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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.

https://www.researchgate.net/figure/Polar-coordinates-with-an-electric-dipole-with-moment-vector-along-the-z-axis-is-the_fig2_1760037

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?

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  • $\begingroup$ 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? $\endgroup$ – hyportnex Sep 12 '19 at 12:21
  • $\begingroup$ @hyportnex Thanks for pointing that out. The charges are actually time-varying in this case. $\endgroup$ – S.D. Sep 12 '19 at 14:28
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    $\begingroup$ 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. $\endgroup$ – hyportnex Sep 12 '19 at 17:13
  • $\begingroup$ @hyportnex Thanks! That should be an answer. :) $\endgroup$ – S.D. Sep 12 '19 at 17:14
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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.

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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.

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  • $\begingroup$ What's the definition of "polar vector" and how do we prove that vector potential is necessarily a polar vector? $\endgroup$ – S.D. Sep 12 '19 at 2:26
  • $\begingroup$ @blue see here for some explanation of the distinction between polar and axial (often called true and pseudo) vectors: en.m.wikipedia.org/wiki/…. $\endgroup$ – Tyberius Sep 12 '19 at 2:35
  • $\begingroup$ 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)... $\endgroup$ – S.D. Sep 12 '19 at 2:43
  • $\begingroup$ @Blue I added an explanation, does it help? $\endgroup$ – Javier Sep 12 '19 at 12:21
  • $\begingroup$ Yes, it does. Thank you! $\endgroup$ – S.D. Sep 12 '19 at 14:23

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