## Tuesday, February 21, 2017

### Geodesic on a Plane in Polar Coordinates

Geodesic on a Plane in Polar Coordinates This is a simple problem in the calculus of variations where we prove that the shortest distance between two points is a straight line. But, instead of the normal Cartesian coordinate system, we will use polar coordinates $$\left( r, \phi \right)$$. Some people use $$\theta$$ for the angle but to be consistent with the cylindrical and spherical coordinate nomenclature, I choose to use $$\phi$$. I was somewhat stumped at the end because there is a difficult integral to evaluate but we'll cross that bridge when we get to it.

In typical fashion, we wish to minimize the functional $$I = \int_a^b ds$$ And the first thing we need to do is figure out what $$ds$$ is. Looking at the figure below,

we can see that the differential length of the curve is the hypotenuse of the right triangle (well, not quite a triangle but it approaches one as $$d\phi \rightarrow 0$$) and is given by $$ds = \sqrt{dr^2 + (rd\phi)^2}$$ In order to get something we can integrate, we have a decision to make: we need to decide which differential to factor out of the radical and integrate by either $$dr$$ or $$d\phi$$. Do we go with $$ds = \sqrt{\left(\frac{dr}{d\phi}\right)^2 + r^2} d\phi$$ or $$ds = \sqrt{1 + (r\frac{d\phi}{dr})^2} dr$$ In order to decide, we need to look at the Euler-Lagrange equation knowing that the $$F$$ in the equation is the integrand in our functional. $$F = F(x,y,y') : \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) - \frac{\partial F}{\partial y} = 0$$ The equation becomes significantly simpler to solve if we can choose an integrand that is not a function of the middle argument, i.e. a function only of the independent variable and the derivative. If this is the case, then $$\frac{\partial F}{\partial y} = 0$$ leaving the first term to be 0 as well. If the first term is 0, then inner partial derivative must be a constant because the derivative of a constant is 0. $$\frac{\partial F}{\partial y'} = K$$ In our case,that leads us to use the second equation $$ds = \sqrt{1 + (r\frac{d\phi}{dr})^2} dr = \sqrt{1 + (r\phi')^2} dr$$ because the first form contains a dependency on $$r$$ as well as $$r'$$.

So we now have to solve $$F = F(r,\phi,\phi') : \frac{\partial F}{\partial \phi'} = \frac{\partial}{\partial \phi'}\left(\sqrt{1 + (r\phi')^2}\right) = K$$ Taking the partial derivative with respect to $$\phi'$$, we have $${{r^2\phi'} \over \sqrt{1 + r^2\phi'^2}} = K$$ Squaring both sides gives $${{r^4\phi'^2} \over {1 + r^2\phi'^2}} = K^2$$ Then $${r^4\phi'^2} = K^2 \left( 1 + r^2\phi'^2 \right)$$ and solving for $$\phi'2$$ gives us $$\phi'^2 = \frac{K^2}{r^2\left( r^2 - K^2 \right)}$$ Taking the square root and moving $$dr$$ to the right side, gives $$d\phi = \frac{K}{r\sqrt{ r^2 - K^2 }}dr$$ So, here is where it got tricky for me. If you type the above integral into Wolfram-Alpha you'll get something ridiculously complicated. If you're smarter than me, you might be able to simplify it but I had to look at this integral for a while before I came up with an answer. The trick with this stuff is knowing what spell to cast.

Whenever you see square-roots with squares and quotients, think right-triangles.

You can see that the $$\frac{K}{\sqrt{r^2 - K^2}}$$ term is $$\tan \alpha$$. So, we make that substitution giving us $$d\phi = \frac{\tan \alpha}{r} dr$$ Now, what do we do about the $$r$$? Well, looking at the triangle, $$\sin \alpha = \frac{K}{r}$$ This also means that $$r = K\csc \alpha$$ so $$dr = -K\cot(\alpha) \csc(\alpha) d\alpha$$ Making those substitutions leaves us with $$d\phi = \frac{1}{K}\frac{K}{r}\tan(\alpha)dr = -\frac{1}{K}\sin(\alpha)\tan(\alpha)K\cot(\alpha)\csc(\alpha)d\alpha$$ And lo and behold, the entire thing becomes $$d\phi = -d\alpha$$ which is something even I can solve. $$\phi = -\alpha + C$$ Solving for $$\alpha$$ in terms of $$r$$, $$\sin(\alpha) = \frac{K}{r} \Rightarrow \alpha = \sin^{-1}(\frac{K}{r})$$ Plugging this back into our result gives us $$\phi = -\sin^{-1}\left(\frac{K}{r}\right) + C$$ $$\sin(C-\phi) = \frac{K}{r}$$ $$r\sin(C-\phi) = K$$ If you don't recognize the above as a line in polar coordinates, we can go one step further using the sin-of-the-sum-of-angles identity $$r\left(\sin(C)\cos(\phi)-\cos(C)\sin(\phi)\right) = K$$ and gathering up constants, we are left with $$Ar\cos(\phi) + Br\sin(\phi) = K$$ and since $$x = r\cos(\phi), y = r\sin(\phi)$$ are the transformations for polar to Cartesian coordinates $$Ax + By = K$$ which is the general form of a line in Cartesian coordinates.

It's always satisfying to see something work out.

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