1
$\begingroup$

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 $[n,T,T'] = 0.$

  • 0
    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?2012-10-10
  • 0
    I am only thinking of defining xsub1, and xsub2, but not getting very far in that2012-10-10
  • 0
    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?).2012-10-10
  • 1
    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.2012-10-10
  • 0
    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 way2012-10-10
  • 0
    Just replace every instance of $\mathbf{x}_u$ with the column vector $(x^1_u, x^2_u, x^3_u)$ and similarly for the other derivatives. It's not so much that the components make things messier (although they do) so much as that they're unnecessary. I think that including the components makes it difficult to see what the next step should be, especially if you try to actually compute the dot and cross products with all of those components. Better to just manipulate the vectors as individual entities instead.2012-10-10
  • 0
    For instance, you _could_ calculate $n = (\mathbf{x}_u \times \mathbf{x}_v)/|\mathbf{x}_u \times \mathbf{x}_v|$ in terms of components, but I don't think that's a necessary step in this problem.2012-10-10

1 Answers 1

2

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.