Question

Inner product under change of coordinates.

Answer 1

In my notes, it has been noted that the inner product is \(V^TGW\) where V and W are two vectors and G is the metric. The question is how do I calculate the metric G.

Answer 2

@Kainui @ikram002p

Answer 3

I am not sure whether I am suppose to know this in my course but the metric of cylindrical coordinates is provided in my notes. The metric is
\[
\begin{pmatrix}
1&0&0\\
0&r^2&0\\
0&0&1
\end{pmatrix}
\]
which does not make any sense to me. What am I supposed to do with r in the matrix?

Answer 4

G is the dot product of all your basis vectors at that point. So:

\[G_{ij}(Z) = \vec Z_i(Z) \cdot \vec Z_j(Z)\]

and to get your basis vectors you can calculate these from the derivative of your position vectors with resepct to the coordinates:

\[\vec Z_i = \frac{\partial \vec R (Z)}{\partial Z^i}\]

Answer 5

So like that's probably not the explanation you want sorry I think I over did it :X

Answer 6

Is R a path of some sort?

Answer 7

V and W are the components of your vectors, but they are not truly your vectors in space since they depend on your choice of basis vectors. actually wait before I go on, Are you familiar with Einstein summation notation?

Answer 8

I know what it is but I always get horribly confused with the summation.

Answer 9

Put it more precisely, I understand what Einstein summation notation is and I know how to convert it to explicit summation form but I do not have an intuitive grasp of it.

Answer 10

Ok since we're working in 3 dimensions I'll write out both side by side. (does your class use contravariant and covariant indices?)

So here's a vector \(\vec V\) written out in some chosen basis:
\[ \vec V = \sum_{i=1}^3 V^i \vec Z_i =V^i \vec Z_i = V^1 \vec Z_1 + V^2 \vec Z_2 + V^3 \vec Z_3 \]

If we want to find the inner product between two vectors we would write it like this:

\[\vec V \cdot \vec W = \left( \sum_{i=1}^3 V^i \vec Z_i \right) \cdot \left( \sum_{j=1}^3 W^j \vec Z_j \right) \]

Now we can distribute these sums together in the same way we would \((a+b)(x+y)=ax+ay+bx+by\) so we write it as:

\[\vec V \cdot \vec W =\sum_{i=1}^3 \sum_{j=1}^3 V^i \vec Z_i \cdot W^j \vec Z_j \]

Now we pull the constants out of the dot product just to make our lives easier, he dot product is really only between our basis vectors:


\[\vec V \cdot \vec W =\sum_{i=1}^3 \sum_{j=1}^3 V^i W^j \vec Z_i \cdot \vec Z_j \]

And we rename specifically this object of 9 dot products the metric tensor, but really it's just a square symmtric matrix.

\[\vec Z_i \cdot \vec Z_j = G_{ij}\]

plugged in:
\[\vec V \cdot \vec W =\sum_{i=1}^3 \sum_{j=1}^3 V^i W^j G_{ij}\]

Everything on the right hand side is now your expression, since it represents \(V^\top G W\). You can remove all the summation signs and that will be what the tensor notation wil be in general. I think it might help to write out the entries of \(G_{ij}\) explicitly. I hope this is kinda what you were looking for however I have a feeling there's more you'd like to know.

I'm trying not to make it confusing but tensors just are so sorry about that :X

Answer 11

I should probably add that the tensor notation is component-wise which basically means matrix multiplication is commutative which is cool.

\[V^i W^j G_{ij} = W^j V^i G_{ij}\]

Also, the transpose operation sort of went away, although usually order of indices matters in this case it doesn't since the metric tensor is symmetric, so it doesn't matter which way you write it. Here's a quick proof using the fact that the dot product is commutative:

\[G_{ij} = \vec Z_i \cdot \vec Z_j = \vec Z_j \cdot \vec Z_i = G_{ji}\]

It might be a good idea to: do a lot of writing to get a hang of this stuff. I'm not sure, typing is cheap and quick for me but I feel like I might be actually missing your question of wanting to know about this metric tensor so I'm just throwing out stuff take it or leave it don't worry abuot wasting my time or anything lol

Answer 12

Yes I am absolutely sure that this is not in my course now. Tensor is definitely not in this module since there are introductions to matrices in my notes. I strongly regret not spending time on reading some book on tensor calculus before entering uni...

Answer 13

R is the length of your radius as you move out, it's the length of your basis vector pointing in the angular direction. And this I know cause of the dot product of all this... So perhaps I took a very roundabout way of saying this

Answer 14

It is not going to stop me thoguh lol.

Answer 15

What are the basis vectors of spherical coordinate system? The worse part is I don't think the basis vectors are "position independent" and there is no way I could visualise the basis vectors.

Answer 16

I happen to know of a good lecture series that might help, he also has linear algebra stuff and I think watching him on 2x speed (with some pausing when necessary) is good haha https://www.youtube.com/playlist?list=PLlXfTHzgMRULkodlIEqfgTS-H1AY_bNtq

It's really strange, but your basis vectors depend on your coordinates in space. This seems kind of odd and was hard to wrap my head around but the reason this happens is because it allows us to create a basis set of vectors intrinsically, in other words we don't have to leave our space to create them. It's a geometric thing.

I'll draw some pictures for cylindrical coordinates and spherical coordinates, actually I think I'll find the exact lectures where he draws the coordinates

Answer 17

Ok so in polar coordinates (really just the same thing as cylindrical coordinates)

I have drawn the origin, a ring of radius 1 and a ring of radius 2 to represent the coordinates and I have these two local bases at two different points. They depend completely on the coordinates.

In spherical coordinates I found a picture where he's drawn them, although technically he's doing something different than us just peek for a second to see how he's moving around with spherical coordinates:

https://www.youtube.com/watch?v=l7coUMHHpQE&index=26&list=PLlXfTHzgMRULkodlIEqfgTS-H1AY_bNtq