- #421
Hans de Vries
Science Advisor
Gold Member
- 1,094
- 30
I'm sure the magnetic moment plays a major role, which reminds me of another electro-
magnetic duality: There are types of two magnetic dipoles with indistinguishable fields,
the vector dipole and the axial dipole. The first is a combination of two opposite magnetic
monopoles and the second is a point like circular electric current.
So, associating the field with a continuous charge/spin distribution, which of the two
types is involved in case of the magnetic moment? Well, the net force in a magnetic
field is quite different:
Force on a Vector dipole in a B field:
[tex]
\begin{array}{l l |l l l| l | l | l}
& & \partial_x \textsf{B}_x & \partial_y \textsf{B}_x & \partial_z \textsf{B}_x & & \mu_x & \\
\vec{F}_{magn} & =\ \ & \partial_x \textsf{B}_y & \partial_y \textsf{B}_y & \partial_z \textsf{B}_y & \cdot & \mu_y & \quad =\ \partial_j B_i\ \mu_i \\
& & \partial_x \textsf{B}_z & \partial_y \textsf{B}_z & \partial_z \textsf{B}_z & & \mu_z
\end{array}
[/tex]
Force on an Axial dipole in a B field:
[tex]
\begin{array}{l l |l l l| l | l | l}
& & \partial_x \textsf{B}_x & \partial_x \textsf{B}_y & \partial_x
\textsf{B}_z & & \mu_x & \\
\vec{F}_{magn} & =\ \ & \partial_y \textsf{B}_x & \partial_y \textsf{B}_y & \partial_y \textsf{B}_z & \cdot & \mu_y & \quad =\ \partial_i B_j\ \mu_i \\
& & \partial_z \textsf{B}_x & \partial_z \textsf{B}_y & \partial_z \textsf{B}_z & & \mu_z
\end{array}
[/tex]
Note the exchange of the indices. The second case is what you would expect:
[tex]\vec{F} ~~=~~\mbox{grad}(\vec{B}\cdot\vec{\mu})[/tex]
So, it should be easy to distinguish between the two cases you would say,
and this rules out the pair of magnetic monopoles. There is an interesting twist
however: The net force between either two vector dipoles or two axial dipoles
turns out to be same:
[tex]
\vec{F}\ =\ \frac{3\mu_o\mu_e^2}{4\pi r^4}\left[ \left\{\left(
\hat{\mu}_1 \cdot \hat{\mu}_2 \right) - 5\left( \hat{r} \cdot
\hat{\mu}_1 \right)\left( \hat{r} \cdot \hat{\mu}_2 \right)\right\}
\mbox{\Large $\hat{r}$} + \left( \hat{r} \cdot \hat{\mu}_2
\right)\mbox{\large $\hat{\mu}_1$} + \left( \hat{r} \cdot
\hat{\mu}_1 \right)\mbox{\large $\hat{\mu}_2$}\ \right]
[/tex]
At least in the case that both are at rest, which reminds me that I should find some
time to work out the general case too really proof the exclusion of the first case.
Regards, Hans
magnetic duality: There are types of two magnetic dipoles with indistinguishable fields,
the vector dipole and the axial dipole. The first is a combination of two opposite magnetic
monopoles and the second is a point like circular electric current.
So, associating the field with a continuous charge/spin distribution, which of the two
types is involved in case of the magnetic moment? Well, the net force in a magnetic
field is quite different:
Force on a Vector dipole in a B field:
[tex]
\begin{array}{l l |l l l| l | l | l}
& & \partial_x \textsf{B}_x & \partial_y \textsf{B}_x & \partial_z \textsf{B}_x & & \mu_x & \\
\vec{F}_{magn} & =\ \ & \partial_x \textsf{B}_y & \partial_y \textsf{B}_y & \partial_z \textsf{B}_y & \cdot & \mu_y & \quad =\ \partial_j B_i\ \mu_i \\
& & \partial_x \textsf{B}_z & \partial_y \textsf{B}_z & \partial_z \textsf{B}_z & & \mu_z
\end{array}
[/tex]
Force on an Axial dipole in a B field:
[tex]
\begin{array}{l l |l l l| l | l | l}
& & \partial_x \textsf{B}_x & \partial_x \textsf{B}_y & \partial_x
\textsf{B}_z & & \mu_x & \\
\vec{F}_{magn} & =\ \ & \partial_y \textsf{B}_x & \partial_y \textsf{B}_y & \partial_y \textsf{B}_z & \cdot & \mu_y & \quad =\ \partial_i B_j\ \mu_i \\
& & \partial_z \textsf{B}_x & \partial_z \textsf{B}_y & \partial_z \textsf{B}_z & & \mu_z
\end{array}
[/tex]
Note the exchange of the indices. The second case is what you would expect:
[tex]\vec{F} ~~=~~\mbox{grad}(\vec{B}\cdot\vec{\mu})[/tex]
So, it should be easy to distinguish between the two cases you would say,
and this rules out the pair of magnetic monopoles. There is an interesting twist
however: The net force between either two vector dipoles or two axial dipoles
turns out to be same:
[tex]
\vec{F}\ =\ \frac{3\mu_o\mu_e^2}{4\pi r^4}\left[ \left\{\left(
\hat{\mu}_1 \cdot \hat{\mu}_2 \right) - 5\left( \hat{r} \cdot
\hat{\mu}_1 \right)\left( \hat{r} \cdot \hat{\mu}_2 \right)\right\}
\mbox{\Large $\hat{r}$} + \left( \hat{r} \cdot \hat{\mu}_2
\right)\mbox{\large $\hat{\mu}_1$} + \left( \hat{r} \cdot
\hat{\mu}_1 \right)\mbox{\large $\hat{\mu}_2$}\ \right]
[/tex]
At least in the case that both are at rest, which reminds me that I should find some
time to work out the general case too really proof the exclusion of the first case.
Regards, Hans