Euler-Lagrange and Christoffel symbols

[LayMuon]

I am pretty much confused with all the algebra of Christoffel symbols:

I have an expression for infinitesimal length: $F= g_{ij} \frac{dx^i dx^j}{du^2}$ and by using Euler-Lagrange equation (basically finding the shortest distance between two points) want to find the equation for geodesics with affine parameter. But getting bogged in the indices and unable to extricate them so as to introduce the Christoffel symbols. Any help or suggestion on how to improve skills with this type of calculations is highly appreciated.

 

[Bill_K]

You don't need any Christoffel symbols, just write down the Euler-Lagrange equations with F as the Lagrangian, and presto, you've got the geodesic equations.

 

[stevendaryl]

I am pretty much confused with all the algebra of Christoffel symbols:

I have an expression for infinitesimal length: $F= g_{ij} \frac{dx^i dx^j}{du^2}$ and by using Euler-Lagrange equation (basically finding the shortest distance between two points) want to find the equation for geodesics with affine parameter. But getting bogged in the indices and unable to extricate them so as to introduce the Christoffel symbols. Any help or suggestion on how to improve skills with this type of calculations is highly appreciated.

You don't need to introduce the Christoffel symbols, they show up in the equations of motion naturally.

The Euler-Lagrange equations for the Lagrangian $g_{ij} U^i U^j$ (where I introduced $U^i$ as a shorthand for $\dfrac{dx^i}{du}$) are:

$\dfrac{d}{dt} \dfrac{\partial L}{\partial U^i} = \dfrac{\partial L}{\partial x^i}$

$2 \dfrac{d}{dt} (g_{ij} U^i) = \dfrac{\partial}{\partial x^i} g_{kj} U^k U^j$

If you solve this for $\dfrac{dU^i}{dt}$, you get the usual geodesic equation:

$\dfrac{dU^i}{dt} = - \Gamma^i_{jk} U^j U^k$

with $\Gamma^i_{jk}$ replaced by terms involving $g_{ij}$ and $g^{ij}$ and $\dfrac{\partial g_{ij}}{\partial x^k}$.

 

[LayMuon]

I am confused right at the point how to solve that equation and rearrange indices in such a manner as to get Christoffel symbols.

 

[stevendaryl]

I am confused right at the point how to solve that equation and rearrange indices in such a manner as to get Christoffel symbols.

I worked it out for myself, and there are a few tricky parts to it. Maybe there's a more direct way to get the answer, but here's how I did it (fair warning: this is extremely long and boring):

Start with the "Lagrangian" $L = g_{ij} U^i U^j$, where $U^i = \dfrac{dx^i}{ds}$ then the Lagrangian equations of motion are:

$2 \dfrac{d}{ds} (g_{ij} U^j) = (\partial_i g_{jk}) U^j U^k$

where $\partial_i$ is shorthand for $\dfrac{\partial}{\partial x^i}$ (To get the right-hand side, I changed the dummy index $i$ to $k$ to keep from clashing with the $i$ in the $\partial_i$)

Now use the product rule: $\dfrac{d}{ds} (g_{ij} U^j) = \dfrac{d g_{ij}}{ds} U^j + g_{ij} \dfrac{dU^j}{ds}$

Then use the chain rule:
$\dfrac{d g_{ij}}{ds} = \dfrac{\partial g_{ij}}{\partial x^k} \dfrac{dx^k}{ds} = \partial_k g_{ij} U^k$

So the equations of motion become:

$2 (\partial_k g_{ij} U^k U^j + g_{ij} \dfrac{dU^j}{dt}) = (\partial_i g_{jk}) U^j U^k$

Now, rearrange to get:
$g_{ij} \dfrac{dU^j}{dt} = -(\partial_k g_{ij}) U^k U^j + \frac{1}{2} (\partial_i g_{jk}) U^j U^k = -\frac{1}{2}(2 \partial_k g_{ij} - \partial_i g_{jk})U^j U^k$

At this point, we can raise the indices on both sides by multiplying by $g^{li}$ and summing over $i$:

$\dfrac{dU^l}{dt} = -\frac{1}{2}g^{li}(2 \partial_k g_{ij} - \partial_i g_{jk})U^j U^k$

This has the form

$\dfrac{dU^l}{dt} = -Q^l_{jk} U^j U^k$

where the coefficients $Q^l_{jk}$ are defined by:

$Q^l_{jk} = \frac{1}{2}g^{li}(2 \partial_k g_{ij} - \partial_i g_{jk})$

$Q^l_{jk}$ is not quite the Christoffel coefficients, because $\Gamma^l_{jk}$ is symmetric in $j$ and $k$, while $Q^l_{jk}$ is not. However, we can split $Q^l_{jk}$ into symmetric and anti-symmetric parts:
$Q^l_{jk} = S^l_{jk} + A^l_{jk}$
where
$S^l_{jk} = \frac{1}{2}(Q^l_{jk} + Q^l_{kj}) = \frac{1}{2}g^{lj}(\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk})$
$A^l_{jk} = \frac{1}{2}(Q^l_{jk} - Q^l_{kj}) = \frac{1}{2}g^{lj}(\partial_k g_{ij} - \partial_j g_{ik})$

