Noether currents in theories with higher derivatives and theories with differential field transformation in action (DFTA)

Abstract

The article is devoted to the investigation of the Noether currents and integrals of motion in the special subclass of theories with higher field derivatives — theories under differential field transformations in action (DFTA). Under fairly general assertions, we derive a simple representation for the integrals as sums of “old” integrals of motion, terms that vanish on the “old” equations of motion and surface integrals. We show that for some cosmological theories of that kind with high-order derivatives the last contribution can violate gauge invariance, specifically, diffeomorphisms. Then we investigate ambiguity of the Noether procedure for higher derivative theories and fix it in the way the problem resolves in quite general setting. The obtained results are discussed for the several extensions of the General Relativity, which are obtained by DFTA, in particular, for mimetic, disformal and Regge–Teitelboim theories of gravity. In particular, we calculate Iyer–Wald black hole entropy from the corrected currents and also discuss the derivation of the integral constraints on Sachs–Wolfe effect.

Appendix

1.1 Explicit form of $L_{(k)a}{}^{\rho \alpha _1..\alpha _k}$

The coefficients $L_{(k)a}{}^{\rho \alpha _1..\alpha _k}$ defined in the decomposition (39) plays key role in the calculations throughout the paper. These coefficients are calculated, for example, in [55]:

$$\begin{aligned} L_{(k)}{}_a{}^{\rho \gamma _1..\gamma _k} ={}&\theta \left( N_d-k+1/2\right) \sum _{l = \max (k-N+1,0)}^{\min (k,M)}\Omega {}^{\rho }{}_a{}^{\gamma _1..\gamma _k}(k-l,l)+ U_a{}^{\rho \gamma _1...\gamma _k}, \end{aligned}$$

(120)

where

$$\begin{aligned} \Omega {}^{\rho }{}_a{}^{\alpha _1..\alpha _m\beta _1..\beta _l}(m,l) \equiv&\frac{1}{(m+l)!}\Bigg \{\sum _{j=m+1}^N\sum _{i = 0}^{j-m-1}(-1)^iC^m_{j-i-1}(\partial _{\alpha _{j-i}..\alpha _{j-1}}Z^{a\rho \alpha _1..\alpha _{j-1}})\nonumber \\&\times \partial _{\alpha _{m+1}..\alpha _{j-i-1}}H_{(l)}{}_{a\mu }{}^{\beta _1..\beta _l}\Bigg \}^{\alpha _1..\alpha _m\beta _1..\beta _l}, \end{aligned}$$

(121)

and $U_\alpha {}^{\rho \gamma _1...\gamma _k}$ are defined as follows:

$$\begin{aligned} K^{\rho } \equiv -\sum _{i = 0}^CU_a{}^{\rho \alpha _1\dots \alpha _i}\partial _{\alpha _1\dots \alpha _i}\xi ^a. \end{aligned}$$

(122)

Here we used $K^{\rho }$ defined in (9).

1.2 Commutators for $N_o = 2$ case

It is clear that the commutation relations (77)–(79) are still valid for the case $N_o = 2$ considered in Sect. 5.2. Aside from these, there is also need for the commutators of $\partial _{\alpha \beta }$ with $\partial /\partial \partial _\rho \varphi _A, \partial /\partial \partial _{\rho \beta }\varphi _A$ and $\partial /\partial \partial _{\rho \beta \gamma }\varphi _A$. All the needed commutation relations are written down below:

$$\begin{aligned}&\left[ \frac{\partial }{\partial \varphi _A},\partial _{\alpha \beta }\right] \lambda _B = 0, \end{aligned}$$

(123)

$$\begin{aligned}&\left[ \frac{\partial }{\partial \partial _\rho \varphi _A},\partial _{\alpha \beta }\right] \lambda _B = \left\{ \delta _\alpha ^{\rho } \partial _\beta \left( \frac{\partial \lambda _B}{\partial \varphi _A}\right) \right\} _{\alpha \beta }, \end{aligned}$$

