Part 12

The _dx′__{σ}’_s_ are expressed as linear and homogeneous function of _dx__{ν}’_s_; we can look upon the differentials of the co-ordinates as the components of a tensor, which we designate specially as a contravariant Four-vector. Everything which is defined by Four quantities A^{σ}, with reference to a co-ordinate system, and transforms according to the same law,

(5a)

$$ A^{\sigma} = \sum_{\nu} \frac{\partial x'_{\sigma}}{\partial x_{\nu}} A^{\nu} $$

we may call a contra-variant Four-vector. From (5. a), it follows at once that the sums (A^{σ} ± B^{σ}) are also components of a four-vector, when A^{σ} and B^{σ} are so; corresponding relations hold also for all systems afterwards introduced as “tensors” (Rule of addition and subtraction of Tensors).

_Co-variant Four-vector._

We call four quantities A_{ν} as the components of a covariant four-vector, when for any choice of the contra-variant four vector B^{ν} (6) ∑_{ν} A_{ν} B^{ν} = _Invariant_. From this definition follows the law of transformation of the co-variant four-vectors. If we substitute in the right hand side of the equation

∑_{σ} A′_{σ} B^{σ′} = ∑_{ν} A_{ν} B^{ν}.

the expressions

$$ \sum_{\sigma} \frac{\partial x_{\nu}}{\partial x_{\sigma'}} B^{\sigma'} $$

for B^{ν} following from the inversion of the equation (5a) we get

$$ \sum_{\sigma} B^{\sigma'} \sum_{\nu} \frac{\partial x_{\nu}}{\partial x_{\sigma'}} A_{\nu} = \sum_{\sigma} B^{\sigma'} A'_{\sigma} $$

As in the above equation B^{σ′} are independent of one another and perfectly arbitrary, it follows that the transformation law is:—

$$ A'_{\sigma} = \sum \frac{\partial x_{\nu}}{\partial x_{\sigma'}} A_{\nu} $$

_Remarks on the simplification of the mode of writing the expressions._ A glance at the equations of this paragraph will show that the indices which appear twice within the sign of summation [for example ν in (5)] are those over which the summation is to be made and that only over the indices which appear twice. It is therefore possible, without loss of clearness, to leave off the summation sign; so that we introduce the rule: wherever the index in any term of an expression appears twice, it is to be summed over all of them except when it is not expressedly said to the contrary.

The difference between the co-variant and the contra-variant four-vector lies in the transformation laws [(7) and (5)]. Both the quantities are tensors according to the above general remarks; in it lies its significance. In accordance with Ricci and Levi-civita, the contravariants and co-variants are designated by the over and under indices.

§ 6. Tensors of the second and higher ranks.

Contravariant tensor:—If we now calculate all the 16 products A^{μν} of the components A^{μ} B^{ν}, of two contravariant four-vectors

(8) A^{μν} = A^{μ}B^{ν}

A^{μν}, will according to (8) and (5 a) satisfy the following transformation law.

(9)

$$ A^{\sigma \tau'} = \frac{\partial x'_{\sigma}}{\partial x_{\mu}} \frac{\partial x'_{\tau}}{\partial x_{\nu}} A^{\mu \nu} $$

We call a thing which, with reference to any reference system is defined by 16 quantities and fulfils the transformation relation (9), a contravariant tensor of the second rank. Not every such tensor can be built from two four-vectors, (according to 8). But it is easy to show that any 16 quantities A^{μν}, can be represented as the sum of A^{μ}B^{ν} of properly chosen four pairs of four-vectors. From it, we can prove in the simplest way all laws which hold true for the tensor of the second rank defined through (9), by proving it only for the special tensor of the type (8).

_Contravariant Tensor of any rank_:—It is clear that corresponding to (8) and (9), we can define contravariant tensors of the 3rd and higher ranks, with 4³, etc. components. Thus it is clear from (8) and (9) that in this sense, we can look upon contravariant four-vectors, as contravariant tensors of the first rank.

_Co-variant tensor._

If on the other hand, we take the 16 products A_{μν} of the components of two co-variant four-vectors A_{μ} and B_{ν},

(10) A_{μν} = A_{μ} B_{ν}.

for them holds the transformation law

(11)

$$ A^{\sigma \tau'} = \frac{\partial x'_{\mu}}{\partial x_{\sigma'}} \frac{\partial x'_{\nu}}{\partial x_{\tau'}} A^{\mu \nu} $$

By means of these transformation laws, the co-variant tensor of the second rank is defined. All re-marks which we have already made concerning the contravariant tensors, hold also for co-variant tensors.

