We now prove a law, which will be often applicable for proving the tensor-character of certain quantities. According to the above representation, Aμν Bμν is a scalar, when Aμν and Bστ are tensors. We also remark that when Aμν Bμν is an invariant for every choice of the tensor Bμν, then Aμν has a tensorial character.
Proof:—According to the above assumption, for any substitution we have
Aστ′ Bστ′ = Aμν Bμν.
From the inversion of (9) we have however
Substitution of this for Bμν in the above equation gives
This can be true, for any choice of Bστ′ only when the term within the bracket vanishes. From which by referring to (11), the theorem at once follows. This law correspondingly holds for tensors of any rank and character. The proof is quite similar. The law can also be put in the following form. If Bμ and Cν are any two vectors, and if for every choice of them the inner product Aμν Bμ Cν is a scalar, then Aμν is a co-variant tensor. The last law holds even when there is the more special formulation, that with any arbitrary choice of the four-vector Bμ alone the scalar product Aμν Bμ Bν is a scalar, in which case we have the additional condition that Aμν satisfies the symmetry condition. According to the method given above, we prove the tensor character of (Aμν + Aνμ), from which on account of symmetry follows the tensor-character of Aμν. This law can easily be generalized in the case of co-variant and contravariant tensors of any rank.
Finally, from what has been proved, we can deduce the following law which can be easily generalized for any kind of tensor: If the quantities Aμν Bν form a tensor of the first rank, when Bν is any arbitrarily chosen four-vector, then Aμν is a tensor of the second rank. If for example, Cμ is any four-vector, then owing to the tensor character of Aμν Bν, the inner product Aμν Cμ Bν is a scalar, both the four-vectors Cμ and Bν being arbitrarily chosen. Hence the proposition follows at once.
A few words about the Fundamental Tensor gμν.