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

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## Retarded gravitational potentials

Already the German physicist and astronomer Karl Schwarzschild (1873–1916) described retarded potentials for elecrodynamic fields (he still used the term "electrokinetic potential") in 1903.[1] These potentials with a delayed effect were adopted one year later with reference to Schwarzschild by the German mathematician Alexander Wilhelm von Brill (1842–1935) in Über zyklische Bewegung (English: "About cyclic movement"), where he coined the term "retardiertes Potential" (English: "retarded potential") in a footnote. Furthermore, he emphasized that these retarded potentials would cause a non-zero divergence in space, i.e. that there would be sources and sinks or that the medium would have the possibility of storing or releasing potential energy.[2] Retarded potentials are a mathematical description of potentials in a field theory in which a field quantity propagates at a finite speed (speed of light) and not instantaneously. They occur in the investigation of time-dependent problems, such as the radiation of electromagnetic waves, but also in the propagation of gravitational waves.

Objects that are moving towards the outer rim of the universe will experience retarded gravitational potentials by the expanding dark matter that can be assumed at the very edge of the universe. Therefore, the appropriate gravitational forces will have delayed effects, and due to the large distances and the finite velocity of gravitational wave propagation they also will be weaker in the direction of the former location of the objects. As a result, the net gravitational force is directed in the direction of movement of these objects, and therefore, all objects that move outwards will be accelerated in the direction of their own movement, which would become observable as an accelerated expansion of the visible universe.

The gravitational potential ${\displaystyle \Phi }$ due to a mass ${\displaystyle M}$ in the distance ${\displaystyle s}$ is given by:

${\displaystyle \Phi (s)=-{\frac {G\cdot M}{s}}}$

The gravitational force ${\displaystyle F}$ to another mass ${\displaystyle m}$ results as follows:

${\displaystyle F=-m\,{\frac {\mathrm {d} \Phi (s)}{\mathrm {d} s}}=G\,{\frac {m\cdot M}{s^{2}}}}$

The time-depending potential ${\displaystyle \Phi ({\vec {x}},t)}$ is the solution of the inhomogeneous wave equation, where ${\displaystyle v({\vec {x}},t)}$ is the inhomogeneity, and ${\displaystyle c}$ is the speed of the propagation of the waves. For gravitational forces we can consider it as equal to the vacuum speed of electromagnetic waves:

${\displaystyle {\frac {1}{c^{2}}}\,{\frac {\partial ^{2}\Phi ({\vec {x}},t)}{\partial t^{2}}}-\Delta \Phi ({\vec {x}},t)=\Box \Phi ({\vec {x}},t)=v({\vec {x}},t)}$,

where ${\displaystyle \Delta =\nabla ^{2}}$ is the Laplace operator, ${\displaystyle \Box }$ is the D’Alembert operator.

The solution of the inhomogeneous wave equation is called retarded potential, and in three dimensions it can be given as:

${\displaystyle \Phi _{\text{ret}}({\vec {x}},t)={\frac {1}{4\pi }}\int {\frac {v({\vec {s}},t_{\text{ret}})}{|{\vec {x}}-{\vec {s}}|}}\,\mathrm {d} ^{3}s}$

The retardation is to be interpreted in such a way that a source element ${\displaystyle v({\vec {x}},t_{\text{ret}})}$ at the point ${\displaystyle {\vec {s}}}$ and at the time ${\displaystyle t_{\text{ret}}}$ only influences the potential at the distant point of impact at ${\displaystyle {\vec {x}}}$ at a later time ${\displaystyle t}$:[3]

${\displaystyle t=t_{\text{ret}}+{\frac {|{\vec {x}}-{\vec {s}}|}{c}}}$
${\displaystyle t_{\text{ret}}=t-{\frac {|{\vec {x}}-{\vec {s}}|}{c}}}$

${\displaystyle t_{\text{ret}}}$ is called the retarded time. At the location ${\displaystyle {\vec {x}}}$ and at the time ${\displaystyle t}$ the retarded potential only depends on the inhomogeneity ${\displaystyle v}$ in the retarding back cone of the location. This inhomogeneity has a retarded effect to the solution, and it is delayed with the wave velocity ${\displaystyle c}$.

