Black Hole time dilation + biological paradox

[moocownarf]

If a spaceship housing humans were to travel near a black hole, time would slow down due to the increased gravity. However, with what we currently know about cellular biology and gravity, humans need to stay around 1g to proliferate properly.

Now if the spaceship generates an artificial gravity of 1g aboard to sustain life, would time dilation only occur on the outside of the ship and time would travel at the same rate as it would on Earth?

Not really a "paradox" but a problem with the thought experiment of using a black hole.

 

[bcrowell]

First off, you're confusing two different things here: the gravitational field g (which is 9.8 m/s2 at the surface of the earth) and the gravitational potential Φ (gravitational potential energy per unit mass). Time dilation relates to Φ.

The second thing is that it's not clear that the presence of the black hole plays any role in your logic.

Third, you haven't discussed by what means you're talking about producing artificial gravity.

 

[moocownarf]

Sorry, perhaps I could have been more clear. The objective of the thought experiment is to travel to the event horizon of a black hole to "travel to the future". A complication arises when humans cannot sustain life in such high gravity conditions. And as such a solution would be to somehow produce gravitational conditions (no idea, perhaps you could help me out here) that are viable ie. similar to Earth.

Now my question is if life were to be sustained on the ship, would time on the ship flow at a slower rate because of the proximity of the black hole, or at a similar rate to that of Earth's due to faux gravitational life support?

 

[yuiop]

If a spaceship housing humans were to travel near a black hole, time would slow down due to the increased gravity. However, with what we currently know about cellular biology and gravity, humans need to stay around 1g to proliferate properly.

Now if the spaceship generates an artificial gravity of 1g aboard to sustain life, would time dilation only occur on the outside of the ship and time would travel at the same rate as it would on Earth?

Not really a "paradox" but a problem with the thought experiment of using a black hole.

I don't think it quite works like that. If a spaceship was near a black hole, one way to reduce the crushing effect of gravity on humans would be to have the spaceship orbit the black hole. In a natural orbit the acceleration due to gravity can be reduced to zero and the humans would be free falling or "weightless" like astronauts in a space station. They would still be subject to the gravitational time dilation of sqrt(1-2m/r) and an additional time dilation of sqrt(1-v^2) due to their local relative velocity. Of course if they needed 1g so that they could walk around their spaceship normally, the speed of the spaceship could be arranged to be slightly less than the natural stable orbital speed and a continuous rocket thrust directed outwards to stop the spaceship spiralling inwards would be required.

 

[bcrowell]

Sorry, perhaps I could have been more clear. The objective of the thought experiment is to travel to the event horizon of a black hole to "travel to the future". A complication arises when humans cannot sustain life in such high gravity conditions. And as such a solution would be to somehow produce gravitational conditions (no idea, perhaps you could help me out here) that are viable ie. similar to Earth.

Now my question is if life were to be sustained on the ship, would time on the ship flow at a slower rate because of the proximity of the black hole, or at a similar rate to that of Earth's due to faux gravitational life support?

OK, that's clearer. The next thing you have to realize is that the gravitational field g does not have any unambiguous meaning in GR. The gravitational field g can have any value you like at a given location. For example, a free-falling observer measures g to be zero. This is the equivalence principle at work.

So you're imagining that the people aboard the ship need to be protected from the black hole's strong gravity by some kind of antigravity, produced by some unknown futuristic technology. Actually all you have to do is let them free-fall, and then they will experience g=0. If you put them in a circular orbit around the black hole, they are in free fall (because orbital motion is a kind of free fall).

You will then have both a kinematic time dilation and a gravitational time dilation. Comparing to an observer at infinity, both of these effects are in the same direction -- they both make the people aboard the ship age less.

IMO your scenario is not scientifically plausible. The amount of energy needed to transport your cancer patients to the black hole, insert them in a circular orbit, and later get them back out is going to be comparable to the energy contained in all the world's present-day nuclear arsenals. And even if you had that amount of energy at your disposal, with roughly the same energy expenditure you could simply have your patients fly around the galaxy in circles and harvest the kinematic time dilation. You don't need the black hole at all. But of course you may be able to sidestep these issues with creative storytelling. A good example that's similar to yours is Pohl's novel Gateway.

