Metric tensor in spherical coordinates
Douglas A. Palmer, PhD
Given that I have a metric, expressed in Cartesian coordinates, g = g(x,t)
with an invariant line element:
ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}.
How do I go about expressing this invariant line element in spherical
coordinates? I have expressions for all of the g's such as g11 etc. in
terms of x, y, z. Is this problem best approached from the standpoint of
converting the components of a vielbein representation for g? The problem
does not appear to be really trivial or it would pop up as a transform in
Maple/GrTensorII. I would appreciate any pointers to references.
Thanks,
John Baez
Douglas A. Palmer, PhD <dpa...@cox.net> wrote:
>Given that I have a metric, expressed in Cartesian coordinates, g = g(x,t)
>with an invariant line element:
>
>ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}.
>
>How do I go about expressing this invariant line element in spherical
>coordinates? I have expressions for all of the g's such as g11 etc. in
>terms of x, y, z.
1) Express x,y,z as functions of the spherical coordinates theta, phi, r.
2) Differentiate to get dx,dy,dz in terms of dtheta, dphi, dr.
3) Plug these formulas for dx,dy,dz into the above formula for ds^2.
4) Also, rewrite each function g_{\mu\nu}(x,y,z) as a function of
spherical coordinates.
WaiteDavid137
>From: "Douglas A. Palmer, PhD" dpa...@cox.net
>Date: 9/9/2003 11:46 AM US Mountain Standard Time
>Message-id: <002401c36b7d$91d2a110$0301a8c0@DOUGLAS>
One way: Make the following replacements in the line element ds^2 and then
simplify
x = rcos(theta)sin(phi)
dx^2 = [cos(theta)sin(phi)dr - rsin(theta)sin(phi)dtheta +
rcos(theta)cos(phi)dphi]^2
y = rsin(theta)sin(phi)
dy^2 = [sin(theta)sin(phi)dr + rcos(theta)sin(phi)dtheta +
rsin(theta)cos(phi)dphi]^2
z = rcos(phi)
dz^2 = [cos(phi)dr - rsin(phi)dphi]^2
Then read the metric tensor off the new expression for the line element.
Second way: Write the transformation matrix for dx^i from the differential
expressions above, then transform the metric tensor directly. Then write any
x,y,z expressions that remain in your metric in terms of the spherical
coordinates as above.
Lubos Motl
transforming the metric into other coordinates is not hard. Believe me.
If you have some new coordinates x'^a - imagine the spherical ones - then
you can calculate the partial derivatives
d x^b / d x'^{a'} = t^b_{a'}
which has an upper index "b" (Cartesian one) and a lower index "a'"
(spherical index). The metric in the spherical coordinates is simply
g_{c'd'} = sum_{a,b} g_{ab} . t^a_{c'} . t^b_{d'}
You sum over all values of the indices a,b; g_{ab} is expressed in the
Cartesian (or, more generally, the old coordinates), and t's are the
transformation matrices defined above. The previous displayed equation is
simply the transformation rule for tensors.
For spherical coordinates you finally get
ds^2 = dr^2 + r^2 (d theta^2 + sin^2(theta) d phi^2)
The mixed terms cancel. I defined t^b_{a'} in such a way that they have an
easy form for spherical coordinates - the derivatives of
r.sin(theta).cos(phi), for example, are always "easy" expressions.
Best wishes
Lubos
______________________________________________________________________________
E-mail: lu...@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
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Superstring/M-theory is the language in which God wrote the world.
Jose B. Almeida
Why not use the striaghtforward replacement
x = r sin(theta) cos(phi)
y = r sin(theta) sin(phi)
z = r cos(theta)
Jose
davidoff404
Quite trivial. If you recall, one can give a simple coordinate
transformation between Cartesian coordinates and spherical polar
coordinates of the form x=rsin(theta)cos(phi), etc. Work out the metric
components as a direct consequence of this transformation and you'll end up
with your line element.
As a starter hint, the line element on a unit two-sphere can be written as
follows:
ds^2 = dr^2 + sin(theta)^2 d(theta)^2