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Self Energy Field

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A Possible Scalar Term Describing Energy Density in the Gravitational Field

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The gravitational field of a point mass and the electric field of a point charge are structurally similar. Each may be represented by a vector field in which the vectors are directed along radial lines emanating from the point, and the vector magnitudes decrease as the inverse square of the [radial] distance from the point. The electric field, $\vec{E}$, when multiplied by the magnitude of a test charge $q$ in the field, gives the force $\vec{f}$ exerted locally on the test charge $$ \vec{f} = q\vec{E} \qquad 1. $$ The gravitational field, $\vec{g}$, when multiplied by the magnitude of a test mass $m$ in the field, gives the force $\vec{f}$ exerted locally on the test mass $$ \vec{f} = m\vec{g} \qquad 2. $$ In each case, the force has the same vector sense as the field. The electric field may point radially toward or away from the source charge, depending on the sign of the source charge. The gravitational field points toward the source mass in all known cases. The electric field has a scalar energy density field (or, following some older texts, a pressure field) associated with it. When the field vector at a point is $\vec{E}$, then the energy density at the same point is $$ u_E = \frac{1}{2}\epsilon_0(\vec{E}\cdot\vec{E}) = \frac{1}{2}\epsilon_0 E^2 \qquad 3. $$ where $\epsilon_0$ is the permittivity of free space. Energy density is a measure of the energy stored in the field per unit volume of space. The electric energy density has been an essential ingredient for our calculation of the Electromagnetic Mass of the electron. Also related is a Question at Mathematics Stack Exchange: Cauchy distribution instead of Coulomb law? Its unit of measure is $J/m^3$ (or $N/m^2$, if it is thought of as a pressure). While eq. 3 represents the energy density for the electric field, and a similar expression represents the energy density for the magnetic field, no such energy density term has ever been defined for the gravitational field. But one suspects that it could be, and possibly even should be. Such in concordance with a Question at PHYSICS Stack Exchange: Is there an energy density associated with a Gravitational Field?, where it is worth noting that he energy density will always be negative, in this case.

Let us use the similarity between the gravitational and electric fields to construct a gravitational energy density term.

We begin by noting how the permittivity of free space enters into the expression for $E$ : $$ E = kQ/r^2 \qquad 4. \\ \mbox{where} \quad k = 1/(4\pi\epsilon_0) \qquad 4a. $$ is a universal constant, $Q$ is the source charge, and $r$ is radial distance from the source.

The gravitational field is given by the expression $$ g = GM/r^2 \qquad 5. $$ where $G$ is a universal constant, $M$ is the source mass, and $r$ is radial distance from the source.

The electrical field energy density may be written in terms of $k$ as $$ u_E = \frac{1}{2}\cdot 1/(4\pi k)\cdot E^2 = E^2/(8\pi k) \qquad 6. $$ By analogy, a candidate gravitational energy density term may now be constructed and written as $$ u_G = \color{red}{-}\, g^2/(8\pi G) \qquad 7. $$ Near the surface of the earth $$ g = 9.807\; m/s^2. \\ \mbox{Also} \quad G = 6.672 \times 10^{-11}\;N m^2/kg^2 \\ \mbox{so that} \quad u_G = \color{red}{-}\, 5.736 \times 10^{10} \; J/m^3. $$ Using eq. 7, it might be possible to construct a classical argument for the rotation of perihelion of a planet around its central body [sun]. [ GR advertizing deleted ] Historically, attempts to modify Newton's law of gravitation to account for the observed motion of Mercury have proved unsatisfactory. So did the introduction of another [hypothetical] planet called Vulcan, with an orbit inside that of Mercury. [ GR advertizing deleted ] Throughout this calculation, we will use Newton's law of gravitation without modification, and incorporate the energy density term described above to render a qualitative description of the rotation of perihelion.

