# Black Hole

1. Dec 18, 2005

### touqra

If my twin were to be in a Schwarzschild black hole, and have passed the event horizon, I would see him as passing the horizon in infinite time. But, what would he see? How would I appear to him? Would my twin ever know that he is in a black hole?

If my twin shines a torchlight towards me, who is outside the black hole, would the light travel to me, ie escape the horizon? If not, what would happen to the light photon?

2. Dec 19, 2005

### pervect

Staff Emeritus
If the black hole is Schwarzschild, I can provide some answers. Real black holes will probably have a different internal geometry (the so-called BKL singularity) so the answers I provide are probably not entirely accurate for a real black hole.

If we take the case of an infalling observer who is falling from rest at infinty, nothing particular special will happen at the event horizon. The outside universe will appear as a single point above him once he reaches the event horizon - he will only be able to see you if you remain on a direct radial path.

The universe will appear "red-shifted" to the observer falling in from infinity, the doppler shift factor will be 50% at the horizon. The red shift will continue to increase, approaching a red-shift to zero freqency at the moment the infalling observer reaches the singularity.

Light emitted from the infalling observer will never reach you outside the horizon. Light emitted _exactly_ at the horizon outwards will "hang" at that location to be seen only by subsequently infalling observers. Light emitted inside the horizon will fall with the observer, hitting the central singularity.

Much of this description is based on the following geodesic using infalling Finklestein coordinates. (This may be worked out in a text somewhere, I wound up calculating it with GRT-II).

For a mass 1 black hole falling from at rest at infinity, the geodesic path for r and the finklestein coordinate v as a function of an infalling mass will be

$$r(\tau) = \frac{1}{2} \sqrt[3]{36 \tau^2}$$
$$v(\tau) = -\sqrt[3]{-48 \tau} + \frac{1}{2} \sqrt[3]{36 \tau^2} + 4 ln(\frac{\sqrt[3]{6 \tau}}{2}+1)$$

v is the ingoing "Finklestein" parameter, defined so that the path of an ingoing photon is given by v=constant. Thus dv/dtau gives you the "redshift factor" for ingoing light.

Note that at $\tau=0$, r=0 and v=0, this is when the object hits the central singularity. $\tau < 0$ is the region of interest when the object is still falling and the equations are valid. The Schwarzschild radius is located at r=2

Last edited: Dec 19, 2005
3. Dec 19, 2005

### Garth

It also depends on the mass of the BB. For a BB a few times the mass of the Sun he would certainly know when he approached the event horizon as the tidal forces would build up so much that he would become 'spaghettified', going in feet first his feet would be torn from his head and he would be squeezed into a sort of tube.

If however it was a super massive BB of > 106Msolar the tidal forces over a human body would be minimal and he may not notice anything at all.

If you had a very large condensed star cluster the gravitational field could reach that needed to create an event horizon ($\frac{2GM}{rc^2}=1$) and you could inadvertently be drawn into it without knowing, only to find that you couldn't get out! (However, whether such a cluster would be stable enough to exist is another matter)

Garth

4. Dec 19, 2005

### touqra

What do you mean by a single point? As in a singularity kind of dimension?
Does it mean that this twin will have something like a tunnel vision?

Why aren't we considering the possibility that the earth is being slowly sucked into a black hole since we are detecting redshifts too?

"will fall with the observer"...does it mean that I can see the photon moving together with me? Wouldn't this violates the equivalence principle, that in a local frame, I shouldn't detect acceleration/gravity? assuming I understand correctly the equivalence principle.

5. Dec 19, 2005

### George Jones

Staff Emeritus
As pervect says, a star directly "above" an observer that freely falls into a black hole will be infinitely redshifted - I give another derivation below. But this is not true for the light from all stars in the universe!

The shadow (black region of the sky) of the black hole grows bigger as the observer falls, but, even at the event horizon take up less that half the observer's field of view. "At" the singularity, the shadow takes up half the field of view.

Consider far way stars that are distributed roughly equally in all (Schwarzschild coordinate) directions. Images of stars that are not directly "above" the observer move towards the shadow region because of stellar aberration. Also, depending on image location, some of the stars are blue shifted and some are red shifted.

Just before the singularity, all the images (except for stars directly above) are concentrated in a brilliant ring around the shadow region, i.e, in a ring perpendicular to "direction" of travel.

All of this can be derived fairly easily without tensors. As an example, consider the image of a star directly above.

