What's Wrong With The Robertson-Walker Metric, and Why a Critical Mass Has No Meaning

The Robertson-Walker metric supposedly describes a homogeneous and isotropic universe.It supposedly gives the criterion of a critical mass of the universe whose deficiency has led to the supposition of "dark matter."

It is commendable that the same criterion also follows from an elementary, nonrelativistic, derivation where the total kinetic and gravitational energies of the universe are conserved at two different times


where the subscript "0" refers to here and now, $t_0$, and the unsubscripted symbols to a future time. If we introduce the present Hubble law, $v_0=H_0R_0$, and the total mass of the universe $M_0=\rho_04\pi R_0^3/3$, then as $t\rightarrow\infty$, $R\rightarrow\infty$ so that

$$v^2\rightarrow R_0^2\left(H_0^2-\frac{8\pi}{3}G\rho_0\right).$$

If, at some point in the evolution of the universe, $v$ should vanish then there will result in a critical density

$$\rho_c\equiv\frac{3H_0^2}{8\pi G}.$$

From this three universes can be delineated: One where $v^2>0$ and the density $\rho_0<\rho_c$, a flat, "finely tuned" universe with $v^2=0$, and a third universe with $"v^2<0"$ with $\rho_0>\rho_c$, resulting in a big "crunch". This is notwithstanding the fact that the critical density was derived by letting $R(t)\rightarrow\infty$ as $t\rightarrow\infty$. 

Be that as it may, it is "amazing enough" in the words of Goldberg and Scradon, in their text Physics of Stellar Evolution and Cosmology, that a detailed analysis of the theory of general relativity leads to the same constraints and values for $\rho_c$ as the classical description given above." But, what about relativistic effects that surely play a role in the evolution of the universe, the transverse forces involved in the evolution of rotational galaxies, and the geometry of spacetime that general relativity holds so precious?

The Tully-Fisher law should dissuade us from believing in a critical mass as given above. That law predicts that the velocity is proportional to the fourth root of the luminosity which is proportional to the mass, $v\propto M^{1/4}$.Instead of the usual gravitational force,

$$\frac{v^2}{r}\propto \frac{GM}{r^2},$$

that law requires the force to be of the form


where $a_0$ has dimensions of an acceleration. This is the modification made by Milgrom to explain the 'flat' rotation curves of spherical galaxies  than being disc shaped. Notice that the force  is proportional to $1/r$ instead of the $1/r^2$. This is just like a radiation term (c.f. the electric force below). As an alternative, it has been suggested that  "dark" matter makes up a good chunk of the matter in the universe. 

$F$ can be rationalized by considering a tangential acceleration. The tangential force is what results when the gravitational force becomes aberrated,

$$F_t\propto \frac{GM}{r^2}\cdot\frac{1}{3}\left(\frac{v}{c}\right)^3,$$

where $v=\omega r$ is the rotational velocity. This can be written as

$$F_t\propto\frac{GM}{r^2}\cdot \frac{r\omega^2}{c^2}\cdot \left(\frac{r\omega}{c}\right)=\frac{GM}{rc^2}a\left(\frac{r\omega}{c}\right),$$

where $a=r\omega^2$ is the tangential acceleration. Not only has an acceleration been introduced, but, also, the inverse square has been reduced to an inverse radius. The first factor shows how small it is, being the ratio of the Schwarzschild radius to the actual radius. All this shows that a simple law, $v^2\propto GM/r$ in determining a critical mass is much too naive--and so are the conclusions.

The supposed "equivalence" with general relativity occurs through the Robertson-Walker metric. To see the skulduggery afoot, it suffices to consider the Landau and Lifshitz derivation in their book Classical Fields. According to them, to "investigate the metric it is convenient to start from a geometrical analogy, by considering the geometry of an isotropic three-dimensional space as the geometry of a hyperspace known to be isotropic, in a fictitious four-dimensional space." They are quick to point out that this has nothing to do with "four-dimensional space-time." The question is "Why not?"

The flat space time metric of the four dimensional space can be written as


The equation of a hypersphere of radius $1/\kappa^2$ in four dimensions has the form


On account of this constraint,



$$-c^2dt^2=dx_4^2=\frac{(x_idx_i)^2}{1/\kappa^2-x_ix_i}=\frac{\kappa^2 rdr}{1-\kappa^2r^2},$$

where the last term follows when spherical coordinates, $x_1=r\sin\theta\cos\phi$, $x_2=r\sin\theta\sin\phi$, and $x_3=r\cos\theta$, are introduced. That is,



$$dx_1dx_1+dx_2dx_2+dx_3dx_3=dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2,$$

so that the metric becomes


where the $+$ and $-$ signs apply to whether $x_4$ just another "space" variable, or $ict$. It does not reduce to the space component of the Robertson-Walker metric

$$dl^2=\frac{dr^2}{1-\kappa^2r^2}+r^2d\theta^2+r^2\sin^2\theta d\phi^2,$$

unless $\kappa=0$, which defeats the whole purpose of the exercise. 

If the Robertson-Walker metric,

$$ds^2=c^2dt^2-\frac{dr^2}{1-\kappa^2r^2}-r^2d\theta^2-r^2\sin^2\theta d\phi^2,$$


$$ds^2=c^2(1-\kappa^2r^2)dt^2-\frac{dr^2}{1-\kappa^2r^2}-r^2d\theta^2-r^2\sin^2\theta d\phi^2,$$

making it look more like the "inner" solution to the Schwarzschild metric, then it would take on the aspect of the Beltrami metric,

$$c^2dt^2=dr^2\left(\frac{1}{(1-\kappa^2r^2)^2}+r^2\frac{\theta^{\prime 2}+\sin^2\theta \phi^{\prime 2}}{1-\kappa^2r^2}\right),$$

where the prime denotes differentiation with respect to $r$. This would mean that the space-time metric has meaning only for null-geodesics that are followed by light rays. Otherwise, it has no meaning.

