Biot-Savart for finite volume currents (Wangsness)

[jack476]

Homework Statement

For a distribution of volume currents $\vec{J}^\prime(\vec{r{^\prime}})$ distributed throughout a finite volume, show that the Biot-Savart law can be written as $\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\int_{V^{\prime}}\frac{\nabla^\prime \times \vec{J}(\vec{r}^\prime)}{R}d\tau^\prime$. The source currents are steady and divergenceless.

Relevant Equations

The vector $\vec{r}$ is a field point, the vector $\vec{r}^\prime$ is a source point, and $\vec{R} = \vec{r}-\vec{r}^\prime$.

When $f(\vec{R})$ and $\vec{F}(\vec{R})$ are functions that depend only on $\vec{R}$ the following identities hold: $\nabla f(\vec{R}) = -\nabla^\prime f(\vec{R}), \nabla\cdot\vec{F}(\vec{R}) = -\nabla^\prime\cdot\vec{F}(\vec{R}), \nabla \times \vec{F}(\vec{R}) = -\nabla^\prime \vec{F}(\vec{R})$

Since $\vec{R}/R^3 = -\nabla(\frac{1}{R}) = \nabla^\prime(\frac{1}{R})$, the standard form of the Biot-Savart law for volume currents can be re-written as: $$\frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\vec{J}^\prime (\vec{r}^\prime)\times\vec{R}}{R^3}d\tau^\prime = \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\vec{J}^\prime (\vec{r}^\prime)\times\nabla^\prime\left(\frac{1}{R}\right)d\tau^\prime$$

By using the vector identity $\nabla \times (f\vec{F}) = (\nabla\times \vec{F})f -\vec{F}\times(\nabla f)$ the RHS can be re-written, resulting in:

$$\frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\vec{J}^\prime \times\vec{R}}{R^3}d\tau^\prime = \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\nabla^\prime \times \vec{J}^\prime}{R}d\tau^\prime - \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\nabla^\prime\times\left(\frac{J^\prime}{R}\right)d\tau^\prime$$

So the problem is solved if:

$$ \int\limits_{V^\prime}\nabla^\prime\times\left(\frac{\vec{J}^\prime}{R}\right)d\tau^\prime = 0$$

And I am at a complete loss as to how to prove that. Is there some reason for why $\vec{J}/R$ should be a conservative field even though $\vec{J}$ isn't?

 

[TSny]

So the problem is solved if:

$$ \int\limits_{V^\prime}\nabla^\prime\times\left(\frac{\vec{J}^\prime}{R}\right)d\tau^\prime = 0$$

And I am at a complete loss as to how to prove that.

Consider the identities in the following list:
https://en.wikipedia.org/wiki/Vector_calculus_identities#Surface–volume_integralsCan you find one that helps?

 

[jack476]

Consider the identities in the following list:
https://en.wikipedia.org/wiki/Vector_calculus_identities#Surface–volume_integralsCan you find one that helps?

The third identity in the list can be used to turn the expression into a surface integral. The current has to be tangent at the surface so at the surface the current be written as $K\hat{t}_1$ where $\hat{t}_1$ is a one of the two tangent vectors at the surface, and since $d\vec{a}$ is normal to the surface the integral can be written as:

$$-\int\limits_{\partial V^\prime}\frac{K\hat{t}_2}{R}da $$

Where $\hat{t}_2$ is the other tangent vector, but how does that help?

 

[Delta2]

The third identity in the list can be used to turn the expression into a surface integral. The current has to be tangent at the surface so at the surface the current be written as $K\hat{t}_1$ where $\hat{t}_1$ is a one of the two tangent vectors at the surface, and since $d\vec{a}$ is normal to the surface the integral can be written as:

$$-\int\limits_{\partial V^\prime}\frac{K\hat{t}_2}{R}da $$

Where $\hat{t}_2$ is the other tangent vector, but how does that help?

My intuition tells me that since the current density is divergenceless this last integral must be equal to zero, but can't exactly find the formal proof right now.

