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Confused Physicist

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

[tex]ds^2=2x^2dx^2+y^4dy^2+2xy^2dxdy[/tex]

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

[tex]\Gamma^x_{xx}=\frac{1}{x}[/tex]

MY PROCEDURE HAS BEEN:

the Lagrangian squared is given by:

[tex]\mathcal{L}^2=2x^2\dot{x}^2+y^4\dot{y}^2+2xy^2\dot{x}\dot{y}[/tex]

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

The Euler-Lagrange equations for this Lagrangian are:

[tex]

\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}

[/tex]

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

[tex]xy^2\ddot{x}+y^2\dot{x}^2=0\quad\longrightarrow\quad \ddot{x}+\frac{1}{x}\dot{x}^2=0[/tex]

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

[tex]xy^2\ddot{y}=0\quad\longrightarrow\quad \ddot{y}=0[/tex]

Knowing that the general expression for the geodesic equation is:

[tex]\ddot{x}^\alpha+\Gamma^\alpha_{\beta\gamma}\dot{x}^\beta \dot{x}^\gamma=0[/tex]

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

[tex]ds^2=2x^2dx^2+y^4dy^2+2xy^2dxdy[/tex]

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

[tex]\Gamma^x_{xx}=\frac{1}{x}[/tex]

**Is this correct?**MY PROCEDURE HAS BEEN:

the Lagrangian squared is given by:

[tex]\mathcal{L}^2=2x^2\dot{x}^2+y^4\dot{y}^2+2xy^2\dot{x}\dot{y}[/tex]

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

The Euler-Lagrange equations for this Lagrangian are:

[tex]

\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}

[/tex]

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

[tex]xy^2\ddot{x}+y^2\dot{x}^2=0\quad\longrightarrow\quad \ddot{x}+\frac{1}{x}\dot{x}^2=0[/tex]

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

[tex]xy^2\ddot{y}=0\quad\longrightarrow\quad \ddot{y}=0[/tex]

Knowing that the general expression for the geodesic equation is:

[tex]\ddot{x}^\alpha+\Gamma^\alpha_{\beta\gamma}\dot{x}^\beta \dot{x}^\gamma=0[/tex]

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

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