(124)

$$\begin{aligned}&\left[ \frac{\partial }{\partial \partial _{\rho \lambda }\varphi _A},\partial _{\alpha \beta }\right] \lambda _B = \frac{1}{2}\left\{ \left\{ \delta ^\lambda _\beta \partial _\alpha \left( \frac{\partial \lambda _B}{\partial \partial _\rho \varphi _A}\right) \right\} _{\alpha \beta }\right\} ^{\lambda \rho }+\frac{1}{2}\frac{\partial \lambda _B}{\partial \varphi _A}\left\{ \delta ^{\rho } _\alpha \delta ^\lambda _\beta \right\} ^{\rho \lambda }, \end{aligned}$$

(125)

$$\begin{aligned}&\left[ \frac{\partial }{\partial \partial _{\rho \varphi \tau }\varphi _A},\partial _{\alpha \beta }\right] \lambda _B = \frac{1}{6}\left\{ \frac{\partial \lambda _B}{\partial \partial _\rho \varphi _A}\delta ^\varphi _\alpha \delta ^\tau _\beta \right\} ^{\rho \varphi \tau }. \end{aligned}$$

(126)

1.3 Derivation of the formula 32

To derive (32) one can use (31) and substitute it to the (23). However, one should first make change $\alpha _{p+1}\rightarrow \rho$. As the index $\alpha _{p+1}$ is present in antisymmetrizer, one needs to be careful and should properly split sums before applying the change:

$$\begin{aligned}&K_{(p)a}{}^{\alpha _{p+1}\alpha _1..\alpha _{p}} = \frac{1}{p+1}\sum _{i = 1}^p\left[ K_{(p)a}{}^{\alpha _{p+1}\alpha _1..\alpha _p}\right] ^{\alpha _{p+1}\alpha _i}\end{aligned}$$

(127)

$$\begin{aligned} \nonumber -&\sum _{k = p+1}^{N_d}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+2}..\alpha _{k+1}}\left[ K_{(k)a}{}^{\alpha _{p+1}\alpha _1..\alpha _p\alpha _{p+2}..\alpha _{k+1}}\right] ^{\alpha _{p+1}\alpha _{k+1}}\\ \nonumber&+\sum _{k=p+1}^{N_d}\sum _{i = 1}^p\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+2}..\alpha _{k+1}}\left[ K_{(k)a}{}^{\alpha _{k+1}\alpha _1..\alpha _p\alpha _{p+1}\alpha _{p+2}..\alpha _{k}}\right] ^{\alpha _{k+1}\alpha _i}\\&+\sum _{k=p+2}^{N_d}\sum _{i = p+2}^k\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+2}..\alpha _{k+1}}\left[ K_{(k)a}{}^{\alpha _{k+1}\alpha _1..\alpha _{k}}\right] ^{\alpha _{k+1}\alpha _i}. \end{aligned}$$

(128)

The last term equals zero due to the symmetry of the operator $\partial _{\alpha _{p+2}..\alpha _{k+1}}$ and antisymmetrization of $K_{(k)a}{}^{\alpha _{k+1}\alpha _1..\alpha _{k}}$ with respect to $\alpha _{k+1}, \alpha _i$ for $i\in [p+2,k+1]$. Hence, one can proceed to the change $\alpha _{p+1}\rightarrow \rho$:

$$\begin{aligned}&K_{(p)a}{}^{\rho \alpha _1..\alpha _{p}} = \frac{1}{p+1}\sum _{i = 1}^p\left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _i}\\ -&\sum _{k = p+1}^{N_d}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+2}..\alpha _{k+1}}\left[ K_{(k)a}{}^{\rho \alpha _1..\alpha _p\alpha _{p+2}..\alpha _{k+1}}\right] ^{\rho \alpha _{k+1}}\\&+\sum _{k=p+1}^{N_d}\sum _{i = 1}^p\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+2}..\alpha _{k+1}}\left[ K_{(k)a}{}^{\alpha _{k+1}\rho \alpha _1..\alpha _p\alpha _{p+2}..\alpha _{k}}\right] ^{\alpha _{k+1}\alpha _i}. \end{aligned}$$

