Astroquizzical: does a black hole have a shape? (Synopsis)

Does a black hole have a shape? Is there a front and back or side view? Does it look the same from all vantage points?

When you think about a black hole, you very likely think about a large amount of mass, pulled towards a central location by the tremendous force of gravity. It's not all that different from our own Sun, which is the largest mass in town. Some 300,000 times as massive as Earth, despite its rotation, the Sun is almost perfectly spherical, differing by less than 0.0001%.

Image credit: Gary Palmer, July 2005, using a violet calcium-K filter. Image credit: Gary Palmer, July 2005, using a violet calcium-K filter.

While black holes themselves may be perfectly spherical (or for rotating black holes, almost perfectly spherical), there are important physical cases that can cause them to look tremendously asymmetrical, including the possession of an accretion disk and, in the most extreme case, a merger with another black hole.

Image credit: Bohn et al 2015, SXS team. Image credit: Bohn et al 2015, SXS team.

Jillian Scudder has the entire story on her latest Astroquizzical column, with some remarkable images and videos to help visualize the whole thing!

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The images of merging black holes all seem to show light traveling between the bodies, which I imagine is the point of the column. It seem almost like a Lagrange Point that pushes back the event horizon. Would it be possible to use a black hole to retrieve something that had fallen through the event horizon of another black hole?

Infalling gasses are compressed and excited by X-Ray emissions and emit light. These gasses are what you're seeing.

You can't see light itself going in to something else: the light has to go to YOU for you to see it.

So black holes have quite the temperature fluctuation over their life span?

By Ragtag Media (not verified) on 30 May 2015 #permalink

@Ragtag Media #3: Their temperature is directly proportional to their surface area. Their surface area scales with mass.

You might want to do some research on the subject of "black hole temperature" and "black hole thermodynamics" (even just the Wikipedia articles) before commenting further.

By Michael Kelsey (not verified) on 30 May 2015 #permalink

Why does a black hole have a bigger than zero observed radius? If it does then what is the observed radius - is it a function of its mass? Isn't it supposed to be a singularity hence having zero radius thus degrading into a point for the obserer? If the radius is non-zero, can we really say that the matter (or what's left of it) density inside is less than infinite?

Are anti-matter black holes possible theoretically? An encounter of a matter black hole with the anti-matter one would probably be the most violent event imaginable

@Alex #6

You are confusing a Black Hole with a singularity. The singularity is at the center of a black hole. The observed radius of the black hole is called the Event Horizon.

Click here for image

The distance between the singularity and the event horizon is determined by the mass of the black hole. Big, heavy, super-massive black holes have a larger event horizon than black holes of lesser mass.

@Alex #6: (Denier quite correctly answered your #5): Anti-matter has exactly the same kind of mass as normal matter. We know this directly, for example, from experiments at CERN in which anti-hydrogen has been created, and observed to fall to the bottom of the apparatus, rather than the top.

Consequently, a black hole, which is observably only distinguishable by _mass_, would look the same whether it had been constructed from matter, anti-matter, or any mixture of the two. You may wish to look up the "no-hair theorem" in Wikipedia for further discussion of what we can possibly know about black holes.

By Michael Kelsey (not verified) on 30 May 2015 #permalink

Suppose you had recently fallen into a very huge black hole ...
From things I've seen online, it sounds like the ratio of the Schwarzschild radius to the infall time (proper time from the event horizon to the singularity) is about 1.5c (is it exactly this?)
Suppose you had recently fallen into a black hole with an infall time of a billion years.
What would the universe look like then?
Tidal forces would probably not be measurable, I'm guessing.
Would there be a redshift or a blueshift for distant galaxies?
How would the Hubble constant be changing over time?
If mass were falling into the black hole fast enough, is it possible you'd never reach the singularity?

By "recently fallen into" in #9, I mean that you're inside the event horizon but not very close to the singularity.

And what would your perceived curvature of space time be, in this very huge black hole?

Here's a partial answer to my question:

From measurements made in the 1920s of the velocities of distant galaxies, it was found that all galaxies are flying away from us at a rate that depends only on their distance from us. No matter which direction we look in the sky this is true. We assume therefore that the Universe is isotropic, meaning rotationally invariant. There is no special or "preferred" direction to space.

Inside a black hole the situation is quite different - all objects are propelled toward the singularity at the center of the black hole. There is a preferred direction in space in the vicinity of a black hole.

