I know that if an observer falls into a Schwarzschild black hole, he can still receive radio signals from the outside even after he passes the event horizon. What I don’t know is how to calculate the frequency he would receive.
Assume the radio transmitter is at a fixed radius R with respect to the black hole and transmits on frequency F in a reference frame stationary with respect to the transmitter and at the same gravitational potential. Assume also that the observer starts falling from the transmitter’s position and has zero initial velocity. How, then, can we calculate the frequency he will receive in his own reference frame?
I don’t have time to work through the details right now, but here’s how I would solve the problem: The Schwarzschild metric has a Killing vector field t[sup]a[/sup] = (1,0,0,0) (those are the t, r, theta, and phi components respectively.) One of the fun properties of Killing vector fields is that the inner product of an observer’s four-velocity with a Killing vector field is a constant along the observer’s worldline. You have two worldlines in question here: the falling observer, and the radio photons. You can use the Killing constant of the motion to figure out what the time component of the observer’s motion is at any value of r (remembering that he’s at rest when he starts.) Since the norm of his four-velocity is always -1, you can then find the radial component of his four-velocity. Similarly, you can use the Killing constant to find the time and radial component of the photon’s four-velocity at any value of r; the value of the constant is set by the fact that the transmitter observes the photons to have frequency F at the initial radius. Once you’ve done that, the frequency observed by the falling observer at any radius r is just the inner product of these two four-velocities.
Hope this isn’t too advanced, but you seem to know the basics; feel free to reply with requests for clarificiation. You might find Eddington-Finkelstein coordinates useful for this purpose, though be careful with the transformations of the four-velocities.
Well MikeS knows a lot more about this than me. But I believe your light cone would both close and tip as you near and pass the horizon. So I don’t believe you could receive anything at all from the outside world.
In Eddington-Finkelstein coordinates the cone tips smoothly while in other coordinates it’s abrubt. Chronos would know for sure. Lets hope he shows up.
It’s true that you can view the future lightcone as “tipping inwards” when you pass the horizon, and that this implies that you can’t send signals to larger values of r once you’re in the black hole. The corresponding statement, though, is that the past lightcone tips “outwards” when you pass the horizon, and therefore you can’t receive signals from smaller values of r once you’re in the black hole. An observer in the black hole can still get signals from outside the black hole, he just can’t send signals in return.
Oh, I should mention in my above derivation that I’m assuming a purely radial infall. If the observer and/or photon have angular momentum, you can still do the same kinds of calculations; you’ll just have another constant of the motion which you can find using the Killing field (0,0,1,0).
Yes, but it would seem that the closing of the LCs would give you at most a view of a dimensionless point. Although I’m not sure if this is true in all coordinate systems.
I just wanted to say here that I’ve seen the thread, and will be giving a full answer in the next day or two. But I want to make sure I have all my ducks in a row before committing to an answer.
I’ve been perusing a number of texts and really couldn’t find anything applicable to the OP. However in Taylor and Wheeler’s Exploring Black Holes page B-18 I found the following:
In case you haven’t noticed this is in direct contradiction to Matt McIrvin’s statement. It appears the experts are either on some level of which I am unfamiliar, or else, God forbid, they actually disagree with each other.
There isn’t a contradiction. He said “faraway events in the arbitrarily distant future” could not reach the observer falling past the horizon. He’s explaining why the infalling observer doesn’t see the outside universe speed up to infinity, even though the outside observer does see the infalling observer slow down to infinity.
Interesting. Maybe you could define how exactly faraway “faraway events” are? Or how distant “arbitrarily distant” is? Or why your definitions of the previous terms cannot wind up on his past light cone but others can. Or what exactly determines whether something can or cannot wind up on his past light cone?
The limit on what the infalling observer can see is set by the fact that the observer eventually hits the singularity. Are you familiar with Penrose diagrams? It’s easy to understand what’s going on if you look at the Penrose diagram for a black hole. Here’s one (this one’s showing stellar collapse to form a BH). The infalling observer moves on a line ending somewhere on the singularity. Because the observer has mass, the slope of his world line is greater than 1 in magnitude, like the line representing the surface of the star. Now everything the observer is able to see, before he hits the singularity, is within the past light cone of the point he hits on the singularity. Draw a 45-degree line backward from the observer’s intersection with the singularity, toward past infinity; this is the observer’s past light cone. Notice that there’s a region of spacetime forward of this light cone (the region above this line). This is what the observer can’t see. You can also draw the past light cone for the point at which the observer crosses the horizon. Notice that this is a different light cone, representing a smaller region of spacetime. So the observer continues to see things after crossing the horizon.
