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The energy of a magnetic dipole in an external magnetic field is

[tex]U = - \mathbf{m} \cdot \mathbf{B}[/tex]

Yet if I try to show this by calculating the energy in the fields, I get the wrong answer. I believe the problem is that I am making some arguements which neglect strange behavior at r=0, and thus am neglecting a factor.

The math is short, so let me just write it out:

magnetic field of a dipole

[tex]\mathbf{B}(\mathbf{r}) = \frac {\mu_0} {4\pi r^3} \left(3(\mathbf{m}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}\right) + \frac{2\mu_0}{3}\mathbf{m}\delta^3(\mathbf{r})[/tex]

energy in a magnetic field

[tex]U = \frac{1}{2} \int d^3 r \ \frac{1}{\mu_0}B^2[/tex]

So, for a magnetic dipole in a constant external field B_0, we have:

[tex]U = \frac{1}{2} \int d^3 r \ \frac{1}{\mu_0}(\mathbf{B}_0 + \mathbf{B}_{dipole})^2[/tex]

[tex]U = \frac{1}{2 \mu_0} \left[ \int d^3 r \ B_0^2 + \int d^3 r \ B_{dipole}^2 + \int d^3 r \ 2 \mathbf{B}_0 \cdot \mathbf{B}_{dipole} \right][/tex]

The first two terms are independent of the orientations, and thus are just constants that can be ignored when calculating the energy dependence on the dipole's orientation with respect to the external field. So continuing...

[tex]U = \frac{1}{\mu_0} \int d^3 r \ \mathbf{B}_0 \cdot

\left( \frac {\mu_0} {4\pi r^3} \left(3(\mathbf{m}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}\right) + \frac{2\mu_0}{3}\mathbf{m}\delta^3(\mathbf{r}) \right)[/tex]

[tex]U = \frac{2}{3} \mathbf{m} \cdot \mathbf{B}_0 + 4 \pi \int d^3 r \

\frac {1} {r^3} \left( 3(\mathbf{m}\cdot\hat{\mathbf{r}})(\mathbf{B}_0 \cdot \hat{\mathbf{r}})-\mathbf{B}_0 \cdot \mathbf{m} \right)[/tex]

Now, chosing spherical coordinates with the "z" axis parallel to B_0 (I'll be using the physics notation where theta is the angle from z and phi is the angle in the x-y plane from the x-axis), we see

[tex](\mathbf{m}\cdot\hat{\mathbf{r}})(\mathbf{B}_0 \cdot \hat{\mathbf{r}}) = (m_x \cos \phi \sin \theta + m_y \sin \phi \sin \theta + m_z \cos \theta) B_0 \cos \theta [/tex]

Doing the integral over [tex]d\phi[/tex] will kill the m_x and m_y terms. So we have

[tex]U = \mathbf{m} \cdot \mathbf{B}_0 \left[ \frac{2}{3} + 4 \pi \int d^3 r \ \frac {3\cos^2 \theta - 1} {r^3} \right][/tex]

However, since

[tex]\int_0^{\pi} (3\cos^2 \theta - 1) \sin \theta d\theta=0[/tex]

this means:

[tex]U = \frac{2}{3} \mathbf{m} \cdot \mathbf{B}_0[/tex]

which is wrong.

My arguements for showing that each "shell" at r doesn't contribute (doing the d\theta and d\phi integral yield zero) look correct to me,

At r=0, theta and phi aren't even valid coordinates. And since I know the answer is wrong, it seems to suggest to me that the integrand is some kind of delta function (since it seems to contribute zero for all r accept at r=0 where is must be infinite such that the integral is a constant).

But how do I go about calculating that contribution at r=0?

[tex]U = - \mathbf{m} \cdot \mathbf{B}[/tex]

Yet if I try to show this by calculating the energy in the fields, I get the wrong answer. I believe the problem is that I am making some arguements which neglect strange behavior at r=0, and thus am neglecting a factor.

The math is short, so let me just write it out:

magnetic field of a dipole

[tex]\mathbf{B}(\mathbf{r}) = \frac {\mu_0} {4\pi r^3} \left(3(\mathbf{m}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}\right) + \frac{2\mu_0}{3}\mathbf{m}\delta^3(\mathbf{r})[/tex]

energy in a magnetic field

[tex]U = \frac{1}{2} \int d^3 r \ \frac{1}{\mu_0}B^2[/tex]

So, for a magnetic dipole in a constant external field B_0, we have:

[tex]U = \frac{1}{2} \int d^3 r \ \frac{1}{\mu_0}(\mathbf{B}_0 + \mathbf{B}_{dipole})^2[/tex]

[tex]U = \frac{1}{2 \mu_0} \left[ \int d^3 r \ B_0^2 + \int d^3 r \ B_{dipole}^2 + \int d^3 r \ 2 \mathbf{B}_0 \cdot \mathbf{B}_{dipole} \right][/tex]

The first two terms are independent of the orientations, and thus are just constants that can be ignored when calculating the energy dependence on the dipole's orientation with respect to the external field. So continuing...

[tex]U = \frac{1}{\mu_0} \int d^3 r \ \mathbf{B}_0 \cdot

\left( \frac {\mu_0} {4\pi r^3} \left(3(\mathbf{m}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}\right) + \frac{2\mu_0}{3}\mathbf{m}\delta^3(\mathbf{r}) \right)[/tex]

[tex]U = \frac{2}{3} \mathbf{m} \cdot \mathbf{B}_0 + 4 \pi \int d^3 r \

\frac {1} {r^3} \left( 3(\mathbf{m}\cdot\hat{\mathbf{r}})(\mathbf{B}_0 \cdot \hat{\mathbf{r}})-\mathbf{B}_0 \cdot \mathbf{m} \right)[/tex]

Now, chosing spherical coordinates with the "z" axis parallel to B_0 (I'll be using the physics notation where theta is the angle from z and phi is the angle in the x-y plane from the x-axis), we see

[tex](\mathbf{m}\cdot\hat{\mathbf{r}})(\mathbf{B}_0 \cdot \hat{\mathbf{r}}) = (m_x \cos \phi \sin \theta + m_y \sin \phi \sin \theta + m_z \cos \theta) B_0 \cos \theta [/tex]

Doing the integral over [tex]d\phi[/tex] will kill the m_x and m_y terms. So we have

[tex]U = \mathbf{m} \cdot \mathbf{B}_0 \left[ \frac{2}{3} + 4 \pi \int d^3 r \ \frac {3\cos^2 \theta - 1} {r^3} \right][/tex]

However, since

[tex]\int_0^{\pi} (3\cos^2 \theta - 1) \sin \theta d\theta=0[/tex]

this means:

[tex]U = \frac{2}{3} \mathbf{m} \cdot \mathbf{B}_0[/tex]

which is wrong.

My arguements for showing that each "shell" at r doesn't contribute (doing the d\theta and d\phi integral yield zero) look correct to me,

__.__*except at r=0*At r=0, theta and phi aren't even valid coordinates. And since I know the answer is wrong, it seems to suggest to me that the integrand is some kind of delta function (since it seems to contribute zero for all r accept at r=0 where is must be infinite such that the integral is a constant).

But how do I go about calculating that contribution at r=0?

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