# Index Notation - Vector Calculus Help

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So I know vaguely how to do it for some basic examples but wanted to know why some of it worked and what are the general rules so i don't get stuck on harder or unfamiliar questions

so when i prove a x (bxc) = b(a.c)-c(a.b) i get to the answer but wanted to know/understand what things are allowed:

after the first few steps you get to

Eijk aj Eklm bl cm

on the next step can you now move the second Eklm to the front, why? because its a constant??...

and also is the following equivalent?

EijkEklm aj bl cm =EijkEklm bl cm aj = EijkEklm cm bl aj

if so why? and can you always interchange the vectors like this

so when i prove a x (bxc) = b(a.c)-c(a.b) i get to the answer but wanted to know/understand what things are allowed:

after the first few steps you get to

Eijk aj Eklm bl cm

on the next step can you now move the second Eklm to the front, why? because its a constant??...

and also is the following equivalent?

EijkEklm aj bl cm =EijkEklm bl cm aj = EijkEklm cm bl aj

if so why? and can you always interchange the vectors like this

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#3

(Original post by

So I know vaguely how to do it for some basic examples but wanted to know why some of it worked and what are the general rules so i don't get stuck on harder or unfamiliar questions

so when i prove a x (bxc) = b(a.c)-c(a.b) i get to the answer but wanted to know/understand what things are allowed:

after the first few steps you get to

Eijk aj Eklm bl cm

on the next step can you now move the second Eklm to the front, why? because its a constant??...

and also is the following equivalent?

EijkEklm aj bl cm =EijkEklm bl cm aj = EijkEklm cm bl aj

if so why? and can you always interchange the vectors like this

**Jammy4410**)So I know vaguely how to do it for some basic examples but wanted to know why some of it worked and what are the general rules so i don't get stuck on harder or unfamiliar questions

so when i prove a x (bxc) = b(a.c)-c(a.b) i get to the answer but wanted to know/understand what things are allowed:

after the first few steps you get to

Eijk aj Eklm bl cm

on the next step can you now move the second Eklm to the front, why? because its a constant??...

and also is the following equivalent?

EijkEklm aj bl cm =EijkEklm bl cm aj = EijkEklm cm bl aj

if so why? and can you always interchange the vectors like this

*components*of vectors (or other tensors) and are, as such, scalars. So yes, you can rearrange their order at will.

All the important vector-y-ness is encoded in the indices, so as long as you maintain those relationships you're fine. Remember, they show the order in which you sum over components. In an ordinary vector equation this is shown by ordering vectors and matrices in a specific way, which is where your confusion comes in.

So yes, the final is equivalent.

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(Original post by

You have to remember that these aren't vector equations. Everything in these equations are

All the important vector-y-ness is encoded in the indices, so as long as you maintain those relationships you're fine. Remember, they show the order in which you sum over components. In an ordinary vector equation this is shown by ordering vectors and matrices in a specific way, which is where your confusion comes in.

So yes, the final is equivalent.

**Rinsed**)You have to remember that these aren't vector equations. Everything in these equations are

*components*of vectors (or other tensors) and are, as such, scalars. So yes, you can rearrange their order at will.All the important vector-y-ness is encoded in the indices, so as long as you maintain those relationships you're fine. Remember, they show the order in which you sum over components. In an ordinary vector equation this is shown by ordering vectors and matrices in a specific way, which is where your confusion comes in.

So yes, the final is equivalent.

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First one is:

a x a

( a x a)i (ith component)

= Eijk aj ak

= -Eikj aj ak (Swapping order in E means It will be minus, so an extra minus sign is needed for equality)

= -Eikj ak aj. (Components of vectors, I.e scalar so can interchange)

= - (a x a)i

And since (a x a)i = -(a x a)i then a x a = 0

Is this right?

a x a

( a x a)i (ith component)

= Eijk aj ak

= -Eikj aj ak (Swapping order in E means It will be minus, so an extra minus sign is needed for equality)

= -Eikj ak aj. (Components of vectors, I.e scalar so can interchange)

= - (a x a)i

And since (a x a)i = -(a x a)i then a x a = 0

Is this right?

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I just realised the above proof must be wrong. But why?

I know this cause I could do the same thing with

a x b where a does not equal b and it would imply a x b = 0

I know this cause I could do the same thing with

a x b where a does not equal b and it would imply a x b = 0

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#7

(Original post by

First one is:

a x a

( a x a)i (ith component)

= Eijk aj ak

= -Eikj aj ak (Swapping order in E means It will be minus, so an extra minus sign is needed for equality)

= -Eikj ak aj. (Components of vectors, I.e scalar so can interchange)

= - (a x a)i

And since (a x a)i = -(a x a)i then a x a = 0

Is this right?