 

[bcrowell]

Hi, yuiop,

Looks like you type faster than I do :-)

Of course if they needed 1g so that they could walk around their spaceship normally, the speed of the spaceship could be arranged to be slightly less than the natural stable orbital speed and a continuous rocket thrust directed outwards to stop the spaceship spiralling inwards would be required.

I think it would be easier just to make a rotating space station. That way you wouldn't have any continuous expenditure of energy.

 

[stevebd1]

You might put a spaceship in orbit around a SM black hole, the time dilation would then be a product of SR (velocity) and GR (gravity)-

[tex]d\tau_{\text{total}}=dt\sqrt{1-\frac{2M}{r}}\cdot\sqrt{1-v_s^2}[/tex]

where [itex]v_s[/itex] is the velocity required for a stable orbit at a specific radius (c=1) and [itex]M=Gm/c^2[/itex]. For a stable orbit [itex]a_g=a_c[/itex] where [itex]a_g[/itex] is gravity and [itex]a_c[/itex] is centripetal acceleration of the orbiting object. It's worth noting that there is some suggestion that while gravity [itex](a_g=M/r^2)[/itex] increases by [itex]1/\sqrt(1-2M/r)[/itex], centripetal acceleration [itex](a_c=v^2/r)[/itex] reduces by [itex]\sqrt(1-2M/r)[/itex] so [itex]v_s[/itex] would increase exponentially the closer you orbit to the BH. [itex]a_g=a_c[/itex] for a stable orbit could be rewritten-

[tex]\sqrt{1-2M/r}\,\frac{v_s^2}{r}=\frac{M}{r^2 \sqrt{1-2M/r}}[/tex]

The equation rewritten relative to [itex]v_s[/itex] matches the Kepler equation for a stable orbit in Schwarzschild metric-

[tex]v_s=\frac{\pm\sqrt{M}\,r^2}{\sqrt{r^2-2Mr}\,r^{3/2}}[/tex]

This applies to a static black hole only, a rotating one would be more complex as the frame dragging would contribute to a ship in prograde orbit's tangential velocity and hence the ship should be able to orbit closer (though radiation from the BH/accretion disk would most likely be an issue)

Gravity on board the ship could be sustained using extended rotating modules, the contribution to the first equation being negligible.EDIT:
In the time it took me to compose my post there were 3 replies!

 

[moocownarf]

Thanks for the answers guys. I saw the scenario on the discovery channel, and being a bio guy, didn't really get how they could survive in such adverse conditions. The orbit explanation clarifies things for me.

 

[PeterDonis]

IMO your scenario is not scientifically plausible. The amount of energy needed to transport your cancer patients to the black hole, insert them in a circular orbit, and later get them back out is going to be comparable to the energy contained in all the world's present-day nuclear arsenals. And even if you had that amount of energy at your disposal, with roughly the same energy expenditure you could simply have your patients fly around the galaxy in circles and harvest the kinematic time dilation. You don't need the black hole at all.

There's actually another issue--there are no stable orbits around a black hole inside of r = 6M (3 times the radial coordinate of the horizon). The orbital velocity of the innermost stable orbit, at r = 6M, is half the speed of light, so the total time dilation factor is [itex]\sqrt{\frac{3}{2}}[/itex] for the gravitational time dilation at r = 6M, times [itex]\frac{2}{\sqrt{3}}[/itex] for the kinematic time dilation at v/c = 1/2, for a total of [itex]\sqrt{2}[/itex].

Of course you could increase the orbital velocity further, and compensate by rocket thrust that accelerated the orbiting station inward (not outward--if you're orbiting at greater than the free-fall orbital velocity, you need to accelerate inward to keep yourself from being thrown out to a greater radius). You could calculate how much additional velocity you would need to have an inward acceleration of 1 g, and see what additional time dilation that gave you, but I think bcrowell is right that you'd be much better off, energy-expenditure-wise, by just having a rotating station traveling around the galaxy in circles at some high velocity.

