# How does the star that has collapsed to form a Schwarschild black hole appear to an observer falling into the black hole?

I understand that to an outside observer, the light from a star that is collapsing into a black hole will become more and more red-shifted as the surface of the star appears to approach the black hole event horizon. The outside observer will never actually see the surface of the star cross the black hole event horizon. This applies to all outside observers: at infinity, in orbit around the star/black hole or those using a rocket to hover above the black hole.

Conversely, I know that for someone on the surface of the star that is collapsing to form a black hole it will appear quite different. The observer on the surface will not see anything unusual happen as they cross the event horizon and in a finite time they will reach the singularity at the center of the black hole where we do not know what will happen since general relativity breaks down in a singularity.

So, now consider an observer that starts at a great distance from the star who is continually falling directly into the star that has formed a black hole. Assume that he is falling in an exactly radial direction with no angular momentum. While the observer is still very far from the black hole, he will see the (original) visible light of the star get red-shifted to infrared, to microwaves and then to longer and longer radio waves. But as he approaches the black hole, he starts to fall faster and faster, so I assume he starts to see that red-shifted photons from the star surface will begin to be blue-shifted by his increasing speed as he falls. So, I think that the red-shift photons will be blue-shifted such that when the observer crosses the event horizon he will see the "visible light" photons that were sitting there at the horizon waiting for him. Since he is falling at the speed of light (in some sense) when he crosses the horizon these photons will become "visible light" again.

So the first question is this true? When he crosses the event horizon will the surface of the star be back to having no net red or blue shift? Will a visible light photon emitted by the star as it crosses the horizon be once again a visible light photon?

The second part of my question is what about the number density of photons? Will it look as "bright" as it would have looked for an observer falling with the surface of the black hole or will it look "dimmer" as if the surface was further away?

Finally, what happens as the falling observer continues his fall past the event horizon of the black hole. Will the photons form the original in-falling star be red or blue shifted. What will the observer see during the short time before he himself hits the singularity?

This is a follow on to my previous question since this was not answered there.

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You need to be a lot more careful when you use the phrase red-shift, due to how frequency is measured in general relativity. Roughly speaking, a photon is characterised by its wave vector $k$, which is a light-like four vector. The frequency measured by an observer is $g(\tau,k)$ where $g$ is the metric tensor, and $\tau$ is the unit vector tangent to the observer's world-line.

A little bit of basic Lorentzian geometry tells you that for any given photon $k$, instantaneously there can be observers seeing that photon with arbitrary high and arbitrarily low frequency.

So: start with your space-time. Fix a point on the event horizon. Fix a photon passing through that space-time event. For any frequency you want to see, you can choose a time-like vector at that space-time event that realises that frequency. Now, since the vector is time-like and the event horizon is null, the geodesic generated by that vector must start from outside the event horizon and crosses inside. Being a geodesic, it represents a free fall. So the conclusion is:

For any frequency you want to see, you can find a free falling observer starting outside of the black hole, such that it crosses the event horizon at the given space-time event and observes the frequency you want him to see.

So you ask, what is this whole business about gravitational red-shift of Schwarzschild black holes? I wrote a longer blog post on this topic some time ago and I won't be as detailed here. But the point is that on the Schwarzschild black holes (and in general, on any spherically symmetric solution of the Einstein's equations), one can break the freedom given by local Lorentz invariance by using the global geometry.

On Schwarzschild we have that the solution is stationary. Hence we can use the time-like Killing vector field* for the time-translation symmetry as a "global ruler" with respect to which to measure the frequency of photons. This is what it is meant by "gravitational redshift" in most textbooks on general relativity (see, e.g. Wald). Note that since we fixed a background ruler, the frequency that is being talked about is different from the frequency "as seen by an arbitrary infalling observer".

(There is another sense in which redshift is often talked about, which involves two infalling observers, one "departing first" with the second "to follow". In this case you again need the time-translation symmetry to make sense of the statement that the second observer "departed from the same spatial point as the first observer, but at a later time.)

It turns out, for general spherically symmetric solutions, there is this thing called a Kodama vector field, which happens to coincide with the Killing vector field on Schwarzschild. Outside of the event horizon, the Kodama vector field is time-like, and hence can be used as a substitute for the global ruler with respect to which to measure red-shift, when the space-time is assumed to be spherically symmetric, but not necessarily stationary. Again, this notion of redshift is observer independent. And it has played important roles (though sometimes manifesting in ways that are not immediately apparently related to red-shift, through choices of coordinates and what-not) in the study of dynamical, spherically symmetric gravitational collapse in the mathematical physics literature.

To summarise:

If you just compare the frequency of light measured (a) at its emission at the surface of the star in the rest frame associated to the collapse and (b) by an arbitrary free-falling observer, you can get basically any values you want. (Basically because the Doppler effect depends on the velocity of the observer, and you can change that to anything you like by choosing appropriate initial data for the free fall.)

