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Quick question. Given Lagrangian density

$$\mathcal{L} = -\frac12 h \Box h + \frac13 \lambda h^3 + Jh ,\tag{3.69}$$

where the scalar $h$ represents the gravitational potential, and given the Euler-Lagrange equation

$$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}h)} - \frac{\partial \mathcal{L}}{\partial h} = 0 \tag{1}$$

we have the equations of motion (according to Schwartz's QFT eq. 3.70)

$$\Box h -\lambda h^2 - J = 0. \tag{3.70}$$

I get $$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}h)} = 0 \tag{2}$$

and

$$\frac{\partial \mathcal{L}}{\partial h} = -\frac12 \Box h + \lambda h^2 + J\tag{3}$$

so where did I go wrong with the factor of $-\frac12$?

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3 Answers 3

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You have to be careful with the way you've defined your Lagrangian. You're treating the second derivative of $h$, $\Box h$, as if it were independent of both $h$ and $\partial_\mu h$. If you integrate by parts then you get a new Lagrangian that is much cleaner: $$ \mathcal L'=\frac{1}{2}\partial_\mu h\partial^\mu h+\frac{1}{3}\lambda h^3+Jh $$ we are, of course, assuming that $h\partial_\mu h$ vanishes on the boundary of spacetime, so that the new Lagrangian gives the same equation of motion.

Applying the Euler-Lagrange equation to this new Lagrangian will give you the correct result. Cheers!

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  • $\begingroup$ It seems that $ \partial_{\mu} \frac{\partial (\partial_\nu h\partial^\nu h)}{\partial(\partial_{\mu}h)} = \partial_\mu \partial^\mu h $, but why is that? when $\mu \neq \nu$, is it 0? $\endgroup$
    – Watanabe.N
    Commented Jan 12 at 1:16
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    $\begingroup$ @Watanabe.N Each $\partial_\mu h$ is an independent degree of freedom, so yes. $\endgroup$
    – Damian Sowinski
    Commented Jan 16 at 0:37
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  1. The correct field-theoretic Euler-Lagrange (EL) equation reads in general $$ 0~\approx~\frac{\delta S}{\delta h} ~=~\frac{\partial {\cal L}}{\partial h} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}h)} + \sum_{\mu\leq \nu} \frac{d}{dx^{\mu}} \frac{d}{dx^{\nu}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\partial_{\nu}h)} - \ldots,\tag{A} $$ where the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher-derivative terms. If you include second-derivative terms, you will get agreement with Schwartz' eom (3.70).

  2. Alternatively, you can start by removing higher-derivative terms in the action via integration by parts. Then the EL equation (A) will not contain higher-derivative terms. That is the strategy of Damian Sowinski's answer.

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Use functional derivation and notations of Schwartz's book $$ \square \equiv \partial _{\mu }^2\equiv \partial _{\mu }\partial ^{\mu } $$ so $$ h \square h=\partial _{\mu }^2h $$

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