Is the Universe Expanding or Are Galaxies Moving Apart?

[yuiop]

First, I would like to get clarification of Kev’s post #44, as I know he has spent some considerable time looking that the issue of relativity. Which observer perceives the conservation of energy, the distant observer or onboard observer?

[1] Distant Observer v_R = c\left( 1+ \frac {Rs}{r}\right)* \sqrt {\frac {Rs}{r}}

[2] Onboard Observer v_R = c\sqrt {\frac {Rs}{r}}

Now equation [1] suggests that the observed radial velocity goes to zero, while the onboard radial velocity goes to [c] at [r=Rs]. I would have thought only [2] was consistent with the conservation of energy?

Hi mysearch,

First I should make it clear that I am comparing distant measurements with local measurements made by non-inertial observers that are stationary in the gravitational field and not "onboard" measurements of free falling observers.

The Total Energy (E_T) equation from Special Relativity relates Rest Energy (R_E) to Momentum Energy (M_E) by the relationship:

E_T = \sqrt{E_R^2 + E_M^2} = \sqrt{m^2c^2 + m^2(1-v^2/c^2)^{-1} v^2 c^2}

To put this into the context of a free falling object in the Schwarzschild metric that falls from infinity with an initial velocity of zero, the relativistic gamma factor of sqrt(1-v^2/c^2) can be replaced by the gravitational gamma factor sqrt(1-R_s/R) because the local falling velocity v/c = sqrt(R_s/R) to give:

E_T = \sqrt{M^2 C^4 + M^2(1-R_s/R)^{-1} V^2 C^2}

I have also taken the liberty of using uppercase M, V and C to indicate measurements made by a distant observer and reserved lower case m, v and c for local measurements .

From other threads we know that

C = c\left( 1 - \frac {Rs}{R}\right)

and this applies to all velocities so

V = v\left( 1 - \frac {Rs}{R}\right)

The total energy equation can now be written as:

E_T = \sqrt{M^2 c^4 (1-R_s/R)^4 + M^2(1-R_s/R)^{-1} (v^2 c^2) (1-R_s/R)^4}

By assuming the Total Energy of the particle at rest at infinity is mc^2 and by further assuming that the Total Energy is constant for a falling particle in coordinate measurements, we can now say

mc^2 = \sqrt{m^2 f^2 c^4 (1-R_s/R)^4 + m^2 f^2 v^2 c^2 (1-R_s/R)^3}

where f is some factor that relates M to m.
By substituting c*sqrt(R_s/R) for v we can solve for f to give:

f = (1-R_s/R)^{-1.5} so that

M^2 = m^2 f^2 = m^2(1-R_s/R)^{-3} and also

mc^2 = \sqrt{m^2 c^4 (1-R_s/R) + m^2 v^2 c^2}

It can be seen that the expression is now in terms of local velocity and that when v=c AND R_s=R the total energy is mc^2 and also when v+0 AND R=infinity that the total energy is still mc^2

Solving the last equation for the local falling velocity gives the result v/c = sqrt(R_s/R) which should not come as too much of a surprise as that was assumed in deriving the equation.

I originally derived the equation without assuming v/c = sqrt(R_s/R) by simply assuming M = m(1-R_s/R)^(-1.5) because in SR parallel relativistic mass is also related by m = mo(1-v^2/c^2)^(-1.5)

Have you noticed that in SR the parallel kinetic transformations

L = Lo (1-v^2/c^2) ^(0.5)
T = To (1-v^2/c^2) ^(-0.5)
M = Mo (1-v^2/c^2) ^(-1.5)

are analogous to the vertical gravitational transformations

L = l (1-R_s/R) ^(0.5)
T = t (1-R_s/R) ^(-0.5)
M = m (1-R_s/R) ^(-1.5)

and that the transverse SR kinetic transformations

L = Lo
T = To (1-v^2/c^2) ^(-0.5)
M = Mo (1-v^2/c^2) ^(-0.5)

are analogous to the horizontal gravitational transformations ?

L = l
T = t (1-R_s/R) ^(-0.5)
M = m (1-R_s/R) ^(-0.5)

My assumption of M = m (1-R_s/R) ^(-1.5) from the above observations happily gives us v/c = sqrt(R_s/R)

Anyway, going back to your original concern about whether we are talking about velocity measured by a local or distant observer, the total coordinate energy equation can be expressed in terms of the velocity V measured by the distant observer as:

E_T = \sqrt{m^2 c^4 (1-R_s/R) + m^2 V^2(1-R_s/R)^{-2} c^2} . . . . . . [3]

Substituting c*sqrt(R_s/R)*(1-R_s/R) for V we get:

E_T = \sqrt{m^2 c^4 (1-R_s/R) + m^2 c^2(R_s/R) c^2}

which simplifies to

E_T = \sqrt{m^2 c^4 (1-R_s/R+R_s/R)}

which is obviously constant for all R.

We can also solve equation [3] for V to obtain

V = c*\sqrt{(R_s/R)}*(1-R_s/R)

which agrees with your equation [1] (assuming you made a typo with the sign)

Please note that I have not proved that coordinate energy is conserved because I assumed that in the derivations, but it seems a reasonable assumption because we already know that the coordinate energy of a falling photon is constant. It should also be clear that energy is not conserved according to measurements by local observers and Garth has also indicated it is not conserved from the point of view of the falling particle. However, I hope I have shown that IF the coordinate total energy of a falling particle is conserved then the rest energy of the particle diminishes as the particle falls.

