Saturday, October 23, 2021

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).


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.


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.


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.

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