# Something is broken about black hole theory

1. Sep 24, 2011

### x_engineer

Imagine something falling into a black hole, starting at infinity. When it gets to the event horizon, its velocity has to be c. However, from the perspective of an observer outside the event horizon, its mass has become infinite, which means it would also have an event horizon and that event horizon swallows the entire universe. Thus if anything actually falls into a blackhole, the entire universe also winds up inside the blackhole.

OK, maybe we never get so see something actually fall in.

2. Sep 25, 2011

### Polyrhythmic

Why would that be the case? That's not true.

3. Sep 25, 2011

Staff Emeritus
That's not true either.

The whole premise of this thread is wrong. Just because you don't understand something that you have not studied does not mean something is broken in the theory.

4. Sep 25, 2011

### skeptic2

Well according to Edwin F. Taylor and John Archibald Wheeler in their book Exploring Black Holes - Introduction to General Relativity, chapter 3, it is true. From Figure 5. p 3-16 [a graph showing the infalling velocities of an object from different frames of reference] "At the horizon, the shell speed rises to the speed of light (equation [24]) while the bookkeeper speed drops to zero (equation [21])."

It is not true however that mass increases as the object approaches c because although the object is accelerating, it is not being accelerated. Even if it were being accelerated, it is the momentum, not the mass that increases.

Last edited: Sep 25, 2011
5. Sep 25, 2011

### WannabeNewton

What is true? That the radial velocity of the particle approaches c? The general equation, in schwarzchild space - time, for radial acceleration is $$\frac{1}{2}(e^{2} - 1) = \frac{1}{2}(\frac{dr}{d\tau })^{2} - \frac{M}{r} + \frac{L^{2}}{2r^{2}} - \frac{ML^{2}}{r^{3}}$$ For a particle starting off at r = infinity falling in radially this reduces to $$\frac{dr}{d\tau } = (\frac{2M}{r})^{1/2}$$ so at r = 2M we have $\frac{dr}{d\tau } = 1$ so yes it does approach c. As you know, the velocity $\mathbf{U}$ of a particle can be written in terms of components in the default coordinate chart on the scwarzchild 4 - manifold (valid outside r = 2M) and of this we have $U^{r} = \frac{dr}{d\tau }$ but keep in mind that this is coordinate velocity. If you transform to eddington coordinates then you can make the singularities at r = 2M disappear so while $\frac{dr}{d\tau }\rightarrow 1$ as $r\rightarrow 2M$ this is a coordinate based result. As stated, a coordinate transformation to something non - singular at r = 2M will get rid of that result. Coordinate singularities tell us nothing about the actual geometry of the 4 - manifold in question.

6. Sep 26, 2011

### twofish-quant

BZZZT.

In 1940, a physicist named George Gamow published a popular book on relativity called Mr. Tompkins In Wonderland. In that book, he made the statement that when objects travel fast, they behave as if their mass increases.

In the subsequent decades, many a physicist wishes that Gamow hadn't said that because, it leads to all sorts of confusion. When something gets close to c, it gets harder to make it go faster, but it's gravitational force does *NOT* increase, and if you kick it from the side, it won't behave differently.

7. Sep 26, 2011

### Phrak

This is true.

8. Sep 26, 2011

### x_engineer

Thanks twofish_quant.

I could see my understanding of what happens to an object moving close to the speed of light is not correct. So, just to be clear, all that energy does not create a gravitational field?

9. Sep 26, 2011

### phyzguy

The answer to your question is, I believe, more subtle than the answers that have been given here, and it is a puzzling aspect of general relativity that was really not clarified until R.C. Tolman explained it in 1930. Imagine a cloud of 'dust' (non-interacting particles) collapsing from infinity under their mutual gravitational attraction. As they accelerate inward, they gain kinetic energy. The right-hand side of Einstein's field equation does not contain just the mass of the particles, but contains the full stress-energy tensor. So the $$T_{00} = \sum (m^2+p^2)$$ term on the right-hand side is the total energy of all of the particles, which clearly increases as the particles accelerate inward. And yet we know that the effective mass of the black hole does not keep increasing as they fall inward, and even simple Newtonian theory tells us that the gravitational field depends only on the total mass inside. So what is going on? The answer is that the energy to accelerate the particles is coming from the gravitational potential energy, which is decreasing as the particles fall inward by the same amount as their kinetic energy is increasing. This needs to be taken into account, and when it is, the total energy on the right-hand side of Einstein's equation is just the mass (or rest mass) of the particles, and it does NOT increase as they fall inward.

