Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $\bar {h}_{\nu \beta}$?

share|improve this question
add comment

1 Answer

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" $h_{\mu\nu} \to h_{\mu\nu} + \partial_\mu \epsilon_\nu+\partial_\nu \epsilon_\mu$

[EDIT]

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)

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.