Nonzero Gravitational Force Exerted by a Spherical Shell on a Body Moving Inside It, and Cosmological Implications

Abstract

The shell theorem, proved by Newton in his Principia (1687), states that the net force exerted by a uniform spherical shell on a body located anywhere inside it is zero, as long as the force is proportional to the inverse square of the distance between the interacting bodies. This null result remains valid whenever the interaction depends only on the distance between the bodies, but not on their relative motion. In this work, I develop a direct closed-form evaluation of the integral of the elements of force to show that Weber-like interactions, which take into account the relative motion between the body and the shell, yield a nonzero force opposite to the acceleration of the body with respect to the shell, whatever be its position and velocity. For gravitational interactions, this nonzero force is relevant in cosmology since it can be identified with the force of inertia, as caused by the celestial sphere (i.e., the set of distant stars), which allows for a full mathematical implementation of Mach’s principle.

Appendices

Appendix A. CLOSED-FORM EXPRESSION FOR THE LAST INTEGRAL IN Eq. (10)

The last integral in Eq. (10) can be evaluated by first expanding it as

$$I_{4}\equiv\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}{\frac{\vec{a}\cdot\vec{R}}{r^{3}}\vec{R}\sin{\theta}d\theta d\varphi}$$

$${}=a_{x}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{x\frac{\sin{\theta}}{r^{3}}\vec{R}d\varphi d\theta}+a_{y}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{y\frac{\sin{\theta}}{r^{3}}\vec{R}d\varphi d\theta}$$

$${}+a_{z}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{z\frac{\sin{\theta}}{r^{3}}\vec{R}d\varphi d\theta}=a_{x}\hat{i}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{x^{2}\frac{\sin{\theta}}{r^{3}}d\varphi d\theta}$$

$${}+a_{y}\hat{j}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{y^{2}\frac{\sin{\theta}}{r^{3}}d\varphi d\theta}$$

$${}+a_{z}\hat{k}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{z^{2}\frac{\sin{\theta}}{r^{3}}d\varphi d\theta},$$

(A.1)

where the zero value of the integrals with crossed terms—i.e., $xy$, $yz$, and $xz$—follows from the zero value of the azimuthal integrals

$$\int\limits_{0}^{2\pi}{\cos{\varphi}d\varphi},\quad\int\limits_{0}^{2\pi}{\sin{\varphi}d\varphi},$$

$$\int\limits_{0}^{2\pi}{\sin{\varphi}\cos{\varphi}d\varphi}=\frac{1}{2}\int\limits_{0}^{2\pi}{\sin{(2\varphi)}d\varphi}.$$

The remaining integrals can be evaluated in a closed form in spherical coordinates as follows:

$$I_{4}=R^{2}\Bigg{(}a_{x}\hat{i}\int\limits_{0}^{\pi}{\frac{\sin^{3}{\theta}}{r^{3}}d\theta}\int\limits_{0}^{2\pi}{\cos^{2}{\varphi}d\varphi}$$

$${}+a_{y}\hat{j}\int\limits_{0}^{\pi}{\frac{\sin^{3}{\theta}}{r^{3}}d\theta}\int\limits_{0}^{2\pi}{\sin^{2}{\varphi}d\varphi}$$

$${}+2\pi a_{z}\hat{k}\int\limits_{0}^{\pi}{\cos^{2}{\theta}\frac{\sin{\theta}}{r^{3}}d\theta}\Bigg{)}=\pi R^{2}\left(a_{x}\hat{i}+a_{y}\hat{j}\right)$$

$${}\times\int\limits_{0}^{\pi}{\frac{\sin^{3}{\theta}}{r^{3}}d\theta}+2\pi R^{2}a_{z}\hat{k}I_{5},$$

(A.2)

where $I_{5}\equiv\int\limits_{0}^{\pi}{\cos^{2}{\theta}\sin{\theta}/r^{3}d\theta}$. Since $\sin^{3}{\theta}=(1-\cos^{2}{\theta})\sin{\theta}$,

$$\int\limits_{0}^{\pi}{\frac{\sin^{3}{\theta}}{r^{3}}}=\int\limits_{0}^{\pi}{\frac{\sin{\theta}}{r^{3}}d\theta}-\int\limits_{0}^{\pi}{\cos^{2}{\theta}\frac{\sin{\theta}}{r^{3}}d\theta}$$

$${}=\frac{2}{R\left(R^{2}-C^{2}\right)}-I_{5},$$

(A.3)

where the first integral is the same which was already evaluated in Eq. (8). Fortunately, $I_{5}$ can be determined using the same substitution as was applied to Eq. (8), which results in

$$I_{5}=\frac{1}{4R^{3}C^{3}}\int\limits_{R-C}^{R+C}{\frac{\left(R^{2}+C^{2}-u^{2}\right)^{2}}{u^{2}}du}$$

$${}=\frac{2\left(R^{2}+2C^{2}\right)}{3R^{3}\left(R^{2}-C^{2}\right)}.$$

(A.4)

On back-substituting this closed-form expression for $I_{5}$ into Eq. (A.3) and then into Eq. (A.2), Eq. (10) becomes

$$\vec{F}=\xi\frac{Gm\sigma R^{2}}{c^{2}}\left(I_{3}-I_{4}\right)$$

$${}=-\xi\frac{4\pi Gm\sigma R}{3c^{2}}\bigg{[}a_{x}\hat{i}+a_{y}\hat{j}$$

$${}+\frac{3C^{2}a\cos{\alpha}+a_{z}\left(R^{2}+2C^{2}\right)}{R^{2}-C^{2}}\hat{k}\bigg{]},$$

(A.5)

where the integral $I_{3}$ was solved in Eq. (11).

