# 2 Metric tensor question~

1. Nov 3, 2005

### yukcream

Q1 If given a 2D Riemannian space, ds^2 = dx^2 + x^2dy^2, do the componets of the metric tensor are these:
g_11 = 1, g_12 = 0
g_21 = o, g_22 = x^2 ?

In addition, I got a question from my lecturer:
Q2. 2 metrics, defined in a Riemannian space, are given by ds^2 = g_ijdx^idy^j
and ds'^2 = g'_ij dx^idy^j = e^z g_ijdx^idy^j , respectively, where z is a function of the coordinates x^i.
Find the relation between the Chritoffel symbols corresponding to the 2 two metrics~~~
I have no ideal how to solve it and what is e here? treat it as a function or is it represents the persudo tensor????

Can any one help me~~

yukyuk

2. Nov 3, 2005

### Jimmy Snyder

Yes, this is correct.

I assume you mean: $ds^2 = g_{ij}dx^idx^j$ and $ds'^2 = g'_{ij}dx^idx^j = e^z g_{ij}dx^idx^j$. (Note that I have replaced references to $y^j$ with references to $x^j$.)

The $e^z$ here is the exponential function. Note that its partial derivatives are (,i is shorthand for $\partial / \partial x^i$):

$e^z{}_{,i} = z_{,i}e^z$

The formula for the Christoffel symbol in terms of the metric tensor is:

$\Gamma^m{}_{ij} = \frac{1}{2}g^{km}(g_{ik,j} + g_{jk,i} - g_{ij,k})$

This should be enough to get you started. If you still have trouble, post again.

Last edited: Nov 3, 2005
3. Nov 3, 2005

### yukcream

To jimmysnyder:

Hope I can understand what you mean~~
I work out the steps, am I correct?

$\Gamma^m{}_{ij} = \frac{1}{2}g^{km} (-g_{ij,k}+g_{jk,i}+g_{ki,j})$
$\Gamma'^m{}_{ij} = \frac{1}{2}g'^{km} (-g'_{ij,k}+g'_{jk,i}+g'_{ki,j})$
$\Gamma'^m{}_{ij} = \frac{1}{2}e^{z(i)}g^{km} (-e^{z(i)}g_{ij,k}+e^{z(i)}g_{jk,i}+e^{z(i)}g_{ki,j}$
$\Gamma'^m{}_{ij} = \frac{1}{2}e^{z(i)}g^km{}[-e^{z(i)}g_{ij,k}+z(i)_{,i}e^{z(i)}g_{jk,i}+e^{z(i)}g_{jk,i}+e^{z(i)}g_{ki,j}]$
$\Gamma'^m{}_{ij} = e^{2z(i)}[\Gamma^m{}_{ij} + \frac{1}{2}g^{km}z(i)_{,i}g_{jk,i}]$

Thx so much and you are so smart that you can correct my mistake as what i want to write is dx^j not dy^j~

yukyuk

Last edited: Nov 3, 2005
4. Nov 3, 2005

### Jimmy Snyder

So far, so good.

The exponential factor that goes with $g^{km}$ should be $e^{-z}$, because you want $g'^{np}g'_{pm} = \delta^n{}_m$. Also in order to make things more clear, I would remove the parameter from z and add parentheses as follows:
$\Gamma'^m{}_{ij} = \frac{1}{2}e^{-z}g^{km} (-(e^{z}g_{ij})_{,k} + (e^{z}g_{jk})_{,i} + (e^{z}g_{ki})_{,j})$

The next steps are not correct, but if you start from the equation I have given you, I think you can get the rest.

Last edited: Nov 3, 2005
5. Nov 3, 2005

### yukcream

Once more question is :

Given a fame S' which is falling along -z-axis with constant acceleration in an inertial frame S. Find a form of metric in the S' frame, assume in Newtonian approximation of the absolute time (t=t').

I just know a definition that
a = sqrt(g_ij dx^idx^j) but how to get the g_ij??

yukyuk

6. Nov 3, 2005

### yukcream

I think my answer is correct~~
as z is only a function of coordinate x^i , derative of z wrt x^j & x^k will be zero ~ right? Do I make the mistake there?

7. Nov 3, 2005

### Jimmy Snyder

Yes, you have made a mistake. In this case, i is an index that takes all 4 values, 0, 1, 2, and 3. In your original post you have written "z is a function of the coordinates x^i". Note coordinates, not coordinate. That means it is a function of all 4 coordinates and partials must be taken with respect to each of them. The letters k and j are just different index letters that also take on the 4 values 0, 1, 2, and 3.

Last edited: Nov 3, 2005