# Flat Space, Curved Spacetime

1. Aug 28, 2008

### mysearch

In another thread I was reminded that there is a clear distinction between curved space and curved spacetime. However, to be honest, I wasn’t sure that I really understood this difference and was pretty sure I couldn’t visualise it. So, by way of a learning process, I wanted to see if there was any way of providing a simplified, but possibly more intuitive visualisation of the nature of both curved space and curved spacetime. As such, the first 3 postings are simply some initial ideas and issues that I am trying to clarify.
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Visualisations:
One of the first visualisation of curvature we often come across is the heavy ball on the rubber sheet. Of course, this image has its limitations in that it is just illustrating the curvature of 2D-space, i.e. time is not really being addressed, unless you assume one axis as time, but then observed time dilates on approach rather than expands.

Next, we might consider the image of a sphere with a triangle on it, accompanied by a description that the angles of this triangle don’t add up to 180 degrees and the observation that 2 people walking due North from the Equator converge at the North Pole, i.e. parallel lines converging. Initially, this appears to be more relevant to spatial curvature, which in the context of the current cosmological model is said to be nearly flat, at least, within the current era.

The next image that is possibly more relevant to cosmology relates to the balloon analogy, which alludes to not only the curvature, but also the expansion of the universe. However, we are usually reminded that this visualisation only applies to the surface of the balloon, not its volume, which might leave the question as to what frame of reference is the surface of the balloon said to be curving around? Still, as often pointed out, it is only an analogy and should not be taken literally.

The following link seems to be a valid attempt to provide an initial visualisation of space and time curvature, which presumably combine to form a concept of spacetime curvature. http://www.math.gatech.edu/~berglund/GR.html [Broken]
However, this presentation seems to be focused on the curvature resulting from the proximity to a large mass-density, which does not necessarily reflect the assumptions of a low mass density homogeneous universe that may or may not have a centre of gravity.

The next link is a much more sophisticated treatment, which includes a brief discussion of cosmological models from which I have extracted a diagram, attached below as Figure-6.jpg by way of quick reference. http://www.rpi.edu/dept/phys/Courses/Astronomy/CurvedSpacetimeAJP.pdf

However, I really wanted to see if I it was possible to provide a far simpler starting point for visualising the specific ideas of flat space and curved spacetime within a k=0 homogeneous expanding cosmological model. I fully accept that this attempt may end in failure. Even so, I will raise some specific issues for clarification in subsequent posts in order to keep each posting to a manageable length. Thanks

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2. Aug 28, 2008

### mysearch

Spacetime Curvature

The attachment ‘spacetime.jpg’ considers the relative path of two photons through spacetime in terms of the basic idea that parallel lines never converge or diverge. In the diagram, 2 photons start off at the same time, in parallel, separated by a given distance, i.e. $$\Delta S_1$$, which it is assumed would be unchanging if both space and spacetime were not curved (?). However, if the universe is spatially flat, but expands with time, (non-linear in the diagram), then the distance $$\Delta S_1$$ would still increase to $$\Delta S_2$$ suggesting that the photons have travelled along a curved geodesic path, irrespective of the spatial flatness of the universe.

So is this example, at least, illustrative of the concept of spacetime curvature being independent of spatial curvature?

Conceptually, the change in the parallel separation suggests the photons take a longer curved path through curved spacetime, but this curvature seems to be a direct consequence of space expansion with time. So some issues for clarification:

- In the context of an expanding cosmology model, Friedmann’s equation might be cited as a description of the expansion. As such, the net energy density is the cause of expansion, while spacetime curvature is the effect?

- Presumably, massive localised concentration of matter, e.g. black holes, can give rise to far more complex curvatures of both space and spacetime, but in a homogeneous model, it seems that the expansion of space does not necessarily imply that space curves, i.e. only spacetime curves?

