# Is time, inside an event horizon, time-like or space-like?

1. Apr 17, 2008

### tiny-tim

(the story so far … I maintain that, inside an event horizon, there is a useful distinction between "space" and "space-like" dimensions, and that in any realistic coordinate system, time is space-like. JesseM maintains that, in any realistic coordinate system, time must be time-like.

JesseM maintains that, "in a local coordinate system constructed out of freefalling rulers and clocks, the laws of physics must look identical to those of SR." I maintain that this is not possible … that inside an event horizon, the laws of physics are the same, but the geometry is different, and therefore the physics must look different.

now read on … )

conservation of mass
Hi JesseM!

a thought experiment …

In Schwarzschild coordinates, imagine a free-falling observer inside an event horizon. His path is the axis of two "cylinders" of free-falling particles, following him, but faster.

He can see them until they hit his backward light-cone ("diagonal" light does go fast enough for that). And the light by which he seems them reaches him at the same time from both "cylinders".

In your SR-physics local coordinate system, what happens to these "cylinders" when they hit the Schwarzschild light-cone?

Doesn't the Schwarzschild light-cone become a coordinate singularity which swallows them up? And isn't that singularity on the same radius as the observer?

So each "cylinder" converges to an event (possibly "at infinity") whose displacement from him is light-like … and either it is ahead of him, so he sees the particles disappearing in his own future … or it is behind him, in which case he literally has a singularity in his wake, and where is there room for the rest of space-time to fit in?

Or is it in his own present (as the Schwarzschild system requires)?

(And we can shrink the inside "cylinder", so that that event can be as "close" to the observer as we like.)

And, if the laws of physics are the same as in SR, doesn't this coordinate singularity have to be an actual singularity … which it obviously isn't?

And, if the laws of physics are the same as in SR (and if he can't see infinitely far into the future):
what happened to the mass?

2. Apr 17, 2008

### Ich

The Schwarzschild-time-basisvector is spacelike in any realistic local coordinate system. The local time-basisvector is different from the Schwarzschild one, and of course timelike.

3. Apr 17, 2008

### JesseM

I don't think I used the phrase "realistic coordinate system". But if an observer inside the horizon uses a network of freefalling clocks to define his time-coordinate, surely you agree this time-coordinate would be timelike? After all, the path of each clock is a timelike geodesic inside the horizon as well as outside, no?
I'm not sure I understand what you mean by "the axis of two cylinders of free-falling particles". How can a single path be the axis of two separate cylinders? Does one cylinder enclose the other, so their axes coincide? And are they "cylinders" in spacetime or cylinders in space?
This would seem to suggest cylinders in spacetime which enclose one another, so that if we have 2 space dimensions, it's more like the guy is sitting at the center of two concentric rings in space which are moving along with him. But I don't know how to make this jibe with your statement that the particles are "following him, but faster". Maybe you could draw a spacetime diagram of what you mean, with either 1 or 2 space dimensions and 1 time dimension? For simplicity we could start out with a diagram in ordinary flat SR spacetime, then we could talk about how this would work if we extended both the observer's worldline and the worldlines of the particles forming the cylinder into a Schwarzschild or Eddington-Finkelstein diagram where they all go into a BH together (Eddington-Finkelstein would be simpler since it avoids a coordinate singularity at the horizon for infalling particles).
What do you mean by "Schwarzschild light-cone"? Do you mean a light cone in a diagram of Schwarzschild coordinates, perhaps one from an event exactly at the event horizon (as suggested by the 'coordinate singularity') comment? If we want to deal with things happening right at the event horizon, then as I said, it would probably be simpler to use Eddington-Finkelstein coordinates or some other system which doesn't blow up there.
I can't picture what you're talking about here. What event are you talking about, physically? An event on the horizon, on the wordline of the singularity, or somewhere else? Why are the cylinders "converging" to this event?

4. Apr 18, 2008

### tiny-tim

Hi JesseM!
No, you didn't … what would you prefer?
In Schwarzschild coordinates, a clock moves faster than light, and so I agree its geodesic (in any coordinates, of course) is timelike.
Yes … as I said, I'm describing everything in Schwarzschild coordinates … in those coordinates, my (massive) particles freefall down the cylinder faster than the observer freefalls down the axis.
But we're only discussing a local coordinate system, so I'm not bothered about the horizon or the singularity.
The cones in that famous Finkelstein diagram of 1956, which I think was in Schwarzschild coordinates.
I can't picture it, either.

