Schwarzschild from Minkowski?

  • #1
Demystifier
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In https://arxiv.org/abs/gr-qc/0309072 Visser starts from Minkowski metric (5), performs a coordinate transformation (6)-(7) and gets Schwarzschild geometry (12). But this should be impossible. Minowski metric has vanishing Riemann curvature tensor, while Schwarzschild geometry hasn't. What do I miss?
 

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  • #2
George Jones
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In https://arxiv.org/abs/gr-qc/0309072 Visser starts from Minkowski metric (5), performs a coordinate transformation (6)-(7) and gets Schwarzschild geometry (12). But this should be impossible. Minowski metric has vanishing Riemann curvature tensor, while Schwarzschild geometry hasn't. What do I miss?
My guess on what Visser has done.

Short version: the coordinates ##\left\{ t_{FF}, x_{FF}, y_{FF}, z_{FF} \right\}## are defined on a open subset of spacetime, but (5) only holds for the metric tensor field evaluated at a particular event ##p## in this open subset. Just as the the number ##f\left(3\right)## says nothing about the derivatives of a function ##f : \mathbb{R} \rightarrow \mathbb{R}##, the metric tensor field evaluated at ##p## says nothing about its derivatives. Visser then does some manipulations and arrives at a new form of the metric field evaluated at ##p##. After this, Visser implicitly asks. "What do we get if we assume this new form of the metric is valid in some (other) open coordinate neighbourhood of ##p##?" Asking this same question of (5) would, as you have noted, give zero spacetime curvature.
 
  • #3
PeterDonis
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In https://arxiv.org/abs/gr-qc/0309072 Visser starts from Minkowski metric (5), performs a coordinate transformation (6)-(7) and gets Schwarzschild geometry (12). But this should be impossible.
Not locally, which is how he's doing it. Globally, the transformation is not constant; it changes when you pick a different event as the "origin" of the local free-fall frame. (In Visser's notation, the ##v## that appears in (7) is a function of ##r##, not a constant.) This is no different from any other curved spacetime.
 
  • #4
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Not locally, which is how he's doing it. Globally, the transformation is not constant; it changes when you pick a different event as the "origin" of the local free-fall frame. (In Visser's notation, the ##v## that appears in (7) is a function of ##r##, not a constant.) This is no different from any other curved spacetime.
I don't see how it helps. No coordinate transformation (local or global) can transform a zero tensor into a non-zero one. Yet his final metric has non-zero Riemann curvature tensor.
 
  • #5
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My guess on what Visser has done.

Short version: the coordinates ##\left\{ t_{FF}, x_{FF}, y_{FF}, z_{FF} \right\}## are defined on a open subset of spacetime, but (5) only holds for the metric tensor field evaluated at a particular event ##p## in this open subset. Just as the the number ##f\left(3\right)## says nothing about the derivatives of a function ##f : \mathbb{R} \rightarrow \mathbb{R}##, the metric tensor field evaluated at ##p## says nothing about its derivatives. Visser then does some manipulations and arrives at a new form of the metric field evaluated at ##p##. After this, Visser implicitly asks. "What do we get if we assume this new form of the metric is valid in some (other) open coordinate neighbourhood of ##p##?" Asking this same question of (5) would, as you have noted, give zero spacetime curvature.
Are you saying that (7) is not integrable, i.e. not an exact differential? If so, that would solve the puzzle.
 
  • #6
PeterDonis
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No coordinate transformation (local or global) can transform a zero tensor into a non-zero one.
This is true for an open neighborhood, yes. But strictly speaking, in the "free fall" coordinates the metric coefficients only take their Minkowski values at a single point; they are not necessarily Minkowski at other points, which means that the Riemann tensor is not necessarily zero. At least, that's how it normally works when transforming from local inertial coordinates to global coordinates in a curved spacetime.

As a heuristic, the way Visser is approaching it, the description @George Jones gives seems reasonable to me.
 
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  • #7
George Jones
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Are you saying that (7) is not integrable, i.e. not an exact differential? If so, that would solve the puzzle.
Yes, I am saying that (7) is not an equation between exact differentials. The 1-forms (cotangent vector fields) ##\mathrm d x_{\mathrm{FF}}##, ##\mathrm d x_{\mathrm{rigid}}##, ##\mathrm d t_{\mathrm{rigid}}## all exist as exact differentials (exterior derivatives of 0-forms that are coordinate functions), but (7) does not hold as a 1-form equation on the intersection of the two coordinate charts, it is an equation that holds when the cotangent vector fields are evaluated at a particular ##p##, i.e., (7) should be
$$\mathrm d \vec x_{\mathrm{FF}} \left(p\right) = \mathrm d \vec x_{\mathrm{rigid}} \left(p\right) - \vec v \left(p\right) \mathrm d t_{\mathrm{rigid}} \left(p\right)$$
for a particular ##p##.
 
