On Tuesday, February 23, 2016 at 7:05:13 PM UTC-6, JanPB wrote:
> This post is a continuation of the thread "Lagranian Method Revisited" [sic]
> which discussed Karl Schwarzschild's 1916 paper "Ueber das Gravitationsfeld
> eines Massenpunktes nach der Einstein'schen Theorie" (1916) (English
> translation "On the gravitational field of a mass point according to
> Einstein's theory" available at
arvix.org in physics/9905030v1).
>
> It is sometimes claimed (by the above English translators, for example) that
> the paper describes a different solution of the Einstein field equations than
> the one described currently in textbooks. Specifically, it is asserted that
> his original solution does not exhibit a black hole but merely a static
> spherically symmetric vacuum region with a pointlike mass in the centre.
>
> In this note I'd like to explain why this conclusion is incorrect and also to
> go in some detail through the relevant parts of Schwarzschild's paper in
> order to find the subtle error he made.
>
> This is in no way to diminish his achievement, in those days hardly anybody
> knew differential geometry anyway, and among physicists it was practically
> unknown.
>
> To appreciate the subtlety of the problem we need to begin with simple
> examples illustrating potential complications.
>
>
> Section I.
> METRICS AND GEOMETRIES IN MANIFOLDS WITH FIXED COORDINATE SYSTEMS
>
> In this section I'll use exclusively manifolds with the _identity map_ as
> their coordinate system. I won't discuss _other_ coordinate systems on
> manifolds until Section II.
>
> First we need to state precisely what we mean by geometry. It's important to
> keep in mind that metric tensor determines geometry but that relation is NOT
> one-to-one. For example, the following tensor describes the flat (Minkowski)
> geometry in 2D:
>
> ds1^2 = -dt^2 + dx^2
>
> But the same geometry is also described by a _different_ tensor:
>
> ds2^2 = 4 dt dx
>
> (Again, I'm stressing here that in this section the coordinate system never
> changes, so ds2^2 is a _different tensor_ than ds1^2.)
>
> The geometry is the same because the Riemann curvature of the second metric
> is identically zero (skipping the calculation here).
>
> So what determines whether two given metrics define the same geometry?
>
> Answer:
> Two metrics ds1^2 and ds^2 determine the same geometry if and only if there
> exists a _diffeomorphism_ F (of the manifold) which induces a transformation
> of one metric tensor into the other, say ds1^2 into ds2^2:
>
> F^*(ds1^2) = ds2^2
>
> The "F with the superscript star" notation stands for the transformation of
> covariant tensors induced by F. I'll define this mapping in a moment.
>
> "Diffeomorphism" here simply means "smooth map which is one-to-one and whose
> inverse is also smooth". "Smooth" here means "infinitely differentiable".
> (A bit later I'll state a useful criterion for checking whether a given map
> is a diffeomorphism that's easier to use than this definition.)
>
> For example:
> (1) f(x) = sinh(x) is a diffeomorphism of the entire real line (all x) since
> it's 1-1, infinitely differentiable for all x, and so is its inverse
> f^(-1)(x) = arsinh(x) (recall the derivative of arsinh(x) is 1/sqrt(1+x^2),
> and similar for higher derivatives, so the denominator never causes any
> trouble).
>
> (2) g(x) = x^3. This is NOT a diffeomorphism of the entire real line.
> Although it is 1-1, smooth, and invertible, its inverse is NOT smooth:
> g^(-1)(x) = x^(1/3), so its derivative is 1/(3x^(2/3)) - and the denominator
> is zero when x = 0. Nevertheless, g(x) _is_ a diffeomorphism of on the
> subset of the real line consisting of _all nonzero_ real numbers. So the
> domain is important for being a diffeomorphism or not.
