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Confused by DeSitter again!
On Thu, 7 Aug 2003, Serenus Zeitblom wrote:
I think it is pretty well-known by now that DeSitter space can be sliced in lots of different ways, eg by flat slices or by spherical ones. Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3 respectively are all present and valuable for various purposes. What I don't understand is this: why does anyone take the flat slicing seriously? I mean, it is obviously going to lead to a geodesically incomplete spacetime---you can have timelike worldlines coming in from "beyond the universe"! That is pretty ridiculous It would indeed be absurd to say that by changing to a different coordinate chart on M, you can alter an intrinsic property (geodesic completeness). Of course the answer is that you -can't- do this; you must be somehow confusing the notion of a coordinate chart on M (which need not cover all of M) with M itself. In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic properties include its topology (M is homeomorphic to the topological manifold RxS^3), the geodesic completeness property, and its conformal structure, which can be diagrammed ______ |\ /| | \ / | | \/ | | /\ | | / \ | |/____\| (This crude ASCII sketch is a stand-in for a similar diagram in which the diagonals have slope +/-1.) For details concerning what I just said, see author = {S. W. Hawking and G. F. R. Ellis}, title = {The Large Scale Structure of Space-Time}, publisher = {Cambridge University Press}, year = 1973} In this book you will also find details of several charts; as an exercise you can find transformations from any of those to any in the following list which are not discussed in HE. 1. Comoving with an irrotational timelike geodesic congruence (integral curves of a vector field X) everywhere orthogonal to a (contracting then re-expanding) family of hyperslices with S^3 geometry: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ], -infty t infty, 0 r pi/2, -pi z,u pi This chart covers almost all of M. Note that the coordinate lines z = z0, r = r0, u = u0 correspond to the world lines of a certain family of inertial observers, i.e. a certain family of timelike geodesics on M. Because the geodesics "fill up an open set without intersection"-- or better yet, because they are the integral curves of a vector field X on M-- we call it a "congruence". Because this congruence of timelike geodesics is irrotational (i.e. the vector field X has vanishing vorticity tensor), it defines an orthogonal family of spatial hyperslices t = t0. (See Hawking & Ellis for congruences and vorticity.) From the form of the line element it is obvious that these hyperslices are three-spheres of "radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a minimal size reached at t = t0, and then expand again. It is very important to understand that this chart actually corresponds to a whole family of different congruences. To understand this, consider the usual embedding as a hyperboloid of one sheet of the two dimensional de Sitter space H^(1,1). Here the "latitude circles" correspond to the slices t = t0, with the "equatorial circle" corresponding to the "minimal size" circle t = 0. Note that as t increases, the latitude circles shrink to radius a and then re-expand. (This is obviously very closely analogous to what we saw for our irrotational congruence of inertial observers orthogonal to S^3 hyperslices above.) But now imagine applying a boost in the embedding space E^(1,3) to our H^(1,1), moving the original latitude circles to a new family of "tilted" circles on H^(1,1). The same boost moves all but two of the world lines of our observers to new hyperbolas, so we have a new irrotational congruence of timelike geodesics with a new family of orthogonal slices (circles). Similarly on H^(1,3). If this seems puzzling, it might help to note that one can define on E^3 not just one "cylindrical coordinate chart" ds^2 = dz^2 + dr^2 + r^2 du^2, -infty z infty, 0 r infty, -pi u pi but a whole -family- of similar charts. To obtain the others: rotate the omitted half plane u = pi, or less trivially, translate/rotate the axis of symmetry r = 0 to a new line. Similarly, by choosing various pairs of antipodal points on S^2 (embedded in E^3 in the usual way) we obtain different families of "latitude circles", so we have a whole family of "polar spherical charts" on S^2. Given any one we can obtain the others by rotating the semicircle ("International Date Line" u = 0) or, less trivially, by choosing a new pair of antipodal points ("North Pole" and "South Pole"). Note that the freedom to define a whole family of charts such that the metric tensor takes the same form in terms of the various coordinates is closely connected to the existence of symmetries (Killing vector fields) for these familiar Riemannian manifolds. Pedantic aside: I said above that the domain of the chart above covers -almost- all of H^(1,3). Only "almost" because we need to delete certain "world sheets" corresponding on t = t0 to a pair of linked great circles (the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the failure of this chart to cover all of M is due only to the failure of the Hopf chart to cover all of S^3, which is closely analogous to the failure of an ordinary polar spherical chart to cover all of S^2 (which omits the two poles), or of a cylindrical chart to cover all of E^3 (which omits a half plane with boundary the axis of cylindrical symmetry). So this failure to cover all of M is as harmless as the failure of a cylindrical coordinate chart to cover all of E^3. Another way to understand this is to observe that our vector field X is defined on all of H^(1,3), and by moving around the omitted surfaces on each S^3 hyperslice we obtain "trivially different" charts which taken together cover M, and are each associated with one and the same vector field X on M. 2. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Y) everywhere orthogonal to an expanding family of spatial hyperslices with E^3 geometry: ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ], -infty t,x,y,z infty (Not the same t coordinate an in (1)!) The domain of this chart covers the region ______ |\*****| | \****| | \***| | \**| | \*| |_____\| Note well: even though the coordinates range over all of R^4, the chart only covers half of M! This doesn't mean that the other half has vanished, only that it is not represented in this chart. In particular, even though our new irrotational timelike geodesic congruence (integral curves of vector field Y) is defined only on half of M, our original vector field X is of course still defined, and as the diagram suggests half of each integral curve of X appears in our new chart. To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2) in the usual way. This is a ruled surface with rulings comprising two families of null geodesics in E^(1,2) (straight lines in the embedding space), which are precisely the null geodesics on H^(1,1). On "the" equatorial circle of H^(1,1) choose two antipodal points, and choose a pair of "parallel" rulings. These define the boundary of the analogous chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using different examples of such charts (change the defining circle and antipodal point pair). Exercise: does each Y define a unique chart of the above form? Compare with X vs. the S^3 slices above. (See the discussion in Hawking and Ellis for a nice picture of the analogous slices in H^(1,1).) Readers familiar with hyperbolic geometry should note that these vanishing (three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to "horospheres" in H^3 and "horocycles" in H^2, which also have vanishing intrinsic curvature. Exercise: Use the method of geodesic Lagrangians to find the geodesic equations for our chart. Use the method of effective potentials to find the following first integrals (asterisk = d/ds, s the parameter of the geodesic to be found, e = -1,0,1 for timelike, null and spacelike geodesics respectively): x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a) t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a) Explain why this shows at once that the coordinate lines mentioned above are indeed timelike geodesics. Can you write down the complete solution of the geodesic equations? Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line element (dual respectively to one timelike and three spacelike vector fields). Take the exterior derivatives of these one-forms and guess the connection one-forms o^1_2 = exp(t/a) dx/a o^1_3 = exp(t/a) dy/a o^1_4 = exp(t/a) dz/a Compute the covariant derivative D_X X and verify again that the coordinate lines above are geodesics. Exercise: Use the connection one-forms above to compute the curvature two-forms and read off the components of the Riemann tensor wrt the given coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to geodesic divergence of our congruence, i.e. uniform negative curvature, as expected. Note that again a appears as a "size" parameter analogous to the "radius" of a sphere.) Exercise: Verify that Y is irrotational and show that it has constant expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute E_(ab) = R_(abcd) Y^b Y^d and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these results in terms of physical observations by the family of observers corresponding to Y. Does the scaling with parameter a make sense? (Hint: think "dilation"!) Exercise: Look up past posts on finding the wave operator on a Lorentzian manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave equation on H^(1,3) in our chart takes the simple form h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ] Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where F(x) = a1 exp(-A x) + a2 exp(A x) G(y) = b1 exp(-B y) + b2 exp(B y) H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E - a^(1/2) exp(a^(1/2) E exp(-t/a)) ] + e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a)) + exp(-(t+a^(3/2) E exp(-t/a))/a) a E ] Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant? Can you write down the general solution of the wave equation on H^(1,3)? Next, look up past posts on computing symmetry groups of PDEs; can you find the symmetry group of the wave equation in the form above? Choose unidimensional subgroups and find the corresponding solutions. Explain how these results are related to what you found using separation of variables. Exercise: Verify that k = exp(-t/a) (e_1 + e_2) defines a null geodesic congruence with expansion scalar exp(-t/a)/a and vanishing shear and twist scalars. Interpret physically. Exercise: Find a transformation to another chart based on Y, ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ], 0 w infty, -infty x,y,z infty Analyze this following the model of the preceding exercises. Exercise: Find a transformation to a third chart based on Y, ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2), -infty T infty, 0 r infty, 0 u pi, -pi v pi Sketch some null geodesics and integral curves of Y. What happens to Y at the sphere r = a? Is this a coordinate singularity? Interpret physically. (Hint: this chart is the "zero mass" case of a Painleve chart for the Schwarzschild-de Sitter spacetime.) 3. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Z) everywhere orthogonal to a family of spatial hyperslices with H^3 geometry: ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ], 0 t, z infty, -infty x, y infty (Not the same t coordinate as in (1),(2)!) The domain of this chart covers the region ______ |\****/| | \**/ | | \/ | | /\ | | / \ | |/____\| Note that the hyperslices t = t0 are each -complete- hyperbolic spaces H^3. IOW, note that in our first slicing, the slices were compact but in the second two they are noncompact. The discussion above, especially if read with reference to the pictures/discussion in Hawking and Ellis, should make it clear why this is not at all a contradiction. We don't normally do that kind of thing; geodesic incompleteness is tolerated only when the curvature diverges or something like that. So why should we tolerate it in the case of DeSitter space? Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This property holds irrespective of what coordinate charts we use to represent this manifold. If we choose a chart whose domain does not cover the entire manifold, we will be able to find some geodesics which reach a "boundary" of that chart (or have coordinates which diverge) after finite parameter lapse, but this just means we can't study the entire geodesic in question using the given chart. To follow it outside the domain of our chart we must switch to another chart. A homely example: consider an ordinary polar coordinate chart ds^2 = du^2 + sin(u)^2 dv^2, 0 u pi, -pi v pi Note that the domain omits the North pole (u = 0), the South pole (u = pi), and the International Date Line (the semicircular arc v = 0, aka v = pi). Now, every latitude circle (including the equator u = pi/2, which is a geodesic) will encounter the Date Line. This doesn't mean that S^2 is geodesically incomplete, only that to follow latitudes across the Date Line, we need to change to another chart, e.g. by setting v' = v + pi/2, which gives a new polar chart in which the Date Line is just the semicircular arc v' = pi/2, which lies in the domain of validity of the new chart. Similarly, to follow longitudes through the North Pole, we need to change to a new polar chart defined by a different pair of antipodal points; e.g. choose such a pair lying on the equator of our original chart. Compare this example with a polar chart for H^2, and then with H^(1,1) as above. It should be clear that whether our chart omits a "set of volume zero" or "half the manifold", in each case we can follow geodesics or other curves outside a given chart simply by changing to another chart partially overlapping the first, but covering at least some of the region omitted by it. The only special thing there is that, being empty of matter, DeSitter is extremely symmetric, so you can hide the incompleteness with a nice choice of coordinates. Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic property which is not affected by the failings of particular charts. This is just like the Milne Universe, a so-called cosmology which is really just a chunk of Minkowski space. The Milne chart is analogous to chart with H^3 hyperslices above. The fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively do not contradict the geodesic completeness of these manifolds. The fact is, however, that the full DeSitter has spheres as slices; the flat slicing is just a mathematical trick of no real importance, ^^^^^^^^^^ How about -utility-? Exercise: use the method of geodesic Lagrangians to find the geodesic equations in a chart you may be more familiar with: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ], -infty t infty, 0 R pi, 0 U pi, -pi V pi Compare with the equations found in the exercises above. Which would you rather try to solve? In short, DeSitter is REALLY a cosmology with finite spherical space. Right? I hope it is now abundantly clear that H^(1,3) has multiple slicings with any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives different geometrical insight into the nature of the geometry of H^(1,3), but in various circumstances one may be more convenient or natural than the others. |
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Confused by DeSitter again!