- In the frame of rest of the photons $$[d\tau=0]$$, which seems to imply that the distance covered by the photon (AB) must also zero, such that the velocity [c] cannot be calculated as a simple function of [s/t] in this frame?

- From all other observer frames of reference, do we assume the constancy of [c], but then infer the distance [AB] from other measurements, e.g. luminosity, to calculate the time taken, i.e. [t2-t1=AB/c]?

- If the distance AB is indirectly measured via receiving photons from a distant object, is it reflective of the geodesic curved path length travelled by the received photons?

- If so, the photons associated with the current luminosity of a given object started off at a distance [ct], but this object must have since receded [vt], where [v] is a time-dependent function of the recessional velocity implicit within Hubble’s constant [H=v/d].

Note, this last point is somewhat tangential to the issue of curvature. However, photons that are now reaching us today must have set off at a much early time in the timeline of an expanding universe. Friedmann’s equation suggests a much higher recession velocity at earlier times and if you add any initial recession velocity [vi] to the increased velocity [ve] that must have occurred due to the subsequent expansion of the universe, whilst the photons were on route, then the current distance of a given source would be [d=ct+(vi+ve)t]. The implication being that this distance could now exceed the Hubble radius [R=c/H]. Does this process go some way to explain the concept of a particle horizon’ being some 3x the Hubble radius?

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3. Aug 28, 2008

### mysearch

Space Curvature:

Based on Friedmann’s equation, if k=0, the energy density of the universe would have to be equal to the critical density and imply that the universe would expand forever, albeit at an ever-decreasing rate, i.e.

$$\Omega = \frac {\rho}{\rho_c} = \frac {8\pi G \rho}{3H^2} = 1$$

Current models, inclusive of matter, cold dark matter and dark energy, suggest that present-day $$[\Omega]$$ is very close to 1, such that [k=0], which in-turn is said to support a spatially flat universe. However, this only gives a mathematical rationale for the definition of spatial curvature without necessarily helping with any visualisation. Of course, we can visualise geometrical 2-D flatness in terms of the angles of a triangle adding up to 180 degrees, but mapping this onto a physical 3-D concept of space appears to be more problematic.

The attachment space.jpg shows an equilateral triangle that is assumed to enclose a vast area of space at 2 different points in time. As a possible futile attempt to consider 3-D space curvature, I have included a triangular pyramid to parallel the 2-D concept of flatness. The problems of simultaneity are ignored and it is simply assumed that the length of the sides and angles are relative to an inertial observer and space-like at each time instance. There are a number issues raised for clarification

- It is recognised that the example is intrinsically flawed in that space is a 3-D concept that cannot really be visualised in terms of a flat 2-D triangle.

- Spatial flatness only requires the angles of this triangle to add up to 180 degrees at any single instance in time, i.e. t1 or t2? However, in the real universe, we would have to first establish a spatially flat reference triangle at [t1], which we might assume remains geometrically flat at [t2]?

- Of course, we have now introduced the effects of time and the possibility of ‘distortion’ due to spacetime curvature (?). However, while each corner of the equilateral triangle might travel along curved spacetime in an expanding universe (?) this does not seem to implicitly change its spatial flatness, i.e. spatial curvature and spacetime curvature can be independent of each other?

- However, if we run time backwards, does space within our triangle always remain flat? If not, what ‘point’ in the universe would this space be curving around in a homogeneous model without any centre of gravity?

- I realise that last point may be meaningless in that the curvature of 3-D space may have no meaningful centre or axis. However, if perpendiculars could be drawn to any tangents of curvature in a higher dimension would these perpendiculars converge to a ‘point’ and what meaning might be inferred on this point?

- Finally, is it possible that the curvature of space has no physical meaning, i.e. it can only be inferred from the motion of some object moving through this space?

