# What would a Lagrangian without a kinetic term represent?

Imagine we write a Lagrangian without kinetic term, for example something like:

$$\mathcal{L} = -\frac{m^2}{2} \phi^2 -\frac{\lambda}{4!} \phi^4. \tag{1}$$

What would that represent? Let's look at the equations of motion:

$$\frac{\delta \mathcal{L}}{\delta \phi} = -\frac{\partial \mathcal{L}}{\partial\phi} = - \left( m^2 + \frac{\lambda}{3!} \phi^2 \right) \phi \overset{!}{=} 0. \tag{2}$$

This means that either $$\phi = 0$$ or $$\phi = \pm\sqrt{\frac{-6}{\lambda} m^2} = \pm i m \sqrt{\frac{6}{\lambda}}$$. Does that then correspond to a useless, constant background field that cannot propagate, or is there more to say about that?

• These solutions are the true and false vacuum ground states of some Higgs-like field, which don't propagate. – Cham Sep 5 '19 at 15:30

FWIW, a Lagrangian with (without) a kinetic term/time derivatives represents a dynamic (static) model, respectively.

There is no dynamical content without kinetic terms, and in fact it makes the theory sick.

Suppose you have

$$\mathcal L = \frac{X}{2}(\partial \phi)^2 -\frac{m^2}{2} \phi^2 -\frac{\lambda}{4!} \phi^4$$

where $$X$$ is some number that must be taken to zero to obtain a kinetic-term-less theory. Canonically normalizing the field by making the transformation $$\displaystyle \phi = \frac{\psi}{\sqrt{X}}$$, we get

$$\mathcal L = \frac{1}{2}(\partial \psi)^2 -\frac{m^2}{2} \frac{\psi^2}{X} -\frac{\lambda}{4!} \frac{\psi^4}{X^2}$$

We then see that taking the limit $$X \to 0$$ implies infinitely strong interactions, rendering the theory sick.

• Your transformation is singular at $X \rightarrow 0$. It cannot be applied for this value. – Cham Sep 5 '19 at 15:33
• @Cham The argument does have some hand-waving but it is one way of seeing that there's a problem with no kinetic terms. See Schwartz pg 134, for instance. I'm doing something which is similar to what is done in regularization: taking limits after computing something. – Avantgarde Sep 5 '19 at 16:21
• @Avantgarde I see what you are trying to show, but I do think that in that case the hand-waviness is actually too strong. To me it still seems that the theory described by a Lagrangian with kinetic terms represents a static model as mentioned by Qmechanic, but not that it makes the theory sick in the way that you describe. Or do you another argument that can justify your claim? – Jxx Sep 5 '19 at 20:18
• @Jxx I've mentioned in the first line that absence of kinetic terms implies no dynamical content, as you say too. The 'sickness' in terms of 'infinitely strong interactions' is maybe too hand-wavy, but I don't know of a way to make the argument concrete. – Avantgarde Sep 5 '19 at 20:32