Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: PhDifferent on 29/06/2021 08:23:39
-
In order to escape from the gravitational field of a large object it's necessary to achieve escape velocity. Since the equations of motion are reversible, an object falling towards a massive object will arrive at the surface travelling at escape velocity. For a black hole at the event horizon, escape velocity is the velocity of light. Under normal circumstances no object with mass can go that fast because the relativistic mass increase means that the faster it's going the heavier it gets, and the harder it is to push it faster, and its mass approaches infinity as its speed approaches light speed. However, in this case the force causing the accelleration is gravity itself, which increases as its mass increases, so the two effects should cancel out.
The implications of this are thoroughly ridiculous, so could somebody please point out the error in the argument?
-
Hi. I hope you are well.
....so could somebody please point out the error in the argument?
Here's some issues:
1. Escape velocity is required to escape to infinity: An object launched from Earth with escape velocity will travel infinitely far away from Earth, reducing velocity as it goes and eventually (as t → ∞) come to rest with respect to the earth when it is infinitely far away. Reversing the equations, an object falling toward Earth only hits earth with escape velocity if it came from infinitely far away where it was initially at rest with respect to the earth. If the object was at rest with respect to earth but much closer to the earth (let's say 100 metres altitude) then it hits earth a velocity much lower than the escape velocity. It takes an infinite amount of time for an object to reach infinity after being launched with escape velocity or conversely to come from infinity and fall toward the earth, so we can argue that it has never happened.
2. There are complications defining velocity ("peculiar velocity") in General relativity. An object in free-fall does not experience any force or acceleration. It's not that the object is being accelerated toward the black hole, it isn't. The object itself doesn't need to have any significant velocity through space. Sometimes people think of the space that the object is in as being dragged into the black hole while the object remains quite still in space. The relativistic effects that you were describing (e.g. relativistic mass) are concepts from special relativity - this a theory that applies locally only. Locally, the object is not moving through space.
It's only under Newtonian physics that Black holes are modelled as something that provides a gravitational force and accelerates objects tword them. There is a Newtonian concept of a "dark star" where an object would need an escape velocity equal to the speed of light to escape from the star's gravitational field but this is not a good model of what a black hole is.
-
In order to escape from the gravitational field of a large object it's necessary to achieve escape velocity.
Not so. Rockets have escaped from Earth without ever having reached the 11.2 m/sec escape velocity at its surface. A slow winch will do if you have one (like a space elevator).
You seem to be applying rules of inertial coordinate systems to non-inertial cases. Yes, given one rock falling in near the event horizon passing an identical rock going the other way at escape velocity, each of them will have near infinite mass relative to the local inertial frame of the other.
Relative to the distant observer, either rock has kinetic energy that cancels out the negative gravitational potential energy, leaving no change to the total energy of the rock. The rock has negligible coodinate speed (it never reaches the EH according to the distant observer), so the mass isn’t a function of relativistic speed as you are painting it. Plus, the rock having more mass than does the black hole doesn’t make sense.
-
In order to escape from the gravitational field of a large object it's necessary to achieve escape velocity.
Not so. Rockets have escaped from Earth without ever having reached the 11.2 m/sec escape velocity at its surface. A slow winch will do if you have one (like a space elevator).
You seem to be applying rules of inertial coordinate systems to non-inertial cases. Yes, given one rock falling in near the event horizon passing an identical rock going the other way at escape velocity, each of them will have near infinite mass relative to the local inertial frame of the other.
Relative to the distant observer, either rock has kinetic energy that cancels out the negative gravitational potential energy, leaving no change to the total energy of the rock. The rock has negligible coodinate speed (it never reaches the EH according to the distant observer), so the mass isn’t a function of relativistic speed as you are painting it. Plus, the rock having more mass than does the black hole doesn’t make sense.
To be fair, in order to escape a body, you eventually have to reach "escape velocity", it's just that it doesn't need to be the surface escape velocity. For instance, if your rocket has reached a distance of 6378 km above the Earth's surface, and is moving at 7.91 km/sec, It has achieved escape velocity for that distance from the Earth. It can cut its engines and coast, never to return.
-
If you mean that its relativistic mass increase to infinity, if we define 'c' to be a sort of infinity, then yes in a Schwarzschild black hole passing the 'real' event horizon. It should also, as I think, reach 'c' as it does it. Can't prove it though.
But that's the relativistic mass. As it is in a gravitational acceleration, locally defined impossible to differ from a 'free fall' (ignoring tidal forces), there is no " local real acceleration " to it, and the relativistic mass will locally defined shine with its absence. It's a interesting idea if we could get it to collide in/at that event horizon. Then the relativistic energy must play a role.
