$$ \chi = \frac{\partial^2 \phi}{\partial x_{\mu} \partial x_{\nu}}
\frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} + \frac{\partial \phi}{\partial
x_{\mu}} \frac{d^2 x_{\mu}}{ds^2} $$
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)
$$ A_{\mu\nu} = \frac{\partial A_{\mu}}{\partial x_{\nu}} -
\begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} A_{\tau} $$
We call the tensor A_{μν} the “extension” of the tensor A_{μ}. Then we
can easily show that this combination also leads to a tensor, when the
vector A_{μ} is not representable as a gradient. In order to see this we
first remark that ψ (_d_φ/∂_x__{μ}) is a co-variant four-vector when ψ
and φ are scalars. This is also the case for a sum of four such terms :—
$$ S_{\mu} = \psi^{(1)} \frac{\partial \phi^{(1)}}{\partial x_{\mu}} +
... + \psi^{(4)} \frac{\partial \phi^{(4)}}{\partial x_{\mu}} $$
when ψ^{(1)}, φ^{(1)} ... ψ^{(4)}, φ^{(4)} are scalars. Now it is
however clear that every co-variant four-vector is representable in the
form of S_{μ}.
If for example, A_{μ} is a four-vector whose components are any given
functions of _x__{ν}, we have, (with reference to the chosen co-ordinate
system) only to put
ψ^{(1)} = A₁ φ^{(1)} = _x₁_
ψ^{(2)} = A₂ φ^{(2)} = _x₂_
ψ^{(3)} = A₃ φ^{(3)} = _x₃_
ψ^{(4)} = A₄ φ^{(4)} = _x₄_.
in order to arrive at the result that S_{μ} is equal to A_{μ}.
In order to prove then that A_{μν} is a tensor when on the right side of
(26) we substitute any co-variant four-vector for A_{μ} we have only to
show that this is true for the four-vector S_{μ}. For this latter case,
however, a glance on the right hand side of (26) will show that we have
only to bring forth the proof for the case when
A_{μ} = ψ ∂φ/∂_x__{μ}.