Electric potential is single value since curl E =0.

According to me magnetic scalar potential should be single valued ( since curl B=0) and the vector potential should be multi valued. I can just not get how scalar potential is multi valued . It should be single valued since curl B=0.

What does it mean intuitively for curl B being 0. I read that in space where there is no current curl B=0. But B is produced due to currents (maybe far off from the point in space where we are measuring B) and B doesn't have any divergence.In the field pictures it always curls.

So What does it mean intuitively for curl B being 0 ?


1 Answer 1


So intuitively $\nabla \cdot B = 0$ is best understood by appeal to the continuity equation $\dot \rho = -\nabla \cdot J,$ in other words if you imagine $\vec B$ as a flow field $\vec v$ for some fluid of constant density, this describes a flow of stuff which does not accumulate at any particular points: it is not "flowing into" or "flowing out of" any of those places.

The expression $\nabla\times B = 0$ is a little harder to understand when you imagine $\vec B$ as a flow field $\vec v$, but if you imagine that you insert a little pinwheel into the flow field at a point, the curl says something about how much torque there will be on that pinwheel due to the fluid trying to drag its tines, with the direction of the curl being the direction that makes the pinwheel spin the most. So $\nabla\times B=0$ means literally "this flow field does not make a pinwheel spin." That helps to understand the strange flow field $\vec v = \alpha ~ \hat \theta / r,$ which seems to "curl" around the origin but has $\nabla \times \vec v = 0$ at all points that are not the origin. If you think about a tiny little area $dr~r~d\theta,$ it has a slightly longer "outside edge" than its "inside edge" and so the flow along that outside edge drags the pinwheel slightly more; but if the flow drops like $1/r$ this slight lengthening is exactly compensated by the slight attenuation in the flow field, meaning that the pinwheel is not rotated at all.

Helmholtz's theorem says that any "nice" vector field $X$ can be "integrated" into a sum of two fields, a curl and a gradient: $$X = \nabla \times Y - \nabla \zeta.$$Since $\nabla \cdot X = -\nabla^2 \zeta$ and $\nabla \times X = \nabla \times (\nabla \times Y)$ due to standard vector identities, whenever we know that $\nabla \cdot X = 0$ we can solve this trivially with $\zeta = 0$ and we know that $X = \nabla \times Y$ for this "vector potential" $Y.$ Similarly when we know that $\nabla\times X = 0$ we know that $Y=0$ works and that $X = -\nabla \zeta$ for this "scalar potential" $\zeta.$

How this plays out in classical E&M

The full Maxwell Equations are (SI units) $$\begin{array}{ll}\nabla \cdot E = \rho/\epsilon_0&~~\nabla \times E = -\dot B\\ \nabla \cdot B = 0&~~\nabla \times B = \mu_0 J + \mu_0\epsilon_0 \dot E. \end{array}$$ Clearly we always have $B = \nabla \times A$ for some vector potential $\vec A$ by the third equation. More subtly, the second equation then works out to $\nabla \times (E + \dot A) = 0$ and therefore $E = -\dot A - \nabla \phi$ for some scalar potential $\phi.$ The choices that you make for these are not unique as you can add any $\nabla \psi$ to $A$ and still preserve $B,$ because the curl of a gradient is zero. If you work out the effect of this on $E$ you'll see that you have to also subtract $\dot \psi$ from $\phi$ to preserve $E$ as well. This is called the "gauge freedom" but it really reflects this freedom to choose your "constants of integration" in the Helmholtz decomposition.

