# Field Momentum of a Dipole in a Static Field

## Homework Statement

Consider an ideal stationary magnetic dipole $\vec{m}$ in a static electric field $\vec{E}$. Show that the fields carry momentum $\vec{p} = -\epsilon_0\mu_0\left(\vec{m}\times\vec{E}\right)$.

## Homework Equations

Momentum stored in electromagnetic fields: $\vec{p} = \epsilon_0\int(\vec{E}\times\vec{B})d\tau$ where $d\tau$ is an infinitesimal volume element.
For static electric fields: $\vec{E} = -\nabla V$ where $V$ is the electric potential.
A product rule I used in my attempt: $\nabla\times(f\vec{A}) = f(\nabla\times\vec{A}) - \vec{A}\times(\nabla f)$ where $f$ is a scalar function while $\vec{A}$ is a vector function.
For static magnetic fields: $\nabla\times\vec{B} = \mu_0\vec{J}$ where $\vec{J}$ is the volume current density.
Also, for steady currents: $\nabla\cdot\vec{J} = 0$.
Another identity I used: $\oint(\vec{c}\cdot\vec{r})d\vec{l} = \vec{a}\times\vec{c}$ where $\vec{c}$ is a constant vector and $\vec{a} = \int_S d\vec{a}$ is the vector area.
Yet another identity: $\int_{\nu}(\nabla\times\vec{v})d\tau = -\oint_S\vec{v}\times d\vec{a}$ where $\nu$ is a volume and $S$ is its surface.
Finally, $\int\vec{J}d\tau = \frac{d\vec{P}}{dt}$ where $\vec{P}$ is the total electric dipole moment (0 in our problem)

## The Attempt at a Solution

First, I substitute $\vec{E} = -\nabla V$ into $\vec{p} = \epsilon_0\int(\vec{E}\times\vec{B})d\tau$ to get $\vec{p} = \epsilon_0\int(\vec{B}\times\nabla V)d\tau$. From here, I use the product rule quoted above to get that $\int\vec{B}\times\nabla V d\tau = \int V(\nabla\times\vec{B})d\tau - \int\nabla\times(V\vec{B})d\tau$. Now, by another identity: $- \int\nabla\times(V\vec{B})d\tau = -\oint_S\vec{v}\times d\vec{a}$. But what is the volume/surface we're integrating over? All of space! Thus, the surface is at infinity and $\vec{B}=0$ there (since it is far away from the dipole) and this integral vanishes. We're thus left with $\vec{p} = \epsilon_0\int V(\nabla\times\vec{B})d\tau = \epsilon_0\mu_0\int V\vec{J}d\tau$. Since the only current is confined to the infinitesimal volume near the origin (i.e. the dipole), this integral is nonzero only there. Thus, $V$ can be expanded with a Taylor series for vectors about the origin: $V ≈ V(0) - \vec{E}(0)\cdot\vec{r}$. With this, our expression becomes: $\vec{p} = \epsilon_0\mu_0\int (V(0) - \vec{E}(0)\cdot\vec{r})\vec{J}d\tau$. Focusing our attention on the first integral, $\int V(0)\vec{J}d\tau = V(0)\int\vec{J}d\tau = 0$ as noted above. Thus, $\vec{p} = -\epsilon_0\mu_0\int (\vec{E}(0)\cdot\vec{r})\vec{J}d\tau$.
Now I am stuck. I wanted to just say that $\vec{J}d\tau → \vec{I}dl = Id\vec{l}$ since the dipole can be treated as a current loop. Then, everything works out by using the other identities involving line integrals I quoted. But my math conscientiousness didn't allow me to. So can anyone give me any hint as to how to proceed with this proof or suggest a simpler method for doing it?

Any suggestions/comments will be greatly appreciated!

• Salazar

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gabbagabbahey
Homework Helper
Gold Member
Hi ELB27

A good attempt, but two things stand out to me:

(1) I don't see why you assume the total electric dipole moment is zero
(2) Your series expansion for $V(0)$ looks dubious in general. In particular, how do you justify throwing away the higher order terms, and how does an expression with just $\vec{E}(0)$ help you reach the desired formula?

