Here however we can not at once deduce the existence of any tensor. If we however take that the curves along which we are differentiating are geodesics, we get from it by replacing d^2x_{ν}/ds^2 according to (22)
χ = [ [part]^2φ/([part]x_{μ}[part]x_{ν}) - {^{μν} _{τ}} ([part]φ/[part]x_{τ})] (dx_{μ}/ds) · (dx_{ν}/ds).
From the interchangeability of the differentiation with regard to μ and ν, and also according to (23) and (21) we see that the bracket {^{μν} _{τ}} is symmetrical with respect to μ and ν.
As we can draw a geodetic line in any direction from any point in the continuum, [part]x_{μ}/ds is thus a four-vector, with an arbitrary ratio of components, so that it follows from the results of §7 that
(25) A_{μν} = [part]^2φ/([part]x_{μ}[part]x_{ν}) - {^{μν} _{τ}} ([part]φ/[part]x_{τ})
is a co-variant tensor of the second rank. We have thus got the result that out of the co-variant tensor of the first rank A_{μ} = [part]φ/[part]x_{μ} we can get by differentiation a co-variant tensor of 2nd rank
(26) A_{μν} = [part]A_{μ}/[part]x_{ν} - {^{μν} _{τ}} A_{τ}.