>
> OK, so what does this "F^*" mean. Let's say that ds1^2 is:
>
> ds1^2 = g_ij dx^i dx^j
>
> Then F^*(ds1^2) (called the _pullback_ of ds1^2 under F) is defined as
> follows:
>
> F^*(ds1^2) = (g_ij . F) d(x^i . F) d(x^j . F)
>
> ...where by the dot "." I mean composing functions. So "g_ij.F" is F followed
> by g_ij, and "x^i.F" means "the i-th component of F" or "F followed by the
> i-th coordinate function".
>
> We then also say that the manifold with the metric ds1^2 is _isometric_ to
> the same manifold with the metric ds2^2. The map F is called the _isometry_.
> Isometric manifolds are indistinguishable geometrically and physically. The
> reason is that isometries preserve the dot product, hence the distances and
> the angles, hence the whole geometry. That's why pullbacks enter the picture
> here. I should also mention (although this is getting ahead of myself a bit)
> that curvature tensors of metrics related by a pullback are themselves
> related by pullback which in our case means that if a ds^2 satisfies the
> Einstein equation, so does F^*(ds^2) for ANY diffeomorphism F.
>
> All this pullback business should become clear by examining the Minkowski
> example from a minute ago:
>
> ds1^2 = -dt^2 + dx^2
> ds2^2 = 4 dt dx
>
> In this case they are isometric because ds2^2 is the pullback of ds1^2 under
> the following diffeomorphism F of the (t,x)-plane:
>
> F(t, x) = (x - t, x + t)
>
> Let's check it using the definition of the pullback (note that the
> compositions are: -1.F = -1 and 1.F = 1):
>
> F^*(ds1^2) =
>
> = F^*(-dt^2 + dx^2) =
>
> = -d("t-component of F")^2 + d("x-component of F")^2 =
>
> = -d(x - t)^2 + d(x + t)^2 =
>
> = -(dx - dt)^2 + (dx + dt)^2 =
>
> = -dx^2 - dt^2 + 2 dt dx + dx^2 + dt^2 + 2 dt dx =
>
> = 4 dt dx =
>
> = ds2^2
>
> The silly factor 4 can be removed by one more pullback via the following
> diffeomorphism G:
>
> G(t, x) = (t/2, x/2)
>
>
> G^*(ds2^2) =
>
> = G^*(4 dt dx) =
>
> = 4 d(t/2) d(x/2) =
>
> = 4 1/2 dt 1/2 dx =
>
> = dt dx
>
> So ds3^2 = dt dx is _also_ the flat Minkowski space (this too can be verified
> directly by computing the curvature of ds3^2 by hand).
>
> OK, so this was the easy part. Now the subtle part begins. Let's imagine a
> slightly trickier diffeomorphism:
>
> H(t, x) = (t, 1/x)
>
> Notice that its domain isn't the whole (t,x)-plane, it's the plane with the
> set of points where x = 0 (i.e., the t-axis) _removed_.
>
> H a diffeomorphism there because it's clearly 1-1 with both F and its inverse
> infinitely differentiable at each x =/= 0.
>
> BY THE WAY: there exists a very good criterion for verifying whether a given
> map is a diffeomorphism that does NOT require computing the inverse of the
> map: it's called "the inverse function theorem" and is one of the two pillars
> of real analysis. (The other pillar of real analysis, incidentally, is
> "Lebesgue dominated convergence theorem" - but I digress.)
>
> That theorem goes like this: in order to check F is a diffeomorphism all you
> need to verify is:
>
> 1. F is 1-1,
> 2. F is smooth,
> 3. the Jacobian of F is nonzero _at every point of the domain of interest_.
>
> No need to invert F, very handy-dandy. It's easy to check that all our
> diffeomorphisms so far satisfy this criterion [exercise].
>
> Let's now calculate the pullback (call it ds4^2) of the Minkowski ds1^2 under
> the above H:
>
> ds4^2 =
>
> = H^*(ds1^2) =
>
> = H^*(-dt^2 + dx^2) =
>
> = -dt^2 + d(1/x)^2 =
>
> = -dt^2 + [ (-1/x^2) dx ]^2 =
>
> = -dt^2 + dx^2/x^4
>
> So this is also the flat Minkowski space (and yet again, one can verify this
> directly by calculating the curvature by hand).
