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About space expansion

by Born2Perform
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mysearch
#73
Aug12-08, 04:19 AM
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#68: Frankly I do not see how physical interpretations of spacetime (curved or not) are going to be relevant here. Fact is gamma does not apply to curved spacetimes except locally.
The relevance is to the questions raised in #65. I am assuming that there is a physical interpretation to the assertion that spacetime is curved. If so, I would appreciate some ‘helpful’ clarification of the physical causes and scale of the effect in-line with current measurements rather than what appears to be a throwaway soundbite.

As a clarification, in #65, I simply tried to list some possible factors, which may or may not contribute to spacetime curvature. Given my statements about the localised effects of gravitation in an expanding k=0 homogeneous universe, without a centre of gravity, I was not asserting anything about the scope of [tex][\gamma][/tex]. The root of my confusion is that most references imply mass-gravity to be the cause of ‘local’ spacetime curvature, but the gravitation effects on the very large scale seem to be more ambiguous.

In part, my questions were triggered by, what I took to be, an outline of the concept of parallel transport by Garth in #61 associated with the discussion of the conservation of energy and momentum.

#61: The problem in GR is introduced, by space-time curvature. Energy is the time component of a particle's energy-momentum vector, how do you transport a vector across curved space time?
Presumably, the complexity of this issue is a direct function of the scale of the curvature and this is why I was trying to get some sort of physical interpretation to its cause and the scale of the curvature.
mysearch
#74
Aug12-08, 04:46 AM
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#72: GR has to be reserved for PhD students only, since the regular physics students are to stupid to understand the curvature of space. My opinion was that people interested in physics need to know all the truth...
In-line with an earlier exchange, while I may well be too stupid, or should I say 'cerebrially challenged' in this politically correct world, to understand all the implications of GR and its maths; I would like to think that it was part of the ethos of the PF to, at least, try to help us, the 'cerebrially challenged' , onto the ‘path of enlightenment’, even if we might subsequently falter along the way.
JimJast
#75
Aug12-08, 06:44 AM
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Quote Quote by mysearch View Post
In part, my questions were triggered by, what I took to be, an outline of the concept of parallel transport by Garth in #61 associated with the discussion of the conservation of energy and momentum.

#61: The problem in GR is introduced, by space-time curvature. Energy is the time component of a particle's energy-momentum vector, how do you transport a vector across curved space time?
Presumably, the complexity of this issue is a direct function of the scale of the curvature and this is why I was trying to get some sort of physical interpretation to its cause and the scale of the curvature.
This problem is solved by an assumption that from the point of view of parallel transport the spacetime is "flat" (no change in direction of energy-momentum vectors). For this type of faltness of spacetime neither space not time needs to be flat (Euclidean), only their combination the spacetime. That's why we say that spacetime is curved or flat depending on which aspect of it we are talking about, and most people understand why it happens though it might be confusing and then you should ask if you see an inconsistence before thinking that all GR people are idiots (even if some might ).
mysearch
#76
Aug12-08, 10:51 AM
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Jim: thanks for the response, but was a little puzzled by the inference at the end of the following quote:
That's why we say that spacetime is curved or flat depending on which aspect of it we are talking about, and most people understand why it happens though it might be confusing and then you should ask if you see an inconsistence before thinking that all GR people are idiots.
I don’t think I ever implied `any` GR people were idiots. In fact, I thought my comments in #76 made it clear that I accepted the limitations of my own knowledge in this area. However, I would confess to having some empathy with the Feynman reference. However, my main issue is still with securing a better overall understanding of the implied spacetime curvature on the scale of the universe.
This problem is solved by an assumption that from the point of view of parallel transport the spacetime is "flat" (no change in direction of energy-momentum vectors). For this type of flatness of spacetime neither space not time needs to be flat (Euclidean), only their combination the spacetime.
Given the inference that SR is associated with Minkowski flat spacetime, I assume the source of curvature lies within GR, which we might summarise in the form of the Wheeler quote:
Matter tells space-time how to curve, and curved space tells matter how to move.
However, this only seems to explain localized spacetime curvature. The point I was really trying to clarify was based on the assumption of a homogeneous k=0 universe with no center of gravity. What causes curvature, if any, on this scale?
MeJennifer
#77
Aug12-08, 11:16 AM
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Note that k is not a quantity pertaining to spacetime but instead a quantity pertaining to a particular coordinate chart.

