Crossing the Event Horizon of a Black Hole

JesseM

There is no physical "change from real to imaginary space and time" at the horizon, it is simply the fact that the Schwarzschild "time" coordinate becomes spacelike past the horizon, and the "radial" coordinate becomes timelike. But this is just a weakness of how Schwarzschild coordinates are defined! If you use Kruskal-Szekeres coordinates, for example, in these coordinates the time coordinate is always timelike both inside and outside the horizon, and the radial coordinate is always spacelike. Also, in many ways Kruskal-Szekeres coordinates are to Schwarzschild coordinates as inertial coordinates are to Rindler coordinates in flat spacetime; just as the Rindler horizon is seen as the diagonal surface of a future light cone when viewed in inertial coordinates, and objects at constant position coordinate in Rindler coordinates are seen as having worldlines that look like hyperbolas in inertial coordinates (see the second diagram on this page), so it is similarly true that the black hole event horizon looks like a diagonal line in Kruskal-Szekeres coordinates (these coordinates have the special property that all lightlike surfaces appear diagonal on a graph), and objects at constant Schwarzschild radius have worldlines that look like hyperbolas as well. See my post #4 on this thread for a quick rundown on the basics of Kruskal-Szekeres coordinates and a few illustrative Kruksal-Szekeres diagrams scanned from the textbook Gravitation.

DiamondGeezer

No they both become imaginary.

No, they both become imaginary as well.

The Kruskal-Szekeres tranformed coordinates are:

[tex] u=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)\cosh\left(\frac{T}{4m^{ *}}\right)[/tex]

and

[tex] v=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)sinh\left(\frac{T}{4m^{* }}\right)[/tex]

which also become imaginary because of

[tex]\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}[/tex]

when r<2M

Neither Kruskal in his original paper, nor Misner/Wheeler/Thorne in Gravitation ever deal with the fact that however the Schwarzschild Metric is changed, the worldlines become imaginary when r<2M. Instead they take great comfort from the fact that the function is continuous, as if that meant anything. If you want to see how waffley some brilliant scientists become when faced with the obvious, then re-read Gravitation and how it deals with the event horizons.

This was something of a surprise when I read Kruskal's paper that he appeared to be oblivious that after so much effort, the result was the same.



In the K-S diagram, the west and east are outside the event horizon and real, north and south are inside the EH and imaginary.

As an example of this phenomenon, consider the simplest possible function:

[tex]y= \sqrt{x}[/tex]

Conventionally, people will say that the function is only real (in both senses of the word) when x > 0 and that's where most people stop. But not so.

When you allow y to be imaginary then the function [tex]y= \sqrt{x}[/tex] is continuous through zero from [tex]- \infty <x< \infty [/tex]

What the K-S diagram does is flatten the real and imaginary parts of the graph onto a 2-d diagram.

George Jones

No, MTW does not state this. Look at the bottom of page 833. Have you deliberately misrepresented MTW? The coordinate transformation which you have given is valid only when [itex]r > 2m[/itex]. For [itex]r < 2m[/itex], the transformation (given in MTW and dozens, if not hundreds, or relativity books) is

[tex] u = \left( 1 - \frac{r}{2m}\right)^{\frac{1}{2}}\exp\left(\frac{r}{4m}\right)\sinh\lef t(\frac{T}{4m}\right)[/tex]

and

[tex] v = \left( 1 - \frac{r}{2m}\right)^{\frac{1}{2}}\exp\left(\frac{r}{4m}\right)\cosh\lef t(\frac{T}{4m}\right).[/tex]
So, you can see the "obvious," but thousands of professional relativists can't?

JesseM

To be clear, DiamondGeezer, are you claiming that any coordinate-invariant quantities determined by the metric, like the proper time between two events on a given worldline, become imaginary once you cross the horizon? (of course if you try to calculate the proper time along a spacelike path you get an imaginary number, but this is equally true outside the horizon)

DrGreg

I thoroughly disagree. In the context being discussed here, the two scenarios are very similar indeed.

(Specifically, we are talking here about near the horizon, above, at and below, and not what happens near the central singularity, which, of course, exists only in the case of the black hole.)

Oh yes we are! Do you not realise that an observer who is hovering at constant height above a black hole is undergoing outward proper acceleration? That is a rather fundamental application of the equivalence principle. The "acceleration due to gravity" that you feel is because you are properly-accelerating upwards. The Schwarzschild coordinate system is a non-inertial accelerated coordinate system.


OK, you've mentioned the "infinite energy" issue before, and I don't think anyone gave an adequate reply.

In the case of a black hole, we have already seen, earlier in this thread, that different observers hovering at different constant heights above a horizon will measure the energy of an infalling particle with a number that gets ever greater the closer the observer is to the horizon. Remember, energy is an observer-dependent quantity. Even in non-relativistic theory, ½mv² depends on what you measure v relative to. So there's no need to ask "where the extra energy comes from" when you switch from one hovering observer's measurement to another, as it's an irrelevant question: energy isn't conserved when you are comparing one observer's measurement against another's.