It might help to think about a small region of space around the observer. Remember that space is locally flat through the event horizon (the event horizon is just a global property; a freely-falling observer can’t detect it locally), all the way until the neighborhood of the singularity. So within a box falling through the horizon, everything behaves normally. The observer inside the box will continue to measure the speed of light at c. In particular, the infalling light from the radio station above him continues to pass him on its journey to the singularity. So he can still listen to the radio all the way to the singularity.
I understand this statement so let me ask a question. Isn’t the time it takes for the infalling observer to hit the singularity his proper time? And isn’t it true that to a faraway observer the coordinate time t really does come to a stop at the event horizon? (If you’re lowered from a rocket to just outside the EH your watch really will run slower than the watches on the rocket.)
If the above is true then it would seem that the infalling observers finite proper time is actually infinite time for the rest of the universe. So given infinite time why can’t all light reach the poor falling in slob before he hits the singularity?
one other question. Is it true that in some coordinate system (Schwarzschild’s?) the light cone completely closes at the Horizon? And if not, doesn’t it close as it reaches the singularity, or at least partially closes?
And still another question. If to a faraway observer time really does comes to a stop at the horizon, then it would seem that to all faraway observers (us) (in their proper time which pretty well matches up with t ) black holes have no other option then to be truly frozen stars. I understand about falling in observers and Finkelstein coordinates, but, still, if time stops, then I can’t see how it’s not a frozen star (as far as we’re concerned)?
Yeah, future experimental GR is sometimes publish-and-perish for the poor grad students.
I’m not sure I understand your argument here, though. It’s true that the hovering observer’s time dilates relative to observers stationary at infinity, but why is that relevant to the time seen by an infalling observer?
The problem is not that the observer’s proper time is finite, it’s that he hits the singularity, which is something in spacetime which prevents the GR equations from giving solutions all the way to future timelike infinity. In the Penrose diagram I linked to, all of the timelike geodesics that make it to the end of the universe end up at a single point (future timelike infinity “i+”), which is the point at the right end of the singularity zigzag and the top-left end of the diagonal “I+” line. The infalling observer ends on the singularity, which is a point causally disconnected from i+: no information can pass in either direction between the singularity and future timelike infinity.
As the FAQ answer points out, the symmetry argument is not done correctly. The argument goes something like this: Tenured professor Alice is hovering stationary far above a black hole, watching as her graduate student Bob (second author on their forthcoming paper) falls freely into the black hole. With careful observations Alice is able to detect redshifted, outward-directed radiation from Bob’s last moments above the horizon, forever after(*). Thus (goes the flawed reasoning) since a few seconds of Bob’s proper time are visible for ever after to Alice, all of Alice’s future should also be visible to Bob. But this argument is just pretending that we can somehow reverse the worldlines of all photons. This is clearly not true in general–as in this situation, Bob may have moved after sending all of his photons to Alice–although in a lot of optics setups the positions of the endpoints and the properties of the optical path change negligibly over the transit time and so it becomes a reasonable approximation.
I think Bob does see an infinity as he passes through the horizon, but it’s in the other direction. At the horizon is a trapped shell of (absurdly-redshifted) outgoing radiation that’s been sitting there since past timelike infinity(*), for a primordial black hole, or at least since the time the black hole formed. Bob passes through this shell, so he sees all of this past history as he passes through the horizon.
I’m sure I’m not explaining this very well, sorry. [Question, again: Are you comfortable with Penrose diagrams? They’re the best way I know for sketching the causal structure of horizons.]
(*) The FAQ points out that “forever” actually is only true assuming continuous radiation and that if we compute the expected times of photon emissions then we get a finite number of photons out.
Yes, in Schwarzschild coordinates the light cone closes at the horizon. Here’s a sketch, which looks like it’s taken from MTW. Note that the light cone closes both at the horizon and at the singularity, but it closes in different directions because t and r change roles at the horizon. Also, the orientation of the light cones has a discontinuous direction change at the horizon–a strong indication that these coordinates are somewhat untrustworthy near the horizon. For a freely-falling observer the Schwarzschild light cones are not terribly relevant.
As I understand it (the FAQ seems to agree), the objection to the notion of a frozen star is just with the idea of the view of the star, and whatever else passes through the horizon, being “frozen”–i.e., with no redshift to blacken the view–at the moment it crosses the horizon. The idea that all of the history of the horizon can be found in the (absurdly-redshifted) shell of nearly-trapped radiation is still OK, apart from the problem of counting individual photons mentioned above.
