# Proving vector calculus identities using summation notation

• lostminty
In summary, we are verifying that a vector field F has a scalar potential and a vector potential, and we do so by showing that the scalar potential is equal to F \cdot r and the vector potential is equal to 1/2F x r. We use the fact that \nabla\cdot F = 0 to simplify our calculations and show that the scalar potential and vector potential are indeed valid.
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.

hi lostminty!

(try using the X2 button just above the Reply box )
lostminty said:
$\frac{∂x_{i}}{∂x_{j}}$ = 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 ∂xi/∂xj = ejek = δij

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

ejekxjxk

= δjkxjxk

= xkxk

(and try the third one again)

Ok, sounds good.

so

3. Attempt at solving

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

$\frac{∂}{∂x_{j}}(δ_{lm}x_{l}x_{m})=$

$\frac{∂}{∂x_{j}}(x_{l}x_{l}) =$

$\frac{∂}{∂x_{j}}(x^{2}) =$

$2x_{j}$

hi lostminty!
lostminty said:
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

lostminty said:

## 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 said:
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 $δ_{jl}=0$

$\frac{∂}{∂x}x^{2}_{j} =$

$2x_{j}$

lostminty said:
$\frac{∂}{∂x}(x_{l}^{2})_{j}$

but that doesn't mean anything!

lostminty said:
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)

$\frac{∂}{∂x}x_{j}x_{j} =$

$2x_{j}$

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

tiny-tim said:
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

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

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

## Homework Statement

$\nabla\cdot \vec{r}=$δii=3

∂xi/∂xjij

## 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

hi lostminty!
lostminty said:
So you do it that way because you can't have x2 instead you do product rule

yes
= 2(∂ xl/∂xj) xl

= δjlxl

if l=j

= 2xj

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" )
lostminty said:
$\nabla\cdot \vec{r}=\nabla_{i}r_{i}$

= ∂/∂xiri

= ∂xi/∂xi = δii

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

yes that's fine

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

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

## The Attempt at a Solution

∅ = F $\cdot$ r

= Firi

F = $\nabla$ ∅

= $\nabla$ (Firi)

= ∂/∂xk (Firi)

= (∂Fi/∂xk)ri + (∂ri/∂xk)Fisince $\nabla\cdot$ F = 0

(∂Fi/∂xk)ri = 0

and

∂xi/∂xj = δij = 1 when j=i

(∂ri/∂xk)Fi = δikFiei = Fiei = F

Last edited:
lostminty said:
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

## The Attempt at a Solution

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

=?

hi lostminty!
lostminty said:

## 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

= Firi

F = $\nabla$ ∅

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

= $\nabla$ (Firi)

= ∂/∂xk (Firi)

no, you need to write either

$(\nabla$ (Firi))_k

= ∂/∂xk (Firi)​

or

$\nabla$ (Firi)

= ∂/∂xk (Firi) e_k​
= (∂Fi/∂xk)ri + (∂ri/∂xk)Fi

fine
since $\nabla\cdot$ F = 0

(∂Fi/∂xk)ri = 0

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

and

∂xi/∂xj = δij = 1 when j=i

you're doing it again!

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

just write "∂xi/∂xj = δij"
(∂ri/∂xk)Fi = δikFiei = Fiei = 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)

## 1. How do you prove vector calculus identities using summation notation?

To prove vector calculus identities using summation notation, you must first understand the fundamental properties of vectors and summation notation. Then, you can manipulate the equations and use algebraic techniques to show that both sides of the equation are equal.

## 2. What are the benefits of using summation notation in proving vector calculus identities?

Using summation notation makes it easier to express and manipulate vector equations, making the proofs more concise and efficient. It also allows for a more general approach, as summation notation can be used for any number of terms in the equation.

## 3. Can you provide an example of proving a vector calculus identity using summation notation?

Yes, for example, to prove the identity ∑(u × v) = u × ∑v, we can use the distributive property and the definition of cross product to show that both sides are equal.

## 4. Are there any special techniques or tips for proving vector calculus identities using summation notation?

One helpful tip is to keep track of the indices used in the summation notation and make sure they match on both sides of the equation. It's also important to fully understand the properties of vectors and summation notation in order to correctly manipulate the equations.

## 5. Is there a limit to the number of terms that can be included in a summation notation for proving vector calculus identities?

No, there is no limit to the number of terms that can be included in a summation notation. However, as the number of terms increases, the proof may become more complex and difficult to follow, so it's important to use good notation and explain each step clearly.

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