=
I don't find it ridiculous at all. But it if it does, it happens outside our observations as you need it to pass that local 'real' event horizon. It must find it locally defined, in its own 'proper time', but redshifts and time dilation relative our 'far away observer' should make it incredibly tricky to define/observe.
Actually I don't think we would be able to see it? The radiation from such a collision should be past that event horizon without any 'paths' except those keeping them inside that event horizon. Although, we should be able to observe some collisions outside that limit. I'm not sure at all what we would be able to see though.
-
Does an object falling into a black hole acquire infinite mass?
If you spiral into a black hole in a looping orbit, you will radiate a considerable amount of energy away in the form of gravitational waves.
So (the total mass of the black hole at the end of the collision) will be less than (the original mass of the black hole) + (the original mass of the unfortunate astronaut).
PS: There have been suggestions that an advanced civilisation could harness a friendly local black hole as a power source, extracting a far greater % of energy out of matter than does hydrogen fusion...
-
In order to escape from the gravitational field of a large object it's necessary to achieve escape velocity.
Not so. Rockets have escaped from Earth without ever having reached the 11.2 m/sec escape velocity at its surface. A slow winch will do if you have one (like a space elevator).
You seem to be applying rules of inertial coordinate systems to non-inertial cases. Yes, given one rock falling in near the event horizon passing an identical rock going the other way at escape velocity, each of them will have near infinite mass relative to the local inertial frame of the other.
Relative to the distant observer, either rock has kinetic energy that cancels out the negative gravitational potential energy, leaving no change to the total energy of the rock. The rock has negligible coodinate speed (it never reaches the EH according to the distant observer), so the mass isn’t a function of relativistic speed as you are painting it. Plus, the rock having more mass than does the black hole doesn’t make sense.
To be fair, in order to escape a body, you eventually have to reach "escape velocity", it's just that it doesn't need to be the surface escape velocity. For instance, if your rocket has reached a distance of 6378 km above the Earth's surface, and is moving at 7.91 km/sec, It has achieved escape velocity for that distance from the Earth. It can cut its engines and coast, never to return.
Technically escape velocity is parallel or greater to the normal at the Earth's surface with no atmosphere to beyond a geosynronous orbit. An escape force is a better description, but then you still have to punch through the troublesome atmosphere which even at the thinness of the upper atmosphere is a problem, as seen in heat shields on spacecraft.
-
Thanks for all the replies, folks.
Yes, I realise that my first couple of sentences on escape velocity were an oversimplification, but I was in a hurry to get on to the real question. Let's say, for the sake of this thought experiment, that I could go to a "safe" distance from the black hole, where the escape velocity is only equal to the muzzle velocity of a rifle, and then fire a shot accurately at the centre of the black hole, managing to avoid the bullet going into orbit, or colliding with the orbiting junk in the accretion disk.
Now the bullet starts and ends its travel at EV, so when it reaches the event horizon it should be travelling at c, with all the weird and wonderful consequences of that. (Oh all right, at some point its mass will become comparable with that of the black hole, which will start to move towards it, so its velocity won't quite reach c :-)
Or is it the case that relativistic mass is somehow different from rest mass, and is not acted on by gravity in the same way?
-
That seems a incredibly tricky question to me, and it depends on how you look at it. You have a momentum as you 'move' in a relative motion. It's easy to prove it existing, just collide with something. That momentum represents a energy and relativity is about energy, the 'stress energy tensor'.
But it's also 'frame, or, observer dependent'. What that state is that this collision you get involved in is a result of a system consisting of you, and what you collide with. And as there is no golden standard for your, or your counterpart 'speed', it means that you can't decide who had what speed in that collision. It means that this frame dependence always exist.
so what are your 'relative motion'?
Your 'speed'?
That we have a limit doesn't tell you a thing about it. And the same should be applicable on accelerations.
-
If you want to look at it as everything is relative, then no, it's not a 'gravitational force' locally defined. If you want a 'global approach' then yes. It must exist, and express itself, globally defined. That global approach versus a local seems more of a abstraction though to me, than a local reality.
=
A easy example is you colliding in your car with a brick wall 'at rest' with earth. Earth has, depending on definitions and choice of comparison, a infinite amount of different relative motions trough space, simultaneously, if you want it to become abstract. With those 'motions' follows different momentum's, and as I think also possible directions. You can pick your choice there. So you can get to different quantities of relative (relativistic) mass existing, simultaneously, although none of them interfering with your calculation of the total energy involved in that crash.
-
Now the bullet starts and ends its travel at EV, so when it reaches the event horizon it should be travelling at c
No. In the coordinate system used by the the guy firing the shot, EV is the speed of light, but speed of light approaches zero, not c, at the EH. So the bullet ends its travels at a stop without ever crossing the EH, at least in that coordinate system. No relativistic mass change. If you want a different speed, you're going to need a different coordinate system, such as say one where the bullet actually goes in.
