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Let $Q_{ab} = (\psi_{;a})(\psi_{;b}) - (1/2)g_{ab}|\nabla \psi|^2$ be the energy-momentum tensor of the wave equation in some space time. I will use semicolons to refer to covariant differentiation and $\partial$s to refer to coordinate differentiation. Let $\pi_{ab}$ be the deformation tensor for some fixed vector field $X$. In deriving "almost conservation laws" one uses the identity.

$(Q_{ab}X^b)^{;a} = (\psi^{;a}_{;a})(X^a\partial_a\psi) + (1/2)Q^{ab}\pi_{ab}$

Does there exist a physical interpretation of the scalar $Q^{ab}\pi_{ab}$ that is not based on the above formula? I am most interested in how one should think about this quantity in relativistic contexts, e.g. black hole geometries.

P.S. Just in case anyone is tempted, I am looking for something more than "$Q^{ab}\pi_{ab}$ vanishes if the flows of $X$ are isometries."

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Just for clarity, by $X \psi$ do you understand the action of the vector field $X$ on the function $\psi$? Your usage of indices is non-standard, to say the least$\ldots$ – Marek Jul 29 '11 at 5:35
Yes, $X\psi$ means the action of $X$ on the function $\psi$. I apologize about the notation, it is standard in the mathematics setting (for example see page 33 of…). Feel free to edit it into more comprehensible physics notation. – Yakov Shlapentokh-Rothman Jul 29 '11 at 13:55
Actually, this is neither mathematical nor physical notation. Anytime Einstein's summation convention is assumed (as it is in your second equation) one expects the "conservation of indices" which doesn't happen here. More importantly, it's very confusing to mix coordinate-free and indexful notation. Either use one or the other. – Marek Jul 29 '11 at 14:07
Where is the lack of conservation of indices? Is it the g in the □g? This just refers to the metric to remind the reader that it is the wave equation with respect to the metric. In terms of mixing notation, would you prefer I write the equation as $(Q_{ab}X^b)^{;a} = (\psi^{;a}_{;a})(X^ae_a(\psi)) + (1/2)Q^{ab}\pi_{ab}$, where the $e_a$s are the tangent vectors associated to a coordinate system $\{x^a\}$? – Yakov Shlapentokh-Rothman Jul 29 '11 at 14:22
In case it is not clear the $;$ refers to a covariant derivative. – Yakov Shlapentokh-Rothman Jul 29 '11 at 14:23
up vote 3 down vote accepted

Observe the following: the Einstein-Hilbert stress-energy is formally given by the first variation of the matter Lagrangian density relative to the inverse metric:

$$ Q = \frac{\delta L_\phi \mathrm{dvol}_g}{\delta g^{-1}} $$

where $L_\phi = g^{-1}(\nabla\phi,\nabla\phi)$ is the Lagrangian function for the scalar field. Further observe that $\mathcal{L}_X\mathrm{Id} = 0$, where $\mathrm{Id}:TM\to TM$ is the identity map and $\mathcal{L}_X$ is the Lie derivative. So $\mathcal{L}_X (g^{-1} g) = (\mathcal{L}_Xg^{-1})g + g^{-1}\pi = 0$. Hence $\mathcal{L}_X g^{-1} = -g^{-1}\pi g^{-1}$. This means that

$$ c(Q g^{-1} \pi g^{-1}) = -c(Q \mathcal{L}_Xg^{-1}) $$

where $c(\cdot)$ maps 1,1-tensors to scalars is the contraction operation. Next, recall that the Lie derivative can be defined by the one parameter family of diffeomorphisms generated by a vector field. More precisely, let $\Psi_t$ be the one parameter family of diffeomorphisms generated by $X$. Then we have a one parameter family of tensor fields $G_t := \Psi_t^* g^{-1}$ defined by the pushforward of the inverse metric. For this family we have that $\mathcal{L}_Xg^{-1} = \frac{d}{dt}G_t |_{t=0}$. Holding the scalar field $\phi$ fixed, we write $\hat{L}_t = G_t(\nabla\phi,\nabla\phi) \mathrm{dvol}_{(G_t)^{-1}}$ for the corresponding one parameter family of Lagrangian densities, then we have that the chain rule gives

$$ \frac{d}{dt} \hat{L}_t |_{t = 0} = c(Q\mathcal{L}_Xg^{-1}) $$

Note that we in fact can split $$\mathcal{L}_X (L_\phi\mathrm{dvol}_g) = \frac{d}{dt}\hat{L}_t |_{t=0} + 2 g^{-1}(d\phi, d\mathcal{L}_X\phi) \mathrm{dvol}_g$$ into the "geometric/gravity" part and the "matter" part using that the exterior derivative commutes with the Lie derivative.

