Suffix Notation Help: Nabla Vector Calculation

[Matt atkinson]

Homework Statement

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}

Homework Equations

above

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).

 

[pasmith]

Homework Statement

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}

Homework Equations

above

The Attempt at a Solution

\nabla \times \vec{p} = \epsilon_{ijk} \epsilon_{klm} \frac{d}{dx_j} B x_m |r|^{-3}

You're missing an l suffix on the B and the derivative is partial:
<br /> (\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
<br /> \frac{\partial r}{\partial x_j} = \frac{x_j}{r} and <br /> \frac{\partial x_m}{\partial x_j} = \delta_{jm}<br />

 

[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.

 

[pasmith]

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}

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

I just don't see how to use \frac{\partial r}{\partial x_j} maybe I'm being a little stupid.

<br /> \frac{\partial}{\partial x_j} \left(\frac{1}{r^3}\right) = \frac{d (r^{-3})}{dr} \frac{\partial r}{\partial x_j}<br />

 

[Matt atkinson]

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

 

[Matt atkinson]

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

<br /> \frac{\partial}{\partial x_j} \left(\frac{1}{r^3}\right) = \frac{d (r^{-3})}{dr} \frac{\partial r}{\partial x_j}<br />

Okay, I am 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 can't 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 I am assuming the first and third term are zero? but I am not sure why, sorry about all the questions this suffix notation is very new to me.

 

[pasmith]

Okay, I am sorry i got stuck again, so i did the math and substituent the two epsilons for deltas uisng the identities i gave before, but i can't 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}

The suffix on B should be an l; try again with
<br /> \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}

 

[Matt atkinson]

The suffix on B should be an l; try again with
<br /> \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}

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 don't see how two of the terms cancel, not sure what I am missing

 

[pasmith]

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 don't see how two of the terms cancel, not sure what I am missing

x_jx_j = r^2.

 

[Matt atkinson]

x_jx_j = r^2.

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