4
$\begingroup$

Imagine you are walking along the xy-plane. There is a landmark at the origin of the plane which distorts time at every point on the plane, such that the distortion is a function of the distance between that point and the origin. What this essentially means is that, if you're trying to get from point A to point B, you'll want to stay a reasonable distance away from the origin, because the cost of moving a given unit of distance is higher the closer you get to the origin.

What I'm trying to do is, given two points A and B, find the path from A to B which will minimize the cost of travel: the "subjective distance".

The first step, then, is to construct an equation to represent the cost for a given path; the second step is to find a path that minimizes that cost, using the Calculus of Variations.


The equation for temporal distortion is given as $\tau = C r^{-n} + 1$, where r is the distance from the origin.

If we define the path as $r(\theta)$, then the cost of travelling from A to B should be $\int_{r(\theta)} C r^{-n} + 1 ds$, or \int_{\alpha}^{\beta} \left( C \left[r(\theta)\right]^{-n} + 1 \right) \sqrt { \left[ r(\theta) \right]^2 + \left[ r'(\theta) \right]^2 } d\theta .

Now, in order to find a $r(\theta)$ such that the integral is minimized, I'm supposed to use the Beltrami Identity (because there's no $\theta$ term in the integral!), which is given as

L - r' \frac{\partial L}{\partial r'} = const

where L is the integrand above.

So I find the partial derivative:

\begin{align} &\frac{\partial}{\partial r'} \left[ \left( C r^{-n} + 1 \right) \sqrt{ r^2 + r'^2 } \right]\\ &{} = \left( C r^{-n} + 1 \right) \frac{\partial}{\partial r'} \sqrt{ r^2 + r'^2 }\\ &{} = \left( C r^{-n} + 1 \right) \cdot \frac{1}{2} \left( r^2 + r'^2 \right)^{-\frac{1}{2}} \cdot 2r'\\ &{} = \frac{r' \cdot \left( C r^{-n} + 1 \right)}{\sqrt{r^2 + r'^2}} \end{align}

And I substitute into the Beltrami Identity:

L - r' \frac{\partial L}{\partial r'} = const\\ \left[ \left( C r^{-n} + 1 \right) \sqrt{ r^2 + r'^2 } \right] - r' \frac{r' \cdot \left( C r^{-n} + 1 \right)}{\sqrt{r^2 + r'^2}} = const\\ \frac{\left( C r^{-n} + 1 \right) \left( r^2 + r'^2 \right)}{\sqrt{r^2 + r'^2}} - \frac{\left( C r^{-n} + 1 \right) r'^2}{\sqrt{r^2 + r'^2}} = const\\ \frac{\left( C r^{-n} + 1 \right) r^2}{\sqrt{r^2 + r'^2}} = const

This is the differential equation I must solve to find the optimum path.


However, this cannot be the correct differential equation. We can observe that $r = const$ is a solution to this differential equation, but if we let C equal 0, then the distortion disappears... which would imply that the shortest distance between two points on a normal plane is a circular arc between them. Which is incorrect.

I can't find a point in my equation where I assumed $C \neq 0$, so I must have done something wrong algebraically... or conceptually. Can anyone spot my error?

1 Answers 1

3

I find this quite a good and well-posed question; I don't know why it's only received $2$ votes and hasn't been answered for hours.

The problem is that the Beltrami identity is a necessary but not sufficient condition for stationarity of the integral. It's implied by the Euler-Lagrange equation, but it doesn't imply the Euler-Lagrange equation.

The situation is directly analogous to that of a pendulum that you displace from its equilibrium position and let go. The Euler-Lagrange equation, which is second-order and expresses $F=ma$, tells you that it will oscillate, whereas the Beltrami identity, which is first-order and expresses energy conservation, allows the further solution that the pendulum just sits there without doing anything, its energy conserved. This spurious solution slips in because to get from $F=ma$ to $E=\text{const}$ you have to multipy through by the velocity before integrating, which isn't an equivalence if the velocity is zero.