How to make the Einstein Ricci curvature definition coordinate independent?

UPDATE November 1, 2021: We do not need to make Ricci curvature coordinate independent. A tensor -  by definition - is coordinate dependent!

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Our blog post yesterday showed that the Einstein definition for Riemann curvature of a manifold (M, g) is coordinate dependent.

On the other hand, Gauss curvature of a manifold is usually defined by the observers living inside the manifold measuring the circumference and the radius of a circle. That definition is, of course, coordinate independent.

https://en.wikipedia.org/wiki/Ricci_curvature

Wikipedia gives the formula for Ricci curvature for new coordinates defined with a mapping:

       y: ℝ⁴ → U,

where U is an open subset of ℝ⁴ and (U, g) is the manifold (M, g).

J(q) is the matrix of the first derivatives of y at the point q ∊ ℝ⁴. That is, J(q) is the Jacobian matrix. In the formula, the matrix on the right side is multiplied by J(q) and the transpose of J(q). The Ricci curvature matrix on the left side is not the same as on the right side. The value of Ricci curvature changes in the coordinate transformation.

Can we make the Einstein Ricci curvature coordinate independent?

How to make the Einstein formula for Ricci curvature coordinate independent?

It is not enough to take the derivatives against the proper distance and proper time. In the M. W. Cook case we had the derivatives against proper time, but the rogue cross term still appeared.

What about using proper time and the proper distance as coordinates? If the surface is curved, one cannot map the measured distances to a euclidean plane. It is not possible to use such coordinates.

Let us define a function:

      s(p, q) = the proper distance of points p ∊ U and q ∊ U.

Ricci curvature should be calculated from the function s. Then it would be coordinate independent since it would be something that observers living inside the manifold measure themselves.

We may introduce polar coordinates around each spacetime point p. Curvature would mean that tangential distances are surprisingly short or long. The radial coordinate would be the proper distance. We will calculate what this is in terms of the metric.

https://en.wikipedia.org/wiki/Theorema_Egregium

Curvature in polar coordinates is like fitting a sphere to the curvature, where proper distances are the framework. We need to look at theorema egregium by Carl Gauss (1827).

Let us have a 2D surface. Let us assume that the metric is

 

       g_x(x, y)         0

  

       0         g_y(x, y)

Furthermore, we assume that

 

       g_x(0, 0) = g_y(0, 0) = 1

and the first derivatives of g_x and g_y are zero at (0, 0).

                           y

                           ^ 

                           |

               •           |           •

                           |

     ----------------------------------------> x

                           |

               •           |           •

                           |

We want to calculate the proper circumference of a square

       (s, s) -- (s, -s) -- (-s, -s) -- (-s, s)

around the origin, where s > 0 is a small real number.

To get an estimate for the proper length of the upper line (-s, s) -- (s, s) of the square we need the second derivative

       D_yy g_x.

To estimate the proper length of the "radius" (0, 0) - (0, s) we need the second derivative

       D_yy g_y.

If there is some distortion in the proper circumference of the square relative to its proper "radius", then the following simple formula calculates it:

       D_yy g_x  -  D_yy g_y  +  D_xx g_y  - D_xx g_x.     (1)

If g_x(0, 0) is arbitrary, we need to "normalize" it to 1 and its first derivatives to 0:

       G_x(x, y) = [ g_x(x, y) - x D_x g_x(0, 0) - y D_y g_x(0, 0) ]

                          / g_x(0, 0).

Furthermore, we need to normalize the second derivatives D_yy and D_xx.

Introduce a preferred coordinate system?

Another option is to take a preferred coordinate system. The obvious candidate is an inertial frame in the Minkowski space which is asymptotic around a mass system. The Schwarzschild external and internal solutions seem to be sensible. Their coordinate system fixes the time to the proper time of a distant observer. The radial coordinate r is defined as the circumference measured by observers divided by 2 π.

In the waterfall analogy of a black hole, the coordinates of the river banks are the natural ones that we should use. Observers swimming in the water measure the geometry with sound signals. Their view is very different from someone standing on the river bank. Using the syrup model of gravity, the water is the syrup which is flowing and taking all frames of internal observers along with it.

A rubber membrane model of gravity is naturally coordinate independent

If we treat spacetime as rubber which is bent and stretched, we will have a coordinate independent theory. The problem with the Einstein equations may be that they do not allow pressure to change the spatial metric.

The name of the stress-energy tensor suggests that it is rubber which is being stretched.

In rubber, it is not curvature itself, but the stretching of the material which takes energy. The Einstein tensor, in some sense, describes the "force" with which spacetime resists being curved. In a rubber model, the force should be calculated by varying the metric and determining the energy it takes to vary the metric.

Calculate the Ricci curvature in the coordinates which minimize it?

Another option to remove coordinate dependency is to demand that the Ricci curvature must be calculated in global coordinates which, in some sense, minimize the curvature.

This would make calculations very difficult. Besides optimizing the metric, we should also optimize the coordinate system.

The covariant derivative

https://en.wikipedia.org/wiki/Covariant_derivative

Around the year 1900 Gregorio Ricci-Curbastro and Tullio Levi-Civita introduced the concept of a covariant derivative.

Wikipedia states:

"This new derivative was covariant in the sense that it satisfied Riemann's requirement that objects in geometry should be independent of their description in a particular coordinate system."

But when we look at the "Informal definition", the formula of the derivative is:

The derivative is against xⁱ, which is a coordinate. The derivative is not against an infinitesimal proper length on the manifold. The derivative does depend on the coordinate system. It is not "intrinsic" in the manifold.

The value of the derivative is easy to transform to new coordinates, though.

Conclusions

The fact that the Einstein equations are coordinate dependent suggests that there might be something wrong with them. In a reasonable physical theory the choice of coordinates has no effect whatsoever.

For a rubber membrane model we have:

1. preferred coordinates: the Minkowski coordinates of the surrounding space;

2. naturally coordinate independent formulation.

We will look at making the Einstein equations coordinate independent.

Riemann and Ricci curvature as defined by Albert Einstein are coordinate dependent

UPDATE November 1, 2021: Of course, the Ricci curvature tensor is coordinate dependent. A tensor is something that you have to transform to new coordinates!

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In his 1916 paper about the foundations of general relativity Albert Einstein defined Ricci curvature with Christoffel symbols.