_Remark_:—

It is convenient to treat the scalar Invariant either as a contravariant or a co-variant tensor of zero rank.

_Mixed tensor._ We can also define a tensor of the second rank of the type

(12) A_{μ}^{ν} = A_{μ}B^{ν}

which is co-variant with reference to μ and contravariant with reference to ν. Its transformation law is

(13)

$$ A^{\tau'}_{\sigma} = \frac{\partial x_{\tau'}}{\partial x_{\beta}} \frac{\partial \alpha}{\partial x_{\sigma'}} A^{\beta}_{\alpha} $$

Naturally there are mixed tensors with any number of co-variant indices, and with any number of contra-variant indices. The co-variant and contra-variant tensors can be looked upon as special cases of mixed tensors.

_Symmetrical tensors_:—

A contravariant or a co-variant tensor of the second or higher rank is called symmetrical when any two components obtained by the mutual interchange of two indices are equal. The tensor A^{μν} or A_{μν} is symmetrical, when we have for any combination of indices

(14) A^{μν} = A^{νμ}

or

(14a) A_{μν} = A_{νμ}.

It must be proved that a symmetry so defined is a property independent of the system of reference. It follows in fact from (9) remembering (14)

$$ A^{\sigma \tau'} = \frac{\partial x_{\sigma'}}{\partial x_{\mu}} \frac{\partial x'_{\tau}}{\partial x_{\nu}} A^{\mu \nu} = \frac{\partial x_{\sigma'}}{\partial x_{\mu}} \frac{\partial x_{\tau'}}{\partial x_{\nu}} A^{\nu \mu} = A^{\tau \sigma'} $$

_Antisymmetrical tensor._

A contravariant or co-variant tensor of the 2nd, 3rd or 4th rank is called _antisymmetrical_ when the two components got by mutually interchanging any two indices are equal and opposite. The tensor or A^{μν} or A_{μν} is thus antisymmetrical when we have

(15) A^{μν} = -A^{νμ}

or

(15a) A_{μν} = -A_{νμ}.

Of the 16 components A^{μν}, the four components A^{μμ} vanish, the rest are equal and opposite in pairs; so that there are only 6 numerically different components present (Six-vector).

Thus we also see that the antisymmetrical tensor A^{μνσ} (3rd rank) has only 4 components numerically different, and the antisymmetrical tensor A^{μνστ} only one. Symmetrical tensors of ranks higher than the fourth, do not exist in a continuum of 4 dimensions.

§ 7. Multiplication of Tensors.

_Outer multiplication of Tensors_:—We get from the components of a tensor of rank _z_, and another of a rank _z′_, the components of a tensor of rank (_z_ + _z′_) for which we multiply all the components of the first with all the components of the second in pairs. For example, we obtain the tensor Τ from the tensors A and B of different kinds:—

Τ_{μνσ} = A_{μν}B_{σ},

Τ^{αβγδ} = A^{αβ}B^{γδ},

Τ_{αβ}^{γδ} = A_{αβ}B^{γδ}.

The proof of the tensor character of Τ, follows immediately from the expressions (8), (10) or (12), or the transformation equations (9), (11), (13); equations (8), (10) and (12) are themselves examples of the outer multiplication of tensors of the first rank.

_Reduction in rank of a mixed Tensor._

From every mixed tensor we can get a tensor which is two ranks lower, when we put an index of co-variant character equal to an index of the contravariant character and sum according to these indices (Reduction). We get for example, out of the mixed tensor of the fourth rank A_{αβ}^{γδ}, the mixed tensor of the second rank

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.

Proof:—According to the above assumption, for any substitution we have

A_{στ′} B^{στ′} = A_{μν} B^{μν}.

From the inversion of (9) we have however

$$ B_{\mu \nu} = \frac{\partial x_{\mu}}{\partial x_{\sigma'}} \frac{\partial x_{\nu}}{\partial \tau'} B^{\sigma \tau'} $$

Substitution of this for B^{μν} in the above equation gives

$$ (A_{\sigma \tau'} - \frac{\partial x_{\mu}}{\partial x_{\sigma'}} \frac{\partial x_{\nu}}{\partial x_{\tau'}}) B^{\sigma \tau'} = 0 $$

This can be true, for any choice of B^{στ′} 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 B^{μ} and C^{ν} are any two vectors, and if for every choice of them the inner product A_{μν} B^{μ} C^{ν} is a scalar, then A_{μν} 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 B^{μ} alone the scalar product A_{μν} B^{μ} B^{ν} is a scalar, in which case we have the additional condition that A_{μν} satisfies the symmetry condition. According to the method given above, we prove the tensor character of (A_{μν} + A_{νμ}), from which on account of symmetry follows the tensor-character of A_{μν}. 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 A_{μν} B^{ν} form a tensor of the first rank, when B^{ν} is any arbitrarily chosen four-vector, then A_{μν} is a tensor of the second rank. If for example, C^{μ} is any four-vector, then owing to the tensor character of A_{μν} B^{ν}, the inner product A_{μν} C^{μ} B^{ν} is a scalar, both the four-vectors C^{μ} and B^{ν} being arbitrarily chosen. Hence the proposition follows at once.

