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Say you were to travel from Paris to Tokyo by digging a tunnel between both cities. If the tunnel is straight, one can easily compute that the time for travelling from one city to the other (independently from where both cities are actually) is a constant (about 42 minutes). It goes without saying : only gravitationnal potential and no friction.

But let's upgrade on the idea. Fix the two cities : what is the optimal shape of the tunnel, so that the time it takes for going from one city to the other is shortest ?

I assume a planar movement, where the plane contains both cities and the center of the earth. With Gauss' law, it is easy to compute that the gravitationnal potential on a mass m inside the earth at any distance $r$ from the center :

$V(r) = m\alpha r^2$ where $\alpha$ is a constant. I parameterize the tunnel by a polar function $r(\theta)$ where the origin is taken at Earth's center. The problem reduces to minimizing the following functional

$\mathcal{T}(\dot{r},r,\theta) = \int \limits_{\theta_1}^{\theta_2} \frac{ds}{|v|} = \int \limits_{\theta_1}^{\theta_2} d\theta \frac{\sqrt{\dot{r}^2 + r^2}}{|v|} = \sqrt{\frac{m}{2}}\int \limits_{\theta_1}^{\theta_2} d\theta \frac{\sqrt{\dot{r}^2 + r^2}}{\sqrt{E - m\alpha r^2}} \equiv \sqrt{\frac{m}{2}}\int \limits_{\theta_1}^{\theta_2} d\theta G(\dot{r},r,\theta)$

$\theta_1$ and $\theta_2$ are the angular coordinates of both cities. $\dot{r}$ is the derivative of $r$ with respect to $\theta$. With this parametrisation, the 'Hamiltonian' is conserved :

$\frac{\partial G}{\partial \dot{r}}\dot{r} - G = c$

Which leads to

$\frac{-r^2}{\sqrt{E-2\alpha r^2}\sqrt{\dot{r}^2+r^2}} = c$

And here I get stuck. Even if I try to play with special initial conditions, I cannot solve this equation. The easiest way is to perform separation of variables, but the integralcannot be carried out.

Any ideas on how to solve this problem ? I am sure the solution is a nice well known symmetric function. I just don't know how to tackle this problem. Would you tink it'd be easier to solve this in cartesian coordinates ?

Qmechanic
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Mathusalem
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  • Possible duplicate: http://physics.stackexchange.com/q/7421/2451 – Qmechanic May 04 '13 at 19:43
  • Paris to Tokyo goes deep. Is constant density a good approximation for that trip? – dmckee --- ex-moderator kitten May 04 '13 at 21:02
  • It is not. But the problem is hard enough already.I've seen your links Qmechanic, thanks. It seems there are no elegant solutions to this problem. – Mathusalem May 04 '13 at 21:09
  • It is a nonlinear differential equation. Nothing guarantees an elegant solution exists. You can start by looking for stationary solutions, those with constant $r$. I don't know if i've done the calculations well, but they should exist for $r=\pm\sqrt{\frac{Ec^2}{1+2\alpha c^2}}$. I don't know how to interpret it because I didn't understand the physical meaning of $c$. – Martino May 04 '13 at 21:46
  • I guess those solutions can only correspond to tunnels which are not tunnels, that is paths that go along the surface of the earth. Otherwise there is no way to connect both cities with an arc (constant r). You can actually sovle the problem exactly. The solution is hypocycloid, but the calculations are very, very tedious. I found it digging through the links of Qmechanic. – Mathusalem May 06 '13 at 07:47
  • keep in mind your tunnel can't go deeper than 100km (edge of lithosphere) because the temperature and density beyond that are high enough to kill you. And I can't think of anything that takes longer to get somewhere than never arriving. – Jim May 07 '13 at 19:59
  • Why is the solution not simply the straight-line arc connecting the two cities? This appears intuitively obvious. I did the calculus on this many moons ago, and I know the solution was closed-form. – Andrew Palfreyman May 07 '13 at 19:47
  • Simply because it is not optimal. Just like above earth, the straight line is not the fastest path from point A to point B. In this case, the closed form is a hypocycloid. Now the thing I'm wondering about is how I can justify a planar movement, that is constant $\phi$. – Mathusalem May 07 '13 at 19:48
  • @Jim For this kind of blue-sky problem we usually simply assume that such issues have been solved. You know? "That's just an engineering detail" kind of thing... – dmckee --- ex-moderator kitten May 13 '13 at 17:50
  • @dmckee You'll have to forgive me, I am an engineer and often consider those details. – Jim May 13 '13 at 18:41
  • @JimOh, I know. As a practicing experimenter I have enough experience designing real stuff to feel your pain. The detector we build is always messier, more complicated and less beautiful than the one we talk about when we're explaining it at the whiteboard. – dmckee --- ex-moderator kitten May 13 '13 at 18:48

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