The remaining equations work out to $$\begin{array}{rl}-\nabla\cdot \dot A -\nabla^2 \phi ~=& \rho/\epsilon_0\\ \nabla(\nabla \cdot A) - \nabla^2 A ~=& \mu_0 J - \mu_0\epsilon_0\ddot A - \mu_0\epsilon_0\nabla \dot \phi,\end{array} $$ once you use this identity that $\nabla \times (\nabla \times X) = \nabla(\nabla\cdot X) - \nabla^2 X.$ These can be put into a very elegant form by recognizing the wave operator $\square X = \mu_0\epsilon_0 \ddot X - \nabla^2 X$ and then incorporating the other spare terms in the second equation into a scalar field $\lambda = \mu_0\epsilon_0\dot \phi + \nabla\cdot A.$

Note that your gauge freedom maps $\lambda \mapsto \lambda - \square \psi$ so you can basically choose any functional form you want for $\lambda$ by solving a wave equation, in theory. Note also that $\lambda$ enters into the first equation via replacing $\nabla\cdot \dot A$ with $\dot \lambda - \mu_0 \epsilon_0 \ddot \phi,$ so you get $$\begin{array}{rl}\square \phi ~=& \rho/\epsilon_0 + \dot \lambda\\ \square A ~=& \mu_0 J - \nabla\lambda.\end{array}$$The "Lorentz gauge" sets $\lambda = 0$ because $(c\rho, J)$ is a 4-vector in relativity where $c = 1/\sqrt{\mu_0\epsilon_0}$ and hence this gives you an equation $\Box(\phi/c, \vec A) = \mu_0 (c\rho, \vec J),$ and since $\Box$ is relativistically covariant you get that $(\phi/c, \vec A)$ is also a 4-vector in relativity. If that doesn't make much sense to you just accept "hey, it makes these equations look exactly alike" and that's good enough.

The "Coulomb gauge" sets $\lambda = \mu_0\epsilon_0 \dot\phi$ so that the first equation reduces to just $-\nabla^2\phi = \rho/\epsilon_0,$ and we can basically assume that the scalar potential $\phi$ updates instantaneously to all of the charges, which makes it super-simple to calculate. If you look up at the definition of $\lambda$ you find out that it also sets $\nabla\cdot A = 0$ so $A$ can be integrated in terms of a vector potential potential, $A = \nabla \times W.$

  • $\begingroup$ Thanks for the answer. I somewhat get it. But why is Scalar potential multi valued when curl B =0. Since curl B =0 Scalar potential should be single valued just like curl E =0 and electric potential is single valued $\endgroup$
    – Shashaank
    Commented Feb 6, 2017 at 16:21
  • $\begingroup$ Also , if I drop a stick at an angle from a certain height , it rotates while falling. But gravitational field is curl less so should rotate ? $\endgroup$
    – Shashaank
    Commented Feb 6, 2017 at 16:36
  • $\begingroup$ @Shashaank: if you're careful to not impart rotation to a falling stick, it will not rotate except by air drag, which will minimize its cross-section that faces downwards. It sounds like you're examining a scalar potential for $\vec B$ in the case where $J=0$ and $\dot E=0$? The only reason for such things to become multivalued is in cases like the above, $\vec v=\alpha\hat\theta/r,$ where there is a "hole" in the space where $\nabla\times\vec v=0,$ in this hole $\nabla\times\vec v\ne0$ and you have a multivalued function based on how many times you "go around the hole." Does that help? $\endgroup$
    – CR Drost
    Commented Feb 6, 2017 at 17:12
  • $\begingroup$ Regarding the stick - Will it not rotate if there is no air drag . Because I had thought its the force of gravity which causes rotation. If that had been true then a curl less field was causing rotation. Is the rotation because of air drag ? Regarding the Scalar potential After reading your answer - Actually it was quite enlightening. The only problem is that I haven't learn't much about the wave operator and gauge transformation though I understood the Vector identities part. $\endgroup$
    – Shashaank
    Commented Feb 6, 2017 at 18:03
  • $\begingroup$ Yet I am unable to understand that curl E =0 causes electric potential to be single valued ( in at statics) but curl B =0 does not cause Scalar potential to be single valued. I suspect from the last line of your comment that is because of the particular expression of B. $\endgroup$
    – Shashaank
    Commented Feb 6, 2017 at 18:06

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