Christoffel symbols are cumbersome to calculate. Let us take the definition of the Einstein tensor from literature for certain metrics.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455

The Oppenheimer and Snyder 1939 paper contains the Einstein tensor which is calculated for the metric above. The component T₁⁴ contains the time derivative with respect to the coordinate time of the radial metric λ. It is marked with the dot on top of λ. If we define a new time coordinate

       T = 2 t,

then the derivative against T has a different value than against t, at the same spacetime point. The value of the Einstein tensor changes through a simple rescaling of time.

https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll8.html

Sean Carroll has calculated the Ricci tensor for the FLRW metric given above. Let us define a new radial coordinate

       R = 2 r.

Then dr = 1/2 dR and r = 1/2 R. The metric, written in terms of the R coordinate is (we take k = 0):

       ds² = -dt² + a²(t) / 4 [ dR² + R² (dθ² + sin²θ dφ²) ]

We see that the metric written in terms of the R coordinate has to be

       A(t) = a(t) / 2.

The Ricci R₁₁ component with the R coordinate then has a value which is only 1/4 of the value it had with the r coordinate.

A solution for the Einstein equations is not coordinate independent

If we had a solution for the FLRW metric, we showed above how to break the solution: simply rescale the radial coordinate r.

The coordinate dependency of the Einstein equations is much worse than for newtonian mechanics. One cannot break a solution of newtonian mechanics by defining new coordinates.

Browsing the internet, one cannot find anyone explicitly claiming that the Ricci curvature tensor is coordinate independent. However, it is easy to get such an impression, since Riemann curvature is sometimes defined in a coordinate free way, with a parallel transport of a vector.

What coordinates we can choose in order to satisfy the Einstein equations?

Karl Schwarzschild chose the standard Minkowski coordinates where a distant observer defines the time coordinate t, and the radial coordinate is determined by the circumference of a circle as measured by observers. He was able to solve the Einstein equations - both externally and also inside the uniform spherical mass.

Question. If one tries to use proper time as the time coordinate, can one find a solution for the metric of a spherical uniform mass?

The time inside the spherical mass lags constantly. The time coordinate lines become more and more bent. Can one solve the Einstein equations then?

https://www.publish.csiro.au/ph/PH750413

A week ago we were suspecting that there might be an error in the M. W. Cook (1975) equations because distorting the time coordinate brought a new cross term which breaks the Friedmann equations. Now we understand that the problem is in the coordinate dependency of solutions for the Einstein equations.

We will investigate what consequences does the coordinate dependency have.

Friedmann equations have a solution for a mix of matter - even for a sudden jump in pressure

We presented a hypothesis in our blog post on October 20, 2021 that a mixture of slow massive matter and radiation has no solution for both Friedmann equations. The hypothesis is false. All mixtures of matter do satisfy the second Friedmann equation if they satisfy the first one. This is because of energy conservation.

Our October 16, 2021 claim that Friedmann equations do not have a solution for a sudden jump of pressure was incorrect.

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

Theorem. Any uniform matter has a solution for both Friedmann equations in the case where the curvature parameter k = 0. We assume energy conservation.

Proof. Let us have any uniform matter. The first Friedmann equation clearly has a solution. Let

        ρ(a)

be the mass-energy density as a function of the scale factor a. If the matter is slow massive matter, then

       ρ(a) ~ 1 / a³,

       D_a ρ(a) = -3 / a * ρ(a),

where we denote the derivative operator against a by D_a

We assume energy conservation. If the derivative D_a ρ(a) has some other value, it has to be because pressure is doing work. If we let a grow to a + da, then the lost mass-energy is

       [-3 / a * ρ(a) - D_a ρ(a)] da * a³ * c²

in a cube whose side is a. The volume growth of the cube is 3 a² da. The pressure has to be

        p(a) = [-ρ(a) - 1/3 a D_a ρ(a)] * c²            (1)

to conserve energy.

We denote the derivative against the time coordinate t with a prime mark '. The first Friedmann equation is

       (a' / a)² = 2/3 κ ρ                                        (F1)

<=>

       a'          = sqrt(2/3 κ) a sqrt(ρ).

We calculate the time derivative of both sides of the equation (we denote C = sqrt(2/3 κ):

=>  a''          = C a' sqrt(ρ)

                       + C a * 1/2 * 1 / sqrt(ρ) * 

                          * (D_a ρ) a'

                    = C² a sqrt(ρ) sqrt(ρ)

                        + 1/2 C² a² sqrt(ρ) / sqrt(ρ)

                           * D_a ρ

                    = 2/3 κ a ρ

                       + 1/3 κ a² D_a ρ

<=>

     a'' / a     = 2/3 κ ρ                                           (2)

                      + 1/3 κ a D_a ρ.

The second Friedmann equation is:

     a'' / a       = -1/3 κ ρ - κ p / c²                      (F2)

Now we subtract (F2) from (2). We assume that we have a solution for (F1), which means that (2) is true. Then (F2) is true if and only if the subtracted equation is true:

    (F2) <=>

     0            = ρ

                      + 1/3 a D_a ρ

                      + p / c².

Our formula for pressure (1) makes the right side of the equation zero, which means that the subtracted equation is true if we assume energy conservation. QED.

Friedmann equations and changing the metric of time

UPDATE October 29, 2021: We updated the text to reflect the fact that the theorem in the October 16, 2021 blog post is erroneous.

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Our previous blog post on October 23, 2021 assumed that the Einstein field equations by M. W. Cook (1975) are correct for an isotropical metric:

https://www.publish.csiro.au/ph/PH750413

We got an astonishing result: by rescaling the time coordinate t and the metric of time b(t) we can always add a new term to the Friedmann equations.

The Einstein tensor is made from the Riemann curvature tensor, and the basic idea of the Riemann tensor is that it is coordinate independent. Simply rescaling the time coordinate should have no effect.

One can derive the Friedmann equations from symmetry arguments in a uniform universe. A solution of Einstein field equations must have a uniform metric of space.

Similarly, we can argue that a solution must have a uniform metric of time.

Assume that we have found such a uniform solution for the Einstein equations, and the global coordinate time is T. Then we can replace our global time coordinate T with the proper time coordinate t of an observer in the universe. We get a solution which has the Friedmann metric:

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

Because of coordinate independence, the solution with the new time coordinate t is a solution of the Einstein equations. It also a solution of the Friedmann equations.