As an alternative approach, you might try working with the magnetic vector potential for the magnetic dipole instead, and expanding its curl a little before plugging it into the definition for EM momentum. A quick BoE (Back of Envelope) calculation gives me the desired equation plus one extra term which presumabely integrates to zero.

Cheers,
gabba

• ELB27
Hi ELB27

A good attempt, but two things stand out to me:

(1) I don't see why you assume the total electric dipole moment is zero
(2) Your series expansion for $V(0)$ looks dubious in general. In particular, how do you justify throwing away the higher order terms, and how does an expression with just $\vec{E}(0)$ help you reach the desired formula?

As an alternative approach, you might try working with the magnetic vector potential for the magnetic dipole instead, and expanding its curl a little before plugging it into the definition for EM momentum. A quick BoE (Back of Envelope) calculation gives me the desired equation plus one extra term which presumabely integrates to zero.

Cheers,
gabba
(1) Your'e right. After looking at the definition, I cannot justify this step. In my original approach, I just thought that since there are no electric dipoles, there is no dipole moment...
(2) I took the expansion from here (scroll all the way down). I justify throwing away all higher order terms by noting that the integral is nonzero only in an infinitesimal volume enclosing the origin (since that's where the dipole is and $\vec{J}$ is nonzero only there). As for $\vec{E}_0$, again your'e probably right. I thought that after the integration, I will be able to get $\vec{E}$ back.

Now to your method. By definition, $\vec{A}_{dip} = \frac{\mu_0}{4\pi}\frac{\vec{m}\times\hat{r}}{r^2}$. Taking the curl: $\vec{B} = \nabla\times\vec{A} = \frac{\mu_0}{4\pi}\nabla\times\left(\vec{m}\times\frac{\hat{r}}{r^2}\right)$. Using a product rule to expand the curl of the cross product: $\nabla\times\left(\vec{m}\times\frac{\hat{r}}{r^2}\right) = \left[(\frac{\hat{r}}{r^2}\cdot\nabla)\vec{m} - (\vec{m}\cdot\nabla)\frac{\hat{r}}{r^2} + \vec{m}(\nabla\cdot\frac{\hat{r}}{r^2}) - \frac{\hat{r}}{r^2}(\nabla\cdot\vec{m})\right]$. Now $\vec{m}$ is a constant vector, therefore all terms involving its various derivatives vanish. Also, $\nabla\cdot\frac{\hat{r}}{r^2} = 4\pi\delta^3(\vec{r})$ where $\delta^3(\vec{r})$ is the 3-dimensional Dirac-Delta function. Thus, we're left with $\left[4\pi\delta^3(\vec{r})\vec{m} - (\vec{m}\cdot\nabla)\frac{\hat{r}}{r^2}\right]$. To simplify the second term even more, let's use another product rule for gradient of a dot product: $\nabla\left(\vec{m}\cdot\frac{\hat{r}}{r^2}\right) = \vec{m}\times(\nabla\times\frac{\hat{r}}{r^2}) + \frac{\hat{r}}{r^2}\times(\nabla\times\vec{m}) + (\vec{m}\cdot\nabla)\frac{\hat{r}}{r^2} + (\frac{\hat{r}}{r^2}\cdot\nabla)\vec{m}$. Noting that, again, $\vec{m}$ is constant and that the curl of $\frac{\hat{r}}{r^2}$ is zero (this vector diverges from the origin without any rotation), we get that $(\vec{m}\cdot\nabla)\frac{\hat{r}}{r^2} = \nabla\left(\vec{m}\cdot\frac{\hat{r}}{r^2}\right)$. Putting everything together: $\vec{B} = \frac{mu_0}{4\pi}\left[4\pi\delta^3(\vec{r})\vec{m} - \nabla(\vec{m}\cdot\frac{\hat{r}}{r^2})\right]$. Now, we substitute this formula into the definition of EM momentum: $\vec{p} = \epsilon_0\int(\vec{E}\times\vec{B})d\tau = \frac{\epsilon_0\mu_0}{4\pi}\left[ \int 4\pi\delta^3(\vec{r})(\vec{E}\times\vec{m})d\tau - \int\vec{E}\times\nabla\left(\vec{m}\cdot\frac{\hat{r}}{r^2}\right)d\tau\right]$ where the integration is over all space. Now, let's split this monster into two. The first term is easy: $\int 4\pi\delta^3(\vec{r})(\vec{E}\times\vec{m})d\tau = 4\pi(\vec{E}\times\vec{m})$. The second term requires another product rule for curl of a scalar and a vector: $\nabla\times\left[(\vec{m}\cdot\frac{\hat{r}}{r^2})\vec{E}\right] = (\vec{m}\cdot\frac{\hat{r}}{r^2})(\nabla\times\vec{E}) - \vec{E}\times\nabla(\vec{m}\cdot\frac{\hat{r}}{r^2})$. Since $\vec{E}$ is static, it curl is zero. Therefore: $-\vec{E}\times\nabla(\vec{m}\cdot\frac{\hat{r}}{r^2}) = \nabla\times\left[(\vec{m}\cdot\frac{\hat{r}}{r^2})\vec{E}\right]$. Now, let's focus again on the integral and use the formula for the volume integral of the curl I quoted in the OP: $\int\nabla\times\left[(\vec{m}\cdot\frac{\hat{r}}{r^2})\vec{E}\right]d\tau = -\oint_S (\vec{m}\cdot\frac{\hat{r}}{r^2})\vec{E}\times d\vec{a}$. But what is this surface of integration? Since the volume integral is all of space, the surface is at infinity. There, $\frac{\hat{r}}{r^2} = 0$ and therefore, assuming $\vec{E}\neq\infty$ at $r=\infty$, $\oint_S (\vec{m}\cdot\frac{\hat{r}}{r^2})\vec{E}\times d\vec{a} = 0$. Thus, we're finally left with: $\vec{p} = \frac{\epsilon_0\mu_0}{4\pi}4\pi(\vec{E}\times\vec{m}) = -\epsilon_0\mu_0(\vec{m}\times\vec{E})$.