To simplify this further it is convenient to shift indices for the operators $\partial _{\alpha _{p+2}..\alpha _{k+1}}$ (as they are dummy) by $-1$. For convenience we also want to increase inner sum’s upper limit to $k-1$ as it does not change anything due to the symmetry of the operator $\partial _{\alpha _{p+2}..\alpha _{k+1}}$ ($\partial _{\alpha _{p+1}..\alpha _{k}}$ after the mentioned index shift):

$$\begin{aligned} K_{(p)a}{}^{\rho \alpha _1..\alpha _{p}} =&\frac{1}{p+1}\sum _{i = 1}^p\left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _i}- \sum _{k = p+1}^{N_d}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+1}..\alpha _{k}}\left[ K_{(k)a}{}^{\rho \alpha _1..\alpha _{k}}\right] ^{\rho \alpha _{k}}\end{aligned}$$

(129)

$$\begin{aligned}&+\sum _{k=p+1}^{N_d}\sum _{i = 1}^{k-1}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+1}..\alpha _{k}}\left[ K_{(k)a}{}^{\alpha _{k}\rho \alpha _1..\alpha _{k-1}}\right] ^{\alpha _{k}\alpha _i}. \end{aligned}$$

(130)

Now one can multiply this expression by $\partial _{\alpha _1..\alpha _p}\xi ^{a}$ and substitute the result into (32):

$$\begin{aligned} -J^{\prime\rho } =&\sum _{p = 1}^{N_d}\sum _{i = 1}^p\frac{1}{p+1}\left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _i}\partial _{\alpha _1..\alpha _p}\xi ^{a}-\sum _{p = 0}^{N_d}\sum _{k = p+1}^{N_d}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+1}..\alpha _{k}}\left[ K_{(k)a}{}^{\rho \alpha _1..\alpha _{k}}\right] ^{\rho \alpha _{k}} \end{aligned}$$

(131)

$$\begin{aligned}&\times \partial _{\alpha _1..\alpha _p}\xi ^{a}+\sum _{p = 0}^{N_d}\sum _{k=p+1}^{N_d}\sum _{i = 1}^{k-1}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+1}..\alpha _{k}}\left[ K_{(k)a}{}^{\alpha _{k}\rho \alpha _1..\alpha _{k-1}}\right] ^{\alpha _{k}\alpha _i}\partial _{\alpha _1..\alpha _p}\xi ^{a}. \end{aligned}$$

(132)

Our next step is to convert r.h.s to the full derivative. We start with the first two terms by changing the summation order in the second term:

$$\begin{aligned} \sum _{p = 0}^{N_d-1}\sum _{k = p+1}^{N_d} = \sum _{k = 1}^{N_d}\sum _{p = 0}^{k - 1}. \end{aligned}$$

(133)

After that one should interchange indices $k\leftrightarrow p$ in the second term and change them$i\rightarrow k$ in the first one. Finally, after applying shift $k\rightarrow k+1$ in the second term the can be properly merged:

$$\begin{aligned} -J^{\prime\rho } =&\sum _{p = 1}^{N_d}\sum _{k = 1}^p\frac{1}{p+1}\left[ \left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _k}\partial _{\alpha _1..\alpha _p}\xi ^{a}+(-1)^{p-k}\partial _{\alpha _k..\alpha _{p}}\left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _p}\partial _{\alpha _1..\alpha _{k-1}}\xi ^{a}\right] \nonumber \\&+\sum _{p = 0}^{N_d}\sum _{k=p+1}^{N_d}\sum _{i = 1}^{k-1}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+1}..\alpha _{k}}\left[ K_{(k)a}{}^{\alpha _{k}\rho \alpha _1..\alpha _{k-1}}\right] ^{\alpha _{k}\alpha _i}\partial _{\alpha _1..\alpha _p}\xi ^{a}. \end{aligned}$$

(134)