From where does the cross term originate?

https://arxiv.org/abs/1210.5269#

Alan R. Parry (2014) has written in the link a very detailed derivation of the Ricci tensor for a varying metric of time.

The cross term comes from the first product term of two Christoffel symbols. It is the term V_t * M_t on the right. V represents the metric of time and M is the spatial metric.

Suppose that we take a history of the universe, and replace the global time coordinate t with

       t' = t        for t < 0,

       t' = 2 t     for t >= 0.

The metric b(t) of time suddenly drops to a half when t' = 0. Obviously the change of the time coordinate should not affect the Riemann tensor or the Ricci tensor.

However, if proper time slows down just in a part of the universe, that does introduce extra curvature.

The cross term probably should be something like

       d² b(t, r) / (dt dr)

to reveal the curvature which comes from non-uniform rate of proper time in the universe. That is, instead of a product of partial derivatives with respect to t and r, we should have a second derivative.

The error probably is in the use of Christoffel symbols. Instead of a product there should be a second derivative.

Riemann curvature is defined through a parallel transport of a vector. Literature claims that the curvature can be calculated through Christoffel symbols using the formula above, but we did not find any proof for this claim.

https://einsteinpapers.press.princeton.edu/vol6-doc/124

Albert Einstein in his paper on the foundations of general relativity in 1916 writes that the curvature can be calculated with that formula from Christoffel symbols. He uses the curvy bracket notation {...} for the symbols.

Almost all known solutions for Einstein equations have the cross term zero, because the metric of time does not change in them. The cross term does appear in the Oppenheimer-Snyder 1939 paper, but it is not relevant there.

https://archive.org/details/raumzeitmateriev00weyl/page/118/mode/2up

Hermann Weyl's 1918 book "Raum. Zeit. Materie." states that the Einstein tensor of general relativity is defined through the above formula of Christoffel symbols. It might be that Albert Einstein introduced the formula, and no one ever checked if it is coordinate independent.

Does the cross term say anything about the Riemann curvature?

If we have an arbitrary function

       f(x, y),

we may imagine that it describes the height of each point on a hill. Then

       d²f / dx²,   d²f / dy²

might measure "curvature", while

       d²f / (dx dy),    d²f / (dy dx)

might measure "torsion". On the other hand, the cross term

       df / dx * df / dy

can be non-zero even if the surface is a plane.

If the hill is circularly symmetric, then

       d²f / dr²

tells us the curvature. If that is zero everywhere but on the top, then the hill is a cone.

On a cone,

       d²f / dx², d²f / dy²

are != 0.

Let us imagine observers living on the surface of a Riemannian manifold

       (M, g).

The observers obey the metric g and measure the angles of triangles, whose sides have to be straight (geodesic). They deduce the curvature of space from the sum of angles, if it is 180 degrees or something else.

This setup is self-evidently coordinate independent. The metric determines what the observers see or measure.

Correcting the Christoffel symbol formula makes general relativity more sensible?

The correct formula for Riemann or Ricci curvature has to be coordinate independent. We will try to correct the Christoffel symbol formula.

A basic idea of general relativity is that mass focuses straight lines drawn to the direction of time, and pressure focuses straight lines drawn to spatial directions. How can we focus lines to spatial directions without stretching the spatial metric? It might be that a corrected Christoffel symbol formula changes general relativity to a "rubber membrane" model where pressure does affect the spatial metric. Then there might exist a solution for a sudden pressure change.

In this blog we have been perplexed by the interior Schwarzschild metric, where pressure does not affect the spatial metric at all. That is very different from a rubber membrane model. We tried to design a perpetuum mobile which utilizes the infinite rigidity of space. If the corrected formula removes this strange feature, then general relativity looks better.

Friedmann equations might have solutions after we have corrected the Christoffel symbols. If pressure can affect the spatial metric, then there is more freedom, and a solution may exist. It is like a rubber membrane which can adapt to all kinds of stresses.

https://en.wikipedia.org/wiki/Ricci_curvature

Wikipedia says that the Ricci tensor is defined in the (unique?) way, such that calculating the Ricci tensor for new coordinates defined by a mapping y, is surprisingly easy, requiring only the Jacobian matrix of first derivatives for the mapping y.

However, this is not what coordinate independence should be. It should be that observers living on the manifold, under the metric, measure triangles, and calculate the curvature. The curvature should not depend on an arbitrary choice of coordinates y at all. It is not enough that the Ricci curvature is easy to calculate for the new coordinates y.

The obvious way to get coordinate independence is to use locally proper distances and proper time at a given point p as the framework. Essentially, it is a local observer on the manifold who measures the curvature. Then the choice of coordinates has no effect whatsoever on the calculated curvature.

If proper time and proper distances are approximately the same as coordinate time and coordinate distances, then the Ricci curvature calculated from the coordinates is approximately the same as the Ricci curvature measured by a local observer.

Unfortunately, using proper distances and times as the framework does not work in a cosmological FLRW model where k = 0. In such a universe all observers measure that curvature is zero, which makes it impossible to satisfy the Einstein field equations.

It may be that we need to introduce a preferred frame where curvature is measured from global coordinates. This is in the spirit of our Minkowski & newtonian model, where inertial Minkowski coordinates are the preferred frame.

Conclusions

Moving to a preferred frame would prevent us from changing coordinates in a way where the rogue cross term can attain arbitrary values.

Spherically symmetric Einstein equations: the second Friedmann equation is always satisfied

UPDATE November 2, 2021: The calculations below are erroneous. The stress energy tensor has to be transformed to the new coordinates.

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UPDATE October 25, 2021: There may be an error in the M. W. Cook formulae. The cross term

       ν_t * λ_t  =  b' / b * a' / a

seems to break the coordinate independence of the Einstein tensor, which is built from the Riemann tensor.

Let us assume that we have a solution of the Friedmann equations.

Then we change the global coordinate t to another coordinate T where T is "stretched" in an interval t = 1 ... t = 2. We keep the metric unchanged relative to t.

The cross term was zero with the original coordinate t. It is strange if it appears after a pure coordinate transformation of t to T. Riemann curvature is about transporting a vector around on the manifold, guided by the metric. An observer moving on the manifold uses the metric as his guide. A pure coordinate transformation should not affect what the observer does on the manifold.