Phew! Thank you very much for suggesting this method. It may be long but much more convincing. Also, I couldn't resist: • Salazar
TSny
Homework Helper
Gold Member
ELB27,

Here is a link to a paper that contains a derivation very similar to your first attempt: http://www.lepp.cornell.edu/~rb532/thesis_final.pdf

There are of course many papers on electromagnetic field momentum. You can look at the references contained in the above link. Also, K. McDonald has written a number of interesting papers on this topic which you can find here: http://www.hep.princeton.edu/~mcdonald/examples/. You will need to sift out the relevant papers from the long list of papers on other topics. This guy has written a lot of very interesting papers.

Also, there is a fairly recent resource letter in the American Journal of Physics by Griffiths:
Am. J. Phys. 80, 7 (2012); http://dx.doi.org/10.1119/1.3641979

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• ELB27
ELB27,

Here is a link to a paper that contains a derivation very similar to your first attempt: http://www.lepp.cornell.edu/~rb532/thesis_final.pdf

There are of course many papers on electromagnetic field momentum. You can look at the references contained in the above link. Also, K. McDonald has written a number of interesting papers on this topic which you can find here: http://www.hep.princeton.edu/~mcdonald/examples/. You will need to sift out the relevant papers from the long list of papers on other topics. This guy has written a lot of very interesting papers.

Also, there is a fairly recent resource letter in the American Journal of Physics by Griffiths:
Am. J. Phys. 80, 7 (2012); http://dx.doi.org/10.1119/1.3641979
Thank you very much for these! I just read the first one and saw that there is another identity I could have used that would solve the problem. For future reference: $\int(\vec{r}\cdot\vec{c})\vec{J}d\tau = -\frac{1}{2}\vec{c}\times\int(\vec{r}\times\vec{J})d\tau$. I was aware of the page of Kirk McDonald before and I have to say that this is a fantastic resource and it came in handy on numerous occasions.

• Salazar