**Jammy4410**)First one is:

a x a

( a x a)i (ith component)

= Eijk aj ak

= -Eikj aj ak (Swapping order in E means It will be minus, so an extra minus sign is needed for equality)

= -Eikj ak aj. (Components of vectors, I.e scalar so can interchange)

= - (a x a)i

And since (a x a)i = -(a x a)i then a x a = 0

Is this right?

(Original post by

I just realised the above proof must be wrong. But why?

I know this cause I could do the same thing with

a x b where a does not equal b and it would imply a x b = 0

**Jammy4410**)I just realised the above proof must be wrong. But why?

I know this cause I could do the same thing with

a x b where a does not equal b and it would imply a x b = 0

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(Original post by

Yep, looks good to me.

Nope, it tells you that a x b= -b x a, which is fine. It's only when the vectors are the same when this swapping doesn't change anything meaning it must be zero. Remember, you swapped the order of the indices, which is equivalent to changing the order of the vector equation.

**Rinsed**)Yep, looks good to me.

Nope, it tells you that a x b= -b x a, which is fine. It's only when the vectors are the same when this swapping doesn't change anything meaning it must be zero. Remember, you swapped the order of the indices, which is equivalent to changing the order of the vector equation.

Problem 2: The divergence of the curl

(Gonna use V for the grad sign, forgotten the name atm, and u for the arbitrary vector) and di for the index notation for V

V . (V x u) (i) Ith component

= di Eijk dj uk

=Eijk dj di uk (orde don't matter)

= -Ejik dj di uk

= -dj (V x u) jth component

= - V. (V x u). Jth company

A property of a scalar is that

a = a transpose

And since this produces a scalar then we have shown

a= a transpose = -a if and only if a=0

Is the reasoning correct?

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Problem 3: involving scalar functions

An arbitrary vector crossed with the grad of a scalar function

(Gonna use a for the scalar function)

Va x u ith component

= Eijk dja uk

= Eijk dj auk

=V x au

I know this is wrong but why? And why cant you move scalars about like this

Is it because you have to treat 'dj a ' together as one thing and cant be separated? An that the way you write it out makes it look like a constant that can be moved around

An arbitrary vector crossed with the grad of a scalar function

(Gonna use a for the scalar function)

Va x u ith component

= Eijk dja uk

= Eijk dj auk

=V x au

I know this is wrong but why? And why cant you move scalars about like this

Is it because you have to treat 'dj a ' together as one thing and cant be separated? An that the way you write it out makes it look like a constant that can be moved around

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#10

(Original post by

Oh yeah, I'm being silly.

Problem 2: The divergence of the curl

(Gonna use V for the grad sign, forgotten the name atm, and u for the arbitrary vector) and di for the index notation for V

V . (V x u) (i) Ith component

= di Eijk dj uk

=Eijk dj di uk (orde don't matter)

= -Ejik dj di uk

= -dj (V x u) jth component

= - V. (V x u). Jth company

A property of a scalar is that

a = a transpose

And since this produces a scalar then we have shown

a= a transpose = -a if and only if a=0

Is the reasoning correct?

**Jammy4410**)Oh yeah, I'm being silly.

Problem 2: The divergence of the curl

(Gonna use V for the grad sign, forgotten the name atm, and u for the arbitrary vector) and di for the index notation for V

V . (V x u) (i) Ith component

= di Eijk dj uk

=Eijk dj di uk (orde don't matter)

= -Ejik dj di uk

= -dj (V x u) jth component

= - V. (V x u). Jth company

A property of a scalar is that

a = a transpose

And since this produces a scalar then we have shown

a= a transpose = -a if and only if a=0

Is the reasoning correct?

(Original post by

Problem 3: involving scalar functions

An arbitrary vector crossed with the grad of a scalar function

(Gonna use a for the scalar function)

Va x u ith component

= Eijk dja uk

= Eijk dj auk

=V x au

I know this is wrong but why? And why cant you move scalars about like this

Is it because you have to treat 'dj a ' together as one thing and cant be separated? An that the way you write it out makes it look like a constant that can be moved around

**Jammy4410**)Problem 3: involving scalar functions

An arbitrary vector crossed with the grad of a scalar function

(Gonna use a for the scalar function)

Va x u ith component

= Eijk dja uk

= Eijk dj auk

=V x au

I know this is wrong but why? And why cant you move scalars about like this

Is it because you have to treat 'dj a ' together as one thing and cant be separated? An that the way you write it out makes it look like a constant that can be moved around

Also you probably want to write it like

To avoid confusion, rather than putting the index on the del.

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(Original post by

Tbh I'm unsure why you're transposing things, is that in the question? But your reasoning looks correct, if you're trying to show the divergence of a curl is 0.

Yea so, grad a is a vector, of which you are looking at the components. You can't just split up the del operator and the scalar function. The scalar function itself does not exist in your equation, only the jth component of its gradient.

