Geodesic equations and christoffel symbols

Question

I want to learn explicitly proof of the proposition 9.2.3.

Which books or lecture notes I can find? Please give me a suggestion. Thank you:)

Answer 1

http://www.math.uh.edu/~minru/4350-11/geodesic.pdf

See pages 3 and 4 of this PDF!

Answer 2

Here's one way to look at this:

Suppose $\nabla$ is the unique covariant derivative operator associated with the metric tensor $g$ on the surface $S$, i.e. it is the unique symmetric connection on $S$ with $\nabla g = 0$. It is well know that the geodesics $\gamma(t)$ of $g$ satisfy

$\nabla_{\mathbf w}\mathbf w = 0, \tag{1}$

where $\mathbf w$ is the tangent vector to the geodesic $\gamma(t) = (u(t), v(t))$; the components of $\mathbf w$ are thus $w^1 = \dot u$ and $w^2 = \dot v$. We also have that

$\nabla_i \mathbf e_j = \Gamma_{ij}^k \mathbf e_k \tag{2}$

for the basis vectors $\mathbf e_1 = \frac{\partial}{\partial u}$ and $\mathbf e_2 = \frac{\partial}{\partial v}$, where we use the shorthand $\nabla_i$ for $\nabla_{\mathbf e_i}$: $\nabla_i \equiv \nabla_{\mathbf e_i}$; furthermore $\Gamma_{ij}^k = \Gamma_{ji}^k$ for $1 \le i, j, k \le 2$; the Einstein summation convention is deployed in writing (2) and throughout this answer. Expanding (1) in terms of the basis $e_i$ we obtain

$0 = \nabla_{\mathbf w}\mathbf w = w^k \nabla_k (w^l \mathbf e_l), \tag{3}$

and we continue with

$0 = w^k \nabla_k (w^l \mathbf e_l) = w^k \nabla_k(w^l) \mathbf e_l + w^l w^k \nabla_k \mathbf e_l = w^k \nabla_k(w^l) \mathbf e_l + w^l w^k \Gamma_{kl}^j \mathbf e_j. \tag{4}$

The coefficient of $\mathbf e_l$ in first term on the right-hand side of (4) is

$w^k \nabla_k(w^l) = \mathbf w[w^l] = \dfrac{dw^l}{dt} = \dot w^l, \tag{5}$

since $\mathbf w \equiv \frac{\partial}{\partial t}$, the derivative, with respect to $t$, along the curve $\gamma(t)$ with

$\dot \gamma(t) = \dot u \frac{\partial}{\partial u} + \dot v \frac{\partial}{\partial v} = w^1\frac{\partial}{\partial u} + w^2 \frac{\partial}{\partial v}; \tag{6}$

we see from (6) that $\dot w^1 = \ddot u$ and $\dot w^2 = \ddot v$. We thus also have, for the second term on the right of (4),

$w^l w^k \Gamma_{kl}^j \mathbf e_j = (\dot u^2 \Gamma_{11}^j + \dot u \dot v \Gamma_{12}^j + \dot v \dot u \Gamma_{21}^j + \dot v^2 \Gamma_{22}^j) \mathbf e_j$ $= (\dot u^2 \Gamma_{11}^j + 2\dot u \dot v \Gamma_{12}^j + \dot v^2 \Gamma_{22}^j) \mathbf e_j, \tag{7}$

where $\Gamma_{kl}^j = \Gamma_{lk}^j$ has been used in (7). If we now substitute (7) and (5) in (4), using $w^1 = \ddot u$ and $w^2 = \ddot v$, and then invoke the linear independence of $\mathbf e_i$ to pick off the individual component equations, we find

$\ddot u + \dot u^2 \Gamma_{11}^1 + 2\dot u \dot v \Gamma_{12}^1 + \dot v^2 \Gamma_{22}^1 = 0, \tag{8}$

with the corresponding, similar equation holding for $v$ with the $\Gamma_{jk}^1$ replaced by the $\Gamma_{jk}^2$.

One can in fact run all this stuff backwards, keeping careful track of the definitions, to derive (1) from (8) and it's companion equation for $v$; that gives you the requested "if and only if" demonstration.

This material is covered in any number of differential geometry books, many of which are mentioned in THW's most helpful comment. I like O'Neill, and I have also found a good source for such things is texts on general relativity, like Misner, Thorne, and Wheeler's Gravitation. They work things out in detail so even the physicists can figure them out! ;-) Very helpful indeed!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!