### In a shell

In the simplified example in the adjacent figure the source term is the linear mass density ${\displaystyle \lambda _{m}}$ that is not time-dependant and only exists in the circle of the outer rim with the constant radius ${\displaystyle r}$ and its centre at ${\displaystyle \bigcirc }$:

${\displaystyle x_{\bigcirc }=0}$
${\displaystyle y_{\bigcirc }=0}$
${\displaystyle x^{2}+y^{2}=r^{2}\rightarrow \lambda _{m}={\text{ constans}}}$

In all other locations within the plane the linear mass density ${\displaystyle \lambda _{m}}$ is zero:

${\displaystyle x^{2}+y^{2}\neq r^{2}\rightarrow \lambda _{m}=0}$

All mass elements outside the regarded plane have an symmetrical effect to the mass, and therefore, in these locations the contribution of the mass elements to the potential can be neglected for the determination of inhomogeneity:

${\displaystyle z\neq 0\rightarrow \lambda _{m}=0}$

Furthermore, the mass element ${\displaystyle \mathrm {d} M}$ on the homogenous circumference ${\displaystyle C}$ of a circle with the radius ${\displaystyle r}$ is given by:

${\displaystyle \mathrm {d} M=\lambda _{m}\cdot \mathrm {d} C=\lambda _{m}\cdot r\cdot \mathrm {d} \alpha }$
${\displaystyle M=\int dM=\int \limits _{0}^{2\pi }\lambda _{m}\cdot r\cdot \mathrm {d} \alpha =\lambda _{m}\cdot 2\pi \cdot r=\lambda _{m}\cdot C}$

The cosine formula gives us the relation between the location of the mass point ${\displaystyle m}$ in the horizontal distance ${\displaystyle x}$ of the origin ${\displaystyle \bigcirc }$ of the circle with the radius ${\displaystyle r}$, when the mass element on the circle ${\displaystyle \mathrm {d} M}$ is in the direction of the angle ${\displaystyle \alpha }$ and the distance ${\displaystyle s}$:

${\displaystyle r^{2}=x^{2}+s^{2}-2xs\,\cos(\pi -\alpha )=x^{2}+s^{2}+2xs\,\cos \alpha }$

In the normalised standard form of the quadratic equation, we get:

${\displaystyle s^{2}+2xs\,\cos \alpha +x^{2}-r^{2}=0}$

The solution for the distance ${\displaystyle s}$ is:

${\displaystyle s={\sqrt {x^{2}\,\cos ^{2}\alpha -x^{2}+r^{2}}}-x\,\cos \alpha }$

It is obvious that the following simplifications are valid:

• ${\displaystyle x=0\rightarrow s=r}$
• ${\displaystyle \alpha =0\rightarrow s=r-x}$
• ${\displaystyle \alpha =\pm {\frac {\pi }{2}}\rightarrow s={\sqrt {x^{2}+r^{2}}}}$
• ${\displaystyle \alpha =\pi \rightarrow s=r+x}$

The origin of the coordinate system can be also shifted to the mass ${\displaystyle m}$:

${\displaystyle x_{\bigcirc }=-x(t)}$
${\displaystyle y_{\bigcirc }=0}$

The retarded time ${\displaystyle t_{\text{ret}}}$ and the retarded potential ${\displaystyle \Phi _{\text{ret}}({\vec {x}},t)}$ are given as follows, where ${\displaystyle c}$ represents the propagation speed of the potentials:

${\displaystyle t_{\text{ret}}=t-{\frac {s}{c}}}$
${\displaystyle \Phi _{\text{ret}}({\vec {x}},t)={\frac {1}{4\pi }}\int \limits _{0}^{2\pi }{\frac {v(s,t_{\text{ret}})}{s}}\,\mathrm {d} \alpha }$

### Illustration

In a simplified example we only consider the infinitesimally small angles ${\displaystyle \mathrm {d} \alpha }$ in the origin of a spherical shell with the radius ${\displaystyle r}$ and the areal density ${\displaystyle \rho _{A}}$.

#### Symmetrical geometry

If the mass ${\displaystyle m}$ is in the centre of a spherical shell with the radius ${\displaystyle r=s_{0}}$ we have the following situation:

• Both angle elements ${\displaystyle \mathrm {d} \alpha }$ are equal.
• Both distances ${\displaystyle s_{0}}$ are equal to ${\displaystyle r}$.
• Both areal elements ${\displaystyle \mathrm {d} A_{0}}$ are equal.
• Both mass elements ${\displaystyle \mathrm {d} M_{0}}$ are equal.
• The retarded times of the gravitational potentials of both mass elements ${\displaystyle t_{ret}={\frac {s_{0}}{c}}={\frac {r}{c}}}$ are equal.

