From https://doi.org/10.1063/1.2155755

he limited himself to second-order differential equations.

Our experience in elementary-particle physics has taught us that any term in the field equations of physics that is allowed by fundamental principles is likely to be there in the equations

I guess the author means from the effective field theory point of view. Namely, effective actions include non-renormalizable terms, which can lead to higher derivatives. I try to see an example beyond second-order differential equations.

Let me start from $\phi^4$. The effective Lagrangian is, e.g., Peskin & Schroeder eq. (12.23) $$ \int d^d x \mathcal{L}_{\mathrm{eff}} = \int d^d x' \left[ \frac{1}{2} \left( \partial'_{\mu} \phi' \right)^2 + \frac{1}{2} m'^2 \phi'^2 + \frac{1}{4} \left( \lambda' \phi'^4 + C \left( \partial'_{\mu} \phi' \right)^4 + D' \phi'^6 +\cdots \right) \right] \tag{1} $$

I suppose $$ \mathcal{L}_{\mathrm{eff}} = \frac{1}{2} \left( \partial'_{\mu} \phi' \right)^2 + \frac{1}{2} m'^2 \phi'^2 + \frac{1}{4} \left( \lambda' \phi'^4 + C \left( \partial'_{\mu} \phi' \right)^4 + D' \phi'^6 +\cdots \right) \tag{2} $$

Try to cook a classical equation of motion. From the Euler-Lagrangian equation, $$ \frac{ \partial \mathcal{L} }{ \partial \phi} - \partial_{\mu} \frac{ \partial \mathcal{L} }{ \partial \left( \partial_{\mu} \phi \right) } = 0\tag{3} $$ plug in the effective lagrangian, we should get some extra terms than the Klein-Gordon equation $$ \square \phi' - m^2 \phi' + C \partial'_{\mu} \left[ \left( \partial'^{\mu} \phi' \right) \left( \partial'_{\mu} \phi' \right)^2 \right] +\cdots = 0.\tag{4} $$

So far the extra term with prefactor $C$ still looks like a second-order differential equation, as one first-order derivative outside the square bracket, $\partial'_{\mu}$, acting on one first-order derivative term $\left( \partial'^{\mu} \phi' \right) $ times the other first-order derivative term (a first-order derivative times itself) $\left( \partial'_{\mu} \phi' \right)^2$, i.e., $(fg)' = f'g + fg'$. If I further organize the inside square bracket part of the extra term by $f' g = (fg)' - f g' $,

$$ C \partial'_{\mu} \left[ \left( \partial'^{\mu} \phi' \right) \left( \partial'_{\mu} \phi' \right)^2 \right] \\ \equiv C \partial'_{\mu} \left\{ \left( \partial'^{\mu} \phi' \right) \left( \partial'_{\mu} \phi' \right)^2 \right\} \\ = C \partial'_{\mu} \left\{ \partial'^{\mu} \left[ \phi' \left( \partial'_{\mu} \phi' \right)^2 \right] - \phi' \partial'^{\mu}\left[ \left( \partial'_{\mu} \phi' \right)^2 \right] \right\} \\ = C \underline{\partial'_{\mu}} \left\{ \partial'^{\mu} \left[ \phi' \left( \partial'_{\mu} \phi' \right)^2 \right] - 2 \phi' \left[ \left( \partial'^{\mu} \phi' \right) \left( \underline{ \partial'^{\mu} \partial'_{\mu}} \phi' \right) \right] \right\}.\tag{5} $$

It seems I get a third-order differential equation from the underline part of the above equation. Is my reasoning right?

I think I did not impose any quantization in getting the equation of motion (except effective action from path integrals), since I think the view in the physics today essay is not much about quantization. Or I am not even wrong?

Or a second-order differential equation should be counted as the total number of the derivatives terms than taking a second-order differentiation on a single term?

  • $\begingroup$ You also get a third order term from the first term in the curly braces. The third-order terms should all cancel if you write everything out. But honestly I think it's simpler just to take the derivative directly from line 1 or line 2, rather than "further organize" things in lines 3 and 4. $\endgroup$
    – Andrew
    Commented Nov 27, 2021 at 5:10
  • $\begingroup$ Okay... then hopefully there is an example for higher-order differential equations $\endgroup$ Commented Nov 27, 2021 at 5:17
  • $\begingroup$ Sure. $(\square \phi)^2$. $\endgroup$
    – Andrew
    Commented Nov 27, 2021 at 14:15

1 Answer 1

  1. OP is right that if the Lagrangian density remains of 1st order, then the Euler-Lagrange (EL) equations will only be of 2nd order. See also e.g. this & this related Phys.SE posts.

  2. However, the Wilsonian effective action
    $$\begin{align} \exp&\left\{ -\frac{1}{\hbar}W_c[J^H,\phi_L] \right\}\cr ~:=~~~&\int \! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\exp\left\{ \frac{1}{\hbar} \left(-S[\phi_L+\phi_H]+J^H_k \phi_H^k\right)\right\} \end{align}$$ is defined by integrating out heavy/high modes $\phi^k_H$ and leaving the light/low modes $\phi^k_L$. Here $J^H_k$ denotes sources for the heavy modes. The (possibly non-local!) Wilsonian effective action $W_c[J^H,\phi_L]$ is the generating functional of connected $\phi_H$ Feynman diagrams in a background $J^H,\phi_L$.

  3. Nevertheless, the heavy propagators are exponentially suppressed, so the non-locality is mild, and can be taking into account by a Taylor expansion, cf. e.g. my Phys.SE answer here.

  4. The upshot is that, in the Wilsonian renormalization group flow, the Wilsonian Lagrangian density will in principle contain all possible terms that are not excluded by symmetry, e.g. $$ \ldots + \ldots +\frac{E}{2} (\partial_{\mu}\partial_{\nu}\phi)(\partial^{\mu}\partial^{\nu}\phi) + \frac{F}{2} (\partial_{\mu}\phi)(\partial^{\mu}\partial^{\nu}\phi)(\partial_{\nu}\phi) + \ldots ,$$ i.e., the Lagrangian density becomes of higher order.

  5. For higher-order Lagrangian theories, the EL equations (3) become $$ 0~\approx~\frac{\delta S}{\delta \phi} ~=~\frac{\partial {\cal L}}{\partial \phi} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi)} + \sum_{\mu\leq \nu} \frac{d}{dx^{\mu}} \frac{d}{dx^{\nu}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\partial_{\nu}\phi)} - \ldots. $$ Here the $\approx$ symbol means equality modulo eoms, and the ellipsis $\ldots$ denotes possible higher-derivative terms.

  6. In general, if the Lagrangian density is of $n$'th order, then the EL equations will be of $2n$'th order.


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