${\mathfrak {t}}_{4}^{'4}={\frac {c}{2\varkappa }}\left\{\nu '^{2}+{\frac {1}{r}}(\lambda -\mu )\left[{\frac {1}{r}}(\lambda -\mu )+2(\lambda '-\mu ')\right]\right\}$ |
(111) |

Thus we see (comp. § 58) that at a distance from the attracting sphere ${\mathfrak {t}}_{4}^{'4}$ decreases proportionally to ${\tfrac {1}{r^{4}}}$. Further it is to be noticed that on account of the indefiniteness pointed out in § 58, there remains some uncertainty as to the distribution of the energy over the space, but that nevertheless the total energy of the gravitation field

$E=4\pi \int \limits _{0}^{\infty }{\mathfrak {t}}_{4}^{'4}r^{2}dr$

has a definite value.

Indeed, by the integration the last terra of (111) vanishes. After multiplication by $r^{2}$ this term becomes namely

$(\lambda -\mu )^{2}+2r(\lambda -\mu )(\lambda '-\mu ')={\frac {d}{dr}}\left[r(\lambda -\mu )^{2}\right]$

The integral of this expression is 0 because (comp. §§ 57 and 58) $r(\lambda -\mu )^{2}$ is continuous at the surface of the sphere and vanishes both for $r=0$ and for $r=\infty$.

We have thus

$E={\frac {\pi c}{\varkappa }}\int \limits _{0}^{\infty }\nu '^{2}r^{2}dr$ |
(112) |

where the value (107) can be substituted for $\nu '$. If e.g. the density ${\overline {\varrho }}$ is everywhere the same all over the sphere, we have at an internal point

$\nu '={\frac {1}{3}}\varkappa {\overline {\varrho }}r$

and at an external point

$\nu '={\frac {1}{3}}\varkappa {\overline {\varrho }}{\frac {a^{3}}{r^{2}}}$

From this we find

$E={\frac {2}{15}}\pi c\varkappa {\overline {\varrho }}^{2}a^{5}$

§ 61. The general equation (99) found for ${\mathfrak {t}}_{4}^{'4}$ can be transformed in a simple way. We have namely

${\begin{array}{c}\sum (abfe){\frac {\partial }{\partial x_{e}}}\left({\frac {\partial Q}{\partial g_{ab,fe}}}\right)g_{ab,f}=\sum (abfe){\frac {\partial }{\partial x_{e}}}\left({\frac {\partial Q}{\partial g_{ab,fe}}}g_{ab,f}\right)-\\\\-\sum (abfe){\frac {\partial Q}{\partial g_{ab,fe}}}g_{ab,fe}\end{array}}$

and we may write $-Q_{2}$ (§ 54) for the last term. Hence