In this situation the mass ${\displaystyle m}$ does not experience any acceleration in the classical approach (see above) or if its velocity is zero.

#### Moving masses

The situation is changed, if the masses are moving starting at ${\displaystyle t=t_{0}}$ within a time span of ${\displaystyle \Delta t}$. The mass ${\displaystyle m}$ moves with the velocity ${\displaystyle v_{m}>0}$ to the right and the two mass elements ${\displaystyle \mathrm {d} M_{1}\propto \mathrm {d} A_{1}}$ and ${\displaystyle \mathrm {d} M'_{1}\propto \mathrm {d} A'_{1}}$ move with the radial velocity ${\displaystyle v_{M}>0}$:

At the time ${\displaystyle t_{1}}$ the mass has moved with the velocity ${\displaystyle v_{m}>0}$ the distance ${\displaystyle \Delta s}$ to the right:

${\displaystyle t_{1}=t_{0}+\Delta t}$
${\displaystyle \Delta s=v_{m}\cdot \Delta t}$

The spherical shell has expanded with the velocity ${\displaystyle v_{M}>0}$ and gained an increased radius:

${\displaystyle \Delta r=v_{M}\cdot \Delta t}$
${\displaystyle s_{1}=s_{0}+\Delta s+\Delta r=s_{0}+v_{m}\cdot \Delta t+v_{M}\cdot \Delta t=s_{0}+(v_{M}+v_{m})\,\Delta t}$
${\displaystyle s'_{1}=s_{0}-\Delta s+\Delta r=s_{0}-v_{m}\cdot \Delta t+v_{M}\cdot \Delta t=s_{0}+(v_{M}-v_{m})\,\Delta t}$

Therefore:

${\displaystyle s_{1}>s'_{1}}$

This means that the distance of mass element ${\displaystyle \mathrm {d} M_{1}}$ to the mass ${\displaystyle m}$ is always greater than the distance of mass element ${\displaystyle \mathrm {d} M'_{1}}$ to the mass ${\displaystyle m}$.

For the two retarded times for these distances to the location of ${\displaystyle m}$ we get:

${\displaystyle t_{{\text{ret}},1}=t_{1}-{\frac {s_{1}}{c}}}$
${\displaystyle t'_{{\text{ret}},1}=t_{1}-{\frac {s'_{1}}{c}}}$

With ${\displaystyle t_{0}=0}$ and therefore ${\displaystyle t_{1}=\Delta t}$:

${\displaystyle t_{{\text{ret}},1}=\left(1-{\frac {v_{M}+v_{m}}{c}}\right)\cdot \Delta t-{\frac {s_{0}}{c}}}$
${\displaystyle t'_{{\text{ret}},1}=\left(1-{\frac {v_{M}-v_{m}}{c}}\right)\cdot \Delta t-{\frac {s_{0}}{c}}}$

Therefore:

${\displaystyle t'_{{\text{ret}},1}>t_{{\text{ret}},1}}$

This means that the retarded time for the mass element ${\displaystyle \mathrm {d} M'_{1}}$ is always later than the retarded time for the mass element ${\displaystyle \mathrm {d} M_{1}}$.

#### Common case

In the adjacent diagram there are three mass points that move in space. Their speed is given as follows:

• Outer mass element top (green): ${\displaystyle v_{M}}$
• Mass point in between (blue): ${\displaystyle \color {Blue}v_{m}}$
• Outer mass element botton (green): ${\displaystyle -v_{M}}$

Their time-depending location is given by these three functions for their x-coordinates:

${\displaystyle \color {Blue}x_{m}(t)=v_{m}\cdot t}$
${\displaystyle \color {OliveGreen}x(t_{r})=-v_{M}\cdot t_{r}}$
${\displaystyle \color {OliveGreen}x'(t'_{r})=v_{M}\cdot t'_{r}}$

The time-depending effective distances ${\displaystyle \color {OliveGreen}s(t)}$ and ${\displaystyle \color {OliveGreen}s'(t)}$ between the outer mass elements and the mass point in between them is the difference of the corresponding x-coordinates and linked to the propagation velocity of the interacting waves ${\displaystyle \color {OliveGreen}c}$ as follows:

