Why the Real Scalar Field lagrangian has this form?

The lagrangian of the Real Scalar Field $$\phi$$ is given by

$$$$\mathcal{L} = \frac{1}{2}\eta^{\mu \nu} \partial _\mu \phi \partial _{\nu} \phi - \frac{1}{2} m^2 \phi^2$$$$ $$$$\eta^{\mu \nu} = diag(+1,-1,-1,-1)$$$$ But what is the real motivation behind this equation?

I've read in several books this equation. In some books it is "defined", in others is given a completely unsatisfactory derivation for this equation , like "let's make an analogy with a coupled oscillators system". I've understood that the natural units makes the Lagrangian have dimension of $$[mass]^4$$ and the fact that the Lagrangian must be a Lorentz Invariant so the derivatives of the field must be $$\partial _\mu \phi$$, but how can i get the "1/2" coefficient in this Lagrangian and where does come from this "$$m^2 \phi^2$$? The "$$m$$" coefficient is the mass of the field or its just a constant with dimension of mass? Please help with this issue.

• What is the motivation behind any lagrangian?
– user87745
Jan 10, 2020 at 0:56

As you say, when building a Lagrangian for a theory, we almost always wish it to be a Lorentz scalar. Obviously, we can take any power of the scalar field, $$\phi$$, that is, we can have terms of the form,
$$\mathcal L = \dots + \sum_{n\geq 0} c_n \phi^n + \dots$$
with coefficients of our choosing. This means we could consider a term like $$\sin \phi$$ since upon Taylor expansion, at least at the level of the Lagrangian, it amounts to simply adding an infinite number of $$\phi^n$$ terms (this results in Sine-Gordon theory, incidentally).
We can also consider derivatives of the field $$\partial_\mu \phi$$ and indeed combining two of them, we can produce the scalar $$\partial^\mu \phi \partial_\mu \phi$$. A Lagrangian with just this term gives us the wave equation, $$\square \phi = 0$$, which as you should be aware is fundamental and ubiquitous in physics. This alone is sufficient motivation to consider such a Lagrangian for the scalar.
Adding a mass term gives the Klein-Gordon equation; that it is a "mass" term (distinguished from other powers of $$\phi$$ one could add) is shown by quantising the theory and considering scattering. Finally the overall factor of $$\frac12$$ is a convenience. We could put any constant in front, it won't change the physics.