 

[Nereid]

[...]

If possible I would like to try and get a better idea as to the scope and scale of the ‘errors’ in Newtonian predictions in comparison with GR. As an initial statement, it was my understanding that the discrepancies between Newtonian physics and Relativity only become apparent under the extremes of gravitational potential, i.e. space curvature, and/or the velocity of an observed frame approaching [c].

[...]

What follows may be a list of things that you already know very well (in which case I have misunderstood your question)...

If by "extremes of gravitational potential" you include the surface of the Earth and the solar system from the Sun's photosphere to the orbit of Jupiter (say), then indeed "the discrepancies between Newtonian physics and Relativity only become apparent" there.

However, such environments are not, I submit, usually so regarded.

And yet it is just these in which "the discrepancies between Newtonian physics and Relativity only become apparent"! (as long as we add "first").

Example: the anomalous advance of the perihelion of Mercury - known for ~half a century before the publication of GR

Example: the bending of light as it passes the Sun's 'surface' - known since the 1919 total eclipse (and now a standard correction in contemporary astrometry, for example, for GAIA this deflection needs to be factored in across the whole sky, and all planets and the Moon will produce a 'weak lensing' signal that will be easily measurable by GAIA)

Example: gravitational redshift (as in Pound-Rebka).

 

[mysearch]

General Response to #48, #49, #51, #52

Thanks for all the helpful comments, especially the detailed explanation in #51. I wanted to initially respond by trying to explain my general approach and then respond to specific points after I have had a chance to work through the details provided.

- First, apologises for the incorrect sign in equation [1] #47, as picked up by kev #51.

- Second, while appreciating Nereid’s comments, my reasons for wanting to compare the approximation of Newtonian physics with GR were mainly in response to the Jim-Jon exchanges. However, as a general comment, while the issues surrounding the perihelion of Mercury and bending of light support GR, the effects in absolute terms are quite small. Therefore, I will try to better explain my rationale below.

From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction. Hence my correlation of GR theory to Newtonian physics, but please note I am not advocating a Newtonian preference. However, I did want to test Jim’s statement that `Newtonian predictions are wrong` rather than the usual implication that they were still a good approximation under most circumstances. Therefore, the point of my original equations (1-4) in #47 was to try to narrow the overall complexity, as alluded to in Garth’s comments in #48, to a very specific, and possibly over-simplistic, example in the hope that some physical interpretation might be more obvious. For example, equation [2] is the free-fall velocity as perceived by both the onboard observer and any local observer at [r].

[2] v_R = c\sqrt {\frac {Rs}{r}}

It is possibly worth highlighting that [2] is also the perspective derived from Newtonian physics. As such, what are the real physical implications of equation [#47:1], if the velocity, as perceived by the distant observer, slows under free-fall gravitational acceleration to zero?

Based on what appears to be the most ‘real’ physical interpretation, and I realize that some may question the applicability of the word ‘real’, I then tried to correlate the implications of equation #2 on equations [3,4]. In this respect, I was focusing on the onboard observer, while Kev in #51 appears to have focused more on the distant observer. Equation [3] being the Newtonian form, while [4] is derived from Schwarzschild’s metric and therefore assumed consistent with GR under the general caveats assumed by this solution. However, the implication on [3] and [4] for an onboard free-falling observer is that orbital velocity [vo=0] and, as such, both equations appear to collapse to the form in [5], i.e.

[3] Newtionian: E_T = 1/2 m\left(v_r^2 + v_o^2\right) +\left (-\frac{GMm}{r} \right)

[4] GR: E_T =1/2 m\left(v_r^2 + v_o^2\right) - \frac{GMm}{r} \left( 1+ \frac{v_o^2}{c^2}\right)

[5] E_T = 1/2 mv_R^2 - \frac{GMm}{r}

Substituting for [v_R] and Rs=2GM/c^2] leads to the form:

[6] E_T = \frac{GMm}{r} - \frac{GMm}{r} = 0

Note, this is only valid for the radially solution, not an orbiting one; also this equation appears not to account for any relativistic effects, i.e. velocity or gravitation as normally associated with [\gamma]. However, the form of equation [6] has transpose the kinetic energy associated with [v_R] into an equivalence amount of potential energy via equation [2]. In this form, the only quantity that would be subject to relativistic effects is [m], so any value of [\gamma] would cancel out in [6].

As such, energy would seem to be conserved based on potential energy being converted to kinetic energy, noting that classically, potential energy is negative ranging from a maximum of zero at an infinite [r] to a minimum, i.e. maximum negative, as the centre of mass [M] is approach. It is noted that free-falling into a black hole raises an anomaly when [v=c] at [r=Rs]. However, generally, the conclusion being forwarded seems to contradict the statement made in #51and so I would like to better understand if I have made a wrong assumption or misinterpreted what was being said in this specific case?

#51: It should also be clear that energy is not conserved according to measurements by local observers and Garth has also indicated it is not conserved from the point of view of the falling particle.

