Curvature of space; curvaure of spacetime

[jonmtkisco]

Hi smallphi and Garth,

Could one of you please confirm whether I am correct in understanding that, in an overall flat FLRW universe, both the deflection of light rays around a gravitating mass and the related Shapiro time delay involve spacetime curvature but not spatial curvature? My understanding is that these phenomena are explained by the combination of the deflection caused by travel through a Newtonian field potential, together with GR grativational time delation. Without requiring any geometrical curvature of a spacelike slice. True?

Does the Schwarzschild vacuum metric calculate any curvature near a point mass that can be identified specifically as spatial curvature rather than spacetime curvature?

Jon

 

[smallphi]

Deflection of light rays and Shapiro time delays about spherical mass are about Schwarzschild metric which is a vacuum solution of Einstein eq - there is vacuum outside the mass - and all components of the Riemann tensor are zero there hence the spacetime curvature, the Ricci scalar is zero too. As a spacetime of zero spacetime curvature, probably it can be foliated in hypersurfaces of different signs of spatial curvature. The original Schwartzschild coordinates are not in FRW form, but I've seen them somewhere transformed to FRW type of metric with scale factor changing in 'time'.

 

[Garth]

Deflection of light rays and Shapiro time delays about spherical mass are about Schwarzschild metric which is a vacuum solution of Einstein eq - there is vacuum outside the mass - and all components of the Riemann tensor are zero there hence the spacetime curvature, the Ricci scalar is zero too.

If the components of the Riemannian are all zero at a certain event then at that event there is no gravitational field, i.e. no tidal forces and no inertial frame acceleration. This is clearly not the case in a true vacuum outside the spherical mass of the Schwarzschild spherical solution.

T_{\mu\nu} = 0 for all \mu , \nu does not mean that R_{\mu\nu} = 0 or R = 0, just that R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R = 0.

Similarly R_{\mu\nu} = 0 does not mean that R_{\mu\nu\omega\sigma} = 0 just that the double trace R^{\mu\nu}_{\mu\nu} = 0

jonmtkisco, even in a cosmologically flat FLRW universe the perturbations of 'Schwarzschild' masses cause local space-time and spatial curvature.

The time dilation difference between two rays of light passing either side of a gravitational lens is one method of probing cosmological parameters.

Garth

 

[MeJennifer]

Spacetime curvature occurs locally in the presence of gravity; i.e. mass-energy. Spacetime curvature causes timelike geodesics to be displaced from a "straight" line, towards a source of mass-energy, but only as function of the passage of time. At any given fixed instant in time, spacetime curvature does not imply any spatial curvature. For example, the spacelike coordinate geometry can be perfectly flat, even if located very close to an intense gravitational source such as a black hole. In that scenario, the circumference of a circle remains 2\pi r.

In reading what you write I suspect you understand something incorrectly about spacetime and its components. It is true that spacetime is curved by mass-energy, one can state that in absolute terms. However, how this curved spacetime is decomposed into "space" and "time" components is entirely observer dependent. Furthermore, with that in mind, how then curvature is assigned to the individual components is entirely coordinate dependent.

 

[smallphi]

Yes sorry, I constantly forget that the Riemann tensor of Schwarzschilld metric is not zero because it's not Minkowski, but the Ricci tensor, the curvature scalar, and the Einstein tensor are all zero.

T_{\mu\nu} = 0 for all \mu , \nu does not mean that R_{\mu\nu} = 0 or R = 0, just that R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R = 0.

Zero energy-momentum tensor does imply zero Ricci tensor and curvature scalar. That follows from the 'trace inverted' version of Einstein eq:

R_{\mu \nu} =8\pi G (T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu})

where T is the trace of energy-mom.

 

[jonmtkisco]

Hi smallphi, Garth & MeJennifer,

I understand that using different coordinates can cause different foliations of spacetime slices.

