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A Event horizon vicinity in isotropic coordinates

  1. Apr 6, 2016 #1

    Jonathan Scott

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    The Schwarzschild radial coordinate ##r## is defined in such a way that the proper circumference of a sphere at radial coordinate ##r## is ##2\pi r##. This simplifies some maths but creates some rather odd side-effects, so to get a more physical picture I like to use isotropic coordinates instead, where the spatial metric scale factor is the same in all directions. The isotropic radial coordinate ##R## is related to the Schwarzschild radial coordinate ##r## as follows:

    $$ r = R (1 + Gm/2Rc^2)^2 $$
    or, turning this round (and choosing one of many possible forms)
    $$ R = \frac{r - Gm/c^2 + r \sqrt{1 - 2Gm/rc^2}}{2} $$

    In isotropic coordinates, unlike in Schwarzschild coordinates, the spatial part of the Schwarzschild metric has a factor of ##(1+Gm/2Rc^2)^2## in all directions and appears to be perfectly well-behaved when approaching the event horizon, where ##R = Gm/2c^2## and proper distance at the event horizon is simply 4 times coordinate distance in any direction.

    If we calculate the proper circumference of a sphere via ##r## as it shrinks towards the event horizon, we find that as we get near the event horizon, it stops decreasing (so if we imagine it made of a ring of rulers, the rulers are now shrinking proportionally to the coordinate radius). What is more, if we pass the event horizon in isotropic coordinates the proper circumference starts increasing again, as the rulers are now shrinking more rapidly than proportionally to the radius, which means that the Schwarzschild radial coordinate defined in terms of that circumference would actually now be increasing again!

    As can be seen from the equations relating ##r## and ##R##, when ##0 < r < 2Gm/c^2## the expression for ##R## would involve the square root of a negative number, so it could not have a real value. However, the isotropic expression for the space factor in the metric goes past the event horizon without anything special happening (although of course the time factor goes to zero there).

    This presumably means that this metric (at least with real coordinates) can only be a valid solution of the Einstein field equations in one of the two coordinate systems and not both. Can anyone point me to the specific way in which one of these coordinate systems does not give a valid solution inside the event horizon? The fact that the isotropic solution seems to behave nicely at the event horizon and implies that the Schwarzschild radial coordinate starts to increase again seems to make sense physically, but is clearly incompatible with the usual assumptions about black holes. Is it simply that no isotropic coordinate system is possible inside the event horizon, and if so, why?
     
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  3. Apr 6, 2016 #2

    PeterDonis

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    No, it just means that these isotropic coordinates cover a different region of the maximally extended spacetime than Schwarzschild coordinates do. If you look at a Kruskal diagram of Schwarzschild spacetime, there are four regions: region I, the "right wedge", the usual exterior region; region II, the "top wedge", the black hole region; region III, the "left wedge", the other exterior region; and region IV, the "bottom wedge", the white hole region.

    The coordinate patches ##2M < r < \infty## in Schwarzschild coordinates and ##M / 2 < R < \infty## in isotropic coordinates both cover region I. But the coordinate patch ##0 < r < 2M## in Schwarzschild coordinates covers region II, while the coordinate patch ##0 < R < M / 2## in isotropic coordinates covers region III. Region III is an exterior region, containing 2-spheres with areal radius larger than ##2M##, so there is no issue with isotropic coordinates covering it.

    The only other technical point here is that, as you note, the spatial part of isotropic coordinates is well-behaved at ##R = M / 2##, i.e., at the event horizon. On a Kruskal diagram, this value of ##R## corresponds to the point at the center of the diagram, where the two horizon lines cross. However, all of the lines of constant time coordinate ##t## also cross there; so isotropic coordinates are in fact not valid at ##R = M / 2##, because there is only a single point instead of an infinite line corresponding to that value of ##R##. (Another way of seeing this is to note that, as you say, the metric coefficient ##g_{00}## is zero at ##R = M / 2##, which means that the metric has zero determinant and therefore has no inverse.)

    This particular transformation from Schwarzschild to isotropic coordinates makes the spatial part of the metric isotropic; but "spatial part" in itself implies a particular choice of spacelike hypersurfaces as surfaces of constant coordinate time, in this case the surfaces of constant ##t## in region I. But those surfaces do not exist inside the horizon (i.e., in region II, or region IV either); they only exist outside it (in region I and region III). More precisely, hypersurfaces of constant Schwarzschild coordinate ##t## exist in region II (and region IV), but they are not spacelike--so the transformation from Schwarzschild to isotropic coordinates cannot be done in those regions since the resulting line element manifestly requires surfaces of constant ##t## to be spacelike.

