Moving objects in retarded gravitational potentials of an expanding spherical shell/Gravitational redshift

Gravitational redshift

Schwarzschild radius

Relation between the Schwarzschild radius $r_{S}$ and a redshift of an emitted photon $z$ in a distance $r$ caused by the gravitation of the massive sphere in the centre.

The Schwarzschild radius $r_{S}$ of any mass $M$ in a Schwarzschild sphere (typically a black hole) is given by:

$r_{S}={\frac {2\,G\,M}{c^{2}}}$

Apart from its mass, the Schwarzschild radius is only depending on two natural constants:

• Gravitational constant: $G=6.67\cdot 10^{-11}{\frac {{\text{m}}^{3}}{{\text{kg}}\,{\text{s}}^{2}}}$
• Speed of light: $c=299792458{\frac {\text{m}}{\text{s}}}$

The gravitational redshift z of photons emitted to the opposite direction of the Schwarzschild sphere and with a distance to the centre of the sphere $r$can be computed by the following formula:

$z={\frac {1}{\sqrt {1-{\frac {2\,G\,M}{r\,c^{2}}}}}}-1={\frac {1}{\sqrt {1-{\frac {r_{S}}{r}}}}}-1$

The redshift of very far objects such as the galaxy JADES.GS.z14-0 has a value of more than 14, and this is much larger than expected. The possible high speed of such a galaxy is not sufficient for such a high value. The distance of this galaxy is given by 13.5 billion light-years, and its age is assumed to be 290 million years after Big Bang.

The cosmic microwave background even has a redshift value of 1089, which is extremely high. It is associated with the first hydrogen atoms that occurred some 380,000 years after the Big Bang.

In a shell

The sum of all gravitational forces between the mass $m$ and a spherical arc (blue) with the linear mass density $\lambda _{M}$ that represents a hollow spherical cap. It has the same effect as the gravitational force between the mass $m$ and an effective spherical mass $M_{eff}$ (black).

However, a gravitational redshift can not only occur outside of spheres, but also within in a hollow spherical cap. To estimate its gravitational redshift, the effective mass $M_{eff}$ of such a cap can be integrated for any point within the cap. The corresponding effect can be described by a Schwarzschild sphere with the Schwarzschild radius.

For very fast-moving objects we can assume that they only experience the retarded gravitational potentials of the mass elements in front of them, since the backward potentials are even much more retarded, and therefore, contribute only weakly to the net gravity. In the following section only the arc of the shell cap is considered. The gravitational forces rectangular to the direction of movement are very small and can be neglected.

This also holds in first approximation for the gravitational forces behind the very fast-moving moving mass $m$. If the distance $d$ between the mass $m$ and the shell is given, we can carry out an estimation for the fraction of the retarded forces from the opposite part of the shell in the distance $2\,R-d$, where $R$ is radius of the universe (cf. previous section "Retarded gravitational potentials"):

$F_{right}\propto {\frac {1}{d^{2}}}$

$F_{left}\propto {\frac {1}{(2\,R-d)^{2}}}$

If we assume that the distance is a fraction of the radius

$d={\frac {R}{n}}$

then the ratio of the two forces has the following relation:

${\frac {F_{right}}{F_{left}}}=\left({\frac {2\,R-{\frac {R}{n}}}{\frac {R}{n}}}\right)^{2}=(2\,n-1)^{2}$

The ratio of the retarded gravitational forces in the distance of oberserved objects versus the corresponding redshifts

This means if the distance $d$ is a tenth of the radius $R$ of the universe, the error by neglecting the left force would be less than 0.3 percent. This distance is corresponding to a redshift of about 7.5.

In the diagram on the left the ratio of the two forces is plotted against the redshift which is observed at corresponding distances. The high values of redshift at far distant objects seem to be dominated by the gravitaional redshift, whereas at shorter distances the redshift is dominated by the Doppler effect.

