A problem with polar coordinates and black hole

1. Sep 26, 2015

Harel

Hey, I know that one doesn't work with polar coordinates (t,r,θ,φ) because they don't behave well in the event horizon. But my problem is with raidal null curves, if we take
ds2=0 and dφ, dθ = 0 so we have

When, if i'm correct, the + sign determine that it's outgoing and the - infalling, so with that logic in r<2GM an infalling photon (minus sign) will become outgoing because the sign will change to +. And the fact that the math tells me that we can have a photon in the r<2GM region that is outgoing(r is increasing) is my problem.
It will help a lot if someone could tell me what is wrong with that logic.

2. Sep 26, 2015

Staff: Mentor

You might find http://arxiv.org/abs/0804.3619 helpful. Note especially the bit about how, because the coordinate singularity at $r=2GM$ means that the coordinate patch with $r<2GM$ cannot be joined smoothly to the patch with $r>2GM$, $r$ and $t$ inside the horizon are not the same coordinates as $r$ and $t$ outside the horizon.

(You said "polar" coordinates but your post suggests that you meant Schwarzschild coordinates).

3. Sep 26, 2015

Staff: Mentor

Which "+ sign" and "- sign" do you mean? There are actually three signs involved here:

(1) The sign of $\left( 1 - 2GM / r \right)$, which is + for $r > 2GM$ and - for $r < 2GM$;

(2) The choice of + or - for the $\pm$ sign in front of $\left( 1 - 2GM / r \right)$ in the formula for $dt/dr$;

(3) The sign of $dt/dr$ that corresponds to "ingoing" or "outgoing".

In the region $r > 2GM$, sign (1) is always +, so that's easy. Sign (3) is also simple: a positive $dt/dr$ means an outgoing geodesic, and a negative $dt/dr$ means an ingoing geodesic. The question is what determines our choice of sign (2); and the answer is that, given (1) and (3), we must choose the + sign for an outgoing geodesic and the - sign for an ingoing geodesic.

In the region $r < 2GM$, sign (1) is always -, so again that's easy. But sign (3) is now reversed: a positive $dt/dr$ means an ingoing geodesic, not an outgoing one (and vice versa). So if we hold our choice of sign (2) constant along the same geodesic (which we must), then everything works out: sign (1) flips inside the horizon, but so does the meaning of the sign of $dt/dr$, so the formula still describes an ingoing geodesic.

Last edited: Sep 26, 2015
4. Sep 27, 2015

Orodruin

Staff Emeritus
All geodesics within the event horizon are ingoing in the sense that the r coordinate is always decreasing. The r coordinate is time-like inside the event horizon and the t-coordinate is space-like.

5. Sep 27, 2015

Harel

Hey first thankes for your help, I ment for the sign of dt/dr and as I understood negative dt/dr means ingoing geodesic anywhere, I didn't understand why dt/dr change his meaning in the r<2GM region, so if you could explain to me why I will be gratefull.
And yet even with that logic, if we consider a photon that been sent radially inside the r<2GM region (meaning the sign (1) is -) by some astronaut and we choose the sign (2) to be + than dt/dr is negative meaning that it's outgoing, which again is a problem. I'm sure that something is wrong with my understanding yet I don't know what.

6. Sep 27, 2015

DrGreg

Maybe this Kruskal spacetime diagram (drawn for 2GM = 1) will help. The pink lines are the Schwarzschild coordinate gridlines outside the event horizon. The blue lines are the Schwarzschild coordinate gridlines inside the event horizon. The black dotted line is the event horizon. Null lines are always at 45° (parallel to the dotted lines) anywhere on the diagram.

Inside blue region II, timelike r is always decreasing, but spacelike t can decrease for a null line going to the left ("ingoing") or increase for a null line going to the right ("outgoing"). But, either way, there is no escape from reaching the darker blue line, the singularity.

Image credit: Kruskal diagram of Schwarzschild chart by Dr Greg at Wikimedia Commons, Creative Commons Attribution-Share Alike 3.0 Unported licence.

7. Sep 27, 2015

Staff: Mentor

Not in the maximally extended spacetime; that includes the white hole region, where all geodesics are outgoing (more precisely, all geodesics going from region IV, the white hole, to region I in DrGreg's diagram). The coordinate assumptions the OP made do not rule out this possibility.

8. Sep 27, 2015

Staff: Mentor

DrGreg's diagram illustrates it, but here is a quick explanation in words: as you approach the horizon from above (i.e., from the region $r > 2GM$), the $t$ coordinate increases without bound; this is sometimes expressed, heuristically, as saying that at the horizon, $t = \infty$. So we have increasing $t$ and decreasing $r$ for an ingoing geodesic, and $dt/dr$ is negative.

But $t$ can't get any larger than $\infty$. So inside the horizon, in the region $r < 2GM$, $t$ must be decreasing as we go inward, not increasing. That means that an ingoing geodesic will have decreasing $t$ and decreasing $r$, so $dt/dr$ is positive.

9. Sep 27, 2015

Staff: Mentor

Just to clarify, the OP was talking about null geodesics coming from region I in your diagram, and all of those must be ingoing if they enter region II (the black hole). The diagram makes this evident.

