How Can Time Dilation Be Achieved Near a Black Hole in Sci-Fi?

[Skyler Hooks]

Super frustrated English major here trying to do calculus! Please help! I'm in way over my head...

So, I'm writing this manuscript that involves humans with genetically inherited teleporting abilities. Near the end of the book, my protagonist wants to fling the antagonist close enough to a black hole for significant time dilation to occur. My first inclination towards a specific black hole was Sagittarius A* as I know it's close and it's huge (four million solar masses right?) so I thought that would make it easy. I tried using the Wolphram Alpha calculator to figure it out but I couldn't get the distance right. Either it was in the swarzchild radius (which means inside the event horizon correct?) or the effect was minimal (like eight times normal speed). I'm trying to get the kind of effect like in Interstellar, where one hour was seven years or 61,320 faster than normal (24x365x7).

Or faster. Much faster if possible. The antagonist will be able to return and probably within a few minutes his time (long enough to open a gate and travel through it in zero-gravity). I would like, if possible, for a few minutes to equal a few years. Or much longer. I'm open to any ideas on how to make this work. A different black hole maybe?

In fact, perhaps a smaller one would be more practical because the distance to experience the effect wouldn't be measured in the millions of kilometers, correct? Like I could just put them a few thousand kilometers away from the event horizon of a smaller one and get the same effect right? Escape velocity isn't an issue; for the sake of the story I'm assuming as long as he doesn't cross the event horizon, he can escape.

So, what I need is:

-A specific candidate for the black hole, preferably one we know about or at least the dimensions of one that would work
-If possible, the distance needed to achieve, say, 525,600x times normal speed (1 minute=a year). Or even faster would be great. I'm not sure what the limits on the effect are, how far I could take it. Best case scenario would be like 1 minute = 100 years (52,560,00x times normal speed). whatever works
-Any other practical information

Thank you so much for your time. I don't know if this is work for you guys or if you can just figure it out in two minutes but I greatly appreciate any time spent working on this problem and will gladly pay you back in haikus or limericks. I can do those all day.

 

[Vanadium 50]

-A specific candidate for the black hole, preferably one we know about or at least the dimensions of one that would work

I don't think you need this. Why not "black hole BH14921776"? It has all the properties you want.

 

[m4r35n357]

So, I'm writing this manuscript that involves humans with genetically inherited teleporting abilities. Near the end of the book, my protagonist wants to fling the antagonist close enough to a black hole for significant time dilation to occur. My first inclination towards a specific black hole was Sagittarius A* as I know it's close and it's huge (four million solar masses right?) so I thought that would make it easy. I tried using the Wolphram Alpha calculator to figure it out but I couldn't get the distance right. Either it was in the swarzchild radius (which means inside the event horizon correct?) or the effect was minimal (like eight times normal speed). I'm trying to get the kind of effect like in Interstellar, where one hour was seven years or 61,320 faster than normal (24x365x7).

As I recall the figures in the film were obtained under the assumption of a prograde orbit very close to a spinning black hole (giving much bigger effect). I think your calculations are probably right for Schwartzschild.

 

[Nik_2213]

Somewhat OT, but the FTL drive in my 'Convention' universe has time running faster aboard a 'City Class' star-ship. Hand-waving 'Inverse Time Dilation' prevented several obscure causality issues...

 

[Janus]

If you assume that he is thrown into a circular orbit around the BH, then you can use
$$ \sqrt{1- \frac{3 R_2}{2r}} $$
for the time dilation factor where
R_s is the Schwartzschild radius of the black hole.
For any given black hole and desired time dilation, solve for r.

 

[m4r35n357]

There is a thread here that discusses time dilation in the film Interstellar, which might be a little advanced for the OP but the discussion may help or inspire . . . and there are a few links too.

 

[Tiran]

I don't really get how using a black hole would work, as the amount of acceleration necessary to get in and out of the black hole's gravity is identical to speeding up that much by anywhere in space. And if you just do it in regular space you don't have the problem of the extreme tidal effects near black holes. I just thought the black hole thing was one of the many extremely flimsy bits of science in Interstellar.

 

[Janus]

I don't really get how using a black hole would work, as the amount of acceleration necessary to get in and out of the black hole's gravity is identical to speeding up that much by anywhere in space.

No. For a clock in orbit, the time dilation factor is
$$ \sqrt{1-\frac{3GM}{rc^2}}$$
which is the same as
$$ \sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}$$
where v is the orbital speed of the clock at orbital radius r, or
$$v = \sqrt{\frac{GM}{r}}$$

Now imagine you are at some distant point from mass M, and you want to enter orbit at distance r. If you just allow yourself to fall inward towards M, when you reach a distance of r you will be moving at escape velocity or
$$v= \sqrt{\frac{2GM}{2}$$

To enter orbit, you just need to shed the difference between your present velocity and the orbital velocity*. This works out to be less than 41% of the orbital velocity. This is by how much you would have to accelerate by. The time dilation factor in that orbit is the equivalent of that of moving at orbital speed compounded by the gravitational time dilation, so the time dilation factor during orbit will be greater than that if you had just accelerated to that same 41% of orbital speed in open space.

If you were comparing an clock at rest with respect to M at distance r to a clock at rest some great distance from M, then yes, the acceleration needed to lift the clock from r to that great distance would be the same as needed to accelerate the distant clock up to a speed where its time dilation matched that of the clock at r, but here we are not dealing with a clock at rest at r, but one in orbit. This not only increases the time dilation factor, but decreases the change of velocity needed to enter and leave that orbit for our distant clock.

* In order to prevent falling directly into M, you will need to start with a slight "not directly at M" velocity, but this doesn't have to be very large and doesn't effect the end conclusion significantly.

 

[PeroK]

I don't really get how using a black hole would work, as the amount of acceleration necessary to get in and out of the black hole's gravity is identical to speeding up that much by anywhere in space. And if you just do it in regular space you don't have the problem of the extreme tidal effects near black holes. I just thought the black hole thing was one of the many extremely flimsy bits of science in Interstellar.

If you are interested there was a long thread about this titled "Using Black holes to time travel into the future" from Feb this year.

 

[Tiran]

No. For a clock in orbit, the time dilation factor is

Hmmm - I hadn't considered that orbiting objects are at escape velocity. Thank you for the correction.

Would the Cliff Notes version of the problem be that a ship could "fall" into the stable and very fast orbit of a planet near a black hole, and that orbit is both inside a dilated gravity well AND at such a high orbital velocity to balance the immense gravity that the two combine to have very high time dilation? And that high orbital speed around the black hole means that the ship only requires the same amount of propellent as leaving the orbit of a similar sized planet around a regular sun?

 

[Janus]

Hmmm - I hadn't considered that orbiting objects are at escape velocity. Thank you for the correction.

Not at escape velocity, but for a circular orbit, a bit under 71% of escape velocity.

Would the Cliff Notes version of the problem be that a ship could "fall" into the stable and very fast orbit of a planet near a black hole, and that orbit is both inside a dilated gravity well AND at such a high orbital velocity to balance the immense gravity that the two combine to have very high time dilation?

Yes, the effective time dilation wold be more than what it would be for either gravitational time dilation or time dilation due to the velocity alone. Though you can't just "drop" the ship into a low circular orbit, you would have to provide some velocity change to remain close to the BH. Otherwise you will just swing aorund the BH and head back out again.( though you will experience a net time differential compared to a clock that stayed at your starting position.)

And that high orbital speed around the black hole means that the ship only requires the same amount of propellent as leaving the orbit of a similar sized planet around a regular sun?

That depends on what you mean. If you're in an orbit at distance r from a mass M, then it doesn't matter if mass M is a regular Sun or a Black hole as long as the mass is the same.
However, if you are orbiting a BH some several hundred times the mass of a regular sun at that same distance r, then your orbital speed and the propellant needed to escape orbit goes up accordingly.

What makes black hole unique is the fact that they are so compact, that they allow orbits with small enough radii for time dilation effect to become significant. With a regular star, you would strike it surface before achieving the type of distance from the center or orbital speeds needed for really significant dilation effects.

Another point concerns the tidal forces you mentioned earlier. The more massive the black hole, the less of a problem this becomes. If you go back and look at the time dilation equation for a orbiting clock, you will note that increasing the mass of the BH and the radius of the orbit by the same factor gives you the same time dilation factor. Tidal forces are proportional to the mass but inversely proportional to the cube of the distance. Thus doubling the mass of the BH and radius of the orbit would give you the same time dilation, but the tidal forces would be 1/4 as strong.

 

[Tiran]

Not at escape velocity, but for a circular orbit, a bit under 71% of escape velocity.

Okay, I see where the 70.71% difference comes from between the parking orbit and escape velocity equations, which basically means that the additional velocity needed to escape is in proportion to the strength of the gravity that the orbit balances (29% above escape velocity).

Which takes me back to my initial reaction to this topic - if the escape velocity acceleration from a normal system (like Voyager leaving from Earth) is X, how much higher would it be to reach escape velocity in a system where the orbital point is under considerable time dilation? In other words, if the BH planet's orbital velocity is a significant percentage of C already, is adding another 29% to that enormous velocity really practical?

I realize that this gets difficult to think about because time dilation means that the crew's perspective of that 29% is different than a non-dilated observer, but the basic question is whether it is really possible to climb out of certain orbits because 29% of velocity 100 units is only 29, but 29% of velocity 100,000 is 29,000 units. The Solar escape velocity from Earth is an additional 12.32km/s over orbital speed, while Mercury's is an additional 13.88km/s over its orbital speed. If Voyager needed a 12.32 boost from Earth, and would have needed a 13.88 boost from Mercury, what kind of boost are we talking about with a highly dilated orbit around a BH?

And is whatever that number might be remotely practical with some sort of propellant based drive? Because that's what I don't have the math to answer for myself.

 

[Tiran]

I followed the link above to the "science of" thread. That lists an orbital velocity of 163399 km/s, which means the additional escape velocity from that orbit is 47,860 km/s. What I don't know is whether that is an external observer velocity, or what the ship itself would have to produce that much delta. If the dilation of 60,000:1 applies, then the velocity needed is virtually nothing - which seems unlikely. If the 47,860 km/s is accurate for those on the landing ship, they would need it to produce the equivalent of a 14 hour 1G burn. And after that burn it is going to take 53 hours per number of AU that the main ship is out parked at.

Without knowing more about it, it appears that the trip is in the realm of possibility.

 

[stevebd1]

It's worth pointing out that the max realistic spin a Kerr black hole can have is a/M=0.998 simply due to the CMB, photons, other radiation and dark matter that falls into a black hole, stopping the bh from approaching too close to maximal. Kip Thorne mentions this in relation to the bh in Interstellar which while theoretically possible, has an unlikely spin parameter of a/M=1-1e-14. Based on a/M=0.998 and orbiting at the MSO with a tangential velocity of ~0.565c, the time dilation would be dτ/dt=0.09267 (interestingly irrespective of mass, v_t and time dilation remain constant for variable M). You could venture to the marginally bound orbit (normally 4M for a static bh) which would require a tangential velocity of ~0.722 c and provides a time dilation of 0.02283 which means 8.333 days in orbit is equivalent to 1 year far outside the gravity well though this would require constant adjustments to stay in orbit which isn't required at the MSO. You'd be pushed to orbit any closer, the next recognised orbit in is the photon sphere which requires a v_t of c.