Referencing `Exploring Black Holes` by Taylor and Wheeler p3-12/section-5:

The fact that E/m is constant for a free particle yields a great simplification in describing the motion of a radially plunging particle

Equation [10] in this reference also gives an equation of the form:

[7] \frac{E}{mc^2} = \left(1-Rs/r\right) \frac{dt}{d\tau} = 1

From which it might be assumed that:

[8] \frac{dt}{d\tau} = \left(1-Rs/r\right)^{-1}

This raised the question in my mind as to whether the relativistic factor [\gamma] could be physically interpreted as follows:

[9] \gamma = \frac{1}{\sqrt(1-v^2/c^2)}*\frac{1}{\sqrt(1-Rs/r)}

However, in the specific case of a free-falling observer, velocity is also proportional to radius such that:

[10] \frac{1}{\sqrt(1-v^2/c^2)} = \frac{1}{\sqrt(1-Rs/r)}

I believe this is consistent to Kev’s statement in #51. However, in #49, Jim presented the following equation and I would therefore like to clarify that the use of the metric tensor [g00] is essentially equivalent to equation [7] given the assumption of this specific case?

E = \frac{m}{\sqrt{1-\frac{v^2}{c^2}}}c^2\sqrt{g_{00}}

This post is already too long, and possibly drifting off the main thrust of this thread, so I will terminate at this point, but will continue to review the detailed points already raised. Many thanks

 

[JimJast]

Why is it necessarily conserved? Are you assuming a time-like Killing vector?

I was reminded already a few times, that such problems can't be discussed in regular threads that may spoil the minds of innocent students who are coming here for knowledge and not for some crazy ideas. Those crazy ideas (like global conservation of energy) have to be placed in IR section of this forum. IR section is provided specially for those who propose (as eg. Feynman did in his Feynmans lectures...) that energy is conserved separately from momentum and not together, as Wheeler's momenergy, as the present form of GR requires. If you want to discuss a time-like Killing vector and consequences of its existence you should open a thread in IR section. We can't discuss it here since this thread is about space expansion which necessarily excludes the global conservation of energy.

My remark about energy necessarily conserved regards Einstein's theory from 1916, and so from before it was assumed that the universe is expanding that could happen only after 1929 when Hubble proposed the dependence of recession velocities of galaxies on distance. Before that it was universally accepted (also by Einstein, the patent office clerk) that energy and momentum are conserved separately, and that no theory that contradicts this principle can be consdered a scientific theory (the patent office didn't even allow applications for devices proposing perpetual motion machines; maybe they allow them now).

But since you mention this subject you might know the answer to the question how energy can be converted into momentum and vv. to keep the both of them being conserved together? I was banned from astronomy forum before learning the answer to this question (and possibly just for asking it) and so I'm still curious.

 

[Garth]

As I have said now several times, if you are talking about GR then you need a time-like Killing vector to conserve energy.

If you are not talking about GR, but some unpublished theory of your own, then you need to post on the IR thread.

Garth

 

[JimJast]

As I have said now several times, if you are talking about GR then you need a time-like Killing vector to conserve energy.

If you are not talking about GR, but some unpublished theory of your own, then you need to post on the IR thread.

You're (obviously) right about a time-like Killing vector and its necessity to conserve energy. You are wrong though that Einstein's theory of 1916 is my own unpublished theory even if it requires posting it in IR thread to be discussed. I wouldn't call this theory neither mine (dispite that I believe it is true) nor unpublished and I would even go as far as calling it a GR theory, distinctive from the mainstream GR by the fact that it has the conservation of energy built in (and therefore it is admitting a time-like Killing vector). Since I don't want to be banned from this forum for discussing Einstein's theory in the mainstream forum I submited my post to IR thread hoping that the discussion of Einstein's theory will be allowed there. We just need to wait for the decision of moderators.

 

[Garth]

But the Einstein's static model as a possible GR cosmological solution has been falsified.

1. By the observation of Hubble red shift, which not be there in a static GR model.

2. By the fact that such a solution would be unstable wrt small perturbations. Any perturbation would increase and kick the model into an ever increasing expanding or contracting solution.

3. Therefore, apart from the other observations and tests it would have to pass that I mentioned in post #28, any resurrection of the model would have to include new physics and therefore be an unpublished new theory.

Garth

 

[yuiop]

Hi,


In General Relativity can we always find a coordinate system or reference frame where the total energy of an evolving system is conserved over time independently of momentum?


I ask, because in Special Relativity the total energy of an evolving system is always conserved independtly of momentum from the point of view of an inertial observer even if the observer is not at rest with respect to the centre of momentum frame of the system. The only time energy appears not to be conserved in Special Relativity is when switching from one reference frame to another.

I am trying to make clear the difference between energy not being conserved over time and energy not being conserved when switching reference frames or coordinate systems.

 

[JimJast]

But the Einstein's static model as a possible GR cosmological solution has been falsified.

1. By the observation of Hubble red shift, which not be there in a static GR model.

2. By the fact that such a solution would be unstable wrt small perturbations. Any perturbation would increase and kick the model into an ever increasing expanding or contracting solution.

3. Therefore, apart from the other observations and tests it would have to pass that I mentioned in post #28, any resurrection of the model would have to include new physics and therefore be an unpublished new theory.