Also you probably want to write it like

To avoid confusion, rather than putting the index on the del.

**Rinsed**)Tbh I'm unsure why you're transposing things, is that in the question? But your reasoning looks correct, if you're trying to show the divergence of a curl is 0.

Yea so, grad a is a vector, of which you are looking at the components. You can't just split up the del operator and the scalar function. The scalar function itself does not exist in your equation, only the jth component of its gradient.

Also you probably want to write it like

To avoid confusion, rather than putting the index on the del.

i mentioned the transpose cause in my working i showed that

the ith component of V.(V x u) was equal to the jth component of - V.(V x u)

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problem 4:

how do i show that

V. (aU) = (Va).U only if a is a constant vector

where

V is the del operator

a is a scalar field

U is an arbitrary vector field

how do i show that

V. (aU) = (Va).U only if a is a constant vector

where

V is the del operator

a is a scalar field

U is an arbitrary vector field

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#13

(Original post by

i mentioned the transpose cause in my working i showed that

the ith component of V.(V x u) was equal to the jth component of - V.(V x u)

**Jammy4410**)i mentioned the transpose cause in my working i showed that

the ith component of V.(V x u) was equal to the jth component of - V.(V x u)

You showed that

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#14

(Original post by

problem 4:

how do i show that

V. (aU) = (Va).U only if a is a constant vector

where

V is the del operator

a is a scalar field

U is an arbitrary vector field

**Jammy4410**)problem 4:

how do i show that

V. (aU) = (Va).U only if a is a constant vector

where

V is the del operator

a is a scalar field

U is an arbitrary vector field

You have, in index notation,

(If you're not familiar with higher and lower indices, just ignore it)

The del is a differential operator, just use the product rule and see what happens.

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(Original post by

So I'll set it up for you.

You have, in index notation,

(If you're not familiar with higher and lower indices, just ignore it)

The del is a differential operator, just use the product rule and see what happens.

**Rinsed**)So I'll set it up for you.

You have, in index notation,

(If you're not familiar with higher and lower indices, just ignore it)

The del is a differential operator, just use the product rule and see what happens.

i got:

V. (aU) ith

= di (aU)i

= di (aUi)

=(dia)Ui + a(diUi)

= (V.a) + a(V.U) ith component

is that right?

and if a is constant then the first term is zero.

so we get

V.(aU) = a(V.U) and then I'm guessing you can re write the RHS as (Va).U why??

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problem 5 (based on problem 3, similar to problem 4):

is there any similar relationship for the vector product i.e.

is there a relationship between:

Va x U and V x aU, and maybe even equivalent if a is a constant vector??

is there any similar relationship for the vector product i.e.

is there a relationship between:

Va x U and V x aU, and maybe even equivalent if a is a constant vector??

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#17

(Original post by

i thought you had to use the product rule but was having trouble knowing when to use it and when you can just interchange, i guess this is 'di' of a product unlike the others:

i got:

V. (aU) ith

= di (aU)i

= di (aUi)

=(dia)Ui + a(diUi)

= (V.a) + a(V.U) ith component

is that right?

and if a is constant then the first term is zero.

so we get

V.(aU) = a(V.U) and then I'm guessing you can re write the RHS as (Va).U why??

**Jammy4410**)i thought you had to use the product rule but was having trouble knowing when to use it and when you can just interchange, i guess this is 'di' of a product unlike the others:

i got:

V. (aU) ith

= di (aU)i

= di (aUi)

=(dia)Ui + a(diUi)

= (V.a) + a(V.U) ith component

is that right?

and if a is constant then the first term is zero.

so we get

V.(aU) = a(V.U) and then I'm guessing you can re write the RHS as (Va).U why??

You actually said a was a constant vector. Because a is a scalar, I assumed you meant

**U**was a constant vector. Worth saying, that gives the right answer. Are you sure the question said a was constant?

Plus, if a were a constant, the fact that the answer includes a differential of a would seem strange.

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(Original post by

Not quite.

You actually said a was a constant vector. Because a is a scalar, I assumed you meant

Plus, if a were a constant, the fact that the answer includes a differential of a would seem strange.

**Rinsed**)Not quite.

You actually said a was a constant vector. Because a is a scalar, I assumed you meant

**U**was a constant vector. Worth saying, that gives the right answer. Are you sure the question said a was constant?Plus, if a were a constant, the fact that the answer includes a differential of a would seem strange.

in what way was it right?

and what way was it wrong? (based on what each letter is defined as)

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i think i get it, if a was a scalar the first term would be automatically zero but with 'a' as a vector the first term will appear but be zero if a is a constant vector

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i just realised i did mean that 'a' was a scalar field and U was a constant vector sorry about that

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