In FRW models (containing matter) spacetime is not flat. In FRW models the Weyl curvature tensor vanishes while the Ricci curvature tensor does not. In FRW models there is actually no such thing as empty space as it models a pressureless perfect fluid. Which makes the often made statements that "space expands" or the "space between objects increases" rather out of place.
mysearch
#78
Aug12-08, 12:49 PM
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Note that k is not a quantity pertaining to spacetime but instead a quantity pertaining to a particular coordinate chart.
Thanks, point duly noted, but is it still correct to say that [k=0] implies a spatially flat universe?

In FRW models (containing matter) spacetime is not flat. In FRW models the Weyl curvature tensor vanishes while the Ricci curvature tensor does not. In FRW models there is actually no such thing as empty space as it models a perfect fluid. Which makes the often made statements that "space expands" or the "space between objects increases" rather strange.
The following bullets are purely to help me infer some physical meaning onto the introduction of Weyl and Ricci tensors. Further clarification of my simplistic definitions welcomed:

- If we equate a unit of 4-dimensional spacetime to a sphere, it has the ability to change its volume and its shape.

- The Ricci tensor describes how the volume changes in any given direction.

- The Weyl tensor describes how it changes its shape.

- Based on your FRW description, if the Weyl tensor vanishes, i.e. goes to 0?, while the Ricci tensor does not, does our sphere only change its volume, but not its shape?

- This suggests an expanding model of the universe, which does not distort in shape as it expands. Is this correct as you seem to be refuting the notion of space expanding?

- However, don’t these tensors only reflect the change, not the reason for the change?

Realise the whole issue of expansion is another ‘bag of worms’ but, at this stage, I still just want to get some physical explanation of what causes spacetime on the scale of the universe to be curved, if this is what LCDM cosmology is actually implying. However, I appreciate the input and would welcome any further clarifications.
MeJennifer
#79
Aug12-08, 12:53 PM
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Quote Quote by mysearch View Post
I still just want to get some physical explanation of what causes spacetime on the scale of the universe to be curved.
Mass and energy curves spacetime. Only spacetimes that have no mass and energy are flat.
mysearch
#80
Aug12-08, 01:20 PM
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Mass and energy curves spacetime. Only spacetimes that have no mass and energy are flat.
OK, that is helpful, as I had really only been focused on mass. So is it correct to infer that the mass-density of the universe, ignoring energy for the moment, cannot appreciably curve spacetime on the scale of the universe given my basic homogeneous, no centre of gravity assumptions?

Classically, we often infer energy associated with mass in the form of kinetic (+) and potential (-) energy, but from the perspective of the conservation of energy these two forms are thought of as adding up to zero?

Therefore, not sure how this is accounted within any total mass-energy density?

While I know there is much speculation about dark energy, possibly accounting for 75% of the total mass-energy density, I thought this form was thought to have negative pressure. It is unclear to me as to what assumptions are being made about its effects on spacetime curvature of the universe as a whole?
yuiop
#81
Aug20-08, 11:49 AM
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Quote Quote by Garth View Post
Hi kev.

You have to define how you measure energy and then how to compare that measurement between different coordinate systems.

The problem in GR is introduced, by space-time curvature.

Energy is the time component of a particle's energy-momentum vector, how do you transport a vector across curved space time?

Consider the surface of a cylinder with a vector in the tangent plane at one point on that surface.

Now transport the vector over the surface.

If you go around the curved surface the tangent plane and the vector being parallel transported in it will turn, so although the vector's overall length does not change a coordinate based measurement of its components will in general change.