>
> BUT there is now a kink: the metric ds4^2 is undefined at x = 0! To be sure,
> that new tensor is flat wherever it's defined but it's _not defined_ at x = 0!
>
> BINGO.
>
> Here we've just encountered the standard phenomenon of the trade: the _same
> geometry_ represented by two tensors _whose domains do not coincide_. In this
> case ds1^2 is defined everywhere in the (t,x)-plane but ds4^2 isn't: one of
> its coefficients (1/x^4) blows up.
>
> In other words: ds4^2 has a singularity at x = 0 but the geometry (and the
> manifold) does NOT. (This ought to sound familiar...)
>
> AGAIN:
>
> The tensor ds4^2 _DOES NOT_ extend over x = 0 (the t-axis), it has a
> _genuine_ singularity there, but it is the pullback of a tensor that DOES
> extend. (Namely, the ds1^2 one.) So the geometry is well-defined, including
> at x = 0. The singular behaviour of ds4^2 is only due to choosing a slightly
> misbehaving isometry to transform one into the other.
>
> So:
> A geometry does not have to stop where a metric tensor defining it stops. It
> may be possible the geometry is simply a restriction of a larger geometry
> which can be seen by selecting an appropriate pullback.
>
> In our case, this geometry extending process would go something like this:
> someone hands you over the metric ds4^2, tells you that it's been written
> with respect to the canonical coordinates, you see it's not defined at x = 0
> but for some reason or other you suspect the singularity is not really
> present in the geometry (say, your reason is that you've calculated the
> curvature of ds4^2 and obtained a value (namely, zero!) which does NOT blow
> up at x = 0). To prove it we need to produce an explicit pullback that
> extends over x = 0. There is no general recipe for doing this other than
> educated guesswork. In our case we know what to do, of course, given that we
> know where ds4^2 came from. We set the following diffeomorphism (the inverse
> of H, which happens to look the same as H):
>
> K(t, x) = (t, 1/x).
>
> Then:
>
> K^*(ds4^2) =
>
> = K^*(-dt^2 + dx^2/x^4) =
>
> = -dt^2 + d(1/x)^2 / (1/x)^4 =
>
> = -dt^2 + (-1/x^2)^2 dx^2 x^4 =
>
> = -dt^2 + dx^2
>
> ...and this is defined on the whole plane, so is the geometry.
>
> Time for two more examples illustrating certain notable patterns.
>
> Example A:
>
> The location of tensor singularity is not fixed, it can change in a
> pullback. Let's look at ds4^2 one more time:
>
> ds4^2 = -dt^2 + dx^2/x^4
>
> It's singular at x = 0. But consider a really basic diffeomorphism:
>
> L(t, x) = (t, x + 1)
>
> Pulling back:
>
> L^*(ds4^2) =
>
> = -dt^2 + [d(x + 1)]^2/(x + 1)^4 =
>
> = -dt^2 + dx^2/(x + 1)^4
>
> Now we got a tensor describing the Minkowski geometry which is singular at
> x = -1. This is borderline trivial but worth remembering: the coordinate
> value of the singularity is not a fixed property of the tensors defining
> the geometry.