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  • #9
strangerep
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In sect 2.4, Visser says that his metric in (11) is an exact solution of Einstein's equations ##R_{ab}=0##, but he doesn't give details of that calculation. If he's using an expression for ##R_{ab}## that doesn't include an object of anholonomicity, then those calculations may be wrong.

(I need to calculate the anholonomicity in the "rigid" frame to check this...)
 
  • #10
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In sect 2.4, Visser says that his metric in (11) is an exact solution of Einstein's equations ##R_{ab}=0##, but he doesn't give details of that calculation.
It is a standard textbook result that the Schwarzschild spacetime metric ##g_{\mu\nu}## satisfies ##R_{\mu\nu}=0## everywhere, except at ##r=0##. However, ##R^{\alpha}_{\beta\mu\nu}\neq 0## everywhere except at ##r\rightarrow \infty##.

Anyway, my suspicion is that Visser's result can be interpreted as a calculation of ##g_{ab}## with certain unusual tetrads, without clarifying the difference between ##g_{ab}## and ##g_{\mu\nu}##.

Additional note. It is one thing to define
$$R^{a}_{bmn}=e^a_{\alpha}e_b^{\beta}e_m^{\mu}e_n^{\nu}R^{\alpha}_{\beta\mu\nu}$$
and another thing to compute ##R^{a}_{bmn}## by pretending that ##g_{ab}## is a spacetime metric (which it is not). Since ##R^{\alpha}_{\beta\mu\nu}=0## in Minkowski spacetime, the first definition above gives ##R^{a}_{bmn}=0## while the second computation may give ##R^{a}_{bmn}\neq0##. In effect, if my interpretation is correct, Visser seems to implicitly perform the second kind of computation.
 
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  • #11
haushofer
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Maybe I'm missing something, but is this question not just "how can we relate flat spacetime to any curved spacetime by vielbeins if the curvature of flat spacetime vanishes, while the curvature of curved spacetime does not?"
 
  • #12
martinbn
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@Demystifier are you saying the following? Suppose you have a metric say

##ds^2=-(vdu+dv)(vdu+dv)+(v^2du+\sin{u} dv)(v^2du+\sin{u} dv)##

One should expanded and write it in the usual form, and check that it isn't flat.

Give names to the one forms ##\omega^1=vdu+dv## and ##\omega^2=v^2du+\sin{u} dv##. Then the metric is ##ds^2=-\omega^1\otimes\omega^1+\omega^2\otimes\omega^2##. Pretend that there are corrdinates so that ##\omega^1=dt## and ##\omega^2=dx##, then the metric is ##ds^2=-dt^2+dx^2##, in other words Minkowski.

Now do it in reverse order. Start with Minkowski and so on. I think that that is exactly what he does. Except that it isn't so arbitrary but motivated by physics, Newtonian gravity.
 
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  • #13
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@Demystifier are you saying the following? Suppose you have a metric say

##ds^2=-(vdu+dv)(vdu+dv)+(v^2du+\sin{u} dv)(v^2du+\sin{u} dv)##

One should expanded and write it in the usual form, and check that it isn't flat.

Give names to the one forms ##\omega^1=vdu+dv## and ##\omega^2=v^2du+\sin{u} dv##. Then the metric is ##ds^2=-\omega^1\otimes\omega^1+\omega^2\otimes\omega^2##. Pretend that there are corrdinates so that ##\omega^1=dt## and ##\omega^2=dx##, then the metric is ##ds^2=-dt^2+dx^2##, in other words Minkowski.

Now do it in reverse order. Start with Minkowski and so on. I think that that is exactly what he does. Except that it isn't so arbitrary but motivated by physics, Newtonian gravity.
Yes.
 
  • #14
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Maybe I'm missing something, but is this question not just "how can we relate flat spacetime to any curved spacetime by vielbeins if the curvature of flat spacetime vanishes, while the curvature of curved spacetime does not?"
When you put it in this form, it sounds as if the answer should be trivial. Do you think it's trivial?
 