I realise that the actual answers to some of these issues may be buried in GR maths or Riemann n-D manifolds, but was trying to focus on whether there was any simple approach that might help visualise the basic concepts. My initial conclusion is that space curvature is harder to visualise than spacetime curvature. However, would welcome any clarification of the issues raised in the first 3 posts. Thanks

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4. Aug 30, 2008

### mysearch

Footnote to spatial curvature [k]:

The symbol [k], and presumably the concept of spatial curvature, seems to be linked to Friedmann’s equation. The derivation of this equation is said to have its roots in GR, but the form of this equation can be derived using ideas associated with the conservation of energy, e.g. http://arxiv.org/abs/astro-ph/0309756, If you follow this logic, an expression for [k] can be derived as follows:

[1] $$-\frac {kc^2}{a^2} = \frac {2E_T}{ma^2}$$ as per 2.4 in reference above

[2] $$k=-\frac {2E_T}{mc^2}$$

In the context of equation [2], [m] appears to be the unit mass subject to expansion in-line with the conservation of energy. Therefore, it is assumed that $$E_T$$ is the total energy associated with the mass [m]. However, it seems that [k] can only equal zero when $$E_T=0$$. As far as I can see, if $$E_T=0$$, it might be suggesting that the expansive (+) positive energy would have to be matched by an equal contracting (-) energy, i.e. net zero.

Would this not suggest the Big Crunch, if [k=0], rather than the asymptotic slow down of an ever-expanding universe?

One other interesting dimension of [k] is raised in Sean Carroll paper on the Cosmological Constant (pdf version: p.11). Cross-referencing equation [1], Carroll seems to suggest that [k] can be associated with an energy density: http://relativity.livingreviews.org/Articles/lrr-2001-1/ [Broken]

$$\rho_k = \frac{3kc^2}{8 \pi Ga^2}$$

This energy density can also be converted into pressure [P] via $$[\omega=-1/3]$$, i.e. expansive according to the Friedmann-Acceleration equation. As such, it is difficult to interpret whether [k] is a cause or an effect’. Of course, based on the Wheeler quote: "Matter tells space-time how to curve, and curved space tells matter how to move” this may seem perfectly natural. However, visualizing spatial curvature still appears to be problematic and possibly a futile undertaking?

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5. Aug 30, 2008

### jonmtkisco

Hi mysearch,

I think you are tilting at windmills to try to visualize spatial curvature. It really can't be completely visualized, only hinted at. If you haven't yet, check out the Wikipedia article on http://en.wikipedia.org/wiki/3-sphere" [Broken] which gets into some detail on this subject.

Keep in mind also that, unlike spacetime curvature, spatial curvature is frame-dependent. Any manifestation of spatial curvature can always be entirely eliminated by shifting the observer's velocity vector to a suitable alternative reference frame. So it is not an invariantly tangible phenomenon.

I think the most important mathematical attribute of non-flat space is that the circumference of a sphere will not measure at $$2\pi$$ times the radius. Thus when space is curved, a sphere is mathematically calculated to enclose more or less volume than would a sphere of the same radius in flat space.

Jon

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6. Aug 31, 2008

### mysearch

Response to #5

Hi Jon,
I cannot totally reject the Don Quixote inference, but in my defence I will quote a Chinese proverb and say that I had, at least, begun to recognise the possible futility of any attempt to visualise space curvature .
However, would still be interested in knowing whether expanding space in time, as described in post #2, is sufficient to describe curved spacetime when k=0. Equally, would like to better understand if space curvature is a ‘cause’ or ‘effect’ based on the associated energy density described in #4. I raise this issue in light of your statement:
When you say, “spatial curvature is frame-dependent” are you implying the value of [k] is dependent on the frame of reference? Would this have any implication on the Friedmann equation?
Actually, this is quite a nice visualisation, which is used by Taylor and Wheeler in their book on black holes. In this context, it almost leads to a ‘visualisation’ of 3-D spatial curvature around extreme gravitational objects, although still usually drawn analogous to the rubber sheet outlined in post #1. Whether you can transpose this concept to the spatial curvature of the universe as a whole presumably depends on where you place the centre of the universe and whether this question is even allowed of the universe

