# What is a fourth rank tensor

summed up according to this index (“rejuvenation”). One wins

so z. B. from the mixed tensor of fourth rank A_{}^{} the

mixed tensor of the second order

and from this, again by rejuvenation, the tensor

zeroth rank A = A_{}^{} = A_{}^{}.

Proof that the result of rejuvenation is effective

Lich has a tensor character, results either from the

Tensor representation according to the generalization of (12) in

Connection with (6) or from the generalization of (13).

Inner and mixed multiplication of tensors. These

consist in the combination of the outer multiplication with

of rejuvenation.

Examples. - From the second-order covariant tensor

A._{} and the contravariant first-order tensor B^{} form

we get the mixed tensor by external multiplication

By tapering according to the indices , the co-

variant quad vector

We also refer to this as the inner product of the tensors

A._{} and B^{}. Analogously, A is formed from the tensors_{} and

B.^{} by external multiplication and tapering twice

the inner product A_{}B.^{}. Through external product formation

and one-time taper is obtained from A_{} and B^{}the

mixed tensor of second rank D_{}^{} = A_{}B.^{}. One can

aptly refer to this operation as a mixed one; because

it is an external one with respect to the indices and , an inner one

regarding the indices and.

We now prove a theorem that proves the

Tensor character is often usable. After the just presented

is A_{}B.^{} a scalar if A_{}and B^{} Tensors

are. But we also claim the following. If A_{}B.^{}For

any choice of tensor B^{} is an invariant, then A has_{} Tensor

character.

Proof. - It is according to requirement for any

substitution

But after reversing (9),

This, inserted into the above equation, yields:

This can be done with any choice of B^{}^{'} only then met

be when the bracket disappears, from which with reverse

Looking at (11) the assertion follows.

This theorem applies to arbitrary tensors

Rank and character; the proof must always be carried out analogously.

The proposition can also be proved in the form: are

B.^{} and C^{} arbitrary vectors, and with every choice the-

same the inner product

a scalar, so is A._{} a covariant tensor. This latter

Theorem still applies even if only the more specific statement

it is true that with any choice of the four-vector B^{} the

scalar product

is a scalar if one also knows that A_{} the sym-

operating condition A._{} = A_{} enough. Because on the earlier

given paths one proves the tensor character of

, from which then because of the symmetry property

the tensor character of A_{} itself follows. This sentence too

can be easily generalized to the case of covariant and

contravariant tensors of any rank.

Finally follows from what has been proven that

dear tensors generalizable proposition: if the sizes

A._{}B.^{} with any choice of the four-vector B^{} a tensor

form first rank, then A_{} a second rank tensor.

Is C^{} an arbitrary four-vector, so is because of the

Tensor character A_{}B.^{} the inner product A_{}C.^{}B.^{} at

any choice of the two four-vectors C^{} and B^{} a

Scalar, from which the claim follows.

§ 8. Some things about the fundamental tensor of g_{}.

The covariant fundamental tensor. In the invariant

Expression of the square of the line element

plays d x_{} the role of a freely selectable contravariant

Vector. Since also_{} = g_{}, it follows after the considerations

of the last paragraph from this that g_{} a covariant tensor

second rank is. We call it the “fundamental tensor”.

In the following we derive some properties of this tensor

ab, which are inherent in every tensor of the second rank; but

the special role of the fundamental tensor in our theory,

which their in the peculiarity of the gravitational effects

has a physical reason, it means that the

winding relations only for the fundamental tensor for

matter to us.

The contravariant fundamental tensor. Is one formed in that

Determinant scheme of g_{} to each g_{} the sub-determi-

nante and divides this by the determinant g = the

G_{}, one obtains certain quantities g^{}(= g^{}) of which we

want to prove that they form a contravariant tensor.

According to a well-known set of determinants is

(16) |

where the sign _{}^{} Means 1 or 0, as the case may be =

or is. Instead of the above expression for d s^{2} can

We also

or according to (16) also

write. But now form according to the multiplication rules

of the previous paragraph the sizes

a covariant four-vector, namely (because of the will-

free choice of d x_{}) any freely selectable

Four-vector. By introducing it into our expression

we receive

Since this with any choice of the vector d _{} a scalar

is and g^{} according to its definition in the indices and sym-

is metric, it follows from the results of the previous para-

graph that g^{} is a contravariant tensor. From (16)

still follows that too _{}^{} is a tensor we mixed the

Can call fundamental tensor.

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