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Is it possible to employ relativity to develop computational technology?

Here is a really basic example:

Build a Computer and Feed it the Problem (say the problem is projected to take 10 years to solve) Start the solution of the problem using algorithm. Now go on to your convenient relativistic jet... accelerate very near the speed of light (utilizing time dilation to your advantage) and make a round trip back to the planet, which takes a reasonable amount of time for the observer within the relativistic jet to complete (say a couple weeks) and allows them to arrive back to their computer with the problem solved.

This sort of system could solve very difficult problems "relatively" quickly in the sense that the user would be able to find the solution to their problem in much less time ,relative to themselves, then normal.

A more user convenient system can be found if research is being conducted in a heavy gravitational field (time speeds up).

For example:

If you have a lab close to the surface of a black hole, and you send your computer far away (much less gravitational attraction) a mere couple of minutes could pass in your lab's frame of reference while the computer had thousands of years to work, in it's frame of reference, so then the computer can be fetched back to the lab with the solution found.

Unfortunately Earth's gravitational time dilation is not nearly enough to take advantage of this sort of system (which is convenient in that it allows the user to remain stationary).

Are there any better ideas? Just for the sake of though experiment and logic.

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    $\begingroup$ Yep, that would work. $\endgroup$ – Loourr May 1 '13 at 16:12
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    $\begingroup$ It would work, but it would not be very energy efficient: it is very hard to accelerate a lab or a person to relativistic velocities! It's probably better spending the energy building a better computer. $\endgroup$ – Michael May 1 '13 at 16:15
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    $\begingroup$ keep in mind that your scenario requires the scientist to miss 10 years of the lives of everyone else back on Earth. And maybe it's just me, but I wouldn't want to miss 10 years of the lives of say my wife and kids just because I didn't want to wait for a computer model. $\endgroup$ – Jim May 1 '13 at 16:23
  • $\begingroup$ Suppose you were based on earth? You launch the CPU out to space, it finds a spot , decelerates and sits very still... In low gravitational fields? How much computation gain would we find from it? $\endgroup$ – frogeyedpeas May 1 '13 at 16:32
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    $\begingroup$ Well, chris. You'd gain nothing from it, because relative to Earth, the cpu would have accelerated, effectively slowing time for it so that when it comes back to us, less time has passed for it than for us. $\endgroup$ – Jim May 1 '13 at 16:58
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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.

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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 $\nabla\phi=4 \pi G \rho$ 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.

  1. 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).

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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.

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