(N.B. the observers at different heights really are in different frames because although they are at fixed distances from each other, each has a different notion of radial distance and time -- gravitational time-dilation. Therefore their definitions of energy disagree.)

If you consider a sequence of hovering observers getting closer to the horizon, yes, in the limit as the distance tends to zero then the "physical speed" (as defined earlier in the thread) of the infalling particle tends to c and its energy tends to infinity. But these are mathematical limits and can never actually be observed by a hovering observer. No observer can actually hover on the horizon itself.

And if you want, instead, to imagine an oberver who slowly descends to the horizon, that observer isn't hovering, so the equations derived earlier are no longer valid. The difference may be tiny at a significant height, but very close to the horizon the difference will be substantial (in fact, diverging to infinity). Any observer will only ever measure a finite energy (and a "physical speed" less than c). George Jones chose his words carefully in post #16 to avoid saying that the particle's physical speed was equal to c.

Everything I've said above also applies to my example of an accelerating rocket. The only difference is that whenever I speak of a "hovering observer" above, that should be interpreted as meaning "stationary relative to the rocket, as measured by the rocket". In the diagram I referred to in my previous post, an apple dropped from the rocket when T=t=0 follows a vertical worldline on the spacetime diagram. As the apple gets closer to the horizon at x=ct, the local Rindler observers are travelling faster and faster relative to the apple, approaching c in the limit. These observers, each at rest in the rocket's frame, measure a kinetic energy of the apple that diverges to infinity in the limit. So ask yourself, where does this energy come from? Does the question make sense?

In both cases, the rocket and the black hole, the reason the falling object "gains infinite energy" (as you put it) is because the observer is accelerating towards the speed of light, relative to the non-accelerating object, not because of something happening to the falling object.

This has already been explained. You need to check the definitions of "spacelike" and "timelike". If you have an equation that seems to be giving you imaginary time, you have made a mistake and what you have actually found is a real distance. If you have an equation that seems to be giving you imaginary distance, you have made a mistake and what you have actually found is a real time. The same issues apply to both the accelerating rocket and the black hole.

I don't really understand what you are getting at here. Everything is measured relative to the rocket, so of course the rocket cannot be a non-zero distance away from itself (which is at R=1).

But this is the point. The construction of the (T,R) coordinate system is such that an object cannot cross the Rindler horizon at finite coordinate values, but it certainly does cross the horizon at finite (t,x) coordinates. This is an artefact of the (T,R) coordinate system. Similarly, an object cannot cross a black hole's event horizon at finite Schwarzschild coordinates, but it certainly does cross the horizon at finite coordinates in other coordinate systems. Again, this is an artefact of the Schwarzschild coordinate system.

Incidentally the "Rindler horizon" has something equivalent to a black hole's Hawking radiation. It is called the Unruh effect.



A final comparison between my rocket example and a black hole.

My inertial coords (t,x) are equivalent to Kruskal-Szekeres coords.

My (T,R) coords are equivalent to Schwarzschild coords.

All of this is discussed in Rindler's book which I referenced in my old post that I previously linked to.

DiamondGeezer

No. What I point out is that there is no mathematical justification for doing so if the line integral is continuous.

It's a fudge, George.

MTW fudges it by rolling out two different equations which avoid the fact that the line integral becomes complex when r<2M. Taylor and Wheeler do the same. Kruskal tried it with a different coordinate transformation but got the same result.

Are you seriously arguing that thousands of professional scientists cannot be wrong and therefore the laws of mathematics can be suspended by popular vote?

If the line integral is continuous then "The coordinate transformation which you have given is valid only when [itex]r > 2m[/itex]" and there's another one for [itex]r < 2M[/itex] then the line integral is discontinuous and you are talking about two different universes.

Or that I cannot argue that this is mathematically invalid lest I get a from George Jones?

DiamondGeezer

But their measurement of 4-energy will be the same because that is invariant and goes to infinity.

This is exactly what I have said, except that I didn't limit myself to "observers on shells near the event horizon". Every calculation of speed and energy from any inertial or non-inertial POV comes to the same answer about the speed of an infalling particle and the energy of same at the event horizon.

The relativistic energy as you know is: [itex]E = \gamma mc^2[/itex] so as [itex]r \rightarrow 2M[/itex] then [itex]\gamma[/itex] reaches [itex]\infty[/itex]

I have never claimed that the infalling particle will be observed to be moving at c. George danced around this point because he knows that the implication of a speed of c implies quite a few unphysically realistic results if the black hole event horizon exists.

I do know the difference. I'm not stupid.

Yes it does. The rocket produces energy. The rocketeers will not "see" the apple reaching c.

Neither will they or anything else reach c (which requires infinite energy that the Universe does not have). They will not reach r=1/2, which isn't on their worldline.

On the other hand, any infalling particle (or apple) apparently gets infinite energy from falling into a black hole. Every free falling particle reaches r=2M.