Computation

Diagram for integrating the effective mass of an outer shell $M_{eff}$ acting on a fast moving mass $m$ by varying the centre angle $\alpha$ from $-\alpha _{R}$ to $\alpha _{R}$.
$R$ = radius of the visible universe
$d$ = distance of mass $m$ to the outer shell
$\lambda _{M}$ = linear mass density of the outer shell

The bold arc on the right-hand side of the diagram is representing all mass elements of the outer shell of the universe with the linear mass density $\lambda _{M}$ in front of the mass $m$ that is moving to the right with high velocity. The outer shell is consisting of dark matter (mainly hydrogen), and in the central area of the universe this dark matter might only be visible due to the cosmic microwave background.

We use the following constant values for the estimates derived from these premises:

• Light-year: $1\,{\text{ly}}=9.46\cdot 10^{15}{\text{m}}$
• Hubble length: $R=1.36\cdot 10^{26}{\text{m}}=14,4\cdot 10^{9}{\text{ly}}$ (14.4 billion light-years)

The mass of this arc $M_{arc}$ can be computed by integrating the arc between the angles $-\alpha _{R}$ and $+\alpha _{R}$ with its linear mass density $\lambda _{M}$:

$\alpha _{R}=\arcsin {\frac {x}{R}}$

$M_{arc}=\int _{-\alpha _{R}}^{+\alpha _{R}}\mathrm {dM} =\int _{-\alpha _{R}}^{+\alpha _{R}}R\,\lambda _{M}\,\mathrm {d\alpha } =2\,\int _{0}^{\alpha _{R}}R\,\lambda _{M}\,\mathrm {d\alpha } =2\,R\,\lambda _{M}\,\alpha _{R}$

The mass of the whole shell $M_{S}$ is given by integrating a complete circle:

$M_{S}=\int _{-\pi }^{+\pi }\mathrm {dM} =2\pi \,R\,\lambda _{M}$

The Pythagorean theorem gives the following result for the half chord of the circle:

$x={\sqrt {2\,R\,d-d^{2}}}$

Detail for integrating the infinitesimal mass element $\mathrm {dM}$ acting on a mass $m$ by varying the angle $\alpha$.
$\alpha$ = angle as seen from origin $\bigcirc$
$R$ = radius of the universe (Hubble length)
$h$ = height og the mass element $\mathrm {dM}$
$d$ = distance of mass $m$ to the outer shell
$s$ = distance between mass $m$ and mass element $\mathrm {dM}$
$e$ = auxiliary sagitta
$\beta$ = angle as seen from $m$

If the distance $d$ between the mass $m$ and the outer shell is given, we can compute the distances of the mass $m$ to any infinitesimal mass element $\mathrm {dM}$ on the arc depending on the angle $\alpha$:

$\alpha =\arcsin {\frac {h}{R}}$

$s(\alpha )={\sqrt {2R^{2}-2\,R\,d+d^{2}-2\,R\,(R-d)\,cos\,\alpha }}\,\geq \,d$

The angle $\beta$ as seen from the moving mass m to the infinitesimal mass element $\mathrm {dM}$ on the arc can be derived by applying the law of sines and is given by the following expression:

$\beta =\arcsin \left({\frac {R}{s}}\sin \,\alpha \right)=\arcsin {\frac {h}{s}}$

With the auxiliary sagitta $e$ we get:

$e=R\,\left(1-\cos \alpha \right)$

$\cos \beta ={\frac {d-e}{s}}$

The vertical components of the gravitational forces are symetrical, and therefore, their net effect is zero. The net horizontal force of this arc $F_{hor}$ that is accelerating the mass $m$ in horizontal direction can be achieved by integrating a semi-circle with the correction factor $\cos \beta$:

$F_{hor}=G\,m\int _{-{\frac {\pi }{2}}}^{+{\frac {\pi }{2}}}{\frac {\cos \beta }{s^{2}}}\,\mathrm {dM} =G\,m\,R\,\lambda _{M}\,\int _{-{\frac {\pi }{2}}}^{+{\frac {\pi }{2}}}{\frac {\cos \beta }{s^{2}}}\,\mathrm {d\alpha }$

