Understanding Metric Tensors in Riemannian Spaces

[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 anyone help me~~

yukyuk

 

[Jimmy Snyder]

Q1 If given a 2D Riemannian space, ds^2 = dx^2 + x^2dy^2, do the componets of the metric tensor are these:
<br /> g_{11} = 1, g_{12} = 0<br />
<br /> g_{21} = 0, g_{22} = x^2<br />

Yes, this is correct.

In addition, I got a question from my lecturer:
Q2. 2 metrics, defined in a Riemannian space, are given by ds^2 = g_{ij}dx^idy^j
and ds&#039;^2 = g&#039;_{ij} dx^idy^j = e^z g_{ij}dx^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 anyone help me~~

I assume you mean: ds^2 = g_{ij}dx^idx^j and ds&#039;^2 = g&#039;_{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.

 

[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&#039;^m{}_{ij} = \frac{1}{2}g&#039;^{km} (-g&#039;_{ij,k}+g&#039;_{jk,i}+g&#039;_{ki,j})
\Gamma&#039;^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&#039;^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&#039;^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

 

[Jimmy Snyder]

\Gamma^m{}_{ij} = \frac{1}{2}g^{km} (-g_{ij,k}+g_{jk,i}+g_{ki,j})
\Gamma&#039;^m{}_{ij} = \frac{1}{2}g&#039;^{km} (-g&#039;_{ij,k}+g&#039;_{jk,i}+g&#039;_{ki,j})

So far, so good.

\Gamma&#039;^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}

The exponential factor that goes with g^{km} should be e^{-z}, because you want g&#039;^{np}g&#039;_{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&#039;^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.

 

[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

 

[yukcream]

So far, so good.
The exponential factor that goes with g^{km} should be e^{-z}. Also in order to make things more clear, I would remove the parameter from z and add parentheses as follows:
\Gamma&#039;^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.

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?

 

[Jimmy Snyder]

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?

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.