However it we move up the surface parallel to the axis of the cylinder, where there is a symmetry, the vector will not turn. In this case the components will not change, they will be conserved under the translation.

If the vector is energy momentum and the axis of the cylinder represents the time axis of a space-time surface then there is a time-like Killing vector and the 'axis' component that is conserved is energy.

I hope this helps.

Garth
Did you mean a sphere?

When a tangent vector is parallel transported on a cylindrical surface keeping it tangent to the surface it does not turn. When tangent vector A is parallel transporting to tangent vector B then if they are parallel the same is true whatever route A is transported via to get to B. There is no problem with defining parallel on a cylindrical surface and essentially it can be thought of as flat spacetime. The cylindrical surface can be cut and rolled out flat with no distortion.

The same is not true for a sperical surface. There is no way to "flatten out" the surface without distortion. Parallel transport on a cylindrical surface does indeed cause the tangent vector to turn and when comparing tangent vector A with B the comparison varies depending upon the route taken by A. That is one way of defining "curved space" and showing that a cylinder has extrinisic curvature but not intrinsic curvature. The sphere has intrinsic curvature and that defines curved space.

By the way, I thought of a method for comparing tangent vectors on a sphere to see if they are parallel. It requires marking out a grid on the sphere in the same way we put lines of longitude and latitude on the global map of the Earth. The latitude lines are great circles or meridians radiating out from the North pole and the longitude lines are parallel to the equator.
The method is applied to a reference vector that is co-transported with vector A and is initially parallel to vector A. Vector A is parallel transported in the normal way and the co-transported reference vector has to obey these 3 simple rules.

1) When transporting the reference vector along a line of longitude for every degree increase in longitude, rotate the reference vector one degree in the opposite direction relative to parallel transported vector A.

2) When transporting the reference vector along a line of latitude do not turn the reference vector relative to parallel transported vector A.

3) The reference vector and parallel transported vector A must remain in the tangent plane of the sphere at all times.

When parallel transported vector A arrives at the location of vector B, then if the reference vector is parallel to vector B, then A and B were initially parallel. If I got it right, then this statement will be true for any route taken by A using these rules.

A simple example that might be easier to visualise would be to imagine looking down from high above the North pole. If a vector being parallel transported around the equator appears to be have gone 90 degrees anti-clockwise around the North pole then its reference vector should be rotated 90 degrees clockwise during the transportation. When the vector is transported from the equator to the North pole then the vector and its reference vector should be parallel transported in the normal way.
jonmtkisco
#82
Aug20-08, 02:04 PM
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Hi mysearch,
Quote Quote by mysearch View Post
So is it correct to infer that the mass-density of the universe, ignoring energy for the moment, cannot appreciably curve spacetime on the scale of the universe given my basic homogeneous, no centre of gravity assumptions?
Are you asking about spacetime curvature or spatial curvature? They are very different things. Anywhere there is gravity (and optionally pressure) there is spacetime curvature, they are defined to mean the same thing. Both gravity and pressure are significant on the scale of our universe, and do not equally offset each other in the current era, so by definition there is appreciable "net" cosmic spacetime curvature. At a particular earlier era (around 7Gy), gravity and pressure were exactly in balance, so at that time the "net" cosmic spacetime curvature was zero. During that brief period, the cosmic expansion essentially coasted, with approximately zero net acceleration.

On the other hand, the question of whether the universe is characterized by cosmic spatial curvature is a subject of much observational and theoretical analysis right now. Currently the error bars in our measurement techniques do not enable us to discern whether the curvature is different from zero. For most analytical purposes, it is reasonable to start with the assumption that the universe is spatially flat.
Quote Quote by mysearch View Post
Classically, we often infer energy associated with mass in the form of kinetic (+) and potential (-) energy, but from the perspective of the conservation of energy these two forms are thought of as adding up to zero?