A few words about the Fundamental Tensor _g__{μν}.

The co-variant fundamental tensor—In the invariant expression of the square of the linear element

_ds²_ = _g__{μν} _dx__{μ} _dx__{ν}

_dx__{μ} plays the rôle of any arbitrarily chosen contravariant vector, since further _g__{μν} = _g__{νμ}, it follows from the considerations of the last paragraph that _g__{μν} is a symmetrical co-variant tensor of the second rank. We call it the “fundamental tensor.” Afterwards we shall deduce some properties of this tensor, which will also be true for any tensor of the second rank. But the special rôle of the fundamental tensor in our Theory, which has its physical basis on the particularly exceptional character of gravitation makes it clear that those relations are to be developed which will be required only in the case of the fundamental tensor.

_The co-variant fundamental tensor._

If we form from the determinant scheme | _g__{μν} | the minors of _g__{μν} and divide them by the determinant _g_ = | _g__{μν} | we get certain quantities _g_^{μν} = _g_^{νμ}, which as we shall prove generates a contravariant tensor.

According to the well-known law of Determinants

(16) _g__{μσ} _g_^{νσ} = δ_{μ}^{ν}

where δ_{μ}^{ν} is 1, or 0, according as μ = ν or not. Instead of the above expression for _ds²_, we can also write

_g__{μσ} δ_{ν}^{σ} _dx__{μ} _dx__{ν}

or according to (16) also in the form

_g__{μσ} _g__{ντ} _g_^{στ} _dx__{μ} _dx__{ν}

Now according to the rules of multiplication, of the fore-going paragraph, the magnitudes

_d_ξ_{σ} = _g__{μσ} _dx__{μ}

forms a co-variant four-vector, and in fact (on account of the arbitrary choice of _dx__{μ}) any arbitrary four-vector.

If we introduce it in our expression, we get

_ds²_ = _g_^{στ} _d_ξ_{σ} _d_ξ_{τ}.

For any choice of the vectors _d_ξ_{σ} _d_ξ_{τ} this is scalar, and _g_^{στ}, according to its definition is a symmetrical thing in σ and τ, so it follows from the above results, that _g_^{στ} is a contravariant tensor. Out of (16) it also follows that δ^{ν}_{μ} is a tensor which we may call the mixed fundamental tensor.

_Determinant of the fundamental tensor._

According to the law of multiplication of determinants, we have

| _g__{μα} _g_^{αν} | = | _g__{μα} | | _g_^{αν} |

On the other hand we have

| _g__{μα} _g_^{αν} | = | δ^{ν}_{μ} | = 1

So that it follows (17) that | _g__{μν} | | _g_^{μν} | = 1.

_Invariant of volume._

We see first the transformation law for the determinant _g_ = | _g__{μν} |. According to (11)

$$ g' = | \frac{\partial x_{\mu}}{\partial x_{\sigma'}} \frac{\partial x_{\nu}}{\partial x_{\tau'}} g_{\mu u} | $$

From this by applying the law of multiplication twice, we obtain

$$ g' = | \frac{\partial x_{\mu}}{\partial x_{\sigma'}} | | \frac{\partial x_{\nu}}{\partial x_{\tau'}} | | g_{\mu \nu} | = | \frac{\partial x_{\mu}}{\alpha_{\sigma'}} | g $$

or

(A)

$$ \sqrt{g'} = | \frac{\partial x_{\mu}}{\partial x_{\sigma'}} | \sqrt{g} $$

On the other hand the law of transformation of the volume element

_d_τ′ = ∫ _dx₁_ _dx₂_ _dx₃_ _dx₄_

is according to the wellknown law of Jacobi.

(B) $$ d\tau' = | \frac{dx'_{\sigma}}{dx_{\mu}} | d\tau $$

by multiplication of the two last equations (A) and (B) we get

(18) = √_g_ _d_τ′ = √_g_ _d_τ.

Instead of √_g_, we shall afterwards introduce √(-_g_) which has a real value on account of the hyperbolic character of the time-space continuum. The invariant √(-_g_)_d_τ, is equal in magnitude to the four-dimensional volume-element measured with solid rods and clocks, in accordance with the special relativity theory.