However, as I understand it, 4-dimensional "spacetime" is not a tangible, physical "geometry" per se, it is merely a particularly convenient abstraction for organizing coordinate systems as a function of the passage of time. In my mind, there has to be some sensible convention for separating spacelike and timelike slices which respects the geometric integrity of space at a chosen instant of time regardless of which particular clock is referred to.

Let's assume we're using the coordinate system of a distant observer whose inertial frame is approximately at rest with respect to the spherical central mass in the Schwarzschild metric. Assume that in this frame, the spherical mass is at rest for a long period of time. At any chosen single instant on the distant observer's clock (which particular instant is irrelevant), the observer measures (takes a "snapshot") of whether C = 2 \pi r over a small set of transverse spatial coordinates very near to the central mass. If it does, then isn't the observer justified in concluding that the spatial curvature is zero?

It seems to me that even if this observer's coordinates are set in an inertial frame which is moving with respect to the rest frame of the mass, or in an accelerating frame (including but not limited to a frame in freefall with respect to the mass), at any arbitrary instant in time the observer will obtain a consistent spatial snapshot of whether C = 2 \pi r over a small set of spatial coordinates near the central mass.

It seems reasonable for the observer to conclude that the spatial curvature is flat, while the spacetime curvature is curved. The curvature of spacetime constituting merely an abstraction of the idea that the timelike or null geodesic of a particle in freefall will be displaced away from a "straight" line as a function of the passage of time. If exactly zero time elapses on any given clock, then exactly zero time must elapse on every possible clock everywhere (even if those clocks do not register that instant in time to be "simultaneous" in their various time metrics), so the instantaneous spacetime curvature everywhere in the Schwarzschild gravitational field must be zero regardless of the coordinate system used. In that sense, I would consider "instantaneous spacetime curvature" to equate to "spatial curvature." Again, I'm assuming that the physical configuration of the central mass remains perfectly constant over a longer time period than is necessary for all possible observers to take their measurement snapshot.

Apparently all of the above is incorrect, so please explain why as explicitly as you can. Generic references to Riemann manifolds and Ricci tensors alone don't seem to provide an intuitive physical explanation.

Jon

 

[MeJennifer]

In my mind, there has to be some sensible convention for separating spacelike and timelike slices which respects the geometric integrity of space at a chosen instant of time regardless of which particular clock is referred to.

Note that there is a distinction between the two terms spacelike and timelike and the space and time decomposition of spacetime.

With regards to your other comments note that coordinate charts do not represent physical realities but instead produce a particular space and time slice or spacetime. This should lead to the conclusion that there is no one physical sense of a separate absolute space and time.

Unfortunately many cosmologists or relativists simply ignore that and talk about characteristics of a given coordinate chart as if they were physical phenomena.

So rather than searching for the actual space and time representation of the universe think of spacetime as the conglomerate of all possible space and time representations that our universe encompasses.

 

[Cuetek]

Hi Mike,... I would like to keep this thread focused on the specific mechanics of spatial curvature, so I encourage you to move your material to a new original thread on this forum... Jon

Sorry about that, Jon. I'll yank it. I'm new and haven't got the etiquette down yet. But indulge me for a moment. In terms of your original question about the difference between locally identifiable gravitational effects on the tensor equations and general vast-scale curvature issues of "the universe," the presumption of extra-Big Bang structures is perhaps pertinent.

In that there is a high (Bayesian) probability of such structures beyond the scale of the Big Bang, but not beyond exerting slight effects in the curvature of Big Bang space, the question becomes not about the curvature of "everything," but about the curvature of mega-regional space. Sure, it's almost philosophical when the hypothetical effects from such extra-Bang structures are near-zero across the scale of the visible universe. But you guys become philosophers by default when you speak about the curvature of "the universe."

What I'm saying is that, the difference between believing that you are talking about mega-regional effects and believing that you are talking about "the universe" is an idea that is constantly presumed in favor of the latter at the expense of what I suggest is the more likely presumption of the former. This distinction makes a great deal of difference in what you expect from the next data set which speaks your aspirations as physicists.

