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#73
Aug1208, 04:19 AM

PF Gold
P: 522

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 massgravity 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. 


#74
Aug1208, 04:46 AM

PF Gold
P: 522




#75
Aug1208, 06:44 AM

P: 44




#76
Aug1208, 10:51 AM

PF Gold
P: 522

Jim: thanks for the response, but was a little puzzled by the inference at the end of the following quote:



#77
Aug1208, 11:16 AM

P: 2,043

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. 


#78
Aug1208, 12:49 PM

PF Gold
P: 522

 If we equate a unit of 4dimensional 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. 


#79
Aug1208, 12:53 PM

P: 2,043




#80
Aug1208, 01:20 PM

PF Gold
P: 522

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 massenergy density? While I know there is much speculation about dark energy, possibly accounting for 75% of the total massenergy 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? 


#81
Aug2008, 11:49 AM

P: 3,967

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 cotransported with vector A and is initially parallel to vector A. Vector A is parallel transported in the normal way and the cotransported 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 anticlockwise 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. 


#82
Aug2008, 02:04 PM

P: 531

Hi mysearch,
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. 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 


#83
Aug2108, 05:08 AM

PF Gold
P: 522

 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 largescale cosmology, was primarily linked to the assumption of a homogeneous model with no centre of gravity. What is spacetime curving around in a massgravity 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 antigravitational 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 reintroducing 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 


#84
Aug2108, 06:54 AM

P: 2,043

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? 


#85
Aug2108, 08:07 PM

P: 531

Hi Jen,
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 


#86
Aug2108, 08:32 PM

P: 531

Hi mysearch,
Jon 


#87
Aug2208, 02:16 PM

PF Gold
P: 522

[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. 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? 


#88
Aug2208, 02:26 PM

Mentor
P: 6,246

For example, see the expression for the curvature scaler given in http://en.wikipedia.org/wiki/Friedmann_equations. If the curvature scalar is nonzero, then the Riemann curvature tensor is nonzero. 


#89
Aug2208, 09:14 PM

P: 531

Hi George,
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 


#90
Aug2308, 10:53 AM

P: 531

Jon 


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