# I Motivation for mass term in Lagrangians

Tags:
1. Apr 28, 2017

### Frank Castle

In field theory a typical Lagrangian (density) for a "free (scalar) field" $\phi(x)$ is of the form $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi -\frac{1}{2}m^{2}\phi^{2}$$ where $m$ is a parameter that we identify with the mass of the field $\phi(x)$.

My question is, what is the motivation for including this mass term? Is it simply that in doing so one can reproduce the Klein-Gordon equation or are there some other physical arguments? Furthermore, is it in some sense a "self-interaction" term? Classically the Lagrangian of a free particle would just have a kinetic term so how does this Lagrangian describe a free field?

Last edited: Apr 28, 2017
2. Apr 29, 2017

### Orodruin

Staff Emeritus
If you do not include the mass term, you cannot describe a massive field and there is nothing forbidding its presence.

You can choose whether to include it in the free lagrangian or not. If you do it becomes part of your propagator, if you don't it becomes a (rather boring) self-energy contribution. Both approaches give the same results in the end.

Also, this is not the Lagrangian of a particle, it is the Lagrangian of a field. There are several classical situations that would lead to such a Lagrangian.

3. Apr 29, 2017

### Frank Castle

I was just wondering if there's any motivation for why one considers a Lagrangian of this form? Is it simply because we know that a relativistic massive field should satisfy the Klein-Gordon equation and so we want to construct a Lagrangian such that the Euler-Lagrange equations reproduce it?!

Also, is the reason why free-field Lagrangians are always constructed to be quadratic because this ensures that the corresponding EOM are linear and as such the solutions satisfy the superposition principle, i.e. they are non-interacting (intuitively they "pass through" each other)?!