Alternatively, Riemann curvature is about measuring triangles on the manifold, guided by the metric. The observer checks if the sum of their angles is 180 degrees. A pure coordinate transformation cannot affect what the observer measures.

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Let us look at the full spherically symmetric Einstein equations.

We use the metric and the equations of M. W. Cook (1975).

https://www.publish.csiro.au/ph/PH750413

The time coordinate is t, and the comoving radial coordinate is r. Let us use a simpler denotation:

       b² = exp(ν),

       a² = exp(λ),

or

       ν = 2 log(b)

       λ = 2 log(a).

That is, a is the scale factor for r, and b tells the proper time per coordinate time.

A spatially uniform metric which does not depend on r at all

We assume that the solution is spatially uniform. The values

       b(t),

       a(t)

depend only on the coordinate time t, not on the radial coordinate r, or other coordinates. All the derivatives marked with a subscript r above are zero.

We denote the derivative with respect to t with a prime mark '. We have

       λ'  =  2 a' / a,

       λ'' =  2 [a'' / a  -  (a' / a)²].

The two equations are then

          -2 κ ρ    =  -3/2 * 1 / b² * (2 a' / a)²,

    -2 κ p / c² r   =   1 / b² * { 4 [a'' / a - (a' / a)²]

                                            + 3/2 (2 a' / a)²

                                             - 2 b' / b * 2 a' / a }.

In a neater form they are

         κ ρ / 3    =  1 / b² * (a' / a)²,                  (1)

        -κ p / c²   =  1 / b² * { 2 a'' / a                 (2)

                                          + (a' / a)²

                                          - 2 b' / b  *  a' / a }.

One may subtract the upper equation from the lower equation to obtain:

  -κ / 6 * (3 p / c² + ρ)   =   1 / b²  *  {a'' / a   -    b' / b * a' / a}.

If we set b(t) identically to 1, these are the Friedmann equations with the spacetime curvature parameter k = 0.

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

In the equations above, it is conspicuous that for each derivative ... / dt relative to the coordinate time t, there is the factor 1 / b. We can remove the factors 1 / b² from the front of the right sides of the equations, and write all the derivatives in the form

       ... / (b dt).

There, b dt is the duration of dt in the proper time of an observer inside the universe. The derivatives are now relative to the proper time, and the equations are:

                            κ ρ / 3  =  (a' / a)²,

       -κ / 6 * (3 p / c² + ρ)  =  a'' / a   -   b' / b * a' / a.

They are the Friedmann equations (k = 0) except for the term b' / b * a' / a in the lower equation.

We may ask why there is so much freedom in the choice of b(t). Since b(t) is uniform in the whole universe, changing the "speed" of time has essentially no effect on observers living in the universe. It is like playing a movie at different speeds.

The theorem also answers our question if pressure changes can explain dark energy: the answer is no in the case k = 0. We can use a pressure change to force a' / a higher or lower (where the derivative is against the coordinate time t), but the first equation (1) above says that the effect is always exactly canceled by a change in b(t). An observer sees no change in the Hubble constant a' / a in his own proper time.

A Robertson-Walker metric: the spatial metric depends on r

Now we allow a(t, r) depend on r, but b(t) only depends on t.

Note that what we call a(t, r), is

       a(t) sqrt(1 / (1 - r² / K²))

in the formula of the Robertson-Walker metric above.

If K² is positive, then r is restricted to values r < K.

http://diposit.ub.edu/dspace/bitstream/2445/59759/1/TFG-Arnau-Romeu-Joan.pdf

J. Arnau Romeu (2014) derives in the link the Friedmann equations from the Robertson-Walker metric, the formula above.

The value K² = ∞ corresponds to the flat spatial space, which we considered in the previous section. If K² > 0, then the spatial space is the surface of a sphere whose radius is K. If K² < 0, then the spatial space is hyberbolic.

Let us analyze what happens when we allow the metric of time to vary and a(t, r) depends on r, too. The formula of the Robertson-Walker metric above would change in a way where the term dt² is replaced with

       b(t)² dt².

The spherically symmetric Einstein equations of M. W. Cook are:

In the previous section we showed that we can absorb the term b(t)² (= exp(ν) in the formula above) into the derivatives with respect to the coordinate time ... / dt. The time derivatives become derivatives with respect to the proper time.

The formulae after that look like the ones where b(t) is identically 1, except that in the second formula we have the extra term

        -λ_t  *  ν_t,

or

       -b(t)' / b(t)  *  a(t)' / a(t).

If k != 0, then the first Friedmann equation contains a term k c² / a² which comes from the term c² exp(-λ) ... of the first M. W. Cook equation.

If λ_t = a' / a suddenly changes, then b(t) still has to compensate the change to satisfy the first equation of M. W. Cook. The observer does not see any change in the expansion rate of the universe.

Gravity can "fool" other forces by manipulating the metric of time b(t) and win them?

We speculated in our October 20, 2021 blog post that gravity can "fool" other forces by manipulating the potential, and always come up with enough energy to satisfy the demands of other forces. Gravity always wins.

Manipulating the potential is manipulating the flow of time.

The observed accelerated expansion of the universe must be a result of increased mass-energy density ρ(t). If there is an unknown non-gravity attractive force of either visible or dark matter, then ρ(t) grows. That would explain the acceleration. We do not need to appeal any strange dark energy field which fills space. An ordinary force is enough.

A summary of the Minkowski & newtonian gravity so far: a photon falls through the horizon in finite global Minkowski time

UPDATE October 29, 2021: We updated the text to reflect that the theorem of our October 16, 2021 blog post is incorrect.

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We are claiming in this blog that curvature of spacetime is just an illusion created by complex effects of newtonian gravity in the global flat Minkowski metric.

Our view is a new interpretation of general relativity. We expect our interpretation to be completely equivalent to the standard geometric interpretation of general relativity.

If our interpretation follows the mathematics of the Einstein-Hilbert action, then it will certainly reproduce the same effects as the standard interpretation. However, we have not proved that our interpretation totally conforms to the Einstein-Hilbert action.

Let us look at some questions which we have raised in the past two weeks.

Lowering down of a mechanical clock in a Schwarzschild potential

https://en.wikipedia.org/wiki/Schwarzschild_metric

We claim that the standard Schwarzschild coordinates r and t are the "true" coordinates of the global flat Minkowski metric outside the event horizon. For an observer in the potential well, proper time has slowed down, and proper radial distances are stretched, relative to the global Minkowski coordinates.

