# 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

## Answers and Replies

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
Thank you for the reply. First, to address your questions:
(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