I'm assuming, for sake of argument, that you are right in saying that an observer inside the horizon using a network of freefalling clocks (and only looking at his immediate neighbourhood) (hereafter called "your observer" ) cannot tell whether he is inside or outside the horizon.

The event I'm talking about, described in Schwarzschild coordinates, is the perfectly ordinary event of the particles crossing the light-cone of the observer.

At that event, your observer says that they disappear, doesn't he?

I was trying to describe how your observer would describe that, assuming that he cannot tell whether he is inside or outside the horizon.

I maintain that he can't … that however he tries, he ends up either with mass not being conserved, or with a coordinate singularity which is not an actual singularity.

In other words, if he assumes the physics is the same inside as outside, then he gets a contradiction.

Now, this is a perfectly simple and valid experiment, isn't it?

So … how do you say your observer describes this event (which I have defined in Schwarzschild coordinates)?

5. Apr 18, 2008

### JesseM

Prefer for what? I never suggested that any coordinate system was better than any other. My main arguments were:

a) in a locally inertial coordinate system constructed out of freefalling rulers and clocks, the laws of physics are identical to those of SR

b) your statement that lightlike geodesics from an event would expand in all directions when projected onto a surface of constant t if the event was outside the horizon, but would only expand in a limited cone of directions if the event was inside the horizon, was just a feature of the coordinate system you were using, it has no real physical significance since we could also find coordinate systems where lightlike geodesics expand in all directions from an event inside the horizon when projected onto a surface of constant t, or even a coordinate system where lightlike geodesics would expand in a limited cone of directions outside the horizon.
Let's think about what's going on far from the horizon, where spacetime is close to flat. If we are talking about a universe with 2 space dimensions, then at any given time do you mean the central observer would be surrounded by a ring of particles, so that the worldlines of particles in this ring forms a cylinder in spacetime? If so, what can it mean to say the particles are falling towards the black hole faster than the central observer? Surely if this were the case, then the central observer would not remain at the center of the ring, instead the center of the ring would be moving in the direction of the BH faster than the observer, so the observer would quickly pass out of the edge of the ring.
I thought that when you talked about a "Schwarzschild light cone" you might mean one on the horizon...if you were talking about an event right on the horizon then we would need to be bothered about the fact that Schwarzschild coordinates blow up there, but maybe that's not what you meant.
I am not familiar with "that famous Finkelstein diagram of 1956", unless you are just talking about a standard Eddington-Finkelstein diagram (but then you presumably wouldn't say it was in Schwarzschild coordinates, since Eddington-Finkelstein diagrams are based on Eddington-Finkelstein coordinates). Could you be a little more specific here?
I don't understand what you mean at all. Only instantaneous events have light-cones, an observer has a worldline made up of a continuum of events. If he is surrounded by a ring of particles, then at any given instant on his worldline there will be events on the ring that he doesn't know about because they're not in the past light cone of that instant, but he'll still learn about those events at some later instant that they are in the past light cone of. There need never be a time when the ring disappears from view, because at any given instant his past light cone will intersect with some set of events on the world-cylinder of the ring around him. But I'm still not sure if an observer surrounded by a ring (in a universe with 2D space) or a sphere (in a universe with 3D space) is really what you were talking about in the first place, you really need to state your scenario more clearly.

6. Apr 18, 2008

### tiny-tim

Hi JesseM!
ok … I define my "cylinder" as θ = constant, in the standard Schwarzschild coordinates (t,r,θ,φ), where θ is latitude.

(I tend to put quotes round it, since strictly it isn't a cylinder, but a cone.)
Yes … at any instant, an observer inside a horizon has a cone (in Schwarzschild coordinates) whose rear half contains all and only the directions from which he can receive photons.

It is a different cone, of course, for different positions on his same world-line. Its half-angle becomes smaller as he falls further.

For any particular particle on the "cylinder", falling faster than the observer (in Schwarzschild coordinates), it is inside the cone up to a certain time, then it is outside it.

