# Landau-Lifschitz gravitational energy-momentum pseudotensor

## Energy conservation principle in general relativity

Two of the most popular definitions of the combined Energy and Momentum Conservation Principle (EMCP) are:

1. the energy and momentum of a closed isolated system are conserved
2. the stress-energy-momentum (SEM) tensor of a closed isolated system is a constant quantity

The second definition is preferred in theoretical physics. The SEM tensor, Tik, which combines energy, mass, momentum, their fluxes and stresses (also called "super energy-momentum tensor", etc.), is the content of the Universe. It is both the matter and the electromagnetic fields. EMCP requires that the SEM tensor is constant. Imagine the closed isolated system as a volume (Universe, black hole), enclosed by an impenetrable envelope such that no matter or radiation can pass through. The SEM tensor within the enclosed volume is constant: there is no tensor flow through the envelope. Tensor flow, or tensor divergence, is the sum of partial derivatives of tensor components with respect to coordinates (similar to vector divergence). Divergence should be zero according to EMCP (LL32,4):

∂Tik/ ∂x k ≡ Tik,k = 0 (EMCP).

Presence of gravitation spoils this clear picture in several ways. Gravitation curves space. In curved space, tensor (and vector) flows are ill-defined because they change direction in each point. Parallel transfer, which is essential for definition of derivatives, is very complicated on curved paths. To cut it short, derivatives are obtained only after making corrections for the curvature – one correction for each tensor index. These corrected derivatives are known as covariant derivatives.  In curved space, ordinary derivatives are replaced by covariant ones and (EMCP) becomes

∂Tik/ ∂x k − Γ likTlk + Γ kklTil ≡ Tik;k = 0. (85,13) .

The Γ lik terms above are the corrections for spacetime curvature: a negative correction for the lower index in Tik and a positive correction for the upper index. Tik is a symmetric tensor and according to 86,11

Tik;k = (√−g) −1 (Tik√−g) , k − ½gkl,iTkl (96,1).

The first term in the right hand side can be made an ordinary derivative of the SEM tensor by pulling out the constant √−g from inside the differential operator. However, the real difficulty lies in the second term −½gkl,iTkl. The metric tensor gkl and its derivatives can be anything depending on the spacetime topology. Taking up the above analogy, the volume that contains the SEM tensor itself stretches, shrinks, and moves in every imaginable way. It cannot be bound in an envelope with zero volume. The spacetime itself generates tensor flows. Do what you want, you can't just ignore or remove the second term. Failure to do so means nothing less than the startling conclusion: in the curved spacetime of general relativity, the EMCP is no longer valid. Whether this is real or apparent remains subject to interpretation. It is easy to see, however, that this result can be expected from what we know from elementary physics. In the heart of the concept of the energy conservation principle is the assumption that space is isotropic while the momentum conservation principle requires symmetric space. Both isotropy and symmetry are no longer assumed in curved spacetime. The energy and momentum conservation principles fail in principle. The very basis on which they are built sinks into a quicksand. Is there a way to salvage something?

## In search of symmetry

Many efforts have been and are being made to sidetrack this important unsolvable problem. Landau-Lifshitz turned their attention to a couple of possible loopholes.

1. the immediate neighborhood of every point is locally flat; this means that the tiny volume around this point can be enclosed in an envelope into which EMCP is valid
2. the identity EMCP does not fix the SEM tensor exactly but leaves a quantity undefined. Namely, according to LL(32,7), one can always find an antisymmetric tensor ηikl = −ηilk such that if T ik = &etaikl,l then Tik,k = ηikl,k,l ≡ 0. 

The possibility to locally fulfil EMCP gives the idea to find a SEM tensor that locally has a zero divergence and then correct it to make it valid globally, in curved spacetime. An inkling to the nature of this correction may be taken from the analogous case with the covariant derivatives. There all corrections contain Christoffel symbols which are composed from first derivatives of the metric tensor. The way to make the required SEM tensor is already obvious: it must be such antisymmetric expression that upon differention becomes a second derivative which is identically zero. When constructing it, we must throw out all first metric derivatives that would prevent the expression from turning to zero.

Landau-Lifshitz start from the Einstein field equations to put Tik into the required form

Tik = ηikl,l (etaform).