7. Aug 31, 2008

### jonmtkisco

Re: Response to #5

Hi Mysearch,
We discussed this topic in the "Expanding Space" thread, the answer is essentially "yes." George Jones said:
As a special case, if one postulates an "empty" universe with no matter but with Lambda at w = -1/3, I think the negative pressure of Lambda would exactly offset its own gravitational mass, such that the curvature scalar would be zero, and therefore the spacetime curvature would also be zero. Even though such universe has gravity and is expanding. This special universe coasts forever with zero acceleration, so it does not evolve as a function of time. Even its density does not vary with time.
In the Schwarzschild metric for central masses (such as a black hole), some degree of spatial curvature is always measured unless the observer is plunging radially toward the mass at exactly the negative of the mass's escape velocity at the then current radius. A mismatch between the plunging speed and the square root of the central mass's M/r means there is localized spatial curvature.

In the FLRW metric, spatial curvature occurs only when the Hubble expansion rate does not equal the escape velocity of the mass-energy contained within the spherical volume being studied (such as our observable universe). This could also be roughly characterized as a mismatch between the universe's expansion speed and the square root of its M/r.

It doesn't make sense to me to suggest the converse, that non-flat spatial curvature "causes" the mismatch between plunge rate or expansion rate on the one hand and M/r on the other. I guess I can't categorically specify the order of causation because the underlying physical causes are unknown.

During the eras in which the cosmic expansion rate is dominated by radiation and matter, if the universe is even the tiniest bit non-flat going in, the spatial curvature will continually increase thereafter during those eras. If the universe eventually enters a Lambda-dominated era, Lambda's exponential expansion rate will conversely cause any existing spatial curvature to diminish continually thereafter. This is a consequence of how the Friedmann equations work.
Interesting question, which occurred to me. I'm not entirely sure, but I think cosmic spatial curvature could be eliminated (flattened out) by adopting a cosmic reference frame whose own comoving spherical expansion rate is either suitably slower or faster than the Hubble expansion rate, depending on whether the spatial curvature is positive or negative, respectively. In that reference frame, the mismatch between the Hubble expansion rate and the density could theoretically be exactly offset and thereby "corrected".
I love their book Exploring Black Holes and I heartily recommend it to anyone interested in how gravity works.
The "corrective" reference frame I suggested above does not require the universe to have a center, nor would the origin of the frame need to be at the cosmic center if there were one. As I've mentioned before, if a very large expanding dust ball universe had an original center, its location becomes meaningless and indistinguishable once the expansion is underway. The Friedmann equations work exactly the same in any spherical, homogeneous large region of a universe whether or not that universe has a center (unless edge effects are observable.)

Jon

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8. Sep 1, 2008

### mysearch

Response to #7

Hi Jon,
Thanks for the food for thought, but struggling a bit to reconcile the equations with your following example, but possibly I have misinterpreted the implications:
When you say ‘empty’ do you mean empty of everything except dark energy ($$\Lambda$$) plus any resulting spatial curvature. However, you seem to imply that both space and spacetime curvatures are zero. If the universe were expanding, albeit with zero acceleration, wouldn’t a photon still follow a geodesic path, as described in #2, in the sense that parallel beams would still diverge?

Based on Friedmann’s equation, each terms must reduce to units of $$1/sec^2$$ and, as such, equivalent energy density equations can be expressed for [k] and $$[\Lambda]$$:

[2] $$\frac {kc^2}{a^2} \equiv \frac {8 \pi G \rho} {3}; \ \frac {\Lambda c^2}{3}\equiv \frac {8 \pi G \rho} {3}$$

[3] $$\rho_k = \frac {3kc^2}{8 \pi Ga^2}; \ \rho_{\Lambda} = \frac {\Lambda c^2}{8 \pi G}$$