Instead of looking for permanent structural unknowns looming vaguely in the periphery of all our models, we look for conformity to a simplified model. That's fine when the data will arrive in due time to correct our thinking. But as new data becomes ever more difficult and expensive to acquire at such scales, our presumptions about the most probable nature of the data we do have as a local manifestation under much wider influences than we can currently detect (just like it has always been for us humans) becomes ever more essential for us to presume.

It is this implication of my otherwise off-topic post that speaks directly to your original question about the difference between local curvature and broad-scale curvature, and I leave it to you guys (and MeJennifer) who have the math chops to actually affect the process to ponder in a way that has any chance at all of taking mere philosophy and making it science.

 

[smallphi]

A local observer can define only a local physical coordinate frame in his/her vicinity. Global coordinates are defined by an infinite family of observers whose world lines thread the whole spacetime - for example the 'fundamental' comoving observers in the case of FRW metric. So a single observer at infinity won't be able to measure anything close to the spherical mass directly.

 

[Garth]

Zero energy-momentum tensor does imply zero Ricci tensor and curvature scalar. That follows from the 'trace inverted' version of Einstein eq:

R_{\mu \nu} =8\pi G (T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu})

where T is the trace of energy-mom.

Yes of course, my mistake; that is what comes of posting quickly before rushing off to work!

Garth

 

[jonmtkisco]

In that there is a high (Bayesian) probability of such structures beyond the scale of the Big Bang, but not beyond exerting slight effects in the curvature of Big Bang space, the question becomes not about the curvature of "everything," but about the curvature of mega-regional space.

Hi Cuetek,

Yes, I agree that a mega region may well be different from "everything".

Jon

 

[jonmtkisco]

... [N]ote that coordinate charts do not represent physical realities but instead produce a particular space and time slice or spacetime. This should lead to the conclusion that there is no one physical sense of a separate absolute space and time.

Hi MeJennifer and smallphi,

I understand the general sense of what you're saying, but I don't understand how it works.

Are you literally saying that observers in various nearby frames who are observing some test particles in freefall within a small area of local space close to an isolated central mass will disagree with each other as to whether a measurement of C = 2 \pi r holds in that region? I don't see how that could be so. Could you please cite a couple examples of nearby frames and explain how they would disagree about this physical observation? I understand that observers in different frames may disagree about clocks, simultaneity, and acceleration measurements, but I do not see how they could disagree about the basic spatial curvature measurement I described.

Jon

 

[MeJennifer]

They do not disagree on the spacetime curvature but they may disagree about how space or time separately is curved. Note the situation in the case of flat spacetime, different observers do not disagree on the spacetime interval but they may disagree about the space and time intervals separately.

 

[smallphi]

"Exploring black holes" by Taylor and Wheeler

contains all the answers. It's written for undergraduates so knowing GR is not required. They discuss different observers around the black hole.

 

[jonmtkisco]

Hi smallphi and MeJennifer,

I'm going to read Taylor & Wheeler when I can get my hands on it.

My understanding is that it is always possible to choose coordinates which generate a metric that is flat (Gallilean) along spatial hypersurfaces at constant time. Therefore, following Hamilton & Lisle, we should be able to interpret any element which appears in other coordinates to be a spatial curvature, as in fact constituting nothing more than a SR Lorentz contraction in the flat Gallilean coordinate system. In that sense, we could be justified in identifying such a spatial geometry as intrinsically flat, even though other interpretations are possible as well.

Jon

 

[MeJennifer]

My understanding is that it is always possible to choose coordinates which generate a metric that is flat (Gallilean) along spatial hypersurfaces at constant time.

All you get with that is an observable reality for a certain class of observers.

The mistake is to think that should be equalized with an universal reality (as some in cosmology do).