The field equations in contravariant form are

Tik = (c4 / 8πk) (Rik − ½gikR). (95,5)

All tensors in 95,5 are decomposed in metric derivatives and only the second derivatives are taken. The detailed transformation of the right hand side of the last expression (explanation of each step is provided in notes) is:

(16πk / c4) Tik note 1 = 2 (Rik − ½gikR) = 2 (gimgkpgln − ½gikglngmp) Rlmnp note 2 ≈ (gimgkpgln − ½gikglngmp)(glp,m,n + gmn,l,p − gln,m,p − gmp,l,n) note 3 = gimgkpglnglp,m,n + gimgkpglngmn,l,p − gimgkpglngln,m,p − gimgkpglngmp,l,n − ½ gikglngmpglp,m,n + ½ gikglngmpgmn,l,p − ½ gikglngmpgln,m,p − ½ gikglngmpgmp,l,n note 4 = gimgkpglngnp,m,l + gipgkmglngnp,m,l − gimgklgnpgnp,m,l − gingkpglmgnp,m,l − ½ gikglngmpgnp,m,l + ½ gikglngmpgnp,m,l − ½ gikgnpgmlgnp,m,l − ½ gikglmgnpgnp,m,l note 5 = gnp,m,l (gimgkpgln + gipgkmgln − gimgklgnp − gingkpglm − ½gikglngmp − ½gikglngmp + ½gikgnpgml + ½gikglmgnp) note 6 = gnp,m,l (gimgkpgln + gipgkmgln − gimgklgnp − gingkpglm − gikglngmp + gikgnpgml) note 7 = [gnp,m(gimgkpgln + gipgkmgln − gimgklgnp − gingkpglm − gikglngmp + gikgnpgml)] , l − gnp,m (gimgkpgln + gipgkmgln − gimgklgnp − gingkpglm − gikglngmp + gikgnpgml),l note 8 ≈ [gnp,m(gimgkpgln + gipgkmgln − gimgklgnp − gingkpglm − gikglngmp + gikgnpgml)],l note 9 = (gimgkpglngnp,m + gipgkmglngnp,m − gimgklgnpgnp,m − gingkpglmgnp,m − gikglngmpgnp,m + gikgnpgmlgnp,m),l note 10 = [−gimgkpgnpgln,m − gipgkmgnpgln,m − gimgkl(g,m/g) + ginglmgnpgkp,m + gikglngnpgmp,m + gikgml(g,m/g)],l note 11 = [−δkngimgln,m − δingkmgln,m − gimgkl(g,m/g) + δipglmgkp,m + δlpgikgmp,m + gikgml(g,m/g)],l note 12 = [− gimgkl,m − gkmgil,m − gimgkl(g,m/g) + glmgik,m + gikglm,m + gikgml(g,m/g)],l note 13 = [−gilgkm,m − gkmgil,m − gilgkm(g,m/g) + glmgik,m + gikglm,m + gikglm(g,m/g)],l note 14 = [(glmgik,m + gikglm,m) − (gilgkm,m + gkmgil,m) + (g,m/g)(gikglm − gilgkm)],l note 15 = [(gikglm),m − (gilgkm),m + (g,m/g)(gikglm − gilgkm)],l note 16 = [(gikglm − gilgkm),m + (g,m/g)(gikglm − gilgkm)], l note 17 = [(−g/− g)(gikglm − gilgkm),m + (−1/−1)(g,m/g)(gikglm − gilgkm)], l note 18 = {1/(−g)[(−g)(gikglm − gilgkm),m + (−g),m(gikglm − gilgkm)]},l note 19 = {1/(−g)[(−g)(gikglm − gilgkm)],m},l note 20 = 1/(−g)[(−g)(gikglm − gilgkm)],m,l note 21.

Note 1. Both sides are multiplied by 16πk/c4 to get rid of the constant in the right hand side. The factor 2 on the right should subsequently neutralise the ½ in the decomposition.

Note 2. The Ricci tensor is

Rik = gimgkpglnRlmnp (92,6).

The term with the scalar curvature, gikR, is

gikR = gikglngmpRlmnp (92,9).

Note 3. The Riemann tensor, Rlmnp, in which all first metric derivatives in Christoffel symbols are ignored, is approximated by

Rlmnp ≈ ½ (glp,m,n + gmn,l,p − gln,m,p − gmp,l,n) (92,1).

Note 4. All parentheses are opened resulting in 2 x 4 = 8 terms.

Note 5. Pairs of dummies are exchanged so as to make all secondary derivatives, gnp,m,l with uniform indexes.

Note 6. The second derivatives gnp,m,l are collected from all terms. Note that the last four terms are pairwise equal when we take into account that the metric tensor is symmetric, e.g., gml = glm. The pairwise equal terms are marked in different colors: red and blue.