Now we can write the generic expressions $$\rho \propto a^{-n}$$ and $$n=3(1+\omega)$$ from which we might infer:

[4] $$\rho_k = a^{-2}; \ \rho_{\Lambda} = a^0$$

[5] $$\omega_k = -1/3; \ \omega_{\Lambda} = -1$$

This seems to align to standard conclusions, but on the basis that $$P=\omega \rho c^2$$ can it be suggested that energy density contributes to a contracting gravitation mass and an expansive pressure, and would these equations be consistent with your example above? Sorry, to go on a bit, but I am really trying to get a better handle on this aspect of the model, see the following thread if you are interested: https://www.physicsforums.com/showthread.php?t=252511
I agree, in the context of the dust ball model, that it appears the centre cannot be distinguished from the homogeneous model, but I not sure that makes it `meaningless’ if gravity is a function of the distance [r] from this point.

9. Sep 1, 2008

### jonmtkisco

Re: Response to #7

Hi mysearch,
I was assuming no matter or free radiation, and no spatial curvature. Just spatially flat with Lambda w = -1/3.
I do not think that parallel photon beams would diverge in this model. If the beams were initially moving apart (not parallel), they would continue to move apart at a constant angle, not an ever increasing angle. If the beams were initially parallel, not moving apart, then in no sense are they partaking in any pre-existing Hubble expansion. Their motion with respect to each other is entirely peculiar. The amount of Lambda between them would remain constant unless something causes the distance between them to increase. The Lambda located between them exerts no net "antigravity" force because it is exactly offset by Lambda's own gravity.
Not meaningless to someone who can observe the outer edge of the dustball, but meaningless to everyone located far away from the edge in the interior. I do not believe that an interior observer could conduct any test which would reveal gravitational directionality.

Jon

10. Sep 2, 2008

### mysearch

Response to #9

Hi Jon,
Thanks for all the clarifications, however, was still interested in understanding how the equations support your conclusions. As a good approximation, Friedmann’s equation can be derived from the conservation of energy, as can the Fluid equation linked to the 1st law of thermodynamics. Finally, the Acceleration equation is derived using both these equations. As such, any assumptions drawn from the Acceleration equation seem to be predicated on the implications of the Friedmann and Fluid equations.

Now based on $$\omega_\Lambda=-1/3$$ and $$P=w \rho c^2$$, I see how the Acceleration equation implies zero acceleration, but I am not so sure what the same values imply with respect to the Friedmann’s and the Fluid equations. In the case of the Fluid equation, it seems to suggest that $$d \rho/dt$$ is negative, i.e. reducing, and the Friedmann equation seems to suggest that contracting gravity will always be greater than expansive pressure, i.e. so how this universe ever get going?
I don’t want to belabour the point, but we might speculate for a moment that we exist within a dustball universe and although we cannot distinguish the gravitational effects of a centre of mass from the accepted assumptions of a homogeneous model, it seemed that the dustball model is still predicated on the existence of an unobserved and possibly unverifiable centre.

11. Sep 2, 2008

### jonmtkisco

Re: Response to #9

The only source of density in this universe is Lambda, and the increase of Lambda is exactly equal to the increase in volume, as a function of time. It is impossible for the density of this model to ever increase or decrease.

What imparted the initial expansion motion to this universe? Well, it's a model, so one has to supply a hypothetical answer. Perhaps inflation imparted the motion. Of course there's nothing to say that a universe based on this model exists or could ever possibly exist, and maybe someday we'll be smart enough to rule out the possibility completely. As a thought experiment, like any other hypothetical model, it is just a way of exploring the math and physics and discovering interesting nuances, which in turn allows a deeper understanding of more conventional models.
Yes, I agree that a spherical expanding dust ball has a physical center. I'm just emphasizing that an interior observer can't find it, and wouldn't notice anything different occuring there even if they stumbled onto it.

Jon

Last edited: Sep 2, 2008