Let be $X$ a Riemannian manifold. I fix a local chart $(U,\varphi)$

If $ \phi : [a,b] \rightarrow U$ is $C^\infty$ I consider $$H(\phi )= \int_a^b k(\phi (t),\dot{\phi(t)})\; dt$$ where

$$ k(\phi(t),\dot{\phi(t)})= \sum_{\alpha, \beta} g_{\alpha,\beta}(\phi(t)) \:\dot{\phi}(t)^\alpha\: \dot{\phi}(t)^\beta$$ and $\phi^1(t),\dots, \phi^n(t)$ are the local coordinates of $\phi(t)$ (i.e. $\varphi \circ \phi=(\phi^1, \dots \phi^n)$)

I have that $k$ satisfies the Euler-Lagrange equations: $$ \frac{d}{dt}\big(\frac{\partial k}{\partial \dot{x}^\gamma}\big)-\frac{\partial k}{\partial x^\gamma}=0.$$ I don't understand why I have that:

$$ \frac{\partial k}{\partial x^\gamma}= \sum_{\alpha, \beta} \frac{\partial g_{\alpha,\beta}} {\partial x^ \gamma} \:\dot{\phi}^\alpha\: \dot{\phi}^\beta$$ and $$ \frac{\partial k}{\partial \dot{x}^\gamma}=2 \sum_{\alpha} g_{\alpha,\gamma}\dot{\phi}^\alpha.$$

Why, when I do these derivatives, have I to consider $\phi^j$ as $x^j$ and the $\dot{\phi}^j$ as $\dot{x}^j$ ?

Thanks for the clarification.

  • $\begingroup$ You will understand why if you understand the derivation of the Euler-Lagrange equation. That's the definition of these notation. $\endgroup$ – user99914 Sep 21 '15 at 18:41

In fact, the mistake comes from the definition of $k$: the one that you gave above is correct only up to an abuse of notation (usually tolerated in books, but not rigorous): since $k$ must be defined on the tangent bundle and since you work in a chart (which induces a local trivialization above $U$ of the tangent bundle), you must tell what the value of $k$ is in a point $(x,v) \in T_x U$. According to your formula above, this should be $k(x,v) = \sum \limits _{\alpha, \beta} g_{\alpha, \beta} (x) v^\alpha v^\beta$.

(Many authors use the notation $\dot x$ instead of $v$, but do not make the mistake of (yet) thinking of it as a time derivative, because so far we haven't yet seen any curve (we shall, briefly). With this notation, $k(x,\dot x) = \sum \limits _{\alpha, \beta} g_{\alpha, \beta} (x) \dot x ^\alpha \dot x ^\beta$.)

You obtain, then, that $\frac {\partial k} {\partial v^i} = \sum \limits _\alpha g_{i, \alpha} (x) v^\alpha + \sum \limits _\alpha g_{\alpha, i} (x) v^\alpha = \sum \limits _\alpha (g_{i, \alpha} + g_{\alpha, i}) (x) v^\alpha$. In most cases $g$ will be the Riemannian structure so the above becomes $2 \sum \limits _\alpha g_{i, \alpha} (x) v^\alpha$.

Similarly, $\frac {\partial k} {\partial x^i} = \sum \limits _{\alpha, \beta} \frac {\partial g_{\alpha, \beta}} {\partial x^i} (x) v^\alpha v^\beta$.

Now, evaluate the above along the curve $\phi$, i.e. apply both quantities computed above in the point $\left( \phi (t), \dot \phi (t) \right) \in T_{\phi(t)} X$ - which means to replace $x$ with $\phi(t)$ and $v$ (or $\dot x$, if you prefer this notation) with $\dot \phi (t)$ (thus replacing the dependence on $(x,v)$ with one on $t$), derive $\frac {\partial k} {\partial v^i} \left( (\phi (t), \dot \phi (t) \right)$ with respect to $t$ and equate the results. These are the E-L equations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.