The apparent slowing down of time deep in the gravity potential well is due to:

1. the increased inertia of a mechanical clock part and

2. the weakening of forces (the gravity between the central mass and the clock does not weaken, though).

The movement of a clock part moves large energy in the gravity field, which causes extra inertia. The extra inertia is many times the original inertia when we are close to the horizon.

The energy which is used to start the clock has done work when we lowered it down, because the energy was pulled by the gravity field. Consequently, there is less energy available to move a clock part.

If the potential is -1% * m c², then the horizontal inertia is 1% larger and forces are 1% weaker. The clock ticks 1% slower.

When a photon dives down in the 1% potential, it retains its energy since the work it did goes to its own (kinetic) energy. The inertia of the photon grows by 1% in horizontal movement, and the horizontal speed of light is 1% lower.

In vertical movement, the inertia is 2% higher, and the measuring rod has contracted 1%. A clock, whose parts do vertical movement, ticks as fast as a clock whose parts do horizontal movement.

A photon falling down to the event horizon: it typically passes the horizon in less than a millisecond of global Minkowski time

In the standard Schwarzschild coordinates, the proper time relative to the Schwarzschild coordinate time has slowed down by a factor

      sqrt(1 - r_s / r),

where r_s is the Schwarzschild radius of a central mass M (the mass is measured from far away), and the observer is at the distance r from the center in Schwarzschild coordinates.

Proper radial distances have stretched by the inverse factor

       1 / sqrt(1 - r_s / r).

Therefore the radial speed of light in Schwarzschild coordinates is slowed down by the factor

        1 - r_s / r.

The coordinate time t (not proper time) that it takes a descending photon to reach the horizon, if it starts from R, is

        

                           R

       t =  1 / c  *  ∫   1 / (1 - r_s / r) dr

                      r_s

                         R

       =  1 / c  *  ∫   r / (r - r_s) dr

                    r_s

                         R - r_s

       =  1 / c  *  ∫   r_s / x dx,

                      0

if R is only slightly larger than r_s. We have denoted x = r - r_s.

The elapsed coordinate time t diverges logarithmically. As if the photon would float forever close to the horizon.

However, in the Minkowski & newtonian model the slowdown of light is due to the extra inertial mass it gains from the field of the black hole, and that inertial mass cannot be infinite.

The smallest possible factor for the slowdown of light is

       m / M,

where m is the energy of the photon divided by c². The largest possible value of the integrand above is M / m. That is reached when x = r_s m / M. We calculate the corrected integral separately for [0, x] and [x, R - r_s].

The corrected value for the integral is

       t = 1 / c * M / m * (r_s * m / M)

            + 1 / c * r_s * ( log(R - r_s) - log(r_s m / M) )

 

         = r_s / c

            + r_s / c * ( log(R / r_s - 1) - log(m / M) ).

Since -log(m / M) is typically a large number, 100 or more, we can ignore smaller terms and write:

       t = r_s / c * log(M / m).

An example: we let a 1 MeV photon fall into a solar mass black hole. The logarithm is then log(10⁶⁰) = 138. It takes at most

       t = 1.38 ms

of global Minkowski time for the photon to fall through the horizon. We cannot really say that the photon stays "floating" at the horizon. Floating is an idealization which would happen at an infinite mass black hole, or if the photon has infinitesimal energy.

The collapse of a uniform ball of dust

According to our calculation in the previous section, a large mass will fall through a horizon in a millisecond or less of the global Minkowski time.

Furthermore, our "syrup" model of gravity suggests that in a large collapsing mass, light is dragged along the mass. Even though light slows to a crawl relative to the mass, the mass itself can move fast as seen in external Minkowski coordinates.

We believe that the combination: external Schwarzschild metric and internal FLRW metric describes the collapse of a uniform ball of dust (an Oppenheimer and Snyder collapse) in our Minkowski & newtonian gravity model. That is, the description of general relativity is correct.

Question. What are the "true" global Minkowski coordinates for the interior of the collapsing ball? That is, if we take into account complex effects of the newtonian gravity field.

In the case of the curvature k = 0, proper distances inside the FLRW part probably are the spatial global Minkowski coordinates. But what is the time coordinate?

In our calculation in the previous section we assumed that a single photon can acquire the entire mass of the black hole as its inertial mass. That probably is an exaggeration.

Singularities

It looks like that the Minkowski & newtonian model creates the same singularities as general relativity.

That is not a big surprise because a uniform dust ball does collapse to a singularity in the classic, year 1687 version of newtonian gravity.

If the dust ball has passed its Schwarzschild radius, gravity, apparently, can "fool" other forces. If another force resists contraction, gravity can lower the potential of the collapsing mass, and in that way provide whatever energy the other force is demanding.

Birkhoff's theorem states that a spherically symmetric collapse has to keep contracting if it has passed the Schwarzschild radius. All light cones point downward. The only way that gravity can prevail over a stronger resisting force is that gravity must fool that other force.

Thus, the Minkowski & newtonian model does not save us from singularities. However, uncertainty principles seem to prevent singularities, according to our previous blog post.

The Big Bang is much like a reverse collapse of a uniform ball of dust, in both general relativity and the Minkowski & newtonian model. Quantum mechanics saves us from a singularity there, too - at least in the Minkowski & newtonian model.

A singularity in a black hole destroys information: quantum mechanics saves it

Stephen Hawking believed that a black hole is characterized just by its mass, charge, and angular momentum.

He also believed that a black hole can evaporate through radiation, and that the radiation does not contain the information that was dumped into the black hole.

This is the famous black hole information paradox: can a black hole destroy information by first devouring it, and then losing the total energy of the black hole in radiation which does not contain that information?

We in this blog have not been able to derive the existence of Hawking radiation. If the radiation would destroy information, that would be strong evidence against its existence.

The information paradox of singularities

However, we have another information paradox in a black hole. If the information which it devours, is squeezed into a zero volume, we can say that it destroyed information. A singularity, whether a point or a ring, is a bad thing. A rotating black hole in the Kerr metric is assumed to have a ring singularity.

Note that an idealized collapse of uniform dust to a point does not destroy much information, since uniform dust contained just a few bits of information. A realistic collapse of some 10⁶⁰ elementary particles into a point or a ring singularity would destroy lots of information.