The event I am talking about is the event on the world-line of that particle when it changes from being inside to being outside.
No … because (in Schwarzschild coordinates) he is going faster than the photons, and that event is the last event on the world-line of that particle from which the photons can catch him.

Photons whose direction is at an angle equal to the half-angle of that cone (in other words: photons from events on his cone) reach him instantaneously (in Schwarzschild coordinates).

Photons from the same event whose direction is at a lesser angle take too long, and can be seen only by another observer following the original observer.

Having defined my event carefully … on the world-line of a particular particle, it is the last event from which your observer can receive photons … again I ask …
… how do you say your observer describes this event?​

In particular, how does he describe the "disappearance" of that particular particle from his view?

7. Apr 18, 2008

### JesseM

Outside the horizon, if we take a snapshot at any given t, that would look like an actual cone which intersects all the concentric spheres of increasing r at a fixed latitude (see the diagram here), but then if you vary t the cone is unchanging...how is anything falling in here? Are you imagining this cone's surface is made up a continuum of infalling particles?

If so, then since your claim is about the infalling particles "falling out of view" for the observer in the horizon, it seems unnecessarily complicated to talk about a continuum of particles, why not just pick a single infalling particle? I assume your claiming that if we select any one infalling particle that the observer can see before crossing the horizon, that particle will disappear from view for him sometime after he crosses it?
If you're talking about things like "half-angles" here, presumably you're picturing something...could you please try to draw it? Your verbal explanations really aren't helping much.
Are you claiming that the worldline of a particular particle can be such that it intersects the edge of the past light cone of some earlier event on the worldline of the observer, but then it doesn't intersect anywhere on the edge of the past light cone of some later event on the worldline of the observer? That seems geometrically impossible to me. Again, a diagram would be useful.
But the fact that he's going "faster" than photons is just an artifact of the coordinate system, where the t in dr/dt is no longer timelike. I don't see how it would have real physical implications like photons from an object that was previously in his past light cone being unable to "catch up with him" after a certain point. And if this physical implication is correct, it should show up in any coordinate system, no? Do you think it would show up in Eddington-Finkelstein coordinates, even though in these coordinates photons on the "underside" of the future light cone of an event on the worldline of an infalling particle always move faster than the particle, while photons on the "upper side" of the future light cone move slower than the particle?

Speaking of coordinate artifacts, I think it's worth pointing out Schwarzschild coordinates look kind of ugly inside the horizon because they make the path of every object appear time-reversed, with the earliest event on their worldline inside the horizon being the collision with the singularity, and the worldline getting closer to the horizon at later values of t, with the distance from the horizon approaching zero as t approaches infinity...see the diagram on p. 848 of Gravitation for example. And if you have that book, perhaps you could draw a similar diagram to illustrate what you mean about photons from a particle that could previously be seen by the infalling observer being unable to "catch up" with any points on the worldline the infalling observer after a certain point. Though again, I would much prefer to see an Eddington-Finkelstein type diagram, which has nice features like the fact that photons emitted on the "underside" of the light cone always move on perfect diagonals, and the fact that it doesn't time-reverse worldlines inside the horizon like Schwarzschild coordinates do (also, I've found Eddington-Finkelstein diagrams which trace the complete worldlines of photons emitted from some event on the worldline of an infalling observer, but I can't find analogous diagrams in Schwarzschild coordinates, although the one on p. 848 does show the immediate direction of the photons after being emitted from an event).

One last thing I'd note is that even if we were talking about faster-than-light "tachyons" in ordinary flat SR spacetime using Minkowski coordinates, as far as I can tell, if some object was in the past light cone of an earlier event on the worldline of the tachyon, it would continue to be in the past light cone of all later events on the worldline of that tachyon. So even at this basic level, I don't see how the fact that something is moving faster than light in a given coordinate system would imply that any object would ever "disappear from view".
I still am inclined to think that the notion of a "disappearance" is a mistake you are making here, because you don't seem to have thought the visual diagrams through very well, and you certainly don't seem to have done any detailed mathematical analysis.

Last edited: Apr 18, 2008
8. Apr 18, 2008

### DrGreg

Let's just examine JesseM's claim about local coordinate systems inside the event horizon. I'm going to explicitly calculate such a coordinate system for you.

