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 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
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.
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.
Ok thanks thats cleared some things up but I've got 5-6 problems I'm still confused by. So il do them one by one
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
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.
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.
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
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
(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?
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.
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
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
(∇a)i
To avoid confusion, rather than putting the index on the del.
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
(∇a)i
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)
(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 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??
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??
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.
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.
im a bit confused now..
in what way was it right? and what way was it wrong? (based on what each letter is defined as)
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