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

## Gravitational redshift

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

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

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

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

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 ${\displaystyle r}$can be computed by the following formula:

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

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

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 ${\displaystyle M_{eff}}$ 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 ${\displaystyle m}$. If the distance ${\displaystyle d}$ between the mass ${\displaystyle m}$ 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 ${\displaystyle 2\,R-d}$, where ${\displaystyle R}$ is radius of the universe (cf. previous section "Retarded gravitational potentials"):

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

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

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

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

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

This means if the distance ${\displaystyle d}$ is a tenth of the radius ${\displaystyle R}$ 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

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 ${\displaystyle \lambda _{M}}$ in front of the mass ${\displaystyle m}$ 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: ${\displaystyle 1\,{\text{ly}}=9.46\cdot 10^{15}{\text{m}}}$
• Hubble length: ${\displaystyle R=1.36\cdot 10^{26}{\text{m}}=14,4\cdot 10^{9}{\text{ly}}}$ (14.4 billion light-years)

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

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

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

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

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

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

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

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

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

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

With the auxiliary sagitta ${\displaystyle e}$ we get:

${\displaystyle e=R\,\left(1-\cos \alpha \right)}$
${\displaystyle \cos \beta ={\frac {d-e}{s}}}$

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

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

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

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

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

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

### Schwarzschild distance

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

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

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

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

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

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

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

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

The Schwarzschild distance ${\displaystyle d_{S}}$ can be interpreted as the distance between the surface of the Hubble sphere with the Hubble radius ${\displaystyle R}$ and the inner Schwarzschild sphere of the black shell, which is the visible limit of the universe. The Hubble radius ${\displaystyle R}$ 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 ${\displaystyle t_{H}}$:

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

### 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 ${\displaystyle q_{M}}$:

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

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

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

In this model the mass ${\displaystyle M_{S}}$ 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.

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 ${\displaystyle M_{S}=2.8769\cdot 10^{53}\,{\text{kg}}=0.96865\cdot M_{universe}}$, Schwarzschild distance ${\displaystyle d_{S}=377\cdot 10^{3}\,{\text{ly}}=3.56\cdot 10^{21}\,{\text{m}}=2.62\cdot 10^{-5}\cdot R}$), 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 ${\displaystyle M_{S}=2.92\cdot 10^{53}\,{\text{kg}}=0,984\cdot M_{universe}}$, Schwarzschild distance ${\displaystyle d_{S}=293\cdot 10^{6}\,{\text{ly}}=2.78\cdot 10^{24}\,{\text{m}}=0.020\cdot R}$), 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 ${\displaystyle M_{S}=2.97\cdot 10^{53}\,{\text{kg}}=M_{universe}}$).

### Schwarzschild radius of the universe

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

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

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 ${\displaystyle M_{universe,visible}}$ or the gravitational constant ${\displaystyle G}$. 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 ${\displaystyle t}$ and reciprocal to the mass of the visible universe ${\displaystyle M_{universe,visible}}$.[1]

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

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

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

Today this value is approximately ${\displaystyle \Lambda =1.7\cdot 10^{-53}{\text{ m}}^{-2}}$, 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:

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

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 ${\displaystyle 2.2\cdot 10^{-5}}$:

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

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:

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

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

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

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:

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

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:

${\displaystyle r_{S,universe}-r_{CMB}\approx 0.9\cdot 10^{9}{\text{ ly}}}$
${\displaystyle r_{H}-r_{V}\approx 0.8\cdot 10^{9}{\text{ ly}}}$

Therefore:

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

This assumption is supported by the fact that the mass of ordinary matter within radius of the visible universe ${\displaystyle r_{visible,universe}}$ has the following value that can be computed by its density ${\displaystyle \rho _{ordinary\,matter}}$ and the corresponding volume ${\displaystyle V_{visible,universe}}$ yielded by the Planck Survey of 2013:[2]

${\displaystyle \rho _{ordinary\,matter}=4.08\cdot 10^{-28}{\frac {\text{kg}}{{\text{m}}^{3}}}}$
${\displaystyle r_{visible,universe}=4.3\cdot 10^{26}{\text{ m}}}$
${\displaystyle V_{visible,universe}={\frac {4}{3}}\,\pi \,r_{visible,universe}^{3}=3.3\cdot 10^{80}{\text{ m}}^{3}}$
${\displaystyle M_{ordinary\,matter}=\rho _{ordinary\,matter}\cdot V_{visible,universe}=1.4\cdot 10^{53}{\text{ kg}}\approx {\frac {1}{2}}\,M_{universe}={\frac {1}{2}}\,2.97\cdot 10^{53}{\text{ kg}}\approx 1.5\cdot 10^{53}{\text{ kg}}}$

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

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

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 ${\displaystyle M_{ordinary\,matter}}$ 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 ${\displaystyle M_{S}}$. 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.

### References

1. Milne, Edward Arthur (1935). "World picture on the simple kinematic model - Comparison with local Newtonian gravitation and dynamics - §§412-418". Relativity, gravitation and world-structure. Oxford, Great Britain: Clarendon Press. pp. 291–294.
2. Tatum, Eugene Terry (2015-06-01). "Could Our Universe have Features of a Giant Black Hole?". Journal of Cosmology. 25: 13063–13072.