>
> Example B:
>
> Minkowski ds1^2 again:
>
> ds1^2 = -dt^2 + dx^2
>
> Now let's look at this easy diffeomorphism:
>
> M(t, x) = (t, 2x)
>
> Therefore,
>
> M^*(ds1^2) =
>
> = -dt^2 + d(2x)^2 =
>
> = -dt^2 + 4 dx^2
>
> Well, nothing deep here, it would work the samy way for any nonzero number
> instead of 2 just the same:
>
> N(t, x) = (t, Cx) where C is any nonzero constant (you should know by
> now why we insist on C =/= 0 here)
>
> Therefore,
>
> N^*(ds1^2) =
>
> = -dt^2 + d(Cx)^2 =
>
> = -dt^2 + C^2 dx^2
>
> Now we got a tensor describing the Minkowski geometry which apparently
> depends on an _extra constant_. This is again borderline trivial but worth
> remembering: a constant appearing in metric tensor coefficients does NOT
> necessarily have ANY physical significance! It can be simply an artifact of
> the method used to calculate the given tensor and in such cases can be
> "pulled back away". Here it can be naturally "pulled back away" by the
> inverse diffeomorphism:
>
> N^(-1)(t, x) = (t, x/C)
>
>
> It would be nice if we could now start discussing Schwarzschild's paper but
> "unfortunately" we need to cover one more degree of freedom: coordinate
> changes. This is necessary because Einstein, Schwarzschild, and modern
> textbooks all use this concept. I said "unfortunately" because the "single
> coordinate system + pullbacks" approach used in the previous section is IMHO
> clearer to understand and to use, esp. when tricky subjects like
> singularities and geometry extensions arise. OTOH coordinate changes
> introduce a certain extra degree of topological complication, you'll see what
> I mean in a minute.
>
>
> Section II.
> METRICS AND GEOMETRIES IN DIFFERENT COORDINATES
>
> This is in a sense a "dual" degree of freedom: given a metric tensor written
> with respect to some coordinates, we can keep it _unchanged_ but instead
> change the linear bases of the tangent spaces of the manifold and expand the
> tensor at each point in those new bases.
>
> Warning: it turns out BOTH approaches produce FAPP same-looking formulas, so
> one must be specific and always state clearly what a given formula
> represents: _two different tensors_ (one before, one after the pullback) in a
> _fixed_ coordinate system OR _one tensor_ written with respect to _two
> different coordinate systems_.
>
> From the point of view of geometry or physics, both approaches are equivalent
> since they produce different but _isometric_ manifolds.
>
> NB: The pullback approach of Section I is sometimes referred to as "active
> view" and the coordinate change approach of this section as "passive view".
> The terminology suggests the former view changes the object in question (the
> tensor) while the latter one does not.
>
> Let's again consider one of the metrics from Section I to see how this
> approach works:
>
> ds1^2 = -dt^2 + dx^2
>
> ...written, as before, in the canonical coordinates (t, x). Let's now
> introduce, for the first time in this posting, _new_ coordinates, call them
> (T, X), defined as:
>
> T = (x - t)/2 (*)
> X = (x + t)/2
>
> i.e.:
>
> t = X - T
> x = X + T
>
> Hence:
>
> dt = dX - dT
> dx = dX + dT
>
> So we obtain:
>
> ds1^2 =
>
> = -(dX - dT)^2 + (dX + dT)^2 =
>
> = -dX^2 -dT^2 + 2 dT dX + dX^2 + dT^2 + 2 dT dX =
>
> = 4 dT dX
>
> This looks almost identical to the calculation in Section I, the mathematical
> transformation process is the same (except for using the upper-case letters)
> but the meaning and the objects involved are very different:
>
> --> in Section I we had _two different tensors_, ds1^2 and F^*(ds1^2), both
> written in terms of _the same coordinate system_. Here we have _one tensor_,
> ds1^2, written in terms of _two different coordinates_: the first system,
> (t,x), being the canonical one, and the second system, (T,X), being defined
> by the formulas (*) above.
>
> In all cases we have ONE geometry: in Section I that's because ds1^2 and
> F^*(ds1^2) are related by a pullback, hence they are isometric, and in this
> section we have just one tensor, so one geometry by definition.
