not the case, i.e., $$\mathbf{J} \ne J_z \hat z$$, when $$\mathbf{J}$$ is confined within a bounded region the integral and thus $$\mathbf{A}$$ must go to zero as $$|\mathbf{x}|\rightarrow \infty$$, and no constant non-zero component allowed. A standard example where $$\mathbf{J}$$ is not confined to a finite region is the problem of an infinite linear current and its field: the vector potential is still parallel with the line current but does not go to zero as $$|\mathbf{x}|\rightarrow \infty$$, in fact it is independent of z.

In a few words, it's because the vector potential is a polar vector, so it can't depend on the right hand rule. But if you give it a $$\varphi$$ component you have to pick one direction or the other, and the right hand rule is the only thing that can give you a preferred direction.

More explicitly, being a polar vector means that it changes sign under inversion of the coordinates. Under an inversion, the point $$M$$ goes to its antipodal opposite, and the dipole flips around. You can then reverse that with a 180º rotation, bringing the point $$M$$ back to its original position, rotating $$\mathbf{A}$$ with it, and making the dipole point upwards again. Since the source (the dipole) is back where it was at first, the potential must be the same. But if $$\mathbf{A}$$ points in the $$\varphi$$ direction, it will flip around after this inversion-rotation process! The only way out is for $$A_\varphi$$ to be zero. And note that this doesn't apply to any $$r$$ or $$\theta$$ components: they stay the same after inverting and rotating.

The fact that the vector potential is a polar vector is crucial. The magnetic field is an axial vector, and if instead of a dipole you had a small piece of current, you could certainly have a $$B_\varphi$$ component. In fact that's all you can have: the above argument implies that we must have $$B_r = B_\theta = 0$$.

Why is $$\mathbf{A}$$ a polar vector and $$\mathbf{B}$$ an axial vector (or pseudovector)? Let's start from the basics: the position vector $$\mathbf{r}$$ of a particle is a vector (i.e. polar vector) basically by definition, so the velocity, acceleration and hence force are also vectors. Also, here comes the crucial point: given two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$, their cross product $$\mathbf{a}\times\mathbf{b}$$ is a pseudovector. You can verify this explicitly by just looking at what happens to everything under an inversion of the coordinates, but the deep reason is the right hand rule: it changes to a left hand rule after inversion. Similarly, the cross product of a vector and a pseudovector is a vector, and the cross product of two pseudovectors is a pseudovector.

In short: a pseudovector depends on the right hand rule, a vector doesn't. Applying the right hand rule an even number of times gives you something that doesn't depend on it.

Now look at the Lorentz force

$$\mathbf{F} = q \mathbf{v} \times \mathbf{B}.$$

Since $$\mathbf{F}$$ and $$\mathbf{v}$$ are vectors, $$\mathbf{B}$$ must be a pseudovector. But $$\mathbf{B} = \nabla \times \mathbf{A}$$, so $$\mathbf{A}$$ must be a vector. You just count the number of cross products.

• What's the definition of "polar vector" and how do we prove that vector potential is necessarily a polar vector? – S.D. Sep 12 '19 at 2:26
• @blue see here for some explanation of the distinction between polar and axial (often called true and pseudo) vectors: en.m.wikipedia.org/wiki/…. – Tyberius Sep 12 '19 at 2:35
• I saw that Wiki page...it doesn't really give the proof for vector potential being a polar vector (from the definition of vector potential)... – S.D. Sep 12 '19 at 2:43
• @Blue I added an explanation, does it help? – Javier Sep 12 '19 at 12:21
• Yes, it does. Thank you! – S.D. Sep 12 '19 at 14:23