$$ \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} = g^{\tau\alpha} \begin{bmatrix}\mu\nu\\\alpha\end{bmatrix} $$ § 10. Formation of Tensors through Differentiation. Relying on the equation of the geodetic line, we can now easily deduce laws according to which new tensors can be formed from given tensors by differentiation. For this purpose, we would first establish the general co-variant differential equations. We achieve this through a repeated application of the following simple law. If a certain curve be given in our continuum whose points are characterised by the arc-distances _s_, measured from a fixed point on the curve, and if further φ, be an invariant space function, then _d_φ/_ds_ is also an invariant. The proof follows from the fact that _d_φ as well as _ds_, are both invariants Since $$ \frac{d \phi}{ds} = \frac{\partial \phi}{\partial x_{\mu}} \frac{\partial x_{\mu}}{\partial s} $$ so that $$ \psi = \frac{\partial \phi}{\partial x_{\mu}} \frac{dx_{\mu}}{ds} $$ is also an invariant for all curves which go out from a point in the continuum, _i.e._, for any choice of the vector _dx__{μ}. From which follows immediately that A_{μ} = ∂φ/∂_x__{μ} is a co-variant four-vector (gradient of φ). According to our law, the differential-quotient χ = ∂ψ/∂_s_ taken along any curve is likewise an invariant. Substituting the value of ψ, we get $$ \chi = \frac{\partial^2 \phi}{\partial x_{\mu} \partial x_{\nu}} \frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} + \frac{\partial \phi}{\partial x_{\mu}} \frac{d^2 x_{\mu}}{ds^2} $$ 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 _d²__x__{ν}/_ds²_ according to (22) $$ \chi = \begin{bmatrix}\frac{\partial^2 \phi}{\partial x_{\mu}\partial x_{\nu}} - \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} \frac{\partial \phi}{\partial x_{\tau}} \end{bmatrix} \frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} $$ From the interchangeability of the differentiation with regard to μ and ν, and also according to (23) and (21) we see that the bracket $$ \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} $$ is symmetrical with respect to μ and ν. As we can draw a geodetic line in any direction from any point in the continuum, ∂_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) $$ A_{\mu\nu} = \frac{\partial^2 \phi}{\partial x_{\mu} \partial x_{\nu}} - \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} \frac{\partial \phi}{\partial x_{\tau}} $$ 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} = ∂φ/∂_x__{μ} we can get by differentiation a co-variant tensor of 2nd rank (26) $$ A_{\mu\nu} = \frac{\partial A_{\mu}}{\partial x_{\nu}} - \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} A_{\tau} $$