>
> Let's do one more example, again similar to something in Section I, to show
> how tensor domains can change in different coordinates. This is where it gets
> a bit hairy. Start with ds1^2 again:
>
> ds1^2 = -dt^2 + dx^2
>
> Change the coordinates like so:
>
> T = t
> X = 1/x (obviously defined for x =/= 0 only)
>
> This means:
>
> t = T
> x = 1/X
>
> therefore (this should look annoyingly familiar by now):
>
> ds1^2 =
>
> = -dT^2 + d(1/X)^2 =
>
> = -dT^2 + [ -1/x^2 dX ]^2 =
>
> = -dT^2 + dX^2/X^4
>
> Again:
>
> 1. the mathematical manipulation looks practically the same as in Section I
> except for the upper-/lower-case distinguishing the coordinates,
>
> 2. the same geometry throughout (the flat Minkowski).
>
> But here comes a subtle point which I never see properly discussed in
> textbooks, even the unusually careful Robert Wald glosses over it. So let's
> look at another, deceptively similar scenario: someone gives you a metric
> like the one above but written instead in the _standard_ coordinates (t, x):
>
> ds^2 = -dt^2 + dx^2/x^4
>
> Remember, these coordinates are well-defined everywhere, so the blowing up of
> the coefficient 1/x^4 at x = 0 means the tensor ds^2 _itself_ blows up there
> too.
>
> No matter, we can change the coordinates anyway, why not try the (T, X)
> again. We get, by the same type of calculation as before:
>
> ds^2 = -dT^2 + dX^2
>
> ...and this expression does extend to X = 0. Wait a minute, didn't we just
> say that ds^2 blows up? We haven't changed ds^2, only the coordinates. So
> how come it suddenly doesn't blow up at zero?
>
> The answer is that we extend it over X = 0 which in the original coordinates
> does NOT correspond to _any point_ in the (t,x)-plane! (since 0 cannot equal
> 1/x) So where on earth does this ds^2 extend _to_? Where _is_ this point X =
> 0 if it's nowhere in the (t,x)-plane?
>
> What really happens is that we are extending the original expression (-dt^2 +
> dx^2/x^4) over _an extra set of manifold points which wasn't there before_.
> Namely we extend -dt^2 + dx^2/x^4 over the extra "line at x = +/- infinity"
> while at the same time removing the "line at x = 0" (the t-axis) over which
> -dt^2 + dx^2/x^4 _does not_ extend.
>
> The resulting manifold consists of a slightly different set of points but
> it's _isometric_ to the Minkowski plane.
>
> You can imagine the original (t,x)-plane wrapped around a cylinder (the
> t-axis pointing in the direction of the cylinder's axis) so that the two
> extremities of the plane where x approaches plus and minus infinity end up
> _just_ shy of the vertical line on the other side of the cylinder, opposite
> the t-axis x = 0. It is THAT vertical line which we add while removing the
> line x = 0. The result is clearly an isometric copy of the entire Minkowski
> plane but sort of "turned inside out".
>
> This sort of thing is precisely what is involved in extending the standard
> modern textbook Schwarzschild metric (typically written in the spherical
> coordinates space) over the horizon by switching to, say, the
> Eddington-Finkelstein coordinates: the metric does not extend over the space
> r = 2M but instead extends over a _copy_ of that space attached _differently_
> (namely, attached as limit points of the infalling (say) geodesics on the
> usual (t,r)-diagram).
>
> (BTW, this type of cutting and pasting parts of manifolds is called "surgery"
> and there is a HUGE field of topology devoted to it. GR textbooks avoid
> discussing this, presumably because this operation isn't much done in GR
> otherwise and ignoring it is of no consequence 99.9% of the time.
> Historically this sort of manifold manipulations and manifold extension arose
> for the first time in the context of complex analysis in the study of Riemann
> surfaces.)
>
> You can see now why I prefer the "active" approach of Section I where
> typically there is no need for surgery, just transform tensors by pullbacks
> as needed which keeps the geometry unchanged and the coordinate systems
> fixed, so less confusion.
>
> Finally:
>
>
> Section III.