If we restrict ourselves to one-dimensional motion along a straight line to the centre of a black hole (so $d\theta = d\phi = 0$) the Schwarzschild metric simplifies to

$$ds^2 = \left( 1 - \frac{2m}{r} \right ) dt^2 - \left ( 1 - \frac{2m}{r} \right )^{-1} dr^2$$​

with the usual convention that units are chosen so that c = G = 1.

Consider the event (t=t0, r=r0) inside the event horizon, so that

$$1 - \frac{2m}{r_0} < 0$$​

Now define new coordinates

$$T = \left ( \frac{2m}{r_0} - 1\right )^{-1/2} (r - r_0)$$
$$R = \left ( \frac{2m}{r_0} - 1\right )^{1/2} (t - t_0)$$​

Substitute these into the equation for the metric and you will get

$$ds^2 = \frac{1 - 2m/r}{2m/r_0 - 1} dR^2 - \frac{ 2m/r_0 - 1}{1 - 2m/r}dT^2$$​

When T is very small, r is approximately r0 and so ds is approximately

$$ds^2 \approx dT^2 - dR^2$$​

But this is just the standard Minkowski metric for flat spacetime i.e. special relativity, so, very close to the event (t=t0, r=r0) i.e. (T=0, R=0), the coordinates (T,R) behave just like (and therefore are) time and space coordinates (in that order) in special relativity.

The local geometry is determined entirely by the local behaviour of the metric. So I've just shown the local geometry approximates to the geometry of special relativity.

Q.E.D.

Outside the event horizon, you would substitute

$$T = \left ( 1 - \frac{2m}{r_0} \right )^{1/2} (t - t_0)$$
$$R = \left ( 1 - \frac{2m}{r_0} \right )^{-1/2} (r - r_0)$$​

instead.

9. Apr 18, 2008

### tiny-tim

Hi DrGreg!

Have I got this right … ?

All you've done is swap over dr and dt, and called them dT and dR, respectively, after applying a linear transformation to make c = 1?

Then the full metric becomes, locally, the standard Minkowski metric, with time and radial distance swapped over.

But … that's obvious!

I'm not disputing you can do that.

I'm disputing the claim that an observer, using local observation only, cannot tell whether he's inside or outside an event horizon.

In particular, I'm claiming that he can tell he's inside by noticing that everything goes faster than light (in the sense that a series of massive particles will overtake light), and also by noticing that faster-freefalling particles disappear from view.

Can you help on this?

10. Apr 18, 2008

### JesseM

Suppose we take a very small patch of spacetime in Schwarzschild coordinates where the worldlines of a photon and a massive particle intersect, and transform into local Minkowski coordinates on that patch. Are you claiming the massive particle will still be moving faster than the photon even in locally inertial Minkowski coordinates? If so I'm sure you're wrong just by the equivalence principle, although I don't have the expertise in GR to do the math here...maybe DrGreg or someone else can help, if indeed you would make the claim that electrons move faster than photons even in locally inertial coordinates.

11. Apr 19, 2008

### MeJennifer

12. Apr 20, 2008

### tiny-tim

Hi Jennifer!

hmm … "Light cones tipping over" … I did a thread-title search on this sub-forum before starting this thread, but I didn't think to look for those words.

-> which reminds me ->​

Whenever I'm at the Restaurant At The End Of The Black Hole …

I never know how much to tip!

13. Apr 20, 2008

### tiny-tim

… I-Can't-Believe-It's-Not-Water coordinates …

Hi JesseM!
No! But it's a coordinate effect, not a physical one.

Since, in Schwarzschild coordinates, a massive particle has dx/dt > 1, it's obvious that if you swap x and t, you get dx/dt < 1!

(btw, do you accept, by the same argument, that massive particles must go faster than light in Schwarzschild coordinates?)

In other words: if an observer inside a horizon insists on using locally inertial Minkowski coordinates, then it's obvious that he will regard massive particles as slower than light.

However, that doen't alter the physical fact that the massive particles will overtake light, and in particular they will reach the singularity first.
Isn't that begging the question?

I keep asking people to comment on how your observer describes the inconvenient physical fact of the disappearance from his view of a faster particle … but nobody answers!

If you start by saying "I believe that no observer using locally inertial Minkowski coordinates can tell whether he's inside a horizon", then obviously you thereby disprove the very existence of inconvenient facts!