Here's a reference,
"Tolman, R.C., "On the Use of the Energy-Momentum Principle in General Relativity", Phys Rev, V35 #8, P.875, Apr. 15, 1930.

10. Sep 27, 2011

### x_engineer

phyzguy: Thanks!

Would the corollary be that energy is stolen from the black hole? I seem to have read some things about the black hole disappearing just before passing the event horizon from the point of view of the infalling object. From a remote observer's point of view the crossing never happens (or rather it happens in the indefinite future).

Also, there is all this talk about tidal forces ripping you apart before you reach the event horizon. Would this necessarly be true for a large enough black hole? I can visualize the math going either way.

11. Sep 27, 2011

### George Jones

Staff Emeritus
The larger the black hole, the smaller the tidal force at the event horizon. For a very large black hole, you are not ripped apart until you are inside the horizon. For a very small black hole, you are ripped apart above the horizon.

12. Sep 27, 2011

### twofish-quant

I don't think so. There are ways of removing rotational energy from a rotating black hole, but no way of removing gravitational energy from a static one.

Probably not the best way of describing what happens. The event horizon is the limit at which once you cross it, you can no longer cross back. However, when you do cross the event horizon, you don't notice anything odd at all.

From a remote observers point of view, you never see someone crossing the event horizon, because the light gets infinitely redshifted. But it does.

What you will see when someone crosses the event horizon is that they will disappear.

It happens with small black holes but not large ones.

13. Sep 28, 2011

### stevebd1

Here are some equations you might find useful regarding an object falling from rest at infinity towards a static black hole (this is often referred to as the rain frame), $M=Gm/c^2$ (equations taken from 'Exploring Black Holes' by Taylor & Wheeler)-

E/m (energy/mass ratio of the object)-

$$\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}\ =\ 1$$
where $d\tau$ is the time dilation for an object in free fall from infinity which in this case is the sum of both the time dilation for gravity and velocity- $d\tau=\sqrt(1-2M/r)\cdot\sqrt(1-v^2)=(1-2M/r)$ where $v=\sqrt(2M/r)$ for an object which has fallen from rest at infinity (see below).

Eshell/m (energy/mass ratio of object relative to shell frame)-

$$\frac{E_{shell}}{m}=\left(1-\frac{2M}{r}\right)^{-1/2}$$
Note- This is energy as observed at a specific radius as the object falls radially towards the BH and while the above tends to infinity at the EH, there would be no static observer at the EH.

vshell (velocity of object relative to shell frame)-

$$v_{shell}=\left(\frac{2M}{r}\right)^{1/2}$$
Note- No stable shell frames with the EH.

v (velocity of in-falling object as observed from infinity, dr/dt)-

$$v=\left(\frac{2M}{r}\right) \left(\frac{2M}{r}\right)^{1/2}$$

Regarding tidal forces-

$$dg=\frac{2Gm}{r^3}dr$$
which can be rewritten in respect of a change of 1g from head to toe (i.e. dr=2m)-

$$r_{ouch}=\left(\frac{2Gm}{g_E}dr\right)^{1/3}$$
and as previously stated, this 'ouch' radius will occur outside the EH of small black holes and inside the EH for very large black holes.

See question 4 in the following link-

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html" [Broken]

Last edited by a moderator: May 5, 2017
14. Sep 29, 2011

### Polyrhythmic

Why does the term on the right hand side survive the r->infinity limit? Sure, it goes to zero slower than the others, but if we assume the ideal point r=infinity, it is also zero. You can basically think about it like this: Schwarzschild geometry is asymptotically flat (i.e. minkowski at infinity). Thus if we drop a particle "at infinity" (which is of course an idealization), it will not move at all since it doesn't feel the gravitational field.

15. Sep 29, 2011

### WannabeNewton

That's not why it survives and the others don't. I said a particle falling in radially from infinity. The other terms vanish because L = 0 for a particle falling radially and e = 1 at r = infinity.

16. Sep 29, 2011

### Polyrhythmic

True, didn't notice that at first, sorry. However, my argument still holds.