Or if you like, as the apple approaches the event horizon, the kinetic energy of the apple rises without limit (even past the total energy of the Universe itself).

I argue that that points to a fundamental flaw in black hole theory, one that cannot be transformed away.

If the observer is accelerating towards the speed of light, then that energy must come from somewhere. Unless the black hole has infinite energy then something is wrong.

No, what I have found is that the classical Schwarzschild Metric has been mathematically fudged to permit an infalling particle to gain infinite energy, achieve light speed and access a Universe at right angles to our own.

I question those fudges that avoid those questions.

I think there IS a transformation of the Schwarzschild Metric which leads to a mathematically consistent non-contradictory solution for the region about a mass which undergoes infinite collapse.

But if so, then black holes do not exist. Something else does.

atyy

While we're discussing energy conservation, can someone tell me please tell me if this is true:

1) Outside the horizon there is a time-like Killing vector so energy is conserved for freely falling particles.

2) Inside the horizon there is no time-like Killing vector, so energy is not conserved for freely falling particles.

?

DrGreg

Tending to infinity in the limit is not the same thing as reaching infinity.

A very, very large energy can be observed only by someone travelling at very, very high velocity relative to the apple being measured. The source of the high velocity is the rocket motor of the observer who has accelerated to that high velocity. No-one measures the apple's energy as infinite because no-one's rocket motors can accelerate the observer to high enough velocity. This applies to both my rocket example and a black hole. When you are close to an event horizon (or an apparent horizon) the natural tendency is to fall into it, and to resist that you need to expend a huge amount of energy.

Or to put it another way, there's no single coordinate system in which energy conservation applies but in which the apple's energy changes from finite to infinite as it falls. The infinite value arising in this thread was a limit of energies in lots of different coordinate systems.

The energy comes from the observer's rocket, resisting the fall into the hole, and never reaches infinity in practice because the rocket eventually runs out of fuel. Shortly afterwards, the rocket falls through the horizon at a non-zero speed, and as it does so it measures the apple's energy as finite (i.e. a "physical speed" less than c).

DiamondGeezer

Clearly that is incorrect. Unlike a rocket motor, the acceleration of any infalling particle to the event horizon involves arbitrarily high energies.

Except in the case of black holes, where any coordinate system regardless of whether it is inertial or not, measures the speed of an infalling particle to reach the speed of light and does so in the proper time of the particle (it would take an infinite amount of time from the perspective of an outside observer).


The problem is that the energy of an infalling particle into a black hole appears to rise without limit.

So either the black hole possesses infinite energy or there's something fundamentally wrong with the theory of black holes because they possess such unphysical properties.

DrGreg

I don't think anyone ever said that in this thread, and it isn't true. In fact no locally inertial system measures the particle's speed as c. As I've said several times before, that value arose by considering the unrealisable limit of a family of different observers.
Er, a freely falling particle's acceleration (i.e. proper acceleration) is zero. The observers are accelerating, the falling particles are not. Maybe this is the cause of your confusion.
All of this almost seems to suggest you think energy is an absolute concept. I'm sure you know it isn't. You have to specify energy relative to something. And, in essence, that "something" has to be a particle with non-zero mass. You can't sensibly measure energy relative to an event horizon (if you try, you get an infinite answer). There is no coordinate system with 3 space axes and 1 time axis in which the event horizon is at rest. (NB. The event horizon lies outside the two Schwarzschild coordinate systems.)

JesseM

This obviously can't be true in Kruskal-Szekeres coordinates, since only null worldlines are depicted as diagonal in a Kruskal-Szekeres diagram, all timelike worldlines have a slope closer to the vertical time axis.

DiamondGeezer

The diagonals in the K-S diagram represent the "event horizon" aka "zero".

JesseM

The event horizon is represented as a diagonal, but any other null geodesic would be represented as a diagonal too (and the event horizon is indeed a null geodesic itself, since a photon sent outward at the moment an object was crossing the horizon would remain on the horizon rather than falling in or escaping). That's just a property of how Kruskal-Szekeres coordinates work.

DiamondGeezer

I think I've repeated myself several times on this thread as to the difference between physical (observed) and coordinate speed.

Er, even a freely falling particle can ascertain that it is accelerating, even if it feels no net forces, by measuring tidal acceleration - a trick which won't work "inside" a black hole. It might also be able to ascertain that the Universe is accelerating relative to it.

I don't think energy is absolute. I do think that 4-energies are invariant.

Unless you think that a black hole is massless, it too possesses an energy which is invariant.

By the way, have you spotted why the fudge of using two different metrics inside and outside the event horizon is incorrect? Because the square root of a negative number [itex]\neq[/itex] minus the square root of a positive number. They might meet at zero, but they are strangers after that.

DrGreg

I have no more time to comment today. But I've just time to ask this:
What do you mean by that? If you'd said "4-momentum is covariant" I would have understood.

DiamondGeezer

http://en.wikipedia.org/wiki/Action_%28physics%29