Quantum mechanics bans singularities

Quantum mechanics comes to our rescue and saves the information. Let us assume that there exists a global frame of reference where we can measure things.

We believe that the true metric of spacetime is the Minkowski metric, and gravity only creates an illusion of curved geometry. Thus, we can take an inertial frame in the Minkowski space, and work in that inertial frame.

Assume that elementary particles, like photons, continue their existence. The particles are very tightly bound together through gravity. They collectively form an "atom" with ~ 10⁶⁰ particles in a solar mass black hole.

Just like in the case of the electron in the hydrogen atom, we cannot know the position of an individual particle too precisely, because then the particle would have more energy than is available.

We cannot know the relative positions of particles precisely, either. There would be infinite energy in the interaction if we did.

When a particle interacts with a large mass, the wavelength of the particle becomes lower. An example is a photon on the surface of a neutron star. A global observer, and also a local observer, see the wavelength shorter. The global observer interprets that the photon has acquired more inertial mass in the interaction, and the speed of light is consequently lower. The local observer thinks that the photon was blueshifted when it came down.

In principle, the global observer might see that a photon inside the black hole has gained the entire mass of the black hole as the inertial mass of the photon. In our syrup model of gravity, that would mean that the black hole is made of perfect syrup.

If that is the case, the wavelength of a photon inside a solar mass black hole is at least

       λ = h c / (M c²)

          = h / (M c)

          = 10⁻⁷² m,

where M is the mass of the Sun. The wavelengh is tiny, but it is not zero.

Thus, a hypothetical singularity cannot be a singularity at all but is a soup of waves (or particles).

Assumptions. In our argument we had to assume:

1. there exists a global frame of reference;

2. energy is conserved in the global frame: particles cannot gain infinite energy;

3. particles do not merge in a way which destroys information;

4. uncertainty relations prevail over classical general relativity.

Are our assumptions self-evident? Assumption 1 is controversial. Assumption 2, conservation of energy, is a rock-solid principle of physics. Assumption 3 is implied by unitarity. Assumption 4 is true in everyday physics - it would be surprising if it does not hold in general relativity.

We are suggesting that uncertainty relations prevent singularities from forming. In a uniform dust collapse, no classical force can resist the contraction, dictated by Birkhoff's theorem. However, quantum mechanics works with waves. Waves cannot be squeezed into a point.

Does a realistic classical collapse produce a singularity?

Uniform dust is not a realistic configuration. If we let a real star to collapse, the spacetime geometry will initially be very complicated behind the event horizon.

Let us assume that the angular momentum of the star is zero.

Particles will initially zigzag behind the horizon. Some particles are still close to the horizon and try to move at the light speed outward. They can stay close to the horizon for a long time.

However, we expect most particles to move downward, because the light cones point down.

Intuitively, we should have a cloud of particles, where the upper layers of the cloud become thinner as time passes. More and more particles come close to the center.

Is there information loss when black holes merge and lose a lot of mass-energy in gravitational radiation?

If we have one million black holes of equal mass M, we can, in principle, release about 99.9% of their mass-energy in gravitational radiation by letting them merge together in pairs of equal mass.

99.9% is a lot, but it does not break the Bekenstein entropy formula, because the area of the horizon is a million times larger in a black hole of a mass 1,000 M than in a black hole of a mass M.

Gravitational radiation is coherent, and does not carry much entropy. The bulk of the entropy must stay in the black hole.

We can probably store a million times more information if we make the number of particles 1,000-fold. There is presumably no information loss.

Minkowski & newtonian gravity allows singularities? The syrup model

In our October 7, 2021 blog post we claimed that the Schwarzschild geometry can be derived from the Minkowski metric and newtonian gravity, if one takes into account all the effects of the newtonian gravity field. Increased inertia and weakening of forces makes clocks to tick slower deep in a gravity potential well.

We claimed that an observer falling into a Schwarzschild black hole is frozen close to the horizon and never reaches the horizon. The speed of light is zero at the horizon. We claimed that the observer cannot move faster than the speed of light.

However, we we not sure what happens inside the horizon. Can matter still move there?

In traditional newtonian gravity (in the style of the year 1687), a collapse of a dust ball does end up in a singularity. Can this happen in our Minkowski & newtonian model?

Our considerations of a collapsing sphere of dust in the blog posts October 16 and 18, 2021 suggest that collective movements of matter can still happen inside the horizon. In our example of the two neutron stars approaching each other, the speed of light inside the star and near it can be close to zero relative to the moving mass, but the neutron stars still move at a great speed.

Syrup and pancakes. Photo https://www.publicdomainpictures.net/en/view-image.php?image=242756

It is like syrup: an observer finds it impossible to move quickly relative to the syrup, but the syrup can still flow fast.

The mass-energy of a collapsing star is huge. Even if an individual test mass inside the horizon would appear to have negative energy, the collective material holds firmly positive energy. Thus, the problem of a negative mass-energy does not really appear in a collapsing star.

Hypothesis. A collapse to a black hole is essentially a traditional newtonian (1687) collapse, where the collapsing matter and the space near it is "syrup": an observer finds it impossible to escape from the collapse. He is stuck in the syrup.

Our Minkowski & newtonian model starts to resemble general relativity more and more.

Question. How does the syrup behave? Can it flow all the way to a singularity? Or does pressure from some other force win gravity?

If we dethrone gravity from position as the geometry of spacetime, and demote it to an ordinary force, then the Einstein-Hilbert action may allow very large pressure from some other force to stop the contraction of a star.

We showed in our October 16, 2021 blog post that a sudden increase of pressure breaks the Friedmann equations. In an Oppenheimer-Snyder collapse, the internal metric of a collapsing uniform ball of dust is the FLRW metric, that is, a solution of the Friedmann equations. But pressure breaks those equations. The road to singularity is not so simple then.

If we again look at the newtonian gravity (1687) analogue, there strong pressure does stop the contraction, though a dust ball would collapse to a singularity.

Question. What happens to a small test mass which we drop to the horizon of an old black hole? Does it "go with the flow", or freeze just above the horizon?

Close to the horizon, a clock ticks extremely slowly because its mechanical parts have huge inertia, and non-gravity forces are extremely weak. However, the inertia probably cannot exceed the mass of the black hole, and the progress of the test mass downward is not dependent on non-gravity forces. The test mass might slip through the horizon in a finite time.

