# geodesics and differential equation satisfied

If $\gamma(s) = \mathbf{x}(\gamma^1(s),\gamma^2(s))$ where $\mathbf{x}$ is a coordinate patch, then what is the differential equations that $\gamma^k$ ($k = 1,2$) must satisfy if $\gamma$ is a geodesic?

Note: A unit speed curve $\gamma(s)$ in $M$ is a geodesic iff $[n,T,T'] = 0.$

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What have you tried? Have you tried writing out $T$, $T'$ and $n$ in terms of $\gamma$ and its derivatives? If so, what did you get? –  Jesse Madnick Oct 10 '12 at 6:06
I am only thinking of defining xsub1, and xsub2, but not getting very far in that –  mary Oct 10 '12 at 6:47
By $\mathbf{x}_1$ and $\mathbf{x}_2$, do you mean the partial derivatives? (If so, you're not "defining" them, but computing them.) Hm, I think you should start by computing $T$ and $T'$ (by using the chain rule) and also computing $n$ (maybe there's a formula for it in terms of a cross product?). –  Jesse Madnick Oct 10 '12 at 6:51
Or-- ah, perhaps you mean the components $\mathbf{x}(u,v) = (x_1(u,v), x_2(u,v), x_3(u,v))$? Yes, those you would define. –  Jesse Madnick Oct 10 '12 at 6:53
I mean the components. Can you illustrate it with the components, I don't see how it is going to get messy. I'm used to thinking in one way –  mary Oct 10 '12 at 8:28

I'll get you started.

I'm going to assume that $\gamma$ is a unit-speed curve. I'm also going to use the notation $\dot{f}$ to mean $\frac{df}{ds}$.

Although I could write $\mathbf{x}(u,v) = (x^1(u,v), x^2(u,v), x^3(u,v))$, I'm going to avoid doing that because it would make our calculations even messier.

By the chain rule:

$$T(s) = \gamma'(s) = \mathbf{x}_u\dot{\gamma}^1 + \mathbf{x}_v\dot{\gamma}^2$$

Again by the chain rule (and product rule):

\begin{align*} T'(s) & = \frac{d}{ds}\left[ \mathbf{x}_u\dot{\gamma}^1 + \mathbf{x}_v\dot{\gamma}^2 \right] \\ & = \left[\mathbf{x}_u\ddot{\gamma}^1 + \mathbf{x}_{uu}(\dot{\gamma}^1)^2 + \mathbf{x}_{uv}\dot{\gamma}^1\dot{\gamma}^2\right] + \left[\mathbf{x}_v\ddot{\gamma}^2 + \mathbf{x}_{vv}(\dot{\gamma}^2)^2 + \mathbf{x}_{uv}\dot{\gamma}^1\dot{\gamma}^2\right] \\ & = \mathbf{x}_u\ddot{\gamma}^1 + \mathbf{x}_{uu}(\dot{\gamma}^1)^2 + 2\mathbf{x}_{uv}\dot{\gamma}^1\dot{\gamma}^2 + \mathbf{x}_v\ddot{\gamma}^2 + \mathbf{x}_{vv}(\dot{\gamma}^2)^2. \\ \end{align*}

By multivariable calculus, $$n = \frac{\mathbf{x}_u \times \mathbf{x}_v}{|\mathbf{x}_u \times \mathbf{x}_v|}.$$

To get the differential equation we want, we're going to use the definition of "geodesic" -- namely $[n, T, T'] = 0$. That is, $$(n \times T) \cdot T' = 0.$$ This means that we have to compute $n \times T$, then take its dot product with $T'$, and finally set all of that equal to zero.

I leave the rest to you.

Possibly useful: When computing the cross product, it might be helpful to use the formula $$\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a}\cdot \mathbf{b})\mathbf{c},$$ and to remember that $g_{11} = \mathbf{x}_u \cdot \mathbf{x}_u$, and so on.

When computing the dot product, it might be helpful to remember the Gauss Equations $$\begin{cases} \mathbf{x}_{uu} = \Gamma^1_{11}\mathbf{x}_u + \Gamma^{2}_{11}\mathbf{x}_v + L_{11}n \\ \mathbf{x}_{uv} = \Gamma^1_{12}\mathbf{x}_u + \Gamma^{2}_{12}\mathbf{x}_v + L_{12}n \\ \mathbf{x}_{vv} = \Gamma^1_{22}\mathbf{x}_u + \Gamma^{2}_{22}\mathbf{x}_v + L_{22}n, \\ \end{cases}$$ where the $L_{11}, L_{12}, L_{22}$ are the coefficients of the second fundamental form.

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