# Intuitive explanation for the free field Lagrangian?

The free field Lagrangian is $$\mathcal{L}=\frac 1 2 \partial^\mu\phi\partial_\mu\phi-\frac 1 2m^2\phi^2$$ with sign convention $$(+,-,-,-)$$. Plugging this into the Euler-Lagrange equations gives the KG equation.

In class we were given the semi-intuitive explanation that this resembles the classical Newtonian Lagrangian of $$L=T-V$$ since the first term resembles $$p^2/2m$$ and the second part resembles the potential.

Is there a more satisfying explanation to this Lagrangian?

• For a scalar field, this lagrangian is the simplest one which is invariant under Lorentz transformations. There is no other explanations for this expression.
– Cham
Dec 19 '18 at 14:35
• What would you consider to be an intuitive explanation? This is the very simplest field Lagrangian that can possibly exist. I could say how the Lagrangian for electromagnetism looks somewhat like this, but that would be explaining something simple in terms of something complicated. And you've also said you don't like the comparison with a particle Lagrangian. So what starting point would you feel is 'intuitive'? Dec 19 '18 at 17:49

In field theory, the degrees of freedom are the field, and its first derivative with respect to some parameter (time $$t$$ in classical theory, a space-time event $$x^\mu$$ in relativistic field theory);

1) Hence, your lagrangian will have to be a function of those ingredients, plus possibly the parameter, if no symmetry prevents it from appearing:

$$L=L(\phi,\partial_\mu\phi;x^\mu)$$

2) Then, as you want your theory to be Poincarè covariant (i.e. under Lorentz rotations and translation in Minkowski space-time), you will require your Lagrangian to be a scalar under such group, that imposes:

• not to have $$x^\mu$$ explicitly, as it breaks translation invariance
• to have no free Lorentz indices, so every occurrence of $$\partial_\mu\phi$$ must be contracted with itself or some other 4-vector in order to yield a Lorentz scalar; as no other 4-vector is available you will require $$\partial_\mu\phi$$ to be contracted with itself:

$$\partial_\mu\phi \partial^\mu\phi$$

• you could in principle add terms with $$(\partial_\mu\phi \partial^\mu\phi)^n$$, with $$n$$ any integer;

3) $$\phi$$ is a scalar field, so you could include any power of it;

4) at the end of the day, you want to reproduce the Klein-Gordon equation which is linear in the field, so all the freedom you have reduces to consider the quadratic terms in $$\phi$$ and $$\partial_\mu\phi$$:

$$L=\alpha \partial_\mu\phi \partial^\mu\phi +\beta \phi^2$$

The Lagrange equations say:

$$\partial_\mu \frac{\partial L}{\partial \partial_\mu\phi}= \frac{\partial L}{\partial \phi}$$

$$\Rightarrow \alpha\partial_\mu\partial^\mu\phi=2\beta\phi$$

From which $$\alpha=1$$ and $$\beta=-\frac12m^2$$

The Lagrangian for a real scalar field $$\phi\left(\mathbf x,t\right)$$ given by OP with the sign convention $$(+,-,-,-)$$ is expressed alternatively as
$$$$\mathcal{L}\left(\phi,\boldsymbol{\nabla}\phi,\overset{\;\centerdot}{\phi}\right)\boldsymbol{=}\frac12\overset{\;\centerdot}{\phi}{}^{\,2}\boldsymbol{-}\frac12\left(\left\Vert\boldsymbol{\nabla}\phi\vphantom{\tfrac12}\right\Vert^2\boldsymbol{+}m^2\phi^2\right) \tag{01}\label{01}$$$$ Comparing \eqref{01} to the usual expression for the Lagrangian $$\;L\boldsymbol{=}T\boldsymbol{-}V$$, we identify the kinetic energy of the field as $$$$T\boldsymbol{=}\frac12\!\int \overset{\;\centerdot}{\phi}{}^{\,2}\mathrm d^3\mathbf x \tag{02}\label{02}$$$$ and the potential energy of the field as $$$$V\boldsymbol{=}\frac12\!\int \left\Vert\boldsymbol{\nabla}\phi\vphantom{\tfrac12}\right\Vert^2\mathrm d^3\mathbf x\boldsymbol{+}\frac12\!\int m^2\phi^2\mathrm d^3\mathbf x \tag{03}\label{03}$$$$ The first term in this expression is called the gradient energy, while the phrase $$^{\prime\prime}$$potential energy$$^{\prime\prime}$$, or just $$^{\prime\prime}$$potential$$^{\prime\prime}$$, is usually reserved for the last term.

Reference : $$^{\prime\prime}$$Quantum Field Theory$$^{\prime\prime}$$ by David Tong, $$\S$$ 1.1.1 An Example: The Klein-Gordon Equation.