Crossing the Event Horizon of a Black Hole

  • #126
3,962
20
I wouldn't say this. I would say this that Schwarzschild coordinates really make up two different coordinates systems. One system is valid outside the black hole, but not on the event horizon or inside the black hole. The other system is valid inside the black hole, but not on the event horizon or outside the black hole. The metric unambiguously determines the nature of each coordinate in each coordinate system.

So, inside the event horizon, space doesn't get rotated into a time direction, it's just that human-chosen labels (coordinates) have names (chosen by humans, not by nature) that are very misleading. See

https://www.physicsforums.com/showthread.php?p=1146536#post1146536.
I agree that there are two systems in Schwarzschild coordinates. The system that has real solutions below the event horizon unambiguosly shows that light can travel from the event horizon to the central singularity and in the opposite direction from the singularity to the event horizon. If an event near the central singularity is connected to a future event just below the event horizon and the two events are within a common (and real) future light cone, then it seems suspect to be able to transform those events to another set of coordinates where the two events happen in the reverse order effectively reversing time.
 
  • #127
JesseM
Science Advisor
8,496
15
I agree that there are two systems in Schwarzschild coordinates. The system that has real solutions below the event horizon unambiguosly shows that light can travel from the event horizon to the central singularity and in the opposite direction from the singularity to the event horizon.
But light doesn't have its own proper time, so how do you distinguish what "direction" the light is traveling in? Perhaps you just mean that in Schwarzschild coordinates, some light beams cross the horizon at a t coordinate of -infinity while others cross at a t coordinate of +infinity? But obviously this wouldn't mean much physically, given that inside the horizon the t coordinate is spacelike rather than timelike anyway. Would agree that all timelike worldlines cross the horizon at an earlier proper time than they hit the singularity, assuming we set the direction of increasing proper time along each timelike worldline such that any timelike worldlines which cross at a single point will agree which is the future light cone of that point and which is the past light cone of that point, and that at least one worldline hits the singularity at a later time than it crossed the horizon?
kev said:
If an event near the central singularity is connected to a future event just below the event horizon and the two events are within a common (and real) future light cone, then it seems suspect to be able to transform those events to another set of coordinates where the two events happen in the reverse order effectively reversing time.
"Suspect" in what sense? The absolute freedom of coordinate systems in GR (diffeomorphism invariance) means that coordinate time can work any way you want it to, it doesn't really mean anything physical (for example, even in flat spacetime you could pick a weird coordinate system where the event of my turning 30 happened at a later coordinate time than the event of my turning 29 and at a later coordinate time than the event of my turning 31). If you're interested in physical statements about time, you want to stick to looking at proper time and forget about coordinate time.
 
  • #128
3,962
20
....
If you're interested in physical statements about time, you want to stick to looking at proper time and forget about coordinate time.
OK, let's stick to proper time. Have at a look at this diagram from mathpages: http://www.mathpages.com/rr/s6-04/6-04.htm


http://www.mathpages.com/rr/s6-04/6-04_files/image043.gif [Broken]

It is the trajectory of a particle in proper time and coordinate time, but unusually the time coordinate is horizontal in this chart. In coordinate time the particle leaves the central singularity and goes up towards the event horizon via time past infinity. This is the "white hole" segment of a KS coordinate system and we do not usually worry about it because it is something that happened in the infinite past and anything could have happened then. However when we look at the particle path in terms of advancing proper time, the particle rises to the event horizon in finite time. In the diagram the apogee is centred on coordinate time zero but there is no reason why the apogee could not be in the future and the particle rises up and out through the event horizon in present time. There is no demarcation between white hole and black hole in this diagram. Although light paths are not shown in this diagram they would all be moving downwards towards below the event horizon and the rising particle is moving in the opposite direction to the future and past light cones. This is a serious problem for the interpretation of what happens below the event horizon and the analysis has been done on your terms of only considering advancing proper time.
 
Last edited by a moderator:
  • #129
JesseM
Science Advisor
8,496
15
OK, let's stick to proper time. Have at a look at this diagram from mathpages: http://www.mathpages.com/rr/s6-07/6-07.htm
Actually that image you posted is from a different section of mathpages: http://www.mathpages.com/rr/s6-04/6-04.htm