 

[stevebd1]

Considering a rotating black hole with a spin parameter of a/M=0.998, the prograde marginally stable orbit (MSO) would be at 1.237M. The equation for the total reduction factor for an object rotating around a Kerr black hole is-

[tex]A=\sqrt{-g_{tt} - 2\Omega g_{\phi t}-\Omega^2 g_{\phi \phi}}[/tex]

where

[tex]g_{tt}=-1+2Mr/\rho^2[/tex]

[tex]g_{t\phi}=-2Mra\sin^2\theta/\rho^2[/tex]

[tex]g_{\phi\phi}=(r^2+a^2+[2Mra^2\sin^2\theta]/\rho^2)\sin^2\theta[/tex]

where [itex]\rho^2=r^2+a^2 \cos^2\theta[/itex], [itex]M=Gm/c^2[/itex] and [itex]a=J/mc[/itex]

For a stable orbit [itex]\Omega[/itex] is replaced with [itex]\Omega_s[/itex] where-

[tex]\Omega_s=\frac{\pm\sqrt{M}}{r^{3/2}\pm a\sqrt{M}}[/tex]

where [itex]\Omega_s[/itex] is the angular velocity and [itex]\pm[/itex] denotes prograde and retrograde orbits.

(the results are synonymous with the results from the equations in post #7 when a=0, i.e. static black hole)

Even with this very tight orbit and rapidly spinning black hole, the reduction factor is no greater than 0.09267 for an object in stable orbit right at the MSO, basically 33.85 days would pass for the orbiting object for every year that passes in the universe outside the gravity field.

 

[michelcolman]

Of course if they needed 1g so that they could walk around their spaceship normally, the speed of the spaceship could be arranged to be slightly less than the natural stable orbital speed and a continuous rocket thrust directed outwards to stop the spaceship spiralling inwards would be required.

I think I would prefer an orbit with slightly more than the natural stable orbital speed, and a continuous rocket thrust directed inwards. Much safer in the event of an engine failure, don't you think?

 

[michelcolman]

There's actually another issue--there are no stable orbits around a black hole inside of r = 6M (3 times the radial coordinate of the horizon). The orbital velocity of the innermost stable orbit, at r = 6M, is half the speed of light

I did not know that. Is there an easy way of explaining why there are no lower stable orbits? What happens when you try to orbit lower? Is it just dynamically unstable? (If you stray from the orbit a little bit downward, the required speed to get back up is more than what you gained from potential energy being converted into kinetic energy). Or is it statically unstable already, some relativistic effect draining energy from you? The former would not be a huge problem if you have engines. I can't imagine what could cause the latter.

 

[TheGuardian]

Considering a rotating black hole with a spin parameter of a/M=0.998, the prograde marginally stable orbit (MSO) would be at 1.237M. The equation for the total reduction factor for an object rotating around a Kerr black hole is-

[tex]A=\sqrt{-g_{tt} - 2\Omega g_{\phi t}-\Omega^2 g_{\phi \phi}}[/tex]

where

[tex]g_{tt}=-1+2Mr/\rho^2[/tex]

[tex]g_{t\phi}=-2Mra\sin^2\theta/\rho^2[/tex]

[tex]g_{\phi\phi}=(r^2+a^2+[2Mra^2\sin^2\theta]/\rho^2)\sin^2\theta[/tex]

where [itex]\rho^2=r^2+a^2 \cos^2\theta[/itex], [itex]M=Gm/c^2[/itex] and [itex]a=J/mc[/itex]

For a stable orbit [itex]\Omega[/itex] is replaced with [itex]\Omega_s[/itex] where-

[tex]\Omega_s=\frac{\pm\sqrt{M}}{r^{3/2}\pm a\sqrt{M}}[/tex]

where [itex]\Omega_s[/itex] is the angular velocity and [itex]\pm[/itex] denotes prograde and retrograde orbits.

(the results are synonymous with the results from the equations in post #7 when a=0, i.e. static black hole)

Even with this very tight orbit and rapidly spinning black hole, the reduction factor is no greater than 0.09267 for an object in stable orbit right at the MSO, basically 33.85 days would pass for the orbiting object for every year that passes in the universe outside the gravity field.