@#4 Mike
"You might want to do some research on the subject of “black hole temperature” and “black hole thermodynamics” (even just the Wikipedia articles) before commenting further."

Haha I did, and got more confused so I thought I would ask here and not get jack booted to the curb.

I read through this whole page:
Saw this part:
"The Question

What is the temperature of a black hole?

The Answer

The temperature of a black hole is determined by the 'black body radiation temperature' of the radiation which comes from it. (e.g., If something is hot enough to give off bright blue light, it is hotter than something that is merely a dim red hot.)

For black holes the mass of our Sun, the radiation coming from it is so weak and so cool that the temperature is only one ten-millionth of a degree above absolute zero. This is colder than scientists could make things on Earth up until just a few years ago (and the invention of a way to get things that cold won the Nobel prize this year). Some black holes are thought to weigh a billion times as much as the Sun, and they would be a billion times colder, far colder than what scientists have achieved on Earth."

Then i read another article/comment, (I can't find it now) that referenced as the black hole shrinks it get hotter, that made me think about the Hawkins radiation process I learned about here (starts with A Bang) a while ago so I did more searching and found this:

"Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius."

But I have also read many other comments like:
"Over time, the continuous addition of negative antiparticles to the black hole adds negative energy, resulting in a gradual decrease in the black hole’s mass. This will in turn cause a black hole’s size to gradually decrease. With the decrease in size, the black hole’s temperature increases to such an extent that the black hole vanishes in an extreme burst of gamma radiation, sometimes including all kinds of energetic particles. This marks the end of the black hole. "
Sophia Nasr

So from what I gather, the Black hole life is VERY cold and it's death VERY hot.

Is this true?

OR does it not really have any temp at all in the very center (singularity).

By Ragtag Media (not verified) on 31 May 2015 #permalink

"So from what I gather, the Black hole life is VERY cold and it’s death VERY hot.

Is this true? "


Temperature, thought, is a bulk measure of average kinetic velocity of a body in internal motion. This is impossible to define when the body is infinitely small, since velocity is undefined. So a singularity cannot have a defined temperature.

@Michael Kelsey #8

Has CERN released data since the ALPHA experiment on antihydrogen falling up or down? The error bars on the 2013 ALPHA results are huge, as in error bars of 100%.

Link to the antihydrogen annihilation map

You can see for yourself the locations of the annihilation (red circles) are all over everywhere. It would be so much fun if antihydrogen fell up but the universe has a way of sucking the joy out of everything so it probably falls down. Is there new data that you know of?

@Ragtag Media #13

The Hawking Radiation you referred to IS the heat of a black hole. Tiny black holes generate a lot of Hawking Radiation, so they are said to be hot. Super-massive black holes generate much less Hawking Radiation proportionately and so are said to be cold.

The energy to create this Hawking Radiation has to come from someplace. Because E=mc^2 says that energy and mass are the same thing, the energy of Hawking Radiation radiating heat away from the black hole subtracts from the mass of the black hole, making the black hole smaller.

As said before, smaller black holes generate proportionality more Hawking Radiation than bigger ones, so as the black hole gets smaller it gets hotter. This is strictly due to size, not age. A solar mass black hole that was 1 year old would give off the same amount of Hawking Radiation as an otherwise identical solar mass black hole that was a kazillion-zillion years old.

Thanks Wow

By Ragtag Media (not verified) on 31 May 2015 #permalink

@Denier #16
Thanks for the response, now I have another question here as I have been reading up on this and the virtual particle/antiparticle re-moving the holes mass by one falling in and the other out.
I was stuck on there because I was thinking they should cancel each other out so I keep reading and this from Ethan:…
A better picture (that I prefer) that’s still a little naive is to imagine that you have these quantum fluctuations, but that every time you have a pair of particle-antiparticles where one falls in, you have a corresponding particle-antiparticle pair where the other falls in. The particle/antiparticle pair on the outside annihilate, emitting real, energetic photons, while the ones that fall in take a corresponding amount of mass (via E = mc^2) away from the black hole.

So is the quantum fluctuation the driving force of the black holes evaporation?

By Ragtag Media (not verified) on 31 May 2015 #permalink

@Ragtag Media (various): I think you've gotten the black hole temperature issue understood already; great work!

At #18, you asked, "So is the quantum fluctuation the driving force of the black holes evaporation?" Essentially, yes. What Hawking realized in his original calculations was that the energy density of the gravitational field near an event horizon could be high enough to support pair production. This happens via quantum fluctuations, but the pairs are _not_ necessarily "virtual" (they can be, but you can do the math without that assumption).