Therefore, not sure how this is accounted within any total mass-energy density?
When mass-energy density is used in calculations such as the Friedmann equations, neither the kinetic energy of expansion nor the potential energy of cosmic position are included, because they are not GR concepts. Only rest mass and pressure are counted. So the calculations count the 3 forms of rest mass: matter, free radiation, and the rest mass of dark energy/cosmological constant; and 2 forms of pressure: the negative pressure of dark energy/cosmological constant, and the positive pressure of free radiation (the latter currently being insignificant).
Quote Quote by mysearch View Post
While I know there is much speculation about dark energy, possibly accounting for 75% of the total mass-energy density, I thought this form was thought to have negative pressure. It is unclear to me as to what assumptions are being made about its effects on spacetime curvature of the universe as a whole?
The negative pressure of dark energy/cosmological constant is accounted for in the Friedmann equations, along with the 3 forms of rest mass. For the cosmological constant, its negative pressure = its own rest mass. The Friedmann equations multiply pressure by 3, so regarding the cosmological constant, in effect 1X of its negative pressure goes to offset the gravitational deceleration effect of own its rest mass, and the remaining 2X of its pressure goes toward overcoming the gravitational deceleration caused by matter and then accelerating the overall expansion rate. At the time in the history of the universe when the acceleration component first exceeded the gravitational deceleration component, the shift in the balance of the homogeneous cosmic spacetime curvature caused the sign of expansionary acceleration to reverse.

The Friedmann equations are cleverly designed so that the cosmological constant itself does not affect the spatial curvature of the universe. If the universe was spatially flat before the cosmological constant became dominant, the cosmological constant will not change that. The same is true for the extra gravity caused by the positive pressure of free radiation, which dominated the very early universe. It caused a very high deceleration rate, but did not affect spatial flatness.

Jon
mysearch
#83
Aug21-08, 05:08 AM
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Are you asking about spacetime curvature or spatial curvature? They are very different things. Anywhere there is gravity (and optionally pressure) there is spacetime curvature, they are defined to mean the same thing. Both gravity and pressure are significant on the scale of our universe, and do not equally offset each other in the current era, so by definition there is appreciable "net" cosmic spacetime curvature. At a particular earlier era (around 7Gy), gravity and pressure were exactly in balance, so at that time the "net" cosmic spacetime curvature was zero. During that brief period, the cosmic expansion essentially coasted, with approximately zero net acceleration. On the other hand, the question of whether the universe is characterized by cosmic spatial curvature is a subject of much observational and theoretical analysis right now. Currently the error bars in our measurement techniques do not enable us to discern whether the curvature is different from zero. For most analytical purposes, it is reasonable to start with the assumption that the universe is spatially flat.
Jon: Thanks, this is the most concise and coherent summary of the issues I seen, hence the replication in quotes. However, I wanted to see if I could distill a few further facts from your summary:

- Gravitation causes spacetime curvature, as does pressure, presumably in the form of dark energy? Gravity is associated with negative potential energy, so is pressure (dark energy) considered as energy per unit volume rather than in terms of force per unit area? If so, is it considered positive in respect to gravitational potential energy? If these are the two significant energy factors operating on the scale of the universe do you think they obey the conservation of energy?

- [k=0] is an approximation of the measured spatial curvature/flatness of the universe. However, what would cause spatial curvature independent of the spatial components of spacetime curvature linked to gravity and pressure? While it could just be said that the geometry of space was curved, we usually like to have a reason, e.g. there are no straight lines in quantum mechanics.

- I believe I understand the general concept of spacetime curvature in GR on a local system, e.g. galaxy, however, my issue, in the context of large-scale cosmology, was primarily linked to the assumption of a homogeneous model with no centre of gravity. What is spacetime curving around in a mass-gravity only model?

- If we add pressure (dark energy) to the mix linked to the assumption that it represents energy per unit volume, do we also conclude that this pressure has an effective mass, e.g. [tex][m=E/c^2][/tex], and therefore an anti-gravitational effect by virtue of an opposite sign? As a side issue, if gravity is associated with negative potential energy and dark energy acts in the opposite direction, would it be correct to say that this energy is positive not negtaive?