_Remarks on the character of the space-time continuum_—Our assumption that in an infinitely small region the special relativity theory holds, leads us to conclude that _ds²_ can always, according to (1) be expressed in real magnitudes _d_X₁ ... _d_X_{_h_}. If we call _d_τ₀ the “_natural_” volume element _d_X₁ _d_X₂ _d_X₃ _d_X₄ we have thus (18a) _d_τ₀ = √(_g_)_i_τ.

Should √(-_g_) vanish at any point of the four-dimensional continuum it would signify that to a finite co-ordinate volume at the place corresponds an infinitely small “natural volume.” This can never be the case; so that _g_ can never change its sign; we would, according to the special relativity theory assume that _g_ has a finite negative value. It is a hypothesis about the physical nature of the continuum considered, and also a pre-established rule for the choice of co-ordinates.

If however (-_g_) remains positive and finite, it is clear that the choice of co-ordinates can be so made that this quantity becomes equal to one. We would afterwards see that such a limitation of the choice of co-ordinates would produce a significant simplification in expressions for laws of nature.

In place of (18) it follows then simply that

_d_τ′ = _d_

from this it follows, remembering the law of Jacobi,

(19)

$$ | \frac{\partial x'_{\sigma}}{dx_{\mu}} | = 1 $$

With this choice of co-ordinates, only substitutions with determinant 1 are allowable.

It would however be erroneous to think that this step signifies a

## partial renunciation of the general relativity postulate. We do not seek

those laws of nature which are co-variants with regard to the transformations having the determinant 1, but we ask: what are the general co-variant laws of nature? First we get the law, and then we simplify its expression by a special choice of the system of reference.

_Building up of new tensors with the help of the fundamental tensor._

Through inner, outer and mixed multiplications of a tensor with the fundamental tensor, tensors of other kinds and of other ranks can be formed.

Example:—

A^{μ} = _g_^{μσ} A_{σ}

A = _g__{μν} A^{μν}

We would point out specially the following combinations:

A^{μν} = _g_^{μα} _g_^{νβ} A_{αβ}

A_{μν} = _g__{μα} _g__{νβ} A^{αβ}

(complement to the co-variant or contravariant tensors)

and B_{μν} = _g__{μν} _g_^{αβ} A_{αβ}

We can call B_{μν} the reduced tensor related to A_{μν}.

Similarly

B^{μν} = _g_^{μν}_g__{αβ}A^{αβ}.

It is to be remarked that _g_^{μν} is no other than the “complement” of _g__{μν} for we have,—

_g_^{μα}_g_^{νβ}_g__{αβ} = _g__{μα}δ^{ν}_{α} = _g_^{μν}.

§ 9. Equation of the geodetic line (or of point-motion).

As the “line element” _ds_ is a definite magnitude independent of the co-ordinate system, we have also between two points P₁ and P₂ of a four dimensional continuum a line for which ∫_ds_ is an extremum (geodetic line), _i.e._, one which has got a significance independent of the choice of co-ordinates.

Its equation is

(20) δ{ ∫^{P₂}_{P₁} _ds_ } = 0

From this equation, we can in a wellknown way deduce 4 total differential equations which define the geodetic line; this deduction is given here for the sake of completeness.

Let λ, be a function of the co-ordinates _x__{ν}; this defines a series of surfaces which cut the geodetic line sought-for as well as all neighbouring lines from P₁ to P₂. We can suppose that all such curves are given when the value of its co-ordinates _x__{ν} are given in terms of λ. The sign δ corresponds to a passage from a point of the geodetic curve sought-for to a point of the contiguous curve, both lying on the same surface λ.

Then (20) can be replaced by

{ λ₃ { ∫δω _d_λ = 0 (20a) { λ₁ { { ω² = _g__{μν}(_dx__{μ}/_d_λ)(_dx__{ν}/_d_λ)

But

δω = (1/ω){½(∂_g__{μν}/∂_x__{σ}) · (_dx__{μ}/_d_λ) · (_dx__{ν}/_d_λ) · δ_x__{σ} + _g__{μν}(_dx__{μ}/_d_λ)δ(_dx__{ν}/_d_λ)}

So we get by the substitution of δω in (20a), remembering that

δ(_dx__{ν}/_d_λ) = (_d_/_d_λ)(δ_x__{ν})

after partial integration,

{ λ₃ { ∫ _d_λ _k__{σ} δ_x__{σ} = 0 (20b) { λ₁ { { where _k__{σ} = (_d_/_d_λ){(_g__{μν}/ω) · (_dx__{μ}/_d_λ)} - (1/(2ω))(∂_g__{μν}/∂_x__{σ}

× (_dx__{μ}/_d_λ) · (_dx__{ν}/_d_λ).

