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I'd like to prove that E-critical curves of the energy functional: $$ E(\eta) = \frac 12 \int_I g(\eta'(t), \eta'(t))dt$$ satisfy the geodesic equation: $$\nabla_{\eta'} \eta' = 0$$


So the geodesic equation is as follows:$$\nabla_{v} u = \nabla_{v^{i}e_{i}} u^{j}e_{j} = v^{i}\nabla_{e_{i}} u^{j}e_{j} = v^{i}u^{j}\nabla_{e_{i}} e_{j} + v^{i}e_{j}\nabla_{e_{i}} u^{j} = v^{i}u^{j}\Gamma^{k}_{ij}e_{k} + v^{i}\frac{du^{j}}{dx^{i}}e_{j}$$ Which I've also seen as: $$\frac{d^{2}x^{k}}{ds^{2}} + \Gamma^{k}_{ij} \frac{dx^{i}}{ds} \frac{dx^{j}}{ds^{2}}$$ with: $\Gamma^{k}_{ij} = \frac 12 g^{kl}(\frac{dg_{jl}}{dx^{i}} + \frac{dg_{il}}{dx^{j}} - \frac{dg_{ij}}{dx^{l}})$


My intuition tells me that...

I. First we need to show the curves, $\eta$, satisfy the Euler-Lagrange equation and thus are critical curves

II. Then we show that those curves satisfy the geodesic equation. Correct?


If this is correct then I just need some help with...

i.) When we say $\eta$ is a critical curve what does that tell us about $\eta$ that is useful in solving the geodesic equation? I mean we can just say $\eta$ is an arbitrary function that solves the EL equations but since we're doing this in general we need not give an explicit form for $\eta$. I'm just confused on how to proceed.

ii. I actually solving the geodesic equation I could use a little assistance too. Since we are on an arbitrary (wrt dimension specifically) manifold we can't say how many components of $\eta$ there are or how many basis vectors there are so there must be other properties endowed to $\eta$ (perhaps based on being a critical curve of the energy functional) which suggest to us that say it's 2nd deriv is equal to zero for all components or the connection on the manifold (cristoffel symbols) are always zero for this reason or that reason. Any help is GREATLY appreciated.

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  • $\begingroup$ When I was TA'ing an undergraduate DG course here I wrote a handout about this. It's in portuguese and I did the computation for surfaces in $\Bbb R^3$ but I think you'll manage to understand the gist of it. Page 6 here $\endgroup$ – Ivo Terek May 7 '17 at 23:21
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    $\begingroup$ I see.. I actually took a bit of Spanish in HS and so I can interpret (with the help of our universal language of math, of course) a bit - it's quite fun actually! Thank you I just wanted some clarification: 1. It looks like you've used $u^{i}(t)$ and $u^{i}(t)$(dot) for $\eta(t)$ and $\eta'(t)$ (in my problem) and the problem doesn't specify anything so I'm guessing it's $\mathbb{R}^{n}$. So would I basically need to perform the derivation for $g_{ij} = g_{ij}(\eta^{1}(t),...,\eta^{n}(t))$ with indices on the sums running from 1 to n? $\endgroup$ – Kyle O'Connell May 8 '17 at 1:15
  • $\begingroup$ And of course I' talking about the part where you derive the geodesic equation from the Euler-Lagrange Equation -- Teorema 3. $\endgroup$ – Kyle O'Connell May 8 '17 at 1:16
  • $\begingroup$ I'm glad it wasn't totally useless! If ${\bf x}$ is a parametrization of the submanifold in which the curve lies, the $u^i$ are the curve's coordinates in that parametrization, as in $\eta(t) = {\bf x}(u^1(t),\cdots,u^k(t))$. Your $\eta$ here would be my $\alpha$ there. $\endgroup$ – Ivo Terek May 8 '17 at 1:20
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    $\begingroup$ Excellent that makes perfect sense. I'll enjoy learning a little Portuguese from the mathematical context, as well! Thank you again sir. $\endgroup$ – Kyle O'Connell May 8 '17 at 1:53
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Let $M^n$ be a differentiable n-manifold and $\Gamma = \{\gamma\in C^\infty([a,b], M^n)| \gamma(a) = p_0 \ \gamma(b) = p_1 \}$ be the family of curves stating from $p_0$ and ending at $p_1$.