Note 7. The last four terms that are pairwise equal are added; now we have a second derivative of the metric multiplied by a combination of 6 metric terms that do not contain the factor ½.

Note 8. Let gnp = u and gimgkpgln + gipgkmgln − gimgklgnp − gingkpglm − gikglngmp + gikgnpgml = v. Using the formula for a derivative of a product: u,m,l v = (u,m v),lu,m v,l  we get 2 terms: the first (red) includes second derivatives while the second (blue) includes only first derivatives

Note 9. The second term is dropped because it contains only first derivatives; the resulting expression is approximation valid in the local point.

Note 10. Brackets are expanded by distributing gnp,m among terms.

Note 11. Indexes shown in red are dummies that were exchanged. We have two cases: first, where we have the same upper and lower dummies, LL(86,7) was used in the form: glngnp,m = −gnpgln,m ; second, where both upper and lower dummies are the same, as in gnpgnp,m , the ubiquitous LL(86,4) was used in the form: g,m = ggnpgnp,m so that gnpgnp,m = (g,m/g).

Note 12. Products of metric tensors that contain the same dummy indexes are equal to Kronecker deltas according to LL(83,10): gkpgnp = δkn, and so on.

Note 13. Kronecker deltas when multiplied with metric tensors change the indexes of the latter in the following way: δkngimgln,m = gimgkl,m; making these changes in the above expression results in some shortening.

Note 14. In some terms, m and l indexes are exchanged, using the symmetry of the differential operator.

Note 15. Terms are grouped conveniently ...

Note 16. ... so that the formula for differentiation of a product: glmgik,m + gikglm,m = (gikglm),m can be used to present the terms in the first two parentheses as derivatives of products.

Note 17. The first two parentheses are factored by the ,m differentiation (associative property).

Note 18. The first term is multiplied by (−g/−g) = 1 and the second term by (−1/−1) = 1

Note 19. 1/(−g) is collected in front of the brackets. Inside the brackets we can distinguish products of two terms: (−g) = u and (gikglm − gilgkm) = v.

Note 20. The property of a derivative of product u v,m + u,m v = (u v),m is again used to factor the second differentiation operator ,m.

Note 21. 1/(−g) is pulled out of derivatives by ,l which is possible because 1/(−g) is constant in the special point. This means that 1/(−g) goes in front of the braces which are then dropped as there is no need for them – the expression that remains in the brackets is doubly differentiated.

To summarize, the final expression for the SEM tensor as a result of the above transformations is:

(16πk/c4)Tik = 1/(−g)[(−g)(gikglm − gilgkm)],m,l.

For the following discussion we need this expression somewhat rearranged. Namely, the constant (c4/16πk) is put under ,l (but not under ,m) and both sides are multiplied by (−g):

(−g)T ik = (c4/16πk)[(−g)(gikglm − gilgkm)],m,l = {(c4/16πk)[(−g)(gikglm − gilgkm)],m},l.

After these "simple transformations" (according to Landau-Lifshitz), we obtained a SEM tensor in a completely antisymmetric form:

(−g)Tik = {(c4/16πk)[(−g)(gikglm − gilgkm)],m},l.

The quantity in the braces corresponds to ηikl in (etaform):

ηikl = (c4/16πk)[(−g)(gikglm − gilgkm)],m (96,3).

According to , (−g)Tik vanishes upon differentiation by ,k (check). Furthermore, (−g)Tik is symmetric: (−g)Tik = (−g)Tki (check).

## Landau-Lifshitz pseudotensor as first metric derivatives

The nice and pliable SEM tensor obtained above whose divergence disappears automatically just by dint of symmetry (or, more precisely, antisymmetry) so that it fulfils EMCP, does not work outside the very small flat locality of the point that we chose. Remember that in its derivation above we threw out first derivatives of the metric tensor at a few steps. First derivatives usually combine to form connections (Christoffel symbols) and these last are directly connected to curvature. So we may expect that outside our flat neighbourhood first derivatives will reappear to play a prominent role as corrections for curvature in the same way as they act in covariant derivatives.

We have every reason to expect, then, that (−g)Tik will no longer be equal to ηikl,l

ηikl,l − (−g)Tik ≠ 0.

There will be some non-zero difference that is the quantity that corrects for the transition from local to global geometry, from flat to curved spacetime. Let's define this difference as (−g)tik, then by definition

(−g)(Tik + tik) = ηikl,l (96,5).

