0
$\begingroup$

In classical field theory, we consider the Lagrangians with single time-derivative of fields whereas double derivative of the field w.r.t. space is allowed sometimes. I understand that the reason of abandoning the 2nd order time-derivative of the fields is that we require two initial conditions, one is that of the field and the second is that if the momentum of the field.

What I don't understand is what is the problem with specifying the two initial conditions?

Also, while moving over to QFT from the classical description, how come the above mentioned discrimination of time derivative over space derivative, does not contradict the notion of putting space and time on equal footing?

$\endgroup$
1
$\begingroup$

Metric signature convention: $(+---)$.

First, note that physical dynamics is ultimately decided by the equations of motion, which you get from the Lagrangian $\mathcal{L}$ after using the least action principle. The kinetic term in a $1$-derivative (before integration by parts) field theory goes like $\mathcal{L} \sim \partial_\mu \phi \partial^\mu \phi \sim -\phi \square \phi$ whose equations of motion are $\square \phi + \cdots = 0$. This is a second order differential equation and so needs two initial conditions if you want to simulate the system.

The reason why people get nervous when they see higher derivatives in Lagrangians is that they typically lead to ghosts: wrong-sign kinetic terms, which typically leads to instabilities of the system. Before going to field theory, in classical mechanics, the Ostrogradsky instability says that non-degenerate Lagrangians with higher than first order time derivatives lead to a Hamiltonian $\mathcal{H}$ with one of the conjugate momenta occurring linearly in $\mathcal{H}$. This makes $\mathcal{H}$ unbounded from below. In field theory, kinetic terms like $\mathcal{L} \sim \square \phi (\square+m^2) \phi$ are bad because they lead to negative energies/vacuum instability/loss of unitarity. It has a propagator that goes like $$ \sim \frac{1}{k^2} - \frac{1}{k^2-m^2}$$

where the massive degree of freedom has a wrong sign. Actually, in a free theory, you can have higher derivatives in $\mathcal{L}$ and be fine with it. You won't 'see' the effect of having unbounded energies until you let your ghost-like system interact with a healthy sector. Then, a ghost system with Hamiltonian unbounded from below will interact with a healthy system with Hamiltonian bounded from below. Energy and momentum conservation do not prevent them from exchanging energy with each other indefinitely, leading to instabilities. In a quantum field theory, things get bad from the get-go because (if your theory has a healthy sector, like our real world) the vacuum is itself unstable and nothing prevents it from decaying into a pair of ghosts and photons, for instance.

This problem of ghosts is in addition to the general consternation one has when they are required to provide many initial conditions to deal with the initial value problem.

Also, in certain effective field theories, you can get wrong-sign spatial gradients $ \mathcal L \sim \dot{\phi}^2 + (\nabla \phi)^2$. (Note that Lorentz invariance is broken here). These lead to gradient instabilities.

$\endgroup$
  • $\begingroup$ I was trying to prove the Ostrogradsky instability for Lagrangian with 2nd order time derivative. I am getting a term which should vanish but I can't understand why it will. The term is: $\int\frac{d^2}{dt^2}(\frac{\partial L}{\partial \ddot{q}} \delta q)$ I don't think this is a surface term which vanishes at the boundary. Maybe this is a silly question but can you help me out? Not that I used 'q' as the canonical co-ordinate. $\endgroup$ – Samapan Bhadury Apr 19 at 17:42
  • 1
    $\begingroup$ @SamapanBhadury See this paper arxiv.org/abs/1506.02210v2 $\endgroup$ – Avantgarde Apr 19 at 17:48
  • $\begingroup$ Okay. Thanks. I will look at it. $\endgroup$ – Samapan Bhadury Apr 19 at 17:52
  • $\begingroup$ Thanks a lot for the very beautiful article. It really solved my conceptual difficulty a lot. However, as I was saying earlier, I am stuck at deriving the equation (13) of the article you mentioned i.e. I was trying to derive the Euler-Lagrangian equation for a Lagrangian that depends on 2nd order time derivative of Lagrangian, $L$ w.r.t. time. I could not eliminate the integral I mentioned above.(I forgot to put the measure there, which is simply, $dt$). $\endgroup$ – Samapan Bhadury Apr 19 at 19:13
  • $\begingroup$ @SamapanBhadury See this question and answers physics.stackexchange.com/q/109518/133418. We remove all surface terms assuming nice boundary conditions. $\endgroup$ – Avantgarde Apr 19 at 19:40
1
$\begingroup$

This isn't true; time and space really are on an equal footing in relativistic quantum field theory. For example, the kinetic term for a real scalar field is $$\frac12 (\partial_\mu \phi) (\partial^\mu \phi)$$ which is first-order in both space and time. If you want, you can integrate it by parts to get $$- \frac12 \phi \partial^2 \phi$$ but this is second-order in both space and time, which we don't like for the reasons you said.

$\endgroup$
  • $\begingroup$ When evaluating the momentum $\Pi$ in QFT, we have to differentiate $\phi$ w.r.t time only. I don't understand how this is not counted as a preferential treatment of time. The spatial derivative of the field i.e. $\nabla_i \phi$ doesn't carry any physically significance. Does it? When thinking about momentum of the field, we think of the former one, not the later one. I am a bit confused here. $\endgroup$ – Samapan Bhadury Apr 19 at 10:48
  • $\begingroup$ @SamapanBhadury The canonical momentum is an object in Hamiltonian mechanics, which explicitly has a preferred time. You are free to use Hamiltonian mechanics in relativistic QFT, but then you should check the results actually are Lorentz invariant, which is done in exhaustive detail in most courses. $\endgroup$ – knzhou Apr 19 at 10:49
  • $\begingroup$ I did say why we don't use double derivative of the fields w.r.t. time (which has to do with initial conditions) but I don't understand what is the problem in defining two initial conditions! $\endgroup$ – Samapan Bhadury Apr 19 at 10:50
  • $\begingroup$ @SamapanBhadury, The derivative with respect to $\nabla_i \phi$ does carry significance. It tells you the spatial components of the momentum density four vector $\endgroup$ – octonion Apr 19 at 10:51
  • $\begingroup$ @SamapanBhadury There's nothing wrong with two initial conditions. You need two for a real scalar field. $\endgroup$ – knzhou Apr 19 at 10:52
1
$\begingroup$

the reason to generally avoid second order (time) derivatives in the Lagrangian is that they bring to third order equations of motion, that would need three constants of integration to be solved and that usually imply the possibility of "run-away" solutions (or Ostrogradski instabilities). These are solutions in which the energy encreases exponentially, a standard example is the Abraham-Lorentz force(see https://en.m.wikipedia.org/wiki/Abraham–Lorentz_force ): $$\ddot {v}\sim \frac {\dot {v}}{\tau}$$ where you see that for instance a initial zero speed does not prevent a solution of the form $v\sim e^{t/\tau}$, which means that the object you are describing undergoes a sort of self-acceleration.

There can be also other problems with higher derivatives (google Ostrogradski instability).

Anyway, you are right in saying that including second order spatial derivatives but only first order time derivatives in the Lagrangian, impedes the covariant formulation in which space and time are on the same page. But that may not alsways be dramatic.

Note however that exist some more "exotic" Lagrangians (for instance Galileons and Horndeski theories https://en.m.wikipedia.org/wiki/Horndeski%27s_theory ) in which even though you include second order time (and space) derivatives you still get second order equations and therefore nothing violates the basic physical requirements. These theories are covariant.

$\endgroup$

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.