Relative to a black hole...

Maniax

Hello.
If you were far away from a small black hole, and shot (for the sake of argument a bullet) with such precision so that it would just miss the event horizon, I have learned that the body falling in will gain speed as it approaches the black hole, and at the event horizon the speed is c.

But if the bullet in this case misses the horizon, it would still have massive speed built up and consequently relative to the black hole, its mass would approach infinity. Right?

Then, wouldn't the black hole feel the gravitational pull from this object, and behave thereafter?

And if this holds, then all objects with mass would make the hole wobble around alot as they fall into the hole, some will just miss it , some don't.

Like a dancing hole in the universe...hehe

Is this a correct assumption?

tiny-tim

Hi Maniax! Welcome to PF!
Yes … as it approaches the event horizon, its speed gets closer and closer to the speed of light.

In local inertial coordinates, that would be c.

And if it went inside the event horizon, then in Schwarzshcild coordinates, its speed would actually be faster than the speed of light.
The bullet only gets a finite amount of energy as it falls, Mass is energy, so its mass remains finite.
The black hole behaves as a single gravitational body, just lke anything else, so yes, technically even a bullet will attract it slightly.

Maniax

Hello!
Thanks for the welcome :)
Ok I see your argumentation, but have a few follow up's

1. The mass of my moving body will approach infinity, not become infinite - distinction there!
So, for the sake of argument again, the mass of my bullet is travelling around 0.99999999 of c when missing the event horizon, by the with of an atom - wouldn´t the bullet relative mass to the hole be huge?
Even larger than the hole - then the hole would jerk around!
This of course would then happen with all matter - even photons?

Am i thinking it wrong?

tiny-tim

Hi Maniax!

From our point of view (watching from a safe distance! ), or from that of any inertial observer, the bullet has almost zero speed, and therefore virtually no additional mass or gravitational attraction. The black hole will feel almost nothing.

The centre of mass of the bullet and black hole together does not move.

Flexo

I'm not actually sure there are paths that you could set the bullet on to make it narrowly miss the black hole. But if you could, you are right, the black hole should experience extreme pull from objects falling near it.

Relativistic mass increases as a function of rest mass and velocity. Photons have no rest mass and a constant velocity, so I don't think it would experience any change in mass.

Excuse my speculation here, but seeing as the mass of a body falling into a black hole increases by so much, wouldn't the Schwarzschild radius increase such that the object itself would turn into a black hole right before it reaches the event horizon?

Tim, why would an inertial observer see a slow bullet?

tiny-tim

Hi Flexo!

Because he would regard time dilation as approaching infinity near the event horizon, and so would regard the bullet as taking almost infinitely long to go an extremely short distance.

MeJennifer

Its rest mass will remain the same and rest mass together with energy and pressure are the only things that influence spacetime curvature.

Jorrie

For non-rotating (Schwarzschild) black holes, free-falling objects cannot pass closer then 1.5 times the event horizon radius without spiraling into the hole. In the case of an inspiral, the object will approach the event horizon at a local speed approaching that of light, but unless it is a massive thing, I do not think the hole will really notice. There is no "extreme pull" from a light-weight object (approaching the speed of light) on the hole; it's the other way around, not so?

shalayka

Would heat transfer (energy flux) and viscosity (momentum diffusion) also count as sources of curvature here?

tiny-tim

Hi Jorrie!

I don't think that's right.

1.5 is the distance at which light orbits a blac k hole.

So if light has "horizontal" velocity closer than 1.5, then is must crash into the black hole.

And even light in orbit at exactly 1.5 will be in unstable orbit, and so will either spiral in or spiral out.

But give it a kick radially inwards, from 1.5, and I think it just reaches the opposite point of the orbit with the same kick outwards, and then escapes.

Jorrie

Hi tiny-tim. You're right about the instability, but I think you've got it wrong with the effect of the 'kick'.

For any particle, including light, to move as you described, it must pass a point with pure horizontal (tangential) velocity and with r < 1.5R_hor. Unless it can move at a local speed exceeding c, it will spiral in, don't you think? Look at the 'impact parameter' of MTW.

tiny-tim

Sorry, I read MTW ages ago but I don't have a copy.

If a particle can come in to a tangential velocity, outside the event horizon, then by symmetry surely it must be able to go the other way?

Jorrie

Yes, but in order to 'come in to a tangential velocity' inside r=1.5R_hor it will need a tangential speed exceeding c. You can only have symmetry by reflecting the incoming photon to have a radially outward velocity component again.

Recall that in the region r < 1.5R_hor, there is no tangential geodesic movement possible, so all in-falling particles spiral in, moving precisely radially at the horizon. It is only possible to move tangentially in the region R_hor < r < 1.5R_hor if a positive radial force is applied to the particle.

I reckon you may be thinking of rotating (Kerr) black holes, where a particle can get a 'slingshot' from the hole and escape from within the ergosphere.