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In MTW book there is one exercise in which there was proposed to discuss linearized tensor gravity, which is represented as $$ g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}, \quad \eta_{\mu \nu} = diag(1, -1, -1, -1). $$ I have the question about lagrangian of fields in this case: in the book it is written in a form $$ L_{f} = -\frac{1}{32 \pi G}\left( \frac{1}{2}(\partial_{\alpha}h_{\nu \beta}) \partial^{\alpha}\bar {h}^{\nu \beta} - (\partial^{\alpha} \bar {h}_{\mu \alpha })\partial_{\beta}\bar {h}^{\mu \beta}\right), \qquad (1) $$ where $$ \bar {h}^{\mu \nu} = h^{\mu \nu} - \frac{1}{2}\eta^{\mu \nu}h^{\alpha}_{\alpha} $$ It is argued that this lagrangian refers to the spin-2 massless field. I understand this expression in general, but I don't understand it in one detail.

Why in the first summand in $(1)$ is $h_{\nu \beta}$, not $\bar {h}_{\nu \beta}$?

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up vote 3 down vote accepted

Just test the invariance of the action (so using integrations by part and neglecting surface terms) under "linear diffeomorphism" $h_{\mu\nu} \to h_{\mu\nu} + \partial_\mu \epsilon_\nu+\partial_\nu \epsilon_\mu$


More precisions :

The linear diffeomorphism is only the linearization of the standard diffeomorphism :

$g^{'\mu\nu} = \frac{\partial x^{'\mu}}{\partial x^{\sigma}} \frac{\partial x^{'\nu}}{\partial x^{\tau}} g^{\sigma\tau}$

for a coordinate transformation $x^{'\mu} = x^\mu + \epsilon^\mu(x)$

where you plug $g^{\sigma\tau} = \eta^{\sigma\tau} + h^{\sigma\tau}$, and $\frac{\partial x^{'\mu}}{\partial x^{\sigma}} = \delta^\mu_\sigma +\partial_\sigma \epsilon^\mu$, and where you keep only terms linear in $\epsilon$.

Due to Lorentz invariance, you have only $4$ possible quadratic terms in first derivative of $h_{\mu\nu}$ , and we are searching the correct combination which leaves the action invariant under the linear diffeomorphism (which is nothing but a gauge invariance for spin $2$ field). So you may write :

$S = \int d^4x (A ~\partial_\alpha h_{\nu\beta}~\partial^\alpha h^{\nu\beta} + B ~\partial_\alpha h_\nu^\nu~\partial^\alpha h_\nu^\nu +C ~\partial_\alpha h^{\alpha\nu}~\partial^\beta h_{\beta\nu}+D ~\partial^\alpha h_\nu^\nu~\partial^\beta h_{\beta\alpha}) $

We may fix a value for $A$, here $A = \frac{-1}{32G} (\frac{1}{2})$, so we have only $3$ variables to be checked.

Now, the variation $\delta S$ of the action is going to give, after integration by parts, terms in $\epsilon \partial^3h$, that is terms linear in $\epsilon$ and $h$, with a third derivative in $h$, and there are only $3$ possible terms :

$\epsilon^\nu(\partial^2\partial^\mu h_{\mu\nu}), \epsilon^\nu (\partial_\nu \partial^2 h_\mu^\mu), \epsilon^\nu(\partial_\nu \partial^\mu \partial^\lambda h_{\mu\lambda})$

The coefficients of these terms must be zero, so you have $3$ equations for $3$ variables $B, C, D$, so you are able to calculate their values.

Integration by parts works like this, take for instance the first term, you have :

$\delta_1 S = A~ \delta(\partial_\alpha h_{\nu\beta}~\partial^\alpha h^{\nu\beta}) = 2A~( \partial_\alpha(2\partial_\nu \epsilon_\beta)\partial^\alpha h^{\nu\beta})$

Now, by applying $2$ times the integration by parts, and neglecting the surface terms (total derivatives), you get :

$\delta_1 S=4\epsilon_\beta \partial^2 \partial_\nu h^{\nu\beta}$

The result of your calculus should be : $B=-A, C=-2A, D=2A$ , and you may check that this corresponds to your original expression (developped in $h$ terms only, so by replacing $\bar h$ terms by their value)

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So lagrangian in form of expression $(1)$ realizes the invariance of lagrangian under these transformations? – Andrew McAddams Dec 25 '13 at 19:56
@AndrewMcAdams : I updated the answer. The action is invariant by the "linear diffeomorphism", not the Lagrangian. – Trimok Dec 26 '13 at 10:45
Yes, I should wrote about action, not about lagrangian. Thank you! – Andrew McAddams Dec 26 '13 at 10:59

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