10. Sep 27, 2015

Orodruin

Staff Emeritus
But region IV is not part of the original Schwarzschild space-time so I doubt this is what the OP intended. Within region II, which I believe is what the OP would consider, all geodesics have a decreasing r-coordinate.
Again, look at region II. There are light-like geodesics with both decreasing and increasing t. Both of them have decreasing r and end up at the singularity at r = 0. The OP does not specify that the geodesics must start in region I.

11. Sep 27, 2015

Staff: Mentor

Fair enough.

I assume you mean null geodesics originating from some event in region II, but outgoing instead of ingoing? Yes, these have increasing $t$ instead of decreasing $t$, but decreasing $r$, so they have negative $dt/dr$. So we can still use the sign of $dt/dr$ to distinguish outgoing from ingoing null geodesics in region II; we just have to reverse the rule we use in region I.

Also, in the part of my post that you quoted, I was (implicitly) talking about an ingoing geodesic that originates in region I and crosses into region II. That's why I talked about the behavior of $t$ as the horizon is crossed. Outgoing geodesics that originate in region II will never cross the horizon, so yes, the reasoning I gave does not apply to them.

12. Sep 27, 2015

Orodruin

Staff Emeritus
But this is exactly my issue. I would not like to refer to these geodesics as "outgoing", they are never going to get out of the event horizon. I am uncomfortable with referring to any direction inside the event horizon as out- or ingoing. There are different spatial directions and regardless of which of these directions you go, you end up at the singularity.

13. Sep 27, 2015

Staff: Mentor

I agree that "outgoing" and "ingoing" may not be the best terms to use inside the horizon, for the reason you give. However, they are the conventional terms; and they do have the advantage of providing a continuous meaning for the terms "outgoing" and "ingoing" along trajectories that pass from region I into region II (which are the ones that are usually of interest).

For example, suppose I am free-falling into a black hole, and I have a telescope pointed at a particular distant star that is directly above me. If I fire a laser along the telescope towards the star, that is the "outgoing" direction; the opposite direction is the "ingoing" direction. That continues to be true even after I cross the horizon and my laser pulses fired towards the star will never reach it, but remain trapped inside the black hole and eventually hit the singularity; even though that's the case, there is still an obvious distinction between the two directions along my telescope, and calling those two directions "outgoing" and "ingoing" is the simplest way to describe it.

14. Sep 27, 2015

Orodruin

Staff Emeritus
I do see your point regarding this, I just think it is a bad nomenclature. Once inside region II, that region is symmetric with respect to the T axis and reflections in that axis would exchange the meaning of ingoing and outgoing.

15. Sep 27, 2015

Staff: Mentor

Yes, agreed. But this corresponds to exchanging regions I and III, and region III is not present in any real black hole spacetime (i.e., one that describes a hole formed by gravitational collapse). So the symmetry that's present in the maximally extended spacetime is not present in a real black hole spacetime. The gravitational collapse breaks the symmetry, so to speak, and therefore picks out a preferred sense of "ingoing" and "outgoing". (It also breaks the symmetry about the X axis of Kruskal coordinates, i.e., the symmetry between past and future; that is why region IV is not present in an actual black hole spacetime, while region II is.)

16. Sep 27, 2015

Harel

Thanks for the answers! I know the kruskal diagrams and why a photon that originated at region 2 must end up in the singularity in those coordinates yet I try to understand how one can know that the same thing will happen using only the Schwarzschild coordinates. I had a problem with Peter's answer because he referd to a one that originated in the r>2Gm region but as I stated with that logic one can have a photon that originated in the r<2Gm region and will increase in r. One possible answer is just that the s.coordinate are not enough when talking about this region. Yet maybe there is something else that I'm missing.

17. Sep 27, 2015

Staff: Mentor

You are taking your coordinates too seriously. The worldline of the infalling particle is what it is no matter what the coordinates you use to label events along it, so if you find it hard to make sense of that worldline using Schwarzschild coordinates.... Don't use them. (It is very easy to fall into the trap of believing that Schwarzschild coordinates are somehow less "artificial" because they work so well for observers outside the event horizon).

Here you have two regions of spacetime covered by two different sets of coordinates. In one region, the timelike coordinate is increasing along the world line of the infalling particle while one of the spacelike coordinates is decreasing (and the other two are constant). In the second region, the timelike coordinate and one of the spacelike coordinates are decreasing (while the other two spacelike coordinates are constant) along the infalling worldline. Naturally the signs of the derivatives will be different in the two regions... And that sign flip tells you more about how your coordinates are defined along the world line than about the actual physics.

18. Sep 28, 2015

Demystifier

You mean spherical. Polar coordinates are for 2 dimension, while spherical coordinates are for 3 dimensions.
(I noted this minor mistakes because recently I made the same mistake by myself. )

Yes, such a possibility exists mathematically. Physicaly it is discarded because it corresponds to a motion backwards in time. This is because, for r<2GM, r is a time-like coordinate, so r-increasing really means time-decreasing.

The others above already said that in different words, but it may be instructive to say the same thing in different words.

19. Sep 28, 2015

Orodruin

Staff Emeritus
Spherical coordinates are sometimes referred to as "spherical polar coordinates". Of course, if you want to be clear you should simply not use "polar coordinates" when referring to something three-dimensional and use "spherical coordinates" and "cylinder coordinates" instead depending on the intended meaning, then it will be unambiguous which three-dimensional coordinate system you are referring to.

20. Sep 28, 2015

Harel

But how can you know that r increasing means time decreasing and not increasing?