Metrics and coordinate transforms

  • Thread starter chroot
  • Start date
  • Tags
    Coordinate
In summary: However, it seems that the original question can be summarized as follows: Given the metric for Euclidean space in spherical coordinates and the transformation matrix from spherical to cylindrical coordinates, how can we deduce the metric for Euclidean space in cylindrical coordinates? In summary, the question asks for the deduction of the metric for Euclidean space in cylindrical coordinates using the metric for Euclidean space in spherical coordinates and the coordinate transformation matrix from spherical to cylindrical coordinates. The approach involves using the transformation rule for tensors and matrix multiplication, but it is important to note that tensors should not be thought of as matrices.
  • #1
chroot
Staff Emeritus
Science Advisor
Gold Member
10,295
41
Hi all, I'm trying to solve Exercise 1.4.3 in Foster & Nightingale's "A Short Course in General Relativity."

The question essentially provides the metric for Euclidean space in spherical coordinates and the matrix representing the coordinate transformation from spherical to cylindrical coordinates. The question prompts the deduction of the metric for Euclidean space in cylindrical coordinates using this information and the rule

[tex]g_{i^\prime j^\prime}
= U^k_{i^\prime}
U^l_{j^\prime}
g_{kl}
[/tex]

where primed coordinates are cylindrical, and unprimed coordinates are spherical.

The matrices involved are

1) the metric of Euclidean space in spherical coordinates

[tex][g_{ij}] =
\left[
\begin{array}{ccc}
1 & 0 & 0\\
0 & r^2 & 0\\
0 & 0 & r^2 \sin^2 \theta
\end{array}
\right]
[/tex]

2) the coordinate transformation matrix between spherical and cylindrical coordinates:

[tex][U^{i^\prime}_j] =
\left[
\begin{array}{ccc}
\sin \theta & r \cos \theta & 0\\
0 & 0 & 1\\
\cos \theta & -r \sin \theta & 0
\end{array}
\right]
[/tex]

Now, it seems to me that, in matrix notation:

[tex]
[g_{i^\prime j^\prime}] =
[g_{ij}] [U^T]
[/tex]

But this doesn't seem to be correct. I should end up with the metric for Euclidean space in cylindrical coordinates:

[tex]
g_{i^\prime j^\prime} =
\left[
\begin{array}{ccc}
1 & 0 & 0\\
0 & r^2 \sin^2 \theta & 0\\
0 & 0 & 1
\end{array}
\right]

=

\left[
\begin{array}{ccc}
1 & 0 & 0\\
0 & \rho^2 & 0\\
0 & 0 & 1
\end{array}
\right]
[/tex]

If my thinking *is* correct, somewhere I'm just getting confused. Can anyone help me figure out what's going on? Thanks in advance.

- Warren
 
Physics news on Phys.org
  • #2
Er wait...

[tex][g_{ij}] = [g_{ji}][/tex]

via matrix transposition, yes?

So then

[tex][U^{i^\prime}_j] = [U^j_{i^\prime}][/tex]

via matrix inversion, not transposition, yes?

Oooops... I still am not getting the correct answer, but I suspect I should have written

[tex]
[g_{i^\prime j^\prime}] =
[g_{ij}] [U^{-1}]
[/tex]

- Warren
 
Last edited:
  • #3
Erm, no one can help me?

- Warren
 
  • #4
Your coordinate change matrix doesn't look right...
 
  • #5
Hurkyl,

Any idea what it should be then? That came right out of my book. :-/

- Warren
 
  • #6
Originally posted by chroot
Hurkyl,

Any idea what it should be then? That came right out of my book. :-/

- Warren

I agree with Hurkyl. the matrix that i got was
[tex]\left (\begin{array}{ccc}
\frac{r}{\sqrt{r^2+z^2}} & \frac{z}{\sqrt{r^2+z^2}} & 0\\
\frac{z}{\sqrt{r^2+z^2}} & \frac{-r}{\sqrt{z^2+r^2}} & 0\\
0 & 0 & 1
\end{array}
\right)
[/tex]
 
  • #7
oh, actually, the thing i wrote is the same thing that you had... its OK

anyway. the point of the story is you don t conjugate to transform the metric. you only conjugate to transform operators
 
  • #8
the transformation rule is
[tex]
g'_{\mu\nu}=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu} g_{\rho\sigma}
[/tex]
 
  • #9
Courtesy of Lillian Lieber, "Co, low, primes below". That is, covariant indices are subscript, and the transformation derivatives are with respect to the new, primed coordinates. And of course contravariant indices are just the opposite.
 