> SCHWARZSCHILD'S PAPER
>
> Here we'll find out why the solution described in this paper is identical to
> the one described in all textbooks, i.e. why it describes the usual black
> hole geometry, including the horizon and the inner, non-static, region.
>
> Let's begin with a quick summary of his method. Then we'll stop at the
> "smoking gun moment" where he jumps to a conclusion too fast.
>
> He begins by writing down the vacuum field equations, Ricci = 0, for any
> coordinate system in which the determinant of the metric is identically -1.
> Under this restriction the field equations simplify quite a bit (to half
> their "native" size), he writes this as his equation (4). He then proceeds to
> construct such coordinate system.
>
> First he quotes Einstein's 1915 paper "Erklaerung der Perihelbewegung des
> Merkur aus der allgemeinen Relativitaetstheorie" ("Explanation of the
> Perihelion Motion of Mercury from General Relativity Theory", a rather bad
> English translation is available here:
>
http://www.gsjournal.net/old/eeuro/vankov.pdf) by stating that "according to
> Mr. Einstein's list" the following conditions must be satisfied:
>
> 1. all metric components are independent of the time x_4 (he denotes the time
> coordinate by x_4),
>
> 2. g_i4 = g_4i = 0 for i = 1, 2, 3
>
> 3. the solution is spatially spherically symmetric wrt to the origin,
>
> 4. boundary conditions at infinity: g_44 = 1, g_11 = g_22 = g_33 = -1.
>
> So right away we are faced with the question: what exactly is the basis for
> claims 1 and 2? The field equations are a complicated system of 10 coupled
> PDEs for 10 unknown functions of four variables, why would merely assuming
> spherical symmetry and staticity imply 1 and 2 just like that? This
> conclusion seems wrong and it's easy to concoct a counterexample in fact:
>
> Consider the 4D vacuum with the following metric written in spherical
> coordinates:
>
> ds^2 = -cosh^2(t) dt^2 - 2 cosh(t) dt dr + r^2 dO^2
>
> ...where "dO^2" is the shorthand for the line element of the unit sphere:
> dtheta^2 + sin^2(theta) dphi^2.
>
> This ds^2 has BOTH time-dependent coefficients AND a cross-term (dt dr), so
> it violates conditions 1 and 2 in Schwarzschild's paper, yet it is the
> Minkowski geometry AGAIN, in particular a spherically symmetric static
> vacuum.
>
> To see this, just calculate its curvature or pull ds^2 back via the following
> map (easily seen to be a diffeomorphism by the inverse function theorem
> quoted earlier):
>
> F(t, r, theta, phi) = (arsinh(t - r), r, theta, phi)
>
> (Or use the equivalent "passive" change of coordinates: T = r + sinh(t), R =
> r, THETA = theta, PHI = phi.)
>
> So just saying "the solution is spherically symmetric and static" does not
> seem to _guarantee_ 1 and 2.
>
> Unfortunately many texbooks continue to sloppily insist that it is so, for
> example Ohanian & Ruffini, 2nd edition, p.392: "The demand for a static
> solution forbids terms of the type dr dt, dtheta dt, dphi dt in the
> expression for the spacetime interval because those terms change sign when we
> perform the time reversal t -> -t".
>
> But of course _nothing_ changes when we perform the time reversal, in fact we
> can "perform" ANY diffeomorphism whatsoever without affecting the geometry at
> all! What the authors really have in mind is that the _tensor itself_ should
> be invariant under such transformations. But this too is fishy because
> spherical symmetry refers to the 3D _spatial_ slices of spacetime. QUESTION:
> which ones? OK, so let's say "the spatial slices t = const. where "t" is one
> of the canonical (identity) coordinates on R^4. Well, actually, not quite,
> because due to spherical symmetry we want to use spherical coordinates in
> place of the canonical (x,y,z). So we really mean the spherical coordinate
> system on each spatial slice t = const.
>
> Well, it's flaky but it kind of works. Now the "static" part. A metric tensor
> in the above coordinates is static if its coefficients are t-independent.
