2 4 0 D O C U M E N T 3 1 0 J U L Y 1 9 2 2 I cannot evade the conclusiveness of your arguments and now see that the con- dition has to be set.[2] I must admit, however, that something unsatisfying remains, because although the formal condition makes the inertia of the masses vanish, it does not account for the relativity of the masses. One would prob- ably have to “prohibit” the use of coordinate systems in which the determinant (g) vanishes—even if only at infinity. As you were so kind as to delve into my considerations, I would like to submit to you another argument that—if another error has not crept in—would at least show that one can make the relativity of masses agree formally with the gravitation equations of the first kind. I proceed from the spherically symmetric solution in the form: 1) (H. Weyl, Ann. d. Phys., 54, p. 132 [3] in the Pauli Enc., page 730, formula 421b, there is a misprint[4] ) and transform it by the substitution: . Thus we get 2) . Now I consider the solution that results for r a with (2) and for r a agrees with the continuously attained result 3) . It describes the field of a hollow massive sphere at resting mass m and radius a, where the velocity of light (measured in natural units of length and cosmic time) has the large value at infinity if ε is small. If one now lets ε go to the limit 0, one obtains in the limiting case pseudo-Euclid- ean conditions within the interior and in the exterior space Euclidean metrics, infi- nite velocity of light, and vanishing inertia of the masses! It seems particularly remarkable to me that at the limiting case ε = 0 the velocity of light rises to imme- g44 = mik 0= ds2 1 m 2r ----- + 4 dx1 2 dx2 2 dx3 2 + + [ ] 1 m 2r ----- 2 1 m 2r ----- + 2 ----------------------dx4- 2 + = xi xi 4 --- - ,= i 1 2 3, , , = t 2 ε -- - t ,= ε 1 2m a ------- 0 –= ds2 1 2m r ------- + 2 ---------------- - 2 dx12 dx22 dx32] + + [ 2 ε -- - 2 1 2m r ------- 2 1 2m r ------- + 2 ------------------------dx42 - + = ds2 2 ε 2 ----------- 2 dx1 2 dx2 2 dx3 2 + + [ ] 2 2 ε ----------- 2 dt4 2 + = 2 ε -- -
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