  • #10
Originally posted by chroot

[tex]
[g_{i^\prime j^\prime}] =
[g_{ij}] [U^{-1}]
[/tex]

- Warren


a minor point, but i would like to correct your notation. [tex]\mbox{$ U^i_j$}[/tex] is a matrix element. to represent the matrix that these components make up you can either put the single component in brackets, getting something like [tex]\mbox{$\left [ U^i_j\right ] $}[/tex] (which is what you have done for the metric) or you can use [tex]\mbox{$ U $}[/tex] to stand for the whole tensor. but to put [tex]\mbox{$ U $}[/tex] inside the brackets is meaningless.
 
Last edited:
  • #11
Originally posted by lethe
the transformation rule is
[tex]
g'_{\mu\nu}=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu} g_{\rho\sigma}
[/tex]

Yeah, I'm aware. And in matrix notation, doesn't that end up being

[tex][g_{i^\prime j^\prime}] =
[U^{i^\prime}_k] [g_{kl}] [U^l_{j^\prime}][/tex]

So that the matrix multiplication is correct? Maybe I'm being naive, but I'm just flipping the U's upside down (inverting the corresponding matrix) and reordering terms so that each multiplication has the same index for the columns of the first matrix as the rows of the second.

Is this not the right way to approach setting up this kind of a multiplication?

Is there some deeper, darker secret I need to know to properly evaluate

[tex]g_{i^\prime j^\prime}
= U^k_{i^\prime}
U^l_{j^\prime}
g_{kl}[/tex]

using matrices? The tensor forms make perfect sense to me, but using them as matrices seems to be a stumbling block to me.

- Warren
 
Last edited:
  • #12
Originally posted by chroot
Yeah, I'm aware. And in matrix notation, doesn't that end up being

[tex][g_{i^\prime j^\prime}] =
[U^{i^\prime}_k] [g_{kl}] [U^l_{j^\prime}][/tex]

So that the matrix multiplication is correct?

no, matrix multiplication is not correct. matrix multiplication is the rule for multiplying a (1,1) tensor with another (1,1) tensor.

when you change coordinates, a (1,1) tensor gets conjugated, but not a (0,2) tensor. the metric is a (0,2) tensor.

forget matrices. thinking of tensors as matrices is a barrier to understanding what a tensor is.

also, the formula you wrote above, i can tell just by looking at it that it is not correct, since the indices don t match on both sides of the equation.
 
  • #13
Originally posted by lethe
when you change coordinates, a (1,1) tensor gets conjugated, but not a (0,2) tensor. the metric is a (0,2) tensor.
Conjugation = matrix transposition? Inversion?
forget matrices. thinking of tensors as matrices is a barrier to understanding what a tensor is.
I can't forget matrices. I'm trying to understand this exercise, which asks me to deal with matrices. Further, how can you actually compute something with tensors if you don't consider their components?

- Warren
 
  • #14
Originally posted by chroot
Conjugation = matrix transposition? Inversion?


conjugation means ABA-1. here, B is your tensor, and A is the change of coordinates matrix. this is the operation you were trying to do to the metric, which i claim is only valid for (1,1) tensors.

I can't forget matrices. I'm trying to understand this exercise, which asks me to deal with matrices. Further, how can you actually compute something with tensors if you don't consider their components?

- Warren

alright, keep your matrices. i don t find them so useful, but perhaps you can get more use from them than i do.
 