And on that page, they later explain that the inner region the particles emerge from cannot be the inner region the particles fall into (assuming no closed timelike curves):
To understand the full set of possible trajectories consistent with the Schwarzschild metric, it’s useful to first note an ambiguity present in all pseudo-Riemannian metrics due to their quadratic character. Consider the Minkowski metric (dt)2 = (dt)2 – (dx)2, which obviously doesn’t constrain the signs of the differentials, because they each appear squared. At constant x the metric requires (dt/dt)2 = 1, but the ratio dt/dt itself can be either +1 or -1. Strictly speaking, we are free to choose whether the proper time along a given path increases or decreases as the coordinate time increases. We might fancifully imagine that the Minkowski metric actually entails two separate universes, with proper time increasing with coordinate time in one, and decreasing with coordinate in the other. Alternatively we could imagine a single universe with two families of particles, whose proper times increase in opposite directions of the coordinate time t. (In fact, John Wheeler once speculated that anti-matter particles might be modeled as particles moving backward in time.) However, with a fixed metric like the Minkowski metric, it’s easy to just arbitrarily stipulate the same sign for dt and dt for every path, and then continuity requires that this always remains true (since timelike paths cannot “turn around” in Minkowski space).

The same quadratic ambiguity arises when considering the Schwarzschild metric, but in this case the various possible signs of the differentials are more inter-related, because the coefficients of the metric change signs at r = 2m. For values of r greater than 2m we have a metric of the form (dt)2 = (dt)2 – (dr)2 neglecting scale factors, whereas for values of r less than 2m the metric takes the form (dt)2 = (dr)2 – (dt)2. In the former case, |dt| must always equal or exceed |dt|, but in the latter case |dr| must equal or exceed |dt|. Thus, outside the Schwarzschild radius we must choose the sign of dt/dt, and inside that radius we must choose the sign of dr/dt. The signs of these ratios cannot change along any particle’s path in their respective regions. In effect, the radius r serves as the “time” coordinate inside the Schwarzschild radius.

Now, by analytic continuation, it can be shown that a path crossing the Schwarzschild radius from an outer region must enter an inner region with negative dr/dt. This is why a particle falling inward through the Schwarzschild must thereafter continue to reach smaller and smaller values of r. It cannot “turn around”, but must continue down to r = 0. However, conversely, it can be shown that a particle passing outward through the Schwarzschild radius must have come from an inner region of positive dr/dt. Hence if we observe objects falling into the inner region, and other object emerging from the inner region, we seem forced to conclude that there are two physically distinct inner regions, or else that there exist closed spacetime loops if we insist on a single interior region. One or the other of these consequences is unavoidable if we take seriously the analytic continuation of all geodesics consistent with the Schwarzschild metric. The existence of two distinct inner regions is perhaps not surprising if we note that an in-falling object requires infinite coordinate time to cross the boundary at r = 2m, and conversely an out-going object requires infinite coordinate time to emerge. Clearly these are two very different classes of objects, one coming from the beginning of coordinate time, and the other departing to the end of coordinate time. The same reasoning leads to the potential existence of a second outer region, with negative dt/dt, so the full extent of the manifold entailed by the Schwarzschild metric, if fully developed, consists of four distinct regions. Thus the consideration of simple radial trajectories in Schwarzschild spacetime leads unavoidably to cosmological issues, which are described more fully in the discussions of “black holes” in Section 7.
So, looking the diagram you posted:

http://www.mathpages.com/rr/s6-04/6-04_files/image043.gif [Broken]

Although all of these paths are valid, I believe it is incorrect to think they all represent segments on the worldline of a single particle. Rather, if the inner region is taken as a black hole region, then the right-hand curved segment in the inner region does represent the worldline of the same particle shown in the outer region as it falls into the black hole (its proper time going in a reverse direction from the t coordinate), but the left-hand curved segment in the inner region is from a different particle that fell into the inner region from the "second outer region" mentioned in the quote above, the second universe seen on the left side of Kruskal-Szekeres diagram. Meanwhile if we wanted to continue the worldline of the particle in the outer region to earlier proper times then the proper time on that worldline when it crossed the horizon at t=-infinity in Schwarzschild coordinates, we'd have to draw the continuation in the diagram for the second "white hole" inner region.
kev said:
However when we look at the particle path in terms of advancing proper time, the particle rises to the event horizon in finite time. In the diagram the apogee is centred on coordinate time zero but there is no reason why the apogee could not be in the future and the particle rises up and out through the event horizon in present time. There is no demarcation between white hole and black hole in this diagram.
Not sure what you mean here. The curve showing coordinate position as a function of proper time is entirely separate from the curve showing coordinate position as a function of coordinate time, there's no implication that the inner region the particle was in at early proper times (before it first crossed the Schwarzschild radius in coordinate position) is the same as the inner region the particle was in at later proper times (after it crossed the Schwarzschild radius a second time). And if you're talking about the curves showing coordinate position as a function of coordinate time (which is what I was talking about before), there is no way that it could be true that "the particle rises up and out through the event horizon in present time", any particles that escape the event horizon of a Schwarzschild black hole must have done so at a coordinate time of -infinity in Schwarzschild coordinates.
 
