Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm in search of examples of Lagrangian, which are at least second order in the derivatives and are covariant, preferable for field theories. Up to now I could only find first-order (such at Klein-Gordon-Lagrangian) or non-covariant (e.g. KdV) ones. Also some pointers to the literature about general properties of such systems are welcome. Thanks

share|cite|improve this question
Klein-Gordon can be rewritten as $(\partial \phi)^2 = - \phi \Box \phi$ + boundary terms! And you can probably write many such Lagrangians yourself. Foe example, you can hit fields with powers of the Laplacian, which is covariant. Nothing prevents you from writing $\Box_x F_{\mu \nu}(x) \Box_y F^{\mu \nu}(y).$ – Vibert Apr 6 '13 at 13:59
@Vibert: Of course, I can build such a Lagrangian by myself. But I'm interested in useful ones, e.g. which describe real physical systems. E.g. general relativity is such an example, but I look for easier one. – Tobias Diez Apr 8 '13 at 13:42
I don't think there are many easier ones. By adding derivatives, you're raising the 'dimension' of the operator in your Lagrangian, and you lose renormalisability. This is used a lot in 'effective field theory' (with applications in flavour physics, higgs physics/EWSB etc.) but not in normal, text-book models. – Vibert Apr 8 '13 at 17:19
There's a no-go theorem ("Ostragradski's theorem") which says higher order actions lead to unbounded energies and unstable systems, see section 2 of arXiv:astrop-ph/0601672 for further details on Ostragradski's theorem... – Alex Nelson Jul 5 '13 at 20:40

I) As user Vibert mentions in a comment, the Euler-Lagrange equations are not modified$^1$ by adding total divergence terms to the Lagrangian density

$$ \tag{1} {\cal L} ~ \longrightarrow ~{\cal L} +d_{\mu}F^{\mu}. $$

Adding total divergence terms leads to an inexhaustible source of higher-order Lagrangians.

II) Generically, without some cancellation mechanism in place [such as, that part of the Lagrangian density is (secretly) a total divergence] an $n$-order action would lead to $2n$-order Euler-Lagrange equations.

III) Example. The Einstein-Hilbert (EH) Lagrangian density

$$\tag{2} {\cal L}_{EH}~\sim~\sqrt{-g} \left\{g^{\mu\nu} R_{\mu\nu}(\Gamma_{LC},\partial\Gamma_{LC})-2\Lambda\right\} $$

depends on both second-order temporal and spatial derivatives of the metric $g_{\mu\nu}$. This is of course an important example. Here $\Gamma_{LC}$ refer to the Levi-Civita (LC) Christoffel symbols, which in turn are first order derivatives of the metric $g_{\mu\nu}$. However, it is possible to add a total divergence term to render the Lagrangian density first order, as user drake mentions in a comment. Thus the Euler-Lagrange equations for the Einstein-Hilbert action $S_{EH}[g_{\mu\nu}]$, i.e. the Einstein field equations (EFE), are not of fourth order, as one may naively have expected, but still of second order.

IV) Higher-order Lagrangians are also discussed in many Phys.SE posts, see e.g. here and here.


$^1$ Note that adding total divergence terms (1) may affect consistent choices of boundary conditions for the theory.

share|cite|improve this answer
There is no higher-order cancellation. The Einstein-Hilbert action depends on second order temporal derivatives through a boundary term, i.e., there exits an action that differs from Einstein-Hilbert's by a boundary term and that gives the Einstein equations. It is much like the Klein-Gordon Lagrangian and $\mathcal L= \phi(\partial^2-m^2)\phi$. So in regard to the question the Einstein-Hilbert action is somehow marginal. – Diego Mazón Jul 6 '13 at 6:54
@drake: Thanks. I updated the answer. – Qmechanic Jul 6 '13 at 15:36

You can look at the Lagrangian for the galileon particles for instance in this paper. It has the property that the equations of motion remains 2nd order in the derivatives and covariant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.