From which it follows, since the choice of δν_{σ} is perfectly arbitrary that _k__{σ}_’s_ should vanish. Then

(20c) _k__{σ} = 0 (σ = 1, 2, 3, 4)

are the equations of geodetic line; since along the geodetic line considered we have _ds_ ≠ 0, we can choose the parameter λ, as the length of the arc measured along the geodetic line. Then _w_ = 1, and we would get in place of (20c)

$$ g_{\mu\nu} \frac{\partial^2 x_{\mu}}{\partial s^2} + \frac{\partial g_{\mu\nu}}{\partial x_{\sigma}} \frac{\partial x_{\sigma}}{\partial s} \frac{\partial x_{\mu}}{\partial s} - \frac{1}{2} \frac{\partial g_{\mu\sigma}}{\partial x_{\nu}} \frac{\partial x_{\mu}}{\partial s} \frac{\partial x_{\sigma}}{\partial s} = 0 $$

Or by merely changing the notation suitably,

(20d) $$ g_{\alpha\sigma} \frac{d^2 x_{\alpha}}{ds^2} + \begin{bmatrix}\mu\nu\\\sigma\end{bmatrix} \frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} = 0 $$

where we have put, following Christoffel,

(21)

$$ \begin{bmatrix}\mu\nu\\\sigma\end{bmatrix} = \frac{1}{2} \begin{bmatrix}\frac{\partial g_{\mu\sigma}}{\partial x_{\nu}} + \frac{\partial g_{\nu\sigma}}{\partial x_{\mu}} - \frac{\partial g_{\mu\nu}}{\partial \sigma}\end{bmatrix} $$

Multiply finally (20d) with _g_^{στ} (outer multiplication with reference to τ, and inner with respect to σ) we get at last the final form of the equation of the geodetic line—

$$ \frac{d^2 x_{\tau}}{ds^2} + \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} \frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} = 0 $$

Here we have put, following Christoffel,

$$ \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} = g^{\tau\alpha} \begin{bmatrix}\mu\nu\\\alpha\end{bmatrix} $$

§ 10. Formation of Tensors through Differentiation.

Relying on the equation of the geodetic line, we can now easily deduce laws according to which new tensors can be formed from given tensors by differentiation. For this purpose, we would first establish the general co-variant differential equations. We achieve this through a repeated application of the following simple law. If a certain curve be given in our continuum whose points are characterised by the arc-distances _s_, measured from a fixed point on the curve, and if further φ, be an invariant space function, then _d_φ/_ds_ is also an invariant. The proof follows from the fact that _d_φ as well as _ds_, are both invariants

Since

$$ \frac{d \phi}{ds} = \frac{\partial \phi}{\partial x_{\mu}} \frac{\partial x_{\mu}}{\partial s} $$

so that

$$ \psi = \frac{\partial \phi}{\partial x_{\mu}} \frac{dx_{\mu}}{ds} $$

is also an invariant for all curves which go out from a point in the continuum, _i.e._, for any choice of the vector _dx__{μ}. From which follows immediately that

A_{μ} = ∂φ/∂_x__{μ}

is a co-variant four-vector (gradient of φ).

According to our law, the differential-quotient χ = ∂ψ/∂_s_ taken along any curve is likewise an invariant.

Substituting the value of ψ, we get

$$ \chi = \frac{\partial^2 \phi}{\partial x_{\mu} \partial x_{\nu}} \frac{dx_{\mu}}{ds} \frac{dx_{\nu}}{ds} + \frac{\partial \phi}{\partial x_{\mu}} \frac{d^2 x_{\mu}}{ds^2} $$

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}.

Now the right hand side of (25) multiplied by ψ is

$$ \psi \frac{\partial^2 \phi}{\partial x_{\mu} \partial x_{\nu}} - \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} \psi \frac{\partial \phi}{\partial x_{\tau}} $$

which has a tensor character. Similarly, (∂φ/∂_x__{μ}) (∂φ/∂_x__{ν}) is also a tensor (outer product of two four-vectors).

Through addition follows the tensor character of

$$ \frac{\partial}{\partial x_{\nu}} (\psi \frac{\partial \phi}{\partial x_{\mu}}) - \begin{Bmatrix}\mu\nu\\\tau\end{Bmatrix} (\psi \frac{\partial \phi}{\partial x_{\tau}}) $$

Thus we get the desired proof for the four-vector, ψ ∂φ/∂_x__{μ} and hence for any four-vectors A_{μ} as shown above.