# Euler-Lagrange and Christoffel symbols

• LayMuon
In summary: 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 k.
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.

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.

LayMuon said:
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 $\dfrac{\partial g_{ij}}{\partial x^k}$.

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

LayMuon said:
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 $S^l_{jk}$:

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

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

stevendaryl said:
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.

elfmotat said:
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.

## 1. What is the Euler-Lagrange equation?

The Euler-Lagrange equation is a fundamental equation in calculus of variations, which is used to determine the stationary points of a functional. It is used to find the function that minimizes or maximizes the functional by setting its derivative to zero.

## 2. What are Christoffel symbols used for?

Christoffel symbols are used in differential geometry to describe the curvature of a space. They are used to calculate the covariant derivative, which is a way to differentiate vector fields on a curved space.

## 3. What is the relationship between Euler-Lagrange and Christoffel symbols?

The Euler-Lagrange equation is derived using the concept of the covariant derivative, which involves the Christoffel symbols. The Christoffel symbols represent the connection coefficients that are used to define the covariant derivative in a curved space.

## 4. How are Euler-Lagrange and Christoffel symbols applied in physics?

Euler-Lagrange and Christoffel symbols are used extensively in theoretical physics, particularly in the fields of general relativity and quantum mechanics. In general relativity, they are used to describe the behavior of matter and energy in curved spacetime. In quantum mechanics, they are used to define the path integral, which is used to calculate the probability amplitude for a particle to move from one point to another.

## 5. Are there any practical applications of Euler-Lagrange and Christoffel symbols?

Yes, there are many practical applications of Euler-Lagrange and Christoffel symbols. They are used in engineering and physics to solve optimization problems, such as finding the shortest path between two points on a curved surface. They are also used in computer graphics to create realistic animations of objects moving in curved spaces.

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