${\displaystyle \color {OliveGreen}s(t)=x_{m}(t)-x(t_{r})=(t_{r}-t)\cdot c}$
${\displaystyle \color {OliveGreen}s'(t)=x'(t'_{r})-x_{m}(t)=(t'_{r}-t)\cdot c}$

As a result, the corresponding retarded times ${\displaystyle t_{r}}$ and ${\displaystyle t'_{r}}$ for the two outer mass elemens are:

${\displaystyle t_{r}={\frac {c-v_{m}}{c+v_{M}}}\cdot t}$
${\displaystyle t'_{r}={\frac {c+v_{m}}{c+v_{M}}}\cdot t}$

And therefore:

${\displaystyle x(t_{r})=-v_{M}\cdot {\frac {c-v_{m}}{c+v_{M}}}\cdot t}$
${\displaystyle x'(t'_{r})=v_{M}\cdot {\frac {c+v_{m}}{c+v_{M}}}\cdot t}$

And:

${\displaystyle \color {OliveGreen}s(t)=\left(v_{M}\cdot {\frac {c-v_{m}}{c+v_{M}}}+v_{m}\right)\cdot t}$
${\displaystyle \color {OliveGreen}s'(t)=\left(v_{M}\cdot {\frac {c+v_{m}}{c+v_{M}}}-v_{m}\right)\cdot t}$

These are the time-depending effective distances for the gravitational potentials of the outer mass elements at their retarded times that have a simultaneous effect to the mass in between them at the time ${\displaystyle t}$.

For ${\displaystyle t>0}$ and ${\displaystyle c>v_{M}>v_{m}>0}$ and according to the diagram we can make the following assumption for the comparison of the effective distances:

${\displaystyle t'_{r}>t_{r}}$
${\displaystyle {\frac {c+v_{m}}{c+v_{M}}}\cdot t>{\frac {c-v_{m}}{c+v_{M}}}\cdot t}$
${\displaystyle v_{m}>-v_{m}}$ quod erat demonstrandum

This means that the retarded time of the upper mass element is always later than the retarded time of the lower mass element.

As well as:

${\displaystyle s(t)>s'(t)}$
${\displaystyle v_{M}\cdot {\frac {c-v_{m}}{c+v_{M}}}+v_{m}>v_{M}\cdot {\frac {c+v_{m}}{c+v_{M}}}-v_{m}}$
${\displaystyle v_{M}\cdot {\frac {-v_{m}}{c+v_{M}}}+v_{m}>v_{M}\cdot {\frac {v_{m}}{c+v_{M}}}-v_{m}}$
${\displaystyle 2\cdot v_{m}>2\cdot v_{M}\cdot {\frac {v_{m}}{c+v_{M}}}}$
${\displaystyle 1>{\frac {v_{M}}{c+v_{M}}}}$ quod erat demonstrandum

Since ${\displaystyle c+v_{M}}$ is always greater than ${\displaystyle v_{M}}$, the assumption is proven.

This means that the effective distance of the moving mass in between them to the lower mass element is always greater than to the moving upper mass element. Finally, it can be stated that the absolute value of the retarded gravitational potential at the location of the moving mass in between them is always greater for the upper mass element than for the lower mass element, if both mass elements have the same value ${\displaystyle M}$:

${\displaystyle \Phi (t_{r})=-{\frac {G\cdot M}{s(t)}}}$
${\displaystyle \Phi '(t'_{r})=-{\frac {G\cdot M}{s'(t)}}}$
${\displaystyle |\Phi (t_{r})|<|\Phi '(t'_{r})|}$

The moving mass ${\displaystyle m}$ experiences the corresponding retarded forces:

${\displaystyle F=-G\,{\frac {m\cdot M}{s(t)^{2}}}}$
${\displaystyle F'=+G\,{\frac {m\cdot M}{s'(t)^{2}}}}$

The net force to the mass ${\displaystyle m}$ is the sum of both:

${\displaystyle F_{net}=F'+F=G\cdot m\cdot M\cdot \left({\frac {1}{s'(t)^{2}}}-{\frac {1}{s(t)^{2}}}\right)}$

For the acceleration ${\displaystyle a}$ of the mass ${\displaystyle m}$ we find:

${\displaystyle a={\frac {F_{net}}{m}}=G\cdot M\cdot \left({\frac {1}{s'(t)^{2}}}-{\frac {1}{s(t)^{2}}}\right)}$

Since ${\displaystyle s'(t) the net force ${\displaystyle F_{net}}$ as well as the acceleration ${\displaystyle a}$ are positive, and the mass experiences an acceleration to positive x-values, i.e. in the direction of its movement.

