# B Black holes and time

1. Dec 22, 2016

### charles watson

I recently was watching a television show and something has lodged in my brain. They claim that as something approaches a black hole, it swirls in thru the accretion disk and speeds up to at times close to light speed before entering. They also claimed that it leaves an impression or image of its self on the black hole such as an observer would see me virtually frozen on its surface or at least for a very long time. If that is so, I am wondering why a black hole would be black or invisible at all? It would seem to me to be very bright. I am assuming photons would do the same thing otherwise how could an observer see me?

2. Dec 22, 2016

### jbriggs444

If you imagine the falling object emitting identical pulses of light at a regular rate then from your vantage point far away, two things will happen. You will receive those pulses at a slower and slower rate and the pulses will be red-shifted to a lower and lower energy. That's gravitational time dilation and gravitational red shift (which are pretty much the same thing). You will only see finitely many such pulses before the faller passes the horizon. In terms of photons, only a finite number will be emitted, only a finite number can be received and you will lose sight of the faller after a finite (and rather short) time.

3. Dec 22, 2016

### sophiecentaur

People talk of 'Spaghettification". It would be interesting to know how far into the process our bodies (and transport) would start to function noticeably wrongly. Signals passing through the ship would be out of sync from one end to the other. Would we actually feel something going on? Hollywood has a number of ways of showing it but that's just Hollywood.

4. Dec 24, 2016

### charles watson

I would imagine a long time before you enter the actual even horizon, your dead. The radiation in the accretion disk would kill you not to mention the heat. You'd also have to think, there would be a lot of electromagnetic energy before you arrived which would fry your brain waves and suck the iron right out of your blood the closer you got.

I often wonder why we always seem to think black holes are matter trapped in a tiny little space. It seems to me that if a black hole can twist space and time, it must be dragging space inside it too. If that's so, they could be larger on the inside than on the outside. If that actually happens, its a great concept for the theory that our universe is actually just that and we live inside a blackhole.

5. Dec 24, 2016

### AstroChris

I don't know about larger on the inside but they will always be denser past the event horizon. And everything inside is brought down to a singularity because of the immense gravity.

6. Dec 24, 2016

### Orodruin

Staff Emeritus
There is nothing dense about the standard Schwarzschild black hole. It is a vacuum solution to the Einstein field equations and therefore the density inside (as well as outside) of the event horizon is zero. Furthermore, the singularity is not a place in space.

Please refrain from personal speculation. The mathematics behind a Schwarzschild black hole are very clear on what is going on (nothing particular) and the space-time inside it is well understood apart from breaking down when you approach the singularity. Of course, there is no way to actually check this without entering the event horizon (not recommended) and how you are affected by other stuff (such as radiation from the accretion disk) is not described by the space-time model itself.

7. Dec 24, 2016

### AstroChris

I'm sure you're coming from an angle on this that I'm not thinking of but Black Holes are most definitely dense. A black hole could be as big as an atom and have the mass of a mountain. And the definition of a singularity is a point or region in spacetime in which gravitational forces cause matter to have an infinite density.

8. Dec 24, 2016

### Orodruin

Staff Emeritus
This is very full of popular misconceptions about what a black hole is.

1. Schwarzschild black holes are not dense. If you look at the space-time for a Schwarzschild black hole, it is a vacuum solution to the Einstein field equations, leading to the conclusion that the density everywhere is equal to zero. The "mass" of the black hole is a property of the space-time, not due to "an amount of mass assembled in a point".
2. You really cannot think about black hole size in this way. That it has a Schwarzschild radius $r$ does not mean that it has a volume that is given by $4\pi r^3/3$.
3. No, that is not the definition of the singularity. The singularity is not part of the Schwarzschild space-time and it is where the mathematical description of the black hole breaks down.

9. Dec 24, 2016

### sophiecentaur

I looked at Wiki about this and the Schwarzschild density varies with the original mass. The Sun would need to collapse to a radius of about 3km and it would then have a density of 1.84×1016g/cm2. This is pretty high - but I guess you would argue that, to form a black hole, you would need a much bigger object to start with. For the Milky Way Galaxy, the density of the black hole would be only 3.72×10−8g/m2, which is pretty low. What would be the likely density of any black hole you'd be likely to encounter up-close be, I wonder?
But I'm obviously looking at this with my conventional hat on.

10. Dec 24, 2016

Staff Emeritus
No, they are not. The density of supermassive black holes can be less than the density of water.

As a learning too, it's really better to ask questions than to make incorrect statements hoping they will be corrected.

11. Dec 24, 2016

### AstroChris

"You can use the Schwarzschild radius to calculate the "density" of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius. This is roughly equal to (1.8x1016 g/cm3) x (Msun / M)2, where M is defined as above." This quote is from a postdoctoral from Cornell.

So, are you saying that they just CAN be less dense than water or they just aren't dense at all, a zero sum?

