An alternative method of finding the stationary path for the...

An alternative method of finding the stationary path for the area from equations 12.42 is to use the arc length, $s$, as the independent variable, which is related to the parameter $t$ by the relation $$ \frac{d s}{d t}=\sqrt{x^2+\dot{y}^2} $$ (a) Show that with $s$ as the independent variable equations 12.42 become $$ \lambda \frac{d x}{d s}=\alpha-y \quad \text { and } \quad \lambda \frac{d y}{d s}=\beta+x $$ Further, show that these equations can be converted to $$ \lambda^2 \frac{d^2 y}{d s^2}+y=\alpha \text { having the general solutions }\left\{\begin{aligned} y & =\alpha+a \cos (s / \lambda+\gamma) \\ x & =-\beta-a \sin (s / \lambda+\gamma) \end{aligned}\right. $$ where $a$ and $\gamma$ are constants. (b) Show that $2 \pi \lambda=L$, derive equations 12.41 and deduce that $2 \pi a=L$.

An alternative method of finding the stationary path for the area from equations 12.42 is to use the arc length, $s$, as the independent variable, which is related to the parameter $t$ by the relation

$$
\frac{d s}{d t}=\sqrt{x^2+\dot{y}^2}
$$

(a) Show that with $s$ as the independent variable equations 12.42 become

$$
\lambda \frac{d x}{d s}=\alpha-y \quad \text { and } \quad \lambda \frac{d y}{d s}=\beta+x
$$

Further, show that these equations can be converted to

$$
\lambda^2 \frac{d^2 y}{d s^2}+y=\alpha \text { having the general solutions }\left\{\begin{aligned}
y & =\alpha+a \cos (s / \lambda+\gamma) \\
x & =-\beta-a \sin (s / \lambda+\gamma)
\end{aligned}\right.
$$

where $a$ and $\gamma$ are constants.
(b) Show that $2 \pi \lambda=L$, derive equations 12.41 and deduce that $2 \pi a=L$.

Transcript

00:01 So this question is about getting the geodesic equation by using calculus of variations to minimize the length of a path.

00:11 So first of all, we're given a curve, r of t, and we want to know its length between two points a and b.

00:24 So we know that ds squared is g -i -j, d -x -i, or d -u -i, d -u -j.

00:39 And so ds, which is the increment of length, is the square root of g -i -j, d -u -i -d -u -j.

00:52 Now, we want to get this, which is in terms of the changes in the coordinates, into the form of using the parameter.

01:03 Because we know that the coordinates are given in terms of the parameter by this map r.

01:08 So we can rewrite this as g -i -j, d -u -i by d -t, d -t -d -t -d -t, because inside here it would be d -t squared, outside the square root is just d -t.

01:26 Now l equals the integral from a to b, d -s, which is the integral from a to b, g -i -j, d -y -d -t, d -t, d -u -j by d -t, d -t, d -u -j by d -t, d t which is what we wanted now we want to use calculus of variations now so remember that the euler -la -la -la -grange equations of a so say we we have a functional l of ui ui dot equals the integral from a to b of this density, then when we vary this, we're varying ui, then what we're going to get is and also vary ui dot.

02:50 Then l is going to go to integral from a to b of l, ui, ui, dot, plus dl by d ui, delta ui, delta ui, ui dot delta ui dot d t which is equal to l plus this stuff but now i'm going to use the uh the integration by parts to move this dot off of the dui and onto this so let's just move this a little bit so we've got room to write it minus d l by du i dot and then this dotted delta u i d t and then on this side we have l plus delta l so now if we in the calculus of variations we now we can identify this delta l with this integral here we want to set delta l equals zero to find the extremer of this functional and since the variation is arbitrary we must have that this is vanishing at every point along the curve.

04:49 So this gives us the oil of the grange equations.

04:54 Dl by d ui minus dl by d ui dot dot equals zero.

05:06 So now we just need to make the identification, which is that our l is equal to the square root of g, i, j, which is a function of the coordinates, ui dot uj, so d l by d u i is 1 over l uh 1 over 2 l in fact and d g i or let's call it j k by d ui uj dot uj dot so this is the d l by d ui part and now we're going to do d l by d ui dui dot and this is going to give us one so d by d t of 1 over 2 l and then uh d by du i dot of uj dot uj dot u k dot uj ui or j ui or jk or jk or jk sorry and i'm going to suppress this for clarity and the g -jk doesn't depend on the dots.

07:14 So this is just going to give us 1 over 2l, dg, j -k by d -u -i, u -d -j, u -j, u -k, u -k, plus d -t by d -t, 1 over 2l, and then this is going to give us a 2, and then the g, because the g is symmetric.

07:47 So we're going to get g -i -j -u dot -j.

08:03 So now we're taking the time derivative of this.

08:27 So when the time derivative hits the l, we're going to get a minus 1 over 2 l squared l dot, g -i -j -u dot j, and then plus the d -d -t hitting the g.

08:46 This is going to give us a 1 over l dg -i -j by d -u -k -u -k -u -k -k -k -k -k -u -d -t will hit the u.

09:11 And this will give us a 1 over l -g -i -j u double dot j.

09:26 So now i'm just going to use this g to lower indices.

09:41 Actually, no, i'm going to leave the g -z in.

09:46 So now, remember that l is nothing but ds by d -t, because we've expanded ds, this is ds -d -d -t, and remember that this l, this is the l here.

10:09 So this is ds -by -d -t.

10:11 So s -dot.

10:13 So we're getting this is 1 over s dot, 2 -s dot.

10:25 And then i'm gonna, yeah, this is fine, d, i, g, j, k, u.

10:42 Dot, j, u dot, k, minus one onto s.

10:51 Squared, but this is s double dot, g, i, j, u dot, and then here we're getting a 1 over s dot, dk, g, g, i, j, j, and then here we're getting a 1 over s dot, dk, g, g, u .k, u .j.

11:30 And then this last term is a 1 over s dot g ij, u double dot.

11:41 So now i'm going to multiply by s dot because this is all equal to zero.

11:48 And i'm going to multiply, yeah, let's do that first.

11:52 So 0 equals a half, di, g, j, k, u.

11:57 Dot j, u dot k, u dot k, minus s double dot on 2s dot g .i .j.

12:06 U .j.

12:11 U .j plus dk .j, u .k...