With the effective gravitational force $F_{hor}$ that acts on the mass $m$ by the gravitational force of the virtual effective mass $M_{eff}$ that is located in the intersection point of the trajectory of the mass $m$ with the outer shell, we have to consider the varying distances $s$ between $m$ and the infinitesimal mass elements $\mathrm {dM}$ along the arc:

$F_{hor}=G\,m{\frac {M_{eff}}{d^{2}}}=G\,m\,R\,\lambda _{M}\,\int _{-{\frac {\pi }{2}}}^{+{\frac {\pi }{2}}}{\frac {\cos \beta }{s^{2}}}\,\mathrm {d\alpha }$

We finally get the effective mass $M_{eff}$ that is acting via gravitational forces on the mass $m$ in the distance $d$:

$M_{eff}=R\,\lambda _{M}\,d^{2}\int _{-{\frac {\pi }{2}}}^{+{\frac {\pi }{2}}}{\frac {\cos \beta }{s^{2}}}\,\mathrm {d\alpha } ={\frac {M_{S}\,d^{2}}{2\pi }}\int _{-{\frac {\pi }{2}}}^{+{\frac {\pi }{2}}}{\frac {\cos \beta }{s^{2}}}\,\mathrm {d\alpha }$

Schwarzschild distance

The Schwarzschild distance $d_{S}$ of this effective mass $M_{eff}$ is equal to the Schwarzschild radius of a sphere with the effective mass:

$d_{S}={\frac {2\,G\,M_{eff}}{c^{2}}}$

The gravitational redshift $z$ of photons emitted to the centre of the universe that is caused by the effective mass in the distance $d$ of the outer shell is:

$z={\frac {1}{\sqrt {1-{\frac {2\,G\,M_{eff}}{d\,c^{2}}}}}}-1={\frac {1}{\sqrt {1-{\frac {d_{S}}{d}}}}}-1$

The equations above can be solved for the Schwarzschild distance $d_{S}$ of every shell with the mass $M_{S}$. With the following condition we can get the solutions with $d_{S}=d$, where $z(d_{S})=\infty$:

$2\,G\,M_{eff}(M_{S},d_{S})=d_{S}\,c^{2}$

Therefore $d_{S}$ can be determined for a given $\lambda _{M}$:

$d_{S}(\lambda _{M})={\frac {2\,G}{c^{2}}}M_{eff}(M_{S},d_{S})={\frac {2\,G\,R\,\lambda _{M}\,d_{S}^{2}}{c^{2}}}\int _{-{\frac {\pi }{2}}}^{+{\frac {\pi }{2}}}{\frac {d_{S}-R\,(1-\cos \alpha )}{{\left(2R^{2}-2\,R\,d_{S}+d_{S}^{2}-2\,R\,(R-d_{S})\,\cos \,\alpha \right)}^{\frac {3}{2}}}}\,\mathrm {d\alpha }$

The Schwarzschild distance $d_{S}$ can be interpreted as the distance between the surface of the Hubble sphere with the Hubble radius $R$ and the inner Schwarzschild sphere of the black shell, which is the visible limit of the universe. The Hubble radius $R$ would be the distance between an observer in the centre of the universe and all objects that are receding from him at the speed of light. It can be expressed by the Hubble time $t_{H}$:

$R=t_{H}\cdot c=14.4\cdot 10^{9}\,{\text{s}}\cdot {c}=14.4\cdot 10^{9}\,{\text{ly}}=1.36\cdot 10^{26}\,{\text{m}}$

• Redshift z as a function of the distance
• 
  Redshift z as a function of the distance of the object from the centre of the universe within the Schwarzschild limit, which asymptotically approaches the Hubble radius for large redshifts.
• 
  Circles with a given redshift z within the Hubble circle (dashed line) with the Hubble radius.