 

[jonmtkisco]

MeJennifer, perhaps you're right to characterize my observation as a "mistake", but my point is that maybe there is no such thing as spatial curvature in the real physical world. In that case, although GR defines an infinite number of mathematical solutions which a suggest a spatial curvature component, all of those solutions would ultimately need to be discarded as being non-physical. The Lorentz contraction solution in Galilean coordinates might turn out to be the only physically valid solution.

I'm not in a position to know whether spatial curvature exists as a physical reality, and I don't think the advocates of mainstream GR are either. So it puzzles me that they refer to it with such certainty. Isn't that getting the cart before the horse? Spatial curvature is a deeply radical physical proposition, seemingly as much so as inflation for example, yet in general cosmologists treat the former with far less skepticism than the latter. Perhaps radical theories just tend to become entrenched with the passage of time.

Jon

 

[MeJennifer]

MeJennifer, perhaps you're right to characterize my observation as a "mistake", but my point is that maybe there is no such thing as spatial curvature in the real physical world. In that case, although GR defines an infinite number of mathematical solutions which a suggest a spatial curvature component, all of those solutions would ultimately need to be discarded as being non-physical. The Lorentz contraction solution in Galilean coordinates might turn out to be the only physically valid solution.

I'm not in a position to know whether spatial curvature exists as a physical reality, and I don't think the advocates of mainstream GR are either. So it puzzles me that they refer to it with such certainty. Isn't that getting the cart before the horse? Spatial curvature is a deeply radical physical proposition, seemingly as much so as inflation for example, yet in general cosmologists treat the former with far less skepticism than the latter. Perhaps radical theories just tend to become entrenched with the passage of time.

Jon

A few posts ago I explained you the difference between spacetime curvature and spatial curvature and why spatial curvature is at most an observer dependent and coordinate dependent quantity. However you seem to insist on using spatial curvature, in that case I honestly don't think I can help you.

 

[jonmtkisco]

Hi MeJennifer,

I am hoping for a two-way dialogue, more than just the teacher marking each of my statements correct or incorrect and repeating a single mantra. (Others are encouraged to join this discussion as well.)

I think I do understand your point that spatial curvature is observer dependent and coordinate dependent. After you made that point, I think my statements and questions totally reflected that advice.

However, if what you are actually trying to say is that because spatial curvature is observer- and coordinate- dependent, it doesn't actually need to embody any physical reality, then you have indeed lost me. If you believe that the relationship we measure here on the Earth's surface between circumfererance and radius on a Euclidian plane is not physically real, then I really do need further explanation of your concept of physical reality. For example, given enough time, an observer on a static 3-sphere can travel far enough in one direction to return to the origin of the trip. An observer in a static flat geometry can never do that. Surely there is a physical reality associated with the difference.

In any event, my point is simply that if the concept of (non-zero) spatial curvature (as a component of spacetime curvature) is not physically valid, then all observer- and coordinate-dependent interpretations which purport to measure a spatial curvature component (of spacetime curvature) would turn out to be wrong, and need to be discarded. In that case, only interpretations which purport to measure a zero spatial curvature component (of spacetime curvature) could possibly be physically valid.

In mainstream GR, the concept of spatial curvature refers to a hypersurface of constant curvature which is therefore either 3-spherical or 3-hyperbolic. As I understand it, these mathematically-conceived hypersurfaces can have physical existence only if there are 4 "real" spatial dimensions (obviously time is not one of the 4 spatial dimension). Is this correct?

Do you believe that a compelling case has been made, either mathematically or through physical observation, that our observable universe is actually characterized by (at least) 4 spatial dimensions?

Do you agree that the physical reality of 4 spatial dimensions should be considered a "radical" theoretical proposition (in the normal sense of those words) based on what we know now?

Yes, I am trying to draw out your personal opinion on these very narrow, specific questions.

Jon

 

[jonmtkisco]

Hey MeJennifer,

C'mon, I want to hear your opinion.

Jon