 

[Charles Link]

I can give a couple of hints= @TSny hopefully I don't give the OP too much info: Take a look at the identity for $\nabla \times (\psi \vec{A})=\nabla \psi \times \vec{A}+\psi \nabla \times \vec{A}$, and also notice that Biot-Savart has $\frac{\vec{J} \times \vec{R}}{|\vec{R}|^3}=\vec{J} \times \nabla (\frac{1}{|\vec{R}|})$, (possibly needs a minus sign=I need to check it) and where $R=|\vec{R}|$, and $\psi=\frac{1}{R}$ and $\vec{A}=\vec{J}$. $\\$ One more vector identity is necessary with $\vec{A}=\frac{\vec{J}}{R}$,(changed the definition of $\vec{A}$ for this second part), where $\int \nabla \times \vec{A} \, d \tau=-\int \vec{A} \times dS$. The surface integral is zero for a very large surface,(because $\vec{J}$ vanishes at large $r$), making the volume integral also zero. Then go back to the first identity above, etc.

 

[TSny]

I was assuming that the integration volume $V'$ is large enough to enclose the current source so that $\vec J = 0$ at every point of the surface of $V'$. Maybe that's cheating.
I'm not sure what $V'$ represents in Wangsness's statement of the problem.

 

[jack476]

I was assuming that the integration volume $V'$ is large enough to enclose the current source so that $\vec J = 0$ at every point of the surface of $V'$. Maybe that's cheating.
I'm not sure what $V'$ represents in Wangsness's statement of the problem.

Wangsness uses $V^\prime$ to represent the region occupied by the field sources (charges or currents). The problem doesn't specify any particular region (sphere, cylinder, torus, etc), just an arbitrary finite region bounded by a closed surface.

But either way, the current source is enclosed, since the statement of the problem says that all of the current elements occupy a finite region.

 

[jack476]

I can give a couple of hints= @TSny hopefully I don't give the OP too much info: Take a look at the identity for $\nabla \times (\psi \vec{A})=\nabla \psi \times \vec{A}+\psi \nabla \times \vec{A}$, and also notice that Biot-Savart has $\frac{\vec{J} \times \vec{R}}{|\vec{R}|^3}=\vec{J} \times \nabla (\frac{1}{|\vec{R}|})$, (possibly needs a minus sign=I need to check it) and where $R=|\vec{R}|$, and $\psi=\frac{1}{R}$ and $\vec{A}=\vec{J}$. $\\$ One more vector identity is necessary with $\vec{A}=\frac{\vec{J}}{R}$,(changed the definition of $\vec{A}$ for this second part), where $\int \nabla \times \vec{A} \, d \tau=-\int \vec{A} \times dS$. The surface integral is zero for a very large surface,(because $\vec{J}$ vanishes at large $r$), making the volume integral also zero. Then go back to the first identity above, etc.

I don't know if I'm allowed to assume that $V^\prime$ has a very large surface.

 

[Charles Link]

You can pick the volume of integration to be as large as you want in those vector identities. The identities always hold, and if you pick the volume large enough, $\vec{J}$ will necessarily vanish on the surface.
This calculation appears to be rather straightforward. I think I outlined about 90% of it in post 5. The OP @jack476 can hopefully see how to proceed from the hints that I supplied=supplying any additional detail would essentially be furnishing the complete solution=something that is not permitted in the PF homework rules.

 

[Charles Link]

One additional comment=I believe it is just a typo, but the OP's last equation in the Homework Equations of post 1 should read $\nabla \times \vec{F}(\vec{R})=-\nabla' \times \vec{F}(\vec{R})$. The second "times" sign is missing.

 

[Delta2]

Suppose we take a volume $V''$ such that at its boundary $\partial V''$ the current density is zero. Then
$$\int_{V''}\nabla'\times \frac{J}{R}d\tau'=\oint_{\partial V''} \frac{J}{R}\times dS=0$$

But also
$$\int_{V''}\nabla'\times\frac{J}{R}d\tau'=\int_{V'}\nabla'\times\frac{J}{R}d\tau'=\oint_{\partial V'}\frac{J}{R}\times dS$$
where V' is our "original" region of current density where the current density is not necessarily zero at the boundary $\partial V'$ but tangential and it is zero in the volume inbetween the boundaries $\partial V'$ and $\partial V''$. So the first equality in the above equation holds because the current density is zero in the volume between the boundaries $\partial V'$ and $\partial V''$ while the second equality is from the vector calculus theorem.