This new theory is a published already theory of Einstein known as general relativity. If you admit the conservation of energy (which is built into Einstein gravitation in a form of vanishing divergence of stres-energy tensor) then for a static space you get the Hubble resdshift roughly as observed for the density of matter in space roughly as observed. You may calculate it easily and exactly with Newtonian approximation of GR since it may be done for space with weak gravitational field and slow matter. If you do you'll see how it simulates the accelerating expansion, with observed acceleration, and that the dynamical friction of photons was never before calculated exactly. It was just assumed to be negligible by inventors of the Big Bang hypothesis to provide for their idea of expanding universe. Similarly the stuff about stability (note that we don't know how to solve 3 body problem, and the universe contains even more than 300 bodies, so how we can tell that this system is unstable?)

Regretable we can't write about Einstein's GR in a mainstream thread because certain calculations were neglected and now we need to pretend that we all believe that they were done and Einstein's universe was falsified. Have you ever seen the proof that there is no Hubble redshift in Einstein's universe (while energy is conserved) and that photons fly in it without any redshift while everything else, including space probes (eg. Pioneers 10 ans 11) have this redshift?

 

[Garth]

This new theory is a published already theory of Einstein known as general relativity. If you admit the conservation of energy (which is built into Einstein gravitation in a form of vanishing divergence of stres-energy tensor)

Have you understood nothing about what we have been saying about GR?

You cannot admit the conservation of energy in GR because GR is an Improper Energy Theory that does not in general conserve energy.

The vanishing covariant divergence of the stress-energy tensor, otherwise known as the energy-momentum tensor, T_{\mu \nu}, conserves energy-momentum, not in general energy, which is different.

This is basic 'Einstenian' GR, which you had better get under your belt if you are going to base a PhD thesis on it. Einstein himself realized that energy was not conserved in general and worried about it until Noether explained that this was how a theory such as GR should behave.

then for a static space you get the Hubble resdshift roughly as observed for the density of matter in space roughly as observed. You may calculate it easily and exactly with Newtonian approximation of GR since it may be done for space with weak gravitational field and slow matter.

No, no, no...

The static GR cosmological solution has a spatial curvature constant k = +1, it has to.
Newtonian approximations do not generally work in curved space over cosmological distances, even if the field is weak and the non-Newtonian effects are locally insignificant nevertheless the accumulation of the effect over cosmological distance is not insignificant.

You have to work consistently in GR.

This is where I find your logic baffling; in the GR static solution, because the field is static, there is a time-like Killing vector and energy is conserved! The galaxies are not moving apart and there is no Doppler shift, the energy of the photon is conserved in the frame of all static representative observers in such a universe and the frequency is constant from emission to observation. There would be no red shift.

If you do you'll see how it simulates the accelerating expansion, with observed acceleration, and that the dynamical friction of photons was never before calculated exactly.

What the heck is "dynamical friction of photons" this is pure crackpottery...

It was just assumed to be negligible by inventors of the Big Bang hypothesis to provide for their idea of expanding universe. Similarly the stuff about stability (note that we don't know how to solve 3 body problem, and the universe contains even more than 300 bodies, so how we can tell that this system is unstable?)

de Sitter showed that the static model was unstable in the 1920's.

If you want to learn about it I suggest you read this recent paper by Barrow, Ellis et al. http://www.iop.org/EJ/article/0264-9381/20/11/102/q311l2.html which looks at the model as possible primordial states for the present universe. The conclusions of that paper:

There is considerable interest in the existence of preferred initial states for the universe and in the existence of stationary cosmological models. So far this interest has focused almost entirely upon the de Sitter universe as a possible initial state, future attractor, or global stationary state for an eternal inflationary universe. Of the other two homogeneous spacetimes, the Einstein static provides an interesting candidate to explore whether it could play any role in the past evolution of our universe. It is important to know whether it can provide a natural initial state for a past eternal universe, whether it allows the universe to evolve away from this state, and whether under any circumstances it can act as an attractor for the very early evolution of the universe. We might also ask whether it is not possible for it to provide the globally static background state for an inhomogeneous eternal universe in which local regions undergo expansion or contraction, manifesting an instability of the Einstein static universe. With these questions in mind we have investigated in detail the situations under which the Einstein static universe is stable and unstable.

We have shown that the Einstein static universe is neutrally stable against inhomogeneous vector and tensor linear perturbations, and against scalar density perturbations if c_s^2 > \frac{1}{5} extending earlier results of Gibbons for purely conformal density perturbations. However, we find that spatially homogeneous gravitational-wave perturbations of the most general type destabilize a static universe. We pointed out the link that can be forged between this homogeneous instability and the behaviour of the inhomogeneous gravitational wave spectrum by choosing modes with imaginary wave number. Our results show that if the universe is in a neighbourhood of the Einstein static solution, it stays in that neighbourhood, but the Einstein static is not an attractor (because the stability is neutral, with non-damped oscillations). Expansion away from the static state can be triggered by a fall in the pressure of the matter. Typically, expansion away from the static solution will lead to inflation. If inflation occurs, then perturbations about a Friedmann geometry will rapidly be driven to zero. The nonlinear effects (which will certainly be important in these models because of the initial infinite timescale envisaged) will be discussed in a further paper, as will other aspects of the spatially homogeneous anisotropic modes.

