I am currently reading "Second-order scalar-tensor field equations in a four-dimensional space" from Horndeski.

As far as I understand, it is a proof of the most general "healthy" Lagrangian we can get in the context of general relativity coupled to a scalar field.$\newcommand{\deriv}[1]{\frac{d}{dx^{#1}}}$ During the proof he introduces the following quantities: $$L = L(g_{ij};g_{ij,i_1};\dots;g_{ij,i_1i_2...\dots i_p};\phi;\phi_{,i_1};\dots;\phi_{,i_1\dots i_q})$$ $$E^{ij} = \sum_{h=0}^p (-1)^{h+1}\deriv{i_1}\deriv{i_2}\dots\deriv{i_h}\frac{\partial L}{\partial g_{ij,i_1\dots i_h}}$$ $$ E = \sum_{h=0}^q(-1)^{h+1}\deriv{i_1}\dots\deriv{i_h}\frac{\partial L}{\partial \phi_{,i_1\dots i_h}},$$ where $A_{,i} = \frac{\partial A}{\partial x^i}$. Note that $E^{ij} = 0; E=0$ are the Euler-Lagrange equations for this Lagrangian.

He then invokes the following property (equation 1.8), which I would like to find a proof for: $$\nabla_j E^{ij}=\frac{1}{2}(\nabla^i\phi) E .\tag{1.8} $$

Where $\nabla_i$ denotes the covariant derivative. I have somewhat tried brute-force, but it doesn't get me very far. Horndeski does cite "Horndeski (1973)" where he claims to prove it, however I was unable to find this paper anywhere on the internet.

So, does anyone know a proof for this identity? Or maybe a paper where it is explained?

  • $\begingroup$ Note : I am actually not sure how general this equation is. Sadly on the paper, except for the reference I gave, there is no further precisions. $\endgroup$ – Frotaur Apr 10 '17 at 10:26
  • $\begingroup$ The technique will be probably similar to proving that the canonical (Noether/Einstein-Rosenfeld-Belinfante) stress-energy tensor is equivalent $\endgroup$ – Void Apr 10 '17 at 10:53
  • 1
    $\begingroup$ ...to the Hilbert stress-energy tensor obtained from the variation with respect to the metric. You just have to make an infinitesimal coordinate transform, compute all the variations induced and put the variation of the action to zero (i.e. assume diffeomorphism invariance). As a result you should get something like your equalities. $\endgroup$ – Void Apr 10 '17 at 10:58

Sketched proof:

  1. Consider infinitesimal transformations in the form of Lie derivatives $$ \delta \phi ~=~ {\cal L}_{\varepsilon}\phi~=~\varepsilon^i \partial_i\phi~=~ \varepsilon_i \partial^i\phi, \qquad \delta g_{ij}~=~{\cal L}_{\varepsilon}g_{ij}~=~\nabla_i\varepsilon_j+(i\leftrightarrow j) \tag{A}, $$ where $\varepsilon$ is an infinitesimal vector field.

  2. The action $$ S[\phi,g] ~=~\int d^nx~{\cal L}, \tag{B}$$ (where ${\cal L}$ denotes the Lagrangian density) is assumed to be diffeomorphism invariant: $$0~=~{\cal L}_{\varepsilon}S~=~ \delta S ~=~\int d^nx \left(E~\delta\phi +E^{ij}~\delta g_{ij}\right) + \text{boundary terms}$$ $$~\stackrel{(A)}{=}~\int d^nx \left(E~\partial^i\phi -2\nabla_jE^{ji}\right)\varepsilon_i + \text{boundary terms}. \tag{C}$$

  3. OP's sought-for identity $$ E~\partial^i\phi ~=~2\nabla_jE^{ji}\tag{1.8}$$ follows since eq. (C) should hold for all vector fields $\varepsilon$.

  • $\begingroup$ But then (1.8) is the Euler-Lagrange eq. And not $E^{ij}=0$ and $E=0$. $\endgroup$ – Walter Apr 10 '17 at 21:42
  • $\begingroup$ Eq. (1.8), $E^{ij}$, $E$ are EL eqs. for the functional ${\cal L}_{\varepsilon}S$, $S$, $S$ wrt. the $\varepsilon_i$, $g_{ij}$, $\phi$, respectively. $\endgroup$ – Qmechanic Apr 10 '17 at 21:50
  • $\begingroup$ Great, that's exactly what I was looking for ! Forgot about the diffeomorphism invariance property ! $\endgroup$ – Frotaur Apr 13 '17 at 9:16

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