Appendix A. CLOSED-FORM EXPRESSIONS FOR THE INTEGRALS IN Eq. (23)

The last integral in Eq. (17), as expanded in Eq. (23), can be solved analytically. To do that, we first write it as

$$R\hat{i}\Bigg{(}v_{0}^{2}C^{2}\cos^{2}{\beta}\int\limits_{0}^{\pi}{\frac{\sin^{2}{\theta}}{r^{5}}d\theta}\int\limits_{0}^{2\pi}{\cos{\varphi}d\varphi}$$

$${}+\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{\frac{v_{0x}^{2}x^{2}+v_{0y}^{2}y^{2}+v_{0z}^{2}z^{2}+2v_{0x}v_{0y}xy+2v_{0x}v_{0z}xz+2v_{0y}v_{0z}yz}{r^{5}}\sin^{2}{\theta}\cos{\varphi}d\varphi d\theta}$$

$${}+2v_{0}C\cos{\beta}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{\frac{v_{0x}x+v_{0y}y+v_{0z}z}{r^{5}}\sin^{2}{\theta}\cos{\varphi}d\varphi d\theta}\Bigg{)}$$

$${}+R\hat{j}\Bigg{(}v_{0}^{2}C^{2}\cos^{2}{\beta}\int\limits_{0}^{\pi}{\frac{\sin^{2}{\theta}}{r^{5}}d\theta}\int\limits_{0}^{2\pi}{\sin{\varphi}d\varphi}$$

$${}+\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{\frac{v_{0x}^{2}x^{2}+v_{0y}^{2}y^{2}+v_{0z}^{2}z^{2}+2v_{0x}v_{0y}xy+2v_{0x}v_{0z}xz+2v_{0y}v_{0z}yz}{r^{5}}\sin^{2}{\theta}\sin{\varphi}d\varphi d\theta}$$

$${}+2v_{0}C\cos{\beta}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{\frac{v_{0x}x+v_{0y}y+v_{0z}z}{r^{5}}\sin^{2}{\theta}\sin{\varphi}d\varphi d\theta}\Bigg{)}+R\hat{k}\int\limits_{0}^{\pi}\int\limits_{0}^{2\pi}{f(\theta,\varphi)\cos{\theta}\sin{\theta}d\varphi d\theta}.$$

(B.1)

Clearly, the first terms of both the $\hat{i}$ and $\hat{j}$ components are zero, as well as all other azimuthal integrals with $\sin{\varphi}$, $\cos{\varphi}$, $\sin{\varphi}\cos{\varphi}$, $\sin^{2}{\varphi}\cos{\varphi}$, $\sin{\varphi}\cos^{2}{\varphi}$, $\sin^{3}{\varphi}$, or $\cos^{3}{\varphi}$ in their integrands. Writing the remaining terms in spherical coordinates, one finds

$$\left(2\pi R^{3}v_{0x}v_{0z}\widetilde{I}_{5}+2\pi R^{2}Cv_{0}\cos{\beta}v_{0x}\widetilde{I}_{3}\right)\hat{i}$$

$${}+\left(2\pi R^{3}v_{0y}v_{0z}\widetilde{I}_{5}+2\pi R^{2}Cv_{0}\cos{\beta}v_{0y}\widetilde{I}_{3}\right)\hat{j}$$

$${}+\Big{[}2\pi RC^{2}v_{0}^{2}\cos^{2}{\beta}\widetilde{I}_{4}+\pi R^{3}\left(v_{0x}^{2}+v_{0y}^{2}\right)\widetilde{I}_{5}$$

$${}+2\pi R^{3}v_{0z}^{2}\left(\widetilde{I}_{4}-\widetilde{I}_{5}\right)$$

$${}+4\pi R^{2}Cv_{0}\cos{\beta}v_{0z}\left(\widetilde{I}_{2}-\widetilde{I}_{3}\right)\Big{]}\hat{k},$$

(B.2)

where

$$\widetilde{I}_{5}\equiv\int\limits_{0}^{\pi}{\frac{\sin^{3}{\theta}\cos{\theta}}{r^{5}}d\theta}=\frac{4}{3}\frac{C}{R^{4}\left(R^{2}-C^{2}\right)}$$

$${}=\frac{C}{R}\widetilde{I}_{3},$$

(B.3)

which has also been solved using the substitution $u^{2}=R^{2}+C^{2}-2RC\cos{\theta}$. Since $v_{0z}=v_{0}\cos{(\pi-\beta)}=-v_{0}\cos{\beta}$, both the $\hat{i}$ and $\hat{j}$ components of Eq. (B.2) simplify to zero, whereas the $\hat{k}$ component simplifies to

$$\pi R^{3}\left(v_{0x}^{2}+v_{0y}^{2}\right)\widetilde{I}_{5}$$

$${}+4\pi RCv_{0z}^{2}\left[C\frac{\widetilde{I}_{4}}{2}+\frac{R^{2}}{C}\frac{\widetilde{I}_{4}-\widetilde{I}_{5}}{2}-R\left(\widetilde{I}_{2}-\widetilde{I}_{3}\right)\right]$$

$${}=\frac{4\pi}{3}\frac{C}{R}\frac{v_{0x}^{2}+v_{0y}^{2}+v_{0z}^{2}}{R^{2}-C^{2}}.$$

(B.4)