The classical scalar field theory Lagrangian is:

$$ \mathcal{L} = \frac{1}{2} \left ( \partial_{\mu} \phi \right )^2 - \frac{1}{2} m^2 \phi^2. $$

I'm comfortable with the fact that the Lagrangian is a linear combination of $ \left ( \partial_{\mu} \phi \right )^2 $ and $ \phi^2 $. However, I'm having a hard time interpreting the fact that the dependence of the second term on $ \phi^2 $ is proportional to $ m^2 $.
From the motivation as a set of classical springs, I would expect the first term to exhibit dependence on $ m $ and the second term to be proportional to $ k $. How intuitively can we justify having an $ m^2 $ in front of $ \phi^2 $? Is there no explanation other than the units working out?

P.S. I should note that I'm using $ \hbar = c = 1 $ units. I also presume that $ \phi $ has units of length (like oscillator displacement).
P.P.S. While googling this question, I found how to use Fourier analysis of the equations of motion to prove that $ m $ represents the mass of the scalar field. While this is somewhat helpful, I'm really looking for an explanation of why we cannot simply extend the dependence on $ m $ from our intuition as a set of interconnected springs.

  • $\begingroup$ You can just as well call this parameter $\zeta$, $\chi$ or whatever you like. But from Fourier analysis you will find that it has interpretation of mass, so it only makes sense to call it that. $\endgroup$
    – Blazej
    Sep 7, 2016 at 19:42

2 Answers 2


I think you're confusing two different masses: the mass of the springs and the mass of the field quanta.

Start with coupled masses-on-springs, of mass $M=1$ and spring constant $k$. Let's imagine that the springs are located on a 2d lattice, and they oscillate only in a 3rd direction.

Quantize and take the continuum limit, and you get a quantum field theory, which describes the values of a field which lives on $\mathbb{R}^2$. This QFT has quanta, which have mass $m = \sqrt{k}$. These particles are not_ the original masses-on-spring though; they're collective oscillations of the springs. Their mass governs how fast they move around in $\mathbb{R}^2$, not how fast they move transverse to the 2d plane.

Note that you don't have to (and shouldn't!) think of the original springs as oscillating in the same space as the lattice points. The springs are part of the example because we wanted a system described by a number at each point in space.


How intuitiviely can we justify having an $m^2$ term.Is there no explanation other than the units working .

At best this is a half answer, and it may not be what you are looking for, or that you already are aware of. But if I am incorrect, my apologies and I hope somebody will tell both of us more regarding the subject.

From memory, mass in scalar fields is always associated with quadratic field terms, at in my studies so far. So once you find a $\theta ^2$, if there is a mass term, it will be associated with that.

I hope you get a more sophisticated answer, because this is just from memory, but Wikipedia seems to support it.

From Scalar Fields Wikipedia

The most basic scalar field theory is the linear theory. Through the Fourier decomposition of the fields, it represents the normal modes of an infinity of coupled oscillators (see phonons). The action for the free relativistic scalar field theory is then

$$ {\begin{aligned}{\mathcal {S}}&=\int \mathrm {d} ^{D-1}x~\mathrm {d} t{\mathcal {L}}\\&=\int \mathrm {d} ^{D-1}x~\mathrm {d} t\left[{\frac {1}{2}}\eta ^{\mu \nu }\partial _{\mu }\phi \partial _{\nu }\phi -{\frac {1}{2}}m^{2}\phi ^{2}\right]\\[6pt]&=\int \mathrm {d} ^{D-1}x~\mathrm {d} t\left[{\frac {1}{2}}(\partial _{t}\phi )^{2}-{\frac {1}{2}}\delta ^{ij}\partial _{i}\phi \partial _{j}\phi -{\frac {1}{2}}m^{2}\phi ^{2}\right],\end{aligned}}$$

where ${\displaystyle {\mathcal {L}}}$ is known as a Lagrangian density; $d4−1x ≡ dx ⋅ dy ⋅ dz ≡ dx1 ⋅ dx2 ⋅ dx3$ for the three spatial coordinates; $\delta_{ij}$ is the Kronecker delta function; and $\partial\rho  = \partial/\partial x\rho $ for the ρ-th coordinate $x\rho$.

This is an example of a quadratic action, since each of the terms is quadratic in the field, $\psi$ The term proportional to $m^2$ is sometimes known as a mass term, due to its subsequent interpretation, in the quantized version of this theory, in terms of particle mass.

The equation of motion for this theory is obtained by extremizing the action above. It takes the following form, linear in $\psi$,

$$ \eta ^{\mu \nu }\partial _{\mu }\partial _{\nu }\phi +m^{2}\phi =\partial _{t}^{2}\phi -\nabla ^{2}\phi +m^{2}\phi =0~,$$

where $∇^2$ is the Laplace operator. This is the Klein–Gordon equation, with the interpretation as a classical field equation, rather than as a quantum-mechanical wave equation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.