- On the assumption of a 4% matter, 23% cold dark matter and 73% dark energy split, do we conclude that 27% of the universe is trying to collapse, while 73% is trying to expand? Is there any assumption about the relative strength of these effects, as it would appear that expansion should win hands down? I don’t understand how dark energy changes with time?

- If I consider these effects independent of time, I simply perceived expansion and contraction linked to pressure and gravity, i.e. no specific implication of curvature. Of course, if I introduce time, then the expansion of each unit volume of space as a function of time does lead to the implication of a curved path, at least, with respect to light. As such, two parallel light beams diverge in an expanding universe and converge in a contracting universe, is this the root of the definition of spacetime curvature, at least, on the scale of the universe?

- As a slight tangential point, if light travels at [c] at any measured instance in space and time, but the path followed by the light beam is subject to expansion over time, does this imply that light never conforms to [c=s/t], if [s] is a curved geodesic resulting from the expansion of space with time? As a result, the measure value of the speed of light will always exceed [c] as during the time it takes to travel any distance, as the space over which its has travelled has expanded?

- While I don’t want to confuse the discussion by re-introducing the concept of a centre of gravity, this issue has been raised in the thread below and would presumably be a significant factor in any model?
http://www.physicsforums.com/showthread.php?t=235046
MeJennifer
#84
Aug21-08, 06:54 AM
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Quote Quote by jonmtkisco View Post
At a particular earlier era (around 7Gy), gravity and pressure were exactly in balance, so at that time the "net" cosmic spacetime curvature was zero.
I have trouble with this statement.

When there is gravity and pressure there is spacetime curvature.

What do you mean by "in balance" and what is cosmic spacetime curvature as opposed to spacetime curvature?
jonmtkisco
#85
Aug21-08, 08:07 PM
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Hi Jen,
Quote Quote by MeJennifer View Post
What do you mean by "in balance" and what is cosmic spacetime curvature as opposed to spacetime curvature?
When the gravitational deceleration is exactly balanced by the negative pressure acceleration of Lambda, then the universe expands at a constant, coasting rate. That is, at the same rate it would if there were NO gravity and NO pressure. The acceleration parameter = 0.

I think that means that the cosmic spacetime curvature is temporarily at 0. "Cosmic" meaning the spacetime curvature above the scale of matter homogeneity, > 100 Mpc.

If you believe there is "net" large scale spacetime curvature at that balance time, please explain.

The exact balance time lasted only an instant, although it lasted longer as an approximation.