Look … if I were to break through the boundary of the bowliverse, I could insist on using I-Can't-Believe-It's-Not-Water coordinates, and thereby prove that I was still in water!

I maintain that an observer crossing a horizon (traditionally described as an unremarkable event) will not suddenly decide to swap his r and t coordinates, and will notice if someone else swaps them!

In particular, he will notice that things start disappearing!

Only an observer with spaghetti for brains, or with the memory and attention-span of a goldfish, could fail to …

erm …
I … er …
d'oh!

14. Apr 20, 2008

### JesseM

Why do you say that? Do you deny that if an observer constructs a local coordinate system out of freefalling rulers and clocks moving alongside them, using the same method that's used to construct inertial coordinate systems in SR, then this will result in the same sort of locally inertial Minkowski coordinates?
Of course.
They will reach it "first" in Schwarzschild coordinate time, which is a spacelike coordinate. As I said, in a coordinate system constructed out of freefalling rulers and clocks, this might be equivalent to something like "hitting the singularity further to the left".
No, it's appealing to a known property of the theory of GR to answer a physical question about a GR scenario.
And I keep telling you I don't believe there will be an "physical fact of the disappearance from his view", that you're confusing yourself using handwavey arguments. It would really help if you would answer some or all of the requests for clarification I made in post #7. For example, if you think the "disappearance from view" is an obvious physical fact, why do you refuse to draw a simple diagram showing how the worldline of the disappearing particle could instersect the past light cone of the infalling observer at one event on his worldline, but fail to intersect the past light cone of the same observer at a later event on his worldline?
I don't believe your inconvenient "facts", which you seemed to have arrived at by vague handwavey arguments that you can't even depict in a spacetime diagram and certainly have not supported using detailed math, are facts at all. But if they are physical facts, then they should show up in any coordinate system, including locally inertial Minkowski coordinates (i.e. if the particle disappears from the past light cone of the infalling observer, it must do so when you look at the past light cone and the particle's worldline in these coordinates, no?)
Any "physical fact" will be true in all coordinate systems.
Who said anything about a sudden swap? If the observer uses freefalling clocks to define his time coordinate, the t coordinate will be timelike both inside and outside the horizon, nothing strange will happen at the horizon. It is in Schwarzschild coordinates where you have a "swap", but the t coordinate in Schwarzschild coordinates is not based on readings of actual physical clocks inside the horizon.
If you have any coherent argument for why you think this will happen, you should really be able to answer my questions in post #7. I think you're just confused.

Last edited: Apr 20, 2008
15. Apr 21, 2008

### DrGreg

It's unfortunate that two of the Schwarzschild coordinates are labelled t and r, and often referred to as "time" and "radius" (or "distance"). That gives the false impression that these coordinates correspond to real i.e. proper time and proper length when really they don't. For a distant observer (hypothetically "at infinity") they do indeed correspond to time and radius, but as you approach the black hole they start to distort (they need to be rescaled using the second coord system I gave at the bottom of my post #8) and when you actually reach the event horizon they just break down completely. However they do work again once you are inside, but with time and space swapped as I showed. If you want to analyse what happens as you cross the event horizon, you need a different coordinate system.

I must confess I have only paddled in the shallow waters of GR, and this discussion is a learning curve for me, but a quick search of Wikipedia found the Lemaitre metric which has no discontinuity at the event horizon and also has the advantage that objects that are at rest in Lemaitre coords are free-falling. The article gives the formula for conversion from Schwarzschild coordinates.

(See also Kruskal-Szekeres coordinates and Gullstrand-Painlevé coordinates, which, at a glance, seem less useful but do contain some useful info. I haven't read any of these 3 articles in depth. yet.)

In GR the local speed of light, measured using a local frame of reference, need not be the same as the coordinate speed of light using the coordinate system of a remote observer extrapolated to the region being measured.

When you say "massive particles will overtake light", that is only from the point of view of a distant observer who interprets velocity as "dr/dt" in Schwarzschild coordinates. A local observer interprets velocity as "dR/dT" where (T,R) is a locally-inertial coord system. (Or using any other locally-inertial system related to the first one via a Lorentz transform -- choose the one in which he/she is at rest i.e. in which the coordinates of the observer satisfy dR'/ds = 0.) From the local observer's perspective, massive local particles still move slower than light. The local speed of light is always given by ds=0, in any coordinates.

16. Apr 21, 2008

### tiny-tim

… a coordinate-free definition of "overtaking" …

Hi DrGreg!
No … I can define "overtaking" in a coordinate-free way:

Imagine a series of 10 photons, freefalling inside the horizon down the same radius, labelled 1 to 10 so that 1 is nearest the singularity (either in r or in t … it makes no difference) in Schwarzschild coordinates.

And imagine a similarly labelled series of 10 massive particles freefalling down an adjacent radius.

"Adjacent" means that there is no doubt as to when one of the photons is next to one of the particles.

In Schwarzschild coordinates, |dr/dt| for a massive particle is greater than for a photon.

So, in Schwarzschild coordinates, if particle 1 starts next to photon 1, then shortly afterwards particle 2 will be next to photon 1 (assuming they're close enough together), then particle 3 …

Furthermore, in Schwarzschild coordinates, particle 1 and photon 1 hit the singularity before all the other particles and photons, respectively.

Therefore, following the world-lines of the particles and photons in any coordinate system, we can identify "particles overtaking photons" as meaning "as a particular photon gets nearer the singularity, it is successively next to freefalling particles which are further from the singularity".

Do you agree that the statement in bold is true in Schwarzschild coordinates, and therefore in any coordinate system?

(Similarly, do you agree that "diagonal" photons will overtake radial photons in any coordinate system? And that photons from particles overtaking a freefalling observer will not be able to reach the observer if they start from too great an angle … in other words, the particles will disappear from view, without any singularity, no matter how small the neighbourhood?)

17. Apr 21, 2008

### JesseM

Look at the diagram of Eddington-Finkelstein coordinates here, which I found on this page. With the r=2M line as the event horizon, it shows the wordlines of photons emitted on "both sides" of various light cones--the two paths bordering the ellipses on either side. The diagram also shows a dark line representing a massive particle. Inside the event horizon, you can see that for the photons emitted on the right side of the light cones (the ones whose worldlines are curved rather than diagonal straight lines), the massive particle is moving faster than they are in these coordinates--in the diagram we see it "passing" two separate photon worldlines and hitting the singularity before they do. Is this the type of thing you're talking about?

If so, just note that the sense in which the dark particle worldline "passes" photon worldlines on the right side of the light cones inside the horizon is identical to the sense in which it passes photons worldlines on the right side of light cones outside the horizon--these are just photons moving to the right, outside the horizon, so of course if I see a series of photons moving to the right while I'm moving to the left, I'll pass a series of them in succession. Inside the horizon, it may look like these photons are moving inward just like the particle, but that's just because of the way light cones have been tilted--these are still the photons that would be moving to the right in a locally inertial coordinate system. And note that the particle does not overtake photons on the left side of the light cones, whose worldlines are always diagonal straight lines in Eddington-Finkelstein coordinates.

Last edited: Apr 21, 2008
18. Apr 21, 2008

### tiny-tim

Hi JesseM!

ok … in your University of Winnipeg diagram, t increases upward (at large r), so my photon 1 is the diagonal line at the bottom, and my photon 8 is the diagonal line at the top.

The bold line is my particle 1 (we don't need the other particles).

The curved lines are "oufalling" photons, which we're not interested in, in this example.

photon 1 hits the singularity first, then photon 2, then … photon 5, then particle 1, then photon 6, …

So particle 1 is next to (to use a neutral expression) photon 1 first, then photon 2 …

So, according to the diagram, the particle is indeed overtaken by the photons, and therefore I am wrong and you are right.

However … all this proves is that you can prove anything with diagrams! :rofl:​

The particle line in the diagram is wrong.

It should reach an angle of 45º at the horizon … the same angle as the infalling photons, and then curve over further, overtaking the photons that had previously overtaken it.

I'll take the equations from http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity#Geodesic_equation:

For a radially falling particle (so dφ/dt = L = 0) of mass m, with c = 1, with wikpedia's 1 - rs/r = 1 - 2GM/r = δ, and with a speed "at infinity" of tanhα (so putting wikpedia's E = m.coshα), we have:

(dr/dτ)² = cosh²α - δ; (dt/dτ)² = cosh²α/δ²;

so dr/dt = ±δ√(1 - δsech²α).