Last edited by a moderator:
  • #130
97
3
Eddington-Finkelstein coordinates have imaginary time coordinates below the event horizon.
Not true.

The ratio of proper time to EF coordinate time [itex](d\tau/dt)[/itex] is [itex]\sqrt{(1-2GM/r)}[/itex] when dr=0 just as in Scharzchild coordinates.
Yes, but differentials of coordinates are not coordinates, they are components of differential intervals, and inside the event horizon an interval with dr = 0 is spacelike rather than timelike, which is to say, it is outside the causal light cone. It's an interval of a locally superluminal path. This is exactly similar to the case of flat Minkowski spacetime, where dtau^2 = dt^2 - dx^2. If we consider a locus with dt = 0, we find that (dtau/dx)^2 is negative, so the ratio of dtau to dx is imaginary, but it does not follow that the x coordinate in Minkowski spacetime is imaginary. It simply means that this is a spacelike interval, and we write ds^2 = dx^2 - dt^2.

There are many excellent text books from which you can learn about coordinate systems on differential manifolds.
 
  • #131
Yes, but differentials of coordinates are not coordinates, they are components of differential intervals, and inside the event horizon an interval with dr = 0 is spacelike rather than timelike, which is to say, it is outside the causal light cone.
Now think about that. How to measure the spacetime differentials between two events each outside the causal light cone of the other?

Think...think...no its gone.

It's an interval of a locally superluminal path. This is exactly similar to the case of flat Minkowski spacetime, where dtau^2 = dt^2 - dx^2. If we consider a locus with dt = 0, we find that (dtau/dx)^2 is negative, so the ratio of dtau to dx is imaginary, but it does not follow that the x coordinate in Minkowski spacetime is imaginary. It simply means that this is a spacelike interval, and we write ds^2 = dx^2 - dt^2.
So the coordinates are not imaginary (and who said that they were?) but the spacetime distance between them is imaginary. Well duh!

And what do you conclude from having imaginary spacetime intervals? No, don't tell me.
 
  • #132
JesseM
Science Advisor
8,496
15
Now think about that. How to measure the spacetime differentials between two events each outside the causal light cone of the other?
How about integrating the line element ds along a spacelike path between those events?
DiamondGeezer said:
So the coordinates are not imaginary (and who said that they were?) but the spacetime distance between them is imaginary. Well duh!

And what do you conclude from having imaginary spacetime intervals? No, don't tell me.
Quiz: if you integrate [tex]ds = \sqrt{dt^2 - (1/c^2)*(dx^2 + dy^2 + dz^2)}[/tex] along a spacelike path in flat minkowski spacetime (note that this is the line element which would give the proper time if you integrate along a timelike path), is the result real or imaginary?
 
  • #133
How about integrating the line element ds along a spacelike path between those events?
There are no spacelike paths inside the EH. Only timelike ones.

Quiz: if you integrate [tex]ds = \sqrt{dt^2 - (1/c^2)*(dx^2 + dy^2 + dz^2)}[/tex] along a spacelike path in flat minkowski spacetime (note that this is the line element which would give the proper time if you integrate along a timelike path), is the result real or imaginary?
But we're not in real spacetime below the event horizon. We're in imaginary spacetime.
 
  • #134
JesseM
Science Advisor
8,496
15
There are no spacelike paths inside the EH. Only timelike ones.
Where'd you get that idea? If you draw a line inside the horizon consisting of events that all have the same radial/angular coordinate but different t coordinates (all in Schwarzschild coordinates), then this is a spacelike path, as demonstrated by the fact that if you integrate ds^2 along this path it'll have a different sign then if you integrate ds^2 along a path of constant t coordinate inside the horizon (a timelike path). On the other hand, if you want to work in KS coordinates you can pick any path whose slope is closer to horizontal than a diagonal line (a path of constant time coordinate will work), and regardless of whether this path is inside or outside the horizon it'll be a spacelike path.
DiamondGeezer said:
But we're not in real spacetime below the event horizon. We're in imaginary spacetime.
Does "imaginary spacetime" refer to some property of how Schwarzschild coordinates are defined inside the horizon, or does it refer to some supposed difference in a coordinate-invariant quantity like proper time (or a coordinate-invariant statement about the nonexistence of spacelike paths in this region) which would still be true even if we switched to a different system like KS coordinates? If the latter, can you please address the questions I asked you about KS coordinates in posts #121 and #122?
 
