# Einstein Tensor - Particle at rest?

Tags:
1. Feb 15, 2015

### unscientific

1. The problem statement, all variables and given/known data

(a)Find Christoffel symbols
(b) Show the particles are at rest, hence $t= \tau$. Find the Ricci tensors
(c) Find zeroth component of Einstein Tensor

2. Relevant equations

3. The attempt at a solution

Part (a)

Let lagrangian be:
$$-c^2 \left( \frac{dt}{d\tau}\right)^2 + a^2 \left[ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2 + \left(\frac{dz}{d\tau}\right)^2 \right]$$

Euler-Lagrangian is given by:
$$\frac{d}{d\tau} \left( \frac{\partial L}{\partial (\frac{\partial x^{\gamma}}{\partial \tau})} \right) = \frac{\partial L}{\partial x^{\gamma}}$$

Applying Euler-Lagrangian to temporal part:
$$\frac{d}{d\tau} \left( -2c^2 \frac{dt}{d\tau} \right) = 2 a \dot a \left[ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2 + \left(\frac{dz}{d\tau}\right)^2 \right]$$

$$\frac{d^2 (ct)}{d\tau^2} + \frac{a \dot a}{c} \left[ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2 + \left(\frac{dz}{d\tau}\right)^2 \right] = 0$$

This implies that $\Gamma_{11}^0 = \Gamma_{22}^0 = \Gamma_{33}^0 = \frac{a \dot a}{c}$.

Applying Euler-Lagrangian to spatial part:
$$\frac{d}{d\tau} \left( 2a^2 \frac{dx^{\gamma}}{d\tau} \right) = 0$$
$$2 a \dot a \frac{dt}{d\tau} \frac{dx^{\gamma}}{d\tau} + a^2 \frac{d^2 x^{\gamma}}{d\tau^2} = 0$$
$$\frac{d^2 x^{\gamma}}{d\tau^2} + \frac{2 \dot a}{a c} \frac{dx^{\gamma}}{d\tau} \frac{d (ct) }{d\tau} = 0$$

This implies that $\Gamma_{10}^1 = \Gamma_{20}^2 = \Gamma_{30}^3 = \frac{2 \dot a}{ac}$. Am I missing a factor of $\frac{1}{2}$ somewhere or is the question wrong?

Part (b)

I can't see why the particle is at rest in this frame. From the transport equation, the change in vector $V^{\mu}$ when transported through length $\delta x^{\beta}$ is:

$$\delta V^{\mu} = -\Gamma_{\alpha \beta}^{\mu} V^{\alpha} \delta x^{\beta}$$

Letting $\delta x^{\beta}$ represent the time component and $\alpha$ represent the spatial component:

$$\frac{\delta V^{\mu}}{\delta x^{\beta}} = - \Gamma_{\alpha \beta}^{\mu} V^{\alpha}$$

The LHS represents 'velocity' while the right hand side represents $\int \hat r dt$. The Christoffel symbols are clearly non-zero: $\Gamma_{10}^1 = \Gamma_{20}^2 = \Gamma_{30}^3 = \frac{2 \dot a}{ac}$.

Last edited: Feb 15, 2015
2. Feb 15, 2015

### TSny

Hint: Note that both $\Gamma_{10}^1$ and $\Gamma_{01}^1$ occur in the equations of motion $\frac{d^2 x^1}{d\tau^2} + \Gamma_{\mu \nu}^1 \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = 0$ .
You need to show that a free particle that is initially at rest will remain at rest in this coordinate system. So, you need to show that when a particle is released at rest it's spatial acceleration $\frac{d^2x^k}{dt^2} = 0$. See if you can show this from the equations of motion $\frac{d^2 x^{\gamma}}{d\tau^2} + \Gamma_{\mu \nu}^\gamma \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = 0$ and the known values of the Christoffel symbols.

3. Feb 15, 2015

### unscientific

Ah I see, so $\frac{d^2 x^1}{d\tau^2} + \Gamma_{\mu \nu}^1 \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = \frac{d^2 x^1}{d\tau^2} + \Gamma_{01}^1 \frac{dx^0}{d\tau} \frac{dx^1}{d\tau} + \Gamma_{10}^1 \frac{dx^1}{d\tau} \frac{dx^0}{d\tau}$

I have to show that the 'acceleration' is 0 by summing the RHS up and show it equals 0. Do I only sum over the spatial parts?

$$\frac{d^2 x^{\gamma}}{d\tau^2} = - \Gamma_{\mu \nu}^\gamma \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}$$

Spatial parts and temporal parts are:
$$\left( 3 \times a^2 \frac{\dot a}{ac} \right) - \left( 3 \times \frac{a \dot a}{c} \right) = 0$$

Why do I need to multiply the metric components in order to make this right?

Last edited: Feb 15, 2015
4. Feb 15, 2015

### TSny

If the particle starts at rest at time t = 0, then the particle will remain at rest if you can show that all derivatives $\frac{d^nx^k}{dt^n} = 0$ at $t = 0$. Here $k =$ 1, 2, or 3.

You can use your equation $\frac{d^2 x^{\gamma}}{d\tau^2} = - \Gamma_{\mu \nu}^\gamma \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}$ to first find an expression for $\frac{d^2 x^k}{d \tau^2}$. Then see if you can argue that $\frac{d^2 x^k}{dt^2} = 0$ at the instant the particle is released. Then try to show $\frac{d^3x^k}{dt^3} = 0$, etc.

To relate derivatives with respect to $\tau$ to derivatives with respect to $t$, it will be helpful to write out $\frac{d^2 x^{\gamma}}{d\tau^2} = - \Gamma_{\mu \nu}^\gamma \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}$ for $\gamma = 0$ and keep in mind that we are considering the instant when the particle is released from rest.

I'm not claiming that the above procedure is the quickest way to the result. It's just a way that I think will get you there.

5. Feb 15, 2015

### unscientific

Analyzing the x-component:

$$\frac{d^2 x}{d\tau^2} = - \frac{\dot a}{ac} \frac{dx}{d\tau} \frac{dt}{d\tau} = - \frac{\dot a}{ac} \dot x \left( \frac{dt}{d\tau} \right)^2$$

The only way this is zero is if either $\frac{dx}{d\tau} = \dot x \frac{dt}{d\tau} = 0$ or $\frac{dt}{d\tau} = 0$.

6. Feb 15, 2015

### TSny

I believe there is a missing factor of 2. Your equation for $\frac{d^2 x^{\gamma}}{d\tau^2}$ in post #1 was correct. (But your conclusion that the $\Gamma_{10}^1$, etc., should have a factor of 2 was incorrect.)

We are considering the instant of time when the particle is released at rest in this coordinate system. So, you know the value of $\dot x$ at this instant.

7. Feb 15, 2015

### unscientific

So we say that if $\dot x = 0$, then $\frac{d^2 x}{dt^2} = 0$. Then to show for higher powers, simply differentiate the RHS which leads to more powers of $\dot (x)^n$ which are all zero.

8. Feb 15, 2015

### TSny

Yes. So, if $\dot{x} = 0$ at $t = 0$, then you get $\frac{d^2x}{d\tau^2} = 0$ at $t = 0$. You then want to check that this implies $\frac{d^2x}{dt^2} = 0$ at $t = 0$.