In other words: "at infinity", δ = 1, and dr/dt = ±tanhα, as expected.

But very near the horizon, δ approaches 0, and dr/dt approaches ±δ … which is the same as dr/dt for photons!

And inside the horizon, δ is negative, and so of course |dr/dt| > δ, and the particle falls faster than photons.

Agreed?

19. Apr 21, 2008

### JesseM

The curved lines are only outgoing photons outside the horizon. The curved lines inside the horizon are falling in towards the singularity just like the diagonal lines. If at a particular moment on its worldline the observer emits photons in both left and right spatial directions as seen in locally inertial coordinates, then (assuming at this moment the observer was already inside the horizon) the left photon goes along a diagonal line until it hits the singularity, while the right photon goes on a curved line until it hits the singularity. And you can see that the infalling observer's coordinate speed is greater than the coordinate speed of the photons following curved worldlines, inside the horizon (the slope of the observer's worldline is closer to horizontal).
Not with correctly-drawn diagrams, no. Any physical conclusions you get in a correctly-drawn diagram should carry over to a correctly-drawn diagram in any other coordinate system.
No, you're wrong here, the diagram matches all other diagrams of Eddington-Finkelstein coordinates I have seen. For example, see the identical diagram here, taken from this page which I referred to earlier. Or the diagram here, taken from this page. Or how about the diagram here from this page, which is a scan from the classic Misner-Thorne-Wheeler GR textbook Gravitation, from p. 829. Do you think Misner-Thorne-Wheeler made an error in their diagram too?
I don't really follow the details of the math here, but these equations are in Schwarzschild coordinates rather than Eddington-Finkelstein coordinates, yes? And Schwarzschild coordinates are badly-behaved at the horizon--both massive particles and photons take an infinite time to reach the horizon in Schwarzschild coordinates, so naturally they must both approach the same coordinate velocity of zero as the time coordinate approaches infinity. And inside the horizon Schwarzschild coordinates are also fairly counterintuitive, since as I mentioned before they reverse the order of events, with the particles emerging from the singularity first and their proper time decreasing with increasing t, so that they get closer and closer to the horizon from the inside as t approaches infinity as well. There's a helpful diagram of the worldline of an infalling object in Schwarzschild coordinates on p. 848 of Gravitation which I've scanned and included as an attachment. You can see that even within the horizon, the object's worldline stays within the future light cone of any event on the worldline.

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20. Apr 22, 2008

### tiny-tim

Hi JesseM!
I didn't call those photons outgoing, I called them "outfalling", with the quotes marks. This has the advantage of distinguishing them from infalling photons, without having to talk about "left and right spatial directions" or such-like.

I accept that it's a bit of an oxymoron … since only friends can literally fall out inside a horizon
usually over map-reading!
ok … your first diagram is from the University of Winnipeg, and I've already commented on it … your second diagram is from the University of Syracuse NY, copied from Geroch's book, and it only shows "outfalling" photons, not infalling ones, so that doesn't help either of us (except that the slope on crossing the horizon, at U, looks like 45º to me) …

and your third diagram is from MTW's Gravitation, but it is described as "edited" … is the original diagram the same?

(And your fourth diagram, in Schwarzschild coordinates, scanned directly from p.848 of MTW, although it doesn't show a series of photons, it does show the particle going faster than infalling light, doesn't it? )
Yes.
Agreed … but they're fine right up to the horizon, and they show clearly that, in Schwarzschild coordinates, dr/dt approaches the local speed of light (my δ) as the particle approaches the horizon.

Do you accept that? If so, surely you accept that in the EF coordinates, photons and particles cross the horizon at the same angle … 45º?
Yes … a particle's "own clock" keeps going forward, but its t coordinate goes to infinity at the horizon, and then starts winding back again.

So … what?

What matters is the absolute value of dr/dt … and the equations show clearly that inside a horizon it's greater than δ.
The Schwarzschild coordinates are a perfectly valid coordinate system.

Although, in Schwarzschild coordinates, r and t both approach infinity at the horizon, dr/dt is still well-defined, and enables a valid comparison of speed both at the horizon and beyond!

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