# Christoffel symbol from Variational Principle

• psimeson
O(\phi^2) In summary, the Christoffel symbols can be calculated using the variational principle, but it requires careful calculation and taking into account the dependence of the gravitational potential on proper time. In this case, the Christoffel symbols end up being dependent on the time derivative of the gravitational potential.

## Homework Statement

It's not exactly a homework question. I can find Christoffel Symbols using general definition of Christoffel symbol. But, when I try to find Christoffel Symbols using variational principle, I end up getting zero.

I have started with the space-time metric in a weak gravitational field (with the assumption of low velocity):
ds2=−(1+2ϕ)dt2+(1−2ϕ)(dx2+dy2+dz2)

Where
ϕ<<1 is the gravitational potential.

## Homework Equations

Euler Lagrange(EL) Equation: $\frac{d}{d\tau}$($\frac{dL}{d\dot{x^{a}}}$) = $\frac{dL}{dx}$

## The Attempt at a Solution

Lagrangian:
L = −(1+2ϕ)$\dot{t}$2+(1−2ϕ)($\dot{x}$2+$\dot{y}$2+$\dot{z}$2)

using EL:$\frac{d}{d\tau}$($\frac{dL}{d\dot{t}}$) = $\frac{dL}{dx}$

−(1+2ϕ)$\ddot{t}$= 0

I repeated the similar process for x, y, and z and I got zero for all. Can someone please help me?

Γtti=ϕ,i

Γi00=ϕ,iijk=δjkϕ,i−δijϕ,k−δikϕ,j

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You are taking phi to be constant in your calculation. The Christoffel symbols are indeed zero if phi is constant, but not if it's not.

I know that phi is function of r and I am having hard time to differentiate it. I just got stuck. Since r = sqrt (x^2 + y^2+ z^2) . How can it depend on t. The way it can depend on t by using the definition of proper time. Sorry, I am just lost.

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So you just use chain rule:

$\frac{d}{d\tau} \frac{\partial L}{\partial \dot{x}_i} = 2\frac{d}{d\tau} [(1+2\phi) \dot{x}_i] = 2((1+2\phi)\ddot{x}_i + 2\dot{x}_i \phi_{,\mu} \dot{x}^{\mu})$

and as usual,

$\frac{\partial L}{\partial x_i}= 2 \phi_{,i} ( \dot{t}^2 +\dot{x}_j \dot{x}^j)$

Note that I can be careless with my index notation here, because the errors I'm making would be proportional to phi squared, which I can drop in first order perturbation theory. You should probably be more careful though, it's very easy to make stupid mistakes with this kind of calculations

I didn't get how you get ${\mu}$...why are you using lower index $\dot{x}_i$

$\frac{d}{d\tau} \frac{\partial L}{\partial \dot{x}_i} = 2\frac{d}{d\tau} [(1+2\phi) \dot{x}_i] = 2((1+2\phi)\ddot{x}_i + 2\dot{x}_i \phi_{,\mu} \dot{x}^{\mu})$

What is $\dot{x}_j \dot{x}^j$?

$\frac{\partial L}{\partial x_i}= 2 \phi_{,i} ( \dot{t}^2 +\dot{x}_j \dot{x}^j)$

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psimeson said:
I didn't get how you get ${\mu}$

What is $\dot{x}_j \dot{x}^j$?

Suppose I have a function f(t,x) and t and x depend on the proper time, t=t(λ) and x=x(λ) along my path. Then

df/dλ = ∂f/∂x dx/dλ + ∂f/∂t dt/dλ.

The Einstein summing convention is just my shorthand way of writing this for all 4 indices. So I would write x,y,z,t as a 4-vector xμ and sum over repeating indices.

It's just a shorthand notation. So latin indices get values 1,2,3 (x,y,z) and greek indices get values 0,1,2,3 (t,x,y,z). I also use Einstein summation convention, so for example
$\dot{x}_j \dot{x}^j = \sum_{j} g_{jj} \dot{x}^j \dot{x}^j = \dot{x}^2+\dot{y}^2+\dot{z}^2 + O(\phi)$

psimeson said:
...why are you using lower index $\dot{x}_i$

You're right, my indices are all over the place in that equation :-) You should really do the calculation more carefully.

For time, Christoffel Symbol should be:
${\Gamma^t}_{it}$ = $x_i\frac{\phi'(r)}{r}$

But by doing the way you suggest I didn't get the answer:

$\frac{d}{d\tau} \frac{\partial L}{\partial \dot{t}} = -2\frac{d}{d\tau} [(1+2\phi) \dot{t}] = -2((1+2\phi)\ddot{t} + 2\dot{t} \phi_{,\mu} \dot{x}^{\mu})$

$\frac{\partial L}{\partial t}= 0$

$(1+2\phi)\ddot{t} + 2\dot{t} \phi_{,\mu} \dot{x}^{\mu}=0$

The μ does not really fit, why do you used μ..can't we use {i,j,k}?

psimeson said:
For time, Christoffel Symbol should be:
${\Gamma^t}_{it}$ = $x_i\frac{\phi'(r)}{r}$

But by doing the way you suggest I didn't get the answer

The μ does not really fit, why do you used μ..can't we use {i,j,k}?

My formula gives the same result, so you must be doing something wrong.

You can use absolutely whatever notation you like, as long as you calculate the derivatives right.

I have my solution in the above post I am getting

$\frac{1}{(1+2ϕ)}$ϕ,μ$\dot{x}$μ=0

I think my derivatives are correct

psimeson said:
I have my solution in the above post I am getting

$\frac{1}{(1+2ϕ)}$ϕ,μ$\dot{x}$μ=0

I think my derivatives are correct

$\frac{1}{1+2\phi} \phi_{,\mu} x^{\mu} = -\frac{\partial \phi}{\partial t} \dot{t} + \frac{\partial \phi}{\partial x} \dot{x} + \frac{\partial \phi}{\partial y} \dot{y} + \frac{\partial \phi}{\partial x} \dot{y} + O(\phi^2)$

## 1. What is the Christoffel symbol from Variational Principle?

The Christoffel symbol from Variational Principle is a mathematical concept used in the field of differential geometry to describe the curvature of a space. It is derived from the variational principle, which states that a physical system will follow the path that minimizes its action.

## 2. How is the Christoffel symbol calculated?

The Christoffel symbol is calculated using the metric tensor, which describes the distance between points in a curved space. It is calculated by taking the partial derivatives of the metric tensor and rearranging them into a specific formula.

## 3. What is the significance of the Christoffel symbol?

The Christoffel symbol is significant because it helps us understand the curvature of a space. It allows us to calculate the path that a physical system will follow in a curved space, as well as the rate of change of vectors in that space.

## 4. How is the Christoffel symbol used in physics?

The Christoffel symbol is used in physics to describe the curvature of space and the motion of particles in that space. It is particularly important in the theory of general relativity, where it is used to describe the gravitational field and the motion of objects under the influence of gravity.

## 5. Can the Christoffel symbol be extended to higher dimensions?

Yes, the Christoffel symbol can be extended to spaces with more than three dimensions. In fact, the concept of the Christoffel symbol is an important tool in higher dimensional geometry and is used in theories such as string theory and Kaluza-Klein theory.