1
$\begingroup$

I was reading: https://en.wikipedia.org/wiki/Metric_tensor#Examples

Is it correct that in the polar coordinate example, just after the euclidean metric example, that distance is measured as:

$$ \int_{a}^{b} \sqrt{dr^2 + r^2 d\theta^2} ?$$

And in general given a metric tensor of a space:

$$ G = \begin{bmatrix} g_{00} & g_{01} & ... & g_{0n} \\ g_{10} & g_{11} & ... & g_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ g_{n0} & g_{n1} & ... &g_{nn} \end{bmatrix}$$

With coordinates: $x_1, x_2 ... x_n$, another coordinate system for this space $y_1 ... y_n$, given as $x_i = f_i(y_1 ... y_n)$ along with a matrix

$$ S = \begin{bmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} & ... & \frac{\partial x_1}{\partial y_n} \\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} & ... & \frac{\partial x_2}{\partial y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial x_n}{\partial y_1} & \frac{\partial x_n}{\partial y_2} & ... & \frac{\partial x_n}{\partial y_n}\end{bmatrix} $$

For coordinates $y_1 ... y_n$ then distance is measured as, let $M = S^T G S$

$$ \int_{a}^{b} \sqrt{ \sum_{i,j}^{n} M_{ij} dy_{i} dy_{j} } ?$$

$\endgroup$
1
$\begingroup$

I like to think of it in the coordinate-free notation to remember how it works. If $\gamma(t)$ is your path, going from $t=a$ to $t=b$ (with velocity/tangent vector $\dot{\gamma}(t)$), and if $g$ is the metric then the length of the curve is

$$ \int_a^b \sqrt{g(\dot{\gamma}(t), \dot{\gamma}(t))} \; dt $$

If you want to compute it in different coordinates, just use the pull-back. That is, if $x = \varphi(y)$ changes from the $y$-coordinates to the $x$-coordinates, then the metric expressed in the $y$-coordinates is given by $\varphi^*g$. Then if you want to do this integral in the new coordinates, just make the appropriate substitutions.

In your polar coordinate example, you'd have $\varphi$ given by

$$ x = r\cos(\theta), ~~~ y = r\sin(\theta) $$

The usual Euclidean metric on the plane (in the "$x$"-coordinates) is

$$ g = dx \otimes dx + dy \otimes dy $$

Doing the pull-back $\varphi^*(g)$ gives

$$ \varphi^*g = dr \otimes dr + r^2 d\theta \otimes d\theta $$

So now if you write $\gamma(t)$ in polar coordinates as $\gamma(t) = (r(t), \theta(t))$, with $\dot{\gamma}(t) = (\dot{r}(t), \dot{\theta}(t))$, you get

$$ \int_a^b \sqrt{\dot{r}(t)^2 + r(t)^2 \dot{\theta}(t)^2} \; dt $$

As for your matrix formula: this is correct, and here is how it relates to the above. The metric is a "symmetric bilinear form" on the tangent spaces, and on a vector space, a symmetric bilinear form $\beta(\cdot, \cdot)$ is given by a symmetric $n$-by-$n$ matrix $B$ such that for any vectors $v$ and $w$, you have $\beta(v,w) = \left< v, Bw \right>$, where $\left< \cdot,\cdot \right>$ is the standard Euclidean dot product. In our case, the metric $g$ is represented in coordinates $(x_1,\dots,x_n)$ at every point by a matrix $G = \left(g_{ij}\right)_{i,j=1}^n$. Now suppose at the same point you look at a different system of coordinates $(y_1,\dots,y_n)$, as above, with transition map $\varphi$ going from the $y$-coordinates to the $x$-coordinates, with Jacobian matrix ("push-forward") $\varphi_*$, which is given by the matrix $S$ as in your post. Also, as in your notation, let $M = \left(M_{ij}\right)$ be matrix corresponding to the metric $g$ in the $y$-coordinates. Then your claim is that $M = S^T G S$ is how the metric changes when switching coordinates. This is because of the definition of how "pull-back" works, along with a property of the Euclidean dot product:

On the one hand, in the $y$-coordinates, if we have two vectors $v$ and $w$, then $g(v,w) = \left< v, M w \right>$. On the other hand, using the definition of the pull-back (remember the metric in $y$-coords is $\varphi^* g$, where $g$ is in the $x$-coords): $$ \begin{eqnarray} (\varphi^*g)(v,w) &=& g(\varphi_* v, \varphi_* w) \\ &=& \left< \varphi_* v, G \varphi_* w \right> \\ &=& \left< S v, GS w \right> \\ &=& \left< v, S^TGS w \right> \end{eqnarray} $$

$\endgroup$
  • $\begingroup$ is $\phi$ literally the system of equations converting the two? Also what is $\phi*$? $\endgroup$ – frogeyedpeas Mar 25 '16 at 20:46
  • $\begingroup$ Yes, $\phi(r,\theta) = (r\cos\theta, r\sin\theta)$. The map $\phi^*$ is what is called the "pull-back" of $\phi$. It takes differential forms on the codomain to differential forms on the domain. What it does is substitute the variables: so $dx$ becomes $d(r\cos\theta) = \cos\theta \, dr - r\sin\theta \, d\theta$, and similarly for the $dy$. $\endgroup$ – Nick Mar 26 '16 at 1:53
  • $\begingroup$ Okay so the generalized matrix formula I gave earlier, would that be correct then? $\endgroup$ – frogeyedpeas Mar 26 '16 at 7:46
  • $\begingroup$ Yes. I edited my post. $\endgroup$ – Nick Mar 26 '16 at 14:47

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.