The waterfall and sound analogy of a black hole and a neutron star

The syrup model reminds us of the waterfall analogy of a black hole. Close to the waterfall, the water flows faster than the speed of sound in water. The event horizon is at the point where the speed of water is equal to the speed of sound. A sound wave behind the horizon cannot escape.

Photo https://www.publicdomainpictures.net/en/view-image.php?image=1309

The corresponding analogy for a neutron star is a waterfall where the speed of water does not exceed the speed of sound. A sound wave can escape, but it moves slowly against the flow of water.

In these analogies, the flow of water is created by the mass itself. A single sound wave cannot escape from behind the horizon, but through a collective effort, the mass itself might be able to escape.

A white hole is a waterfall with time reversed. We may imagine that it is the movement of the great mass upward which forces the water to flow up in the waterfall.

Turning dark matter into radiation would explain dark energy in cosmology?

UPDATE October 29, 2021: We updated the blog post to reflect that the theorem of the October 16, 2021 blog post is erroneous.

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UPDATE October 24, 2021: Our blog post on October 23, 2021 proved that pressure changes cannot explain dark energy.

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Let us use the Friedmann equations and consider a phase transition where massive matter turns entirely into radiation.

The metric of time does not change in a Friedmann solution. Only the mapping of spatial, comoving, coordinates to proper distances changes.

For massive matter, the pressure p = 0, while for radiation,

       p = 1/3 ρ / c².

In the second equation, the acceleration suddenly doubles when massive matter is converted into radiation.

Let us analyze what happens.

Assumptions. 1. We assume that prior to the phase transition, the universe is dominated by uniform massive matter, and the metric obeys the Friedmann solution

       a(t)  ~  t^(2/3),

that is, a(t) is "almost" linear on t, if we regard 2/3 as almost 1. The expansion slows down at a moderate pace.

2. We assume that k = 0 and Λ = 0.

When the pressure suddenly goes up, we believe that the expansion starts to slow down faster.

A sudden jump in the pressure in an Oppenheimer-Snyder collapse

Robert Oppenheimer and Hartland Snyder in their famous 1939 paper were only able to calculate the pressureless p = 0 case, that is, the collapse of a uniform dust ball.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455

https://arxiv.org/abs/2006.07418

Valerio Faraoni and Farah Atieh (2020) mention the Mashhoon and Partovi result (1980) that the metric inside a collapsing uniform dust ball must be FLRW.

What would happen if uniform large pressure would suddenly appear at a late stage of the collapse?

Since a Friedmann solution uses comoving coordinates, an observer inside the star knows what are the Friedmann coordinates for each particle. The observer can measure the Hubble constant H = da / dt * 1 / a, as well as the matter density ρ. The first Friedmann equation is strictly about observations.

Large pressure lowers the gravitational potential of an observer inside the star. In the previous blog post we argued that an external observer would see the star to continue contracting at roughly the same pace after the pressure is raised. The time of an internal observer slows down. He observes a sudden rise in the Hubble constant.

In this blog we believe that the "true" geometry of spacetime is the flat Minkowski metric, and curved spacetime is just an illusion created by complex effects of newtonian gravity. The Big Bang might be a reverse Oppenheimer-Snyder collapse.

Calculation of the observed increase of the expansion rate in an expanding ball

The expansion rate of the universe appears to have increased by some 20% relative to the t^(2/3) solution in the past 4 billion years.

Let us assume that we live inside an expanding uniform ball in an asymptotic Minkowski space.

The assumed amount of dark matter in the visible universe is so large that the expanding ball might even be inside its own Schwarzschild radius.

How much does pressure lower the potential of a test mass m inside the ball?

The Schwarzschild metric around the mass m stretches the spatial metric of radial distances. If we consider a sphere of a thickness dr at the distance r, its volume grows by

       dV = r_s / r * r² dr

             = r_s r dr

             = 2 G m / c² * r dr,

where r_s is the Schwarzschild radius of the 1 kg mass. Integrating over r gives:

       ΔV = G m / c² * r².

The radius of the observable universe is 46.5 billion light-years, which is 4.65 * 10²⁶ meters. A 1 kg mass increases the volume of the observable universe by

       ΔV = 3 * 10²⁵ cubic meters.

That is a cube whose side is 300,000 km.

Let us assume that the observable universe holds a pressure p. The potential of the test mass m is

       -p ΔV.

If the potential is -0.2 m c², then time would be slowed down by 20% for the test mass. Let us calculate the required pressure. It is

       p = 7 * 10⁻¹⁰ Pa,

which is the opposite number to the assumed negative pressure from dark energy in ΛCDM.

To create the positive pressure p, the radiation must have three times the energy density of the assumed dark energy density in ΛCDM.

In our calculation we assumed that the test mass m only increases the volume of the observable universe. Is this right? Probably not. The test mass affects the metric of the entire expanding ball. If we take the radius r in the above calculation to be much larger, then the required pressure is much lower.

Conclusions

It looks like that we can explain the observed acceleration in the expansion of the universe through massive dark matter turning into dark radiation, but the details are not yet clear.

We do not yet understand what happens if the expanding ball is inside its Schwarzschild radius.

We have not analyzed the effect of large pressure during the first 380,000 years of the universe.

Friedmann equations have no solution for a sudden pressure jump? Yes they do!

UPDATE October 29, 2021: There is an error in the proof of the Theorem. In a phase change, the exponent n changes. The formula (1) is incorrect and the Theorem is incorrect.

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The standard FLRW model of the universe is built on the Friedmann equations:

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

The formula of the metric has a constant c² factor before the time element dt². A static observer has ds₃ = 0. Then we can define that dt is the proper time interval measured by a static observer anywhere in the universe, and t is his proper time.

The Friedmann equations let pressure only affect the second time derivative of the scale factor a(t):

The time derivative is denoted by the dot over the symbol a (the Newton notation of the derivative).

In principle, we can change the pressure rapidly inside the universe. The FLRW universe should have uniform pressure, but let us relax that requirement, and increase the pressure quickly in a half of the universe, and keep it constant in another half.

Static clocks suddenly tick slower in a half of the universe. Static observers in those zones see the time derivative of the scale factor a(t) to jump suddenly. That is, those observers see the expansion to accelerate rapidly.

The rest of the static observers do not see any jump? Do they see a sudden slowdown?

Let us then raise the pressure also in that half which did not get a pressure lift. Now the pressure is uniform again.

