Calculate Christoffel Symbols of 2D Metric

[Ryomega]

Homework Statement

Consider metric ds² = dx² + x³ dy² for 2D space.
Calculate all non-zero christoffel symbols of metric.

Homework Equations

\Gamma^j_ik = \partiale_i / \partial x^k \times e^j

The Attempt at a Solution

Christoffel symbols, by definition, takes the partial of each vector basis with respect to component (in this case x^k).

So my instinct tells me to differentiate the metric with respect to x and y. Which would give me:

metric differentiated with respect to (x) = 3x² x¹

metric differentiated with respect to (y) = 0 x²

Did I do this right or did I get this completely wrong? If I have done it wrong, please explain.

Thank you very much in advance.

 

[Chestermiller]

Do you know the equation for calculating the christoffel symbols from the partial derivatives of the components of the metric tensor? Alternatively, for this problem,

e_x=i_x

e_y=x^(3/2)i_y

and

e^x=i_x

e^y=x^(-3/2)i_y

where i_x and i_y are unit vectors in the x- and y-directions, respectively.

 

[Ryomega]

Hello! Thank you for the quick reply!

The equation you have mentioned, the only equation I am aware of for relating christoffel symbols to the metric is:

\Gamma^L_ab = \frac{1}{2} g^Lc (g_ac;b + g_cb;a - g_ba;c)

I'm having a hard time understanding the indices, thus just how to use this equation correctly. I'll give it a go and see if I can't arrive at the same answer.

Thanks again!

 

[Chestermiller]

Hello! Thank you for the quick reply!

The equation you have mentioned, the only equation I am aware of for relating christoffel symbols to the metric is:

\Gamma^L_ab = \frac{1}{2} g^Lc (g_ac;b + g_cb;a - g_ba;c)

I'm having a hard time understanding the indices, thus just how to use this equation correctly. I'll give it a go and see if I can't arrive at the same answer.

Thanks again!

Those should be commas (indicating partial differentiation), not semicolons.
For this problem, it would probably be easier to use your original formula. There are only two dimensions, so there are going to be only 8 gammas to evaluate. Either way, it shouldn't be much work.

chet