>
> As you can see, the above definition is sloppy. But it's good enough
> for us here. Good modern texts like Wald or d'Inverno or Hawking & Ellis
> define these things properly, in a way that captures the physics and the
> geometry in a way that does not presume any coordinates. I won't dwell
> more on that point.
>
> So now Schwarzschild begins solving the Einstein equation by positing his
> condition 1 and 2 as an Ansatz: seek the solution in the following form,
> written in the above system consisting of spherical coordinates on slices
> t = const (he sometimes denotes time by "t", sometimes by "x_4"). This is
> Schwarzschild's equation (6):
>
> ds^2 = A(r) dt^2 - B(r) dr^2 - C(r) (dtheta^2 + sin^2(theta) dphi^2)
>
> ...where I took the liberty of denoting by A(r), B(r), C(r) certain functions
> of r he momentarily keeps written out explicitly and which arise from his
> Cartesian-to-spherical switch (their exact form is unimportant here).
>
> He finally makes one more coordinate change: from (t, r, theta, phi) to
> (x_1, x_2, x_3, x_4) [he reuses the letters which at the beginning of his
> paper denote the Cartesians] like so:
>
> x_1 = r^3/3, x_2 = -cos(theta), x_3 = phi, x_4 = t
>
> ...with the obvious ranges inherited from the spherical coordinates:
>
> x_1 > 0, -1 < x_2 < 1, 0 < x_3 < 2pi, -infinity < x_4 < +infinity
>
> In these new coordinates he writes the final general form of the desired
> solution and gets on with the actual solving the equation. That final general
> form is (his equation (9)):
>
> ds^2 =
>
> = f_4 dx_4^2 -
>
> - f_1 dx_1^2 -
>
> - f_2/(1 - x_2^2) dx_2^2 -
>
> - f_3 (1 - x_2^2) dx_3^2
>
> f_1, f_2, f_3, f_4 are the unknowns to solve for.
>
> All this means that _I'm retracting_ my previous claim that an extra
> coordinate change is needed beyond those mentioned above _in the static
> case_. When I was discussing this with Koobee, I had in mind the general case
> of the spherically symmetric geometry (i.e., no assumption of static). In
> that case one cannot posit that _all_ cross-terms disappear: the dt dr term
> has to stay BUT one can then construct an ADDITIONAL change of coordinates
> which removes it. This is the coordinate change I was talking about.
> Nevertheless, this additional coordinate change is not relevant anyway
> because as it turns out it is a _further_ change down the road that's
> responsible for moving the points surrounding the r = 0 set _away_
> from the origin.
>
> Remember this is the reason for this post: to show r = 0 ceases to be
> "pointlike" in Schwarzschild.
>
> OK, so so far we still have: r is the polar coordinate with r = 0 denoting
> the origin, and x_1 = r^3/3, so 1_4 has the same property.
>
> It looks like we're stuck with the "central pointlike mass at r = 0"
> but hang on.
>
> Let's continue following Schwarzschild. He plugs the above ds^2 in the field
> equations and solves them for the above unknown metric coefficients f_1, f_2,
> f_3, f_4. These solutions are his equations (10), (11), (12).
>
>
> THE SMOKING GUN
> And at this point, right after his equation (12), he stumbles. He notes that
> one of the metric coefficients (f_1) is discontinuous and he _immediately_
> assumes this discontinuity _is_ the pointlike mass at the origin! Well, in
> 1916 it must have seemed perfectly sensible: what OTHER singularity IS there??
>
> But now we know, and we've seen examples of this earlier, that _a singularity
> of a metric tensor does not necessarily imply the singularity of the
> geometry_! There may exist a pullback which will change the tensor to
> another one with a different domain which may well cover the supposed
> geometric singularity. (Equivalently: there may exist a coordinate change...
> etc. etc., you know the rest).
