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

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. 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 out rim 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 $$\Phi$$ due to a mass $$M$$ in the distance $$s$$ is given by:


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

The gravitational force $$F$$ to another mass $$m$$ results as follows:


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

The time-depending potential $$\Phi(\vec x,t) $$ is the solution of the inhomogeneous wave equation, where $$v(\vec x, t)$$ is the inhomogeneity, and $$c$$ 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:


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

where $$\Delta = \nabla^2$$ is the Laplace operator, $$\Box$$ 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:


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

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


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


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

$$t_{\text{ret}}$$ is called the retarded time. At the location $$\vec x$$ and at the time $$t$$ the retarded potential only depends on the inhomogeneity $$v$$ 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 $$c$$.

In a shell


In the simplified example in the adjacent figure the source term is the linear mass density $$\lambda_m$$ that is not time-dependant and only exists in the circle of the outer rim with the constant radius $$r$$ and its centre at $$O$$:


 * $$x_O = 0$$


 * $$y_O = 0$$


 * $$x^2 + y^2 = r^2 \rarr \lambda_m = \text{ constans}$$

In all other locations within the plane the linear mass density $$\lambda_m$$ is zero:


 * $$x^2 + y^2 \neq r^2 \rarr \lambda_m = 0$$

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:


 * $$z \ne 0 \rarr \lambda_m = 0$$

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


 * $$\mathrm d M = \lambda_m \cdot \mathrm d C = \lambda_m \cdot r \cdot \mathrm d \alpha$$


 * $$M = \int d M = \int\limits_{0}^{2\pi} \lambda_m \cdot r \cdot \mathrm d \alpha = \lambda_m \cdot 2 \pi \cdot r = \lambda_m \cdot C$$

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


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

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


 * $$s^2 + 2xs \, \cos \alpha + x^2 - r^2 = 0$$

The solution for the distance $$s$$ is:


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

It is obvious that the following simplifications are valid:


 * $$x = 0 \rarr s = r$$
 * $$\alpha = 0 \rarr s = r - x$$
 * $$\alpha = \pm \frac {\pi} {2} \rarr s = \sqrt {x^2 + r^2}$$
 * $$\alpha = \pi \rarr s = r + x$$



The origin of the coordinate system can be also shifted to the mass $$m$$:


 * $$x_O = - x(t)$$


 * $$y_O = 0$$

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


 * $$t_{\text{ret}} = t - \frac {s} {c}$$


 * $$\Phi_{\text{ret}}(\vec x, t) = \frac {1} {4 \pi} \int\limits_{0}^{2\pi} \frac {v (s, t_{\text{ret}})} {s} \, \mathrm d \alpha$$

Illustration
In a simplified example we only consider the infinitesimally small angles $$\mathrm d \alpha$$ in the origin of a spherical shell with the radius $$r$$ and the areal density $$\rho_A$$.

Symmetrical geometry
If the mass $$m$$ is in the centre of a spherical shell with the radius $$r = s_0$$ we have the following situation:


 * Both angle elements $$\mathrm d \alpha$$ are equal.
 * Both distances $$s_0$$ are equal to $$r$$.
 * Both areal elements $$\mathrm d A_0$$ are equal.
 * Both mass elements $$\mathrm d M_0$$ are equal.
 * The retarded times of the gravitational potentials of both mass elements $$t_{ret} = \frac {s_0} {c} = \frac {r} {c}$$ are equal.

In this situation the mass $$m$$ 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 $$t = t_0$$ within a time span of $$\Delta t$$. The mass $$m$$ moves with the velocity $$v_m > 0$$ to the right and the two mass elements $$\mathrm d M_1 \propto \mathrm d A_1$$ and $$\mathrm d M'_1 \propto \mathrm d A'_1$$ with the radial velocity $$v_M > 0$$:

At the time $$t_1$$ the mass has moved with the velocity $$v_m > 0$$ the distance $$\Delta s$$ to the right:


 * $$t_1 = t_0 + \Delta t$$


 * $$\Delta s = v_m \cdot \Delta t$$

The spherical shell has expanded with the velocity $$v_M > 0$$ and gained an increased radius:


 * $$\Delta r = v_M \cdot \Delta t$$


 * $$s_1 = s_0 + \Delta s + \Delta r = s_0 + v_m \cdot \Delta t + v_M \cdot \Delta t = s_0 + (v_M + v_m) \, \Delta t$$


 * $$s'_1 = s_0 - \Delta s + \Delta r = s_0 - v_m \cdot \Delta t + v_M \cdot \Delta t = s_0 + (v_M - v_m) \, \Delta t$$

Therefore:


 * $$s_1 > s'_1$$

This means that the distance of mass element $$\mathrm d M_1$$ to the mass $$m$$ is always greater than the distance of mass element $$\mathrm d M'_1$$ to the mass $$m$$.