The proper time of an observer hovering with constant $r$, $\theta$, and $\phi$ above a black hole is related to Schwarzschild coordinate time by

$$d\tau_S = \left(1 - \frac{2M}{r} \right)^{\frac{1}{2}} dt,$$

and coordinate time is proper time for a far observer. Frequency is the inverse of time, so if light with frequency $f$ starts far from the black hole with frequency and fall radially towards the hole, the hovering observer will see frequency

$$f_S = \left(1 - \frac{2M}{r} \right)^{-\frac{1}{2}} f.$$

Suppose an observer freely falling from rest far away passes the hovering observer at the instant the light is observed. Since the two observers are coincident, the special relativistic Doppler effect relates the frequencies seen by the two observers. The relative speed between the two observers is given by

$$v = \sqrt{\frac{2M}{r}}.$$

For a derivation, see

https://www.physicsforums.com/showpost.php?p=621802&postcount=32.

Consequently, the freely falling observer sees a frequency

$$\begin{equation*} \begin{split} f' &= f_S \sqrt{\frac{1 - v}{1 + v}} \\ &= f \left(1 - \frac{2M}{r} \right)^{-\frac{1}{2}} \sqrt{\frac{1 - \sqrt{\frac{2M}{r}}}{1 + \sqrt{\frac{2M}{r}}}} \\ &= \frac{f}{1 + \sqrt{\frac{2M}{r}}} \end{split} \end{equation*}$$

As pervect calculated, the frequency seen by the freely falling observer as he crosses the event horizon will be half the frequency that the light had far away.

This result can also be calculated using Lorentz transformations.

Even though this derivation is only valid outside the event horizon, I think (but haven't checked) that the result is valid inside the horizon, where is $r$ a timelike coordinate.

Regards,
George

Last edited: Oct 12, 2010
6. Dec 20, 2005

### pervect

Staff Emeritus
I have to make a correction. As George states, and is also pointed out in

(which has some movies of what people see falling into a black hole)

the universe does not contract to a point for an observer freely falling through a black hole. The universe does contract to a point for an observer "holding station" with a rocket, but that wasn't what was asked.

The site above has some interesting graphics and descriptions of what a person _does_ see.

George on I agree on the doppler shift at the event horizon, but I have to look at the general formula more closely - currently I'm getting

$$\frac{dv}{d\tau} = \frac{\sqrt{2r}}{\sqrt{2r}+2}$$

(I've assumed that M=1, thus at the Schwazschild radius, r=2M = 2, and dv/dtau = 1/2 as previously stated).

You can think of the 'v' coordinate as a number that is assigned to flashes of light emitted from an observer at infinity, a number that stays constant along an infalling flash of light, a number that increases by 1 every second (as measured by the outside observer at infinity). The coordinates of an infalling object are given by the light flash number seen at that time (the v coordiante) and the value of the r coordinate.

The rate of change of the v coordinate with proper time should then be the observered red-shift.

At the moment I'm not sure how to explain the discrepancy, I have to look over my math and George's more closely. As I mentioned earlier, though, this part of the math is suspect inside the black hole - not just because of the coordinate singularity there, but because a real black hole is probably not going to have a nice well-behaved Schwarzschild metric in the interior, but is going to be described by a more complex, chaotic metric known as a BKL singularity.

Last edited: Dec 20, 2005
7. Dec 20, 2005

### George Jones

Staff Emeritus
I think I forgot a square root.

$$f'= \frac{f}{1+\sqrt{\frac{2M}{r}}}.$$

Now my answer agrees with pervect's.

Regards,
George

Last edited: Oct 12, 2010
8. Dec 21, 2005

### pervect

Staff Emeritus
Forget I said that, it's wrong (see my previous retraction). :-(

Huh? I don't understand the motivation for this question.

The photon and the observer will take different paths, sorry if I implied otherwise. However, both of them will wind up at the central singularity if the photon is emitted inside the event horizon.

9. Dec 29, 2005

### blackmama

huh?

i really dont get all this math and stuff, but besides that what do u all mean by being in a black hole? i thoght u become speghtee once u got sucked in.:rofl:

10. Dec 29, 2005

### Garth

Hi blackmama and welcome to these Forums!

See my post #3 above.

Garth

11. Dec 29, 2005

### pervect

Staff Emeritus
Eventually the tidal forces will turn you into spaghetti (by stretching you in one direction and compressing you in another), however exactly when this happens depends on the size of the black hole.

Black holes have an "event horizon", which is a region around the black hole from which light cannot escape.

After you cross the horizon of a non-rotating black hole, you are doomed to eventually be turned into spaghetti before you hit the central singularity. However, for a large enough black hole, you will not be turned into spaghetti as soon as you cross the horion, this will happen at a later point in your journey.