First, we rewrite the term $u_G$ by substituting for $g$ in eq. 7.: $$ g = GM/r^2 \quad \mbox{so that} \quad u_G = \color{red}{-}\, GM^2/(8\pi r^4) \qquad 8. $$ Another scalar field may be obtained from the energy density field by using the mass-energy equivalence from special relativity; i.e., a scalar mass density field of the form $$ u_G/c^2 = \color{red}{-}\, GM^2/(8\pi r^4 c^2) \qquad 9. $$ We next assume that the mass due to the term $u_G/c^2$, integrated over a suitable volume of space, behaves, gravitationally, like ordinary matter. For an extended body like the sun (rather than an ideal point mass), the ramifications of this assumption need to be explored

The sun may be considered a sphere of radius $r_0$, and constant (average) density $\rho$. The classical solar mass is then $$ M_0 = 4\pi r_0^3\rho/3 $$ To find the additional mass contribution due to the interior field, we must first rewrite eq. 9 for the sun's interior. We set $$ M = 4\pi r^3\rho/3 $$ which is a function of the radial distance r. We next substitute this result into eq. 9 to obtain $$ u_G/c^2 = \color{red}{-}\, 2G\pi r^2\rho^2/(9c^2) \qquad 10. $$ The corresponding mass contribution $M_f$ is then the volume integral of eq. 10 throughout the sun's interior. The integrand is $$ dM_f = [\color{red}{-}\, 2G\pi r^2\rho^2/(9c^2)] . 4\pi r^2\,dr $$ and the limits of integration are from $r = 0$ to $r = r_0$. Thus $$ M_f = \color{red}{-}\, 8\pi^2 G \rho^2 r_0^5/(45c^2) \qquad 11. $$ The total mass of the sun must now be $$ M = M_0 + M_f \qquad 12. $$ For the sun, we know that $$ M_0 = 1.99 \times 10^{30}\; kg \\ \rho = 1,410\; kg/m^3\;(avg.) \\ r_0 = 6.912 \times 10^8\;m \\ \mbox{We calculate} \quad M_f = \color{red}{-}\, 4.08 \times 10^{23}\; kg $$ As should be expected, $M_f$ is very small compared to $M_0$. In fact, $$ M_f = \color{red}{-}\, (2.05 \times 10^{-7}) . M_0 $$ and its absolute value is, therefore, only slightly greater than one ten-millionth of the classical solar mass.

Outside the sun, the local field experienced by an orbiting planet, asteroid, etc., has contributions from

Let a planet be located a distance $r$ from the center of the sun. Associate with this distance a sphere, concentric with the sun, on whose surface the planet is always to be found. As $r$ increases, the sphere expands. As $r$ decreases, the sphere contracts. The motion of the planet is determined by the total mass inside the sphere. This total mass is made up of the mass contained in the sun plus the mass due to the external field contained inside the sphere. For the external field $$ dM_f' = [\color{red}{-}\, GM^2/(8\pi r^4 c^2)] . 4\pi r^2\; dr $$ The volume integral is to be taken throughout the region between the surface of the sun and the sphere of radius $r$. Thus $$ M_f' = \color{red}{-}\, GM^2(1/r_0 - 1/r)/(2c^2) \qquad 13. $$ This result is in concordance with formula (7.13) in the booklet by Léon Brillouin, Relativity Reexamined, where he writes: This atmosphere surrounding $M_0$ is always negative, whatever the sign of $M_0$ might be. The total mass $M_g$ distributed in the field is directly obtained by integration for the whole space, with $M_g = M_f'$ , $M_0 = M$ , $a = r_0$ , $r \to \infty$ : $$ M_g = -\frac{G}{c^2}\frac{M_0^2}{2a} $$ It is the total mass $M + M_f' (= M_0 + M_f + M_f')$ that attracts the planet, and influences its motion around the sun. The gravitational field $g$ acting on the planet is $$ g = G(M + M_f')/r^2 \qquad 14. \\ = G(M \color{red}{-}\, GM^2(1/r_0 - 1/r)/(2c^2))/r^2 \qquad 14a. $$ This field represents the acceleration of the planet in its orbit around the sun. The classical contribution of the sun is represented by $$ GM/r^2 \qquad 15. $$ and results in a stationary, elliptical orbit, as expected. The additional contribution of the sun, due to its external gravitational field, is represented by $$ \color{red}{-}\, G^2 M^2(1/r_0 - 1/r)/(2c^2 r^2) \qquad 16. $$ This field varies with radial distance $r$, the more so the smaller the value of $r$. Therefore, any planet, near the sun and in a noncircular orbit, will experience a sideways perturbation from a strict classical orbit. If, without this perturbation, the planet would follow a classical ellipse, the perturbation must be such as to divert it sunwards from the ellipse as it travels from perihelion to aphelion. The reverse must be true for the other half of the orbit, with the result that the line of apsides must turn [slowly] with the same directional sense as the orbital motion of the planet (i.e., clockwise or counterclockwise, viewed from a suitable vantage point).