On Thu, 7 Aug 2003, Serenus Zeitblom wrote:
I think it is pretty well-known by now that DeSitter space can be sliced in lots of different ways, eg by flat slices or by spherical ones. Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3 respectively are all present and valuable for various purposes. What I don't understand is this: why does anyone take the flat slicing seriously? I mean, it is obviously going to lead to a geodesically incomplete spacetime---you can have timelike worldlines coming in from "beyond the universe"! That is pretty ridiculous It would indeed be absurd to say that by changing to a different coordinate chart on M, you can alter an intrinsic property (geodesic completeness). Of course the answer is that you -can't- do this; you must be somehow confusing the notion of a coordinate chart on M (which need not cover all of M) with M itself. In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic properties include its topology (M is homeomorphic to the topological manifold RxS^3), the geodesic completeness property, and its conformal structure, which can be diagrammed ______ |\ /| | \ / | | \/ | | /\ | | / \ | |/____\| (This crude ASCII sketch is a stand-in for a similar diagram in which the diagonals have slope +/-1.) For details concerning what I just said, see author = {S. W. Hawking and G. F. R. Ellis}, title = {The Large Scale Structure of Space-Time}, publisher = {Cambridge University Press}, year = 1973} In this book you will also find details of several charts; as an exercise you can find transformations from any of those to any in the following list which are not discussed in HE. 1. Comoving with an irrotational timelike geodesic congruence (integral curves of a vector field X) everywhere orthogonal to a (contracting then re-expanding) family of hyperslices with S^3 geometry: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ], -infty t infty, 0 r pi/2, -pi z,u pi This chart covers almost all of M. Note that the coordinate lines z = z0, r = r0, u = u0 correspond to the world lines of a certain family of inertial observers, i.e. a certain family of timelike geodesics on M. Because the geodesics "fill up an open set without intersection"-- or better yet, because they are the integral curves of a vector field X on M-- we call it a "congruence". Because this congruence of timelike geodesics is irrotational (i.e. the vector field X has vanishing vorticity tensor), it defines an orthogonal family of spatial hyperslices t = t0. (See Hawking & Ellis for congruences and vorticity.) From the form of the line element it is obvious that these hyperslices are three-spheres of "radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a minimal size reached at t = t0, and then expand again. It is very important to understand that this chart actually corresponds to a whole family of different congruences. To understand this, consider the usual embedding as a hyperboloid of one sheet of the two dimensional de Sitter space H^(1,1). Here the "latitude circles" correspond to the slices t = t0, with the "equatorial circle" corresponding to the "minimal size" circle t = 0. Note that as t increases, the latitude circles shrink to radius a and then re-expand. (This is obviously very closely analogous to what we saw for our irrotational congruence of inertial observers orthogonal to S^3 hyperslices above.) But now imagine applying a boost in the embedding space E^(1,3) to our H^(1,1), moving the original latitude circles to a new family of "tilted" circles on H^(1,1). The same boost moves all but two of the world lines of our observers to new hyperbolas, so we have a new irrotational congruence of timelike geodesics with a new family of orthogonal slices (circles). Similarly on H^(1,3). If this seems puzzling, it might help to note that one can define on E^3 not just one "cylindrical coordinate chart" ds^2 = dz^2 + dr^2 + r^2 du^2, -infty z infty, 0 r infty, -pi u pi but a whole -family- of similar charts. To obtain the others: rotate the omitted half plane u = pi, or less trivially, translate/rotate the axis of symmetry r = 0 to a new line. Similarly, by choosing various pairs of antipodal points on S^2 (embedded in E^3 in the usual way) we obtain different families of "latitude circles", so we have a whole family of "polar spherical charts" on S^2. Given any one we can obtain the others by rotating the semicircle ("International Date Line" u = 0) or, less trivially, by choosing a new pair of antipodal points ("North Pole" and "South Pole"). Note that the freedom to define a whole family of charts such that the metric tensor takes the same form in terms of the various coordinates is closely connected to the existence of symmetries (Killing vector fields) for these familiar Riemannian manifolds. Pedantic aside: I said above that the domain of the chart above covers -almost- all of H^(1,3). Only "almost" because we need to delete certain "world sheets" corresponding on t = t0 to a pair of linked great circles (the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the failure of this chart to cover all of M is due only to the failure of the Hopf chart to cover all of S^3, which is closely analogous to the failure of an ordinary polar spherical chart to cover all of S^2 (which omits the two poles), or of a cylindrical chart to cover all of E^3 (which omits a half plane with boundary the axis of cylindrical symmetry). So this failure to cover all of M is as harmless as the failure of a cylindrical coordinate chart to cover all of E^3. Another way to understand this is to observe that our vector field X is defined on all of H^(1,3), and by moving around the omitted surfaces on each S^3 hyperslice we obtain "trivially different" charts which taken together cover M, and are each associated with one and the same vector field X on M. 2. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Y) everywhere orthogonal to an expanding family of spatial hyperslices with E^3 geometry: ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ], -infty t,x,y,z infty (Not the same t coordinate an in (1)!) The domain of this chart covers the region ______ |\*****| | \****| | \***| | \**| | \*| |_____\| Note well: even though the coordinates range over all of R^4, the chart only covers half of M! This doesn't mean that the other half has vanished, only that it is not represented in this chart. In particular, even though our new irrotational timelike geodesic congruence (integral curves of vector field Y) is defined only on half of M, our original vector field X is of course still defined, and as the diagram suggests half of each integral curve of X appears in our new chart. To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2) in the usual way. This is a ruled surface with rulings comprising two families of null geodesics in E^(1,2) (straight lines in the embedding space), which are precisely the null geodesics on H^(1,1). On "the" equatorial circle of H^(1,1) choose two antipodal points, and choose a pair of "parallel" rulings. These define the boundary of the analogous chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using different examples of such charts (change the defining circle and antipodal point pair). Exercise: does each Y define a unique chart of the above form? Compare with X vs. the S^3 slices above. (See the discussion in Hawking and Ellis for a nice picture of the analogous slices in H^(1,1).) Readers familiar with hyperbolic geometry should note that these vanishing (three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to "horospheres" in H^3 and "horocycles" in H^2, which also have vanishing intrinsic curvature. Exercise: Use the method of geodesic Lagrangians to find the geodesic equations for our chart. Use the method of effective potentials to find the following first integrals (asterisk = d/ds, s the parameter of the geodesic to be found, e = -1,0,1 for timelike, null and spacelike geodesics respectively): x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a) t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a) Explain why this shows at once that the coordinate lines mentioned above are indeed timelike geodesics. Can you write down the complete solution of the geodesic equations? Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line element (dual respectively to one timelike and three spacelike vector fields). Take the exterior derivatives of these one-forms and guess the connection one-forms o^1_2 = exp(t/a) dx/a o^1_3 = exp(t/a) dy/a o^1_4 = exp(t/a) dz/a Compute the covariant derivative D_X X and verify again that the coordinate lines above are geodesics. Exercise: Use the connection one-forms above to compute the curvature two-forms and read off the components of the Riemann tensor wrt the given coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to geodesic divergence of our congruence, i.e. uniform negative curvature, as expected. Note that again a appears as a "size" parameter analogous to the "radius" of a sphere.) Exercise: Verify that Y is irrotational and show that it has constant expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute E_(ab) = R_(abcd) Y^b Y^d and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these results in terms of physical observations by the family of observers corresponding to Y. Does the scaling with parameter a make sense? (Hint: think "dilation"!) Exercise: Look up past posts on finding the wave operator on a Lorentzian manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave equation on H^(1,3) in our chart takes the simple form h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ] Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where F(x) = a1 exp(-A x) + a2 exp(A x) G(y) = b1 exp(-B y) + b2 exp(B y) H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E - a^(1/2) exp(a^(1/2) E exp(-t/a)) ] + e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a)) + exp(-(t+a^(3/2) E exp(-t/a))/a) a E ] Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant? Can you write down the general solution of the wave equation on H^(1,3)? Next, look up past posts on computing symmetry groups of PDEs; can you find the symmetry group of the wave equation in the form above? Choose unidimensional subgroups and find the corresponding solutions. Explain how these results are related to what you found using separation of variables. Exercise: Verify that k = exp(-t/a) (e_1 + e_2) defines a null geodesic congruence with expansion scalar exp(-t/a)/a and vanishing shear and twist scalars. Interpret physically. Exercise: Find a transformation to another chart based on Y, ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ], 0 w infty, -infty x,y,z infty Analyze this following the model of the preceding exercises. Exercise: Find a transformation to a third chart based on Y, ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2), -infty T infty, 0 r infty, 0 u pi, -pi v pi Sketch some null geodesics and integral curves of Y. What happens to Y at the sphere r = a? Is this a coordinate singularity? Interpret physically. (Hint: this chart is the "zero mass" case of a Painleve chart for the Schwarzschild-de Sitter spacetime.) 3. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Z) everywhere orthogonal to a family of spatial hyperslices with H^3 geometry: ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ], 0 t, z infty, -infty x, y infty (Not the same t coordinate as in (1),(2)!) The domain of this chart covers the region ______ |\****/| | \**/ | | \/ | | /\ | | / \ | |/____\| Note that the hyperslices t = t0 are each -complete- hyperbolic spaces H^3. IOW, note that in our first slicing, the slices were compact but in the second two they are noncompact. The discussion above, especially if read with reference to the pictures/discussion in Hawking and Ellis, should make it clear why this is not at all a contradiction. We don't normally do that kind of thing; geodesic incompleteness is tolerated only when the curvature diverges or something like that. So why should we tolerate it in the case of DeSitter space? Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This property holds irrespective of what coordinate charts we use to represent this manifold. If we choose a chart whose domain does not cover the entire manifold, we will be able to find some geodesics which reach a "boundary" of that chart (or have coordinates which diverge) after finite parameter lapse, but this just means we can't study the entire geodesic in question using the given chart. To follow it outside the domain of our chart we must switch to another chart. A homely example: consider an ordinary polar coordinate chart ds^2 = du^2 + sin(u)^2 dv^2, 0 u pi, -pi v pi Note that the domain omits the North pole (u = 0), the South pole (u = pi), and the International Date Line (the semicircular arc v = 0, aka v = pi). Now, every latitude circle (including the equator u = pi/2, which is a geodesic) will encounter the Date Line. This doesn't mean that S^2 is geodesically incomplete, only that to follow latitudes across the Date Line, we need to change to another chart, e.g. by setting v' = v + pi/2, which gives a new polar chart in which the Date Line is just the semicircular arc v' = pi/2, which lies in the domain of validity of the new chart. Similarly, to follow longitudes through the North Pole, we need to change to a new polar chart defined by a different pair of antipodal points; e.g. choose such a pair lying on the equator of our original chart. Compare this example with a polar chart for H^2, and then with H^(1,1) as above. It should be clear that whether our chart omits a "set of volume zero" or "half the manifold", in each case we can follow geodesics or other curves outside a given chart simply by changing to another chart partially overlapping the first, but covering at least some of the region omitted by it. The only special thing there is that, being empty of matter, DeSitter is extremely symmetric, so you can hide the incompleteness with a nice choice of coordinates. Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic property which is not affected by the failings of particular charts. This is just like the Milne Universe, a so-called cosmology which is really just a chunk of Minkowski space. The Milne chart is analogous to chart with H^3 hyperslices above. The fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively do not contradict the geodesic completeness of these manifolds. The fact is, however, that the full DeSitter has spheres as slices; the flat slicing is just a mathematical trick of no real importance, ^^^^^^^^^^ How about -utility-? Exercise: use the method of geodesic Lagrangians to find the geodesic equations in a chart you may be more familiar with: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ], -infty t infty, 0 R pi, 0 U pi, -pi V pi Compare with the equations found in the exercises above. Which would you rather try to solve? In short, DeSitter is REALLY a cosmology with finite spherical space. Right? I hope it is now abundantly clear that H^(1,3) has multiple slicings with any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives different geometrical insight into the nature of the geometry of H^(1,3), but in various circumstances one may be more convenient or natural than the others. |
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Confused by DeSitter again!
On Thu, 7 Aug 2003, Serenus Zeitblom wrote:
I think it is pretty well-known by now that DeSitter space can be sliced in lots of different ways, eg by flat slices or by spherical ones. Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3 respectively are all present and valuable for various purposes. What I don't understand is this: why does anyone take the flat slicing seriously? I mean, it is obviously going to lead to a geodesically incomplete spacetime---you can have timelike worldlines coming in from "beyond the universe"! That is pretty ridiculous It would indeed be absurd to say that by changing to a different coordinate chart on M, you can alter an intrinsic property (geodesic completeness). Of course the answer is that you -can't- do this; you must be somehow confusing the notion of a coordinate chart on M (which need not cover all of M) with M itself. In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic properties include its topology (M is homeomorphic to the topological manifold RxS^3), the geodesic completeness property, and its conformal structure, which can be diagrammed ______ |\ /| | \ / | | \/ | | /\ | | / \ | |/____\| (This crude ASCII sketch is a stand-in for a similar diagram in which the diagonals have slope +/-1.) For details concerning what I just said, see author = {S. W. Hawking and G. F. R. Ellis}, title = {The Large Scale Structure of Space-Time}, publisher = {Cambridge University Press}, year = 1973} In this book you will also find details of several charts; as an exercise you can find transformations from any of those to any in the following list which are not discussed in HE. 1. Comoving with an irrotational timelike geodesic congruence (integral curves of a vector field X) everywhere orthogonal to a (contracting then re-expanding) family of hyperslices with S^3 geometry: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ], -infty t infty, 0 r pi/2, -pi z,u pi This chart covers almost all of M. Note that the coordinate lines z = z0, r = r0, u = u0 correspond to the world lines of a certain family of inertial observers, i.e. a certain family of timelike geodesics on M. Because the geodesics "fill up an open set without intersection"-- or better yet, because they are the integral curves of a vector field X on M-- we call it a "congruence". Because this congruence of timelike geodesics is irrotational (i.e. the vector field X has vanishing vorticity tensor), it defines an orthogonal family of spatial hyperslices t = t0. (See Hawking & Ellis for congruences and vorticity.) From the form of the line element it is obvious that these hyperslices are three-spheres of "radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a minimal size reached at t = t0, and then expand again. It is very important to understand that this chart actually corresponds to a whole family of different congruences. To understand this, consider the usual embedding as a hyperboloid of one sheet of the two dimensional de Sitter space H^(1,1). Here the "latitude circles" correspond to the slices t = t0, with the "equatorial circle" corresponding to the "minimal size" circle t = 0. Note that as t increases, the latitude circles shrink to radius a and then re-expand. (This is obviously very closely analogous to what we saw for our irrotational congruence of inertial observers orthogonal to S^3 hyperslices above.) But now imagine applying a boost in the embedding space E^(1,3) to our H^(1,1), moving the original latitude circles to a new family of "tilted" circles on H^(1,1). The same boost moves all but two of the world lines of our observers to new hyperbolas, so we have a new irrotational congruence of timelike geodesics with a new family of orthogonal slices (circles). Similarly on H^(1,3). If this seems puzzling, it might help to note that one can define on E^3 not just one "cylindrical coordinate chart" ds^2 = dz^2 + dr^2 + r^2 du^2, -infty z infty, 0 r infty, -pi u pi but a whole -family- of similar charts. To obtain the others: rotate the omitted half plane u = pi, or less trivially, translate/rotate the axis of symmetry r = 0 to a new line. Similarly, by choosing various pairs of antipodal points on S^2 (embedded in E^3 in the usual way) we obtain different families of "latitude circles", so we have a whole family of "polar spherical charts" on S^2. Given any one we can obtain the others by rotating the semicircle ("International Date Line" u = 0) or, less trivially, by choosing a new pair of antipodal points ("North Pole" and "South Pole"). Note that the freedom to define a whole family of charts such that the metric tensor takes the same form in terms of the various coordinates is closely connected to the existence of symmetries (Killing vector fields) for these familiar Riemannian manifolds. Pedantic aside: I said above that the domain of the chart above covers -almost- all of H^(1,3). Only "almost" because we need to delete certain "world sheets" corresponding on t = t0 to a pair of linked great circles (the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the failure of this chart to cover all of M is due only to the failure of the Hopf chart to cover all of S^3, which is closely analogous to the failure of an ordinary polar spherical chart to cover all of S^2 (which omits the two poles), or of a cylindrical chart to cover all of E^3 (which omits a half plane with boundary the axis of cylindrical symmetry). So this failure to cover all of M is as harmless as the failure of a cylindrical coordinate chart to cover all of E^3. Another way to understand this is to observe that our vector field X is defined on all of H^(1,3), and by moving around the omitted surfaces on each S^3 hyperslice we obtain "trivially different" charts which taken together cover M, and are each associated with one and the same vector field X on M. 2. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Y) everywhere orthogonal to an expanding family of spatial hyperslices with E^3 geometry: ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ], -infty t,x,y,z infty (Not the same t coordinate an in (1)!) The domain of this chart covers the region ______ |\*****| | \****| | \***| | \**| | \*| |_____\| Note well: even though the coordinates range over all of R^4, the chart only covers half of M! This doesn't mean that the other half has vanished, only that it is not represented in this chart. In particular, even though our new irrotational timelike geodesic congruence (integral curves of vector field Y) is defined only on half of M, our original vector field X is of course still defined, and as the diagram suggests half of each integral curve of X appears in our new chart. To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2) in the usual way. This is a ruled surface with rulings comprising two families of null geodesics in E^(1,2) (straight lines in the embedding space), which are precisely the null geodesics on H^(1,1). On "the" equatorial circle of H^(1,1) choose two antipodal points, and choose a pair of "parallel" rulings. These define the boundary of the analogous chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using different examples of such charts (change the defining circle and antipodal point pair). Exercise: does each Y define a unique chart of the above form? Compare with X vs. the S^3 slices above. (See the discussion in Hawking and Ellis for a nice picture of the analogous slices in H^(1,1).) Readers familiar with hyperbolic geometry should note that these vanishing (three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to "horospheres" in H^3 and "horocycles" in H^2, which also have vanishing intrinsic curvature. Exercise: Use the method of geodesic Lagrangians to find the geodesic equations for our chart. Use the method of effective potentials to find the following first integrals (asterisk = d/ds, s the parameter of the geodesic to be found, e = -1,0,1 for timelike, null and spacelike geodesics respectively): x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a) t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a) Explain why this shows at once that the coordinate lines mentioned above are indeed timelike geodesics. Can you write down the complete solution of the geodesic equations? Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line element (dual respectively to one timelike and three spacelike vector fields). Take the exterior derivatives of these one-forms and guess the connection one-forms o^1_2 = exp(t/a) dx/a o^1_3 = exp(t/a) dy/a o^1_4 = exp(t/a) dz/a Compute the covariant derivative D_X X and verify again that the coordinate lines above are geodesics. Exercise: Use the connection one-forms above to compute the curvature two-forms and read off the components of the Riemann tensor wrt the given coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to geodesic divergence of our congruence, i.e. uniform negative curvature, as expected. Note that again a appears as a "size" parameter analogous to the "radius" of a sphere.) Exercise: Verify that Y is irrotational and show that it has constant expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute E_(ab) = R_(abcd) Y^b Y^d and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these results in terms of physical observations by the family of observers corresponding to Y. Does the scaling with parameter a make sense? (Hint: think "dilation"!) Exercise: Look up past posts on finding the wave operator on a Lorentzian manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave equation on H^(1,3) in our chart takes the simple form h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ] Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where F(x) = a1 exp(-A x) + a2 exp(A x) G(y) = b1 exp(-B y) + b2 exp(B y) H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E - a^(1/2) exp(a^(1/2) E exp(-t/a)) ] + e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a)) + exp(-(t+a^(3/2) E exp(-t/a))/a) a E ] Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant? Can you write down the general solution of the wave equation on H^(1,3)? Next, look up past posts on computing symmetry groups of PDEs; can you find the symmetry group of the wave equation in the form above? Choose unidimensional subgroups and find the corresponding solutions. Explain how these results are related to what you found using separation of variables. Exercise: Verify that k = exp(-t/a) (e_1 + e_2) defines a null geodesic congruence with expansion scalar exp(-t/a)/a and vanishing shear and twist scalars. Interpret physically. Exercise: Find a transformation to another chart based on Y, ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ], 0 w infty, -infty x,y,z infty Analyze this following the model of the preceding exercises. Exercise: Find a transformation to a third chart based on Y, ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2), -infty T infty, 0 r infty, 0 u pi, -pi v pi Sketch some null geodesics and integral curves of Y. What happens to Y at the sphere r = a? Is this a coordinate singularity? Interpret physically. (Hint: this chart is the "zero mass" case of a Painleve chart for the Schwarzschild-de Sitter spacetime.) 3. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Z) everywhere orthogonal to a family of spatial hyperslices with H^3 geometry: ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ], 0 t, z infty, -infty x, y infty (Not the same t coordinate as in (1),(2)!) The domain of this chart covers the region ______ |\****/| | \**/ | | \/ | | /\ | | / \ | |/____\| Note that the hyperslices t = t0 are each -complete- hyperbolic spaces H^3. IOW, note that in our first slicing, the slices were compact but in the second two they are noncompact. The discussion above, especially if read with reference to the pictures/discussion in Hawking and Ellis, should make it clear why this is not at all a contradiction. We don't normally do that kind of thing; geodesic incompleteness is tolerated only when the curvature diverges or something like that. So why should we tolerate it in the case of DeSitter space? Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This property holds irrespective of what coordinate charts we use to represent this manifold. If we choose a chart whose domain does not cover the entire manifold, we will be able to find some geodesics which reach a "boundary" of that chart (or have coordinates which diverge) after finite parameter lapse, but this just means we can't study the entire geodesic in question using the given chart. To follow it outside the domain of our chart we must switch to another chart. A homely example: consider an ordinary polar coordinate chart ds^2 = du^2 + sin(u)^2 dv^2, 0 u pi, -pi v pi Note that the domain omits the North pole (u = 0), the South pole (u = pi), and the International Date Line (the semicircular arc v = 0, aka v = pi). Now, every latitude circle (including the equator u = pi/2, which is a geodesic) will encounter the Date Line. This doesn't mean that S^2 is geodesically incomplete, only that to follow latitudes across the Date Line, we need to change to another chart, e.g. by setting v' = v + pi/2, which gives a new polar chart in which the Date Line is just the semicircular arc v' = pi/2, which lies in the domain of validity of the new chart. Similarly, to follow longitudes through the North Pole, we need to change to a new polar chart defined by a different pair of antipodal points; e.g. choose such a pair lying on the equator of our original chart. Compare this example with a polar chart for H^2, and then with H^(1,1) as above. It should be clear that whether our chart omits a "set of volume zero" or "half the manifold", in each case we can follow geodesics or other curves outside a given chart simply by changing to another chart partially overlapping the first, but covering at least some of the region omitted by it. The only special thing there is that, being empty of matter, DeSitter is extremely symmetric, so you can hide the incompleteness with a nice choice of coordinates. Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic property which is not affected by the failings of particular charts. This is just like the Milne Universe, a so-called cosmology which is really just a chunk of Minkowski space. The Milne chart is analogous to chart with H^3 hyperslices above. The fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively do not contradict the geodesic completeness of these manifolds. The fact is, however, that the full DeSitter has spheres as slices; the flat slicing is just a mathematical trick of no real importance, ^^^^^^^^^^ How about -utility-? Exercise: use the method of geodesic Lagrangians to find the geodesic equations in a chart you may be more familiar with: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ], -infty t infty, 0 R pi, 0 U pi, -pi V pi Compare with the equations found in the exercises above. Which would you rather try to solve? In short, DeSitter is REALLY a cosmology with finite spherical space. Right? I hope it is now abundantly clear that H^(1,3) has multiple slicings with any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives different geometrical insight into the nature of the geometry of H^(1,3), but in various circumstances one may be more convenient or natural than the others. |
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Confused by DeSitter again!
