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 [part]φ/[part]x_{μ}.
Now the right hand side of (25) multiplied by ψ is
ψ [part]^2φ/([part]x_{μ}[part]x_{ν}) - {^{μν} _{τ}}ψ [part]φ/[part]x_{τ}
which has a tensor character. Similarly, ([part]φ/[part]x_{μ}) ([part]φ/[part]x_{ν}) is also a tensor (outer product of two four-vectors).
Through addition follows the tensor character of
[part]/[part]x_{ν} (ψ [part]φ/[part]x_{μ}) - {μν/τ}(ψ [part]φ/[part]x_{τ})
Thus we get the desired proof for the four-vector, ψ [part]φ/[part]x_{μ} and hence for any four-vectors A_{μ} as shown above.
With the help of the extension of the four-vector, we can easily define "extension" of a co-variant tensor of any rank. This is a generalisation of the extension of the four-vector. We confine ourselves to the case of the extension of the tensors of the 2nd rank for which the law of formation can be clearly seen.
As already remarked every co-variant tensor of the 2nd rank can be represented as a sum of the tensors of the type A_{μ} B_{ν}.