Question. Do static observers in the whole universe see a sudden jump in the expansion rate?

Alexander Friedmann (1888 - 1925). https://commons.wikimedia.org/wiki/File:Aleksandr_Fridman.png#mw-jump-to-license

Two stars approach each other, and we change their internal pressure

                   star                         star

                    ___                           ___

                  /       \                       /        \

                  \____/                      \____/

                    ---->                        <----

                              ● external

                                   observer

The question above is related to a thought experiment where two stars approach each other and there are observers inside the stars. It is a kind of a "contracting universe". By changing the pressure inside the stars, we can make the clocks of the internal observers to tick faster or slower.

Clearly, the internal pressure inside the stars cannot affect much the speed seen by an external observer.

However, observers inside the stars see a sudden jump or slowdown in the contraction rate of their mini-universe.

If we put the stars inside a huge pressurized vessel, then increasing the pressure suddenly makes the inertia of the stars to grow, because then they are carrying the energy of the pressure field around. In that case, the an external observer sees the speed of the stars to slow down suddenly.

However, if the stars themselves create the pressure, the pressure cannot affect their inertia much.

The FLRW metric inside a newtonian expanding ball

In literature several sources say that a newtonian expanding ball with uniform density has the Schwarzschild metric outside. Birkhoff's theorem dictates the external metric.

They also claim that the interior metric is the FLRW metric.

Let us suddenly increase the pressure inside the ball.

If we can ignore the acceleration of the matter at the edge of the ball, then the FLRW metric claims that there is no sudden jump in the expansion rate, as seen by internal observers.

The pressure is created by the matter of the ball itself. Can the pressure increase enough the inertia of, say, the left half and the right half of the ball, so that the expansion rate would suddenly slow down in the eyes of an external observer, and internal observers would see the expansion rate to stay constant?

Each half does increase the volume of the other half through its gravity field. Thus, each half does get some extra inertia from the pressure force field of the other half. Does this extra inertia slow down the halves enough, so that observers inside the halves see the expansion to continue at the old rate?

The Friedmann equations do not have a solution at all for a sudden pressure jump

Once we have chosen for the universe some scale factor value a(t), and the time derivative of a(t), at a certain initial time t₀, then the second Friedmann equation determines the scale factor a(t) uniquely for all times t.

However, there is no guarantee that the first equation is satisfied. Maybe in most cases there is no solution at all for the Friedmann equations?

Theorem. If we have a solution for the Friedmann equations, but change the setup so that we add a sudden increase in pressure at a certain time, then there is no solution for the Friedmann equations. We assume here that the time derivative of the scale factor a(t) is not zero at the time of the pressure increase. We assume that a(t) is a continuous function and that its second time derivative is defined.

NOTE October 29, 2021: the proof is incorrect. The value of n changes in the phase change, which spoils the calculation of the perturbation in (1).

Proof. Let us assume that we have a solution a(t) for the Friedmann equations. We assume that the time derivative of a(t) is not zero at t = 0.

We change the original setup so that we suddenly increase the pressure p at the time t = 0 by some fixed amount.

Let us assume that b(t) is the new solution for t > 0.

The second Friedmann equation says that the second time derivative of a(t) jumps up suddenly. Then we can write:

       b'(t) = a'(t) - C t

for some constant C, for small t > 0. We denoted the time derivative with the prime mark '.

Then we have

       b(t) = a(t) - C t² / 2,

for small t > 0, and

       ρ_b(t) = ρ(t) / [ 1  -  C t² / 2  *  1 / a(t) ]ⁿ,   (1)

where n is a real number in the range [3 ... 4], and we denote by ρ_b(t) the new solution for small t > 0.

We have the square of t in the formulae for b(t) and ρ_b(t). If t > 0 is very small, we may assume that b(t) = a(t) and ρ_b(t) = ρ(t).

Then we immediately see that the first Friedmann equation is not satisfied for very small t > 0. QED.

Corollary. The metric of time has to change in a spherically symmetric universe if there is a sudden uniform change in pressure. We assume here that the Einstein equations have a solution for such a setup - otherwise we can no longer talk about a metric.

Proof. If there is no change in the metric of time, then there is a solution of the Friedmann equations. But we showed above that there is no Friedmann solution. QED.

Corollary. Assume that the universe satisfies the Friedmann equations before a sudden pressure change. After the pressure change, the universe cannot return to a standard massive matter & radiation solution of the Friedmann equations.

Proof. If the universe would return to a standard solution a(t) after a pressure change at a time t₀, then we could continue that solution all the way down to the time t₀, because the universe would obey the standard pressure rules of the matter content.

Then we can show just like in the proof of Theorem that for a very short time interval before t₀, the Friedmann equations cannot be satisfied. But that contradicts the assumption that the equations were satisfied before t₀. QED.

It looks like the Friedmann equations do not have a solution in most cases. Thus, in most universes the metric of time has to change.

The standard cosmological model ΛCDM and changes of pressure

The standard ΛCDM model of cosmology assumes that the universe was radiation-dominated until the age of 380,000 years.

Then it became matter-dominated.

At about 4 billion years ago, it became dominated with dark energy, which is expressed in the cosmological constant Λ.

Does ΛCDM assume that the time of a static observer has the same metric in all these phases? That is probably wrong. Friedmann equations probably have no solution for the pressure changes.

If pressure can affect the metric of time, then increasing pressure from dark matter (dark radiation) might slow down time, and the expansion of the universe would appear to accelerate, even though it keeps decelerating from the viewpoint of an observer outside the universe. It might be that most of dark matter was massive particles until some 4 billion years ago, and then decayed into radiation.

The dark energy field which creates energy from nothing is a very ugly idea in ΛCDM. There is no such field in the physics that we know. All known fields conserve energy.

A much better hypothesis is pressure from dark radiation.

A newtonian ball cannot contain a dark energy field because that would break energy conservation. But there is no problem in assuming an increase of pressure when matter is turned into radiation.

What is the history of dark matter?

From astronomy we can deduce the history of visible matter: it switched from radiation-dominated to matter-dominated about 380,000 years after the Big Bang.

But we do not know the history of dark matter. It might be that dark matter has switched between matter and radiation phases many times in the past 13.5 billion years.

If dark matted switched to radiation 4 billion years ago, that may explain the illusion of an accelerated expansion of the universe.

There may be surprises in the first 380,000 years, too.