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Assuming a miraculous arrangement of a string of black holes for slingshot purpose, and a solid ball slinging by these black holes one after the other, gaining speed with each slingshot, is it possible in theory, that the ball reaches/exceeds the speed of light at some point? If not, then assuming the string of black holes extends for ever, at what point will the ball stop accelerating and why?

I got couple of answers that basically limit the speed to c. If that is the case, then it is obvious that the acceleration has to be less and less with each sling.

What is the reason for that reduction in acceleration? Is it because ball itself will turn into a black hole and it will start imparting as much energy to other black holes as it gains from them? Or because, there is some other relativistic mechanism involved here? Like, it will start emitting GWs to protect c.

The question was closed sighting another question. That is not the right question for this purpose. Because the other question is about light. Light can gain energy without gaining speed, just by gaining frequency. So, it does not need to gain speed.

kpv
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    Related: https://physics.stackexchange.com/q/36500/2451 , https://physics.stackexchange.com/q/24319/2451 , https://physics.stackexchange.com/q/73279/2451 and links therein. – Qmechanic May 30 '21 at 09:08
  • No, the ball won't turn into a black hole. See https://physics.stackexchange.com/q/3436/123208 OTOH, two speeding balls could turn into a black hole if they smash into each other, but that requires them to have huge $\gamma$ in the collision centre of mass frame. – PM 2Ring May 30 '21 at 15:20
  • Also, SR only places a restriction on a body's velocity, not its acceleration, momentum, or KE. However, space isn't a perfect vacuum, so a body travelling at insane speed faces resistance from that stuff. Even the CMB acts as a source of friction. – PM 2Ring May 30 '21 at 15:24
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    Re "...it is obvious that the acceleration has to be less and less...", it may be obvious, but it just ain't so. The speed is limited to (asymptotically approching) c because that's the way the universe works. That is, you could (given some handwavium fuel source) keep accelerating at 1 G forever, but would never reach c. Doesn't matter HOW you do the accelerating. – jamesqf May 30 '21 at 17:32
  • @jamesqf: Shouldn't there be a difference between some other source (e.g.) rocket, and gravity. Gravity is curvature of space, so would the limit of c still apply when gravity is accelerating something, as compared to something accelerating the body through space? – kpv May 30 '21 at 17:45
  • @jamesqf What you're saying is true of proper acceleration, but OP clearly isn't talking about proper acceleration since motion under gravity alone has no proper acceleration. I expect they're referring to some sense of SR-esque coordinate acceleration, which cannot remain 1 G indefinitely. – jawheele May 30 '21 at 19:30
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    ''is it possible in theory, that the ball exceeds the speed of light at some point'' - I think you already know the answer to that question – Tom May 30 '21 at 20:03
  • There’s less total acceleration each time because the ball is moving faster and spends less time being accelerated by each succeeding black hole. With relativistic time shortening, this would be even less time from a relative perspective. – RBarryYoung May 30 '21 at 20:26
  • Also, slingshotting doesn’t really work like this. – RBarryYoung May 30 '21 at 20:28
  • @kpv: No. See Einstein's equivalence principle: https://en.wikipedia.org/wiki/Equivalence_principle – jamesqf May 31 '21 at 16:10

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Since, the ball keeps experiencing gravitational force, The ball will never stop accelerating. But as its velocity increases, the magnitude of acceleration will keep getting smaller and smaller .

Velocity will approach c, but the closer it gets to c, the smaller will be the acceleration, and hence the smaller will be the subsequent increases in velocity. Hence, velocity will keep approaching c, but never reach c, because acceleration will keep getting smaller and smaller

silverrahul
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No, it is not possible in theory. The short reasoning is: we model the motion of a test particle under gravity via timelike geodesic motion in general relativity, "faster than the speed of light" means a spacelike tangent vector, and geodesics preserve their timelike vs. spacelike vs. lightlike nature.

