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?
 A: 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 $\beta=-\frac12m^2$
A: The Lagrangian for a real scalar field $\phi\left(\mathbf x,t\right)$ given by OP with the sign convention $(+,-,-,-)$ is expressed alternatively as
\begin{equation}
\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}    
\end{equation}
Comparing \eqref{01} to the usual expression for the Lagrangian $\;L\boldsymbol{=}T\boldsymbol{-}V$, we identify the kinetic energy of the field as
\begin{equation}
T\boldsymbol{=}\frac12\!\int \overset{\;\centerdot}{\phi}{}^{\,2}\mathrm d^3\mathbf x  
\tag{02}\label{02}    
\end{equation}
and the potential energy of the field as
\begin{equation}
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}     
\end{equation}
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.

