$$ - \frac{1}{2} \sqrt{-g} \frac{\partial g_{\rho\sigma}}{\partial
x_{\mu}} A^{\rho\sigma} $$
If instead of A^{ρσ}, we introduce in a similar way the symmetrical
co-variant tensor A_{ρσ} = _g__{ρα} _g__{σβ} A^{αβ}, then owing to (31)
the last member can take the form
$$ \frac{1}{2} \sqrt{-g} \frac{\partial g_{\rho\sigma}}{\partial
x_{\mu}} A_{\rho\sigma} $$
In the symmetrical case treated, (41) can be replaced by either of the
forms
(41a)
$$ \sqrt{-g} A{\mu} = \frac{\partial (\sqrt{-g}
A^{\sigma}_{\mu})}{\partial x_{\sigma}} - \frac{1}{2} \frac{\partial
g_{\rho\sigma}}{\partial x_{\mu}} \sqrt{-g} A^{\rho\sigma} $$
or
(41b)
$$ \sqrt{-g} A{\mu} = \frac{\partial (\sqrt{-g}
A^{\sigma}_{\mu})}{\partial x_{\sigma}} + \frac{1}{2} \frac{\partial
g_{\rho\sigma}}{\partial x_{\mu}} \sqrt{-g} A_{\rho\sigma} $$
which we shall have to make use of afterwards.
§12. The Riemann-Christoffel Tensor.
We now seek only those tensors, which can be obtained from the
fundamental tensor _g_^{μν} by differentiation alone. It is found
easily. We put in (27) instead of any tensor A^{μν} the fundamental
tensor _g_^{μν} and get from it a new tensor, namely the extension of
the fundamental tensor. We can easily convince ourselves that this
vanishes identically. We prove it in the following way; we substitute in
(27)
$$ A_{\mu\nu} = \frac{\partial A_{\mu}}{\partial x_{\nu}} -
\begin{Bmatrix}\mu & & \nu\ \rho&\end{Bmatrix} A_{\rho} $$
_i.e._, the extension of a four-vector.
Thus we get (by slightly changing the indices) the tensor of the third
rank
$$ A_{\mu\sigma\tau} = \frac{\partial^2 A_{\mu}}{\partial x_{\sigma}
\partial x_{\tau}} - \begin{Bmatrix}\mu & & \sigma\ \rho&\end{Bmatrix}
\frac{\partial A_{\rho}}{\partial x_{\tau}} - \begin{Bmatrix}\mu & &
\tau\ \rho&\end{Bmatrix} \frac{\partial A_{\rho}}{\partial x_{\sigma}} -
\begin{Bmatrix}\sigma & & \tau\ \rho&\end{Bmatrix} \frac{\partial
A_{\mu}}{\partial x_{\rho}} + \begin{bmatrix} - \frac{\partial}{\partial
x_{\tau}} \begin{Bmatrix}\mu&&\sigma\ \rho&\end{Bmatrix} +
\begin{Bmatrix}\mu&&\tau\ \alpha\end{Bmatrix}
\begin{Bmatrix}\alpha&&\sigma\ \rho&\end{Bmatrix} +
\begin{Bmatrix}\sigma&&\tau\ \alpha\end{Bmatrix}
\begin{Bmatrix}\alpha&&\mu\ \rho&\end{Bmatrix} \end{bmatrix} A_{\rho} $$
We use these expressions for the formation of the tensor A_{μστ} -
A_{μτσ}. Thereby the following terms in A_{μστ} cancel the corresponding
terms in A_{μτσ}; the first member, the fourth member, as well as the
member corresponding to the last term within the square bracket. These
are all symmetrical in σ, and τ. The same is true for the sum of the
second and third members. We thus get
(43)
$$ A_{\mu\sigma\tau} - A_{\mu\tau\sigma} = B^{\rho}_{\mu\sigma\tau}
A_{\rho} $$
$$ B^{\rho}_{\mu\sigma\tau} = - \frac{\partial}{\partial x_{\tau}}
\begin{Bmatrix}\mu & & \sigma\ \rho&\end{Bmatrix} +
\frac{\partial}{\partial x_{\sigma}} \begin{Bmatrix}\mu & &
\tau\ \rho&\end{Bmatrix} - \begin{Bmatrix}\mu & &
\sigma\ \alpha&\end{Bmatrix} \begin{Bmatrix}\alpha & &
\tau\ \rho&\end{Bmatrix} + \begin{Bmatrix}\mu & &
\tau\ \alpha&\end{Bmatrix} \begin{Bmatrix}\alpha & &
\sigma\ \rho&\end{Bmatrix} $$
The essential thing in this result is that on the right hand side of
(42) we have only A_{ρ}, but not its differential co-efficients. From
the tensor-character of A_{μστ} - A_{μτσ}, and from the fact that A_{ρ}
is an arbitrary four vector, it follows, on account of the result of §7,
that B^{ρ}_{μστ} is a tensor (Riemann-Christoffel Tensor).
The mathematical significance of this tensor is as follows; when the
continuum is so shaped, that there is a co-ordinate system for which
_g__{μν}_’s_ are constants, B^{ρ}_{μστ} all vanish.
If we choose instead of the original co-ordinate system any new one, so
would the _g__{μν}’s referred to this last system be no longer
constants. The tensor character of B^{ρ}_{μστ} shows us, however, that
these components vanish collectively also in any other chosen system of
reference. The vanishing of the Riemann Tensor is thus a necessary
condition that for some choice of the axis-system _g__{μν}’s can be
taken as constants. In our problem it corresponds to the case when by a
suitable choice of the co-ordinate system, the special relativity theory
holds throughout any finite region. By the reduction of (43) with
reference to indices to τ and ρ, we get the covariant tensor of the
second rank
(44)
$$ B_{\mu\nu} = R_{\mu\nu} + S_{\mu\nu} $$
$$ R_{\mu\nu} = - \frac{\partial}{\partial x_{\alpha}}
\begin{Bmatrix}\mu & & \nu\ \alpha&\end{Bmatrix} + \begin{Bmatrix}\mu
& & \alpha\ \beta&\end{Bmatrix} \begin{Bmatrix}\nu & &
\beta\ \alpha&\end{Bmatrix} $$
$$ S_{\mu\nu} = \frac{\partial \log \sqrt{-g}}{\partial x_{\mu} \partial
x_{\nu}} - \begin{Bmatrix}\mu & & \nu\ \alpha&\end{Bmatrix}
\frac{\partial \log \sqrt{-g}}{\partial x_{\alpha}} $$
_Remarks upon the choice of co-ordinates._—It has already been remarked
in §8, with reference to the equation (18a), that the co-ordinates can
with advantage be so chosen that √(-_g_) = 1. A glance at the equations
got in the last two paragraphs shows that, through such a choice, the
law of formation of the tensors suffers a significant simplification. It
is specially true for the tensor B_{μν}, which plays a fundamental rôle
in the theory. By this simplification, S_{μν} vanishes of itself so that
tensor B_{μν} reduces to R_{μν}.
I shall give in the following pages all relations in the simplified
form, with the above-named specialisation of the co-ordinates. It is
then very easy to go back to the general covariant equations, if it
appears desirable in any special case.