Regretable we can't write about Einstein's GR in a mainstream thread because certain calculations were neglected and now we need to pretend that we all believe that they were done and Einstein's universe was falsified. Have you ever seen the proof that there is no Hubble redshift in Einstein's universe (while energy is conserved) and that photons fly in it without any redshift while everything else, including space probes (eg. Pioneers 10 ans 11) have this redshift?

Yes I have given the 'proof' above; energy is conserved in a static GR universe there is no motion and therefore in a GR theory there is no red shift in a theory where particle masses are constant.
This is because red shift measurement is the comparison the ratio (energy of the photon)/(mass of the atom) at emission with the ratio (energy of the photon)/(mass of the atom) at observation. In the static GR model it would be the same.

As I have said you have to do it consistently - I have published in peer reviewed journals a theory that does what you seem to be getting at, but it is a modification of GR, not GR itself.

Garth

 

[Garth]

Hi,In General Relativity can we always find a coordinate system or reference frame where the total energy of an evolving system is conserved over time independently of momentum?I ask, because in Special Relativity the total energy of an evolving system is always conserved independtly of momentum from the point of view of an inertial observer even if the observer is not at rest with respect to the centre of momentum frame of the system. The only time energy appears not to be conserved in Special Relativity is when switching from one reference frame to another.

I am trying to make clear the difference between energy not being conserved over time and energy not being conserved when switching reference frames or coordinate systems.

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

 

[JimJast]

What the heck is "dynamical friction of photons" this is pure crackpottery...

not to astronomers for whom this effect is an observable and who are able to calculate it easily in clouds of dust. Eg. in a static cloud of dust of density 6\times10^{-27}kg/m^3 it produces a redshift corresponding to the recession velocity of the source of light 70km/s/Mpc (looks familiar?). Those astronomers are always surprised that the cosmologists despite their sophistication and ease of handling of tensors of GR never heard about the effect that necessarily follows the simple Newtonian physics with its hypothetical conservation of energy, the same effect that is described in popular articles on http://en.wikipedia.org/wiki/Dynamical_friction" .

 

[Garth]

not to astronomers for whom this effect is an observable and who are able to calculate it easily in clouds of dust. Eg. in a static cloud of dust of density 6\times10^{-27}kg/m^3 it produces a redshift corresponding to the recession velocity of the source of light 70km/s/Mpc (looks familiar?). Those astronomers are always surprised that the cosmologists despite their sophistication and ease of handling of tensors of GR never heard about the effect that necessarily follows the simple Newtonian physics with its hypothetical conservation of energy, the same effect that is described in popular articles on http://en.wikipedia.org/wiki/Dynamical_friction" .

I asked what you meant by the "dynamical friction of photons".

From the same wikipedia article that you referred to: (Note, Wikipedia articles have to be used with much caution and are not always a good reference)

Fritz Zwicky proposed in 1929 that a gravitational drag effect on photons could be used to explain cosmological redshift as a form of tired light.[6] However, his analysis had a mathematical error, and his approximation to the magnitude of the effect should actually have been zero, as pointed out in the same year by Arthur Stanley Eddington. Zwicky promptly acknowledged the correction,[7] although he continued to hope that a full treatment would be able to show the effect.

It is now known that the effect of dynamical friction on photons or other particles moving at relativistic speeds is negligible, since the magnitude of the drag is inversely proportional to the square of velocity.

Any attempt to resurrect this discredited idea must therefore be 'new' physics, which must either be tested and published, in which case please give the reference to the published paper, or simply falls into the category of 'crackpottery'.

Garth

 

[Nereid]

Thanks for all the helpful comments, especially the detailed explanation in #51. I wanted to initially respond by trying to explain my general approach and then respond to specific points after I have had a chance to work through the details provided.

- First, apologises for the incorrect sign in equation [1] #47, as picked up by kev #51.

- Second, while appreciating Nereid’s comments, my reasons for wanting to compare the approximation of Newtonian physics with GR were mainly in response to the Jim-Jon exchanges. However, as a general comment, while the issues surrounding the perihelion of Mercury and bending of light support GR, the effects in absolute terms are quite small. Therefore, I will try to better explain my rationale below.

From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction. Hence my correlation of GR theory to Newtonian physics, but please note I am not advocating a Newtonian preference. However, I did want to test Jim’s statement that `Newtonian predictions are wrong` rather than the usual implication that they were still a good approximation under most circumstances. Therefore, the point of my original equations (1-4) in #47 was to try to narrow the overall complexity, as alluded to in Garth’s comments in #48, to a very specific, and possibly over-simplistic, example in the hope that some physical interpretation might be more obvious.

[...]

Thanks for the clarification.

Not to derail the line of discussion you wish to pursue, I think there are some important clarifications to be made.

However, as a general comment, while the issues surrounding the perihelion of Mercury and bending of light support GR, the effects in absolute terms are quite small.

If, for purposes of teaching and learning, you choose to start with something like Newton (and history) and work your way to Einstein and GR, then this is fine. However, as GR is more general and as it has passed all the experimental and observational tests (to date), it is just as sensible to start with GR and derive Newton as a good approximation in the appropriate limit.