On Thu, 7 Aug 2003, Serenus Zeitblom wrote:
I think it is pretty well-known by now that DeSitter space can be sliced in lots of different ways, eg by flat slices or by spherical ones. Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3 respectively are all present and valuable for various purposes. What I don't understand is this: why does anyone take the flat slicing seriously? I mean, it is obviously going to lead to a geodesically incomplete spacetime---you can have timelike worldlines coming in from "beyond the universe"! That is pretty ridiculous It would indeed be absurd to say that by changing to a different coordinate chart on M, you can alter an intrinsic property (geodesic completeness). Of course the answer is that you -can't- do this; you must be somehow confusing the notion of a coordinate chart on M (which need not cover all of M) with M itself. In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic properties include its topology (M is homeomorphic to the topological manifold RxS^3), the geodesic completeness property, and its conformal structure, which can be diagrammed ______ |\ /| | \ / | | \/ | | /\ | | / \ | |/____\| (This crude ASCII sketch is a stand-in for a similar diagram in which the diagonals have slope +/-1.) For details concerning what I just said, see author = {S. W. Hawking and G. F. R. Ellis}, title = {The Large Scale Structure of Space-Time}, publisher = {Cambridge University Press}, year = 1973} In this book you will also find details of several charts; as an exercise you can find transformations from any of those to any in the following list which are not discussed in HE. 1. Comoving with an irrotational timelike geodesic congruence (integral curves of a vector field X) everywhere orthogonal to a (contracting then re-expanding) family of hyperslices with S^3 geometry: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ], -infty t infty, 0 r pi/2, -pi z,u pi This chart covers almost all of M. Note that the coordinate lines z = z0, r = r0, u = u0 correspond to the world lines of a certain family of inertial observers, i.e. a certain family of timelike geodesics on M. Because the geodesics "fill up an open set without intersection"-- or better yet, because they are the integral curves of a vector field X on M-- we call it a "congruence". Because this congruence of timelike geodesics is irrotational (i.e. the vector field X has vanishing vorticity tensor), it defines an orthogonal family of spatial hyperslices t = t0. (See Hawking & Ellis for congruences and vorticity.) From the form of the line element it is obvious that these hyperslices are three-spheres of "radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a minimal size reached at t = t0, and then expand again. It is very important to understand that this chart actually corresponds to a whole family of different congruences. To understand this, consider the usual embedding as a hyperboloid of one sheet of the two dimensional de Sitter space H^(1,1). Here the "latitude circles" correspond to the slices t = t0, with the "equatorial circle" corresponding to the "minimal size" circle t = 0. Note that as t increases, the latitude circles shrink to radius a and then re-expand. (This is obviously very closely analogous to what we saw for our irrotational congruence of inertial observers orthogonal to S^3 hyperslices above.) But now imagine applying a boost in the embedding space E^(1,3) to our H^(1,1), moving the original latitude circles to a new family of "tilted" circles on H^(1,1). The same boost moves all but two of the world lines of our observers to new hyperbolas, so we have a new irrotational congruence of timelike geodesics with a new family of orthogonal slices (circles). Similarly on H^(1,3). If this seems puzzling, it might help to note that one can define on E^3 not just one "cylindrical coordinate chart" ds^2 = dz^2 + dr^2 + r^2 du^2, -infty z infty, 0 r infty, -pi u pi but a whole -family- of similar charts. To obtain the others: rotate the omitted half plane u = pi, or less trivially, translate/rotate the axis of symmetry r = 0 to a new line. Similarly, by choosing various pairs of antipodal points on S^2 (embedded in E^3 in the usual way) we obtain different families of "latitude circles", so we have a whole family of "polar spherical charts" on S^2. Given any one we can obtain the others by rotating the semicircle ("International Date Line" u = 0) or, less trivially, by choosing a new pair of antipodal points ("North Pole" and "South Pole"). Note that the freedom to define a whole family of charts such that the metric tensor takes the same form in terms of the various coordinates is closely connected to the existence of symmetries (Killing vector fields) for these familiar Riemannian manifolds. Pedantic aside: I said above that the domain of the chart above covers -almost- all of H^(1,3). Only "almost" because we need to delete certain "world sheets" corresponding on t = t0 to a pair of linked great circles (the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the failure of this chart to cover all of M is due only to the failure of the Hopf chart to cover all of S^3, which is closely analogous to the failure of an ordinary polar spherical chart to cover all of S^2 (which omits the two poles), or of a cylindrical chart to cover all of E^3 (which omits a half plane with boundary the axis of cylindrical symmetry). So this failure to cover all of M is as harmless as the failure of a cylindrical coordinate chart to cover all of E^3. Another way to understand this is to observe that our vector field X is defined on all of H^(1,3), and by moving around the omitted surfaces on each S^3 hyperslice we obtain "trivially different" charts which taken together cover M, and are each associated with one and the same vector field X on M. 2. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Y) everywhere orthogonal to an expanding family of spatial hyperslices with E^3 geometry: ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ], -infty t,x,y,z infty (Not the same t coordinate an in (1)!) The domain of this chart covers the region ______ |\*****| | \****| | \***| | \**| | \*| |_____\| Note well: even though the coordinates range over all of R^4, the chart only covers half of M! This doesn't mean that the other half has vanished, only that it is not represented in this chart. In particular, even though our new irrotational timelike geodesic congruence (integral curves of vector field Y) is defined only on half of M, our original vector field X is of course still defined, and as the diagram suggests half of each integral curve of X appears in our new chart. To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2) in the usual way. This is a ruled surface with rulings comprising two families of null geodesics in E^(1,2) (straight lines in the embedding space), which are precisely the null geodesics on H^(1,1). On "the" equatorial circle of H^(1,1) choose two antipodal points, and choose a pair of "parallel" rulings. These define the boundary of the analogous chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using different examples of such charts (change the defining circle and antipodal point pair). Exercise: does each Y define a unique chart of the above form? Compare with X vs. the S^3 slices above. (See the discussion in Hawking and Ellis for a nice picture of the analogous slices in H^(1,1).) Readers familiar with hyperbolic geometry should note that these vanishing (three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to "horospheres" in H^3 and "horocycles" in H^2, which also have vanishing intrinsic curvature. Exercise: Use the method of geodesic Lagrangians to find the geodesic equations for our chart. Use the method of effective potentials to find the following first integrals (asterisk = d/ds, s the parameter of the geodesic to be found, e = -1,0,1 for timelike, null and spacelike geodesics respectively): x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a) t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a) Explain why this shows at once that the coordinate lines mentioned above are indeed timelike geodesics. Can you write down the complete solution of the geodesic equations? Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line element (dual respectively to one timelike and three spacelike vector fields). Take the exterior derivatives of these one-forms and guess the connection one-forms o^1_2 = exp(t/a) dx/a o^1_3 = exp(t/a) dy/a o^1_4 = exp(t/a) dz/a Compute the covariant derivative D_X X and verify again that the coordinate lines above are geodesics. Exercise: Use the connection one-forms above to compute the curvature two-forms and read off the components of the Riemann tensor wrt the given coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to geodesic divergence of our congruence, i.e. uniform negative curvature, as expected. Note that again a appears as a "size" parameter analogous to the "radius" of a sphere.) Exercise: Verify that Y is irrotational and show that it has constant expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute E_(ab) = R_(abcd) Y^b Y^d and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these results in terms of physical observations by the family of observers corresponding to Y. Does the scaling with parameter a make sense? (Hint: think "dilation"!) Exercise: Look up past posts on finding the wave operator on a Lorentzian manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave equation on H^(1,3) in our chart takes the simple form h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ] Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where F(x) = a1 exp(-A x) + a2 exp(A x) G(y) = b1 exp(-B y) + b2 exp(B y) H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E - a^(1/2) exp(a^(1/2) E exp(-t/a)) ] + e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a)) + exp(-(t+a^(3/2) E exp(-t/a))/a) a E ] Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant? Can you write down the general solution of the wave equation on H^(1,3)? Next, look up past posts on computing symmetry groups of PDEs; can you find the symmetry group of the wave equation in the form above? Choose unidimensional subgroups and find the corresponding solutions. Explain how these results are related to what you found using separation of variables. Exercise: Verify that k = exp(-t/a) (e_1 + e_2) defines a null geodesic congruence with expansion scalar exp(-t/a)/a and vanishing shear and twist scalars. Interpret physically. Exercise: Find a transformation to another chart based on Y, ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ], 0 w infty, -infty x,y,z infty Analyze this following the model of the preceding exercises. Exercise: Find a transformation to a third chart based on Y, ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2), -infty T infty, 0 r infty, 0 u pi, -pi v pi Sketch some null geodesics and integral curves of Y. What happens to Y at the sphere r = a? Is this a coordinate singularity? Interpret physically. (Hint: this chart is the "zero mass" case of a Painleve chart for the Schwarzschild-de Sitter spacetime.) 3. Comoving with irrotational timelike geodesic congruence (integral curves of vector field Z) everywhere orthogonal to a family of spatial hyperslices with H^3 geometry: ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ], 0 t, z infty, -infty x, y infty (Not the same t coordinate as in (1),(2)!) The domain of this chart covers the region ______ |\****/| | \**/ | | \/ | | /\ | | / \ | |/____\| Note that the hyperslices t = t0 are each -complete- hyperbolic spaces H^3. IOW, note that in our first slicing, the slices were compact but in the second two they are noncompact. The discussion above, especially if read with reference to the pictures/discussion in Hawking and Ellis, should make it clear why this is not at all a contradiction. We don't normally do that kind of thing; geodesic incompleteness is tolerated only when the curvature diverges or something like that. So why should we tolerate it in the case of DeSitter space? Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This property holds irrespective of what coordinate charts we use to represent this manifold. If we choose a chart whose domain does not cover the entire manifold, we will be able to find some geodesics which reach a "boundary" of that chart (or have coordinates which diverge) after finite parameter lapse, but this just means we can't study the entire geodesic in question using the given chart. To follow it outside the domain of our chart we must switch to another chart. A homely example: consider an ordinary polar coordinate chart ds^2 = du^2 + sin(u)^2 dv^2, 0 u pi, -pi v pi Note that the domain omits the North pole (u = 0), the South pole (u = pi), and the International Date Line (the semicircular arc v = 0, aka v = pi). Now, every latitude circle (including the equator u = pi/2, which is a geodesic) will encounter the Date Line. This doesn't mean that S^2 is geodesically incomplete, only that to follow latitudes across the Date Line, we need to change to another chart, e.g. by setting v' = v + pi/2, which gives a new polar chart in which the Date Line is just the semicircular arc v' = pi/2, which lies in the domain of validity of the new chart. Similarly, to follow longitudes through the North Pole, we need to change to a new polar chart defined by a different pair of antipodal points; e.g. choose such a pair lying on the equator of our original chart. Compare this example with a polar chart for H^2, and then with H^(1,1) as above. It should be clear that whether our chart omits a "set of volume zero" or "half the manifold", in each case we can follow geodesics or other curves outside a given chart simply by changing to another chart partially overlapping the first, but covering at least some of the region omitted by it. The only special thing there is that, being empty of matter, DeSitter is extremely symmetric, so you can hide the incompleteness with a nice choice of coordinates. Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic property which is not affected by the failings of particular charts. This is just like the Milne Universe, a so-called cosmology which is really just a chunk of Minkowski space. The Milne chart is analogous to chart with H^3 hyperslices above. The fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively do not contradict the geodesic completeness of these manifolds. The fact is, however, that the full DeSitter has spheres as slices; the flat slicing is just a mathematical trick of no real importance, ^^^^^^^^^^ How about -utility-? Exercise: use the method of geodesic Lagrangians to find the geodesic equations in a chart you may be more familiar with: ds^2 = -dt^2 + a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ], -infty t infty, 0 R pi, 0 U pi, -pi V pi Compare with the equations found in the exercises above. Which would you rather try to solve? In short, DeSitter is REALLY a cosmology with finite spherical space. Right? I hope it is now abundantly clear that H^(1,3) has multiple slicings with any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives different geometrical insight into the nature of the geometry of H^(1,3), but in various circumstances one may be more convenient or natural than the others. |
#5
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Confused by DeSitter again!