Jon
jonmtkisco
#86
Aug21-08, 08:32 PM
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Hi mysearch,
Quote Quote by mysearch View Post
- is associated with negative potential energy, so is pressure (dark energy) considered as energy per unit volume rather than in terms of force per unit area?
As I said, potential energy is not a GR concept, so it is not considered at all in GR. The negative pressure of the cosmological constant is considered as energy per unit volume.
Quote Quote by mysearch View Post
If these are the two significant energy factors operating on the scale of the universe do you think they obey the conservation of energy?
Potential energy is not a GR concept. And as I said, the Friedmann universe does not conserve energy.
Quote Quote by mysearch View Post
- [k=0] is an approximation of the measured spatial curvature/flatness of the universe. However, what would cause spatial curvature independent of the spatial components of spacetime curvature linked to gravity and pressure?
Spatial curvature occurs only when a large region of the universe does not expand at exactly the Newtonian escape velocity of its mass/energy contents. In other words, it's a mismatch between expansion speed and the mass-energy density. As I said, the negative pressure of Lambda and the positive pressure of free radiation do not cause a change in spatial curvature.
Quote Quote by mysearch View Post
- I believe I understand the general concept of spacetime curvature in GR on a local system, e.g. galaxy, however, my issue, in the context of large-scale cosmology, was primarily linked to the assumption of a homogeneous model with no centre of gravity. What is spacetime curving around in a mass-gravity only model?
I don't understand the question.
Quote Quote by mysearch View Post
- If we add pressure (dark energy) to the mix linked to the assumption that it represents energy per unit volume, do we also conclude that this pressure has an effective mass, e.g. [tex][m=E/c^2][/tex], and therefore an anti-gravitational effect by virtue of an opposite sign?
As I already said, the cosmological constant has a gravitational mass-energy equal to the negative of its negative pressure.
Quote Quote by mysearch View Post
As a side issue, if gravity is associated with negative potential energy and dark energy acts in the opposite direction, would it be correct to say that this energy is positive not negtaive?
For the last time, potential energy is not a GR concept.
Quote Quote by mysearch View Post
- On the assumption of a 4% matter, 23% cold dark matter and 73% dark energy split, do we conclude that 27% of the universe is trying to collapse, while 73% is trying to expand? Is there any assumption about the relative strength of these effects, as it would appear that expansion should win hands down?
The ratio between matter and DE changes as a function of time. Right now it is believed to be .27:.73. In addition, the cosmological constant has negative pressure equal to the negative of 3x its gravitational mass-energy. So at the present time, DE dominates the expansion rate. But this was not always so.
Quote Quote by mysearch View Post
I don’t understand how dark energy changes with time?
Dark energy (in the form of the cosmological constant) is believed to be a fixed characteristic of the vacuum. Each cubic meter of vacuum contains a cosmological constant of about 6.7E-27 kg/cubic meter. So as the number of cubic meters in the observable universe has increased over its history, the total kg of cosmological constant has increased. Meanwhile, the amount of matter in the observable universe remained fixed. Eventually (at about 7Gy), the "weight" of cosmological constant in the observable universe began to exceed the weight of matter. By about 9Gy, the cosmological constant was the dominant expansion force, and the deceleration caused by matter had become insignificant by comparison.
Quote Quote by mysearch View Post
- While I don’t want to confuse the discussion by re-introducing the concept of a centre of gravity, this issue has been raised in the thread below and would presumably be a significant factor in any model?
The existence of a center of gravity would cause the cosmic gravitational force to vary in proportion to distance from the center. But since initial expansion velocity would also vary in proportion to distance from the center, all comoving galaxies would decelerate at the same proportion of their current velocity. As a result, I believe that the varying effect of gravity at different distances from the center would not be detectible, and would not cause a change in the matter homogeneity.

Jon
mysearch
#87
Aug22-08, 02:16 PM
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For the last time, potential energy is not a GR concept.
OK, I got the message loud and clear. I understand that GR does not used PE to explain gravity, however, I was not aware that it banned its use as a way of trying to visualise the conservation of energy issues. I am sure you can appreciate that trying to assimilate all the somewhat speculative issues being raised for anybody new to this subject requires a bit of time and some reflection from more than one perspective. However, if you feel you have already addressed my question, then please simply ignore it.
And as I said, the Friedmann universe does not conserve energy.
Sorry, but I could not find where you explicitly made this point before. Is it related to the implication that GR does not always conserve energy? I am raising this issue because there are derivations of the Friedmann and Fluid equations that seem to have the conservation of energy as a root assumption.
Dark energy (in the form of the cosmological constant) is believed to be a fixed characteristic of the vacuum. Each cubic meter of vacuum contains a cosmological constant of about 6.7E-27 kg/cubic meter. So as the number of cubic meters in the observable universe has increased over its history, the total kg of cosmological constant has increased. Meanwhile, the amount of matter in the observable universe remained fixed.
This was a helpful clarification. However, is the value [tex]6.7E-27kg/m^3[/tex] quoted associated with the critical density [tex][\rho_c][/tex] normally inferred from Friedmann’s equation, i.e. [tex]\rho_c = 3H^2/8 \pi G[/tex]?
So the calculations count the 3 forms of rest mass:
- matter, free radiation & dark energy
and 2 forms of pressure:
- negative dark energy & positive free radiation
I have paraphrased the quote above from #82, but wanted to check whether matter included CDM? However, as you highlighted, dark energy also has an associated mass with corresponding gravitational implications, as well as being the pressure (energy per unit volume) that drives expansion. However, could I clarify the use of ‘pressure [P]’ and the ‘cosmological constant [itex][\Lambda][/itex]’ in the following form of the acceleration equation?