  • #15
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In the attached picture I sketch the idea of post #8. Usually one starts with a curved spacetime M, and then on each point of M one defines a flat tangent space T. Alternatively, one can start with a flat spacetime M, and then on each point of M one can define a curved tangent space T.
Photo0588.jpg
 

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  • #16
haushofer
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When you put it in this form, it sounds as if the answer should be trivial. Do you think it's trivial?
Well, I wouldn't use the word "trivial" when it comes down to GR, and I always have to think about this stuff for a while :P

But as I understand it, is that the Vielbein is a local (!) coordinate transformation which transforms the metric to the Minkowski-metric and its first derivatives to zero. The corresponding coordinates correspond to inertial observers, i.e. "tangent space coordinates" . However, the second derivatives of the metric will (in general) not be zero; if they would, I think your manifold would be flat globally (!). See also page 91 and onward of Blau's notes on GR,

www.blau.itp.unibe.ch/newlecturesGR.pdf

Your question is directly adressed in the notes

Tensor Calculus, Part 2 - [email protected]


of Edmund Bertschinger, from page 2 onwards ("If spacetime is not flat, how can we reduce the metric at every point to the Minkowski form? Doesn’t that require a globally flat, Minkowski spacetime? How can one have the Minkowski metric without having Minkowski spacetime?") He's using different language then I'm used to, but maybe it helps.

-edit So the Christoffel components would be zero in inertial coordinates, but the Riemann tensor would not be (in general). In flat coordinates, one has the spin connection $$\omega_{\mu}{}^{ab}$$ and the corresponding curvature $$R_{\mu\nu}{}^{ab}(\omega)$$, which is also non-zero in general. Because this curvature is merely the Riemann curvature contracted with two Vielbeins, this works out OK.
 
  • #17
strangerep
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After sleeping on it, I think I've now got this sorted out. It's essentially as @haushofer described, but here's a little more detail specific to Visser's case...

Visser's FF "coordinates" are not well-defined as coordinates. To see this, start with Visser's (6), (7), with the spatial part transformed to spherical coordinates. We have $$ dt_{FF} ~=~ dt~~;~~~~~~ dr_{FF} ~=~ dr - v(r)dt ~.$$Thus far, the ##dt_{FF}## and ##dr_{FF}## are only vector fields. We don't know whether they can be integrated to give ##t_{FF}, r_{FF}## coords. This is only possible if $$d^2 t_{FF} = 0 = d^2r_{FF} ~.$$However, $$d^2r_{FF} ~=~ - dv \wedge dt ~=~ - v'(r) \, dr \wedge dt ~\ne~ 0 ~.$$
Alternatively, we could determine this via the dual operators. If I haven't made a mistake, these are (with an abuse of the "##\partial##" notation), $$\partial_{t_{FF}} ~=~ \partial_t + v(r) \partial_r ~~;~~~~~~ \partial_{r_{FF}} ~=~ \partial_r ~~.$$ Then evaluate the commutator: $$\left[\partial_{r_{FF}} \;,\; \partial_{t_{FF}} \right] ~=~ v'(r) \partial_r ~\ne~ 0 ~.$$ I.e., there is a nonzero anholonomicity ##\Omega## associated with the FF frame field. (This means we cannot integrate the FF vector fields to give good FF coordinates.)

Hence, even though the metric components are Minkowskian, the curvature is nonzero because of the skew-symmetric part of the connection corresponding to the anholonomicity ##\Omega##.

(The relationship between the spin connection and the anholonomicity is explained here.)

(Edit: @Demystifier, I think the idea in post #15 cannot be correct -- you can't flatten out a curved manifold in general.)
 
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  • #18
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(Edit: @Demystifier, I think the idea in post #15 cannot be correct -- you can't flatten out a curved manifold in general.)
That's not the idea of post #15. One can associate a flat tangent space with a point on a curved manifold, which does not change the fact that the manifold is curved. Note that with each point on the manifold one associates another flat tangent space. The whole construction is a tangent bundle.
 
  • #19
haushofer
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What is a "curved tangent space"? Isn't a tangent space flat per definition?
 
  • #20
strangerep
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What is a "curved tangent space"? Isn't a tangent space flat per definition?
Well, there is Finsler geometry, which is a generalization of Riemannian geometry where the metric depends on direction as well as position. One works with "horizontal--vertical" decompositions to obtain Ehresmann connection(s), one of which is the metric-compatible "Cartan connection" wherein the tangent spaces can indeed be curved.

But I'm sure that's not what Demystifier had in mind. :oldbiggrin:
 
  • #21
Demystifier
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What is a "curved tangent space"? Isn't a tangent space flat per definition?
Tangent space is usually defined as flat, but I am proposing here a generalization. For instance, I can take a flat 2-dimensional base space and construct a fibre bundle by attaching a 2-sphere with each point on the base space.

Sorry for replying after so much time, but I was reading some of my old threads and noticed this.
 
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  • #22
haushofer
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In my mothertongue we say "better let as net". 😋
 
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  • #23
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In my mothertongue we say "better let as net". 😋
In mine it would be "bolje ikad nego nikad".
 

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