From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction.

Hear hear!

For some, the math and the abstractions are the side-show; the experimental results and observations must always been seen as the main show.

 

[mysearch]

Response to #64

Nereid: Thanks for the response. You raise some useful points against which I would like to make some general comments, as I assume that many, like me, must come to the PF to get some help in understanding the science that underpins the standard models.

#64: If, for purposes of teaching and learning, you choose to start with something like Newton (and history) and work your way to Einstein and GR, then this is fine.

In practical terms, most people become acquainted with the principles of physics through Newton’s laws of motion and gravitation. Therefore, this is an implicit starting point, for most people, which normally precedes any introduction to SR/GR. I also assume that it is an implicit postulate of relativity that in the absence of any near light speeds or massive gravitational presence, its equations should reduce to the Newtonian approximations. Finally, there is also a sense of Pareto’s (80/20) rule that suggests that you can learn 80% of something with 20% effort. By this rule GR plus all the maths will demand 80% of my time, in order, to come closer to some overall understanding of the detailed arguments supporting the standard model(s) of cosmology.

#64: However, as GR is more general and as it has passed all the experimental and observational tests (to date), it is just as sensible to start with GR and derive Newton as a good approximation in the appropriate limit.

As stated, I am not suggesting a preference for a Newtonian model and fully accept the weight of authority that supports GR. However, for the reasons above, I not sure that getting to grips with GR and its apparent dependency on differential geometry and tensors always seems the most “sensible” way forward, although I now accept the necessity and have started to work my way into these topics.

#53: From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction

#64ear hear! For some, the math and the abstractions are the side-show; the experimental results and observations must always been seen as the main show.

I fully concur with your statement, i.e. #64, in the following sense. I accept the need that has led to the abstraction of the mathematical notation used in GR, however, the conclusions via this approach should still be expressible in plain English and correlated to “experimental results and observations”. By way of example, I have tried to follow many of the threads discussing the curvature and expansion of spacetime. However, I am still struggling to get a clear physical interpretation of the various arguments being put forward, i.e.

- Measurements suggest that k = 0 implying little to zero space curvature. Does this not suggest that space is flat? If so, what other curvature is implied?

- On the very large scale, the universe is said to be homogeneous. As such, it is assumed that the universe has no overall centre of gravity. If so, does this not suggest that any net gravitational effects must be localised anomalies of mass-density, e.g. galaxies?

- As a very simplistic reduction of relativity, might we look at the effects of the relativistic factor [\gamma] due to velocity and/or gravity? Do the following equations not suggest little overall effect until the relative velocity approaches [c] or a mass object approaches the Schwarzschild radius of mass-density corresponding to a black hole?

[1] velocity \gamma_v = \frac{1}{\sqrt(1-v^2/c^2)}

[2] gravity \gamma_g = \frac{1}{\sqrt(1-Rs/r)}

- Again, if the homogenous mass-density, inclusive any dark matter, is aligned to current estimates and there is no overall centre of gravity, how does gravity, and therefore GR, affects the net curvature of the universe?

- There appears to be conflicting comments over the real nature of any expansion of space. I have include a quote from Garth from another thread purely as any example, as he is also contributing to this thread and may wish to explain the level of “contention”:

The only reason for saying that "space expands", which is itself a contentious expression, is that the FRW predicts it to do so, that is cosmic expansion is one solution of that metric when GR is used to determine the scale factor a(t) and curvature parameter k.

- If the expansion of space is not to violate SR, how does [\gamma_v] have any overall affect on the curvature of space or spacetime?

- On the basis of k=0, the FRW metric appears to reduce to the form below. However, while this metric alludes to an expansion, does it make any other statement about curvature other than [k]?

- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2}\right) = - c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2

However, I fully accept that I still have a lot to learn on this subject, so will get back to my teach-yourself maths courses. However, would welcome any other comments not encoded in tensors

 

[MeJennifer]

Mysearch you cannot apply gamma to curved spacetimes except locally. Note also that special relativity is not compatible with spacetimes that have a non-zero cosmological constant.

 

[mysearch]

Mysearch you cannot apply gamma to curved spacetimes except locally.

Now if you could just explain the physical interpretation of your definition of curved spacetime. I am really not tying to be facetious.

 

[MeJennifer]

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.

 

[jonmtkisco]

Hi Kev,

Are you suggesting that the rest mass energy of a falling particle changes as it falls so that the potential energy of the gravitational field does not have to be invoked to conserve energy?

GR treats a particle's energy as a unified whole, and it is not possible to divide that energy into the Newtonian components of rest mass, kinetic and potential energy. If the particle begins at rest at an infinite distance from a planet, its total GR energy is simply equal to its rest mass. Then as the particle begins to accelerate gravitationally toward the planet, that total energy does not change, as judged by a distant observer. The falling particle does not gain energy from the gravitational source. Energy of mass and motion is an invariant in GR. When the particle collides with the planet, its rest mass is simply added to the planet's rest mass.

As you mention, it is important to recognize that a stationary local observer (say, standing on the surface of the gravitating planet, or orbiting around it), does NOT measure the plunging particle's energy to be invariant. This stationary observer's clock runs at a different rate from both the distant observer and an observer falling alongside the particle, so the calculated energy of the particle is different. If she doesn't adjust for the clock difference, the stationary observer will judge that the total energy of the particle increases as it falls.

