Proving vector calculus identities using summation notation

[lostminty]

Homework Statement

$\frac{∂x_{i}}{∂x_{j}}$ = δ$_{ij}$

Homework Equations

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

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$_{ij}$

Homework Statement

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

Homework Equations

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

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

Homework Statement

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

Homework Equations

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

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.

 

[tiny-tim]

hi lostminty!

(try using the X₂ button just above the Reply box )

$\frac{∂x_{i}}{∂x_{j}}$ = 1 iff i=j

δ$_{ij}$ = 1 iff i=j

therefore

$\frac{∂x_{i}}{∂x_{j}}$ = δ$_{ij}$

fine

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

no, the whole point of knowing that ∂x_i/∂x_j = e_je_k = δ_ij

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

e_je_kx_jx_k

= δ_jkx_jx_k

= x_kx_k ​

(and try the third one again)

 

[lostminty]

Ok, sounds good.

so

3. Attempt at solving

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

$\frac{∂}{∂x_{j}}(δ_{lm}x_{l}x_{m})=$

$\frac{∂}{∂x_{j}}(x_{l}x_{l}) =$

$\frac{∂}{∂x_{j}}(x^{2}) =$

$2x_{j}$

 

[tiny-tim]

hi lostminty!

so

3. Attempt at solving

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

$\frac{∂}{∂x_{j}}(δ_{lm}x_{l}x_{m})=$

$\frac{∂}{∂x_{j}}(x_{l}x_{l}) =$

ok so far

(though you could miss out the line with δ, it's not necessary)

$\frac{∂}{∂x_{j}}(x^{2}) =$

no, you need to turn $\frac{∂}{∂x_{j}}(x_{l})$ into a δ

 

[HallsofIvy]

Homework Statement

$\frac{∂x_{i}}{∂x_{j}}$ = δ$_{ij}$

Homework Equations

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

The Attempt at a Solution

$\frac{∂x_{i}}{∂x_{j}}$ = 1 iff i=j

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

δ$_{ij}$ = 1 iff i=j

therefore

$\frac{∂x_{i}}{∂x_{j}}$ = δ$_{ij}$

Homework Statement

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

Homework Equations

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

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

Homework Statement

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

Homework Equations

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

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.

 

[lostminty]

Ok, sounds good.

so

3. Attempt at solving

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

$\frac{∂}{∂x_{j}}(δ_{lm}x_{l}x_{m})=$

$\frac{∂}{∂x_{j}}(x_{l}x_{l}) =$

$\frac{∂}{∂x_{j}}(x^{2}) =$

$2x_{j}$

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 $δ_{jl}=0$

$\frac{∂}{∂x}x^{2}_{j} =$

$2x_{j}$

 

[tiny-tim]

$\frac{∂}{∂x}(x_{l}^{2})_{j}$

but that doesn't mean anything!

 

[lostminty]

hmmm



$\frac{∂}{∂x}(x_{l}x_{l})_{j} =$

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)

$\frac{∂}{∂x}x_{j}x_{j} =$

$2x_{j}$

 

[tiny-tim]

hi lostminty!

(just got up :zzz:)

you need a push-start …

∂/∂x_j (x_lx_l)

= (∂/∂x_j x_l) x_l + x_l (∂/∂x_j x_l)

= 2(∂/∂x_j x_l) x_l …

carry on from there ​

 

[lostminty]

hi lostminty!

(just got up :zzz:)

you need a push-start …

∂/∂x_j (x_lx_l)

= (∂/∂x_j x_l) x_l + x_l (∂/∂x_j x_l)

= 2(∂/∂x_j x_l) x_l …

carry on from there ​

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

= 2(∂ x_l/∂x_j) x_l

= δ_jlx_l

if l=j

= 2x_j

 

[lostminty]

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

Homework Statement

$\nabla\cdot \vec{r}=$δ_ii=3

Homework Equations

∂x_i/∂x_j=δ_ij

The Attempt at a Solution

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

= ∂/∂x_ir_i

= ∂x_i/∂x_i = δ_ii

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

 

[tiny-tim]

hi lostminty!

So you do it that way because you can't have x² instead you do product rule

yes

= 2(∂ x_l/∂x_j) x_l

= δ_jlx_l

if l=j

= 2x_j

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" )

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

= ∂/∂x_ir_i

= ∂x_i/∂x_i = δ_ii

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

yes that's fine

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

 

[lostminty]

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

Homework Statement

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

Homework Equations

The Attempt at a Solution

∅ = F $\cdot$ r

= F_ir_i

F = $\nabla$ ∅

= $\nabla$ (F_ir_i)

= ∂/∂x_k (F_ir_i)

= (∂F_i/∂x_k)r_i + (∂r_i/∂x_k)F_isince $\nabla\cdot$ F = 0

(∂F_i/∂x_k)r_i = 0

and

∂x_i/∂x_j = δ_ij = 1 when j=i

(∂r_i/∂x_k)F_i = δ_ikF_ie_i = F_ie_i = F

 

[lostminty]

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

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

1/2F x r = 1/2ε_ijkF_jr_k

F= $\nabla\times$ 1/2ε_ijkF_jr_k

= ε_imn∂/∂x_i(1/2ε_ijkF_jr_k)_m

=∂/∂x_i(δ_mjδ_nk-δ_njδ_mk)(1/2F_jr_k)_m

=(1/2)∂/∂x_i(F_mr_n - F_nr_m)_m

=?

 

[tiny-tim]

hi lostminty!

Homework Statement

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

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

verify

∅ = F $\cdot$ r

…


∅ = F $\cdot$ r

= F_ir_i

F = $\nabla$ ∅

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

= $\nabla$ (F_ir_i)

= ∂/∂x_k (F_ir_i)

no, you need to write either

$(\nabla$ (F_ir_i))_k

= ∂/∂x_k (F_ir_i)​

or

$\nabla$ (F_ir_i)

= ∂/∂x_k (F_ir_i) e_k​

= (∂F_i/∂x_k)r_i + (∂r_i/∂x_k)F_i

fine

since $\nabla\cdot$ F = 0

(∂F_i/∂x_k)r_i = 0

no, that doesn't follow at all, does it?

((∂F_i/∂x_k) has nothing to do with $\nabla\cdot$ F)

you need the stronger condition, that F is constant

and

∂x_i/∂x_j = δ_ij = 1 when j=i

you're doing it again!

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

just write "∂x_i/∂x_j = δ_ij"

(∂r_i/∂x_k)F_i = δ_ikF_ie_i = F_ie_i = F

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)