How are field theory Langrangians treated when some terms have 2 derivatives but others have only 1? Because the number of derivatives in a Lagrangian term is more easily even than odd, the discussions for newcomers to physics of breaking up Lagrangians into free theories and perturbations often does not give clear instruction on how to conceptualize and handle terms with a single derivative.

How should a mathematician speaking to physicists refer to the role played by terms with only one derivative (e.g. Chern Simons like terms) in the presence of a Yang Mills kinetic term? Would it be a kinetic term of lower order? A velocity dependent potential term? Would the interpretation change if the second order Yang - Mills like term were discarded? Is an m D m -like term with 1-derivative of a field m treated as part of the perturbation in perturbation theory or as generating a lower order summand of the 2nd order 'free' operator to be inverted?

Please feel free to rephrase the question if you understand what is being asked and the confusion is complicating the inquiry. Thanks in advance.

  • 2
    $\begingroup$ You should probably try to clarify what your asking, since its not clear if you're asking a terminological question or something else... I usually see the term "free theory" to mean a theory whose Lagrangian is quadratic in fields, regardless of the number of derivatives (i. e. its Gaussian). As to what is treated as unperturbed and what as treated as perturbation, this just depends on what you're doing, no? In the usual perturbation theory around the Gaussian theory you would include quadratic terms in the unperturbed Lagrangian regardless of the number of derivatives. $\endgroup$ – BebopButUnsteady Jul 13 '11 at 5:15
  • $\begingroup$ This is exactly what I am confused by. Some descriptions emphasize that all quadratic terms appear in the free field theory. Other descriptions emphasize that free field theory includes only quadratic terms with 2 derivatives (kinetic) and 0 derivatives (mass). The importance of the terminological issue is that much of what is specified is described in prose and references 'free field theory', 'kinetic terms', 'potential terms'. I would like to carry around a clear sense of how physicists treat Lorentz invariant 1 derivative terms in terms of recipes, terminology, and theory. $\endgroup$ – Chet Marone Jul 13 '11 at 5:49

The OP is asking several questions.

1) The kinetic terms are, roughly speaking, the terms with temporal derivatives, e.g., $\frac{m}{2} \dot{q}^2$; or $p_i\dot{q}^i$; or a symplectic potential term $\vartheta_I(z)\dot{z}^I$, where $z^{I}$ is a phase space variable. (I'm using notation from point mechanics for the recognition value, however it works in field theory as well.)

2) Physicists often prefer a Hamiltonian formulation (also known as a first order formulation) withtime-derivatives (=velocities) appearing at most linearly, cf. the symplectic potential term $\vartheta_I(z)\dot{z}^I$. It is in principle possible to achieve a Hamiltonian form of the Lagrangian density ${\cal L}$ by Legendre transformation in the pertinent sectors at the expense of sometimes introducing first and second constraints via a Dirac-Bergmann analysis. Similarly to the Darboux theorem in finite dimensions, it is often possible to locally achieve a symplectic potential term of the standard form $p_i\dot{q}^i$ by a change of variables. See also the Faddeev-Jackiw method.

3) Concerning theories with both quadratic and linear dependencies of the same velocity, it is a good idea to first look at the simple example of a charged particle in an electro-magnetic field. Yang-Mills action plus a Chern-Simons action is, a bit oversimplified, just a field-theoretic version of this.

4) A fourth question is related to how to divide a Lagrangian density $\cal L={\cal L}_0+{\cal V}$ into a free part $\cal L_0$ and an interaction part ${\cal V}$. As a rule of thumb, we put as much as we can solve exactly in the free part, which usually restrict us to quadratic (and linear) terms only. In particular quadratic terms that are linear in velocities belong to the free part. Finally, the Hessian of the free part has to be invertible so that we can find the propagator. Zero-modes may cause extra subtleties. We set up the perturbative expansion around a stationary value of the free part.

  • $\begingroup$ Correction to the answer (v4): The word withtime should be with time. $\endgroup$ – Qmechanic Jun 14 '17 at 20:11

It is practically unknown that the perturbation term is also of kinetic nature; that is why it gives increments (shifts, "corrections") to the fundamental constants. Renormalizations are discarding those increments, and the remaining interaction is the true interaction. It is done perturbatively so it is not "visible". I have this explained in my detailed article.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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