A_{β}^{δ} = A_{αβ}^{αδ} = (∑_{α} A_{αβ}^{αδ})

and from it again by “reduction” the tensor of the zero rank

A = A_{β}^{β} = A_{αβ}^{αβ}.

The proof that the result of reduction retains a truly tensorial

character, follows either from the representation of tensor according to

the generalisation of (12) in combination with (6) or out of the

generalisation of (13).

_Inner and mixed multiplication of Tensors._

This consists in the combination of outer multiplication with reduction.

Examples:—From the co-variant tensor of the second rank A_{μν} and the

contravariant tensor of the first rank B^{σ} we get by outer

multiplication the mixed tensor

D^{σ}_{μν} = A_{μν} B^{σ} .

Through reduction according to indices ν and σ (_i.e._, putting ν = σ),

the co-variant four vector

D_{μ} = D^{ν}_{μν} = A_{μν} B^{ν} is generated.

These we denote as the inner product of the tensor A_{μν} and B^{σ}.

Similarly we get from the tensors A_{μν} and B^{στ} through outer

multiplication and two-fold reduction the inner product A_{μν} B^{μν}.

Through outer multiplication and one-fold reduction we get out of A_{μν}

and B^{στ}, the mixed tensor of the second rank D^{τ}_{μ} = A_{μν}

B^{τν}. We can fitly call this operation a mixed one; for it is outer

with reference to the indices μ and τ and inner with respect to the

indices ν and σ.

We now prove a law, which will be often applicable for proving the

tensor-character of certain quantities. According to the above

representation, A_{μν} B^{μν} is a scalar, when A_{μν} and B^{στ} are

tensors. We also remark that when A_{μν} B^{μν} is an invariant for

every choice of the tensor B^{μν}, then A_{μν} has a tensorial

character.

The Principle of Relativity · 第 111 / 203 题