Therefore the scalar quantity you wrote down is the portion of the change of the matter Lagrangian density in the direction of the vector field $X$ that is caused by the change of the (inverse) metric in that direction. In you are willing to consider the Lagrangian density as a physical quantity, this gives a "physical" meaning to your scalar quantity as the "geometric/gravity" portion of the flow of the Lagrangian density under the vector field $X$. In a not very precise sense, if the vector field $X$ is time-like, you can associate to it a time-like congruence. So this scalar measures the infinitesimal (fictitious) work "done by gravity/acceleration of frame". (I use the word fictitious here to indicate that the $X$ dependence of this scalar is somewhat analogous to the frame-dependence of the "centrifugal force" in classical mechanics.)

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Here we will not give a physical interpretation per se, but just derive an alternative covariant expression for the sought-for scalar.

There are given a scalar field $\psi$ and a metric

$$ds^2~=~g_{ab} ~dx^a dx^b.$$

Let $\nabla$ be the corresponding Levi-Civita connection. The Lagrangian scalar function is:

$$ \Lambda~:=~ \frac{1}{2}\langle \nabla\psi , \nabla\psi \rangle ~=~ \frac{1}{2}\nabla_a\psi ~ g^{ab}~ \nabla_b\psi ~=~ \frac{1}{2}\partial_a\psi ~ g^{ab}~ \partial_b\psi . $$

The energy-momentum tensor reads:

$$ Q_{ab} ~:=~ \nabla_a \psi ~\nabla_b \psi - g_{ab}~\Lambda. $$

We also have a vector field $X$. Let us make a co-vector (=one form) by lowering the index:

$$ \eta_a = g_{ab}~ X^b. $$

The deformation tensor $\pi$ wrt. the vector field $X$ is defined as the Lie derivative of the metric:

$$\pi_{ab} ~:=~ ({\cal L}^{}_X g)_{ab}~=~ (\nabla_a \eta)_b + (\nabla_b \eta)_a, $$ $$\pi^{ab} ~:=~ g^{ac} ~\pi_{cd}~g^{db}~=~ -({\cal L}^{}_X g)^{ab}~=~ g^{ac}\partial_c X^b +g^{bc}\partial_c X^a - X^c\partial_c g^{ab} . $$

The sought-for scalar is:

$$S~:=~ \frac{1}{2}\pi^{ab} Q_{ab}. $$

Another vector field $Y$ is defined as:

$$ Y^a~:=~g^{ab}~ Q_{bc}~X^c~=~g^{ab}~\nabla_b \psi ~X[\psi] - \Lambda X^a, $$ $$ Y[f]~=~X[\psi]~\langle \nabla\psi , \nabla f \rangle - \Lambda X[f].$$

Then it is a bit tedious but straightforward to reproduce the formula in the question formulation (v3):

$$S~=~\mathrm{div}_g Y - \Box_g \psi ~ X[\psi], $$

and also to derive another covariant expression

$$S ~=~ \langle \nabla\psi , \nabla[ X[\psi]] \rangle - \mathrm{div}_g( \Lambda X).$$

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Thanks, that is an interesting alternative formulation. I will have to meditate on it. – Yakov Shlapentokh-Rothman Jul 29 '11 at 19:58
All but the last equation are essentially just definitions written out in full carrying no more information than the OP itself, while the last one follows from simply plugging-in the definition of $Y$ into the next-to-last equation. In particular, where is any physical interpretation in this answer? – Marek Jul 30 '11 at 5:58
Please note that I stressed from the very beginning that the answer does not contain any physical interpretation. If the answer is not useful enough, I can delete it. – Qmechanic Jul 30 '11 at 9:24

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