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


 * $$t_{\text{ret},1} = t_1 - \frac {s_1} {c}$$


 * $$t'_{\text{ret},1} = t_1 - \frac {s'_1} {c}$$

With $$t_0= 0$$ and therefore $$t_1 = \Delta t$$:


 * $$t_{\text{ret},1} = \left( 1 - \frac {v_M + v_m} {c} \right) \cdot \Delta t - \frac {s_0} {c}$$


 * $$t'_{\text{ret},1} = \left( 1 - \frac {v_M - v_m} {c} \right) \cdot \Delta t - \frac {s_0} {c}$$

Therefore:


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

This means that the retarded time of mass element $$\mathrm d M'_1$$ is always later than the retarded time of mass element $$\mathrm d M_1$$.

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): $$v_M$$
 * Mass point in between (blue): $$\color{Blue}v_m$$
 * Outer mass element botton (green): $$-v_M$$

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


 * $$\color{Blue}x_m(t) = v_m \cdot t$$


 * $$\color{OliveGreen}x(t_r) = -v_M \cdot t_r$$


 * $$\color{OliveGreen}x'(t'_r) = v_M \cdot t'_r$$

The time-depending effective distances $$\color{OliveGreen}s(t)$$ and $$\color{OliveGreen}s'(t)$$ 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 $$\color{OliveGreen}c$$ as follows:


 * $$\color{OliveGreen}s(t) = x_m(t) - x(t_r) = (t_r - t) \cdot c$$


 * $$\color{OliveGreen}s'(t) = x'(t'_r) - x_m(t) = (t'_r - t) \cdot c$$

As a result, the corresponding retarded times $$t_r$$ and $$t'_r$$ for the two outer mass elemens are:


 * $$t_r = \frac {c - v_m} {c + v_M} \cdot t$$


 * $$t'_r = \frac {c + v_m} {c + v_M} \cdot t$$

And therefore:


 * $$x(t_r) = -v_M \cdot \frac {c - v_m} {c + v_M} \cdot t$$


 * $$x'(t'_r) = v_M \cdot \frac {c + v_m} {c + v_M} \cdot t$$

And:


 * $$\color{OliveGreen}s(t) = \left( v_M \cdot \frac {c - v_m} {c + v_M} + v_m \right) \cdot t$$


 * $$\color{OliveGreen}s'(t) = \left( v_M \cdot \frac {c + v_m} {c + v_M} - v_m \right) \cdot t$$

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 $$t$$.

For $$t > 0$$ and $$c > v_M > v_m > 0$$ and according to the diagram we can make the following assumption for the comparison of the effective distances:


 * $$t'_r > t_r$$


 * $$\frac {c + v_m} {c + v_M} \cdot t > \frac {c - v_m} {c + v_M} \cdot t$$


 * $$v_m > -v_m$$ 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:


 * $$s(t) > s'(t)$$


 * $$v_M \cdot \frac {c - v_m} {c + v_M} + v_m > v_M \cdot \frac {c + v_m} {c + v_M} - v_m$$


 * $$v_M \cdot \frac {- v_m} {c + v_M} + v_m > v_M \cdot \frac {v_m} {c + v_M} - v_m$$


 * $$2 \cdot v_m > 2 \cdot v_M \cdot \frac {v_m} {c + v_M}$$


 * $$1 > \frac {v_M} {c + v_M}$$ quod erat demonstrandum

Since $$c + v_M$$ is always greater than $$v_M$$, 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 $$M$$:


 * $$\Phi(t_r) = - \frac {G \cdot M} {s(t)}$$


 * $$\Phi'(t'_r) = - \frac {G \cdot M} {s'(t)}$$


 * $$|\Phi(t_r)| < |\Phi'(t'_r)|$$

The moving mass $$m$$ experiences the corresponding retarded forces:


 * $$F = - G \, \frac {m \cdot M} {s(t)^2}$$


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

The net force to the mass $$m$$ is the sum of both:


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

For the acceleration $$a$$ of the mass $$m$$ we find:


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

Since $$s'(t) < s(t)$$ the net force $$F_{net}$$ as well as the acceleration $$a$$ 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:


