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Metrics and coordinate transforms

  1. Nov 23, 2003 #1

    chroot

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    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
     
  2. jcsd
  3. Nov 23, 2003 #2

    chroot

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    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: Nov 23, 2003
  4. Nov 24, 2003 #3

    chroot

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    Erm, no one can help me?

    - Warren
     
  5. Nov 24, 2003 #4

    Hurkyl

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    Your coordinate change matrix doesn't look right...
     
  6. Nov 24, 2003 #5

    chroot

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    Hurkyl,

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

    - Warren
     
  7. Nov 24, 2003 #6
    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]
     
  8. Nov 24, 2003 #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
     
  9. Nov 24, 2003 #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]
     
  10. Nov 24, 2003 #9

    selfAdjoint

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    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.
     
  11. Nov 25, 2003 #10


    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: Nov 25, 2003
  12. Nov 25, 2003 #11

    chroot

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    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: Nov 25, 2003
  13. Nov 25, 2003 #12
    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.
     
  14. Nov 25, 2003 #13

    chroot

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    Conjugation = matrix transposition? Inversion?
    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
     
  15. Nov 25, 2003 #14

    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.

    alright, keep your matrices. i don t find them so useful, but perhaps you can get more use from them than i do.
     
  16. Nov 25, 2003 #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: Dec 8, 2003
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