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Equation of Motion from Lagrangian: Expand the Lagrangian by orders of perturbation to get higher order equation of motion?

Physics Asked on June 20, 2021

Just to state this first: I am not asking about higher order derivatives, so no jerk, snap, crackle or pop is involved here. I am asking about higher order perturbative equations of motion from an expanded Lagrangian.

Say, I have a Lagrangian, depending on a scalar field $phi$ and a metric $g_{ij}$, which I perturb to get
$$
L(phi,g_{ij}) = L_0 + L_1 + L_2 +…
phirightarrowphi(t)+deltaphi(t,x,y,z)
g_{ij}rightarrow g_{ij}+delta g_{ij}
$$

where $L_1$ is of order $epsilon$, $L_2$ is of order $epsilon^2$, and so forth, with $epsilon$ the order of the perturbation.

Now, I would like to derive equations of motion. The background equation of motion is given by the Euler-Lagrange equation
$$
frac{partial L_0}{partial phi}-frac{mathrm{d}}{mathrm{d}t}frac{partial L_0}{partial dot{phi}}=0.
$$

But how do I get, say, the first order equations of motion?

One Answer

I really like this question! Let us define $E[cdot]$ as $$ E[cdot] = left[frac{partial cdot}{partial q} -frac{d}{dt} left(frac{partial cdot}{partial dot q}right)right] $$ Then, $$ E[L] = E[L_0 + L_1+dots] $$ I'm pretty sure its linear operator so I think it will be $$ E[L] = E[L_0] + E[L_1]+dots $$ Separately, your perturbation in $phi$ will lead to Noether's theorem if you assume it leads to an equivalent action.

Correct answer by AngusTheMan on June 20, 2021

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