The naive model which Hawking initially developed has problems with entanglement, information loss, etc., which is why a great deal of active research is going into understanding it better, and refining it.

By Michael Kelsey (not verified) on 31 May 2015 #permalink

@Ragtag Media #18

The quantum fluctuation virtual particles are made up of energy borrowed from space itself. When they annihilate, their energy is returned to space, so in the end you get absolutely nothing. Your model of an outside virtual pair annihilating to create a real photon doesn't work because the annihilating pair already owes everything back to space.

The naive model, as Michael Kelsey puts it, is the model you will most commonly find describing the process. It says that virtual particle-antiparticle pairs cannot exist longer than a very short measurement of time called Planck time. The pair is created and must annihilate in under Plank time or they become real particles. At the Event Horizon of a black hole it is possible for the pair to be created and have one of the pair get stuck on the wrong side of the horizon. Plank time for the particle on our side of the horizon is reached without annihilation ever happening and it becomes real radiation.

There are some problems with that model. Michael Kelsey has a much better model for how it works but it is not as easy to explain so what you'll find out there on the internet is the naive virtual particle version. The math is the same, but if you're just trying to wrap your head around how it works, the naive virtual particle version is good enough to do the trick.

I notice that when one black hole captures another it rings for a long period of time. This seems to imply that the structure is composed of some sort of elastic material. Is it really just an effect of gravity? Seems like finer resolution of the properties of the ringing should reveal interesting properties of spacetime and the event horizon.

That's not part of the black hole, Joel. It's matter around the black hole.

@Joel #21
There's frame dragging when spinning black holes collide.
Maybe that's related to what you're seeing.

@Joel #21 and Wow #22: Joel is right. The "ringing" is due to the fact that two merging black holes of different masses (and hence different circumference event horizons) have a quadrupole distribution of their masses. A quadrupole is the lowest order for which it's possible to generate and emit gravitational radiation.

As I understand it (and I am *NOT* an astrophysicist, nor a theorist, by any means) the angular momentum of the system plus backreaction from gravitation radiation, means that the merging system remains in a non-spherical configuration for longer than you would otherwise expect. If the two black holes were brought together adiabatically, without "spiraling together", then you wouldn't see very much ringdown.

The event horizion is non-spherical because the two singularities (or whatever the heck they are) of the black holes still have a finite separation, as seen by a distant observer. Once the two meet up, so to speak, the event horizon will become simply a sphere, with a circumference determined by the combined mass, and the ringdown will be complete.

By Michael Kelsey (not verified) on 03 Jun 2015 #permalink

I was presuming that, like others, he was looking at the warp of the accretion disk.

As to your understanding (IAAAstrophysicist, just not much practiced), it's quite correct. There's a whole shedload of explanation of what's "really" going on, but the metric gets really twisted. This combines with the chaotic system of the accretion disks between the two masses to produce a far more dispersed and contorted pattern of light around a BH double, most of which is not the distortion of the frames but the accretion disk being flung about in an unstable situation.

The claim that it takes longer than expected isn't really true, except in so far as many (but not really a majority) thought it would fall in quicker and lose energy quicker.

Please note that there will still be lagrangian points around the double BH system. As long as they remain outside the event horizon, you can see their effect too.

In a tripe, it's entirely possible that the lagrangian points will be able to move in and out of the event horizon of one of the masses.

In my opinion, that may be fine: you're not really pulling matter/energy out of the black hole event horizon, but moving the space outside. Matter is coming along for the ride. Then again, this may be excluded from happening and that *could* be a mechanism by which spiralling inward is retarded in a double system too.

A bit like the theories about Warp Drive being possible, if unlikely, with reality as we know it to be at the moment.

Ethan, you say a black hole only has three kinds of information: mass/energy, charge, and angular momentum.

What about regular momentum-momentum? If you fired a grapefruit into a stationary black hole, would the black hole begin to drift gently in the direction the grapefruit was going?

There's no such thing as momentum-momentum, only the force needed to move an object into a different reference frame.

momentum = mass x velocity

and velocity is only defined with respect to an arbitrary reference v=0 frame.

So it doesn't get conserved, since it can be made merely by deciding on a different frame where v=0.

Rotation *does* have a frame of reference: the center of mass and no centripetal forces.