Results

It is assumed that the outer shell is expanding, spherical and consists of dark matter. The effective mass of the outer shell is computed only considering the geometrical region in front of the object. This assumption is based on the fact that the retarded gravitational potentials from the opposite border of the black shell can be neglected. The following diagrams show the solution in two different representations over the relation of the mass of the invisible universe surrounding the visible universe in a spherical shell in units of the mass of the visible universe $q_{M}$:

• The Schwarzschild distance $d_{S}$ between the visible border of the universe and the invisible outer shell of the universe.
• The radius of the visible universe $R_{v}=R-d_{S}$.
• Mass of the universe: $M_{universe}=2.97\cdot 10^{53}{\text{kg}}$:

The overall mass of the invisible black shell $M_{S}$ can be expressed in relation to the mass of the visible universe $M_{universe}$:

$q_{M}={\frac {M_{S}}{M_{universe}}}$

→ See appendix: table with results for different values of $\lambda _{M}$ numerically computed by a Java program.

In this model the mass $M_{S}$ of the invisible outer black shell continuously increases by absorbing mass from the visible universe, due to the net forces of the retarded gravitational potentials and the corresponding acceleration in direction of the black shell. More and more matter is moving from the visible universe behind the event horizon, where it becomes invisible and unaccessible, but leaves its gravitational action to the visible part of the universe. Furthermore, the mass of the black shell is in the same magnitude as the mass of the visible universe even in the earlier cosmical ages.

• Diagrams for the early evolvement of the universe considering the effects of the retarded gravitational potentials of an invisible outer shell
• 
  Scheme with the outer black shell of the invisible universe with dark matter (right), the shell of the origin of the cosmic microwave background (CMB, dark red) as well as a far galaxy (light red). The matter in the vicinity of the black shell is accelerated towards the shell due to the effect of retarded gravitational forces and may disappear into it by crossing the event horizon at the Schwarzschild distance. During the era of transforming dark matter into "dark energy" the Schwarzschild distance was fastly increasing.
• 
  Evolvement of the effective mass of the black shell of the universe with the decreasing radii of the visible universe and the increasing Schwarzschild distances as well as the increasing (effective) masses of the black shell.
• 
  Schwarzschild distance $d_{S}$ between the visible border of the universe and the invisible outer sphere of the universe as a function of the mass of the black shell in units of the mass of the visible universe. The Schwarzschild distance was grewing much faster than the mass of the black shell around $M_{S}\approx 0.96865\cdot M_{universe}$.
• 
  Radius of the visible universe $R_{v}=R-d_{S}$ as a function of the mass of the invisible universe surrounding it in a spherical black shell in units of the mass of the visible universe. The size of the visible universe began to shrink very suddenly at the time around $M_{S}\approx 0.96865\cdot M_{universe}$.

It is very noteworthy to recapitulate that the age of the cosmic microwave background with its afterglow light pattern is about 380,000 years (the corresponding Schwarzschild distance belongs to a shell mass $M_{S}=2.8769\cdot 10^{53}\,{\text{kg}}=0.96865\cdot M_{universe}$, Schwarzschild distance $d_{S}=377\cdot 10^{3}\,{\text{ly}}=3.56\cdot 10^{21}\,{\text{m}}=2.62\cdot 10^{-5}\cdot R$), exactly where the radius of the visible universe had significantly begun to decrease in relation to the Hubble radius (see right diagram). The so-called "dark ages" begun, and they lasted for several hundreds of million years.

The age of the oldest known galaxy JADES.GS.z14-0 represents the youngest stars that emit light, and therefore, the end of the dark ages. Its age is about 290 million years (the corresponding Schwarzschild distance belongs to a shell mass $M_{S}=2.92\cdot 10^{53}\,{\text{kg}}=0,984\cdot M_{universe}$, Schwarzschild distance $d_{S}=293\cdot 10^{6}\,{\text{ly}}=2.78\cdot 10^{24}\,{\text{m}}=0.020\cdot R$), where the conversion from dark matter to "dark energy" during the so-called "dark ages" was finished.