  • #15
OK, so let my spherical coordinates be [tex]r\mbox{, }\theta\mbox{, and } \phi[/tex], and my cylindrical coordinates be [tex]\rho, \psi, z[/tex]

[tex]\frac{\partial x^\text{sph}}{\partial x^\text{cyl}} = \left(\begin{array}{ccc}
\frac{\rho}{\sqrt{\rho^2+z^2}} & \frac{z}{\sqrt{\rho^2+z^2}} & 0\\
\frac{z}{\rho^2+z^2} & \frac{-\rho}{\rho^2+z^2} & 0\\
0 & 0 & 1
\end{array}\right)
[/tex]
where the first column is differentiation with respect to ρ, the second with respect to z, and the third with respect to ψ

the metric in spherical is

[tex]
g^\text{sph} = \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & r^2 & 0\\
0 & 0 & r^2\sin^2\theta
\end{array}\right)
[/tex]

so using the transformation formula, i can get the elements of gcyl

i will only calculate the diagonal components, since the basis is orthogonal.

[tex]
\begin{align*}
g^\text{cyl}_{\rho\rho}&=\frac{\partial r}{\partial \rho}\frac{\partial r}{\partial \rho}g^\text{sph}_{rr}+\frac{\partial\theta}{\partial\rho}\frac{\partial\theta}{\partial\rho}g^\text{sph}_{\theta\theta}+\frac{\partial\phi}{\partial\rho}\frac{\partial \phi}{\partial\rho}g^\text{sph}_{\phi\phi} \\
=& \frac{\rho^2}{\rho^2+z^2} + \frac{z^2}{(\rho^2+z^2)^2}(\rho^2+z^2)
= 1
\end{align*}
[/tex]

ok...

[tex]
g^\text{cyl}_{\psi\psi}=g^\text{sph}_{\phi\phi}=\rho^2
[/tex]

and
[tex]
g^\text{cyl}_{zz}=\frac{z^2}{\rho^2+z^2}+\frac{\rho^2}{(\rho^2+z^2)^2}(\rho^2+z^2)=1
[/tex]

and thus, in cylindrical coordinates, i have

[tex]
ds^2=d\rho^2+\rho^2d\psi^2+dz^2
[/tex]

voilà
 
Last edited:

1. What are metrics and coordinate transforms?

Metrics and coordinate transforms are mathematical tools used to measure and describe the relationships between different objects and points in a given space. They are commonly used in fields such as mathematics, physics, and engineering.

2. How are metrics and coordinate transforms related?

Metrics and coordinate transforms are closely related, as metrics provide the basis for coordinate transforms. A metric is a set of rules for measuring distances and angles in a space, and a coordinate transform is a way of mapping points from one coordinate system to another using these rules.

3. What is the purpose of using metrics and coordinate transforms?

The main purpose of using metrics and coordinate transforms is to simplify the description and analysis of complex systems and spaces. By using a standard set of rules and coordinates, we can better understand and manipulate these systems and make accurate predictions about their behavior.

4. What are some common examples of metrics and coordinate transforms?

Some common examples of metrics and coordinate transforms include Euclidean geometry, which uses a standard set of rules to measure distances and angles in 2D and 3D space, and polar coordinates, which use a different set of rules to describe points in a circular or spherical coordinate system.

5. How do metrics and coordinate transforms impact scientific research?

Metrics and coordinate transforms play a crucial role in scientific research, as they allow us to accurately measure and describe the physical world and make meaningful comparisons between different systems and phenomena. They are used extensively in fields such as physics, astronomy, and geology to study everything from the behavior of subatomic particles to the movement of planets and galaxies.

Similar threads

Replies
6
Views
859
Replies
1
Views
1K
Replies
1
Views
1K
  • Differential Geometry
Replies
21
Views
586
  • Differential Geometry
Replies
10
Views
4K
  • Differential Geometry
Replies
1
Views
945
  • Differential Geometry
Replies
7
Views
2K
  • Differential Geometry
Replies
12
Views
3K
Replies
6
Views
2K
  • Differential Geometry
Replies
2
Views
2K
Back
Top