# Suffix notation help

1. Mar 13, 2014

### Matt atkinson

1. The problem statement, all variables and given/known data
Show that the equation $$\nabla \times \vec{p} = -\frac{\vec{B}}{r^3} + 3\frac{\vec{B} \bullet \vec{r}}{r^5}\vec{r}$$
Where ;
$$\vec{p} = \frac{\vec{B} \times \vec{r}}{r^3}$$
$$\vec{r}=(x_1 ,x_2 ,x_3)$$
and $$\vec{B}$$ is a constant vector.
and r is the magnitude of $$\vec{r}$$
2. Relevant equations
above

3. The attempt at a solution
$$\nabla \times \vec{p} = \epsilon_{ijk} \epsilon_{klm} \frac{d}{dx_j} B x_m |r|^{-3}$$
$$= \epsilon_{ijk} \epsilon_{klj} B |r|^{-3} - 3 \epsilon_{ijk} \epsilon_{klm} B x_m x_j |r|^{-5}$$
I've tried expanding and using various identities such as;
$$\epsilon_{ijk} \epsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$$
If someone could give me a push in the right direction or let me know if i went wrong somewhere (i know i skipped a few steps but it took me 20 mins to right out the latex code for this adk if you don't see what I did).

Last edited by a moderator: Mar 13, 2014
2. Mar 13, 2014

### pasmith

You're missing an $l$ suffix on the $B$ and the derivative is partial:
$$(\nabla \times \vec p)_i = \epsilon_{ijk} \epsilon_{klm} B_l \frac{\partial}{\partial x_j} \left(\frac{x_m}{r^3}\right)$$

You will also need
$$\frac{\partial r}{\partial x_j} = \frac{x_j}{r}$$ and $$\frac{\partial x_m}{\partial x_j} = \delta_{jm}$$

3. Mar 13, 2014

### Matt atkinson

ah okay, hmm.
so I should use the product rule and get
$$\epsilon_{ijk} \epsilon_{klm} B_l x_m \frac{\partial r^{-3}}{\partial x_j} + r^{-3} \delta_{jm}$$
I just don't see how to use $$\frac{\partial r}{\partial x_j}$$ maybe i'm being a little stupid.

Last edited by a moderator: Mar 13, 2014
4. Mar 13, 2014

### pasmith

You are missing some brackets: you should have
$$\epsilon_{ijk} \epsilon_{klm} B_l \left(x_m \frac{\partial r^{-3}}{\partial x_j} + r^{-3} \delta_{jm}\right)$$

$$\frac{\partial}{\partial x_j} \left(\frac{1}{r^3}\right) = \frac{d (r^{-3})}{dr} \frac{\partial r}{\partial x_j}$$

5. Mar 13, 2014

### Matt atkinson

Thankyou so much, ive been trying to do this for a while now, can't believe it was chain rule i forgot

6. Mar 13, 2014

### Matt atkinson

Okay, im sorry i got stuck again, so i did the math and substitute the two epsilons for deltas uisng the identities i gave before, but i cant seem to cancel;
$$\delta_{il}\delta_{jj} B_jr^{-3} -\delta_{ij}\delta_{jl} B_jr^{-3} -3\delta_{il}\delta_{jm}B_j x_m x_j r^{-5}+3\delta_{im}\delta_{jl}B_j x_m x_j r^{-5}$$
to the show that answer im assuming the first and third term are zero? but im not sure why, sorry about all the questions this suffix notation is very new to me.

Last edited: Mar 13, 2014
7. Mar 13, 2014

### pasmith

The suffix on $B$ should be an $l$; try again with
$$\delta_{il}\delta_{jj} B_l r^{-3} -\delta_{ij}\delta_{jl} B_l r^{-3} -3\delta_{il}\delta_{jm}B_l x_m x_j r^{-5}+3\delta_{im}\delta_{jl}B_l x_m x_j r^{-5}$$

8. Mar 14, 2014

### Matt atkinson

okay, I did that and got;
$$3B_i r^{-3} - B_i r^{-3} -3B_i x_j x_j r^{-5}+3B_j x_i x_j r^{-5}$$
I just dont see how two of the terms cancel, not sure what im missing

Last edited: Mar 14, 2014
9. Mar 14, 2014

### pasmith

$x_jx_j = r^2$.

10. Mar 14, 2014

### Matt atkinson

Wow thanks I feel kinda silly now.
Thanks so much for your help.