# Higher order derivatives in field theories

1. Aug 14, 2010

### lasm2000

It is common lore to write lagrangians in field theories in the form

$$L(t)=\int d^{3}x\mathcal{L}(\phi_{a},\partial_{\mu}\phi_{a})$$.

Nonetheless, is there any particular reason for doing that? Why do we neglect higher order derivatives? Does it mess around with Lorentz invariance or something like that? I have heard that higher order lagrangians give origin to a spectra that is not bounded from below (then we can always have radiative transitions to a lower level).

Can someone confirm that or knows of a better reason?

2. Aug 14, 2010

### bcrowell

Staff Emeritus
I'll take a shot at this one, but I could be totally wrong. You seem to have quantum field theory in mind, but the question seems to me like it could just as well be posed in a classical context rather than a quantum-mechanical one, and for discrete particles rather than continuous fields.

For a classical description of particles, it seems to me that higher-order derivatives lead to problems when you want to define boundary-value problems. If I shoot a particle out of a gun, the initial conditions are the position and velocity. I'm not even sure what it would mean to specify the initial acceleration. I'm not sure that I could even do measurements that could distinguish initial conditions that differed only in their acceleration. Acceleration (unlike velocity) can change discontinuously, so it's not clear that it can even be measured at an instant in time.

In the context of relativity, we expect everything to be describable in geometric terms. Geometrically, two points are supposed to determine a line. In other words, if I specify an event, and another event infinitesimally close to it, then I've essentially determined a geodesic, which is the motion of a particle whose world-line passed through those two events. In a theory where initial conditions depended on higher-order derivatives, I'd have to specify more than two points to determine a line. I don't know of a geometry where that makes sense.

3. Aug 15, 2010

### genneth

4. Aug 15, 2010

### lasm2000

Thanks for your answers. Thats interesting, so it gives raise to a linear dependece of H with P1 (one of the canonical momentums that arise due to the fact that we now need four quantities as input). Then we can just decrease the value of H by decreasing the one of P1, so then, no bound from below.