NB> An intrusive question, maybe, but does the gravitational field on Earth caused by it's orbital motion dilate the time dimension for those on Earth? It would have too. Futhermore, would time dilate upon slower-moving planets? Would this mean that each individual celestial body would have a differing time-state from it's neighbour?

 

[stevebd1]

I did not know that. Is there an easy way of explaining why there are no lower stable orbits? What happens when you try to orbit lower? Is it just dynamically unstable? (If you stray from the orbit a little bit downward, the required speed to get back up is more than what you gained from potential energy being converted into kinetic energy). Or is it statically unstable already, some relativistic effect draining energy from you? The former would not be a huge problem if you have engines. I can't imagine what could cause the latter.

You might find this thread of interest-

https://www.physicsforums.com/showthread.php?p=966943#post966943

 

[michelcolman]

You might find this thread of interest-

https://www.physicsforums.com/showthread.php?p=966943#post966943

A, OK, so orbits lower than 3 times the Schwartzschild radius would still be possible as long as you use thrusters to stay in orbit. You just can't have a "free fall" orbit there. So in the scenario where you would be using 1g thrusters to create gravity for the occupants of the ship, there really would not be a problem. You just have to modulate the thrusters a little bit to stay in the orbit.

 

[yuiop]

A, OK, so orbits lower than 3 times the Schwartzschild radius would still be possible as long as you use thrusters to stay in orbit. You just can't have a "free fall" orbit there. So in the scenario where you would be using 1g thrusters to create gravity for the occupants of the ship, there really would not be a problem. You just have to modulate the thrusters a little bit to stay in the orbit.

I think that is about right. There is another webpage here http://www.fourmilab.ch/gravitation/orbits/ that is helpful in describing where the stable orbits are and how the are derived. In a landscape defined by the effective potential, stable orbits are at the deepest parts of the valleys and unstable orbits are delicately balanced on the the peeks.

An interesting aspect of the animated applet on that page is that if you assume the grey circle at the centre of the animation has a radius of r=2m then the test particle seems to dwell below r=3m for some time (several orbits) before popping out again. Not sure if that is possible and it may just be a problem with the way the animation appears to be scaled.

NB> An intrusive question, maybe, but does the gravitational field on Earth caused by it's orbital motion dilate the time dimension for those on Earth? It would have too. Futhermore, would time dilate upon slower-moving planets? Would this mean that each individual celestial body would have a differing time-state from it's neighbour?

You have the last part right but your statement that "the gravitational field on Earth caused by it's orbital motion" is not. The gravitational field of the Earth is not caused by its orbital motion, but is caused by the mass of the Earth. The time dilation on the Earth is product of its gravitational time dilation (dues to its mass) and kinematic time dilation (due to its relative velocity). Different planets with different masses and different velocities relative to the Sun centred frame will have clocks ticking at different rates on their surfaces relative to each other.

 

[PeterDonis]

A, OK, so orbits lower than 3 times the Schwartzschild radius would still be possible as long as you use thrusters to stay in orbit. You just can't have a "free fall" orbit there. So in the scenario where you would be using 1g thrusters to create gravity for the occupants of the ship, there really would not be a problem. You just have to modulate the thrusters a little bit to stay in the orbit.

True, but as bcrowell (I think) said, from an energy usage perspective (or perhaps an "energy usage per unit of time dilation" perspective), it would be a lot more efficient to just accelerate up to some high velocity, set the ship spinning to provide "artificial gravity", and then coast, than to have to constantly provide rocket thrust to maintain orbit.

 

[TheGuardian]

I think that is about right. There is another webpage here http://www.fourmilab.ch/gravitation/orbits/ that is helpful in describing where the stable orbits are and how the are derived. In a landscape defined by the effective potential, stable orbits are at the deepest parts of the valleys and unstable orbits are delicately balanced on the the peeks.

An interesting aspect of the animated applet on that page is that if you assume the grey circle at the centre of the animation has a radius of r=2m then the test particle seems to dwell below r=3m for some time (several orbits) before popping out again. Not sure if that is possible and it may just be a problem with the way the animation appears to be scaled.