You asked about what happens in the interior of the black hole. Again, any frequency can be realised by time-like observers locally. The question then boils down to whether you can construct such time-like observers to have come from free fall starting outside the black hole. By basic causality considerations, if you start with a time-like vector at a space-time event inside the black hole formed from gravitational collapse, going backwards along the time-like geodesic generated by the vector you will either hit the surface of your star, or exit the black hole. Though precisely how the two are divided depends on the precise nature of the gravitational collapse.

I should add that if you use the "global ruler" point of view, arguments have been put forth that analogous to how one expects red shift near the event horizon, one should also expect blue shifts near any Cauchy horizon that should exist. This has been demonstrated (mathematically) in the Reissner-Nordstrom (and similar) black holes. But as even the red-shift can sometimes run into problems (extreme charged black holes), one should not expect the statement about blue shifts near the Cauchy horizons to be true for all space-times.

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I can't seem to find an "add comment on reply" button; but with reference to comments of the previous reply, go for it Ron, and no doubt someone else will comment on what you're unsure of. That's how we all learn.

As the star approached the event horizon Wouldn't "almost radial" (to the hole) light be bent round the hole and be visible, both near and far, as a halo?

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The last stable orbit for photons is at 1.5 Schwarzchild radii so any photons emitted near the event horizon that are at any angle except very nearly perpendicular to the horizon will fall into the black hole -AFAIK. –  FrankH Dec 3 '11 at 22:15
you cannot comment because you are new and do not have enough points. I gave you a +1 so you get some. –  anna v Dec 4 '11 at 7:45
@FrankH: this is true, but there is a subtlety for nearly outward light--- it will stay put (almost) and slowly wind around the horizon. Will it leak in (or out) before making a full turn? This is hard to determine by pure order of magnitude. –  Ron Maimon Jan 9 '12 at 7:31

This is a subtle question, because what you see depends on competing effects. The answer depends on exactly how you fall into the black hole, since the appearence of objects depends very strongly on your boost. If you boost away from any object, the object will redshift, spread out in your field of vision, and dim, and if you boost toward it it will blueshift, compress into your forward field of vision and brighten. This means that what you see depends on whether you speed up toward the black hole to crash in fast, or whether you accelerate away from the black hole so that you are highly boosted outward when you cross (the typical situation for a late-falling observer)

The light-rays at the moment of crossing the horizon from the star stay on the horizon forever, so you can always (classically) collect some radiation from the star no matter how late you cross. But the image size, shape brightness and reshift depends on your boost in such a way that you never see too much of the star at late times. If you go in moving very fast toward the center of the black hole, you see a small bright image of the star directly ahead, toward the direction of the singularity whose size is inversely proportional to the time at which you cross (the affine parameter of your crossing location). If you go in naturally, meaning you spend a while accelerating near the horizon to keep from falling in, and then let yourself go, then you see a spread out dim redshifted image (the same image as before in a different frame), which is redshifted to oblivion if you come in at late times after spending much time near the horizon.

### Near Horizon Solution

The near horizon form of the Schwartschild solution can be found by writing $r=2M+u^2$ in the usual r coordinate, as described here: Why is spacetime near a quantum black hole approximately AdS? . You get (choosing units so that 2M=1, and calling the Schwartschild time $\theta$):

$$ds^2 = - u^2 d\theta^2 + du^2 + (1 + {u^2\over 4} ) d\Omega^2$$

This is a Rindler space cross a sphere, so that you can transform it into Minkowksi space cross $S_2$ by using the coordinates $t=u \sinh(\theta)$ $x=u \cosh(\theta)$

$$ds^2 = - dt^2 + dx^2 + (1 + {x^2 - t^2\over 4} ) d\Omega^2$$

This is the near horizon form, including the leading order variation in the sphere radius with distance from the horizon. The horizon is the light path $x=t$. The region $t>x$ inside the forward lightcone of the origin is the region in which the sphere radius contracts, and this is the interior of the black hole, while the region $t<x$ spacelike and to the right of the origin is the exterior of the black hole

The problem is ray-tracing in a Schwarzschild geometry, so one has to consider light rays starting at a crossing point $x=t=t_0$ on the horizon. The backward light cone from this point can be parametrized by choosing a past-pointing vector in M_2, and adding the appropriate length component along the sphere.

### No winding

The main issue in the solution of the problem is whether you see multiple images. The light rays near the horizon going close to outward slowly travel around the black hole surface, and you might thing that you can see many images of the star, due to rays that slowly crawl around the black hole to reach you from the star after a winding.

This is not so, because the time for one winding is always comparable to the time it takes the light to get away from the surface of the black hole. This is easiest to see in the product near-horizon solution.

Given a light ray coming into your eya at small angle $\theta$ from directly toward the center of the black hole, the failure of the ray to be the horizon generator is proportional to $\theta^2$, while the component along the sphere factor of the near horizon solution is proportional to $\theta$.

