Schwarzschild from Minkowski?

In summary: Visser is assuming that a metric transformation (6)-(7) is valid for a particular coordinate neighbourhood ##p## of the origin. If this assumption is not true, then (12) is not a valid Schwarzschild geometry. However, if the assumption is true, then (7) becomes an equation between exact differentials, and (12) is a valid Schwarzschild geometry.
  • #1

Demystifier

Science Advisor
Insights Author
Gold Member
14,118
6,616
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?
 
Physics news on Phys.org
  • #2
Demystifier said:
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
Demystifier said:
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
PeterDonis said:
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
George Jones said:
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
Demystifier said:
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.
 
  • Like
Likes Demystifier
  • #7
Demystifier said:
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##.
 
  • Like
Likes Demystifier
  • #9
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
strangerep said:
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.
 
Last edited:
  • #11
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
@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.
 
  • Like
Likes Demystifier
  • #13
martinbn said:
@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
haushofer said:
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
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
 

Attachments

  • Photo0588.jpg
    Photo0588.jpg
    24.1 KB · Views: 525
  • #16
Demystifier said:
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 - DSpace@MIT


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
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.)
 
Last edited:
  • Like
Likes Demystifier
  • #18
strangerep said:
(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
What is a "curved tangent space"? Isn't a tangent space flat per definition?
 
  • #20
haushofer said:
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
haushofer said:
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.
 
  • Like
Likes haushofer
  • #22
In my mothertongue we say "better let as net". 😋
 
  • Like
Likes Demystifier
  • #23
haushofer said:
In my mothertongue we say "better let as net". 😋
In mine it would be "bolje ikad nego nikad".
 

1. What is the Schwarzschild metric?

The Schwarzschild metric is a solution to Einstein's field equations in general relativity that describes the geometry of spacetime outside of a non-rotating, spherically symmetric mass. It is named after Karl Schwarzschild, who first derived the solution in 1916.

2. How is the Schwarzschild metric related to Minkowski spacetime?

The Schwarzschild metric is a generalization of the Minkowski metric, which describes flat spacetime in special relativity. In the limit of no mass, the Schwarzschild metric reduces to the Minkowski metric. This means that the Schwarzschild metric describes the curvature of spacetime due to the presence of a massive object, while the Minkowski metric describes the absence of any massive objects.

3. What is the significance of the Schwarzschild radius?

The Schwarzschild radius is a characteristic radius associated with the Schwarzschild metric. It represents the distance from the center of a non-rotating, spherically symmetric mass at which the escape velocity is equal to the speed of light. This is also known as the event horizon, and marks the point of no return for anything falling into a black hole.

4. How does the Schwarzschild metric affect the motion of objects?

The Schwarzschild metric introduces a curvature of spacetime, which affects the motion of objects by altering their trajectories. In this metric, the paths of objects are bent towards the center of the mass, similar to how objects on a curved surface will follow a curved path due to the curvature of the surface. This effect is known as gravitational lensing.

5. Can the Schwarzschild metric be used to describe all massive objects?

No, the Schwarzschild metric is only applicable to non-rotating, spherically symmetric masses. It does not take into account the effects of rotation or other asymmetries in the mass distribution. For these cases, more complex solutions to Einstein's field equations, such as the Kerr metric, are required.

Suggested for: Schwarzschild from Minkowski?

Replies
20
Views
594
Replies
11
Views
415
Replies
14
Views
722
Replies
18
Views
1K
Replies
18
Views
2K
Replies
4
Views
565
Replies
63
Views
5K
Back
Top