    OTOH, it is certainly possible to have other coordinates inside the horizon with the spatial part of the metric being isotropic: Painleve coordinates are the obvious example. But the surfaces of constant Painleve coordinate time are different from the surfaces of constant Schwarzschild coordinate time.
     
  4. Apr 6, 2016 #3

    Jonathan Scott

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    Thanks; I think I understand that, at least mathematically speaking. However, it gives me the distinctly uneasy impression that if one falls through the event horizon in isotropic coordinates, one lands up somewhere different from where one would land up falling through it in Schwarzschild coordinates! :smile:
     
  5. Apr 6, 2016 #4

    PAllen

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    I've never seen coordinates outside my window, only visualized them in my head. I don't expect this have any impact on nature. To answer what happens to an infaller you must construct timelike world line crossing the horizon (geodesic if you want free fall). Then, parts of this line are simply outside the coordinate patches covered by some coordinate systems. It is as simple as a great circle through the poles has two points missing coordinate values in Mercator coordinates.
     
  6. Apr 6, 2016 #5

    PeterDonis

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    You don't fall through the horizon "in" any coordinates. Your worldline through the horizon is an invariant, independent of coordinates.

    However, it is true that isotropic coordinates cannot describe your worldline at or below the horizon; your worldline simply goes off the edge of the chart, just as it does in exterior Schwarzschild coordinates. (Note that, although Schwarzschild coordinates do exist inside the horizon, they are a separate coordinate patch, discontinuous with the Schwarzschild coordinate patch outside the horizon. So there is really no difference between Schwarzschild and isotropic coordinates here.)

    If you look at curves which go from ##R = 0## to ##R = \infty## in isotropic coordinates, you will find that they are all spacelike curves; there are no timelike curves that are covered by the chart that span that entire coordinate range. There are timelike curves that span ##R = \infty## to ##R \rightarrow M / 2##, and there are timelike curves that span ##R = 0## to ##R \rightarrow M / 2##; but these are separate families of curves and are disconnected from each other. Any curve in this chart that passes through ##R = M / 2## must be spacelike. (Looking at the regions covered by isotropic coordinates on a Kruskal diagram makes this obvious.) (Also, strictly speaking, isotropic coordinates in the two regions are two separate patches, since the metric is not invertible at ##R = M / 2##. But we can still restrict ourselves to a single spacelike hypersurface and look at curves going from ##R = 0## to ##R = \infty## in that hypersurface.)
     
  7. Jan 24, 2017 #6
    I think it is an artefact : Kruskal diagram is build from the classical r variable (peripheral radius) and when you obtain r > 2M in this diagram, you automatically refer to the exterior... that is wrong since r > 2M also applies to the "interior" (as Jonathan Scott has judiciously noted).

    If one considers a spacial plane containing the origine, and builds an immersion of it as a curved surface into ℝ3, in such a way that the metric induced by euclidian metric in ℝ3 on the obtained surface is the Schwarzschild one ; then this surface is a paraboloid symmetric with respect to the singularity (with r > 2M on both side).

    The main problem is that, without remaking the reasoning carefully, some physicists of reputation have reproduced the bad hypothesis for r < 2M in some of their works, which now are used as reference ... then other physicists are led to trust that without more checking…
     
  8. Jan 24, 2017 #7

    PeterDonis

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    That's not quite what he said, and it is incorrect. He said (correctly) that the proper circumference (which is just ##2 \pi## times the ##r## you are using here) reaches a minimum at ##R = M / 2## in isotropic coordinates (where the ##R## here is the radial coordinate, not the "peripheral radius"), and then increases again as ##R## gets smaller than ##M / 2##. What he (incorrectly) supposed was that the coordinate region ##R < M/2## in isotropic coordinates referred to the interior (the area below the horizon). It doesn't; it refers to a second exterior region (the left wedge of the Kruskal diagram).

    You are misdescribing the Flamm paraboloid construction. It is only valid for ##r > 2M## in the first place, and in terms of ##r## there is only one "side", the "side" where ##r > 2M##. The "circular" aspect of the paraboloid is due to the angular coordinates (actually only one of them, ##\phi##, since the paraboloid is usually drawn with one angular coordinate suppressed by working in the "equatorial plane" ##\theta = \pi / 2##); it is not because there are "two sides" both having ##r > 2M##. (Btw, this property is not unique to this case; it is a basic fact about spherical coordinates and is just as true in ordinary Euclidean geometry.)