A more physical argument uses conservation of energy in the rest frame of each black hole individually. For simplicity of the argument, I'll assume the black holes are very far apart, so we can consider them separately and meaningfully discuss the speed of the test mass in a special-relativistic manner between the slingshots. The way a slingshot maneuver works is that the test mass' initial and final speeds are the same in the rest frame of the black hole due to conservation of energy (this holds exactly in GR if one models the black hole as Schwarzschild or Kerr in its rest frame), but its velocity gets redirected. If this velocity is redirected in such a way that it's more aligned with the motion of the black hole in your lab frame, then in the lab frame its speed has increased. See the still taken from a gif of an example scenario on the gravity assist Wikipedia article below:

enter image description here

The "planet" here would be our black hole and the "sun frame" would be our lab frame, but the essentials of the dynamics are the same. The blue trajectory is that of the test mass, while the black trajectory is that of the black hole. To be quantitative: if $v$ is both the initial (nonrelativistic) speed of the test mass in the black hole's frame and the speed of the black hole in the lab frame, then in the pictured scenario the test mass' speed increases from $\sqrt{2} v$ to $2v$ in the lab frame.

So, what about passing the speed of light? The point is that, in each maneuver, the test mass' initial speed is the same as its final speed in the rest frame of the appropriate black hole, so its speed in the lab frame after each maneuver is always some Lorentz transformation of a speed less than $c$. Since it was going less than the speed of light in the black hole frame, it will still be going less than the speed of light in the lab frame-- this is a constraint set by the relativistic velocity-addition formula.

As a demonstration: in the example above, if the speed $v$ were relativistic, the initial and final speeds $u_i,\, u_f$ of the test mass in the lab from would be (using units with $c=1$)

$$u_i = v\sqrt{2-v^2}, \;\; u_f = \frac{2v}{1+v^2}.$$

Notice $u_f < 1$. The test mass need not ever stop accelerating, as it can always change its direction, but it will eventually be unable to meaningfully gain speed due to the above constraint of the addition formula.

jawheele
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You would just create a series of accellerations, getting smaller and smaller so you would gain a series that converges in it´s velocity to lightspeed for a series of infinitely many black holes you use as slings. The object you shoot would gain mass according to E = mc^2 with each acceleration so that your projectile would run according to the earlier mentioned converging series.

Maximilian A.
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Firstly, the ball will never transform in a black hole. It gets accelerated by the black holes but a mass that's not a black hole initially can never turn into a black hole by accelerating it. Only when it passes the event horizon of one of the black holes you have in mind will it become a part of a black hole, thereby increasing the size of the black hole it has fallen into.
Depending on the configuration of holes you have in mind, and the initial position and velocity of the ball, the ball will escape the row after having interacted with it, the ball will continue its travel in a stationary way, zig-zagging around the holes (though eventually, due to the emission of gravitational waves, it will enter a black hole), or it will fall in a hole somewhere in the row.
When the ball acquires linear momentum, it will introduce linear frame-dragging (no Schwarzschild radius will ever form, because the mass-momentum tensor involves only restmass, the momentum part will introduce frame-dragging). so the black holes can gain momentum in the direction of the velocity of the ball, leaving them in motion. This will eventually stop the ball, if it zig-zags. On top of GWs.

To answer your question, the ball will only reach the speed of light if it enters a hole and if it started its travel at an infinite distance from the holes. If not, the velocity will be less than the speed of light at the exit of the row, but higher than the initial velocity. A problem might be that in the neighborhood of the hole time is slowed down. Your ball seems to slow down close to the Schwarzschild radius of a hole. so depending on how close your ball gets, you might have to wait a considerable time for your ball to be slung out with high speed.

Deschele Schilder
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  • actually, I'm not sure ... what will happen, once the Schwartzschild radius of the object surpasses it's "Euklid" radius, because of relativistic mass increase? – Apfelsaft May 30 '21 at 20:04
  • @CarlBerger i don't think the ball will ever acquire a Schwarzschild radius. When it obtains linear momentum only the momentum part of the mass-momentum tensor contributes to the curvature of the ball (only rest-mass is involved in this tensor). so a linear motion only results in frame-dragging. this has no significant impact on this problem though. The black holes might acquire a little velocity in the velocity direction of the ball. I'll edit for completeness. – Deschele Schilder May 30 '21 at 20:14
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enter image description here

(Don't mind the v^2 thing to the right, it's not important, I just copied it from wikipedia)

enter image description here

These two explain that the speed (v for velocity, but it can be seen as speed really) can never surpass that of light because gamma will become infinitely large at c (1/0), meaning that the energy needed will become infinitely large.

Kelvin Bouma
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