Last edited:
  • #135
Where'd you get that idea? If you draw a line inside the horizon consisting of events that all have the same radial/angular coordinate but different t coordinates (all in Schwarzschild coordinates), then this is a spacelike path, as demonstrated by the fact that if you integrate ds^2 along this path it'll have a different sign then if you integrate ds^2 along a path of constant t coordinate inside the horizon (a timelike path). On the other hand, if you want to work in KS coordinates you can pick any path whose slope is closer to horizontal than a diagonal line (a path of constant time coordinate will work), and regardless of whether this path is inside or outside the horizon it'll be a spacelike path.
How do you draw a line inside the horizon? In order to do that, the line must be composed of real spacelike integrals of non-zero length. But there are no spacelike integrals at all.

Once again you've moved from the continuous integral outside the EH to a different integral "inside" the EH by reversing the sign, a move of mathematical dubiousness that I challenge.

For KS coordinates, the line integral bends through 90 degrees at the event horizon and heads in a timelike direction towards the supposed centre of the coordinate system which is no longer a point but is a line integral itself.

Does "imaginary spacetime" refer to some property of how Schwarzschild coordinates are defined inside the horizon, or does it refer to some supposed difference in a coordinate-invariant quantity like proper time (or a coordinate-invariant statement about the nonexistence of spacelike paths in this region) which would still be true even if we switched to a different system like KS coordinates? If the latter, can you please address the questions I asked you about KS coordinates in posts #121 and #122?
Yes it would be invariant under transformation. The coordinates U and V become imaginary when r<2M. The spacelike separations also become complex when r<2M because cosh(x) is always positive, while sinh(x)<0 when x<0 ie when r<2M.
 
Last edited:
  • #136
JesseM
Science Advisor
8,496
15
How do you draw a line inside the horizon? In order to do that, the line must be composed of real spacelike integrals of non-zero length.
A line is just an arbitrary parametrized path, it can be spacelike or timelike depending on what happens when you integrate ds^2 along this path using the metric. Do you deny that we can draw a path of constant radial and angular coordinate and varying t coordinate inside the horizon? Do you deny that if we use the usual definition of the metric inside the horizon, integrating ds^2 along this path shows it has the opposite sign from a line of constant t coordinate?
DiamondGeezer said:
Once again you've moved from the continuous integral outside the EH to a different integral "inside" the EH by reversing the sign, a move of mathematical dubiousness that I challenge.
Not sure how you distinguish between a "continuous integral" and a "differential integral". The metric is always defined in a differential way, giving ds^2 as a function of dR and dt and dtheta and dphi, no? In any case, if your critique is based on the fact that the metric is defined differently inside the horizon than outside when we use Schwarzschild coordinates, then let's switch to starting from KS coordinates, where the definition of the metric as as a function of the coordinates is the same inside and outside the horizon.
DiamondGeezer said:
For KS coordinates, the line integral bends through 90 degrees at the event horizon and heads in a timelike direction towards the supposed centre of the coordinate system which is no longer a point but is a line integral itself.
I don't know what you mean when you say the "line integral" bends. A line can bend, an integral is not a geometric object. Perhaps you mean that if we want the sign of ds^2 to remain the same both inside and outside the horizon, then that implies the line must bend 90 degrees at the horizon in KS coordinates? But this is wrong, in KS coordinates all line segments closer to vertical than 45 degrees have the same sign for ds^2 along them, while all line segments closer to horizontal than 45 degrees have the opposite sign for ds^2 when compared to more vertical lines. This is true even if you pick one vertical segment outside the horizon and another horizontal segment inside the horizon, they will have opposite signs for ds^2, so if the vertical segment is outside the horizon and timelike, the horizontal segment inside the horizon must be spacelike. An object with a timelike worldline will continue to remain closer to vertical both inside and outside the horizon, look for example at the worldlines A and F which are labeled as "timelike" geodesics in the right side of this diagram, which is from p. 835 of Gravitation by Misner, Thorne and Wheeler:

p835Gravitation.jpg


If you think that a horizontal path inside the horizon would be timelike despite the fact that a horizontal path outside the horizon would be spacelike, you're obviously disagreeing with the diagram above; please present either the equations that you use to justify this claim so that people can critique it, or else present a reference to some textbook or paper which presents this math, otherwise you're just making unjustified assertions.
DiamondGeezer said:
Yes it would be invariant under transformation. The coordinates U and V become imaginary when r<2M.
What does it mean for "coordinates" to become imaginary? Do you just mean that if we look at ds for an incremental variation in one of the coordinates (while the other coordinate is held constant), ds is imaginary? If so I'm sure your statement above is wrong, if you look at the equation for ds given by the metric in KS coordinates, you should find that incrementing dV while setting dU=0 gives the opposite result from incrementing dU while setting dV=0 (i.e. if one is real than the other is imaginary), regardless of whether you pick your original U,V to be inside the horizon or outside the horizon (so if you pick a U,V outside the horizon and increment dV while keeping dU=0, then you'll get the same type of ds as if you had picked a U,V inside the horizon and incremented dV while keeping dU=0). This is just like with the metric in 2D Minkowski coordinates, where if you calculate ds by incrementing dt while setting dx=0, you get the opposite result from if you had incremented dx while setting dt=0 (time is like imaginary distance while distance is like imaginary time). If you disagree, then again, please present the math behind your claim.
 
  • #137
A line is just an arbitrary parametrized path, it can be spacelike or timelike depending on what happens when you integrate ds^2 along this path using the metric. Do you deny that we can draw a path of constant radial and angular coordinate and varying t coordinate inside the horizon? Do you deny that if we use the usual definition of the metric inside the horizon, integrating ds^2 along this path shows it has the opposite sign from a line of constant t coordinate?
There's a fundamental difference between the line integral of a geodesic and an "arbitrary parametrized path". Perhaps you should stop and think before making such an obvious mistake.

The continuous line integral of a geodesic is always normal to the event horizon and there is no function which allows the geodesic to bend again to allow an infalling particle to reach the "singularity".

Its at this point that MTW gets waffley and produces a new function for the "interior".

Not sure how you distinguish between a "continuous integral" and a "differential integral". The metric is always defined in a differential way, giving ds^2 as a function of dR and dt and dtheta and dphi, no? In any case, if your critique is based on the fact that the metric is defined differently inside the horizon than outside when we use Schwarzschild coordinates, then let's switch to starting from KS coordinates, where the definition of the metric as as a function of the coordinates is the same inside and outside the horizon.
Whatever the metric, the line integral of a geodesic is invariant and is normal to the event horizon. And "inside" the EH, there are no spacelike paths.

Its remarkable that you change coordinate systems expecting to get a different result for the same object.

Why, for example, are there so many different coordinate systems for the same object? And yet, the same object produces radically different results depending on which coordinate system is used?

I don't know what you mean when you say the "line integral" bends. A line can bend, an integral is not a geometric object. Perhaps you mean that if we want the sign of ds^2 to remain the same both inside and outside the horizon, then that implies the line must bend 90 degrees at the horizon in KS coordinates?
No. It means that the geodesic must be normal to the event horizon, regardless of the coordinate system. The line integral of a geodesic is invariant under transformation.

But this is wrong, in KS coordinates all line segments closer to vertical than 45 degrees have the same sign for ds^2 along them, while all line segments closer to horizontal than 45 degrees have the opposite sign for ds^2 when compared to more vertical lines. This is true even if you pick one vertical segment outside the horizon and another horizontal segment inside the horizon, they will have opposite signs for ds^2, so if the vertical segment is outside the horizon and timelike, the horizontal segment inside the horizon must be spacelike. An object with a timelike worldline will continue to remain closer to vertical both inside and outside the horizon, look for example at the worldlines A and F which are labeled as "timelike" geodesics in the right side of this diagram, which is from p. 835 of Gravitation by Misner, Thorne and Wheeler:

p835Gravitation.jpg
This depends on whether the worldlines shown are real or imaginary and whether MTW shuffles the imaginary line into a real one and ignores the "i".

If you think that a horizontal path inside the horizon would be timelike despite the fact that a horizontal path outside the horizon would be spacelike, you're obviously disagreeing with the diagram above; please present either the equations that you use to justify this claim so that people can critique it, or else present a reference to some textbook or paper which presents this math, otherwise you're just making unjustified assertions.
I'm making unjustified assertions? Really? What about this next bit?

What does it mean for "coordinates" to become imaginary? Do you just mean that if we look at ds for an incremental variation in one of the coordinates (while the other coordinate is held constant), ds is imaginary?
Erm no. I've lost track of the number of time YOU have made this straw man argument. I have never made it, so why am I being lambasted to prove a statement I never made? I have never made such a stupid statement as to claim that "coordinates become imaginary".

If so I'm sure your statement above is wrong, if you look at the equation for ds given by the metric in KS coordinates, you should find that incrementing dV while setting dU=0 gives the opposite result from incrementing dU while setting dV=0 (i.e. if one is real than the other is imaginary), regardless of whether you pick your original U,V to be inside the horizon or outside the horizon (so if you pick a U,V outside the horizon and increment dV while keeping dU=0, then you'll get the same type of ds as if you had picked a U,V inside the horizon and incremented dV while keeping dU=0).
In other words I was correct to state that the KS coordinates produce an imaginary spacelike intervals inside of the EH.

This is just like with the metric in 2D Minkowski coordinates, where if you calculate ds by incrementing dt while setting dx=0, you get the opposite result from if you had incremented dx while setting dt=0 (time is like imaginary distance while distance is like imaginary time). If you disagree, then again, please present the math behind your claim.
No it isn't. In the 2D Minkowski coordinates events are separated by real timelike or real spacelike separations. I can accept that in flat spacetime one is the imaginary counterpart of the other.

But in the case of the interior of a black hole, spacelike separations are never real. Which means that a geodesic can never reach the singularity. To reach the singularity, a geodesic must have real spacelike length, but there ain't any space unless you produce an interior solution. KS produces the same result inside the EH, where spacelike separations along geodesics are imaginary and not real.
 
  • #138
JesseM
Science Advisor
8,496
15
There's a fundamental difference between the line integral of a geodesic and an "arbitrary parametrized path". Perhaps you should stop and think before making such an obvious mistake.

The continuous line integral of a geodesic is always normal to the event horizon and there is no function which allows the geodesic to bend again to allow an infalling particle to reach the "singularity".
You're either confused or expressing yourself in a garbled way--a "line integral" is an integral, not a geometric object like a geodesic, so it's meaningless to say the line integral is "normal to the event horizon", only the geodesic itself can have geometric properties like being "normal" to anything. The line integral along a geodesic is an integral which gives the "length" of that geodesic in spacetime (and this length is locally maximized for a geodesic path), but it's conceptually distinct from the geodesic itself. And you can compute the line integral along any arbitrary parametrized path, it doesn't have to be a geodesic one (for example, you can calculate the proper time along the worldline of a rocket which is moving on a non-geodesic path as it fires its engines).
Diamond Geezer said:
Its at this point that MTW gets waffley and produces a new function for the "interior".
Yes, in Schwarzschild coordinates the function is different outside vs. inside, but in both cases the function is a differential one giving ds in terms of dR and dt and so forth, you haven't explained your comment that the outside is a "continuous integral" rather than a "differential integral".

And in KS coordinates, the function is exactly the same inside and outside. And your claims that things work differently inside the horizon than outside in KS coordinates are complete nonsense, which you never present any math to back up.
Diamond Geezer said:
Whatever the metric, the line integral of a geodesic is invariant and is normal to the event horizon. And "inside" the EH, there are no spacelike paths.
Do you understand that determining whether an arbitrary path is "spacelike" or "timelike" just means doing a line integral of the infinitesimal line element ds^2 (given by the metric) along that path, and seeing whether the integral is positive or negative? You can certainly find paths inside the horizon such that the integral of ds^2 along them is positive, and other paths such that the integral of ds^2 along them is negative.
Diamond Geezer said:
Why, for example, are there so many different coordinate systems for the same object?
General relativity is diffeomorphism-invariant, which means you can use an infinite number of different possible coordinate systems in the same physical spacetime, and the Einstein field equations will work in all of them. Read http://www.aei.mpg.de/einsteinOnline/en/spotlights/background_independence/index.html [Broken] for example.
Diffeomorphism invariance said:
And yet, the same object produces radically different results depending on which coordinate system is used?
Nonsense, if you take a given path expressed in the coordinates of one coordinate system, and apply the appropriate coordinate transformation to find the coordinates of the same path in a different coordinate system, and use the equations for the metric in each coordinate system to integrate the line element ds^2 along the coordinates of the path in each system, you will get the exact same answer for the integral both times. If this wasn't true, then there wouldn't be a coordinate-invariant answer for the proper time between two events along a given worldline.
JesseM said:
I don't know what you mean when you say the "line integral" bends. A line can bend, an integral is not a geometric object. Perhaps you mean that if we want the sign of ds^2 to remain the same both inside and outside the horizon, then that implies the line must bend 90 degrees at the horizon in KS coordinates?
Diamond Geezer said:
No. It means that the geodesic must be normal to the event horizon, regardless of the coordinate system. The line integral of a geodesic is invariant under transformation.
It's certainly not true that the geodesic is always normal to the event horizon in KS coordinates, another totally unsupported statement from you. I of course agree that the line integral of a geodesic is coordinate-invariant (that's exactly what I was just saying in the previous paragraph about transforming the coordinates of a path and doing the line integral in each coordinate system for the path expressed in those coordinates) but I'm not sure what you think the relevance of this is to all your other bizarre statements. Can you explain what you meant when you said that in KS coordinates, "the line integral bends through 90 degrees at the event horizon and heads in a timelike direction towards the supposed centre of the coordinate system which is no longer a point but is a line integral itself"?
DiamondGeezer said:
This depends on whether the worldlines shown are real or imaginary and whether MTW shuffles the imaginary line into a real one and ignores the "i".
Calling worldlines "real" or "imaginary" seems to be your own made-up terminology, can you explain what you mean? Certainly the line integral of ds^2 along an arbitrary path can be positive or negative, which means the integral of ds can be real or imaginary. But that's just what physicists mean when they distinguish between "timelike" and "spacelike" paths, do you not understand this?
JesseM said:
If you think that a horizontal path inside the horizon would be timelike despite the fact that a horizontal path outside the horizon would be spacelike, you're obviously disagreeing with the diagram above; please present either the equations that you use to justify this claim so that people can critique it, or else present a reference to some textbook or paper which presents this math, otherwise you're just making unjustified assertions.
DiamondGeezer said:
I'm making unjustified assertions? Really? What about this next bit?
Please don't engage in the childish "but you did it too!" game (known formally as the tu quoque fallacy). Even if I had made some unjustified assertions, that wouldn't be a good reason for you to be evasive about providing the slightest bit of mathematical argument to justify your own claims. Again, are you claiming anything unusual happens to a timelike geodesic in KS coordinates when it crosses the event horizon? Do you disagree that a line which appears vertical in KS coordinates, both inside and outside the horizon, represents a valid timelike worldline (though probably not a geodesic) where the line integral of ds^2 between any two points on this line (whether both points are outside the horizon, both are inside, or one is inside and one is outside) will always have the same sign? Do you disagree that physicists define "timelike" vs. "spacelike" path in terms of the sign of the integral of ds^2 along that path (in arbitrary spacetimes including Minkowski spacetime)? If your answer to any of these questions is "yes", then please provide either a reference to a trustworthy source or your own mathematical equations, otherwise you're just making unjustified assertions which conflict with what is believed by mainstream physicists.
JesseM said:
What does it mean for "coordinates" to become imaginary? Do you just mean that if we look at ds for an incremental variation in one of the coordinates (while the other coordinate is held constant), ds is imaginary?
DiamondGeezer said:
Erm no. I've lost track of the number of time YOU have made this straw man argument. I have never made it, so why am I being lambasted to prove a statement I never made? I have never made such a stupid statement as to claim that "coordinates become imaginary".
Um, except you did so in the very statement I was responding to above, when you said "The coordinates U and V become imaginary when r<2M." If you misspoke then please clarify, but don't go lambasting me for reading what you actually write.
JesseM said:
If so I'm sure your statement above is wrong, if you look at the equation for ds given by the metric in KS coordinates, you should find that incrementing dV while setting dU=0 gives the opposite result from incrementing dU while setting dV=0 (i.e. if one is real than the other is imaginary), regardless of whether you pick your original U,V to be inside the horizon or outside the horizon (so if you pick a U,V outside the horizon and increment dV while keeping dU=0, then you'll get the same type of ds as if you had picked a U,V inside the horizon and incremented dV while keeping dU=0).
DiamondGeezer said:
In other words I was correct to state that the KS coordinates produce an imaginary spacelike intervals inside of the EH.
Perhaps you misunderstood me, what I was saying was that KS coordinates produce both timelike and spacelike intervals, depending on whether you increment the timelike coordinate dV while holding dU=0 (in which case you get a timelike interval) or if you increment the spacelike coordinate dU while holding dV=0 (in which case you get a spacelike interval). This is true regardless of whether you pick your points U,V to be inside the horizon or outside it--if you calculate the interval between (U,V) and (U+dU,V) then it will be spacelike if U,V is a point inside the horizon, and it will be spacelike if U,V is a point outside the horizon. Likewise, if you calculate the interval between (U,V) and (U, V+dV) it will be timelike if U,V is a point inside the horizon, and it will be timelike if U,V is a point outside the horizon. Nothing changes when you cross the horizon. Do you disagree?
DiamondGeezer said:
No it isn't. In the 2D Minkowski coordinates events are separated by real timelike or real spacelike separations. I can accept that in flat spacetime one is the imaginary counterpart of the other.
What does "real spacelike" and "real timelike" even mean? I suppose we can say that the metric in an inertial coordinate system in Minkowski spacetime can be written as one of two possible line elements (assuming units where c=1):

1. [tex]ds^2 = dt^2 - dx^2 - dy^2 - dz^2[/tex]

2. [tex]ds^2 = -dt^2 + dx^2 + dy^2 + dz^2[/tex]

If we choose option #1, then the integral of ds^2 along a timelike path will be positive, while the integral of ds^2 along a spacelike path will be negative, so the integral of ds along a spacelike path is imaginary. Under option 1, how can the separation between spacelike-separated points possibly be "real"? By definition, if you choose this form for the line element, all spacelike-separated points have an imaginary separation. Of course, you could also choose option #2, in which case now spacelike-separated points would have a real separation while timelike-separated ones would have an imaginary separation. But if you stick to one or the other, which physicists normally do, then one type of separation will be treated as real while the other will be treated as imaginary.

