TransWikia.com

Magnetic Moment MIT Bag Model

Physics Asked by cmmigl on May 16, 2021

I’m reading the book "Advances in Nuclear Physics vol 13" by J. W. Negele and Erich Vogt

In chapter 3, one wants to calculate the magnetic moment for a current loop.
In page 29 how does one go from equation 3.8 to 3.9:

equation 38:
$$mu=frac{N^2}{2}sum_iQ_iint_{bag}dtextbf{r} textbf{r}^2[j_0(omega r/R),-isigma_ihat{r}j_1(omega r/R)]
begin{pmatrix}
0 & textbf{r}timessigma_i
textbf{r}timessigma_i & 0
end{pmatrix}
begin{pmatrix}
j_0(omega r/R)
isigma_ihat{r}j_1(omega r/R)
end{pmatrix}
$$

equation 39:

$$mu=mu_0sum_i sigma_iQ_i$$

Manipulating expression 38 I arrive at:

$$sum_iQ_ifrac{N^2}{2}int_0^R dtextbf{r} textbf{r}^2left(ij_0left(frac{omega r}{R}right)j_1left(frac{omega r}{R}right)[textbf{r}timessigma_i,sigma_ihat{r}]right)$$

Where $[textbf{r}timessigma_i,sigma_ihat{r}])$ is the commutator.

How should I proceed? I know I should manage to isolate $sigma_i$, but I don’t know how to go further.

One Answer

I ended up expanding the commutator, using the levi-civita definition of the cross product. Arriving at the result:

$ [textbf{r}times sigma_i, sigma_ihat{r}]=2i(textbf{r}sigma_ihat{r}-textbf{r}sigma_i)$

From there I computed the integral and arrived at 3.9

Correct answer by cmmigl on May 16, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP