I Using Black Holes to Time Travel Into the Future

[stevendaryl]

This is a possible science-fiction scenario, and I'm wondering if it is scientifically plausible.

If someone wanted to take a one-way trip into future, say 1000 years from now, then SR gives you a possible way to do it without dying of old age: Just hop in a rocket ship, accelerate to nearly the speed of light, travel for a year or so, and then turn around and return. If you work things right, you could perhaps age only a year or so while the rest of the Earth ages 1000 years.

The problem with this approach is that it would require an enormous amount of energy to get a ship going that fast. It's not really practical for time travel.

So if you want to make time travel more affordable for middle-class consumers, here's another approach:

Launch yourself into an eccentric orbit around a black hole. Just let gravity take you, so no energy required except for getting into the right orbit.

The question is: Is there a black hole of the appropriate size, and an orbit around that black hole such that

• the proper time for the orbit is reasonable (let's say, less than 20 years)
• the coordinate time for the orbit (Schwarzschild coordinates) is 1000 years.
• at no time are tidal forces great enough to pose a threat to living creatures

My feeling is that if the orbit gets close enough to the event horizon, then the ratio of coordinate time to proper time can be made arbitrarily large, but I'm not sure whether there are limits.

 

[Ibix]

I think you can do this by getting arbitrarily close to an event horizon, yes. You could certainly do it by hovering above the event horizon, and it occurs to me that as you pass a hovering observer (or a sequence thereof) on a ballistic trajectory your clock would be further slowed by SR type effects.

The tidal forces are problematic, though. I know you can reduce them by using a larger black hole.

But I think there's one more issue which is that we aren't aware of any black holes around here, much less any large ones. So you essentially have to solve your SR travel problem to get to a position where you can use your GR solution, no?

 

[PeroK]

I think you can do this by getting arbitrarily close to an event horizon, yes. You could certainly do it by hovering above the event horizon,

That uses a lot of energy, which is what was to be avoided.

 

[PeterDonis]

Just let gravity take you, so no energy required except for getting into the right orbit.

For a non-rotating black hole, no such orbits exist closer than 3/2 of the horizon radius (and any orbits inside 3 times the horizon radius are unstable, so they would require frequent small corrections using rockets). The time dilation factor at that altitude is not very large (just $1 / \sqrt{3}$ at 3/2 of the horizon radius).

For a rotating black hole, if its rate of rotation is close to the maximum possible one, then there will be stable orbits very close to the horizon, provided they are prograde (i.e., going in the same direction the hole is rotating), and the time dilation factors in these orbits can be very large. This is the scenario portrayed in the movie Interstellar, for which Kip Thorne did the physics calculations. They are online and should be findable with a Google search.

 

[PeterDonis]

an eccentric orbit

If by this you mean a hyperbolic orbit, not bound to the hole, but passing very close to the horizon, this won't be very helpful for your goal, since the craft won't spend enough time close to the horizon to accumulate much time dilation. Most of its time will be spent at higher altitudes, where the time dilation factor is much smaller.

 

[stevendaryl]

If by this you mean a hyperbolic orbit, not bound to the hole, but passing very close to the horizon, this won't be very helpful for your goal, since the craft won't spend enough time close to the horizon to accumulate much time dilation. Most of its time will be spent at higher altitudes, where the time dilation factor is much smaller.

Yes, that's the idea---you dive toward the black hole from far away, loop around it once skirting near the event horizon, and then escape again. I know that in terms of proper time, such an orbit would only briefly be near the black hole, but the issue is coordinate time (or the time of a distant stationary observer). When you're closest to the black hole, the ratio (coordinate time)/(proper time) is largest.

 

[PeterDonis]

When you're closest to the black hole, the ratio (coordinate time)/(proper time) is largest.

Yes, but I don't think there will be enough coordinate time elapsed there for this to make much difference, at least not if it's a simple hyperbolic orbit (see below). The point about either hovering static at a low altitude above the horizon, or being in a stable bound orbit, is that you can stay there indefinitely (barring fuel requirements for hovering).

The reason I think there won't be enough coordinate time elapsed for time dilation to make much difference is that, because of the extra attractive term in the effective potential (as compared with the Newtonian case), there is a tradeoff between angular momentum and the minimum distance of closest approach you can have without falling into the hole. The lower the angular momentum, the further that minimal distance is; but the higher the angular momentum (which allows a closer approach), the higher the angular velocity, and the less coordinate time will be spent close to the hole. I haven't done the math, but I suspect this sets an upper bound on how much effective time dilation you can have for this case.

 

[DaveC426913]

Yes, but I don't think there will be enough coordinate time elapsed there for this to make much difference, at least not if it's a simple hyperbolic orbit (see below).

When dealing with SR and GR, one is frequently pushing some value to within a very tiny infinitesimal fraction of its limit to see the really intersting results.

I think the idea is that, while you only spend a short time near the EH, if you are close enough (like infinitesimally close) the dilation factor can become arbitrarily large. Theoretically.

 

[DaveC426913]

Of course, one must ask about practicalities.
Compare
- "SR time travel": how much energy it might take to accelerate a ship to a very large fraction of c while still keeping your subjective travel time low. You can do it right in the solar system.
to
- "GR time travel": how much energy it would take just to get to the nearest BH, let alone one one large enough to have a shallow gradient so as not to squish you.

i.e. SR time travel has the extremely convenient - and fuel-saving - advantage of being able to do it anywhere, whereas GR requires you to be at the right place (which is not here).

 

[PeterDonis]

I think the idea is that, while you only spend a short time near the EH, if you are close enough (like infinitesimally close) the dilation factor can become arbitrarily large. Theoretically.

It is true that if you can get arbitrarily close to the horizon, the time dilation factor can become arbitrarily large. My question is whether it is possible to get arbitrarily close to the horizon on a hyperbolic orbit without falling into the hole. In this regime Newtonian intuitions are no longer valid. The only way to resolve it would be to actually do the math, which I have not done.

 

[DaveC426913]

My question is whether it is possible to get arbitrarily close to the horizon on a hyperbolic orbit without falling into the hole.

A sufficiently large BH would have a safe "corridor" as wide as a spacecraft outside the EH that is also sufficiently shallow in gradient to not tear the ship apart.
Need a higher dilation factor? Find a bigger BH.

 

[PeterDonis]

A sufficiently large BH would have a safe "corridor" as wide as a spacecraft outside the EH that is also sufficiently shallow in gradient to not tear the ship apart.

I'm not talking about tidal gravity near the horizon; you're correct that that can be made negligible by having a large enough hole.

I'm talking about the fact that a free-fall trajectory passing close enough to the horizon that has insufficient angular momentum will fall in. And if it has sufficient angular momentum not to, its angular velocity will be too high for it to spend enough time close to the horizon to accumulate significant time dilation. And because of the extra term in the effective potential in GR that is not there in Newtonian mechanics, "sufficient angular momentum" gets much larger very close to the hole than Newtonian intuitions would suggest. At least, that's the possibility that I think would need to be checked by actually doing the math.

 

[Jorrie]

I'm not talking about tidal gravity near the horizon; you're correct that that can be made negligible by having a large enough hole.

I'm talking about the fact that a free-fall trajectory passing close enough to the horizon that has insufficient angular momentum will fall in. And if it has sufficient angular momentum not to, its angular velocity will be too high for it to spend enough time close to the horizon to accumulate significant time dilation. And because of the extra term in the effective potential in GR that is not there in Newtonian mechanics, "sufficient angular momentum" gets much larger very close to the hole than Newtonian intuitions would suggest. At least, that's the possibility that I think would need to be checked by actually doing the math.

I recall some results from previous simulations, using MTW's orbital eq. (25.16) for a Schwarzschild hole: $$\left(\frac{dr}{d\tau}\right)^2+\tilde{V^2} = \tilde{E^2}$$ with $\tilde{E}$ and $\tilde{V}$ the total energy and effective potential parameters respectively, eq's (25.17) and (25.18).

AFAIK, the equation has no direct analytical solution, but it can be readily solved by most integrator software. Playing around with it I recall finding that the closest periapsis is just outside Schwarzschild parameter $r > 4GM/c^2$, i.e. twice the horizon coordinate. There the local orbital speed equals the local escape speed ($v = 0.707c$), so the orbital solution must stay just outside of that (closer and it spirals in). The propertime for a complete orbit can be obtained out of the solution, but I can't recall the value.

Interestingly, there are analytical solutions for "multiple leaf clover" orbits, that actually circle the hole a number of times before going out around a next leaf, later to close the orbit again. I don't have that link atm, but can possibly find it. This could potentially give you a substantial amount of orbital time near $4GM/c^2$ and bring the clock back for a direct comparison to the more distant one.

Found the link: https://arxiv.org/abs/0802.0459 "A Periodic Table for Black Hole Orbits" by
Janna Levin, Gabe Perez-Giz

These orbits are also referred to as "zoom-whirl" cases (not 'zoom-twirl', as I originally typed).

 

[Ibix]

That uses a lot of energy, which is what was to be avoided.

I wasn't suggesting it as a solution, just noting it as a step on the way. Peter seems to disagree, though. I need to read the paper Jorrie linked and break out a numerical integrator and do some investigating.

I'm still wondering how we actually get to the hole without spending too much energy, though.

 

[stevendaryl]

I wasn't suggesting it as a solution, just noting it as a step on the way. Peter seems to disagree, though. I need to read the paper Jorrie linked and break out a numerical integrator and do some investigating.

I'm still wondering how we actually get to the hole without spending too much energy, though.

Well, for a science fiction story, you can just set the action near a black hole, you don't actually have to get there from here.

 

[Jorrie]

I agree with Steven - to be practical, start the sim in circular orbit at say $r=100GM/c^2$. Then give it the correct deceleration burn and let it fall in and (hopefully) come back from near the BH, to compare clocks. I think it will still require a lot of energy, but you will probably do better on energy budget than for a simple flat spacetime 'twin-paradox' scenario. Especially if the hole is spinning.

 

[stevendaryl]

AFAIK, the equation has no direct analytical solution, but it can be readily solved by most integrator software. Playing around with it I recall finding that the closest periapsis is just outside Schwarzschild parameter $r > 4GM/c^2$, i.e. twice the horizon coordinate.

For a given value of angular momentum, isn't the closest periapsis just the point that maximizes V_{eff}?

According to the paper https://arxiv.org/pdf/0802.0459.pdf,

V_{eff} = \frac{1}{2} - \frac{1}{r} + \frac{L^2}{2r^2} - \frac{L^2}{r^3}

I'm not sure what the units are: Letting r_s = 1?

Then for a given L the minimum r is at:

r = \frac{L^2}{2} \pm \sqrt{\frac{L^2}{4} + 3 L^2}

 

[Jorrie]

According to the paper https://arxiv.org/pdf/0802.0459.pdf,

V_{eff} = \frac{1}{2} - \frac{1}{r} + \frac{L^2}{2r^2} - \frac{L^2}{r^3}

I'm not sure what the units are: Letting r_s = 1?

I think the units are for c=G=M=1, so $r_s$ would be 2.
Yes, the closest Schwarzschild periapsis is at the point that maximizes $V_{eff}$ for a given orbital energy parameter. At escape energy it lies at r=4. As the energy is lowered, r must go up, until at circular orbit it lies at r=6 (where the min and max values meet). Graphs at page 18.

PS: I think the above equation for $V_{eff}$ is the 3PN approximation, good enough for this sort of numerical simulation. The exact equation from MTW in the same units as above is
$$V_{eff} = \sqrt{(1-2/r)(1+L^2/r^2)}$$

 

[Ibix]

break out a numerical integrator and do some investigating

Thinking along these lines, how do I set the conserved quantity E (7.43 in Carroll's lecture notes)? It's the energy per unit mass - can I just set this to $(\gamma-1)c^2$ using velocity relative to a co-located hovering observer at the 100M radius Jorrie suggested?

 

[Jorrie]

Thinking along these lines, how do I set the conserved quantity E (7.43 in Carroll's lecture notes)? It's the energy per unit mass - can I just set this to $(\gamma-1)c^2$ using velocity relative to a co-located hovering observer at the 100M radius Jorrie suggested?

I rather now suggest that you use some of the values from the linked paper, e.g. fig. 11 at page 20. Then it is easy to check your simulation.

I did not check what coordinates Carrol's lecture notes use.

 

[Ibix]

I think the units are for c=G=M=1, so $r_s$ would be 2.
Yes, the closest Schwarzschild periapsis is at the point that maximizes $V_{eff}$ for a given orbital energy parameter. At escape energy it lies at r=4. As the energy is lowered, r must go up, until at circular orbit it lies at r=6 (where the min and max values meet). Graphs at page 18.

PS: I think the above equation for $V_{eff}$ is the 3PN approximation, good enough for this sort of numerical simulation. The exact equation from MTW in the same units as above is
$$V_{eff} = \sqrt{(1-2/r)(1+L^2/r^2)}$$

Apart from a square root and a factor of 2 that's the same as Steven's expression, which is the same as Carroll's, give or take G, M and c.

I rather now suggest that you use some of the values from the linked paper, e.g. fig. 11 at page 20. Then it is easy to check your simulation.

I did not check what coordinates Carrol's lecture notes use.

Thanks - I'll take a look.

 

[Jorrie]

Apart from a square root and a factor of 2 that's the same as Steven's expression, which is the same as Carroll's, give or take G, M and c.

Yes, 3PN (Post-Newtonian) is essentially a truncated 3rd order Taylor series expansion of the full equation. Larger orders are however required to accurately go right to the horizon, like with in-spirals, but it becomes very costly in computing time. Especially when it is not just a primary mass and a particle, but rather two masses, like in mergers.
https://arxiv.org/abs/0907.0671

 

[Ibix]

Yes, 3PN (Post-Newtonian) is essentially a truncated 3rd order Taylor series expansion of the full equation. Larger orders are however required to accurately go right to the horizon, like with in-spirals, but it becomes very costly in computing time. Especially when it is not just a primary mass and a particle, but rather two masses, like in mergers.
https://arxiv.org/abs/0907.0671

Carroll says otherwise - see https://preposterousuniverse.com/wp-content/uploads/grnotes-seven.pdf. The effective potential is given in equation 7.48 at the bottom of the eleventh page (p174) and explicitly described as "not a truncated series" in the second or third paragraph (depends whether you count the paragraph split round the figure as one or two) on the following page.

I haven't read your link yet, and I'm out of time to do the cloverleaf orbits today. I have managed to generate an overall time dilation factor of 11.44 for an infall and return from 100M, basically by trial and error - nothing systematic. Uncommented code below if anyone fancies playing around with it (or checking it!) themselves. You'll need python with scipy and matplotlib - I'm using python 2.7.9, so I make no guarantees that this'll work in python 3. You can set initial conditions M, E, R and psi. The function f is the differential equations. The list t contains the proper time and the list y contains $dr/d\tau, r, \psi,t$ coordinates (all motion is in the equatorial plane). The print statement prints the final t, $\tau$, and the ratio.

from __future__ import division

import math,scipy.integrate,matplotlib.pyplot

G=1
c=1
M=1

def f(t,y,args=None):
    return [-GM/y[1]**2+Lc2/y[1]**3-3*GML2/y[1]**4, \
            y[0], \
            Lc/y[1]**2, \
            E/(csq-2*GM/y[1])]

E=0.079
R=100*G*M/c**2
psi=3*math.pi/4

GM=G*M
csq=c*c
L=R*math.sin(psi)*E/math.sqrt(1-2*GM/(csq*R))
Lc2=L*L*c*c
GML2=GM*L*L
Lc=L/c

t=[0]
y=[[math.cos(psi)*math.sqrt(1-2*GM/(csq*R)),R,0,0]]

r=scipy.integrate.ode(f).set_integrator("dopri5")
r.set_initial_value(y[0],t[0])
dt=0.1

while r.successful() and r.y[1]>2.01*GM/csq and r.y[1]<=1.01*y[0][1]:
    r.integrate(r.t+dt)
    t.append(r.t)
    y.append(r.y)

print t[-1],y[-1][-1],t[-1]/y[-1][-1]

fig=matplotlib.pyplot.figure()
ax=fig.add_subplot(111, projection='polar')
c=ax.scatter([p[2] for p in y],[p[1] for p in y])
matplotlib.pyplot.show()

 

[stevendaryl]

Yes, 3PN (Post-Newtonian) is essentially a truncated 3rd order Taylor series expansion of the full equation. Larger orders are however required to accurately go right to the horizon, like with in-spirals, but it becomes very costly in computing time. Especially when it is not just a primary mass and a particle, but rather two masses, like in mergers.
https://arxiv.org/abs/0907.0671

I don't think the effective potential is a truncated power series. I think it's exact for a Schwarzschild black hole.

Here's how I would get to it:

The equation for proper time in the Schwarzschild metric is:

d\tau^2 = Q dt^2 - \frac{1}{Q} dr^2 - r^2 (d\theta^2 + sin^2(\theta) d\phi^2) where Q = 1 - \frac{2GM}{r}

For a geodesic, we can view t, r, \theta, \phi as functions of proper time \tau. So we can write:

1 = Q (\frac{dt}{d\tau})^2 - \frac{1}{Q} (\frac{dr}{d\tau})^2 - r^2 ((\frac{d\theta}{d\tau})^2 + sin^2(\theta) (\frac{d\phi}{d\tau})^2)

Now, we can choose our coordinates so that the orbit is in the plane \theta = \frac{\pi}{2}, so this becomes:

1 = Q (\frac{dt}{d\tau})^2 - \frac{1}{Q} (\frac{dr}{d\tau})^2 - r^2 (\frac{d\phi}{d\tau})^2

Letting m be the mass of the test particle, we can define:

E = m Q \frac{dt}{d\tau}
L = m r^2 \frac{d\phi}{d\tau}

These are constants of the motion. In terms of them, we can rewrite our equation:

1 =(\frac{E^2}{m^2} - (\frac{dr}{d\tau})^2) \frac{1}{Q} - \frac{L^2}{m^2 r^2}

Now, multiplying through by Q = 1 - \frac{2GM}{r} gives:

1 - \frac{2GM}{r} = \frac{E^2}{m^2} - (\frac{dr}{d\tau})^2 - \frac{L^2}{m^2 r^2} (1 - \frac{2GM}{r})

Getting the constants to one side:

\frac{E^2}{m^2} - 1 = (\frac{dr}{d\tau})^2 - \frac{2GM}{r} + \frac{L^2}{m^2 r^2} (1 - \frac{2GM}{r})

To make it look a little more Newtonian, we can multiply by \frac{m}{2} to get:

\frac{E^2}{2m} - \frac{m}{2} = \frac{m}{2} (\frac{dr}{d\tau})^2 - \frac{GmM}{r} + \frac{L^2}{2m r^2} (1 - \frac{2GM}{r})

So this is just like a one-dimensional Newtonian problem with an effective energy

\mathcal{E}_{eff} = \frac{E^2}{2m} - \frac{m}{2}
V_{eff}(r) = -\frac{GmM}{r} + \frac{L^2}{2mr^2} - \frac{L^2 GM}{mr^3}

and giving r as a function of proper time instead of coordinate time.

 

[PeterDonis]

I don't think the effective potential is a truncated power series. I think it's exact for a Schwarzschild black hole.

This is my understanding as well.

 

[PeterDonis]

Apart from a square root and a factor of 2 that's the same as Steven's expression

There might be an issue of nomenclature here. Some sources do appear to define the "effective potential" $V(r)$ using the square root; but often the equation that actually gets used has $V^2$ in it, and other sources just call that expression the "effective potential". See, for example, the discussion in the caption of Figure 25.2 in MTW, in particular the following:

"Note that one could equally well regard $\tilde{V}^2(r)$ as the effective potential, and define a turning point by the condition $\tilde{V}^2 = E^2$. Which definition one chooses depends on convenience, on the intended application, on the tie to the archetypal differential equation $\frac{1}{2} \dot{x}^2 + V(x) = E$, and on the stress one wishes to put on the correspondence with the effective potential in Newtonian theory."

 

[PeterDonis]

the closest Schwarzschild periapsis is at the point that maximizes $V_{eff}$ for a given orbital energy parameter. At escape energy it lies at r=4.

We're considering hyperbolic orbits, though, so the orbital energy could be larger than the escape energy. That should allow orbits with a closest periapsis all the way down to $r = 3M$. I don't think the periapsis of any escape orbit can be closer than that because at that radius tangentially moving light is in a circular orbit, so any timelike object moving tangentially there (as it would have to be at either of the apsides) would be moving slower than circular orbital speed and would be on an orbit that would end up falling into the hole (i.e., it would have to be at apoapsis, not periapsis).

The above seems to me to make it highly unlikely that a Schwarzschild hole could be used for the purpose intended with any orbit having a single passage. For that it would have to be a Kerr hole, and the closer to maximal spin, the better (since the closer to maximal spin, the closer the photon sphere is to the horizon). I'm not sure the above reasoning rules out the "zoom-whirl" orbits around a Schwarzschild hole for the intended purpose, though; it would depend on how many zooms and whirls could be done.

 

[Jorrie]

Yes, there seems to be some non-clarity on the V_eff definition. The follow-on paper that Janna Levin coauthored (Zoom-whirl behavior) definitely speaks about using the 3PN approximation. It may be because they are definitely out of the Schwarzschild regime there. It is also not clear from MTW whether that equation is precise only at the turning points, or everywhere in Schwarzschild coordinates.

I'm not sure the above reasoning rules out the "zoom-whirl" orbits around a Schwarzschild hole for the intended purpose, though; it would depend on how many zooms and whirls could be done.

Plenty of whirls can be done, I guess. They are the important ones, because there $d\tau/dt$ becomes very small. The big reason for suggesting closed orbits is that some later zoom will close the orbit at the apoapsis for physical clock comparison. We obviously expect that to agree with the simulation to within some error band.

 

[PeroK]

Yes, there seems to be some non-clarity on the V_eff definition. The follow-on paper that Janna Levin coauthored (Zoom-whirl behavior) definitely speaks about using the 3PN approximation. It may be because they are definitely out of the Schwarzschild regime there. It is also not clear from MTW whether that equation is precise only at the turning points, or everywhere in Schwarzschild coordinates.

Plenty of whirls can be done, I guess. They are the important ones, because there $d\tau/dt$ becomes very small. The big reason for suggesting closed orbits is that some later zoom will close the orbit at the apoapsis for physical clock comparison. We obviously expect that to agree with the simulation to within some error band.

How are you getting $d\tau/dt$ to be very small?

 

[Jorrie]

How are you getting $d\tau/dt$ to be very small?

Straight from the Schwarzschild metric, which can be written like this:
$(d\tau/dt)^2 = Q-v_r^2/Q-v_\phi^2$, with c=G=1, Q=1-2M/r, $v_r=dr/dt$ and $v_\phi=rd\phi/dt$, the Schwarzschild radial and tangential speeds respectively.
During the whirl phase, $r \approx 4M$, $v_r \approx 0$ and $v_\phi^2 \approx 2M/r$ (escape speed), so $d\tau/dt$ tends towards zero.

 

[Jorrie]

We're considering hyperbolic orbits, though, so the orbital energy could be larger than the escape energy. That should allow orbits with a closest periapsis all the way down to $r = 3M$.

I do not quite think any periapsis is possible below r=4M in Schwarzschild. Any geodesic that is even slightly 'in-going' at 4M should spiral in, because there is no maximum effective potential below that. Outgoing particles should be possible, provided that $v_r^2/Q + v_t^2 < 1$. Or am I missing something?

I recall this from simulations done a long time ago, but I'm not sure and I don't have access to that software atm. It will be interesting to see what the other members' software efforts come up with.

 

[PeroK]

Straight from the Schwarzschild metric, which can be written like this:
$(d\tau/dt)^2 = Q-v_r^2/Q-v_\phi^2$, with c=G=1, Q=1-2M/r, $v_r=dr/dt$ and $v_\phi=rd\phi/dt$, the Schwarzschild radial and tangential speeds respectively.
During the whirl phase, $r \approx 4M$, $v_r \approx 0$ and $v_\phi^2 \approx 2M/r$ (escape speed), so $d\tau/dt$ tends towards zero.

That looks like the local speed of light to me, not the escape velocity. If that speed is less than light, then you could increase $v_{\phi}$ and have $(d\tau/dt)^2$ negative.

PS the formula for escape velocity is the speed measured by a local observer. This translates to an escape velocity of:

$v_\phi^2 = Q\frac{2M}{r}$

 

[Jorrie]

PS the formula for escape velocity is the speed measured by a local observer. This translates to an escape velocity of:
$v_\phi^2 = Q\frac{2M}{r}$

Oops, yes! What I've used is the radial escape velocity in Schwarzschild. So the metric will give, using Steven's equation, dropping radial components and with $\theta=\pi/2$

$dτ^2 = Q dt^2 - r^2 d\phi^2$

or $(dτ/dt)^2 = Q - r^2 d\phi^2/dt^2 = Q-v_\phi^2 = Q-Q\frac{2M}{r} =Q(1-\frac{2M}{r}) = Q^2$

For periapsis just above r=4M we get $d\tau \approx 0.5\, dt$. So yea, we will need many swirls for Steven to go significantly into the future; or many leafs on his clover...

 

[stevendaryl]

To confirm, if anyone needs it besides me, what @PeterDonis said,

If I computed it correctly, the formula for r_{min}, the closest approach for a hyperbolic orbit is:

r_{min} = L^2 \pm \sqrt{L^4 - 3L^2}

(in units where c=1, 2GM = 1)

There is no solution unless L^4 - 3L^2 &gt;0, so L &gt; \sqrt{3}. So the smallest r_{min} can be is 3.

 

[Jorrie]

To confirm, if anyone needs it besides me, what @PeterDonis said,

If I computed it correctly, the formula for r_{min}, the closest approach for a hyperbolic orbit is:

r_{min} = L^2 \pm \sqrt{L^4 - 3L^2}

(in units where c=1, 2GM = 1)

There is no solution unless L^4 - 3L^2 &gt;0, so L &gt; \sqrt{3}. So the smallest r_{min} can be is 3.

If 2GM = 1, doesn't this mean that r_{min} = 6M in the other common units (c=G=1)? Does not sound right somehow...

 

[PeterDonis]

It is also not clear from MTW whether that equation is precise only at the turning points, or everywhere in Schwarzschild coordinates.

It's exact everywhere. It's straightforward to derive that equation from the geodesic equation. I don't have my copy of MTW handy right now, but I'm pretty sure they discuss this.

Any geodesic that is even slightly 'in-going' at 4M should spiral in, because there is no maximum effective potential below that.

Huh? The maximum effective potential continues to increase all the way down to the (unstable) circular photon orbit at $r = 3M$.

in units where $c=1$, $2GM = 1$

I think it should be $GM = 1$. I don't like units where $M$ disappears from the equations for this reason; $M$ is not a physical constant like $G$ and $c$, it's a solution parameter, so to me it needs to remain explicit in the equation.

 

[Ibix]

I'll post the maths and some revised code later (the code is mathematically the same as above, but better laid out to allow multiple trajectories and plots). But here are some results - tangential launches in the equatorial plane from 3M and 4M. Since a trajectory grazing these radii must pass through this point, I think this points to the notion that 4M is escapable but 3M isn't:

 

[PeterDonis]

tangential launches in the equatorial plane from 3M and 4M

I would try values in between. I agree $3M$ shouldn't be escapable, since it's the photon sphere; but $3.001M$ should be, although with a value of $E$ much larger than $1$.

Also, what exactly does $E$ signify in your code? In the usual meaning of that symbol in the effective potential equation, a marginal escape trajectory (i.e., just coming to rest at infinity) should have $E = 1$. But it looks like your $E = 0.9$ trajectory at $r = 4M$ escapes.

 

[Ibix]

I would try values in between. I agree $3M$ shouldn't be escapable, since it's the photon sphere; but $3.001M$ should be, although with a value of $E$ much larger than $1$.

Here's a launch at 3.001M. E=100 does escape or, at least, passes 50M on a nearly-radial path:

Also, what exactly does $E$ signify in your code? In the usual meaning of that symbol in the effective potential equation, a marginal escape trajectory (i.e., just coming to rest at infinity) should have $E = 1$. But it looks like your $E = 0.9$ trajectory at $r = 4M$ escapes.

It does escape (rightly or wrongly). Assuming I didn't muck anything up, I'm following Sean Carroll's GR notes chapter 7, equations 7.43, 7.44, 7.47 and 7.48 with $\epsilon=c^2$ (=1 in his notation) for a timelike path. What I've actually implemented is$$\begin{eqnarray}
\frac{dp}{d\tau}&=&-\frac{GM}{r^2}+\frac{L^2c^2}{r^3}-\frac{3GML^2}{r^4}\\
p&=&\frac{dr}{d\tau}\\
\frac{d\phi}{d\tau}&=&\frac{L}{cr^2}\\
\frac{dt}{d\tau}&=&\frac{E}{c^2-2GM/r}
\end{eqnarray}$$
For initial values you set E, the radial coordinate at launch, R, and the angle $\psi$, which is what an observer hovering at R measures as the launch angle (0 is "straight up"). In this context L is a derived quantity:$$L=\frac{ER\sin\psi}{\sqrt{1-2GM/c^2R}}$$The initial value of $p=dr/d\tau=\left(\cos\psi\right)\sqrt{1-2GM/c^2R}$. See also George Jones' and your comments in an old thread of mine on Schwarzschild null geodesics.

 

[trainman2001]

How much energy would it take to break away from your orbit once you've reached your targeted future time? If you can't escape, what have you gained? I love how you guys can communicate using the math, but it's unintelligible to me. I do well thinking in terms of images and thought experiments, but don't have the math background to understand the language you use to relate your points. It's my problem not yours.

 

[PeterDonis]

What I've actually implemented is

Yes, this looks ok to me.

 

[Jorrie]

For initial values you set E, the radial coordinate at launch, R, and the angle ψ\psi, which is what an observer hovering at R measures as the launch angle (0 is "straight up"). In this context L is a derived quantity:

Looks good, but like Peter, I'm uncomfortable with your E=1 that plunges in. What E are you using for escape energy?
The large E looks like a particle that is very relativistic 'at infinity'. I would like to see what would happen to a particle that would be nearly at rest at infinity, i.e. just above escape energy.

 

[Jorrie]

How much energy would it take to break away from your orbit once you've reached your targeted future time? If you can't escape, what have you gained?

That's where the "zoom-whirl" scenario comes in. The breakaway is "free".

 

[stevendaryl]

Anyway, I'm pretty convinced that using black holes is a pretty wimpy sort of time travel, if you want to do it on the cheap (not spend much energy). It looks like the best you can do is get time dilation factors that are less than a factor of 2 or 3. So you're not going to get 1000 years into the future using relativity.

So broadening the question a little:

Suppose we have two events e_1 and e_2 that have a timelike separation, and we have two different timelike paths, \mathcal{P}_1(s) and \mathcal{P}_2(s). Let \tau_1 be the proper time for the first path and \tau_2 be the proper time for the second path.

Question for geodesics: If the two paths are geodesics, is there a limit to how large the ratio \frac{\tau_1}{\tau_2} can be? My original thought was that if one of the geodesics got close enough to a black hole event horizon, while the other stayed away, then the ratio could become arbitrarily large. But it seems that that doesn't work. So is there a limit to the ratio imposed by GR? If there is no limit, then what kind of spacetime geometry could have arbitrarily large ratios?

Now that I think about it, the usual twin paradox, which is non-inertial, could probably be done with geodesics if you had stars arranged in the right pattern. You have the stay-at-home twin, who stays on Earth (or orbiting Earth, to make it geodesic), and you have the traveling twin who uses the gravitational slingshot effect again and again to get to higher and higher velocities, all the while remaining inertial.This would require a very specific arrangement of planets and stars, so it's not really what I was looking for.

Question for noninertial paths: Broadening the question a little: Let's let one of the geodesics be noninertial. Then the ratio can become arbitrarily large, if you're willing to expend energy to force an object to follow one of the paths. The sort of open-ended question is: Is there a "cheapest" way to achieve a given ratio, in terms of energy?

 

[checksix]

Slightly off topic for a physics forum, but an interesting sf novel written back in 1970 by Poul Anderson explored the issues of forward time travel using relativistic speeds: https://en.wikipedia.org/wiki/Tau_Zero

 

[Jorrie]

Maybe you could visit Sir Roger Penrose's "black hole energy extraction city" and hitch a ride on one of the garbage containers! See the figure below that I scanned from Gravitation, Misner, Thorne, Wheeler (MTW), fig.33.2.

It shows a mega-city built on a rigid structure around the equator of a massive, spinning black hole. The city is at just the right distance from the hole so that the local gravity is at a comfortable 1g and that tidal forces and inertial frame dragging do not affect the structures adversely.

The city's garbage is processed at the top left by dumping it into suitable containers and dropping the containers at a carefully chosen angle towards the black hole. Due to frame dragging, the containers swing around the black hole in its ergosphere, and so "steal" some of the black hole's angular momentum. Just before the container starts to leave the ergosphere, the garbage is ejected towards the hole, to be "swallowed" by it.

As I understand it, the garbage essentially attains negative energy as it enters the black hole, causing the hole's rest mass to decrease. The "lost" energy (or most of it) is added to the container as kinetic energy and it can become highly relativistic. I'm not sure what the time dilation ratio would be for a "realistic scenario", but at least it might solve your energy requirements problem. I suggest that you jettison your capsule from the container before the city catches it for converting the extra kinetic energy into electricity. The g-forces might be severe in the catching action!

Then all you need to do is find other black holes, stars, whatever, to shape your trajectory so that you can inertially return to the city. Perhaps then traveling around the city's black hole again, but in a retrograde sling-shot, can decelerate your capsule so that that you can land at the spaceport, much younger than your own generation.

Beautiful sci-fi, but all calculable with GR.

 

[PeterDonis]

I would like to see what would happen to a particle that would be nearly at rest at infinity, i.e. just above escape energy.

I think it's possible that a particle that has just the escape energy will fall in if its periapsis is less than $r = 4M$. For any $r$ less than that, the effective potential is positive for an unstable circular orbit; that translates to energy having to be greater than escape energy for a trajectory that is tangent at periapsis to the same circular orbit.

In other words, I think it's possible that your earlier argument does apply to the case of a particle with energy just equal to escape energy.

 

[PeterDonis]

like Peter, I'm uncomfortable with your E=1 that plunges in. What E are you using for escape energy?

I think the resolution of the apparent "paradox" here is that in the GR Schwarzschild geometry it is not case (as it is in Newtonian gravity) that a trajectory with $E = 1$ will escape regardless of its direction. Heuristically, the range of angles in which an $E = 1$ trajectory will escape gets narrower as the $r$ coordinate decreases. $r = 4M$ marks the point at which a tangential $E = 1$ trajectory no longer escapes, and as $r \rightarrow 2M$, the range of angles at which an $E = 1$ trajectory escapes gets narrower and narrower around the radial outward direction.

I have seen this presented in terms of outgoing light trajectories (for the lightlike case, $r = 3M$ represents the point at which a tangential trajectory no longer escapes), but it would have to work for timelike trajectories as well. I have not worked out the details of how the math shows this, though.

 

[Ibix]

Looks good, but like Peter, I'm uncomfortable with your E=1 that plunges in. What E are you using for escape energy?

I have not worked out the details of how the math shows this, though.

I'm slightly suspicious of my own results, I must say. For example, I agree $E <1$ shouldn't escape because as $r\rightarrow\infty$ $E\rightarrow dt/d\tau$, which in the nearly flat spacetime far from the hole should be $\gamma$. So an escape with $E<1$ implies $\gamma <1$. But my E=0.9 trajectory looks like it escapes. I think I need to do some checking.

Edit: I've either messed something up, the integrator isn't appropriate, or the turnaround on that E=0.9 trajectory is much further from the hole than I guessed (aware I can actually check, not guess!)

 

[Jorrie]

What Peter said makes sense, so your program is probably correct. You can set $\psi$ to zero (i.e. radially or near it) and it should escape with E=1.