 * $$c = 1$$


 * $$v_M = \frac {1} {2}$$


 * $$v_m = \frac {1} {4}$$

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


 * $$\color{DarkBlue}x_m(t) = v_m \cdot t = \frac {t} {4}$$


 * $$\color{Red}x(t) = -v_M \cdot t = - \frac {t} {2}$$


 * $$\color{Gray}x'(t) = +v_M \cdot t = + \frac {t} {2}$$

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


 * $$\color{Green}s(t) = x_m(t) - x(t_r) = \frac {t} {4} + \frac {t_r} {2}\color{Black}$$


 * $$\color{Green}s'(t) = x'(t'_r) - x_m(t) = \frac {t'_r} {2} - \frac {t} {4}\color{Black}$$

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


 * $$\color{Dandelion}t_r = \frac {\frac {3} {4}} {\frac {3} {2}} \cdot t = \frac {1} {2} \cdot t$$


 * $$\color{CornflowerBlue}t'_r = \frac {\frac {5} {4}} {\frac {3} {2}} \cdot t = \frac {5} {6} \cdot t$$

Therefore:


 * $$\color{Dandelion}x(t_r) = -\frac {1} {2} \cdot \frac {\frac {3} {4}} {\frac {3} {2}} \cdot t = -\frac {1} {4} \cdot t$$


 * $$\color{CornflowerBlue}x'(t'_r) = \frac {1} {2} \cdot \frac {\frac {5} {4}} {\frac {3} {2}} \cdot t = \frac {5} {12} \cdot t$$


 * $$\color{Dandelion}s(t) = \left( \frac {1} {2} \cdot \frac {\frac {3} {4}} {\frac {3} {2}} + \frac {1} {4} \right) \cdot t = \frac {1} {2} \cdot t$$


 * $$\color{CornflowerBlue}s'(t) = \left( \frac {1} {2} \cdot \frac {\frac {5} {4}} {\frac {3} {2}} - \frac {1} {4} \right) \cdot t = \frac {1} {6} \cdot t$$

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


 * $$v_M = c$$


 * $$v_m = \frac {v_M} {3}$$


 * $$s = c \cdot t$$

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


 * $$c = 1$$

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


 * $$s = t$$

The effective distances $$s_0$$ for the gravitational potential at $$t_0 = 8$$ are both equal:


 * $$s_0 = t_0 - t_{\text{ret,0}} = 8$$

At the time $$t' = 11$$ the mass experiences the retarded potentials on the left-hand side (distance to mass element is $$s_1$$) and at the right-hand side (distance to mass element is $$s_2$$):


 * $$s_1 = t' - t_{\text{ret,1}} = 10$$


 * $$s_2 = t' - t_{\text{ret,2}} = 9$$

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


 * $$t_{\text{ret,0}} = t_0 - \frac {s_0} {c} = 8 - \frac {8} {1} = 0$$, this corresponds to the effective time of the two area elements $$\mathrm d A_0$$.


 * $$t_{\text{ret,1}} = t' - \frac {s_1} {c} = 11 - \frac {10} {1} = 1$$, this corresponds to the effective time of the area element $$\mathrm d A_1$$ on the left.


 * $$t_{\text{ret,2}} = t' - \frac {s_2} {c} = 11 - \frac {9} {1} = 2$$, this corresponds to the effective time of the area element $$\mathrm d A_2$$ on the right.

For a mass $$m$$ moving from the centre of a shell to the right the angle element $$\mathrm d \alpha_1$$ to the left becomes smaller than the original angle element $$\mathrm d \alpha_0$$, and the angle element $$\mathrm d \alpha_2$$ to the right becomes greater than the original angle element $$\mathrm d \alpha_0$$:


 * $$\mathrm d \alpha_2 > \mathrm d \alpha_0 > \mathrm d \alpha_1$$

The following applies to the appropriate mass elements:


 * $$\mathrm d M_0 = \mathrm d M_1 = \mathrm d M_2 = \rho_a \cdot \mathrm d A_0 = \rho_a \cdot 4\pi \cdot s_0^2 \cdot \mathrm d \alpha_0^2$$

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


 * $$\mathrm d F = \mathrm d F_2 + \mathrm d F_1 = G \cdot m \cdot \mathrm d M_0 \cdot \left( \frac {1} {s_2^2} - \frac {1} {s_1^2} \right)$$

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


 * $$\mathrm d a = \frac {\mathrm d F} {m} = G \cdot \mathrm d M_0 \cdot \left( \frac {1} {s_2^2} - \frac {1} {s_1^2} \right)$$

Since $$s_2 < s_1$$ the net force $$\mathrm d F$$ is positive, and the mass experiences an acceleration $$\mathrm d a$$ 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 $$\mathrm d a$$ is proportional to the areal density $$\rho_a$$ of the expanding shell:


 * $$\mathrm d a \propto \rho_a$$

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