Do you see a mini paradox here, where the same integral is zero and not zero at the same time??

 

[Charles Link]

Do you see a mini paradox here, where the same integral is zero and not zero at the same time??

There is no paradox. For a minor detail, the vector identity has a minus sign, but that's a finer point. A uniform tangential $\vec{J}$ will have a $\nabla \times \vec{J}$ that behaves like a delta function, which I believe will resolve this mini paradox. The best example I can give which shows this type of thing is if you take a magnetization vector $\vec{M}$ for a uniform magnetization of a cylinder. When you take $\nabla \times \hat{M}$, this derivative vanishes except where $\vec{M}$ is tangential, where the result is $\nabla \times \vec{M}= \vec{M} \times \hat{n} \, \delta(r)$. $\\$ (This result is a surface current result, but I really don't want to introduce that part here, because we are simply working with pure vectors and $\vec{M}$ corresponds to $\vec{J}$, but let's introduce it: In the E&M, $\vec{J}_m=\nabla \times \vec{M}$, and surface current per unit length that occurs as a surface current at the boundary is $\vec{K}_m=\vec{M} \times \hat{n}$. You should be familiar with this curl operation and the delta function that can arise). $\\$ Note that for the volume integral with the smaller volume, $\int \nabla \times \vec{J} \, d \tau$ does not vanish because of the delta function behavior of $\nabla \times \vec{J}$ at the surface. Likewise, the integral $-\int \vec{J} \times dS$ does not vanish for this smaller volume with the tangential $\vec{J}$.
Edit: Additional note: Something still doesn't seem quite right about this vector identity=either the minus sign, or the identity itself. I need to study it further. Additional edit: The integral I believe vanishes also for the smaller volume, at least in some cases, e.g. for the cylinder with uniform $\vec{M}$, the integral $\int \vec{M} \times \hat{n} dS$ vanishes, even though the integrand is non-zero. The surface current $\vec{K}_m=\vec{M} \times \hat{n}$ with a plus sign. I don't yet have good explanation for the minus sign of the vector identity, although I have to believe it is correct as it also that way in an older E&M textbook that I have (Schwarz). $\\$ And one final edit: I believe I can explain the minus sign: Choose as the surface a small Gaussian pillbox that encompasses a small portion of the tangential interface where $\vec{M}$ exists on one portion and there is vacuum on the other portion. The contribution of $\vec{M} \times dS$ comes from the portion that has $dS$ pointing into the material, and $\hat{n}$ points outward in the surface current definition. Thereby the minus sign. Everything now seems to be consistent. :)

 

[jack476]

Okay, I think I understand now.

It makes no difference whether we integrate the distribution $\vec{J}^\prime$ over $V^\prime$ or if we integrate the following current distribution over a finite region $V^{\prime \prime}$ in which $V^\prime$ is properly contained:
$$
\vec{J}^{\prime \prime}(\vec{r}^\prime) = \begin{cases}\vec{J}^\prime &\text{ if } \vec{r}^\prime \in V^\prime\\
0 &\text{ otherwise } \end{cases}
$$
Then the surface current is zero on $\partial V^{\prime \prime}$ so the integral vanishes:
$$
\int\limits_{V^\prime}\nabla^\prime \times\left(\frac{\vec{J}^\prime(\vec{r}^{\prime})}{R}\right)d\tau^\prime = \int\limits_{V^{\prime \prime}}\nabla^\prime \times\left(\frac{\vec{J}^{\prime \prime}(\vec{r}^{\prime})}{R}\right)d\tau^\prime = \int\limits_{\partial V^{\prime \prime}}\frac{\vec{J}^{\prime \prime} (\vec{r}^{\prime})}{R}dA^\prime =0
$$