12. Dec 24, 2016

### Orodruin

Staff Emeritus
This still has nothing to do with the density of the resulting black hole.

You can of course compute 3M/(4piR^3) and it will have the correct dimensions, but as I have already pointed out, this is not "dividing the black hole mass by the black hole volume".

Your "conventional hat" does not work in curved space-time.

Note the "" in the quote. He knows he is lying, but he is taking a liberty to do so in order to make something popularised - ending up with something people take too litterally.

13. Dec 24, 2016

### Orodruin

Staff Emeritus
Anywhere you are in the Schwarzschild black hole you will observe zero density.

14. Dec 24, 2016

### AstroChris

Ok, so inside there is zero density observed. As we are looking at it mathematically, knowing that a black hole could be measured to be 10 million x our Sun do we calculate density then or is it irrelevant or is it still zero observed density? Getting correct information can be hard I suppose.

15. Dec 24, 2016

### sophiecentaur

I realise that but consider a region containing masses. That region will have a density and an object in orbit round it will follow Newton's Laws and, if it's distant enough, the masses can be regarded as a point mass. The volume of that region will be what it is and can shrink a bit, with the mass staying the same. The orbit of the object will not change, once the masses have condensed far enough to form a black hole. (I think that's right) So, the black hole will occupy a certain volume, observable (after a fashion) from the orbiting object. The density, calculated by mass (remembered) / volume (observed) has a meaning to the observer on the object although it is not the same for a reference frame inside the black hole. So it seems that, along with all the other quantities, relativity gives them different values, depending on the reference frame. That's fair enough and it not as unthinkable as talking about how things look from the point of view of a photon, which is just not valid at all.

16. Dec 24, 2016

### Orodruin

Staff Emeritus
This is not correct. Again, you are trying to impose a Euclidean structure on a space-time where curvature is clearly non-negligible.

While the components of the stress-energy tensor (of which energy density is one) generally depend on the observer - all the components are identically zero in the Schwarzshild space-time.

17. Dec 24, 2016

### sophiecentaur

But objects will orbit around a black hole so they must 'see' a mass that's non-zero or they would carry on in a straight line. That mass and the extent of the black hole will give a finite (possibly high) density result. At the distance I am considering, the space time behaves as if it were Euclidian. I am not questioning the theory of what goes on near the singularity.

18. Dec 24, 2016

### Orodruin

Staff Emeritus
This is a global property of the space-time. Not a result of there being a non-zero density anywhere.

Density is a local concept, not a global one. You also implicitly assume a volume of the black hole that is not well defined. I am sorry, but what you are trying simply does not make sense.

19. Dec 24, 2016

### sophiecentaur

I know that what I am saying is going against your detailed knowledge of GR but, imagine we were in orbit around a black hole, instead of a dead Sun and that it was surrounded by a disc of debris that meant we couldn't actually see it. (Ignore our human vision etc). By looking at our surrounding bright astro objects, we could deduce our orbit and we would identify that there was a mass (or equivalent) in there, keeping us in place. There would be an Equivalent Mass in there and we could hazard a guess about how much room it takes up. As far as we are concerned, that would correspond to a density.
By sticking to "what's really going on" in there, you are invalidating every model of the Universe that existed before GR and Black Holes got going because the old model don't tell the whole story. That's not the way Science works.
You are 'right', if you choose your definition of density in your way. Can you really say that the definition that we all know and love is not a possible way of viewing this newly found phenomenon? It's no surprise that density is frame dependent. Most other things seem to be. But we still use the schoolboy formulae for most of our lives. KE is still KE and Mass is still Mass.
The actual volume of a black hole may not be "well defined" but we can surely assign some upper limit on it by occultation and lensing effects.

20. Dec 24, 2016

### Orodruin

Staff Emeritus
Again, this is a global property of the space-time which a priori tells you nothing about the local stress-energy tensor. Furthermore, your assumption on "how much room it takes up" is also predicated on an assumption about a Euclidean space. This is (very) far from the situation in the case of a black hole and your hazarded guess will be wildly speculative at best.
No it would not. It would have the physical dimension of a density, but it would tell you nothing of the actual density you would find there.

Good! No model before GR gave a proper description of a black hole! Yes you can compute your "density", but it will not be a density in any usual sense of the word. You are refusing to accept that your classical view of space and time does not apply to the situation of a black hole. That is not how science works. Your model tells you something about the global property of the space-time only and therefore has little to do with the local properties, e.g., the local density.

The problem of your density definition here is that you are implicitly defining a "volume" of the black hole. Defining this volume requires a simultaneity convention. This is not trivial to do in a black hole as, e.g., the radial coordinate is time-like. You cannot simply refer to "the frame in which the black hole is at rest", there is no good definition of this that extends into the black hole region - the volume of which you want to compute!

No, this is what you do not seem to grasp. The Schwarzschild radius of the black hole does not give you a volume in any reasonable sense of the word. You are trying to squeeze a square block through a round hole.