I'm not going to comment on your equation because math is not a strong suit for me!

Jon

 

[MeJennifer]

If the particle begins at rest at an infinite distance from a planet, its total GR energy is simply equal to its rest mass. Then as the particle begins to accelerate gravitationally toward the planet, that total energy does not change, as judged by a distant observer. The falling particle does not gain energy from the gravitational source. Energy of mass and motion is an invariant in GR. When the particle collides with the planet, its rest mass is simply added to the planet's rest mass.

That is actually not true under GR. Two masses of 1kg closing in on each other due to gravitation will slowly decrease their total rest mass. Except when they are separated infinitely far from each other their total mass will be less than 2kg.

Also the "stacking order" or matter matters. For instance if you have a set of Lego blocks in space far removed from any gravitational influences then their total mass depends on how they are stacked.

 

[jonmtkisco]

Hi Jennifer,

That is actually not true under GR. Two masses of 1kg closing in on each other due to gravitation will slowly decrease their total rest mass. Except when they are separated infinitely far from each other their total mass will be less than 2kg.

Let's be very careful about what you say isn't true. In the example you describe, I did not say that the combined masses did or did not decrease as they approach each other. I said only that their total energy in GR is invariant as they approach each other and after they collide, as judged by a far distant observer. No net energy is gained or lost in this (initially) two-object system.

Jon

 

[JimJast]

That is actually not true under GR... Two masses of 1kg closing in on each other due to gravitation will slowly decrease their total rest mass. Except when they are separated infinitely far from each other their total mass will be less than 2kg.

So it is true under GR since this is about what jonmtkisco said The mystery (if there is any in it) is solved by noticing that while the total energy (or inertial mass) is conserved in GR (contrary to opinions of some mathematicians who because of their exact minds can't stomach this fact and the expansion of the universe at the same time, in which they are of course right), the "missing" mass is there in a form of kinetic energy of those masses. And all of it might be easily verified by just taking a derivative with respec to displacement, which will then show, white on black (if it is done with a chalk on a blackboard), all parts of total energy of the particle, and also the so called "gravitational force", since "force" is -d(mc^2)/dx, in which case mc^2 (actually it is mc^2\sqrt{g_{00}} to be exact but \qrt{g_{00} is almost 1) must be the potential energy that some people call "gravitational".

As an off topic remark I'd like to add that apparently I was right in my discussion with the head of our Gravitation and Cosmology Dept. who maintained that 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 (that is available at the time) from the very beginning since only then they might be able to discover new things without necessity of wasting many years on unlearning what they "know" but it ain't so, as I had to, believing in things that now I know were not physics but math only and therefore not true. Einstein, who might have had some experience in those matters, said: "As far as the laws of mathematics refer to reality, they are not certain, as far as they are certain, they do not refer to reality" which roughly means that one can get only a trivial thruths (2+2=4) from math but non trivial truths can be found only through experiments, and, "If you are out to describe the truth, leave elegance to the tailor" (quote by the same author). Yet the GR has been constructed from "elegant" math and this elegant math said that the universe is expanding. The point of all of this is that a physicist must never believe in math that is not supported experimentally or observationally. And another point, which I see in this forum that some people interested in physics can think, which was my point in my discussion when I proposed to teach physics of GR from the first year of physics since it is even simpler that Newtonian math that assumes a mysterious and unobserved so far "gavitational attraction". Unfortunately the proposal was rejected by professors who teach the math of GR.

 

[mysearch]

Response to #68

#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 [\gamma]. 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]

Quick response to #72

#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]

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 http://geocities.com/jim_jastrzebski/sci/feynman.htm" ).

 

[mysearch]

Response to #75

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]

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]

Response to #77

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]

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]

Response to #79

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]

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]

Hi mysearch,

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.

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).

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]

Response to #82

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. [m=E/c^2], 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 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 traveled 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?
https://www.physicsforums.com/showthread.php?t=235046

 

[MeJennifer]

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]

Hi Jen,

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]

Hi mysearch,

- 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.

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.

- [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.

- 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.

- 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. [m=E/c^2], 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.

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.

- 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.

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.

- 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]

Response to #86:

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 6.7E-27kg/m^3 quoted associated with the critical density [\rho_c] normally inferred from Friedmann’s equation, i.e. \rho_c = 3H^2/8 \pi G?

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 [\Lambda]’ in the following form of the acceleration equation?

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

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

\frac {\ddot a}{a} = - \rho - (3P)

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. https://www.physicsforums.com/showthread.php?p=1844506#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]

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]

Hi George,

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]

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 \omega = -1/3, 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

 

[jonmtkisco]

Hi myseach,

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.

The Friedmann equations do conserve energy at least for a spatially flat model with zero Lambda.

This was a helpful clarification. However, is the value 6.7E-27kg/m^3 quoted associated with the critical density [\rho_c] normally inferred from Friedmann’s equation, i.e. \rho_c = 3H^2/8 \pi G?

Yes, you use the Friedmann equations to calculate it, using the estimated figures for matter and Lambda density, as a function of time. An easy way to calculate it is to run the Friedmann equations into the arbitrarily distant future when Lambda becomes the only significant component of the universe's mass-energy.