First term can be further simplified by subsequently using Leibniz’ rule in the form similar to (13):

$$\begin{aligned} {\begin{matrix} &{}(-1)^{p-k}\partial _{\alpha _k..\alpha _{p}}(Y^{\alpha _1..\alpha _p})\partial _{\alpha _1..\alpha _{k-1}}Q\\ &{}=\sum _{i = k}^p(-1)^{p-i}\partial _{\alpha _i}\left( \partial _{\alpha _{i+1}..\alpha _p}Y^{\alpha _1..\alpha _p}\partial _{\alpha _1..\alpha _{i-1}}\xi ^{\mu }\right) - Y^{\alpha _1\dots \alpha _p}\partial _{\alpha _1..\alpha _p}Q, \end{matrix}} \end{aligned}$$

(135)

which is true for the arbitrary functions $Y^{\alpha _1..\alpha _p}$ (even without any index symmetry) and the arbitrary function Q. Then by using the simple relation:

$$\begin{aligned} \left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _p}\partial _{\alpha _1..\alpha _{p}}\xi ^{a} = \left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _k}\partial _{\alpha _1..\alpha _{p}}\xi ^{a},\;\; k\le p,\;\; p\ge 1, \end{aligned}$$

(136)

one may rewrite (134) in the following form:

$$\begin{aligned} -J^{\prime\rho } =&\sum _{p = 1}^{N_d}\sum _{k = 1}^p\sum _{i= k}^p\frac{(-1)^{p-i}}{p+1}\partial _{\alpha _i}\left( \partial _{\alpha _{i+1}..\alpha _p}\left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _p}\partial _{\alpha _1..\alpha _{i-1}}\xi ^{a}\right) \nonumber \\&+\sum _{p = 0}^{N_d}\sum _{k=p+1}^{N_d}\sum _{i = 1}^{k-1}\frac{(-1)^{k-p}}{k+1}\partial _{\alpha _{p+1}..\alpha _{k}}\left[ K_{(k)a}{}^{\alpha _{k}\rho \alpha _1..\alpha _{k-1}}\right] ^{\alpha _{k}\alpha _i}\partial _{\alpha _1..\alpha _p}\xi ^{a}. \end{aligned}$$

(137)

The same trick with Leibniz’ rule can be applied to the second term of (137), which will lead to the terms that are full derivatives and the following contribution:

$$\begin{aligned} \sum _{p = 0}^{N_d}\sum _{k=p+1}^{N_d}\sum _{i = 1}^{k-1}\frac{(-1)^{k-p}}{k+1}\left[ K_{(k)a}{}^{\alpha _{k}\rho \alpha _1..\alpha _{k-1}}\right] ^{\alpha _{k}\alpha _i}\partial _{\alpha _1..\alpha _{k}}\xi ^{a}. \end{aligned}$$

(138)

Note that the minimal number of derivatives in $\partial _{\alpha _1..\alpha _k}\xi ^a$ is greater or equal than 2. It becomes obvious then that this expression is zero if one takes into account antisymmetrizer and the symmetry of $\partial _{\alpha _1..\alpha _{k}}\xi ^{a}$. Thus, the formula (137) can be written as follows:

$$\begin{aligned}&-J^{\prime\rho } = \sum _{p = 1}^{N_d}\sum _{k = 1}^p\sum _{i= k}^p\frac{(-1)^{p-i}}{p+1}\partial _{\alpha _i}\left( \partial _{\alpha _{i+1}..\alpha _p}\left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _p}\partial _{\alpha _1..\alpha _{i-1}}\xi ^{a}\right) \nonumber \\&-\sum _{p = 0}^{N_d}\sum _{k=p+1}^{N_d}\sum _{i = 1}^{k-1}\sum _{s = p+1}^k\frac{(-1)^{k+s}}{k+1}\partial _{\alpha _s}\left( \partial _{\alpha _{s+1}..\alpha _k}\left[ K_{(k)a}{}^{\alpha _{k}\rho \alpha _1..\alpha _{k-1}}\right] ^{\alpha _{k}\alpha _i}\partial _{\alpha _1..\alpha _{s-1}}\xi ^{a}\right) . \end{aligned}$$

