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

Schwarzschild radius


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.

For comparison:

The Schwarzschild radius of the universe can be computed by its mass, too:


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

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 $$M_{eff}$$ 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 for the gravitational forces behind the moving mass $$m$$. If the distance $$d$$ between the mass $$m$$ and the shell is given, we can estimate 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:


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



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


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: $$R = 1.36 \cdot 10^{26} \text {m}$$ (equals 14,4 billion light-years)
 * Mass of the universe: $$M_{universe} = 2.97 \cdot 10^{53} \text {kg}$$:

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 vertical components of the gravitational forces are symetrical, and therefore, their net effect is zero.

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


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



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 $$\alpha$$:


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


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

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 horizontal force of this arc $$F_{hor}$$ that is accelerating the mass $$m$$ in horizontal direction can be achieved by integrating the arc with the correction factor $$\cos \beta$$:


 * $$F_{hor} = G \, m \int_{-\alpha_R}^{+\alpha_R} \frac {\cos \beta} {s^2} \, \mathrm {dM} = G \, m \, R \, \lambda_M \, \int_{-\alpha_R}^{+\alpha_R} \frac {\cos \beta} {s^2} \, \mathrm {d\alpha}$$

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:


 * $$F_{hor} = G \, m \frac {M_{eff}} {d^2} = G \, m \, R \, \lambda_M \, \int_{-\alpha_R}^{+\alpha_R} \frac {\cos \beta} {s^2} \, \mathrm {d\alpha}$$

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


 * $$M_{eff} = R \, \lambda_M \, d^2 \int_{-\alpha_R}^{+\alpha_R} \frac {\cos \beta} {s^2} \, \mathrm {d\alpha} = \frac {M_S \, d^2} {2 \pi} \int_{-\alpha_R}^{+\alpha_R} \frac {\cos \beta} {s^2} \, \mathrm {d\alpha}$$

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 $$\lambda_M$$:


 * $$d_S (\lambda_M) = \frac {2 \, G} {c^2} M_{eff} (M_S, d_S) = \frac {2 \, G \, R \, \lambda_M \, d_S^2} {c^2} \int_{-\alpha_R}^{+\alpha_R} \frac {d_S - R \, (1 - \cos \alpha)} {{\left( 2R^2 - 2 \, R \, d_S + d_S^2 - 2 \, R \, (R - d_S) \, \cos \, \alpha \right)}^{\frac {3} {2}}} \, \mathrm {d\alpha}$$

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 $$t_H$$:


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

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 $$R_v = R - d_S$$.

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.

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 $$M_S = 2.8769 \cdot 10^{53} \, \text {kg} = 0.96865 \cdot M$$, 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 $$M_S = 2.92 \cdot 10^{53} \, \text {kg} = 0,984 \cdot M$$, 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.