From this equation, we can in a wellknown way deduce 4 total differential equations which define the geodetic line; this deduction is given here for the sake of completeness. Let λ, be a function of the co-ordinates _x__{ν}; this defines a series of surfaces which cut the geodetic line sought-for as well as all neighbouring lines from P₁ to P₂. We can suppose that all such curves are given when the value of its co-ordinates _x__{ν} are given in terms of λ. The sign δ corresponds to a passage from a point of the geodetic curve sought-for to a point of the contiguous curve, both lying on the same surface λ. Then (20) can be replaced by { λ₃ { ∫δω _d_λ = 0 (20a) { λ₁ { { ω² = _g__{μν}(_dx__{μ}/_d_λ)(_dx__{ν}/_d_λ) But δω = (1/ω){½(∂_g__{μν}/∂_x__{σ}) · (_dx__{μ}/_d_λ) · (_dx__{ν}/_d_λ) · δ_x__{σ} + _g__{μν}(_dx__{μ}/_d_λ)δ(_dx__{ν}/_d_λ)} So we get by the substitution of δω in (20a), remembering that δ(_dx__{ν}/_d_λ) = (_d_/_d_λ)(δ_x__{ν}) after partial integration, { λ₃ { ∫ _d_λ _k__{σ} δ_x__{σ} = 0 (20b) { λ₁ { { where _k__{σ} = (_d_/_d_λ){(_g__{μν}/ω) · (_dx__{μ}/_d_λ)} - (1/(2ω))(∂_g__{μν}/∂_x__{σ} × (_dx__{μ}/_d_λ) · (_dx__{ν}/_d_λ). From which it follows, since the choice of δν_{σ} is perfectly arbitrary that _k__{σ}_’s_ should vanish. Then (20c) _k__{σ} = 0 (σ = 1, 2, 3, 4) are the equations of geodetic line; since along the geodetic line considered we have _ds_ ≠ 0, we can choose the parameter λ, as the length of the arc measured along the geodetic line. Then _w_ = 1, and we would get in place of (20c) $$ g_{\mu\nu} \frac{\partial^2 x_{\mu}}{\partial s^2} + \frac{\partial g_{\mu\nu}}{\partial x_{\sigma}} \frac{\partial x_{\sigma}}{\partial s} \frac{\partial x_{\mu}}{\partial s} - \frac{1}{2} \frac{\partial g_{\mu\sigma}}{\partial x_{\nu}} \frac{\partial x_{\mu}}{\partial s} \frac{\partial x_{\sigma}}{\partial s} = 0 $$ Or by merely changing the notation suitably, (20d) $$ g_{\alpha\sigma} \frac{d^2 x_{\alpha}}{ds^2} + \begin{bmatrix}\mu\nu\\\sigma\end{bmatrix} \frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} = 0 $$ where we have put, following Christoffel, (21) $$ \begin{bmatrix}\mu\nu\\\sigma\end{bmatrix} = \frac{1}{2} \begin{bmatrix}\frac{\partial g_{\mu\sigma}}{\partial x_{\nu}} + \frac{\partial g_{\nu\sigma}}{\partial x_{\mu}} - \frac{\partial g_{\mu\nu}}{\partial \sigma}\end{bmatrix} $$ Multiply finally (20d) with _g_^{στ} (outer multiplication with reference to τ, and inner with respect to σ) we get at last the final form of the equation of the geodetic line— $$ \frac{d^2 x_{\tau}}{ds^2} + \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} \frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} = 0 $$ Here we have put, following Christoffel,