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Let be $(M,g)$ a connected Riemannian manifold and $p,q \in M$.

If $ \phi : [a,b] \rightarrow M$ is $C^\infty$ we define the arc-length of the curve $\phi$ as the quantity:

$$J(\phi )= \int_a^b f(\phi (t),\dot{\phi(t)})\; dt$$ where

$$ f(\phi(t),\dot{\phi(t)})^2= \sum_{\alpha, \beta} g_{\alpha,\beta}(\phi(t)) \:\dot{\phi}(t)^\alpha\: \dot{\phi}(t)^\beta$$ and $\phi^1(t),\dots, \phi^n(t)$ are the local coordinates of $\phi(t)$.

I want to prove that $f(\phi(t),\dot{\phi(t)})$is independent from the choice of the coordinates in a neighborhood of $\phi(t)$. In my text book it is said that this is an obvious fact. Why?

I try to do that substituting in the summation the relation between $g_{j,\, \alpha, \beta} $ and $g_{k,\, \alpha, \beta}$ and the relations between $\frac{\partial}{\partial x_j}$ and $\frac{\partial}{\partial X_k}$ involving the Jacobian matrix (the rules of changing of coordinates for tensors in components) but I didn't succeed in proving the above claim .

Is there a more direct way to check this? Thanks!

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    $\begingroup$ Are you aware that vector fields and the metric are tensors and therefore have special rules for how they change under a change of coordinates? $\endgroup$ – muaddib Sep 3 '15 at 12:41
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That's a bad definition of the length, and it is its badness which is causing you problems.

Given $\sigma:[a,b]\to M$, for each $t\in[a,b]$ essentially by definition $\sigma'(t)$ is an element of the tangent space $T_{\sigma(t)}M$ and therefore we can compute $g(\sigma'(t),\sigma'(t))$, which is a number. This defines a function $f:t\in[a,b]\mapsto g(\sigma'(t),\sigma'(t))\in\mathbb R$. This function is obviously well-defined in the sense that it depends only on $(M,g)$ and on the curve~$\sigma$.

Next check, using coodinates, that it is continuous (smooth, even; in fact, you have almost already done this) and define the length of $\sigma$ as its integral. When done in this way your question does not even arise.

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  • $\begingroup$ Thank you, I've now understood better the definition and I have found the mistake in the calculations that I had done before. I try to prove that my definition is well defined using the rules for vectors. If this relation is true I have finished: $$\sum_{\alpha,\beta} \frac{\partial x_j^\gamma}{\partial x_k^\alpha}\frac{\partial x_j^\delta}{\partial x_k^\beta}= \sum_{\alpha,\beta} \frac{\partial x_k^\alpha}{\partial x_j^\gamma}\frac{\partial x_k^\beta}{\partial x_j^\delta}$$. Is it true? Thank you very much. $\endgroup$ – Gggl. Sep 15 '15 at 8:41
  • $\begingroup$ You should explain what the notation means, as it is otherwise impossible to know if it os correct or not! $\endgroup$ – Mariano Suárez-Álvarez Sep 15 '15 at 9:25
  • $\begingroup$ I consider the two local systems of coordinates $\varphi_j=(x_j^1,...,x_j^n)$ and $\varphi_k=(x_k^1,...,x_k^n)$. $\endgroup$ – Gggl. Sep 15 '15 at 9:32
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    $\begingroup$ Consider the example where $n=1$. Does your equality hold there? $\endgroup$ – Mariano Suárez-Álvarez Sep 15 '15 at 9:45
  • $\begingroup$ no, it is not true... $\endgroup$ – Gggl. Sep 15 '15 at 9:47
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Consider two coordinates $\phi,\ \psi $ where $$ d\phi e_i= E_i,\ d\psi e_i =F_i $$

We have a map $$ E_i = t_{ji} F_j $$

If $c$ is a curve on $M$, let $$ c'(t) =\sum_i A_i E_i =\sum_\alpha B_\alpha F_\alpha $$ so that $$ A_i t_{\alpha i} =B_\alpha $$

Hence $$ |c'(t)|^2 = g_{ij} A_i A_j = g( t_{\alpha i} F_\alpha, t_{\beta j} F_\beta) A_iA_j = g_{\alpha\beta} t_{\alpha i } A_i t_{\beta j} A_j =g_{\alpha\beta} B_\alpha B_\beta$$

Hence $$ L = \int_a^b (g_{ij} A_i A_j)^\frac{1}{2} dt = \int_a^b ( g_{\alpha\beta } B_\alpha B_\beta )^\frac{1}{2} dt $$

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