cup said:
Well, please show me the finite derivation you speak of.
Okay, I'll try to type something up.
cup said:
Why do you ignore Uoutside?
That integral is zero. It is a multiplication of two different spherical harmonics (legendre polynomials) which are orthogonal, so the result is zero.
Also I notice you are using an infinite external source, if you do this you have left out one term (a surface term at infinity). You can either include this term, or just use a finite source.
Okay, let me try to write out the calculations.
----------------------
Consider a spherical shell of radius R and of uniform charge spinning such that it has a magnetic dipole of \mathbf{m}. The field outside the sphere is that of a perfect magnetic dipole, and inside the magnetic field is a constant.
<br />
\mathbf{B}_{dip} = <br />
\begin{cases}<br />
\frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] & \text{if } r \geq R \\<br />
\frac{2 \mu_0}{3} \frac{\mathbf{m}}{\frac{4}{3} \pi R^3} & \text{if } r < R<br />
\end{cases}<br />
The limit R \rightarrow 0 yields an ideal magnetic dipole:
<br />
\mathbf{B}_{dip} = \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] + \frac{2 \mu_0}{3} \mathbf{m} \delta^3(\mathbf{r})<br />
The energy in electromagnetic fields is given by:
<br />
U_{em} = \frac{1}{2} \int (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) d^3r<br />
where d^3r is the volume element and the integration is over all space.
Using an infinite external source:
If we consider an external field \mathbf{B} and the field due to a magnetic dipole \mathbf{B}_{dip}, we have:
<br />
U = \frac{1}{2\mu_0} \int (\mathbf{B} + \mathbf{B}_{dip})^2 \ d^3r = \frac{1}{2\mu_0} \int (B^2 + B^2_{dip} + 2 \mathbf{B} \cdot \mathbf{B}_{dip}) \ d^3r<br />
the B^2 and B^2_{dip} terms are independent of the orientation and so are just constants we will ignore.
Choosing axes such that the external field is in the z direction,
<br />
U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \mathbf{B}_{dip} \ d^3r =<br />
\frac{1}{\mu_0} \int B \hat{\mathbf{z}} \cdot \left[ \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] + \frac{2 \mu_0}{3} \mathbf{m} \delta^3(\mathbf{r}) \right] d^3r<br />
The first term does not contribute, for
<br />
\begin{align*}<br />
\int & \hat{\mathbf{z}} \cdot [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] d^3r \\<br />
& = \int [ 3(m_x \sin\theta\cos\phi + m_y \sin\theta\sin\phi + m_z \cos\theta)\cos\theta - m_z ] r^2 \sin\theta \ d\phi d\theta dr <br />
\end{align*}<br />
of which the m_x and m_y terms vanish after the \phi integration, and the m_z term will vanish after the \theta integration for it contains
<br />
\int [ 3\cos^2\theta - 1] \sin\theta \ d\theta = 0.<br />
Therefore only the delta function term will contribute to the interaction energy.
<br />
U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \frac{2 \mu_0}{3} \mathbf{m} \delta^3(\mathbf{r}) \ d^3r = (\mathbf{B} \cdot \mathbf{m}) \int \frac{2}{3} \delta^3(\mathbf{r}) d^3r<br />
<br />
U = + \frac{2}{3} \mathbf{m} \cdot \mathbf{B}<br />
This has the wrong magnitude and the wrong sign. The reason the magnitude is wrong is because we used an infinite source. When doing so, the energy in the electromagnetic fields also requires a surface term
\frac{1}{2\mu_0} \int (\mathbf{A}\times\mathbf{B}) \cdot d\mathbf{a}
If you work this out, it gives another \frac{1}{3} \mathbf{m} \cdot \mathbf{B}. Or you can just use a finite source, which I'll do here as it is more satisfying for we avoid any limits to infinity.
Another possible complaint is that while doing the \theta and \phi integrals show that the non-delta function terms don't contribute, it is unclear if this argument holds for the point right at r=0 where this coordinate system is undefined. Also the delta function term can be unsettling. However, since the dipole is in an external field, and the finite sized dipole just has a constant field inside (where the external field is also constant) it is trivial to do this with the finite dipole and get the same result (before even taking the limit it approaches an ideal 'point' dipole). The math results in the same answer, so that is not where the problem resides.
Using a finite external source:
In the last attempt, even though the integration was over "all space", the source of the external field was not including anywhere in space. While this satisfies Maxwell's equations, a finite source would be preferred. Since it has already been noted that a spinning shell of uniform charge produces a constant magnetic field inside, this can be used as the source and the limit R \rightarrow \infty be taken if the equivalent of a constant external magnetic field "everywhere" is needed.
