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Field Momentum of a Dipole in a Static Field

  1. Jan 7, 2015 #1
    1. The problem statement, all variables and given/known data
    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)##.

    2. Relevant 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)

    3. 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!
     
  2. jcsd
  3. Jan 7, 2015 #2

    gabbagabbahey

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    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 [itex]V(0)[/itex] looks dubious in general. In particular, how do you justify throwing away the higher order terms, and how does an expression with just [itex]\vec{E}(0)[/itex] 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
     
  4. Jan 8, 2015 #3
    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:

    woody and buzz.jpg
     
  5. Jan 8, 2015 #4

    TSny

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    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
     
    Last edited: Jan 8, 2015
  6. Jan 8, 2015 #5
    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.
     
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