    No, the main problem is that you do not understand how the actual geometry is represented in the coordinates.
     
  9. Jan 25, 2017 #8

    stevendaryl

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    That's something that is strange and maybe misleading about Schwarzschild coordinates. Since they have the same form on both sides of the horizon, you're tempted to identify the coordinates on the two sides. People say things like: when you cross the horizon, your value for "r" switches from being spacelike to being timelike. But that is misleading if you think of it as two coordinate patches--there is no unique way to identify coordinates in different patches.
     
  10. Jan 25, 2017 #9

    PeterDonis

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    That's not quite the way I see it put; a "value" for a coordinate can't be spacelike or timelike. The way I see it put is that the coordinate ##r## (not its value at a particular event) becomes timelike (which is true, but coordinate-dependent--there are other charts in which ##r##, though still defined as the areal radius, remains spacelike inside the horizon), or that "space and time switch roles" inside the horizon (which is misleading, and IMO "not even wrong").

    Yes, there is. The ##r## coordinate is the areal radius of 2-spheres in both cases; that is an invariant which is present in both patches. The angular coordinates can be matched up by using them to label radially infalling geodesics that appear in both patches. And the ##t## coordinate can be defined by labeling hypersurfaces orthogonal to the integral curves of the fourth Killing vector field, which appears in both patches (it's timelike in one and spacelike in the other).

    The reason statements like "space and time switch roles inside the horizon" is misleading is that the wrong definition of "space and time" is being used. In terms of proper time along an infalling worldline, and spacelike hypersurfaces locally orthogonal to that worldline, there is no "switching of roles" at all.
     
  11. Jan 25, 2017 #10

    stevendaryl

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    Thanks. I stand corrected.
     
  12. Jan 25, 2017 #11
    Thank you for all the comments.
    The problem with the Kruskal diagram is that it is not bijective at the level of the singularity: the whole diagonals u = ± v corresponds to only two points r = 2M with t = ±∞. It is a little like angular coordinates on the surface of the Earth: if you represent θ and φ as orthonormal cartesian coordinates, the whole segment θ = 0 with 0 ≤ φ ≤ 2π corresponds to only one point at the north pole. A by side consequence is there seems to exist two "external" disjoined zones in the Kruskal diagram.
    Another interpretation is therefore that r < 2M is not physical (in the vacuum surrounding the star; it may exists inside the star, but with another expression of the metric); if the region I at the right describes interior, then region III at the left describes interior (with r ≥ 2M, just as said JonathanScott). Because of the symmetry, in this region (or at less the part of it that is vacuum) the gravitation is repulsive (attractive towards the singularity).
    I did not want to say that your interpretation is wrong but only I wanted to tell to JonathanScott that there are physicist thinking as him that there exist another possible interpretation.
    In any way this discussion cannot have full efficiency since in the vicinity of the singularity the quantity A(r)'/A(r) tends to infinite (where A(r) is the coefficient of dt2 in the metric). Now this quantity has a role rather analogous to a gravitational field in the newtonian theory, therefore near the singularity necessarily exists an important quantic vacuum polarization: the crossing of the singularity cannot be precisely studied without a quantum theory.
     
  13. Jan 25, 2017 #12

    PeterDonis

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    First, you are talking about the horizon, not the singularity. Second, you have things backwards: the lack of bijectiivity you describe is an issue with Schwarzschild coordinates, not Kruskal coordinates. The way to see this is to look at invariants; for example, looking at infalling geodesics that pass through the horizon. These show that the horizon ##r = 2M## is not just two points as you claim; it is in fact two intersecting lines (or, putting back the angular coordinates, two intersecting infinite families of 2-spheres).

    This is incorrect as well.

    The region I at the right is an exterior region, not an interior region. So is region III at the left. They both have ##r > 2M##. The interior regions are II, at the top, and IV, at the bottom. They both have ##r < 2M##.

    I have no idea what you mean by this; gravity everywhere in this geometry is "attractive towards the singularity", but that does not mean "repulsive" at all.

    What you are describing is not "another possible interpretation"; it is a misunderstanding.

    Once again, ##r = 2M## is the horizon, not the singularity. The quantity ##A'(r) / A(r)## is coordinate dependent; it is easy to find charts in which there is no such "infinite" quantity at the horizon. (Kruskal coordinates are one such chart.) The spacetime geometry at the horizon is perfectly smooth and nonsingular.
     
