Relativistic Computation?

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:

1. 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 $nablaphi=4 pi G rho$ with

$g_{00}approx - left(1+ dfrac{2phi}{c^2}right)$

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:

$dfrac{nu_{obs}- nu_{em}}{nu_{em}}=dfrac{lambda_{em}-lambda_{obs}}{lambda_{obs}}=sqrt{dfrac{1+2phi_{em}/c^2}{1+2phi_{obs}/c^2}} - 1 approx (...) approx left( phi_{em} - phi_{obs} right)/c^2$

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.

2. 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).

You may want to check references regarding Malament-Hogarth spacetimes. They are basically defined as spacetimes which allow hypercomputations, that is infinitely long computations in finite proper time. Famous Malament-Hogarth spacetimes are the rotating and charged black holes and Anti De-Sitter.