[[Mod. note -- lines re-wrapped -- jt]]
"Chris Hillman (remoev -animal to reply)" wrote in message ... On Thu, 7 Aug 2003, Serenus Zeitblom wrote: I think it is pretty well-known by now that DeSitter space can be sliced in lots of different ways, eg by flat slices or by spherical ones. Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3 respectively are all present and valuable for various purposes. What I don't understand is this: why does anyone take the flat slicing seriously? I mean, it is obviously going to lead to a geodesically incomplete spacetime---you can have timelike worldlines coming in from "beyond the universe"! That is pretty ridiculous It would indeed be absurd to say that by changing to a different coordinate chart on M, you can alter an intrinsic property (geodesic completeness). Of course the answer is that you -can't- do this; you must be somehow confusing the notion of a coordinate chart on M (which need not cover all of M) with M itself. I wasn't, of course, suggesting that ALL OF deSitter can be made geodesically incomplete just by changing coordinates. But of course any space can be made geodesically incomplete by cutting parts of it out and throwing them away. Sometimes, you can however do this and cover things up by a clever choice of coordinates on the remaining piece. What I mean is this: I frequently hear and see people saying [A] the WMAP data show that the spatial sections of our universe are flat; and [b] this, combined with the fact that the universe is accelerating, means that, to a first approximation [ignoring matter and radiation], the metric of the universe can be written as ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ] What I am objecting to is the fact that this is not a metric on "deSitter space" ---it is a metric on a PART of deSItter space. If we cut away the other half, then what is left over is of course geodesically incomplete. And normally we don't do that kind of thing! Let me put it to you this way. Suppose you believe that the spatial sections of the universe are flat and simply connected. Then you might observe that the above metric is a metric on R x R^3, and that it solves the Einstein equations with a positive cosmological constant. It looks ok in these coordinates, but IN FACT it is geodesically incomplete. The easiest way to show this is to do as you did, draw the Penrose diagram, and see that what you really have here is just half of the diagram. Cutting away the other half is of course going to produce a geodesically incomplete spacetime. R x R^3 endowed with the above metric is geodesically incomplete, true or false? What I'm driving at is this. If we believe that our accelerating spacetime would be deSitter space if there were no matter or radiation in it, then we are implicitly claiming that the topology of spacetime is really R x S^3. Of course we can do things like take stereographic projections, but only by cutting out and throwing away parts of the space. Agreed? In short, DeSitter is REALLY a cosmology with finite spherical space. Right? I hope it is now abundantly clear that H^(1,3) has multiple slicings with any of S^3, E^3 or H^3 geometry. REALLY! :-/ I agree. I wasn't objecting to these well-known mathematical facts. I was really objecting to the idea that "deSitter spacetime" can have flat spatial sections. It can, but only if you cut half of it off and throw it away. By the way, I am told that the Steady-State people were quite willing to do this, precisely because they didn't believe in matter conservation and actually welcomed the idea of stuff coming from "outside the universe" [ie, really, from the half of deSitter they were throwing away!] Not coincidentally, they always used the flat slicing! Each of these slicings gives different geometrical insight into the nature of the geometry of H^(1,3), but in various circumstances one may be more convenient or natural than the others. As long as nobody tries to tell me that PART OF deSitter space (either the half covered by the flat slicing or the quarter covered by the hyperbolic one) is a viable cosmology BY ITSELF, then of course I have no objections. |
#6
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Confused by DeSitter again!
[[Mod. note -- lines re-wrapped -- jt]]
"Chris Hillman (remoev -animal to reply)" wrote in message ... On Thu, 7 Aug 2003, Serenus Zeitblom wrote: I think it is pretty well-known by now that DeSitter space can be sliced in lots of different ways, eg by flat slices or by spherical ones. Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3 respectively are all present and valuable for various purposes. What I don't understand is this: why does anyone take the flat slicing seriously? I mean, it is obviously going to lead to a geodesically incomplete spacetime---you can have timelike worldlines coming in from "beyond the universe"! That is pretty ridiculous It would indeed be absurd to say that by changing to a different coordinate chart on M, you can alter an intrinsic property (geodesic completeness). Of course the answer is that you -can't- do this; you must be somehow confusing the notion of a coordinate chart on M (which need not cover all of M) with M itself. I wasn't, of course, suggesting that ALL OF deSitter can be made geodesically incomplete just by changing coordinates. But of course any space can be made geodesically incomplete by cutting parts of it out and throwing them away. Sometimes, you can however do this and cover things up by a clever choice of coordinates on the remaining piece. What I mean is this: I frequently hear and see people saying [A] the WMAP data show that the spatial sections of our universe are flat; and [b] this, combined with the fact that the universe is accelerating, means that, to a first approximation [ignoring matter and radiation], the metric of the universe can be written as ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ] What I am objecting to is the fact that this is not a metric on "deSitter space" ---it is a metric on a PART of deSItter space. If we cut away the other half, then what is left over is of course geodesically incomplete. And normally we don't do that kind of thing! Let me put it to you this way. Suppose you believe that the spatial sections of the universe are flat and simply connected. Then you might observe that the above metric is a metric on R x R^3, and that it solves the Einstein equations with a positive cosmological constant. It looks ok in these coordinates, but IN FACT it is geodesically incomplete. The easiest way to show this is to do as you did, draw the Penrose diagram, and see that what you really have here is just half of the diagram. Cutting away the other half is of course going to produce a geodesically incomplete spacetime. R x R^3 endowed with the above metric is geodesically incomplete, true or false? What I'm driving at is this. If we believe that our accelerating spacetime would be deSitter space if there were no matter or radiation in it, then we are implicitly claiming that the topology of spacetime is really R x S^3. Of course we can do things like take stereographic projections, but only by cutting out and throwing away parts of the space. Agreed? In short, DeSitter is REALLY a cosmology with finite spherical space. Right? I hope it is now abundantly clear that H^(1,3) has multiple slicings with any of S^3, E^3 or H^3 geometry. REALLY! :-/ I agree. I wasn't objecting to these well-known mathematical facts. I was really objecting to the idea that "deSitter spacetime" can have flat spatial sections. It can, but only if you cut half of it off and throw it away. By the way, I am told that the Steady-State people were quite willing to do this, precisely because they didn't believe in matter conservation and actually welcomed the idea of stuff coming from "outside the universe" [ie, really, from the half of deSitter they were throwing away!] Not coincidentally, they always used the flat slicing! Each of these slicings gives different geometrical insight into the nature of the geometry of H^(1,3), but in various circumstances one may be more convenient or natural than the others. As long as nobody tries to tell me that PART OF deSitter space (either the half covered by the flat slicing or the quarter covered by the hyperbolic one) is a viable cosmology BY ITSELF, then of course I have no objections. |
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