Relativistic Computation?
To quote a comment Scott Aaronson made on his blog:
Can you perform an arbitrarily long computation with minimal effort, by leaving your computer on Earth, boarding a spaceship that accelerates to close to the speed of light, then turning around and returning to Earth, where you find civilization collapsed, your friends long dead, and the Sun going cold, but your important computation finished?
Here, as I like to point out in talks, the crucial problem is the energy needed to accelerate to relativistic speed. Indeed, if you want to get a superpolynomial speedup by the above means, it’s not hard to show that you need to accelerate your spaceship to faster than c-1/p(n) for any polynomial p. But that, in turn, requires a superpolynomial expenditure of energy (assuming, of course, that you have nonzero mass!). So, as long as we make the reasonable (and separately justifiable) assumption that in T seconds you can only collect poly(T) joules of energy, the Extended Church-Turing Thesis once again seems safe.
To summarize: Yes, physically possible. No, probably not more useful than regular computation.
Black hole computers are often lumped in with the rest of relativistic computers, but I am not so clear about what is the problem with them.
If you want to use gravitational effects to compute something you have to consider the fact that the result of the computational process must be sent to your position using a signal. For this reason the redshift caused by the mass (e.g. a star) changes the emitted signal i.e. the associated emitted frequency(ies) $\nu_{em}$. Let's suppose we have a calculator on the surface of a star. We have to study two cases:
- Weak field limit-> we consider a calculator on the Sun surface, the 00-component of metric tensor is related to the Newtonian potential $\phi$ which is the solution of equation $\nabla\phi=4 \pi G \rho$ with
$g_{00}\approx - \left(1+ \dfrac{2\phi}{c^2}\right)$
from the definition of proper time we also know that:
$\dfrac{\nu_{obs}}{\nu_{em}}=\dfrac{\lambda_{em}}{\lambda_{obs}}= \sqrt{\dfrac{g_{00}(x^\mu_{em})}{g_{00}(x^\mu_{obs})}}$
from the previous equations we can obtain:
$\dfrac{\nu_{obs}- \nu_{em}}{\nu_{em}}=\dfrac{\lambda_{em}-\lambda_{obs}}{\lambda_{obs}}=\sqrt{\dfrac{1+2\phi_{em}/c^2}{1+2\phi_{obs}/c^2}} - 1 \approx (...) \approx \left( \phi_{em} - \phi_{obs} \right)/c^2$
we know that $\phi= - G \dfrac{M_{sun}}{r}$ where r is the distance from the Sun center; if we set $r_{em}=R_{Sun\,surface}$ and $r_{obs}= r_{Sun - Earth}$ we have:
$\dfrac{\Delta \nu}{\nu} \approx \frac{G M_{Sun}}{c^2}\left( - \dfrac{1}{R_{Sun\,surface}} + \dfrac{1}{R_{Sun - Earth} } \right)$.
We also know that $R_{Sun - Earth}= 149.6\, \mbox{x}\, 10^6 km$. Using this value we obtain that:
$\dfrac{\Delta \nu}{\nu} \approx - 0.21 \, \mbox{x} \, 10^{-5} $ $\Rightarrow$ $\Delta \nu$ is negative so we have a redshift.
- Strong gravitational field: let's suppose that our mass is a neutron star described by Schwarzchild metric and $r_{obs} >> r_{em}$ i.e. the observer is located very far from the source emitting signal. $r_{em} = R_{NS}$ i.e. the distance from the center and and the surface of the neutron star, tipically we have $R_{NS} \approx 10 km$ and $M_{NS}=1.4 M_{Sun}$.
$\dfrac{\nu_{obs}}{\nu_{em}}= \sqrt{\dfrac{g_{00}(x^\mu_{em})}{g_{00}(x^\mu_{obs})}}=\sqrt{ \dfrac{1- \dfrac{2G M /c^2}{r_{em}} }{1-\dfrac{2G M /c^2}{r_{obs} } }} \approx \sqrt{1 - \dfrac{2G M /c^2}{r_{em}}}$
for a neutron star we usually have $\dfrac{G M_{NS}}{R_{NS}c^2} \approx 0.21$.
Then $\dfrac{\nu_{obs}}{\nu_{em}} \approx \sqrt{1- 2\, \mbox{x}\, 0.21} \approx 0.76 $ $\Rightarrow$ $\dfrac{\Delta \nu}{\nu}= \dfrac{\nu_{obs} - \nu_{em}}{\nu_{em}} \approx -0.24$ which is related to a greater red-shift than the previous case.
In addition, if we consider a (Schwarzchild) black hole with our calculator orbiting around it we see that as our computer approaches the radius of emission $r_{obs}=2GM/c^2$:
$\nu_{obs} \approx \sqrt{1 - \dfrac{2G M /c^2}{r_{em}}} \nu_{em} \rightarrow 0$.
The observed frequency tends to zero i.e. the result of our computation is lost.
More gravity $\Rightarrow$ More time $\Rightarrow$ more the frequency will tend to zero.
These considerations come from "Lecture notes on General Relativity, Black Holes and Gravitational Waves"(V. Ferrari, L. Gualtieri).