Christoffel symbols knowing Line Element (check my result)

[Confused Physicist]

Hi! I'm asked to find all the non-zero Christoffel symbols given the following line element:

$$ds^2=2x^2dx^2+y^4dy^2+2xy^2dxdy$$

The result I have obtained is that the only non-zero component of the Christoffel symbols is:

$$\Gamma^x_{xx}=\frac{1}{x}$$

Is this correct?

MY PROCEDURE HAS BEEN:

the Lagrangian squared is given by:
$$\mathcal{L}^2=2x^2\dot{x}^2+y^4\dot{y}^2+2xy^2\dot{x}\dot{y}$$

where $$\dot{\quad}\equiv d/d\tau$$, with $$\tau$$ being the proper time.

The Euler-Lagrange equations for this Lagrangian are:
$$\begin{split} \frac{d}{d\tau}\Bigg(\frac{\partial\mathcal{L}^2}{\partial{\dot{x}}}\Bigg)-\frac{\partial\mathcal{L}^2}{\partial x}=0\quad\longrightarrow\quad&\frac{d}{d\tau}\Big(4x^2\dot{x}+2xy^2\dot{y}\Big)-4x\dot{x}^2-2y^2\dot{x}\dot{y}=0\\ &8x\dot{x}^2+4x^2\ddot{x}+2y^2\dot{x}\dot{y}+4xy\dot{y}^2+2xy^2\ddot{y}-4x\dot{x}^2-2y^2\dot{x}\dot{y}=0\\ \text{(1)}\quad&2x^2\ddot{x}+xy^2\ddot{y}+2x\dot{x}^2+2xy\dot{y}^2=0\\ \frac{d}{d\tau}\Bigg(\frac{\partial\mathcal{L}^2}{\partial{\dot{y}}}\Bigg)-\frac{\partial\mathcal{L}^2}{\partial y}=0\quad\longrightarrow\quad&\frac{d}{d\tau}\Big(2y^4\dot{y}+2xy^2\dot{x}\Big)-4y^3\dot{y}^2-4xy\dot{x}\dot{y}=0\\ &8y^3\dot{y}^2+2y^4\ddot{y}+2y^2\dot{x}^2+4xy\dot{x}\dot{y}+2xy^2\ddot{x}-4y^3\dot{y}^2-4xy\dot{x}\dot{y}=0\\ \text{(2)}\quad&xy^2\ddot{x}+y^4\ddot{y}+y^2\dot{x}^2+2y^3\dot{y}^2=0 \end{split}$$

Multiplying equation (1) by $$y^2/x$$ we obtain: $$2xy^2\ddot{x}+y^4\ddot{y}+2y^2\dot{x}^2+2y^3\dot{y}^2=0$$. Subtracting equation (2) from this result we obtain:
$$xy^2\ddot{x}+y^2\dot{x}^2=0\quad\longrightarrow\quad \ddot{x}+\frac{1}{x}\dot{x}^2=0$$

Multiplying equation (2) by $$2x/y^2$$ we obtain: $$2x^2\ddot{x}+2xy^2\ddot{y}+2x\dot{x}^2+2xy\dot{y}^2=0$$.Subtracting equation (1) from this result we obtain:
$$xy^2\ddot{y}=0\quad\longrightarrow\quad \ddot{y}=0$$

Knowing that the general expression for the geodesic equation is:
$$\ddot{x}^\alpha+\Gamma^\alpha_{\beta\gamma}\dot{x}^\beta \dot{x}^\gamma=0$$

$$\boxed{\text{the only non-zero Christoffel symbol is therefore: } \Gamma^x_{xx}=\frac{1}{x}}$$

 

[Orodruin]

Please show your work on homework-like questions. Otherwise it is impossible to know where you go wrong (if you do) or confirm that you did things correctly (if you did) and did not come to the correct answer by chance.

 

[Orodruin]

Your approach is fine. However, your final equation for $\ddot y$ is not. Also, in this particular case, there is a shortcut by a simple change of variables that turns your line element to one with constant coefficients.

 

[Confused Physicist]

Your approach is fine. However, your final equation for $\ddot y$ is not. Also, in this particular case, there is a shortcut by a simple change of variables that turns your line element to one with constant coefficients.

Oh, why is my final equation for $\ddot{y}$ not correct?

And what is that change of variables that simplifies the calculation?

Is the solution I have obtained incorrect then?

 

[Orodruin]

You have an arithmetic error in your multiplication of (2).

The change of variables should be quite evident from the form of the line element. Hint: You want to make a change to variables $\chi(x)$ and $\eta(y)$ so that the line element takes the form $ds^2 = A d\chi^2 + B d\chi\, d\eta + C d\eta^2$, where A B and C are constant.

 

[Confused Physicist]

You have an arithmetic error in your multiplication of (2).

You mean when I multiply (2) by $2x/y^2$? I'm repeating the process again, but I can't find the mistake...

Ohhhhh okay I've seen the mistake now. The mistake is on equation (2). The $2y^3\dot{y}^2$ should be $4y^3\dot{y}^2$.

Oh no. That wasn't the mistake. I still can't see what you're pointing out.

FINAL EDIT: Okay, I've seen the mistake. The equation for $\ddot{y}$ should be: $xy^2\ddot{y}+2xy\dot{y}^2=0$, which gives another non-zero Christoffel symbol: $\Gamma^y_{yy}=\frac{2}{y}$.

 

[Orodruin]

Right. Now try out the coordinate change $\chi = x^2$, $\eta = y^3$ to see what it does to the metric.