Magnetic field from vector potential function using tensor notation

[thatguy14]

Homework Statement

We will see (in Chap. 5) that the magnetic field can be derived from a vector potential function as
follows:
B = ∇×A
Show that, in the special case of a uniform magnetic field B_{0} , one possible
vector potential function is A = \frac{1}{2}B_{0}×r

MUST USE TENSOR NOTATIONm also B0 is constant (uniform magnetic field)

Homework Equations

ε_{ijk}ε_{klm} = δ_{il}δ_{jm} - δ_{im}δ_{jl}

The Attempt at a Solution

I have tried a bunch of different things but I am missing something near the end.
Here is what I have

B = (∇×A)_{i}
B = ε_{ijk}∂_{j}A_{k}
B = ε_{ijk}∂_{j} (\frac{1}{2}ε_{klm}B_{0l}r_{m})
B = \frac{1}{2}ε_{ijk}ε_{klm}∂_{j}B_{0l}r_{m})

where ε_{ijk}ε_{klm} = δ_{il}δ_{jm} - δ_{im}δ_{jl}

So B = \frac{1}{2}[δ_{il}δ_{jm} - δ_{im}δ_{jl}]∂_{j}B_{0l}r_{m}

Changing indicies gives (noting that the derivative of constant = 0 and using the product rule)

B = \frac{1}{2}[B_{0i}∂_{m}r_{m} - B_{0l}∂_{l}r_{i}]

And that's where I am stuck. What comes next? I am assuming that with the last term there, l and i have to be equal (because if they aren't then it equals 0) and I think I have to introduct the krockner delta somewhere but I am unsure. Any help would be greatly appreciated.

 

[TSny]

In the expression $\partial_m r_m$ you are summing over $m$. Consider one term of that sum, say $\partial_2 r_2$. Can you see what that is equal to?

You are right that you can express $\partial_l r_i$ in terms of a Kronecker delta.

 

[thatguy14]

Would ∂_{m}r_{m} just be equal to the divergence of r i.e. ∇°r?

and for expressing it as a krockner delta is it just ∂_{l}r_{i} = δ_{li}∂r?

 

[TSny]

Would ∂_{m}r_{m} just be equal to the divergence of r i.e. ∇°r?

Yes. You should be able to reduce ∂_{m}r_{m} to a number.

and for expressing it as a krockner delta is it just ∂_{l}r_{i} = δ_{li}∂r?

Not quite. It's simpler than you wrote.

 

[thatguy14]

For the first thing, its just 1 right? And that's because the derivative of r with respect to r is just 1?

For the second... l has to equal i is what I am going to go with so does that mean that that portion is equal to 0? Or is that incorrect?

 

[tiny-tim]

For the first thing, its just 1 right? And that's because the derivative of r with respect to r is just 1?

no, you can't differentiate wrt a vector

(and if you mean |r|, that isn't in your equation)

instead of ∂/∂r, try ∂/∂x or ∂/∂y or ∂/∂z ​

 

[TSny]

What does the symbol $\partial_m$ stand for? It is a derivative with respect to something. What is the "something"?

 

[thatguy14]

oh... Okay then let me try again:

since r is x i + y j + z k when we take the divergence we get 3?

 

[TSny]

since r is x i + y j + z k when we take the divergence we get 3?

Yes. Good.

 

[thatguy14]

Okay awesome (rather silly of me though)! Now for the second portion I am still lost. I am going to take a guess that it just equals 1 but I can't justify it to myself. Is that correct? and if it is why is it correct?

 

[tiny-tim]

if you get confused, remember you can always write ∂_mr_n as ∂x_n/∂x_m …

which equals … ? ​

 

[thatguy14]

well it would equal 1 only if the indices are equal so then is it equal to 3 also? (because of the y and z components)?

edit: looking only at the x component the indicies are summed to 3 so for the x component is it equal to 3?

 

[tiny-tim]

well it would equal 1 only if the indices are equal …

hold it there!

so (for fixed m and n) ∂x_n/∂x_m = … ?

 

[thatguy14]

Right we don't have a repeated index

that would then = 0 because you are looking at different components correct?

 

[tiny-tim]

∂x_n/∂x_m = δ_mn

 

[thatguy14]

hm. I can see how that works now but do you think you can explain that a little further if possible?

Also that means then I have this left with the appropriate substituitons:
\frac{1}{2}[3B_{0i} - B_{0l}δ_{li}]

so then δ_{li}] = 1 only if l = i or 0 if l ≠ i

changing indicies

\frac{1}{2}[3B_{0i} - B_{0i}δ(1)]

so then B = B_{0}

and that's it correct?

Thank you by the way guys for your help. I am very glad you didn't just give me the answers and helped me along even if it was a little silly in hindsight!

 

[tiny-tim]

… B_{0l}δ_{li}
…

changing indicies

…B_{0i}δ(1)]

once you've used the δ to change the indices, the δ disappears …

B_{0l}δ_{li} = B_0i ​

(oh, and i forgot to mention: your "B =" on the LHS of all your equations should have been "B_i = " )

 

[thatguy14]

Oh that was a mistake I knew that but thank you. You have been a wonderful help!