Can we determine the order of the equations of motion simply by looking at the action?

Question

Naively, one would expect the EL equations arising from an action to contain derivatives (of the dynamics field) of an order that is twice the order of the highest-order derivative (of the dynamic field) present in the action. However, from the Einstein-Hilbert action, we know that this expectation is not always true.

To recap the story, the highest-order derivatives of the metric in the EH action are the second-order derivatives of the metric, coming from the $\partial \Gamma$ terms of the Ricci scalar. However, the Einstein equations do not contain third-order derivatives of the metric, but only the second-order (or lower) derivatives of the metric. This is easily seen by noticing that the highest-order derivative of the metric present in the Ricci tensor/scalar is the second-order derivative of the metric, again, coming from the $\partial\Gamma$ terms. This magic happens because one can write the EH Lagrangian density as a part containing only the first-order (or lower) derivatives of the metric and a part containing the second-order (or lower) derivatives of the metric, and the second part turns out to be a pure divergence term which would not contribute to the EL equations. Thus, we are left with the EL equations which contain the second-order derivatives of the metric as the highest order derivative of the metric.

I didn't know this until quite recently, and I remember using the reasoning of "the action is of the $k$-th order in derivatives of the dynamic fields so we would get $2k$-th order derivatives of the dynamic field in the equations of motion" multiple situations in physics. Now, I am weirded out about the equations of motion being one order lower than expected in multiple situations for it doesn't seem straightforward to spot as to whether there exists a decomposition of the Lagrangian density into non-pure-divergence and pure-divergence parts in such a way that all the highest order derivative terms are in the pure-divergence parts. Is there a way to examine the action to tell whether this is the case or not without deriving the EL equations explicitly?

Answer 1

1. If the Lagrangian density ${\cal L}$ has at most $k\in\mathbb{N}_0$ spacetime derivatives, then the Euler-Lagrange (EL) equations will have at most $2k$ spacetime derivatives.

2. Total divergence terms $d_{\mu}F^{\mu}$ in the Lagrangian density ${\cal L}$ does not affect the EL equations. As OP already mentions, one might be able to rewrite ${\cal L}=\tilde{\cal L}+d_{\mu}F^{\mu}$ into a piece $\tilde{\cal L}$ with, say, only $\tilde{k}$ spacetime derivatives, so that EL equation will have at most $2\tilde{k}$ spacetime derivatives.

3. The mechanism in pt. 2 is what happens for the 2nd-order Einstein-Hilbert action, cf. e.g. my Phys.SE answer here.

4. In a given theory it could be tedious/difficult to identify all possible hidden total divergence terms. In practice, it is simpler to just derive EL equations to begin with, and if the derivative order is less than expected, try to identify a hidden total divergence term in $\tilde{\cal L}$ that causes the mismatch.