(139)

As terms in the first and the second sequences of sums do not depend on k and p respectively, one can change summation order to calculate sums over these indices. This can be easily done for the first sequence of sums by using the following relation:

$$\begin{aligned} \sum _{k = 1}^p\sum _{i = k}^p = \sum _{i = 1}^p\sum _{k = 1}^i \end{aligned}$$

(140)

For the second term in (139) one should set the upper bound of sum over p to $N_d-1$ (because the sum over k will give zero for $p = N_d$) and use (133), then perform index shift $p\rightarrow p+1$ and use (140). The result will be the following:

$$\begin{aligned} -J^{\prime\rho } =&\sum _{p = 1}^{N_d}\sum _{k= 1}^p\frac{k(-1)^{p-k}}{p+1}\partial _{\alpha _k}\left( \partial _{\alpha _{k+1}..\alpha _p}\left[ K_{(p)a}{}^{\rho \alpha _1..\alpha _p}\right] ^{\rho \alpha _p}\partial _{\alpha _1..\alpha _{k-1}}\xi ^{a}\right) \nonumber \\&-\sum _{k=1}^{N_d}\sum _{i = 1}^{k-1}\sum _{s = 1}^k\frac{s(-1)^{k-s}}{k+1}\partial _{\alpha _{s}}\left( \partial _{\alpha _{s+1}..\alpha _k}\left[ K_{(k)a}{}^{\alpha _{k}\rho \alpha _1..\alpha _{k-1}}\right] ^{\alpha _{k}\alpha _i}\partial _{\alpha _1..\alpha _{s-1}}\xi ^{a}\right) . \end{aligned}$$

(141)

One can now put the derviatives $\partial _{\alpha _k}$ and $\partial _{\alpha _s}$ in the first and the second terms respectively out of the sums by renaming these indices in each term of the sums to $\beta$. As it was for (130), one should be careful when doing this because the indices $\alpha _k$ and $\alpha _s$ can be encountered as the indices of the antisymmetrizer. The final answer is given by the formula:

$$\begin{aligned} -J^{\prime\rho } =&\partial _{\beta }\Bigg [\sum _{p = 2}^{N_d}\sum _{k= 1}^{p-1}\frac{k(-1)^{p-k}}{p+1}\partial _{\alpha _k..\alpha _{p-1}}\left( \left[ K_{(p)a}{}^{\rho \beta \alpha _1..\alpha _{p-1}}\right] ^{\rho \alpha _{p-1}}+\left[ K_{(p)a}{}^{\beta \rho \alpha _1..\alpha _{p-1}}\right] ^{\beta \alpha _{p-1}}\right) \nonumber \\&\times \partial _{\alpha _1..\alpha _{k-1}}\xi ^{a}+\sum _{p = 1}^{N_d}\frac{p}{p+1}\left[ K_{(p)a}{}^{\rho \beta \alpha _1..\alpha _{p-1}}\right] ^{\rho \beta }\partial _{\alpha _1..\alpha _{p-1}}\xi ^{a}\nonumber \\&-\sum _{p=3}^{N_d}\sum _{i = 1}^{p-2}\sum _{s = 1}^{p-1}\frac{s(-1)^{p-s}}{p+1}\partial _{\alpha _{s}..\alpha _{p-1}}\left[ K_{(p)a}{}^{\alpha _{p-1}\rho \beta \alpha _1..\alpha _{p-2}}\right] ^{\alpha _{p-1}\alpha _i}\partial _{\alpha _1..\alpha _{s-1}}\xi ^{a}\nonumber \\&-\sum _{p=2}^{N_d}\frac{p(p-1)}{p+1}\left[ K_{(p)a}{}^{\beta \rho \alpha _1..\alpha _{p-1}}\right] ^{\beta \alpha _{p-1}}\partial _{\alpha _1..\alpha _{p-1}}\xi ^{a}\Bigg ]. \end{aligned}$$

(142)

One can easily compare this result with (32) and (33) to ensure that they coincide.