# Homework Help: Proving vector calculus identities using summation notation

1. Mar 15, 2012

### lostminty

1. The problem statement, all variables and given/known data

$\frac{∂x_{i}}{∂x_{j}}$ = δ$_{ij}$

2. Relevant equations

$\vec{r}$ = x$_{i}$e$_{i}$

3. The attempt at a solution
$\frac{∂x_{i}}{∂x_{j}}$ = 1 iff i=j

δ$_{ij}$ = 1 iff i=j

therefore

$\frac{∂x_{i}}{∂x_{j}}$ = δ$_{ij}$

1. The problem statement, all variables and given/known data

r$^{2}$ = x$_{k}$x$_{k}$

2. Relevant equations

$\vec{r}$ = x$_{k}$e$_{k}$
$\vec{r}$ = x$_{j}$e$_{j}$

3. The attempt at a solution

r$^{2}$ = x$_{k}$e$_{k}$$\bullet$x$_{j}$e$_{j}$

= e$_{k}$e$_{j}$x$_{k}$x$_{j}$

e$_{k}$e$_{j}$ = δ$_{jk}$ = 1 iff j=k

r$^{2}$ = x$_{k}$x$_{k}$ iff j=k

1. The problem statement, all variables and given/known data

($\nabla$r$^{2}$)$_{j}$= $\frac{∂}{∂x_{j}}$($x_{l}$$x_{l}$)= 2x$_{j}$

2. Relevant equations

r$^{2}$ = $x_{k}$$x_{k}$

3. The attempt at a solution

pretty confused by now, so far I've guessed my way through.

is the j index communative? if so where did the l index come from. there is only 2 l's so maybe they cancel and you get x$^{2}$ which differentiation gives 2x. and it inherits the i index from the d/dxi.

2. Mar 15, 2012

### tiny-tim

hi lostminty!

(try using the X2 button just above the Reply box )
fine
no, the whole point of knowing that ∂xi/∂xj = ejek = δij

is that this notation enables you to avoid using those "iffs" …

ejekxjxk

= δjkxjxk

= xkxk

(and try the third one again)

3. Mar 15, 2012

### lostminty

Ok, sounds good.

so

3. Attempt at solving

($\nabla$r$^{2}$)$_{j}=$
$\frac{∂r^{2}}{∂x_{j}}=$

$\frac{∂}{∂x_{j}}(δ_{lm}x_{l}x_{m})=$

$\frac{∂}{∂x_{j}}(x_{l}x_{l}) =$

$\frac{∂}{∂x_{j}}(x^{2}) =$

$2x_{j}$

4. Mar 15, 2012

### tiny-tim

hi lostminty!
ok so far

(though you could miss out the line with δ, it's not necessary)
no, you need to turn $\frac{∂}{∂x_{j}}(x_{l})$ into a δ

5. Mar 15, 2012

### HallsofIvy

What does it equal if $i\ne j$? Don't you think you should say that?

6. Mar 15, 2012

### lostminty

hmmm

so

$\frac{∂}{∂x}(x_{l}x_{l})_{j} =$

$\frac{∂}{∂x}(x_{l}^{2})_{j} =$

$\frac{∂}{∂x}x^{2}_{l}δ_{jl} =$

if l=j $δ_{jl}=1$ else $δ_{jl}=0$

$\frac{∂}{∂x}x^{2}_{j} =$

$2x_{j}$

7. Mar 15, 2012

### tiny-tim

but that doesn't mean anything!

8. Mar 15, 2012

### lostminty

Still struggling with the concept. So I can use a kronecker delta to have the condition of index l being j...

$\frac{∂}{∂x}(δ_{jl}x_{l}x_{l})_{j} =$

( $δ_{jl}=1$ if l=j else 0)

$\frac{∂}{∂x}x_{j}x_{j} =$

$2x_{j}$

9. Mar 16, 2012

### tiny-tim

hi lostminty!

(just got up :zzz:)

you need a push-start

∂/∂xj (xlxl)

= (∂/∂xj xl) xl + xl (∂/∂xj xl)

= 2(∂/∂xj xl) xl

carry on from there

10. Mar 16, 2012

### lostminty

That makes sense! So you do it that way because you can't have x2 instead you do product rule

= 2(∂ xl/∂xj) xl

= δjlxl

if l=j

= 2xj

11. Mar 16, 2012

### lostminty

well I'll assume thats close to right and move onto the next problem which seems to make sense

1. The problem statement, all variables and given/known data

$\nabla\cdot \vec{r}=$δii=3

2. Relevant equations

∂xi/∂xjij

3. The attempt at a solution

$\nabla\cdot \vec{r}=\nabla_{i}r_{i}$

= ∂/∂xiri

= ∂xi/∂xi = δii

=$\sum$ between i=1 and 3 of δii
=$\sum$ between i=1 and 3 of 1
= 1+ 1 + 1
=3

12. Mar 17, 2012

### tiny-tim

hi lostminty!
yes
that "if" line is unnecessary and wrong

stop putting "ifs" into your proofs! (what is it with you and "if"? do you keep saying "if" in ordinary speech?)

the δ takes care of that!

(and you missed a "2" )
yes that's fine

(though i'd be inclined to shorten the ending to just δii = 3 or δii = trace(δ) = 3)

13. Mar 20, 2012

### lostminty

This is another problem I'm a bit stuck on with similar content

1. The problem statement, all variables and given/known data

F is a constant vector field. hence $\nabla\cdot$ F = 0

this means there is a vector potential F = $\nabla\times$ A

also $\nabla\times$ F = 0

this means there is a scalar potential F = $\nabla$ ∅

verify

∅ = F $\cdot$ r

and

A = 1/2F x r

2. Relevant equations

3. The attempt at a solution

∅ = F $\cdot$ r

= Firi

F = $\nabla$ ∅

= $\nabla$ (Firi)

= ∂/∂xk (Firi)

= (∂Fi/∂xk)ri + (∂ri/∂xk)Fi

since $\nabla\cdot$ F = 0

(∂Fi/∂xk)ri = 0

and

∂xi/∂xj = δij = 1 when j=i

(∂ri/∂xk)Fi = δikFiei = Fiei = F

Last edited: Mar 20, 2012
14. Mar 21, 2012

### lostminty

F = $\nabla\times$ A = $\nabla\times$1/2F x r

1/2F x r = 1/2εijkFjrk

F= $\nabla\times$ 1/2εijkFjrk

= εimn∂/∂xi(1/2εijkFjrk)m

=∂/∂ximjδnknjδmk)(1/2Fjrk)m

=(1/2)∂/∂xi(Fmrn - Fnrm)m

=?

15. Mar 21, 2012

### tiny-tim

hi lostminty!
no, leave out "F = " … that's what you're trying to prove!

no, you need to write either

$(\nabla$ (Firi))_k

= ∂/∂xk (Firi)​

or

$\nabla$ (Firi)

= ∂/∂xk (Firi) e_k​
fine
no, that doesn't follow at all, does it?

((∂Fi/∂xk) has nothing to do with $\nabla\cdot$ F)

you need the stronger condition, that F is constant

you're doing it again!!

using "when" is the same as using "if"!!

just write "∂xi/∂xj = δij"
where did those e's suddenly come from?

if you were going to use them, they should have been there from the beginning

(i'll look at the other one later)