Then the equations of motion become:
$\dfrac{dU^l}{dt} = -S^l_{jk} U^j U^k - A^l_{jk} U^j U^k$

The term $A^l_{jk} U^j U^k$ is 0, so only the symmetric part matters. So we have:

$\dfrac{dU^l}{dt} = -S^l_{jk} U^j U^k$

At this point, we can identify $\Gamma^l_{jk}$ with $S^l_{jk}$:

$\Gamma^l_{jk} = \frac{1}{2}g^{li}(\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk})$

 

[LayMuon]

Thank you very much. I was stuck at the point where you split Q into syummetric and antisymmetric parts.

 

[elfmotat]

I worked it out for myself, and there are a few tricky parts to it. Maybe there's a more direct way to get the answer, but here's how I did it (fair warning: this is extremely long and boring):

Start with the "Lagrangian" $L = g_{ij} U^i U^j$, where $U^i = \dfrac{dx^i}{ds}$ then the Lagrangian equations of motion are:

$2 \dfrac{d}{ds} (g_{ij} U^j) = (\partial_i g_{jk}) U^j U^k$

where $\partial_i$ is shorthand for $\dfrac{\partial}{\partial x^i}$ (To get the right-hand side, I changed the dummy index $i$ to $k$ to keep from clashing with the $i$ in the $\partial_i$)

Now use the product rule: $\dfrac{d}{ds} (g_{ij} U^j) = \dfrac{d g_{ij}}{ds} U^j + g_{ij} \dfrac{dU^j}{ds}$

Then use the chain rule:
$\dfrac{d g_{ij}}{ds} = \dfrac{\partial g_{ij}}{\partial x^k} \dfrac{dx^k}{ds} = \partial_k g_{ij} U^k$

So the equations of motion become:

$2 (\partial_k g_{ij} U^k U^j + g_{ij} \dfrac{dU^j}{dt}) = (\partial_i g_{jk}) U^j U^k$

Now, rearrange to get:
$g_{ij} \dfrac{dU^j}{dt} = -(\partial_k g_{ij}) U^k U^j + \frac{1}{2} (\partial_i g_{jk}) U^j U^k = -\frac{1}{2}(2 \partial_k g_{ij} - \partial_i g_{jk})U^j U^k$

At this point, we can raise the indices on both sides by multiplying by $g^{li}$ and summing over $i$:

$\dfrac{dU^l}{dt} = -\frac{1}{2}g^{li}(2 \partial_k g_{ij} - \partial_i g_{jk})U^j U^k$

This has the form

$\dfrac{dU^l}{dt} = -Q^l_{jk} U^j U^k$

where the coefficients $Q^l_{jk}$ are defined by:

$Q^l_{jk} = \frac{1}{2}g^{li}(2 \partial_k g_{ij} - \partial_i g_{jk})$

$Q^l_{jk}$ is not quite the Christoffel coefficients, because $\Gamma^l_{jk}$ is symmetric in $j$ and $k$, while $Q^l_{jk}$ is not. However, we can split $Q^l_{jk}$ into symmetric and anti-symmetric parts:
$Q^l_{jk} = S^l_{jk} + A^l_{jk}$
where
$S^l_{jk} = \frac{1}{2}(Q^l_{jk} + Q^l_{kj}) = \frac{1}{2}g^{lj}(\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk})$
$A^l_{jk} = \frac{1}{2}(Q^l_{jk} - Q^l_{kj}) = \frac{1}{2}g^{lj}(\partial_k g_{ij} - \partial_j g_{ik})$

Then the equations of motion become:
$\dfrac{dU^l}{dt} = -S^l_{jk} U^j U^k - A^l_{jk} U^j U^k$

The term $A^l_{jk} U^j U^k$ is 0, so only the symmetric part matters. So we have:

$\dfrac{dU^l}{dt} = -S^l_{jk} U^j U^k$

At this point, we can identify $\Gamma^l_{jk}$ with $S^l_{jk}$:

$\Gamma^l_{jk} = \frac{1}{2}g^{lj}(\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk})$

You mixed up a few indices towards the end. For example in $\Gamma^l_{jk} = \frac{1}{2}g^{lj}(\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk})$ you should be summing over i, not j.

 

[stevendaryl]

You mixed up a few indices towards the end. For example in $\Gamma^l_{jk} = \frac{1}{2}g^{lj}(\partial_k g_{ij} + \partial_j g_{ik} - \partial_i g_{jk})$ you should be summing over i, not j.

Thanks.