At the break-even point, where the masses of the visible universe and the black shell have the same value, the Schwarzschild distance is about 860 million light-years (it belongs to a shell mass $M_{S}=2.97\cdot 10^{53}\,{\text{kg}}=M_{universe}$).

Schwarzschild radius of the universe

Simplified diagram of the radius $r$ of the universe over time from origin at $t=0$ to time of universe $t_{U}$. The particle horizon is constantly expanding with the speed of light $c$:

Radius of end of inflationary period (Schwarzschild radius of universe):
$r_{S,universe}=t_{U}\cdot c={\frac {2\,G\,M_{universe}}{c^{2}}}\approx 14283{\text{ Mpc}}\approx 46,6\cdot 10^{9}{\text{ ly}}$

Co-moving radius of the cosmic microwave background (CMB):
$r_{CMB}\approx 14000{\text{ Mpc}}\approx 45,7\cdot 10^{9}{\text{ ly}}$

Hubble radius at Hubble time $t_{H}$:
$r_{H}=t_{H}\cdot c\approx 4480{\text{ Mpc}}\approx 14,6\cdot 10^{9}{\text{ ly}}$

Radius of visible universe as seen from observer:
$r_{V}\approx 4230{\text{ Mpc}}\approx 13,8\cdot 10^{9}{\text{ ly}}$

Time of the observer: $t_{observer}=2\cdot t_{H}$
Time of the visible event horizon: $t_{v}=2\cdot t_{H}-{\frac {r_{V}}{c}}>t_{H}$

The Schwarzschild radius of the visible universe $r_{S,universe}$ can be computed by its mass, too:

$r_{S,universe}={\frac {2\,G\,M_{universe}}{c^{2}}}\approx 4.4\cdot 10^{26}{\text{ m}}\approx 46.6\cdot 10^{9}{\text{ ly}}$

This value fits very well with the radius of the invisible particle horizon at the end of inflationary period of the universe today. The question is whether this coincidence is the case by accident or due to varying values of the mass of the visible universe $M_{universe,visible}$ or the gravitational constant $G$. Of course it is disputable, whether the mass of the visible universe and the gravitational constant are constant on large time scales. However, the British mathematician and astrophysicist Edward Arthur Milne (1896–1950) published already in 1935 in his book on "Relativity, Gravitation and World-Structure" the possibility that the gravitational constant could be proportional to the time since the creation of the universe $t$ and reciprocal to the mass of the visible universe $M_{universe,visible}$.

$G={\frac {c^{3}}{M_{universe,visible}}}\,t$

According to Edward Arthur Milne the cosmological constant $\Lambda$ is in this model:

$\Lambda ={\frac {3}{c^{2}\,t^{2}}}$

Today this value is approximately $\Lambda =1.7\cdot 10^{-53}{\text{ m}}^{-2}$, which is about 6.6 times smaller than the current value of the standard model of cosmology known as the ΛCDM model (CDM = cold dark matter).

The time-dependent relation for the gravitational constant would mean that the gravitational constant is grewing with the following rate, if the mass of the universe keeps constant:

${\frac {\mathrm {d} G}{\mathrm {d} t}}={\frac {c^{3}}{M_{universe,visible}}}\approx 9.1\cdot 10^{-29}{\frac {{\text{m}}^{3}}{{\text{kg s}}^{3}}}$

As a result the gravitational constant would change in the following magnitudes, which are much smaller than the uncertainty of the measurements of the gravitational constant of around $2.2\cdot 10^{-5}$:

${\frac {\frac {\mathrm {d} G}{\mathrm {d} t}}{G}}\approx 1.36\cdot 10^{-18}{\frac {1}{\text{s}}}\approx 4.3\cdot 10^{-11}{\frac {1}{\text{a}}}$

Only after half a million of years the corresponding change would be in the magnitude of the uncertainty.