But if you look at the quantity $x^2-t^2$, which is $u^2$,the squared difference of the radial Schwarzschild coordinate from 2M, along the approximate geodesic, it is

$$(t-s)^2 - (t - s\cos(\theta))^2 = s(t-s)\theta^2$$

This quantity increases to a maximum of $(t\theta/2)^2$. When this maximum is comparable to 1, the light ray traced back escapes from the gravitational product region. This heuristic shows that the escape time is for $\theta\propto 1/t$, up to small factors of order unity.

Since the winding time is also $t\propto 1/\theta$, there are no windings--- the light ray escapes the near-horizon product region before it can go once around.

### Size of the stellar image

The angular size of the stellar image is determined by the affine-parameter to escape the horizon region for a back-traced ray at angle $\theta$ away from the line toward the center of the black hole.

Since the solution looks (nearly) like a product right by the horizon, the geodesics are simply straight lines in the Minkowski space, which simultaneously wind around the sphere. No-winding shows that the angular spread of the image of the star (assuming you come in in the same boost frame as the star when the star crosses the horizon) is less than the angle which will wind one full turn in the affine parameter $t$ where you cross.

This means that the angular spread of the star falls as $1/t$. This is the image in the unboosted frame, which is defined by translating the frame of the infalling star with no boosting in the Minkowksi space factor of the product

### Boosting effects

The effect of boosting is a conformal transformation of the sphere of incoming light rays. This is described very nicely in Penrose's Spinors and Space-Time Vol.1. The qualitative effect is clear--- if you go very fast in a certain direction, light in your frame has additional momentum in the opposite direction, concentrating the light into your forward field of vision and blueshifting it. Light behind you is redshifted and spread into oblivion.

For a black hole, time translation along the horizon is a boost, since the external time parameter $\theta$ is a boost parameter. This means that if you wait a long time in the external coordinates, and look at the same velocity translated into the future using the time Killing vector, this velocity has been boosted away from the black hole center by an amount proprotional to the time.

The visual image of the star is only undimmed and shrunk by an amount proportional to t in the "rest frame" of the collapsing star (I put rest frame in quotes because it is a reference frame of the near horizon M_2 x S_2 metric). It is undimmed because the product nature of the solution does not let the light rays spread out, but it is shrunk to an angular size of 1/t because most of the rays miss the star.

But when you are falling in with the same velocity at late times, t, you are boosting by an amount proportional to t. A boost by an amount proprtional to t will spread the angular region behind by an exponentially growing amount in t. This leads the image to dim exponentially, so that you really won't see anything at all if you fall in at late times in a natural way.

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The redshift turns into a blue shift as the infalling observer approaches the event horizon. I don't know for sure if the photons he sees as he crosses the event horizon will be back to no shift at all or if there would still be a net redshift or even a blueshift - does anyone know?

But the surface of the star will look dimmer and dimmer as the infalling observer approaches the event horizon. Also the observer will only be able to see a smaller and smaller patch of the surface of the star directly below him in the radial direction as he approaches the event horizon.

The reason for this is that as the original star approached the event horizon, only photons emitted in an increasingly narrow cone along the radial direction will be able to avoid falling into the black hole. At the event horizon, only photons emitted exactly along the radial direction will remain frozen at the horizon waiting for the infalling observer to see.

Once the observer enters past the event horizon he will not see anything at all since the forward light cone points inward in a "spatial" direction towards the singularity.

Finally, these consideration affect what the observer at infinity sees - as the star surface approaches the event horizon, not only will the photons become more and more redshifted, but in addition, the patch of the star that the observer at infinity can see will be a smaller and smaller disk that will shrink towards a point at infinite time.

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Gave up on getting help from anybody so I just accepted my own answer.... –  FrankH Nov 10 '11 at 8:26
I was thinking about it since you posted the bounty, on and off. I wrote an answer, but there was one sticking point I wasn't 100% sure about, so I didn't post it. It's interesting. Your answer is only partly right. I'll finish it at some point, and post it. –  Ron Maimon Dec 1 '11 at 18:24
Hi @RonMaimon, Please do post your answer! I posted the bounty but it automatically expired in 1 week - I did not take it away. I would have been happy to pay it to you... especially since the bounty goes into the bit bucket when it expires :^(! To make up for the lack of bounty, when you post your answer, I will go and add 10 upvotes on 10 of your many wonderful answers! –  FrankH Dec 1 '11 at 20:10
Hi @FrankH: thanks! I didn't care about the bounty, I just got confused on something stupid (the issue in question was whether the light could wind around the black hole, leading to multiple image of the star, or whether there is only one image--- I sorted it out, there is only one image). I didn't want to post a half-assed answer, but it's done now, and I'll post it in a few minutes. –  Ron Maimon Jan 9 '12 at 7:32