δρσ is 1 or 0, according as ρ = σ or not and r is the quantity
+√(x₁² + x₂² + x₃²).
On account of (68a) we have
(70a) α = κM/4π
where M denotes the mass generating the field. It is easy to verify that this solution satisfies approximately the field-equation outside the mass M.
Let us now investigate the influences which the field of mass M will have upon the metrical properties of the field. Between the lengths and times measured locally on the one hand, and the differences in co-ordinates dxν on the other, we have the relation
ds² = gμν dxμ dxν.
For a unit measuring rod, for example, placed parallel to the x axis, we have to put
ds² = -1, dx₂ = dx₃ = dx₄ = 0
then -1 = g₁₁dx₁².