# Another christoffel symbols from the metric question

• Ai52487963
Rightarrow \frac{d}{d \eta} ( -2A(r) \dot{t} ) = -A'(r) \dot{t}^2 + B'(r) \dot{r}^2\Rightarrow -2A'(r) \dot{t} \dot{r} -2A(r) \ddot{t} = -A'(r) \dot{t}^2 + B'(r) \dot{r}^2\Rightarrow -2A(r) \ddot{t} + 2A'(r) \dot{t} \dot{r} + A'(r) \dot{t}^2 - B'(
Ai52487963
Another "christoffel symbols from the metric" question

## Homework Statement

Find the Christoffel symbols from the metric:

$$ds^2 = -A(r)dt^2 + B(r)dr^2$$

## Homework Equations

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x^a}} \right) = \frac{\partial L}{\partial x^a}$$

## The Attempt at a Solution

Pretty straight forward, but I'm not seeing where one of the cross-terms comes from. I used the euler-lagrange eqns to rearrange for the equation of motion so I could just read the christoffel symbols off:

the t-component of the E-L eqns is zero, so moving on to the r...

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = \frac{\partial L}{\partial r}$$

$$\Rightarrow \frac{d}{dt} \left( 2B(r)\dot{r} \right) = -A'(r)\dot{t}^2 + B'(r)\dot{r}^2$$

$$\Rightarrow 2B'(r)\dot{r} + 2B(r)\ddot{r} = -A'(r)\dot{t}^2 + B'(r)\dot{r}^2$$

$$\Rightarrow 2B(r)\ddot{r} + 2B'(r)\dot{r} - B'(r)\dot{r}^2 + A'(r)\dot{t}^2 = 0$$

$$\Rightarrow \ddot{r} + \frac{B'(r)}{B(r)}\dot{r} - \frac{B'(r)}{2B(r)}\dot{r}^2 + \frac{A'(r)}{2B(r)}\dot{t}^2 = 0$$

But I'm not seeing how the answer given tells us that

$$\Gamma^r_{tt} = \frac{A'(r)}{2B(r)}$$
$$\Gamma^r_{tr} = \frac{A'(r)}{2A(r)}$$
$$\Gamma^r_{rr} = \frac{B'(r)}{2B(r)}$$

From my EOM's it looks like I have the r_rr and r_tt symbols, but I'm not seeing where this r_tr symbol is supposed to come from. I have an extraneous rdot floating around that doesn't have a corresponding tdot , but even if it did, there has to be an A' in the numerator of that factor, so what gives? Where'd I go wrong?

t is a coordinate. Just like r. Don't presume that the "dot" means derivative with respect to t. Write your Euler-Lagrange equation with "dot" meaning derivative with respect to some arbitrary parameter $\eta$:

$\frac{d}{d \eta} \left( \frac{\partial L}{\partial \dot{x^a}} \right) = \frac{\partial L}{\partial x^a}$

Last edited:

Dick said:
t is a coordinate. Just like r. Don't presume that the "dot" means derivative with respect to t. Write your Euler-Lagrange equation with "dot" meaning derivative with respect to some arbitrary parameter $\eta$:

$\frac{d}{d \eta} \left( \frac{\partial L}{\partial \dot{x^a}} \right) = \frac{\partial L}{\partial x^a}$

I agree and it makes sense that I should have a tdot * rdot cross-term, but I'm not seeing where this phantom A' is coming from in that cross-term.

If the partial of L with respect to rdot is really a partial, then it shouldn't act on the -A(r) term in the first part of the lagrangian:

$$L = -A\dot{t}^2 + B\dot{r}^2$$

applying partial wrt rdot:

$$\frac{\partial L}{\partial \dot{r}} = 2B\dot{r}$$

and lacking any A's in the numerator or denominator of the soon-to-be cross term

$$\frac{d}{dt} (2B \dot{r} ) = -A'\dot{t}^2 + B'\dot{r}^2$$

$$\Rightarrow 2B' \dot{t}\dot{r} + 2B\ddot{r}= -A'\dot{t}^2 + B'\dot{r}^2$$

$$\Rightarrow 2B \ddot{r} + 2B'\dot{t}\dot{r} + A'\dot{t}^2 - B'\dot{r}^2 = 0$$

$$\Rightarrow \ddot{r} + \frac{B'}{2B}\dot{t}\dot{r} + \frac{A'}{2B} \dot{t}^2 - \frac{B'}{2B}\dot{r}^2 = 0$$

gives us the Christoffel symbols:

$$\Gamma^r_{tr} = \Gamma^r_{rt} = \frac{B'}{2B}$$
$$\Gamma^r_{tt} = \frac{A'}{2B}$$
$$\Gamma^r_{rr} = \frac{B'}{2B}$$

I'm thinking the mixed term has to be a typo, I can't see what I did wrong otherwise.

You are making some mistakes there. E.g. $\frac{d}{d \eta} (2B \dot{r} ) = 2B(r) \ddot{r} + 2 B'(r) {\dot r}^2$ There is no single $\dot r$ term. And the mixed term is coming from the other EL equation:
$$\frac{d}{d \eta} \left( \frac{\partial L}{\partial \dot{t}} \right) = \frac{\partial L}{\partial t}$$

I appreciate your attempt at finding the Christoffel symbols using the Euler-Lagrange equations. However, I believe you may have made a mistake in your calculation. The Christoffel symbols can be derived from the metric using the formula:
\Gamma^a_{bc} = \frac{1}{2}g^{ad}\left(\frac{\partial g_{db}}{\partial x^c} + \frac{\partial g_{dc}}{\partial x^b} - \frac{\partial g_{bc}}{\partial x^d}\right)

In your case, plugging in the given metric, we get:

\Gamma^t_{tt} = \frac{1}{2}\left(\frac{-A'(r)}{A(r)} + \frac{-A'(r)}{A(r)} - \frac{0}{A(r)}\right) = \frac{-A'(r)}{A(r)}

\Gamma^r_{tr} = \frac{1}{2}\left(\frac{B'(r)}{B(r)} + \frac{0}{B(r)} - \frac{A'(r)}{B(r)}\right) = \frac{A'(r)}{2B(r)}

\Gamma^r_{rr} = \frac{1}{2}\left(\frac{0}{B(r)} + \frac{B'(r)}{B(r)} - \frac{B'(r)}{B(r)}\right) = 0

I hope this clarifies where the r_tr symbol comes from and helps you in your future calculations. Good luck!

## 1. What are Christoffel symbols?

Christoffel symbols are mathematical objects that help describe the curvature of a surface or space. They are used in the study of differential geometry and general relativity.

## 2. How are Christoffel symbols related to the metric?

The Christoffel symbols are derived from the metric tensor, which describes the distance between points in a space. The metric tensor contains information about the curvature of the space and the Christoffel symbols help quantify this curvature.

## 3. Can Christoffel symbols be calculated for any metric?

Yes, Christoffel symbols can be calculated for any metric. However, the calculation can be complex and may require advanced mathematical techniques such as tensor calculus.

## 4. What is the significance of Christoffel symbols in physics?

Christoffel symbols are used in the field of general relativity to describe the curvature of space-time and the motion of objects in a gravitational field. They are also used in other areas of physics, such as in the study of black holes and cosmology.

## 5. How are Christoffel symbols used in practical applications?

Christoffel symbols are used in many practical applications, such as in GPS technology, where they help calculate the shortest path between two points on a curved surface. They are also used in computer graphics to create realistic 3D images and in engineering to model the stress and strain in materials.

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