A 4D Black Hole?

One feature of a universe undergoing Flat Universal Granular Expansion is that the density of the universe is critical and that density is equal to the density of a Schwarzschild black hole with the radius equal to the age of the universe multiplied by the speed of light.

An obscured assumption there is of a 3D black hole, but the universe has at least one more dimension, since time and space are interchangeable under general relativity (see On Time for an example of this interchangeability).  So … the question arises, what happens with a black hole in more than 3 dimensions?

There’s some disagreement as to whether black holes are already 4D, because they exist in a 4D universe (with three dimensions of space and one of time) and they persist over time.  While most say yes, of course, I have seen the occasional person say no, of course not – basically because there is an inherent assumption that we only talk about spatial dimensions in relation to objects.  We can’t imagine a .  That’s not really the question I was pondering though.  I was thinking about how mass would be distributed in a spatially 4D black hole, working from the intuition that because there is more volume in the 3D surface of a 4D glome than in a sphere, then there would need to be more mass.

This is going about things the wrong way though.

Consider a standard Schwarzschild black hole, a mass (M_S) that is neither charged nor rotating, in a volume defined by its radius (r_S=2GM/c²).  This is the point at which the escape velocity is the speed of light, meaning that nothing, not even light can escape.

We can show this easily – to escape, a body must have more kinetic energy than its gravitational potential energy, or

[K+U(r)=½mv²-GMm/r]>0

 

So, at the limit where the kinetic energy is not quite enough,

½v²=GM/r

 

Since we are talking about where the limit is the speed of light at r_S, and rearranging we get:

r_S=2GM/c²

 

Note that we are effectively talking about vectors here, a body must have a velocity, not just a speed, in order to escape.  In other words, the body must have a velocity greater than v=c in the direction of the separation from the mass M.  Every point on the sphere defined by the radius r_S around the centre of mass M_S defines a vector from the locus of the sphere.  The sphere, however, is not really relevant to each body that is able or is not able to escape.  All that matters is the vector from the centre of the mass and the body’s own centre of mass.

The same applies to a 4D black hole and also a hypothetical 2D black hole.  The limit will always be given by r_S=2GM/c², meaning that mass does not change.  A 4D (along with a hypothetical 2D) black hole has the same mass as a 3D black hole.

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Note that if we talk in terms of natural Planck units, the equation resolves to r_S=2M, because both G and c are effectively just exchange rates between units, and it can be seen that the radius is two units of Planck length for every unit of Planck mass.  This could have been intuited from the Mathematics for Imagining a Universe, where the conclusion that “mass … is being added to the universe at the rate of half a Planck mass … per Planck time” is reached, noting that in the scenario, that universe is expanding at the speed of light.

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Some may note that there is a problem.  In On Time, I noted that the equation for kinetic energy is an approximation.  It is noted elsewhere that E_k=½mv² is a first (or second) order approximation.

As v->c, the approximation becomes less valid, as E_k=mc².(1/√(1-v²/c²)-1) tends towards infinity.  The derivation above explicitly uses v=c, so … it doesn’t work.  (Other than the fact that, obviously, it does.)

So, perhaps it’s better come at it from the other side.  From outside of the black hole, using the relativistic effects of gravitation.

Time is dilated and space is contracted (according to normal parlance) in proportion to the force of gravity.  The relevant equation is this:

t₀=t_f.√(1-(2GM/c²)/r)

 

Note that t₀ is “proper time” between two events observed from a distance of r from the centre of the mass M, and t_f is the “coordinate time” between the same two events observed at distance of r_f such that the gravitation acceleration experienced (a_f=GM/r_f²)≈0 (and assuming that no other effects are in play).  This can be made even more complicated by considering that the events must be either collocated or equidistant from both observation locations, but we could just be a little less stringent, and say that t_f is “normal time” and that t₀ is “affected time” (that is, affected by gravitation due to proximity to the mass M).