It is interesting to calculate the approximate magnitude of the mass contribution in eq. 13 for the planet Mercury. Accordingly, we need to know that for Mercury, $$ r = 5.75 \times 10^{10}\; m \\ \mbox{Also, for the sun,} \quad M = 1.99 \times 10^{30}\; kg \\ r_0 = 6.91 \times 10^8\; m \\ \mbox{And} \quad G = 6.67 \times 10^{-11}\; N\,m^2/kg^2. \\ \mbox{Then,} \quad M_f' = \color{red}{-}\, 2.10 \times 10^{24}\; kg \qquad 17. $$ It is interesting that the mass calculated in eq. 17 is roughly equivalent to the mass of the earth. It was commented earlier that when Mercury's rotation of perihelion was first observed, astronomers attempted to account for it by postulating another planet called Vulcan, whose orbit was inside that of Mercury and whose gravitational influence perturbed Mercury's orbit. These observations were made late in the nineteenth century, before the advent of special relativity. [ GR advertizing deleted ]

It may be shown that motion, under a law of central attraction of the general form $$ k/r^2+k_1/r^3 \qquad 18. $$ results in the type of motion we have been describing here; i.e., that the perihelion of a planet, moving under such a law of attraction, will rotate at a rate proportional to the constant, $k_1$ (Levi-Civita, pg. 396). [ stuff deleted ] We may rewrite eq. 14a. as $$ G(M \color{red}{-}\, GM^2(1/r_0-1/r)/(2c^2))/r^2 = \color{red}{GM\left[1 - GM/(2 c^2 r_0)\right]/r^2 + G^2 M^2/(2c^2)/r^3} \qquad 14b. $$ In this representation, comparing with eq. 18: $$ k = GM\left[1 -\, GM/(2 c^2 r_0)\right] \\ k_1 = +\, G^2 M^2/(2c^2) $$ [rest deleted] We are skipping in the book by Amitabha Ghosh, Origin of Inertia for the purpose of finding a formula similar to 18. : $$ F = \frac{G m_s m_p}{r^2} + \frac{C}{r^3} \qquad (2.5) $$ It takes a few pages to find an expression for $C$ . Via $$ m = m_0\,\exp\left[\frac{GM}{c^2r}\right] \qquad (2.23) $$ we arrive at $$ F = \frac{G M m_0}{r^2}\left[1 + \frac{3\,GM}{c^2r}\right] \qquad (2.24) $$ and the resulting excess perihelion rotation is the same as given by Einstein. However, a series expansion of $\,\exp[GM/(c^2r)]\,$ does of course not lead to the desired result. If we anyway assume that (2.24) is correct, then we have at last $$ C = 3\,G^2 M^2/c^2 = 6 \times G^2 M^2/(2c^2) $$ At least the order of magnitude of our constant $\,k_1\,$ can be claimed to be alright.