For r<R, the result is identical to attempt 1, so the energy is:
<br />
U = \frac{1}{\mu_0} \int \mathbf{B} \cdot \mathbf{B}_{dip} \ d^3r = \frac{2}{3} (\mathbf{m} \cdot \mathbf{B}) + U_2<br />
where
<br />
U_2 = \frac{1}{\mu_0} \int_R^\infty \int_0^\pi \int_0^{2\pi} \mathbf{B} \cdot \mathbf{B}_{dip} \ r^2 \sin\theta \ d\phi d\theta dr<br />
This new term, where r > R, both the source and the dipole under investigation have perfect dipole fields. The magnetic dipole of the source will be referred to as \mathbf{m_1} while that of the perfect dipole is \mathbf{m_2}. The axes are chosen such that the external field (and therefore \mathbf{m_1}) is in the z direction.
<br />
U_2 = \frac{1}{\mu_0} \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{\mu_0}{4 \pi r^3} \left[ 3 m_1 \cos\theta \hat{\mathbf{r}} - m_1 \hat{\mathbf{z}} \right] \cdot \mathbf{B}_{dip} \ r^2 \sin\theta \ d\phi d\theta dr<br />
Using the same logic as in the last attempt, the term in the z direction with no angular dependence will not contribute to the integral here. Expanding out the remaining terms results in
<br />
\begin{align*}<br />
U_2 = & \frac{1}{\mu_0} \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{\mu_0}{4 \pi r^3} 3 m_1 \cos\theta \ \hat{\mathbf{r}} \\<br />
& \cdot \ \frac{\mu_0}{4 \pi r^3} [ 3(m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta)\hat{\mathbf{r}} - \mathbf{m_2} ] <br />
r^2 \sin\theta \ d\phi d\theta dr \\<br />
U_2 = & \frac{3 \mu_0}{16 \pi^2} m_1 \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{1}{r^4} \cos\theta [ 3(m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta) \\<br />
& - (m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta)] \sin\theta \ d\phi d\theta dr \\<br />
U_2 = & \frac{3 \mu_0}{8 \pi^2} m_1 \int_R^\infty \int_0^\pi \int_0^{2\pi} \frac{1}{r^4} \cos\theta (m_{2x} \sin\theta\cos\phi + m_{2y} \sin\theta\sin\phi + m_{2z} \cos\theta) \sin\theta \ d\phi d\theta dr<br />
\end{align*}<br />
Performing the \phi integration yields
<br />
\begin{align*}<br />
U_2 &= \frac{3 \mu_0}{4 \pi} m_1 \int_R^\infty \int_0^\pi \frac{1}{r^4} \cos\theta m_{2z} \cos\theta \sin\theta \ d\theta dr \\<br />
&= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \int_R^\infty \int_0^\pi \frac{1}{r^4} \cos^2 \theta \ \sin\theta \ d\theta dr.<br />
\end{align*}<br />
[/itex]<br />
<br />
Do a change of variables u=\cos\theta and complete the integral.<br />
<br />
\begin{align*}<br />
U_2 &amp;= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \int_R^\infty \int_{-1}^1 \frac{1}{r^4} u^2 \ du dr \\<br />
&amp;= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \frac{2}{3} \int_R^\infty \frac{1}{r^4} \ dr \\<br />
&amp;= \frac{3 \mu_0}{4 \pi} (\mathbf{m_1} \cdot \mathbf{m_2}) \frac{2}{3} \frac{1}{3R^3} \\<br />
&amp;= \frac{1}{3} (\frac{2 \mu_0}{3} \frac{\mathbf{m_1}}{\frac{4}{3}\pi R^3} \cdot \mathbf{m_2})<br />
\end{align*}<br /><br />
<br />
Referring to the source equations, the relationship between \mathbf{m_1} and the constant external field \mathbf{B} inside the source simplifies the result to<br />
<br />
\begin{align*}<br />
U_2 = \frac{1}{3} \mathbf{m_2} \cdot \mathbf{B}<br />
\end{align*}<br /><br />
<br />
The final result is therefore<br />
<br />
\begin{align*}<br />
U = + \mathbf{m} \cdot \mathbf{B}<br />
\end{align*}<br /><br />
<br />
While the magnitude is now correct, the sign is still wrong. This calculation seems to indicate the state in which the dipole is aligned with the external field is <i>unstable</i>, and the minimum energy is with the dipole aligned <i>opposing</i> the field.<br />
<br />
While it is hard to do such integral in your head, one can see that it seems intuitively reasonable that a magnetic dipole alligned <i>against</i> a field would reduce the total magnetic field, while one alligned <i>with</i> the field would increase the magnetic field ... therefore from the "energy in the fields" standpoint, this result seems to make sense. However, we know from experiment that this must be wrong as a dipole left to rotate on its own will try to allign with the field. <br />
<br />
... Hence the "paradox".