The point in the above expression is that $t$ and $r$ are interchanged in the Beltrami metric, which is the Lobachevsky metric of velocity space. The square of the relative velocity $v^{\prime}$ of the velocities $\vec{u}$ and $\vec{v}$ in two different planes is

$$v^{\prime 2}=\frac{(\vec{u}-\vec{v})^2-(\vec{u}\wedge\vec{v})^2/c^2}{(1-\vec{u}\cdot\vec{v}/c^2)^2}.$$

For infinitesimal differences, $\vec{u}=\vec{v}+d\vec{v}$, this becomes

$$d\vec{v}^{\prime 2}=\frac{(1-\beta^2)(d\vec{v})^2+(\vec{\beta}\cdot d\vec{v})^2}{(1-\beta^2)^2},$$

where $\beta=v/c$ is the relative velocity. Note that the aberration term, $\vec{u}\cdot\vec{v}$ has become $\beta^2$. A further simplification allows us to write the above expression as

$$d\vec{v}^{\prime 2}=(d\vec{v})^2\frac{(1-\beta^2+\beta^2\cos^2\omega)}{(1-\beta^2)^2}\equiv(d\vec{v})^2\frac{K^2(\omega)}{(1-\beta^2)^2},$$

where our old friend,


Recalling our previous blog,


$s'=c(t-t')$ is the distance from the observer to the light source at the retarded time $t'$, and $s$ is its distance at the present epoch. Hence, the above expression is the inverse velocity. The aberration term does not appear since in forward and backward paths it vanishes since $\beta$ changes sign, like in an interferometer. So depending on which velocity is proportional to $cdt$, 


represents a velocity or its inverse.

The two velocities, $c$ and $c'$ are depicted in the diagram, and occur when two circles intersect orthogonally. The angle $A$ has been dubbed the "banana" angle by Thurston. Physically, it is the angle formed by $s$ and $s'$ at the observer.

The two velocities are inverses of one another, resulting when two circles intersect orthogonally. The angle A is the "banana" angle.
The velocities c and c' are inverses of one another. 

If the distances $s$ and $s'$ are interpreted as current elements, $ds$ and $ds'$, as in Ampere's theory, the angle $A$ represents induction. It tells us how much $ds$ and $ds'$ interact. The distance of separation between $ds$ and $ds'$ is $r$, which is a function of both. For retarded times, the source will have moved $\beta s'$.

The electric field is composed of  velocity and acceleration fields


where $\vec{\gamma}'\equiv\vec{a}'/c$ is the relative acceleration at the retarded time. The interesting point, which will allow us to draw the analogy between Grassmann-Ampere electrodynamics and radiation is that in the Grassmann force, $s$ and $s'$ are replaced by the current elements $ids$ and $i'ds'$, and the relative acceleration by the unit vector, $\hat{r}_{s,s'}$, pointing either in the direction of $ds$ or $ds'$, depending on which component of the force is being considered. If the force is directed to the element $d\vec{s}$, the Grassmann force is


for currents $i$ and $i'$, where $\hat{r}_s$ is the unit vector along $r$ connecting the elements and directed toward $d\vec{s}$ at which the force is being applied. This has the same form as the radiative term in the electric force. However, whereas there is a complete symmetry between the elements, $ds$ and $ds'$, there is no symmetry between $s$ and $s'$ because they are causally related. That is, the above force would be pointing "backward" in time, and not forward in time as it should. The development of the analogy will be saved for a separate blog.

As a final point, we want to comment on the derivation of the critical density from the virial theorem that we discussed in "Weber's Electromagnetic Theory as an Optical Gravitational Theory." Recall that Ampere's law was written as

$$ r^2\frac{d^2F}{ids\cdot i'ds'}=\frac{1}{2}\frac{d^2{\mathcal{I}}}{dsds'}-\frac{3}{2}\frac{dr}{ds}\frac{dr}{ds'},$$

where ${\mathcal{I}}$ is the 'moment of inertia' per unit mass.

To convert it into a dynamical law, we wrote $ids$ as $edt/c^2$, where $c$ is the ratio of the electromagnetic to electrostatic units of charge. In Ampere's law it cancels out,


but not in Weber's law since is an amalgam of a static law, Coulomb's, and kinetic law, Ampere's. The introduction of a static law, like Coulomb's or Newton's, therefore requires the conversion factor $c^2$. It is therefore desirable to all the left-hand side of the above equation, the 'virial' (of Clausius), and set it equal to 


The virial is said to be the potential energy of a cloud of non-interacting particles. From the first equality, we find


where $1/\surd(G\rho)$ is the Newtonian free-fall time. This essentially rends the force $F$ adimensional.

Moreover, in a steady state ${\mathcal{I}}$ is constant so that the virial reduces to


Introducing Hubble's constant, $H$, according to $dr/dt=v=Hr$, the above relation gives three times the critical density found above: $\rho_c=9H^2/8\pi G$, or equivalently, a critical Newtonian free-fall time, $\tau_c=8\pi/9H^2$. This can hardly be considered as the critical density of the universe, and underscores the limitations of such an expression. Even if we were to use the mechanical virial for $k=0$, we would get


which would lead to twice the critical velocity, $\rho_c=3H^2/4\pi G$, found from the conservation of energy above.

 Rather than being an "amazing" coincidence between  classical, non-relativistic and general relativistic results, it points to its complete inadequacy of being considered as a serious candidate for a critical velocity of the universe.