Given a smooth function $L: TM \to \mathbb{R} $ we can define the Lagrangian Action functional $A_L:\Gamma \to \mathbb{R}$ as follows $A_L(\gamma) = \int_a^b L(\gamma(t), \dot{\gamma}(t))dt $. If we put $L(q,\dot{q})= \frac 1 2 g_{i j}(q) \dot{q}^i\dot{q}^j $ we have that $A_L = E$ the Energy ($g $ is the Riemmanian metric).

I) A critical point of the Energy satisfies the Euler-Lagrange equations.

For simplicity assume that $M = \mathbb{R}^n$, (in general it is not sufficient to consider a chart since the curves can "exit" from the chart, but I prefer to do this way since the Euclidean situation is quicker to understand) $\gamma \in \Gamma$ is critical for $A_L$ if

$\forall \eta \in \Gamma_0 = \{\eta \in C^\infty([a,b], \mathbb{R}^n)| \eta(a) = 0 \ \eta(b) = 0 \}$
$\frac d {ds}A_L(\gamma+ s\eta)|_{s=0} = 0$ (In general one should replace $\Gamma_0$ with the tangent space at $\gamma$ in $\Gamma$ which is a Banach manifold and require that $D_\gamma A_L(\eta) = 0\ \forall \eta \ \in T_\gamma \Gamma $).

This condition is equivalent to the Euler Lagrange equations. $\frac d {ds}A_L(\gamma+ s\eta)|_{s=0} = \int_a^b \frac d {ds}L(\gamma(t) + s \eta(t), \dot{\gamma} + s \dot{\eta}(t))|_{s=0}dt $ $=\int_a^b L_q(\gamma(t)) \cdot \eta(t) + L_\dot{q}(\gamma(t))\cdot \dot{\eta}(t) dt $ and integrating by parts $=\int_a^b (L_q(\gamma(t)) - \frac d {dt} L_\dot{q}(\gamma(t))) \cdot {\eta}(t) dt $

By the fundamental Lemma of Calculus of Variations, the latter expression is zero for any $\eta \in \Gamma_0$ iif $L_q(\gamma(t)) - \frac d {dt} L_\dot{q}(\gamma(t)) = 0$ which is exactly the Euler-Lagrange equation.

II) If $\gamma$ satisfies the Euler-Lagrange equation for $L(q,\dot{q})= \frac 1 2 g_{i j}(q) \dot{q}^i\dot{q}^j $ then $\gamma$ is a geodesic (using the Levi-Cìvita connection).

First we rewrite the E-L equation.

$\frac d {dt} L_{\dot{q}^i} = \frac d {dt} (g_{ij} \dot{q}^j) = g_{ij} \ddot{q}^j + \frac {\partial g_{ij}}{\partial q^k} \dot{q}^k\dot{q}^j$

$\frac{\partial L}{\partial q^i} = \frac 1 2 \frac {\partial g_{j k}}{\partial q^i}\dot{q}^j \dot{q}^k$

thus the E-L equation is equivalent to

$g_{ij} \ddot{q}^j = (\frac 1 2 \frac {\partial g_{j k}}{\partial q^i} - \frac {\partial g_{ij}}{\partial q^k})\dot{q}^j \dot{q}^k$

Now multiply the above equation by $g^{r i}$ and relabel the indexes to obtain:

$(1)\quad \ddot{q}^i = g^{r i}(\frac 1 2 \frac {\partial g_{j k}}{\partial q^r} - \frac {\partial g_{rj}}{\partial q^k})\dot{q}^j \dot{q}^k$.

On the other hand we have the geodesics equations:

$(2)\quad \ddot{q}^i = - \Gamma_{j k}^i \dot{q}^j \dot{q}^k $ where $\Gamma_{j k}^i = \frac 1 2 g^{i r}(\partial_j g_{r k} + \partial_k g_{r j} - \partial_r g_{j k})$ for the Levi-Civita connection.

Therefore $\Gamma_{j k}^i \dot{q}^j \dot{q}^k = \frac 1 2 g^{i r}(\partial_j g_{r k} + \partial_k g_{r j} - \partial_r g_{j k})\dot{q}^j \dot{q}^k = g^{i r}(\partial_j g_{r k} - \frac 1 2 \partial_r g_{j k} ) \dot{q}^j \dot{q}^k$

(Since we are summing over $i,j$ then by the symmetry of $\dot{q}^j \dot{q}^k$ we have $(\partial_j g_{r k} + \partial_k g_{r j})\dot{q}^j \dot{q}^k = 2 \partial_j g_{r k}\dot{q}^j \dot{q}^k$ )

Thus the E-L equations $(1)$ is the same as the geodesics equations $(2)$ (remember that $g^{i r} = g^{r i}$ since the metric tensor is symmetric).

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