0
$\begingroup$

I tried to find an answer to this but haven't yet. Going through Jackson's Electrodynamics, the following steps are used to determine the vector potential as the curl of $\mathbf{B}$. The context is magnetostatics, where $\nabla \cdot \mathbf{J} = 0$.

The basic law (5.4) for the magnetic induction can be written down in general form for a current density $\mathbf{J}(\mathbf{x})$: $$ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \int \mathbf{J}(\mathbf{x'}) \times \frac{(\mathbf{x} - \mathbf{x'})}{|\mathbf{x}-\mathbf{x'}|^3} d^3 x' $$ ... In order to obtain the differential equations equivalent to (5.14) we transform (5.14) into the form: $$ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \nabla \times \int \frac{\mathbf{J}(\mathbf{x'})}{|\mathbf{x}-\mathbf{x'}|} d^3 x' $$

This is all that is said about it, taken from the 1st edition. Note the units aren't SI, hence the $c$.

The issue I have with this is the following. The curl of $\mathbf{J}$ divided by $|\mathbf{x}-\mathbf{x'}|$ can be expanded as: $$ \nabla \times \left ( \frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} \right ) = \frac{1}{|\mathbf{x}-\mathbf{x'}|}\nabla \times \mathbf{J} + \nabla \left ( \frac{1}{|\mathbf{x}-\mathbf{x'}|} \right ) \times \mathbf{J} $$ Which is the well-known identity, $$ \nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) + (\nabla f) \times \mathbf{A}$$ Now we can use the fact that $$ \nabla \left ( \frac{1}{|\mathbf{x}-\mathbf{x'}|} \right ) = \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3}$$

[EDIT: This is a mistake. There should be a negative sign, which explains the wrong sign later on.]

Along with the anti-commutativity of the cross product to get: $$ \nabla \times \left ( \frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} \right ) = \frac{\nabla \times \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} - \mathbf{J} \times \nabla \left ( \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} \right ) $$ Or, rearranging, $$ \mathbf{J} \times \left ( \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} \right ) = \frac{\nabla \times \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} - \nabla \times \left ( \frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} \right ) $$

Now we can substitute this in for the quantity in the integral in the first equation, for $\mathbf{B}$, $$ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \int \frac{\nabla \times \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|}d^3x' - \frac{1}{c} \int\nabla \times \left ( \frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} \right )d^3x' $$ Which is to say, $$ \mathbf{B}(\mathbf{x}) = \frac{1}{c} \int \frac{\nabla \times \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|}d^3x' - \frac{1}{c} \nabla \times \int \frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} d^3x' $$ Now, if the first integral is zero, we get the equality Jackson jumps to so quickly, but without the negative sign. But where did the negative sign go? And could that first integral be zero without $\nabla \times \mathbf{J} = 0$? (Which cannot be true unless $\mathbf{J} = 0$ because in we assumed $\nabla \cdot \mathbf{J} = 0$).

I would be more inclined to think that I made a mistake somewhere, but there doesn't seem to be much going on with the derivation, it's relatively straightforward. What am I missing?

$\endgroup$
4
  • $\begingroup$ Update: I'm thinking that the reason is $\mathbf{J}$ is a function of $\mathbf{x'}$, not $\mathbf{x}$, but the curl is taken with respect to $\mathbf{x}$, pretty subtle, it comes up when using the curl identity. Still unsure $\endgroup$ – Sam Gallagher May 28 '20 at 22:07
  • $\begingroup$ Isn't $\nabla(\frac{1}{|x-x'|}) = \frac{x'-x}{|x'-x|^3}$? $\endgroup$ – SamuraiMelon May 29 '20 at 0:49
  • $\begingroup$ @SamuraiMelon Yes you're right, that fixes the minus sign, thanks! $\endgroup$ – Sam Gallagher May 29 '20 at 14:38
  • $\begingroup$ Note that $\nabla\cdot\mathbf J=0$ and $\nabla\times\mathbf J=0$ does not imply that $\mathbf J=0$ $\endgroup$ – BioPhysicist May 29 '20 at 18:43
1
$\begingroup$

The issue here is keeping track of your derivatives - there's a big difference between differentiating with respect to $\mathbf x$ inside the integral and differentiating with respect to $\mathbf x'$ inside the integral.

In what follows, I use the notation $\frac{\partial}{\partial \mathbf x}$ rather than $\nabla$, to emphasize which variables we are differentiating with respect to.