You have the last part right but your statement that "the gravitational field on Earth caused by it's orbital motion" is not. The gravitational field of the Earth is not caused by its orbital motion, but is caused by the mass of the Earth. The time dilation on the Earth is product of its gravitational time dilation (dues to its mass) and kinematic time dilation (due to its relative velocity). Different planets with different masses and different velocities relative to the Sun centred frame will have clocks ticking at different rates on their surfaces relative to each other.

I understand so is it possible to calculate the dilation of time upon the Earth using these formulae? I do recall a media report about the dilation of Time, from NASA I think, and they reported it as a 1/10, 000 subliminal effect. Is this true?

 

[stevebd1]

A, OK, so orbits lower than 3 times the Schwartzschild radius would still be possible as long as you use thrusters to stay in orbit. You just can't have a "free fall" orbit there. So in the scenario where you would be using 1g thrusters to create gravity for the occupants of the ship, there really would not be a problem. You just have to modulate the thrusters a little bit to stay in the orbit.

You could risk the marginally bound orbit (MBO) which for a static black hole is at 4M, here an object is supposed to manage one decent orbit before spiralling into the black hole, using engines pushing you away from the event horizon, you might be able to maintain a stable orbit. For a rotating black hole, the MBO is at-

[tex]R_{\text{mb}}=2M\mp a+2\sqrt{M(M \mp a)}[/tex]

again for a rotating black hole with a spin parameter of a/M=0.998, the prograde MBO is at 1.091M (prograde photon sphere at 1.074M and event horizon at 1.063M). The total reduction factor would be 0.02283 and the prograde tangential velocity would be 0.7218663c based on-

[tex]v_\pm=(\Omega_s-\omega)\frac{R}{\alpha}[/tex]

where [itex]\Omega_s[/itex] is the stable orbit from post #10, [itex]\omega=2Mra/\Sigma^2[/itex] is the frame dragging rate, [itex]R=(\Sigma/\rho) \sin \theta[/itex] is the reduced circumference, [itex]\alpha=(\sqrt{(\Delta)}\rho/\Sigma)[/itex] is the reduction factor relative to the frame dragging, [itex]\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta[/itex] and [itex]\Delta= r^{2}+a^{2}-2Mr[/tex].

at this orbit, 1 year would pass in 8.33 days, this is probably about the maximum reduction you could achieve from orbiting a black hole, you might be able to risk a closer orbit or even have engines pushing you inwards as you increased in speed but as it has already been said, your probably better of taking a round trip around the galaxy slowly approaching c.

 

[TheGuardian]

You could risk the marginally bound orbit (MBO) which for a static black hole is at 4M, here an object is supposed to manage one decent orbit before spiralling into the black hole, using engines pushing you away from the event horizon, you might be able to maintain a stable orbit. For a rotating black hole, the MBO is at-

[tex]R_{\text{mb}}=2M\mp a+2\sqrt{M(M \mp a)}[/tex]

again for a rotating black hole with a spin parameter of a/M=0.998, the prograde MBO is at 1.091M (prograde photon sphere at 1.074M and event horizon at 1.063M). The total reduction factor would be 0.02283 and the prograde tangential velocity would be 0.7218663c based on-

[tex]v_\pm=(\Omega_s-\omega)\frac{R}{\alpha}[/tex]

where [itex]\Omega_s[/itex] is the stable orbit from post #10, [itex]\omega=2Mra/\Sigma^2[/itex] is the frame dragging rate, [itex]R=(\Sigma/\rho) \sin \theta[/itex] is the reduced circumference, [itex]\alpha=(\sqrt{(\Delta)}\rho/\Sigma)[/itex] is the reduction factor relative to the frame dragging, [itex]\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta[/itex] and [itex]\Delta= r^{2}+a^{2}-2Mr[/tex].

at this orbit, 1 year would pass in 8.33 days, this is probably about the maximum reduction you could achieve from orbiting a black hole, you might be able to risk a closer orbit or even have engines pushing you inwards as you increased in speed but as it has already been said, your probably better of taking a round trip around the galaxy slowly approaching c.

Excellent, time dilation is certainly not a subliminal effect. Deducing that it would be 8.3 days per single year, how would it fit in with our planet Earth? The planet is traveling 1/3 of the speed of light so could you calculate how much dilation is occurring? Or have I got it wrong again?