I have paraphrased the quote above from #82, but wanted to check whether matter included CDM?

Yes it is; as you know the great majority of matter mass is believed to be dark energy.

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

\frac {\ddot a}{a} = - \rho - (3P)

No, Lambda is in the equation for a reason, so you cannot simply exclude it.

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 simplest concept for dark energy is the cosmological constant, and every cubic meter of vacuum comes with its own cosmological constant, so by definition it can never be diluted. On the contrary it helps to dilute and eventually dominate everything else.

The ongoing creation of additional space filled with its own (additional) Lambda seems to be a characteristic of both the "dust ball model" and standard cosmology, which suggests that indeed new energy is being constantly created. Perhaps one could also posit a model where the empty vacuum "outside" the dust ball has Lambda but does not expand in accordance with the de Sitter model. In which case the existing Lambda was always there and is not being newly created. But any such theory is entirely speculative. In any event, little satisfaction is gained by saying that infinite energy was "always there" instead of incremental energy being "newly created" with the passage of time.

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?

Standard cosmology readily admits that the lack of a center of gravity is a simplifying assumption, rather than a fact or theory which has been demonstrated to be highly likely. I believe it is a widely accepted assumption because of its mathematical simplicity (no "edge" effects) and philosophical elegance (cosmological principle), and because as a practical matter the observable characteristics of the universe so far have not depended on whether the assumption is correct.

Jon

 

[mysearch]

Response to #91:

Hi Jon,
Given the excessive length of my response in another thread, I will keep this one brief. Thanks for all the help, you given me a lot to think about, however, I now recognise the need do some more detailed reading based on what I hope is now a more ‘expansive’ perspective. Much appreciated.

P.S. Your last statement is very interesting, especially if nobody challenges it!

Standard cosmology readily admits that the lack of a center of gravity is a simplifying assumption, rather than a fact or theory which has been demonstrated to be highly likely. I believe it is a widely accepted assumption because of its mathematical simplicity (no "edge" effects) and philosophical elegance (cosmological principle), and because as a practical matter the observable characteristics of the universe so far have not depended on whether the assumption is correct.

 

[Cuetek]

Standard cosmology readily admits that the lack of a center of gravity is a simplifying assumption, rather than a fact or theory which has been demonstrated to be highly likely. I believe it is a widely accepted assumption because of its mathematical simplicity (no "edge" effects) and philosophical elegance (cosmological principle), and because as a practical matter the observable characteristics of the universe so far have not depended on whether the assumption is correct. Jon

We base a great deal of the Big Bang on the cosmological principle. And it is indeed an excellent model by which to hash out the local nature of the universe. But it is based on limited data in pursuit of an complete description of a universe that always seems to have more in store than our descriptions can convey. To attempt to contrive a complete and sufficient view of the universe, rather than presuming our models to be local characterizations, is ill-advised in my view.

Across 40 orders of spatial magnitude the material universe has proven to be uniformly hierarchical, yet in all our cosmologies we humans like to terminate that hierarchy at the limits of our data (flat earth, crystal sphere, island universe, steady state, etc). Currently we choose to presume the the mere 2 orders of magnitude of homogeneous galactic clustering we see at the limits of our view are sufficient to completely characterize a potentially infinite universe. We do this even though we can easily find in excess of 14 orders of magnitude of homogeneity (water molecule in 10 cubic km of ocean) nested within the 40 orders of hierarchy we have already fully examined. To invest all our efforts in a model based on such a selective application of the data is poor science in my estimate.

The Big Bang is an idealized model of how the universe could work if the material hierarchy did somehow stop at the limits of our ability to examine it. Like the 19th century "Island Universe" model of the Milky Way, the Big Bang is a beautiful, comprehensive, ostensibly accurate model of how the universe would behave if the hierarchy did stop. But like every cosmology we ever devised, the limits we habitually place on the hierarchy have always proven to be false and the hierarchy has always persisted beyond them.

The advantage to presuming an ongoing hierarchy is to better address the anomalous data in the current model. Influences from a greater context often show up as anomalous data in the current model (Al Sufi's nebula, Hubble's red shift). It is just as likely that the questions of dark matter, dark energy and curvature of the universe are related to the material and energetic effects of the surrounding hierarchical context rather than strictly local variable adjustments in the current model.

Sure the Big Bang will be BIG, but it won't be "everything." There will always be more to the hierarchy than we can see from here, just as has always been the case. That's what the material and historical data both consistently indicate. If the cosmological principle is good science, then the hierarchical principle is better science.

-Mike

 

[mysearch]

Response to #93:

There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy.
Hamlet by William Shakespeare: 16th Century

Hi Mike,
In many respects, I think you have eloquently summarised a valid view of, not only, the Big Bang model, but possibly science as a whole. This is not an anti-science or anti-establishment view, far from it, simply a recogniition of the reality of the endeavour and we all need to be reminded of this reality occasionally.

Concepts that have proven useful in ordering things can easily attain an authority over us such that we forget their worldly origin and take them as immutably truths. They are then rubber-stamped as a "sine-qua-non of thinking" and an "a priori given". Such errors often make the road of scientific progress impassable for a long time.
Albert Einstein: 20th Century