We can't tell much about the quantities tik except that they have absorbed all the first derivatives which we threw out before, and also that they are symmetric:

tik = t ki (96,6).

The latter stems from the definition of tik, since they are composed of Tik and ηikl,l which are symmetric as we proved in .

The method to find tik that Landau-Lifshitz propose, is direct but excessively long and tedious: " ... after a rather lenghty calculation we find the following expression ... " One seldom finds in their book such a phrase at the place of the gory detail of the intermediate calculations though it is an understatement as we see in the derivation of LL96,8 that follows.

To be more specific, we shall articulate the plan of calculations in several steps:

1. tik is expressed according to (96,5) as a difference between the Landau-Lifsitz pseudotensor in the special point and the SEM tensor
2. the two differentiations in the Landau-Lifsitz pseudotensor in the special point are carried out
3. derivatives and double derivatives of the metric determinant are transformed into derivatives and double derivatives of the metric tensor according to (86,4)
4. all double derivatives are written as covariant metric tensors rather than as contravariant ones using when necessary (86,4) in the covariant form
5. the SEM tensor is expressed by the Einstein equations and transformed sequentially into Riemann tensor (through (92,6) and (92,9)), Christoffel symbols (through (92,1)), and metric tensor derivatives (through (86,3))
6. the SEM tensor in metric derivatives is subtracted from the Landau-Lifshitz pseudotensor in the special point, expressed also in metric derivatives
7. the second metric tensor derivatives in the resultant expression are presented in such a manner as to cancel and derive an expression containing only first derivatives of the metric tensor (base form of the Landau-Lifshitz pseudotensor)
8. the base form (first metric derivatives) is transformed into (96,8) (Christoffel symbols) using (86,8) and (86,5) or
9. the base form (first metric derivatives) is transformed into (96,9) (metric tensor densities) using (86,6)
10. As before, the intermediate calculations are provided with detailed notes.