[tex] \frac {\ddot a}{a} = - \frac{4 \pi G}{3} \left ( \rho + \frac {3P}{c^2} \right ) + \frac {\Lambda c^2}{3}[/tex]

If I describe dark energy in terms of a negative pressure [P] does this effectively negate the need for the [tex][\Lambda][/tex] term, as its units, i.e. [tex]m^{-2}[/tex], do not really seem indicative of pressure? The following form of the equation above simply removes all constant values and [tex][\Lambda][/tex] to highlight the dependency on just [tex][\rho][/tex] and [P]:

[tex] \frac {\ddot a}{a} = - \rho - (3P) [/tex]

Clearly, for acceleration to be expansive, pressure has to be negative and as you pointed out some of the negative pressure must overcome the effective gravitational mass of dark energy itself. However, what seems even stranger is the fact that the dark energy, which is the energy per unit volume that expands space, does not get ‘diluted’ in the process, i.e. the suggestion appear to be that it remains constant. If I have interpreted this correctly, it gives the impression that mass and energy are being `created` in the process or, at least, tapping into some other source, i.e. zero point energy/vacuum energy.
The existence of a center of gravity would cause the cosmic gravitational force to vary in proportion to distance from the center. But since initial expansion velocity would also vary in proportion to distance from the center, all comoving galaxies would decelerate at the same proportion of their current velocity. As a result, I believe that the varying effect of gravity at different distances from the center would not be detectible, and would not cause a change in the matter homogeneity.
In part, this issue is also being discussed in another thread. I have posted a query against this position, which can be accessed via the following link. http://www.physicsforums.com/showthr...06#post1844506

However, one point that I would like to raise in this thread is simply a level of surprise that nobody has challenged the assumption of a centre of gravity, as I thought this was not normally accepted as part of the standard model?
George Jones
#88
Aug22-08, 02:26 PM
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Quote Quote by jonmtkisco View Post
At a particular earlier era (around 7Gy), gravity and pressure were exactly in balance, so at that time the "net" cosmic spacetime curvature was zero. During that brief period, the cosmic expansion essentially coasted, with approximately zero net acceleration.
This isn't correct. Zero acceleration is not sufficient for zero spacetime curvature. For a spatially flat universe that has zero scale factor acceleration, zero scale factor velocity, (i.e., zero expansion) is a necessary condition for zero spacetime curvature.

For example, see the expression for the curvature scaler given in

http://en.wikipedia.org/wiki/Friedmann_equations.

If the curvature scalar is non-zero, then the Riemann curvature tensor is non-zero.
jonmtkisco
#89
Aug22-08, 09:14 PM
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Hi George,
Quote Quote by George Jones View Post
If the curvature scalar is non-zero, then the Riemann curvature tensor is non-zero.
Good catch. I was thinking about it as if the balance point could be held constant, but of course it cannot. If mass or pressure is present, spacetime curvature changes continuously, so as a function of moving time it is never zero.

Still, there is an instant when the deceleration parameter = 0.

How about a universe where the Lambda equation of state is -1/3, which coasts in perpetuity? In that case is the spacetime curvature zero?

Jon
jonmtkisco
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Aug23-08, 10:53 AM
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Quote Quote by jonmtkisco View Post
How about a universe where the Lambda equation of state is -1/3, which coasts in perpetuity? In that case is the spacetime curvature zero?
No, this one can't be exactly zero spacetime curvature either. Even with an equation of state of [tex]\omega = -1/3[/tex], the universe will evolve over time, as expansion dilutes the density of matter, and Lambda comes to dominate over matter. It will approach true "coasting" only at late times. I believe it will asymptotically approach zero spacetime curvature at late times despite retaining a positive expansion velocity.

Jon


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