Christoffel symbol from Variational Principle

[psimeson]

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?

Answer should be:

Γ^t_ti=ϕ,_i

Γⁱ₀₀=ϕ,_i,Γⁱ_jk=δjkϕ,i−δijϕ,k−δikϕ,j

 

[clamtrox]

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.

 

[psimeson]

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.

 

[clamtrox]

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

 

[psimeson]

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)

 

[clamtrox]

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 x^μ 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)

...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.

 

[psimeson]

For time, Christoffel Symbol should be:
{\Gamma^t}_{it} = x_i\frac{\phi&#039;(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}?

 

[clamtrox]

For time, Christoffel Symbol should be:
{\Gamma^t}_{it} = x_i\frac{\phi&#039;(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.

 

[psimeson]

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

 

[clamtrox]

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)