tik = 1/(−g)ηikl,l − Tik note 22 = (c4/16πk){1/(−g)[(−g)(gikglm − gilgkm)]},m,l − (c4/8πk)(Rik − ½gikR) note 23 = (c4/16πk){(1/g)[g(gikglm − gilgkm)],m,l − (2Rik − gikR) note 24 = (c4/16πk){(1/g)[(gikglm − gilgkm)g,m,l + g,m (glmgik,l − gkmgil,l − gilgkm,l + gikglm,l ) + g,l (glmgik,m − gkmgil,m − gilgkm,m + gikglm,m ) + g(−gil,m gkm,l − gil,l gkm,m + gik,m glm,l + gik,l glm,m + glmgik,m,l − gkmgil,m,l &minus gilgkm,m,l + gikglm,m,l )] − (2gilgkm − gikglm)Rlm} note 25 = (c4/16πk){(1/g)[(gikglm − gilgkm)g,m,l + g,l ( glmgik ,m − gklgim,m − gimgkl,m + gikglm,m ) + g,l (glmgik,m − gkmgil,m − gilgkm,m + gikglm,m ) + g(−gil,m gkm,l − gil,l gkm,m + 2gik,l glm,m + glmgik,m,l − gkmgil,m,l &minus gilgkm,m,l + gikglm,m,l )] − (2gilgkm − gikglm) nlm,n − Γ nln,m + Γ nlmΓ pnp − Γ plnΓ nmp)} note 26 = (c4/16πk){(1/g)[ 2g(gikglm − gilgkm)(2Γ nln Γ pmp + Γ nln,m ) + 2gΓ nln ( 2glmgik,m − gklgim,m − gkmgil,m − gimgkl,m − gilgkm,m + 2gikglm,m ) + g(−gil,m gkm,l − gil,l gkm,m + 2gik,l glm,m + glmgik,m,l − gkmgil,m,l &minus gilgkm,m,l + gikglm,m,l )] − (2gilgkm − gikglm) (Γ nlm,n − Γ nln,m + Γ nlmΓ pnp − Γ plnΓ nmp)} note 27 = (c4/16πk) [−gil,m gkm,l − gil,l gkm,m + 2gik,l glm,m + glmgik,m,l − gkmgil,m,l &minus gilgkm,m,l + gikglm,m,l + 4glmgik,m Γ nln − 2gklgim,m Γ nln − 2gkmgil,m Γ nln − 2gimgkl,m Γ nln − 2gilgkm,m Γ nln + 4gikglm,m Γ nln + 4gikglmΓ nln Γ pmp + 2gikglmΓ nln,m − 4gilgkm Γ nln Γ pmp2gilgkm Γ nln,m + gikglmΓ nlm,ngikglmΓ nln,m + gikglmΓ nlmΓ pnp − gikglmΓ plnΓ nmp − 2gilgkmΓ nlm,n + 2gilgkmΓ nln,m − 2gilgkmΓ nlmΓ pnp + 2gilgkmΓ plnΓ nmp ] note 28 = (c4/16πk) [− (−glnΓ imn −ginΓ lmn) (−gmpΓ klp −gkpΓ mlp) (−glpΓ ilp −gipΓ llp) (−gmnΓ kmn −gknΓ mmn) + 2 (−gknΓ iln −ginΓ kln) (−gmpΓ lmp −glpΓ mmp) + glmgik,m,l − gkmgil,m,l &minus gilgkm,m,l + gikglm,m,l + 4glm (−gkpΓ imp −gipΓ kmp) Γ nln − 2gkl (−gmpΓ imp −gipΓ mmp) Γ nln − 2gkm (−glpΓ imp −gipΓ lmp) Γ nln − 2gim (−glpΓ kmp −gkpΓ lmp) Γ nln − 2gil (−gmpΓ kmp −gkpΓ mmp) Γ nln + 4gik (−gmpΓ lmp −glpΓ mmp) Γ nln + 4gikglmΓ nln Γ pmp + gikglmΓ nln,m − 4gilgkm Γ nln Γ pmp + gikglmΓ nlm,n + gikglmΓ nlmΓ pnp − gikglmΓ plnΓ nmp − 2gilgkmΓ nlm,n − 2gilgkmΓ nlmΓ pnp + 2gilgkmΓ plnΓ nmp ] note 29 = (c4/16πk) [− (glm gnpΓ ilnΓkmp + gilgmnΓ knp Γ plm + gklgmnΓ inp Γ plm + gilgkmΓ nlpΓ pmn) − (glmgnpΓ iln Γ kmp + gklgmnΓ imnΓ plp + gilgmnΓ kmnΓ plp + gilgkmΓ nmnΓ plp) + 2 (−gknΓ iln −ginΓ kln) (−gmpΓ lmp −glpΓ mmp) + glmgik,m,l − gkmgil,m,l &minus gilgkm,m,l + gikglm,m,l + 4glm (−gkpΓ imp −gipΓ kmp) Γ nln − 2gkl (−gmpΓ imp −gipΓ mmp) Γ nln − 2gkm (−glpΓ imp −gipΓ lmp) Γ nln − 2gim (−glpΓ kmp −gkpΓ lmp) Γ nln − 2gil (−gmpΓ kmp −gkpΓ mmp) Γ nln + 4gik (−gmpΓ lmp −glpΓ mmp) Γ nln + 4gikglmΓ nln Γ pmp + gikglmΓ nln,m − 4gilgkm Γ nln Γ pmp + gikglmΓ nlm,n + gikglmΓ nlmΓ pnp − gikglmΓ plnΓ nmp − 2gilgkmΓ nlm,n − 2gilgkmΓ nlmΓ pnp + 2gilgkmΓ plnΓ nmp ] note 30 =

Note 22. tik is obtained by isolating it in (96,5).

Note 23. ηikl,l is obtained from (96,3) ηikl,l = (c4/16πk) [(−g)(gikglm − gilgkm)],m,l and Tik is expressed with Rik and gikR through the Einstein field equations (95,5).

Note 24. (c4/16πk) is pulled out in front; as a result, the Rik and gikR terms are multiplied by 2. The minus sign from 1/(−g) is put under the derivative where it cancels with the minus sign of (−g).

Note 25. The expression in brackets is doubly differentiated by ,m and ,l. Terms that become identical on exchange of l and m indexes are summed (in red). Now in curved spacetime, the determinant g is no longer constant so that its derivatives are generally not zeros. This fact greatly contributes to the complexity of the resulting expressions. The SEM is expressed entirely through the covariant Ricci tensor Rlm according to LL(92,6) Rik = gilgkmRlm and LL(92,9) gikR = gikglmRlm.

Note 24. Differention of the expression in the first brackets (former ηikl,l ) is carried out to give a mixture of first and second derivatives. Christoffel symbols are not decomposed in metric tensors here because our first aim is to obtain LL(96,8) which is the Landau-Lifshitz pseudotensor expressed in Christoffel symbols.

