# Arclength formula under the metric tensor on polar coordinates

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 $M = S^T G S$

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

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): $$\begin{eqnarray} (\varphi^*g)(v,w) &=& g(\varphi_* v, \varphi_* w) \\ &=& \left< \varphi_* v, G \varphi_* w \right> \\ &=& \left< S v, GS w \right> \\ &=& \left< v, S^TGS w \right> \end{eqnarray}$$

• is $\phi$ literally the system of equations converting the two? Also what is $\phi*$? – frogeyedpeas Mar 25 '16 at 20:46
• 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 $dy$. – Nick Mar 26 '16 at 1:53
• Okay so the generalized matrix formula I gave earlier, would that be correct then? – frogeyedpeas Mar 26 '16 at 7:46
• Yes. I edited my post. – Nick Mar 26 '16 at 14:47