#### Special case

Let us have a look at the following special case:

${\displaystyle c=1}$
${\displaystyle v_{M}={\frac {1}{2}}}$
${\displaystyle v_{m}={\frac {1}{4}}}$

Their time-depending location is given by these three functions for their x-coordinates:

${\displaystyle \color {DarkBlue}x_{m}(t)=v_{m}\cdot t={\frac {t}{4}}}$
${\displaystyle \color {Red}x(t)=-v_{M}\cdot t=-{\frac {t}{2}}}$
${\displaystyle \color {Gray}x'(t)=+v_{M}\cdot t=+{\frac {t}{2}}}$

The time-depending effective distances ${\displaystyle \color {Green}s(t)\color {Black}}$ and ${\displaystyle \color {Green}s'(t)\color {Black}}$ between the outer mass elements and the mass point in between them is the difference of the corresponding x-coordinates:

${\displaystyle \color {Green}s(t)=x_{m}(t)-x(t_{r})={\frac {t}{4}}+{\frac {t_{r}}{2}}\color {Black}}$
${\displaystyle \color {Green}s'(t)=x'(t'_{r})-x_{m}(t)={\frac {t'_{r}}{2}}-{\frac {t}{4}}\color {Black}}$

As a result, the corresponding retarded times ${\displaystyle \color {Dandelion}t_{r}}$ and ${\displaystyle \color {CornflowerBlue}t'_{r}}$ for the two outer mass elemens are:

${\displaystyle \color {Dandelion}t_{r}={\frac {\frac {3}{4}}{\frac {3}{2}}}\cdot t={\frac {1}{2}}\cdot t}$
${\displaystyle \color {CornflowerBlue}t'_{r}={\frac {\frac {5}{4}}{\frac {3}{2}}}\cdot t={\frac {5}{6}}\cdot t}$

Therefore:

${\displaystyle \color {Dandelion}x(t_{r})=-{\frac {1}{2}}\cdot {\frac {\frac {3}{4}}{\frac {3}{2}}}\cdot t=-{\frac {1}{4}}\cdot t}$
${\displaystyle \color {CornflowerBlue}x'(t'_{r})={\frac {1}{2}}\cdot {\frac {\frac {5}{4}}{\frac {3}{2}}}\cdot t={\frac {5}{12}}\cdot t}$
${\displaystyle \color {Dandelion}s(t)=\left({\frac {1}{2}}\cdot {\frac {\frac {3}{4}}{\frac {3}{2}}}+{\frac {1}{4}}\right)\cdot t={\frac {1}{2}}\cdot t}$
${\displaystyle \color {CornflowerBlue}s'(t)=\left({\frac {1}{2}}\cdot {\frac {\frac {5}{4}}{\frac {3}{2}}}-{\frac {1}{4}}\right)\cdot t={\frac {1}{6}}\cdot t}$

#### Thought experiment

In a thought experiment we look at the following situation, where ${\displaystyle c}$ is the propagation speed of the gravitational waves. The time line starts at ${\displaystyle t=0}$, and the effect of the retarded potentials is synchronised with the occurence of the moving mass. The left moving mass element is regarded at ${\displaystyle t=0}$ and ${\displaystyle t=1}$, the right moving mass element is regarded at ${\displaystyle t=0}$ and ${\displaystyle t=2}$, whilst the moving mass ${\displaystyle m}$ is regarded at ${\displaystyle t=8}$ and ${\displaystyle t=11}$, when the retarded gravitational potentials have their effect to the mass.

${\displaystyle v_{M}=c}$
${\displaystyle v_{m}={\frac {v_{M}}{3}}}$
${\displaystyle s=c\cdot t}$

In the following diagram the velocity of the waves is normalised and used without unit:

${\displaystyle c=1}$

And therefore, only for simplification and without any units, too:

${\displaystyle s=t}$

The effective distances ${\displaystyle s_{0}}$ for the gravitational potential at ${\displaystyle t_{0}=8}$ are both equal:

${\displaystyle s_{0}=t_{0}-t_{\text{ret,0}}=8}$

At the time ${\displaystyle t'=11}$ the mass experiences the retarded potentials on the left-hand side (distance to mass element is ${\displaystyle s_{1}}$) and at the right-hand side (distance to mass element is ${\displaystyle s_{2}}$):