12. Dec 29, 2005

### Labguy

And how many non-rotating black holes do you, or anyone, think actually exist in nature? Probably none at all.
Just my opinion, but we all spend a lot of time talking about "point singularities" and the Schwartzchild solutions and seem to have forgotten about Kerr-Newman BH's, the ergosphere and ring singularity. That's, of course, for any of those who still believe that a singularity, defined as infinite density, exists at all.

13. Dec 30, 2005

### SpaceTiger

Staff Emeritus
Isn't it standard practice to consider the idealized cases when doing theory? I mean, there aren't any truly spherical stars either. Also, is it really that unreasonable to think that there might be black holes out there with J << M?

I don't see any problem with discussing rotating black holes, but for understanding GR, I would say the Schwarzschild solution is much more instructive.

14. Dec 30, 2005

### pervect

Staff Emeritus
The actual interior structure of a rotating black hole is a very interesting problem that AFAIK has not been totally solved. The interior solution is unlikely to be the Kerr-Newmann solution, however.

The best paper I'm aware of on the topic is

http://lanl.arxiv.org/abs/gr-qc/9902008

and I would be very interested in hearing about any more recent work on the topic. Note that while much work needs to be done, the inner structure of rotating black holes is felt to be more like the traditional Schwarzschild singularity than the traditonal Kerr-Newman singularity, at least as of the time of the writing of this paper, as per the following quote from the abstract.

The interior structure of a non-rotating black hole has probably been solved, and is probably not the traditional symmetric Schwarzschild singularity but a BKL singularity. Most of my information here comes from Thorne's book "Black Holes and Time Warps: Einstein's Outrageous Legacy" rather than the technical literature.

These questions about the interior structure of black holes are separate from those raised by quantum gravity - i.e. they are open questions within the framework of GR, not questions whose answers depend on unkown quantum gravity effects. While quantum gravity effects almost certainly exist, trying to speculate about what these effects might be is difficult due to the lack of any experimental data.

Meanwhile, since posters like "blackmama" are apparently looking for simple, easy answers to complex problems "without math", I try to accomodate them as much as I can while remaining technically accurate. As nearly as I can tell, Blackmama believes that one turns into spaghetti as soon as one reaches the event horizon - I was mainly attempting to explain in very simple terms that this is not the case, while also noting that "spaghetification" was something that probably did happen eventually.

15. Dec 30, 2005

### Labguy

Yes, I suppose it is, but that would cause the string theory proponents a lot of trouble since there are over 20 "idealized cases" to consider. But, I really can't make a case for doing much different than you describe.
You're right about no spherical stars. There are also no circular orbits, no straight-line paths between any two light sources and certainly no parabolic paths followed by a thrown or shot object, etc. Sphere, circle, parabolic and straight are as defined mathematically. My biggest pet-peeve is so many who state simply that: "if a white dwarf accumulates matter from a companion to where it exceeds the Chandrasekhar limit, it becomes a Type Ia supernova". That has been posted a hundred times and isn't even close without ten pounds of caveats. Generalized statements are fine, but not that general! BTW, did you mean J << M or J <= M?
I agree; that's fine, but when you say "for understanding GR" don't you actually mean "more simple? Not that it is a problem, I don't understand any of it anyway...
It isn't too new, but I like the four pages starting here for a geometry description of the differences between static and rotating black holes. I don't see any GR problems presented in either case.
I totally agree with your approach and with trying to explain the "spaghetification" (nice word coining) at the EH. Lots of misunderstanding out there about EH gravitational strengths depending on mass. I am "mathematically challenged", so the text explanations are easier for me to read in under an hour.
To all: Please don't bring up the Type Ia supernova complaint (mine) on this thread since it has been discussed several times before on PF..

16. Dec 30, 2005

### SpaceTiger

Staff Emeritus
Of the ones I've seen, the Schwarzschild solution is certainly one of the simplest. I suspect it's popular because of some sort of unspoken optimization between simplicity and astrophysical significance. As pervect pointed out, the Kerr solution is extremely difficult to work with.

The former. It represents a black hole for which rotation is negligible and the Schwarzschild solution is a close description of reality.

Pretty much, yeah. Do you not agree that simplicity can be valuable for understanding a theory?

17. Dec 30, 2005

### Labguy

Yes, for sure. But, I do have a problem with broad generalzations like the one I mentioned about Type Ia supernova. Some things just can't be described in a simple fashon; at least not accurately.

18. Dec 31, 2005

### SpaceTiger

Staff Emeritus
It's a good point, and I'm not sure I have a strong position on the issue. It could be argued that, in pedagogy, it is sometimes worthwhile to sacrifice accuracy for simplicity. In the case of Type Ia supernovae, I wonder how instructive the caveats are for a layman.