>
> In fact, his "stumble" appears for the first time a bit earlier, on page 3
> [English translation] where he lists conditions his f-coefficients should
> fulfil, specifically his condition 4 which insists on:
>
> "Continuity of the f, except for x_1 = 0".
>
> This is incorrect as matric tensor singularities do not necessarily arise
> from geometry singularities and therefore are NOT restricted to these
> locations.
>
> So his conclusion that rho = alpha^3 is likewise jumping the gun. Instead, he
> should have noticed that one of the two constants (rho) was _redundant_, just
> like the constant C in our Example B at the end of Section I. In other words,
> with the right diffeomorphism the constant rho could be absorbed by the
> pullback, with the tensor domain adjusted accordingly. This is the freedom
> allowed by the fact that Einstein's equation is a PDE for _unknown tensors_,
> not just a regular PDE for _unknown functions_. This would have left him with
> only one constant (alpha) which then could have been shown tied to the mass
> of the central "body". Of course Schwarzschild was hampered in his
> pullback/coordinate freedom by that constraint of the unit determinant.
>
> Let's see how this constant absorbing would've worked had Schwarzschild done
> it in the paper. Schwarzschild's metric ds^2 is hard to write down in ASCII,
> so I'm going to provide here a link to a properly typeset PDF page, but the
> point is that one can transform rho away by using the following simple
> diffeomorphism:
>
> F(x_1, x_2, x_3, x_4) = ((x_1 - rho)/3, x_2, x_3, x_4)
>
> Here is the link to the PDF which is much easier on the eyes:
>
https://drive.google.com/file/d/0B2N1X7SgQnLRY1ExWUczZldYZXM
>
> Summing up in ASCII: pulling Schwarzschild's ds^2 back by F yields:
>
> F^*(ds^2) =
>
> = (1 - alpha*x_1^(-1/3)) dx_4^2 -
>
> x_1^(-4/3)/(1 - alpha*x_1^(-1/3)) * (1/3)^2 dx_1^2 -
>
> (x_1^(2/3))/(1 - x_2^2) dx_2^2 -
>
> (x_1^(2/3))*(1 - x_2^2) dx_3^2
>
> ...with the x_1 coordinate of the pullback satisfying: x_1 > rho (recall
> Schwarzschild's original x_1 was defined as the polar r^3/3, hence his x_1
> satisfied x_1 > 0).
>
> His original singularity of ds^2 was at (his) x_1 = (alpha^3 - rho)/3.
>
> Thus the above pullback F^*(ds^2) has the same singularity (unless it happens
> to extend, which it does not) at:
>
> x_1 = 3((the old singularity location) + rho =
>
> = 3((alpha^3 - rho)/3 + rho) =
>
> = alpha^3
>
> ...which is a strictly positive number. THIS is the coordinate change that
> moves the "r = 0" away from the origin.
>
> Since x_1 > rho, where rho is any constant, we are free to extend the metric
> as far "down" as we please but another glance at F^*(ds^2) shows that the
> metric has another singularity at x_1 = 0 (because of the cube roots of x_1
> in the denominators). A curvature calculation then shows it's actually a
> genuine singularity of the geometry (and you know the rest).
>
> As usual, a completely analogous reasoning can be made using the "passive"
> (coordinate change) viewpoint.
>
> PS:
> Another reason why Schwarzschild's set r = 0 is not a point is that if one
> surrounds it by a sphere, it's surface area does NOT go to zero as r
> approaches 0 (it approaches 4pi alpha^2 which is a nonzero number since alpha
> is nonzero whenever the mass is nonzero). For a proof check this link:
>
https://drive.google.com/open?id=0B2N1X7SgQnLRV1otWnE2T2F6aEk
>
> --
> Jan
MHPF is called the isometry.
Isometric manifolds are indistinguishable geometrically and physically.
Pullbacks, dot products, diffeomorphisms, inverse function theorem. 4pi alpha ^2 which is a nonzero number- which is 10,000.