1080 



Dividing by c* we then obtain the mass of tlie sjslem and in § 4 

 we shall (ind that this mass is identical with the one we obtain by 

 considering the gravitation field at points at a very great distance. 

 As has been said, the integral in (15) must be extended over the 

 whole infinite space. It is however desirable to express the mass of 

 a material system by an integral taken over the material system 

 itself and we shall now show how this can be done. According to 

 a law of v. Laue we have for the energy E also this expression 



■M 



E=c \\\^ (^V^ + r./) dx, d.v^ dx„ . . . . . (16) 



integrated over the whole three-dimensional space. We subtract this 

 equation from equation (15) after having multiplied the latter by 2. 

 As in a stationary field (//~ =: 0, we have because of (10) and (11) 



xr/ = i6*=ix:S'tA (17) 



and we obtain 



E = cJJj {^^^-X,^-l,^-^,')d.v,d.v,d,v, . . . (18) 



The integrand being zero at every point outside the material system, 

 the integral here has only to be extended over the material 

 system itself. 



By means of formula (18) we have expressed the mass of a 

 material system by a space-integral extended over the material 

 system. This space-integral can again be transformed into a surface 

 integral extended over a surface enclosing the material system. This 

 may be made evident in the following way. From formula (14) 

 we see that x (X/ -|- r/) can be expressed as the divergency of a 

 three-dimensional quasivector ^: 



»«(2/ + r/) = ^^\ (19) 



where 



^r=-:S^—~gM (I9a) 



Multiplying (19) by 2 and subtracting (13) from this product 

 while also (17) is taken into account, we obtain 



>«(Ï4-'^/-^.'-3:3^) = ^v"(2 55r-^31.) = ^T^\ . (20) 



According to this equation the application of Gauss's law to 

 (18) gives 



E = ^J^,^df. ........ (-21) 



