# Higher-order derivatives than second-order differential equations

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?

• 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. Nov 27, 2021 at 5:10
• Okay... then hopefully there is an example for higher-order differential equations Nov 27, 2021 at 5:17
• Sure. $(\square \phi)^2$. Nov 27, 2021 at 14:15

\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 $$J^H,\phi_L$$.
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.