What is the mass of a black hole made of?

[Stella.Physics]

It is possible to extrapolate some classical behaviors into the boundary of a black hole. In the case of supermassive black holes, the event horizon is so far from the center of mass that the gradient of the gravitational field is relatively small, A human astronaut could cross that boundary without being aware of the difference of gravitational attraction between his head and his feet. Near a much smaller black hole, the tidal effects could pull him apart. The same processes would hold for smaller objects, even down to the size of atoms. Eventually, the gradient approaching the singularity would pull everything apart. The question remains: how long does an astronaut have until that happens? With his speed of approach to the singularity increasing all the time, it should be expected that his perception of time, relative to an outside observer, should be decreasing. The details may be impossible to work out with confidence, but he could quite possibly spend a subjective eternity approaching the singularity. This is essentially the premise of Tipler's book, The Physics of Immortality.

lets say we have three observers, a) one in moving around a static (Schwarzschild) black hole at a radius let's say r=10M b) one that is at infinity and gets light signals from the first observer and c) an observer that is somehow standing still at radius r=10M and also sees the light signals from the first observer.

Now for the first observer if we say that ##\theta=\pi /2## at a certain level and also his r is fixed we get

##-d\tau^2=ds^2 = - \bigg(1- \frac{2M}{r} \bigg) dt^2 + r^2 d\phi^2## from the metric element

and that would be ##\Delta\tau=20\sqrt{7}\pi M##

for the second observer the time that he would get between those signals are ##\Delta t=20\sqrt{10} \pi M##

and the third and static observer the proper time between those signals would be ##\Delta \tau' = 20\sqrt{8}\pi M##

So we get ##\Delta \tau < \Delta \tau' < \Delta t ## because for the observer far away measuring time t, the static observer feels only the gravitational expansion of time whereas the one going around the black hole feels both the gravitational (GR) and the kinetic (SR) contraction of time.

I've left out many math steps but that would give you an idea although this case is different .

Another thing is while approaching the black hole and for r<2GM time t from timelike becomes spacelike. so when you are traveling towards a Black Hole from a point and on it's like moving forward in time, like going to the future, yes I say what happens when you travel in space by say what will happen from the time point of view.
"Thus you can no more stop moving toward the singularity than you can stop getting older. Since proper time is maximized along a geodesic, you will live the longest if you don't struggle. As you fall to the singularity your feet and head will be pulled apart from each other while your torso is squeezed to infinitesimal thinness" as detailed in the book "Gravitation" by Misner, Thorne and Wheeler.

 

[Nugatory]

We know that black hole temperatures are on the order of a millionth of a degree. There seem to be different laws of physics for things this cold. Anyone want to elaborate?

Temperatures much colder than that have been achieved in labs without any "different laws of physics" being seen. In fact, it often works the other way: random thermal motion overwhelms some of the more subtle effects we're looking for, so many physics experiments must be done at the coldest possible temperatures just so that we can see the laws of physics that we already know at work.

But all of this is a complete red herring because the low temperatures that you're talking about are the equilibrium temperatures at the event horizon, while this thread is about conditions well inside the horizon.

 

[Phil Lawless]

I think you've misunderstood something, or I've misunderstood you. The infalling astronaut has a very short lifetime after crossing the horizon - less than a second, if memory serves. Viewed from a distance, though, the astronaut never reaches the horizon. She can exploit this to return in the far future by dropping arbitrarily close to the horizon and spending a short (to her) time there then returning (assuming a impossibly powerful rocket to hover and return).

Edit: also, you appear to be invoking special relativity's concept of kinematic time dilation, which isn't appropriate here. You need to look at spacetime intervals to get the elapsed times for two observers.

Yes, I was invoking time dilation, but fairly reasonably so. Gravity also makes clocks run slower. As for the astronaut's lifetime, it all does depend on your frame of reference, doesn't it?

 

[phinds]

As for the astronaut's lifetime, it all does depend on your frame of reference, doesn't it?

Yeah, but the astronaut isn't going to care for very long about anyone else's FOR.

 

[kent davidge]

Do the possibilities include the matter being converted to massless particles such as photons? Or perhaps that the energy is stored in a field with no particles at all? Perhaps other types of conservation (charge?, baryon number? ...) forbids those scenarios.

It does not seem to make sense to have several laws of nature being violated inside a black hole. If that was the case, then such a object would be unlike everything physicists ever dealt with, either theoretically or experimentally. To me, it is more probable that something familiar is happening there, that is, something one can explain in part using the laws already known, perhaps with the help of a Quantum Gravity theory.

Does anyone think so?

 

[Ibix]

Yes, I was invoking time dilation, but fairly reasonably so. Gravity also makes clocks run slower.

But gravitational time dilation doesn't have anything to do with speed, which was what you were talking about. Particularly not speed with respect to a black hole singularity, which is problematic in a number of ways as PeterDonis pointed out.

As for the astronaut's lifetime, it all does depend on your frame of reference, doesn't it?

Indeed. But you said that the infalling astronaut could quite possibly spend a subjective eternity approaching the singularity. I took "subjective" in this context to mean that you were referring to the astronaut's proper time - which is very short, and in no way a subjective eternity.

 

[PeterDonis]

for the first observer if we say that ##\theta=\pi /2## at a certain level and also his r is fixed we get

$$
-d\tau^2=ds^2 = - \bigg(1- \frac{2M}{r} \bigg) dt^2 + r^2 d\phi^2
$$

from the metric element

This is correct as far as it goes, but it's not sufficient by itself to calculate ##\Delta \tau##. You need to know what ##dt## and ##d\phi## are. You haven't said anything about them at all. Can you give more details about what specific orbit you are assuming and how you are calculating your numbers?

 

[Stella.Physics]

This is correct as far as it goes, but it's not sufficient by itself to calculate ##\Delta \tau##. You need to know what ##dt## and ##d\phi## are. You haven't said anything about them at all. Can you give more details about what specific orbit you are assuming and how you are calculating your numbers?

##-d \tau^2=ds^2=-(1-\frac{2M}{r})dt^2+
r^2 d\phi ^2## →##d \tau^2 = [(1-\frac{2M}{r})(\frac{dt}{d \phi})^2-r^2]dφ^2## and for circular orbits like the one assumed around the static black hole ##Ω^2=(\frac{d φ}{dt})^2=\frac{M}{r^3}## so you have

##Δτ=\int_{0}^{2π} \sqrt{\frac{r}{M}-3} rdφ=2πr\sqrt{ \frac{r}{M}-3}## and for r=10M you get ##Δτ=20√7πM##

 

[PeterDonis]

for circular orbits like the one assumed around the static black hole ##Ω^2=(\frac{d φ}{dt})^2=\frac{M}{r^3}##

Ah, ok, so you were assuming a free-fall orbit.

 

[smodak]

I was thinking of buying this book. Is it good? Has anyone of you read it?

I read it and like it.