# Inside a black hole

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1. Nov 9, 2014

### JollyOlly

A number of recent threads have discussed what happens when an observer falling into a massive black hole passes the event horizon. What I would like to know is this. For a massive BH of mass M, Schwartzchild Radius Rs, how long would it take for such an observer (who, presumably crosses the event horizon at the speed of light) to reach the singularity? And during this brief (?) time, would he actually see anything?

2. Nov 9, 2014

### Staff: Mentor

First, the observer does not cross the horizon at the speed of light. It's the horizon that is moving (outward) at the speed of light, not the observer.

The time it takes the observer to reach the singularity from the horizon--I assume that by "time" you mean the time by the observer's own clock--depends on the initial conditions--how far above the horizon the observer was when he started falling from rest--and on the mass of the hole. For the idealized case of an observer who falls in from rest "at infinity", it takes him a time $2M$ by his own clock to reach the singularity from the horizon, where $M$ is the mass of the hole. This is in geometric units, where mass, distance, and time all have the same units; in conventional units the time is $2 G M / c^3$, where $G$ is Newton's gravitational constant and $c$ is the speed of light. If you work out the numbers, this turns out to be about 10 microseconds for a black hole of one solar mass, and goes up in proportion to the mass (so for a black hole of a million solar masses, about the size of the one at the center of our galaxy, the time would be 10 seconds).

Yes. Some good animations of what the observer might see are here:

3. Nov 9, 2014

### zoki85

4. Nov 10, 2014

### JollyOlly

Your formula, Peter, reduces to Rs/c which seems to suggest that, from the point of view of the falling observer, the radius of the BH really is something like Rs the Schwartzchild radius. Now I had understood that while the circumference of the BH was equal to 2πRs, in reality the radius inside the BH was much larger than this - possibly infinitely large. (Kip S Thorne: Black Holes and Time Warps p11). I imagine that the confusion arises (as usual) by not stating clearly enough the frame of reference you are viewing the situation from. Would it be true to say that from the point of view of the falling observer the radius is Rs but observed from outside the radius is much larger?

I am also curious to know the apparent angular size of a BH when viewed from a distance. Would the event horizon subtend the same angle as a star whose radius was equal to Rs?

5. Nov 10, 2014

### Staff: Mentor

Not really. First of all, you can't observe the interior of the BH from the outside, so an observer outside the horizon has no way of measuring any distances inside it.

More importantly, though, the black hole doesn't really have a "radius" in the ordinary sense, because it's not an ordinary object. The singularity at the center of the BH is not a place in space; it's a moment in time. To an observer who has just crossed the horizon, the singularity is in the future; it's not some distance away. By that I mean, not just that the observer will reach the singularity some time in his future; I mean it is in the future in the same sense as tomorrow is in the future. There is no meaning to the question "what is the spatial distance from now to tomorrow?"

When Thorne talks about the "radius" inside the BH being possibly infinite, he means something a bit different: he means that the spacetime geometry inside the BH is such that there are spacelike curves that are infinitely long! But none of these curves intersect the singularity, so none of them can be used to say that there is an infinite distance to the singularity. What their presence means is that, as an observer inside the hole is falling inward, there is a sense in which the space he is falling through is infinite--an infinite space that, for him, will come to an end in a finite time.

If you ignore the effects of gravitational lensing, yes. But taking gravitational lensing into account (i.e., the fact that light that passes close to the horizon has its path bent, similar to the way the Sun bends light passing close to it but much more intensely), the horizon will appear to occupy a larger area, as judged by the area in which no objects behind the hole can be seen (you can't actually see the horizon itself).

6. Nov 10, 2014

### JollyOlly

Thank you. I am getting a better picture now. I have also found Amndrew Hamiltons website very interesting (http://casa.colorado.edu/~ajsh/)

With reference to your reply as to the apparent size of a BH from a distance, I have learned that inside 1.5 Schwartzchild radii, no orbits exist, stable or otherwise. If a photon were to enter this sphere, it would be unable to get out again so the apparent disc will be at least this big. Between 1.5 and 3 radii, no stable orbits exist but I am unsure what this means in terms of a passing photon. Can you enlighten me? Outside 3 radii, photons are simply deflected.

I am also interested in the size of the Einstein ring. According to my calculations, the ring should subtend a semi-angle of √(2 Rs / r) where r is the distance between the observer and the BH (assumed much larger than Rs). Would you agree with this?

One more question! Would the Einstein ring around a spinning black hole appear distorted into an ellipse?

7. Nov 10, 2014

### Staff: Mentor

No, that's not correct. There are no stable closed orbits inside 1.5 times the Schwarzschild radius, but there are still trajectories that come in from infinity, pass closer than that to the horizon, and go back out to infinity.