  14. Jan 26, 2017 #13
    You are right ! About this subject, I would have check my language more carefully (what you call the singularity is in the center).

    I think this is wrong. In ... relativity... the change from some coordinates to others changes the way to do the calculus, but the physical result is the same. The unique reason to change is to make the calculus easier. If it is simple with t and r, I can use these coordinates. If two maps are bijective with respect to space-time, they must be bijective between them (in their common domain if it is not the whole).

    Yes, but these lines are exactly what is a problem. It is just a graphic artefact.
    It is analogous to polar coordinates in ℝ2 when you represent them graphically as cartesian coordinates: the distance r to origin horizontally and the angle θ vertically. The origin is only one point in the plane but is represented by the whole segment r = 0 with 0 ≤ θ ≤ 2π. For a curve passing by the origin, the point of this segment gives only information about the way it passes here, but there are not an infinity of origines (this is only a graphic artefact).
    If all the points of this graphic segment are "equivalent" of the same origine, this is associated to the fact that θ is not defined at the origine, and to the fact that the length dl = r dθ = 0 in this case.
    In an analogous way, on the horizon r = 2M the time variable cannot be defined since dtloc = √A(r) dt = 0 in this case (the time does not "flow" here, so you can use use any value for t, this changes nothing). In the Kruskal diagram, for a trajectory coming from region I and "crossing" the horizon in the side of the origin, at a point corresponding to t = +∞, there is an apparent discontinuity and the continuation is in the region III, at a point corresponding to t = -∞. In the interior part, the mobile seems to come from the infinite past, just as it has seemed to disappear in the infinite future.
    Such an discontinuity is only apparent since time is not defined on the horizon. It is just analogous to the apparent π discontinuity of θ for a mobile crossing the origin in ℝ2.

    When the Schwarzschild metric has been studied, all the physicists have considered that the choice of the coordinates is "free". The main part of them choose the peripheral radius r as radial coordinate, but nobody imagine that in a space very curved, there is no reason for r to be an increasing function of the distance from center.
    When the Kruskal diagram has been studied, they (except a few ones) search hardly how to introduce r < 2M since they thought that this is needed to represent the interior. A mathematical extension of Kruskal coordinates, corresponding to regions II and IV, has been found, but it is only an mathematical artefact: interior does correspond to the region III.

    With the interpretation I try to explain, the symmetry of the regions I and III implies a symmetry of the gravitational effects (symmetry for A'/2A with the same values for symmetric points with same r > 2M).

    However, this is only a consequence in the vacuum part of the interior. It must be studied separately inside the star which is in the center.

    When writing that, you knew probably already what I was going to answer: if there is only one correct interpretation, it is mine and it is yours that is a misunderstanding. The fact that great physicists give it their approval (among them Penrose...) is a hard social argument, but it is not a valuable scientific argument.

    You are partly right: apart from the fact that I forgot a factor two, it is ambiguously written. I obviously would have mean [dA/dl]/A where dl = ±√B(r) dr (and B = 1/A) is the length of the radial move (the ± is here needed since r is not everywhere an increasing function of the position.

    The static space has no singularity at the horizon, but A′(r)/2A(r) = ∞ where A(r) = 0 is a gravitational "singularity".

    even if I am in clearly in disagreement with you, I thank you for your comments
     
  15. Jan 26, 2017 #14

    PeterDonis

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    Not if one set of coordinates is singular. Schwarzschild coordinates are singular at the horizon. So you can't calculate anything physical from them at ##r = 2M##.

    No, it isn't. The actual, physical horizon in the maximally extended Schwarzschild geometry is two infinite families of 2-spheres, which intersect at one of them (the one designated by the center point on the Kruskal diagram). As I said, this can be seen by looking at invariants, such as families of ingoing radial geodesics. Different ingoing radial geodesics on the same radial line cross the horizon at different events. If your claim were true, they would all have to cross at the same event--i.e., if you dropped a rock radially towards the horizon, and then dropped another one directly above it a minute later, the two rocks would have to collide at the horizon. They don't. So your claim is false.

    I don't have time right now to respond to the errors in the rest of your post, but you should think very, very carefully before posting again. This is an "A" level thread, which means graduate level knowledge of the subject matter is assumed. You obviously don't have that. Even explaining to you why your claims are false is really off topic in an "A" level thread; if you really want to pursue this discussion, it should go in a separate thread at a lower level.
     
    Last edited: Jan 26, 2017
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