It's exactly the same with the line element for a particular coordinate system in curved spacetime, like the spacetime around a black hole; physicists normally adopt a single convention for the equation for the line element, even though they could choose a different convention (by multiplying the right side of the first equation's convention by -1) which would reverse which types of paths had a real value and which had an imaginary value. For example, when dealing with the line element in Schwarzschild coordinates for the exterior region, there would be two possible conventions:

1. [tex]ds^2 = (1 - \frac{r_s}{r})dt^2 - \frac{dr^2}{1 - \frac{r_s}{r}} - r^2 (d\theta^2 + sin^2 \theta d\phi^2 )[/tex]

2. [tex]ds^2 = -(1 - \frac{r_s}{r})dt^2 + \frac{dr^2}{1 - \frac{r_s}{r}} + r^2 (d\theta^2 + sin^2 \theta d\phi^2 )[/tex]

If you adopt the first convention, then all timelike paths outside the horizon will have real ds while all spacelike paths outside the horizon will have imaginary ds, while if you adopt the second convention, the reverse will be true.
DiamondGeezer said:
But in the case of the interior of a black hole, spacelike separations are never real.
Again, I have no idea why you think any statement like this applies to "the interior" but not equally to the exterior. If you adopt a single convention for the line element in a given coordinate system like KS coordinates, then if that convention treats timelike paths as having a real ds, then automatically it will treat spacelike paths as having imaginary ds, this will be just as true outside the horizon as inside.
DiamondGeezer said:
Which means that a geodesic can never reach the singularity. To reach the singularity, a geodesic must have real spacelike length
What are you talking about? Only a tachyon could have a spacelike path. A slower-than-light particle has a timelike path both outside and inside the horizon, and it has no problem reaching the singularity as shown by paths A and F in that KS diagram (where the singularity is the jagged line).
 
Last edited by a moderator:
  • #139
I'm not going to stretch this thread to infinity.

I'd like to pick up one comment:

General relativity is diffeomorphism-invariant, which means you can use an infinite number of different possible coordinate systems in the same physical spacetime, and the Einstein field equations will work in all of them. Read this article for example.
No, I'll take that as read. The question remains as to why so many coordinate systems have been constructed.

And yet, the same object produces radically different results depending on which coordinate system is used?

Nonsense, if you take a given path expressed in the coordinates of one coordinate system, and apply the appropriate coordinate transformation to find the coordinates of the same path in a different coordinate system, and use the equations for the metric in each coordinate system to integrate the line element ds^2 along the coordinates of the path in each system, you will get the exact same answer for the integral both times. If this wasn't true, then there wouldn't be a coordinate-invariant answer for the proper time between two events along a given worldline.
Unfortunately black holes are not coordinate invariant. They are a product of a bad choice in the construction of the coordinate systems.

It is possible to construct a valid coordinate system where black holes are not the result of gravitational collapse, which produce the same answers for line integral calculations of arbitrarily parametrized path.
 
  • #140
atyy
Science Advisor
14,207
2,466
No, I'll take that as read. The question remains as to why so many coordinate systems have been constructed.
Generically, curved manifolds must be coordinatized with several coordinate patches. The maximally extended Schwarzschild solution given in Kruskal-Szekeres coordinates is an unusual case where a curved spacetime can be covered by a single coordinate patch.

Unfortunately black holes are not coordinate invariant. They are a product of a bad choice in the construction of the coordinate systems.

It is possible to construct a valid coordinate system where black holes are not the result of gravitational collapse, which produce the same answers for line integral calculations of arbitrarily parametrized path.
No.
 
  • #143
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,412
1,051
This thread is done. I have kept it open in the interests of pedagogy, but it now is clear (and has been for some time) that nothing is being accomplished.
 

Related Threads on Crossing the Event Horizon of a Black Hole

Replies
99
Views
20K
Replies
28
Views
4K
Replies
24
Views
2K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
11
Views
6K
Replies
6
Views
2K
Replies
12
Views
2K
Replies
3
Views
916
Top