${\displaystyle s_{1}=t'-t_{\text{ret,1}}=10}$
${\displaystyle s_{2}=t'-t_{\text{ret,2}}=9}$

All inhomogeneities contribute to the retarded potential at the location of the mass with the value they had at the retarded times ${\displaystyle t_{\text{ret,0}}}$, ${\displaystyle t_{\text{ret,1}}}$ and ${\displaystyle t_{\text{ret,2}}}$:

${\displaystyle t_{\text{ret,0}}=t_{0}-{\frac {s_{0}}{c}}=8-{\frac {8}{1}}=0}$, this corresponds to the effective time of the two area elements ${\displaystyle \mathrm {d} A_{0}}$.
${\displaystyle t_{\text{ret,1}}=t'-{\frac {s_{1}}{c}}=11-{\frac {10}{1}}=1}$, this corresponds to the effective time of the area element ${\displaystyle \mathrm {d} A_{1}}$ on the left.
${\displaystyle t_{\text{ret,2}}=t'-{\frac {s_{2}}{c}}=11-{\frac {9}{1}}=2}$, this corresponds to the effective time of the area element ${\displaystyle \mathrm {d} A_{2}}$ on the right.

For a mass ${\displaystyle m}$ moving from the centre of a shell to the right the angle element ${\displaystyle \mathrm {d} \alpha _{1}}$ to the left becomes smaller than the original angle element ${\displaystyle \mathrm {d} \alpha _{0}}$, and the angle element ${\displaystyle \mathrm {d} \alpha _{2}}$ to the right becomes greater than the original angle element ${\displaystyle \mathrm {d} \alpha _{0}}$:

${\displaystyle \mathrm {d} \alpha _{2}>\mathrm {d} \alpha _{0}>\mathrm {d} \alpha _{1}}$

The following applies to the appropriate mass elements:

${\displaystyle \mathrm {d} M_{0}=\mathrm {d} M_{1}=\mathrm {d} M_{2}=\rho _{a}\cdot \mathrm {d} A_{0}=\rho _{a}\cdot 4\pi \cdot s_{0}^{2}\cdot \mathrm {d} \alpha _{0}^{2}}$

For the net force ${\displaystyle \mathrm {d} F}$ to the mass ${\displaystyle m}$:

${\displaystyle \mathrm {d} F=\mathrm {d} F_{2}+\mathrm {d} F_{1}=G\cdot m\cdot \mathrm {d} M_{0}\cdot \left({\frac {1}{s_{2}^{2}}}-{\frac {1}{s_{1}^{2}}}\right)}$

For the acceleration ${\displaystyle \mathrm {d} a}$ of the mass ${\displaystyle m}$ we find:

${\displaystyle \mathrm {d} a={\frac {\mathrm {d} F}{m}}=G\cdot \mathrm {d} M_{0}\cdot \left({\frac {1}{s_{2}^{2}}}-{\frac {1}{s_{1}^{2}}}\right)}$

Since ${\displaystyle s_{2} the net force ${\displaystyle \mathrm {d} F}$ is positive, and the mass experiences an acceleration ${\displaystyle \mathrm {d} a}$ to the right, i.e. in the direction of its movement. This result is absolutely inline with the findings above in the section "Common case" above.

Furthermore, it is noteworthy to state that the acceleration ${\displaystyle \mathrm {d} a}$ is proportional to the areal density ${\displaystyle \rho _{a}}$ of the expanding shell:

${\displaystyle \mathrm {d} a\propto \rho _{a}}$

Nevertheless it should be noted that the areal density is decreasing with the expansion and the increasing radius of the outer shell.

### References

1. Schwarzschild, Karl (1903). "Zur Elektrodynamik" [On electrodynamics]. Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse. Göttingen: 127.
2. von Brill, Alexander (1904). Written at Tübingen. Clebsch, Alfred; Neumann, Carl Gottfried; Klein, Felix; van Dyck, Walther; Hilbert, David (eds.). "Über zyklische Bewegung" [About cyclic movement]. Mathematische Annalen (in German). Leipzig: Benedictus Gotthelf Teubner. 58: 473.
3. Dragon, Norbert (2016-09-26). "Stichworte und Ergänzungen zu Mathematische Methoden der Physik" [Keywords and additions to Mathematical Methods in Physics] (PDF) (in German). pp. 222–224. Retrieved 2024-02-25.