As r decreases with increasing proximity to the mass M, affected time between standard events (in the normal frame) gets shorter until, when 1-(2GM/c²)/r=0, all normal events have no separation at all.  This represents a limit as you can’t get less than no separation (and what you get instead if something were to approach more closely to the mass M is the square root of a negative number).

The value of r at this limit is, of course, r=r_S=2GM/c².  Note that while we have used the value r, we were not specifically talking about a radius, merely a separation.  Therefore the same principles are mentioned above apply.

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But, of course, that equation is predicated on “the Schwarzschild metric, which describes spacetime in the vicinity of a non-rotating massive spherically symmetric object”, so it already assumes r_S=2GM/c².

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So … we have to go deeper.  Looking at relativistic kinetic energy, we can see that it’s not quite so simple.  At relativistic speeds, kinetic energy (K, to remain consistent with above) is given by:

K=√(p²c²+m²c⁴)-mc²

 

Where p is the linear momentum, p=mγv, where γ=1/√(1-v²/c²) which approaches infinity as v->c.  Substituting and rearranging the above then, we get:

K=√((mγv)²c²+m²c⁴)-mc²=√(m²γ²v²c²+m²c⁴)-mc²

K=mc²(√(γ²v²/c²+1)-1)

But since γ=1/√(1-v²/c²), and just focussing on the term within the square root for the sake of clarity:

γ²v²/c²+1=v²/c²/(1-v²/c²)+1=v²/(c²-v²)+1

=v²/(c²-v²)+(c²-v²)/(c²-v²)=c²/(c²-v²)=1/(1-v²/c²)

And so:

K=mc²(√(1/(1-v²/c²))-1)= mc².(γ-1)=mγc²-mc²

 

Which is the same equation as shown in On Time (just the terminology is different), which puts us back to where we were before.

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So, while the intuitive solution has some procedural difficulties, we really need to go via the derivation of the Schwarzschild metric to see that r_S is the radius at which the metric is singular (in this case meaning that there is a division by zero).

It is interesting that, in this case, using an approximation that is based on the assumption that v<<c works perfectly in a situation where v=c.  Why that is the case is currently unclear to me.

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Update: Having thought about it a little more and looking at some other views on why the 2 factor is in the Schwarzschild radius, my suspicion is that the curvature of space in the region around a black hole is such that the escape velocity at any given distance from the black hole’s centre of mass (where r>=r_S) is given by v=√(2GM/r), even if that is the result you might reach simplistically (and by making incorrect assumptions about the value of kinetic energy at relativistic speeds).

Here's my thinking – the escape velocity at the Schwarzschild radius is, due to completely different factors, of a value such that the following equation is true: ½mv²=GMm/r, where m is irrelevant due to being on both sides of the equation, M is the mass of a black hole, v is the escape velocity (in this case c) and r is the distance between the body in question and the centre of the mass M (in this case the Schwarzschild radius).  At a nominal distance r very far from the centre of the mass M, the kinetic energy of a mass m at escape velocity v is indistinguishable from ½mv² and is equal in value to the gravitational potential energy GMm/r, so the equation ½mv²=GMm/r is true (noting that, again, m is irrelevant).

Since we have picked a nominal distance, we could conceivably pick another distance which is slightly closer to the centre of the black hole and have no reason to think that ½mv²=GMm/r is no longer true.  As we inch closer and closer to the black hole, there is no point at which we should expect the equation to suddenly no longer hold, particularly since we know that it holds at the very distance from the black hole at which escape is simply no longer possible because we bump up against the universal speed limit of c.

I accept that it’s not impossible that the escape velocity might gradually depart from the classical equation (becoming lower in magnitude due to the fact that we must consider relativity), only to pop back up to agreement with classical notions right at the last moment.  I don’t have a theory to account for that, and have no intention to create one, but my not having a theory for something doesn’t make it untrue.  It just seems unlikely.  To me.