Note 25. Derivatives of metric determinant are replaced with derivatives of metric tensor with the help of the following variants of LL(86,4): g,m = ggnpgnp,m ; g,l = ggnpgnp,l and g,m,l = (ggnpgnp,m),l = ggnpgqrgnp,mgqr,l + ggnp,lgnp,m + ggnpgnp,m,l (replacements are in red).

Note 26. Expression is fully expanded to obtain 32 terms overall. All metric tensor determinants cancel and disappear. Some terms can be made equal by exchanging pairs of dummy indexes l, m, n, p. Pairs of equal terms (shown in various colors) are added together, reducing the number of terms to 27.

## Footnotes and References

1. All naming and mathematical conventions as well as equation numbering are according to The classical theory of fields. By Lev Davidovich Landau, Evgeniy Mikhaylovich Lifshitz, Morton Hamermesh. Translated by Morton Hamermesh. Edition: 4, illustrated, revised. Published by Butterworth-Heinemann, 1975. ISBN 0750627689, 9780750627689. 402 pages
2. In the short notation, ordinary derivatives are written with a comma "," in front of the indexes of the coordinate arguments, covariant derivatives are written with semicolon ";" instead of comma.
3. Proof that ηikl,k,l is identically zero:

ηikl,k,l = ½(ηikl + ηikl),k,l = ½(ηikl ηilk),k,l note 3.1 = ½ηikl,k,l − ½ηilk,k,l = ½ηikl,k,l − ½ηikl,l,k note 3.2 = ½ηikl,k,l − ½ηikl,k,l note 3.3 ≡ 0.

Note 3.1 Exchange of k and l indexes in the antisymmetric tensor ηikl makes it negative (operations in red).

Note 3.2 k and l are "dummy" indexes, so they can be exchanged without having to worry about consistency with indexes in other terms.

Note 3.3 The differentiation operator is also symmetric, so that, for example, ,k,l is the same as ,l,k.

4. Check that (−g)Tik,k is zero:

(−g)Tik,k = ηikl,l,k = (c4/16πk) [(−g)(gikglm − gilgkm)],m,l,k = (c4/16πk){½[(−g)(gikglm − gilgkm)],m,l,k + ½[(−g)(gikglm − gilgkm)],m,l,k} note 4.1 = (c4/16πk) {½[(−g)(gikglm − gilgkm)],m,l,k ½[(−g)(gilg km gikglm)],m,l,k} note 4.2 = (c4/16πk) {½[(−(gikglm − gilgkm)],m,l,k − ½[(−g)(gikglm − gilgkm)],m,k,l} note 4.3 = (c4/16πk){½[(−g)(gikglm − gilgkm)],m,l,k − ½[(−g)(gikglm − gilgkm)],m,l,k} note 4.4 ≡ 0.

Note 4.1 The calculation procedure is very similar to the one outlined in .

Note 4.2 Change of the sign from plus to minus in front of the second parenthesis makes the terms inside exchange places (shown in red).

Note 4.3 l and k are dummies here, so we exchange them in some terms (in red).

Note 4.4 ,m,k,l = ,m,l,k because the differential operator is symmetric for all indexes: the result of a multiple differentiation does not depend on the order in which the arguments are taken.

Third derivatives of the metric tensor combination, and with them (−g)Tik,k vanish identically as expected.

5. Check that Tik,k is symmetric tensor:

(16πk/c4)(−g)Tik = [(−g)(gikglm − gilgkm)],m,l = ½[(−g)(gikglm − gilgkm)],m,l + ½[(−g)(gikglm − gilgkm)],m,l = ½[(−g)(gki glmgkmgil)],m,l + ½[(−g)(gkiglmgkmgil)],m,l note 5.1 = ½[(−g)(gkiglm − gklgim)],m,l + ½[(−g)(gkiglm − gklgim)],m,l note 5.2 = [(−g)(gkiglm − gklgim)],m,l note 5.3 = (16πk/c4)(−g)Tki.

Note 5.1 Because of the symmetry of the metric tensor gik = gki and the commutativity of tensor multiplication gilgkm = gkmgil.

Note 5.2 The m and l dummies in the second and fourth terms were exchanged but their order was preserved in the differential operator because the latter is symmetric.

Note 5.3 Addition of terms inside the brackets gives a combination of metric tensors in which indexes i and k have exchanged places compared to the original pseudotensor.

The above shows that the SEM tensor thus obtained is symmetric. Note that Tik wouldn't be symmetric if we had left 1/(−g) under the ,l differentiating operator and had not made the note 21 transformation because then we wouldn't have symmetric ,m,l double-differentiating operator.