Meanwhile, in the coordinate system of the guy with the gun, the bullet approaches a contracted length of zero as it approaches the EH, so it is still traveling at X many bullet-lengths per second, which is a different way of measuring speed, a sort of proper velocity, which can exceed c even in flat spacetime just like the speeds used to express recession rates of distant galaxies.
-
Yeah, that's another interesting effect from it :)
frame dependencies
-
This one describes it from the frame of earth, versus the frame of a infalling muon. From earth it gets defined as a time dilation, from the frame of the muon as a length contraction. http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html
and that is also what I call a 'global approach' to relativity
-
And no, locally defined it can't become a 'black hole', and neither should it be able to become one from any other point of view. Don't remember where it is today, but (John?) Baez wrote a piece on that one using two spaceships passing each other (in opposite directions) at relativistic speeds, aka both being in a uniform motion.
=
If you're curious about it, it could be here. https://math.ucr.edu/home/baez/physics/
Sorry, couldn't find the one I was thinking of, it's old but it should be somewhere on the Internet.
This might suffice although the mathematics seems to be missing.
" If you go too fast, do you become a black hole?
When an object approaches the speed of light, its mass increases without limit, and its length contracts towards zero. Thus its density increases without limit. Sometimes people think that this implies it should form a black hole; and yet, they reason, since its mass and volume haven't changed in its rest frame, it should not form a black hole in that frame—and therefore not in any other frame either. So does a black hole form or not?
The answer is that a black hole does not form. The idea that "if enough mass is squeezed into a sufficiently small space it will form a black hole" is rather vague. Crudely speaking, we might say that if an amount of mass, M, is contained within a sphere of radius 2GM/c2 (the Schwarzschild radius), then it must be a black hole. But this is based on a particular static solution to the Einstein field equations of general relativity, and ignores momentum and angular momentum as well as the dynamics of spacetime itself. In general relativity, gravity does not only couple to mass as it does in the newtonian theory of gravity.
Gravity also couples to momentum and momentum flow; the gravitational field is even coupled to itself. It is actually quite difficult to determine the correct conditions for a black hole to form. Hawking and Penrose proved a number of useful singularity theorems about the formation of black holes. But even these theorems do assume certain conditions which we cannot be sure are true "out there". "
and 'mass' in this terminology avoids the term relativistic mass, instead referring to it as momentum.
-
Well, a collision of such falling object and a hovering object is like collision of infinite mass object and a normal mass object.
I mean like fast proton hitting a still standing proton in a particle accelerator. That kind of collision.
-
The notion of the black holes is as follows:
A black hole is a region of spacetime where gravity ... The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.
After a black hole has formed, it can continue to grow by absorbing mass from its surroundings. By absorbing other stars and merging with other black holes, supermassive black holes of millions of solar masses (M☉) may form. There is consensus that supermassive black holes exist in the centers of most galaxies.
https://en.wikipedia.org/wiki/Black_hole
-
Hi.
That's a reasonable or workable definition of a black hole. Thanks for stating the source of information.
That definition is not without problems:
1. The singularity that General Relativity (GR) shows at what we could call the centre of the black hole is not necessarily a region of spacetime. If it exists then it exists outside any manifold we can find as a solution to Einstein's Field Equations. So it is not a region of any spacetime that we can model. Leading to two possibilities:
(i) It is not a region of our spacetime.
or, (ii) The model (GR) is insufficient.
In either case, this definition or model of a black hole does not include the most important singularity. That's OK and probably unavoidable.
2. You have described a black hole by it's origin or formation. Your definition stated that a black hole was caused by a "compact mass" that deformed spacetime. This is OK but it isn't just ordinary mass that could deform spacetime sufficiently. Any source of mass-energy influences spacetime. It is possible to imagine a shell of photons (radiation) being created and directed toward a single point in space. As the shell of radiation converges on that point the mass-energy density becomes large enough to deform spacetime locally. An event horizon is thought to form and appears to grow, coming out to meet the inbound shell of radiation. A conventional black hole seems to have been created using only radiation (energy) as it's seed. More interestingly, the radiation creates an event horizon before it actually enters the interior of the event horizon. It seems that there is some time when there was no matter or energy at or near the centre of the black hole but a black hole has formed anyway.
Alternative definitions of a black hole could be presented. I tend to think of a black hole as being a region of spacetime bounded by a surface that is called the event horizon. That surface is defined by r=2GM for Schwarzschild black holes and I don't worry about anything else that might technically be a black hole.
Best wishes to everyone.
-
That one was fun PHD :)
rereading you " For a black hole at the event horizon, escape velocity is the velocity of light. "
Not really, light doesn't get out.