If we substitute the expression of Milne for the gravitational constant in the formula for the Schwarzschild radius of the universe, we get:

$r_{S,universe}={\frac {2\,c^{3}\,M_{universe}}{c^{2}\,M_{universe,visible}}}\,t$

Assuming that the whole universe is visible, and therefore, $M_{universe}=M_{universe,visible}$, this value would be two times larger than the value of the Schwarzschild radius computed by the mass of the universe:

$r_{S,universe}=2\,c\,t$

It is noteworthy to mention that according to the results the following estimates apply for Schwarzschild distances of around 858 million light-years, where the ratio of the mass of black shell to the mass of visible universe was equal to one:

$M_{universe,visible}\approx M_{universe,invisible}=M_{S}$

This Schwarzschild distance is similar to the differences of the particle horizon and the plane of the cosmic microwave background (CMB) that is determined by the ΛCDM model that is preferred by most cosmologists today:

$r_{S,universe}-r_{CMB}\approx 0.9\cdot 10^{9}{\text{ ly}}$

$r_{H}-r_{V}\approx 0.8\cdot 10^{9}{\text{ ly}}$

Therefore:

$M_{universe,total}=M_{universe,visible}+M_{universe,invisible}\approx 2\,M_{universe,visible}\approx 2\,M_{S}$

This assumption is supported by the fact that the mass of ordinary matter within radius of the visible universe $r_{visible,universe}$ has the following value that can be computed by its density $\rho _{ordinary\,matter}$ and the corresponding volume $V_{visible,universe}$ yielded by the Planck Survey of 2013:

$\rho _{ordinary\,matter}=4.08\cdot 10^{-28}{\frac {\text{kg}}{{\text{m}}^{3}}}$

$r_{visible,universe}=4.3\cdot 10^{26}{\text{ m}}$

$V_{visible,universe}={\frac {4}{3}}\,\pi \,r_{visible,universe}^{3}=3.3\cdot 10^{80}{\text{ m}}^{3}$

$M_{ordinary\,matter}=\rho _{ordinary\,matter}\cdot V_{visible,universe}=1.4\cdot 10^{53}{\text{ kg}}\approx {\frac {1}{2}}\,M_{universe}={\frac {1}{2}}\,2.97\cdot 10^{53}{\text{ kg}}\approx 1.5\cdot 10^{53}{\text{ kg}}$

However, if the gravitational constant would be given by the following relation using the total mass of the universe $M_{universe,total}$ including the black shell, the values would match much better:

$G={\frac {c^{3}}{M_{universe,total}}}\,t\approx {\frac {c^{3}}{2\,M_{universe,visible}}}\,t$

$r_{S,universe}\approx {\frac {2\,c^{3}\,M_{universe,visible}}{2\,c^{2}\,M_{universe,visible}}}\,t=c\,t$

This value is identical to the radius of the particle horizon of our universe. In this case the change of the gravitational constant would be in the magnitude of the uncertainty only after a million years. The total mass of the universe would not change or influence the gravitational constant even if any mass flow from the visible part of the universe to the invisible outer black shell would decrease the mass of the visible part of the universe.

The following applies for the hypothetical case that the Schwarzschild radius of the universe is only determined by the ordinary mass of the visible part of the universe, which is only half of the total mass of the universe. The Schwarzschild radius of the visible universe with the mass $M_{ordinary\,matter}$ that is completely visible to an observer in the centre of the universe, causes any light originating from the centre of the universe being invisible to any observer outside of this sphere, i.e. in the outer black shell with the similar mass $M_{S}$. Vice versa, any light originating from the outer black shell is invisible to any observer in the centre of the universe. If the these two regions share the very same spherical boundary surface, this surface would separate two worlds that obviously cannot exchange any information. According to this assumption, the Schwarzschild radius of our universe would define this spherical boundary surface and would be around 860 million light-years smaller than the radius of the particle horizon respectively the outer radius of the black shell.