Page:LorentzGravitation1916.djvu/54

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\nu=\varkappa\int\limits _{\infty}^{r}\frac{dr}{r^{2}}\int\limits _{0}^{r}\varrho dr (108)

The quantities \lambda and \mu on the contrary are not completely determined by the differential equations. If namely equations (105) and (106) are added to (104) after having been multiplied by -\tfrac{1}{2} and +\tfrac{1}{2} respectively, we find

\lambda+r\lambda'-\mu+r\nu'=0 (109)

and it is clear that (104) and (105) are satisfied as soon as this is the case with this condition (109) and with (106). So we have only to attend to (108) and (109). The indefiniteness remaining in \lambda and \mu is inevitable on account of the covariancy of the field equations. It does not give rise to any difficulties.

Equation (107) teaches us that near the centre

\nu'=\frac{1}{3}\varkappa\overline{\varrho_{0}}r

if \overline{\varrho_{0}} is the density at the centre, whereas from (108) we find a finite value for \nu itself. This confirms what has been said above about the values at the centre. We shall assume that at that point \lambda,\mu and their derivatives have likewise finite values. Moreover we suppose (and this agrees with (109)) that \lambda,\mu,\lambda' and \mu' are continuous at the surface of the sphere.

If a is the radius of the sphere we find from (108) for an external point

\nu=-\frac{\varkappa}{r}\int\limits _{0}^{a}\varrho dr

Without contradicting (109) we may assume that at a great distance from the centre \lambda and \mu are likewise proportional to \tfrac{1}{r}, so that \lambda' and \mu' decrease proportionally to \tfrac{1}{r^{2}}.


§ 59. We can now continue the calculation of \mathfrak{t}_{4}^{'4} (§ 56). Substituting (101) in (99) and using polar coordinates we find

\mathfrak{t}_{4}^{'4}=-\frac{1}{2\varkappa}u\sqrt{\frac{w}{v}}\left(\frac{1}{2}\frac{u'^{2}}{u^{2}}+\frac{u'w'}{uw}\right)

whence by substituting (102) we derive for a field without gravitation

\mathfrak{t}_{4}^{'4}=-\frac{c}{\varkappa}

This equation shows that, working with polar coordinates, we