Starting from

$$\mathbf{B}(\mathbf{x}) = \frac{1}{c} \int \mathbf{J}(\mathbf{x'}) \times \frac{(\mathbf{x} - \mathbf{x'})}{|\mathbf{x}-\mathbf{x'}|^3} d^3 x'$$

we can write

$$\frac{(\mathbf x - \mathbf x')}{|\mathbf x - \mathbf x'|^3} = -\frac{\partial}{\partial \mathbf x} \frac{1}{|\mathbf x-\mathbf x'|}$$

which gives us

$$\mathbf B(\mathbf x) = -\frac{1}{c}\int \mathbf J(\mathbf x') \times \frac{\partial}{\partial \mathbf x}\frac{1}{|\mathbf x - \mathbf x'|}$$

Noting that $\frac{\partial}{\partial \mathbf x} \times (\mathbf A f(\mathbf x))= \frac{\partial f}{\partial \mathbf x} \times \mathbf A$ for a constant vector $\mathbf A$, we can rearrange the above (since $\mathbf J(\mathbf x')$ is constant as far as differentiation with respect to $\mathbf x$ is concerned):

$$\mathbf J(\mathbf x') \times \frac{\partial}{\partial \mathbf x}\frac{1}{|\mathbf x - \mathbf x'|} = - \left(\frac{\partial}{\partial \mathbf x} \frac{1}{|\mathbf x - \mathbf x'|} \right)\times \mathbf J(\mathbf x') = - \frac{\partial}{\partial \mathbf x} \times \left(\frac{\mathbf J(\mathbf x')}{|\mathbf x - \mathbf x'|}\right) $$

which yields that

$$\mathbf B(\mathbf x) = \frac{1}{c} \int \frac{\partial}{\partial \mathbf x} \times \left(\frac{\mathbf J(\mathbf x')}{|\mathbf x - \mathbf x'|}\right) d^3x'$$ $$ = \frac{1}{c}\frac{\partial}{\partial \mathbf x} \times \int \frac{\mathbf J(\mathbf x')}{|\mathbf x - \mathbf x'|} d^3 x'$$


Explicitly, you write

$$\nabla \times \left ( \frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} \right ) = \frac{\nabla \times \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} - \mathbf{J} \times \nabla \left ( \frac{1}{|\mathbf{x}-\mathbf{x'}|} \right )$$

but up to this point, $\nabla$ has denoted differentiation with respect to $\mathbf x$, not $\mathbf x'$, which means that $\nabla \times \mathbf J(\mathbf x') = 0$. If instead you mean $\nabla$ to denote differentiation with respect to $\mathbf x'$, then your derivation works just fine, but in the last line

$$\mathbf{B}(\mathbf{x}) = \frac{1}{c} \int \frac{\nabla \times \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|}d^3x' - \frac{1}{c} \int\nabla \times \left ( \frac{\mathbf{J}}{|\mathbf{x}-\mathbf{x'}|} \right )d^3x'$$

you can't simply pull the $\nabla$ out of the second integral.

$\endgroup$
2
$\begingroup$

$\nabla \left ( \frac{1}{|\mathbf{x}-\mathbf{x'}|} \right ) = \color{red}{-}\frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3},$ which fixes the minus sign.

The first term is usually taken to be zero because of the physical boundary condition of localised charges and currents. I.e. you assume that any current or charge distribution exists over a finite volume or area. Because physically there are no such things as infinite solenoids or wires.

For large volumes, you can take $|\mathbf{x} - \mathbf{x}'| \sim R \rightarrow \infty$, and split the term as: $$\int \frac{\nabla \times \mathbf{J}}{|\mathbf{x}-\mathbf{x'}|}\mathrm{d}^3x' \sim \frac{1}{R}\int \nabla \times \mathbf{J}\,\mathrm{d}^3x' , $$ then you can write it out as: $$\ \frac{1}{R}\int \nabla \times \mathbf{J}\,\mathrm{d}^3x' = \ \frac{1}{R}\int [ \mathbf{J} \times \hat{\mathbf{n}}] \big |_R\cdot \mathrm{d}^2\mathbf{x}', $$ where $\mathbf{J}$ in the last term is integrated at the surface of a sphere of radius $R$.

But because realistic charges/current distributions are localised, $\mathbf{J}(R) \rightarrow 0$ as $R \rightarrow \infty$.

$\endgroup$
2
  • $\begingroup$ If $\nabla$ denotes differentiation with respect to $\mathbf x$, then $\nabla \times \mathbf J(\mathbf x') = 0$. If $\nabla$ denotes differentiation with respect to $\mathbf x'$, then there should be no minus sign in the beginning. $\endgroup$ – J. Murray May 29 '20 at 20:54
  • $\begingroup$ Yeah you are right. I think my physics is correct but I may have adapted it to the maths wrongly. I think your answer is correct. $\endgroup$ – SuperCiocia May 30 '20 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.