Why is mass the quadratic term in a Lagrangian?

Question

Why is mass the quadratic term in a Lagrangian?

Answer 1

I'm not sure if this is the sort of answer you want, but here goes.

Mass, by definition, is the thing in $F=ma$. That is, it's the coefficient in front of the acceleration in the equation of motion. If you build a Lagrangian out of terms involving a coordinate $x$ and its derivative $\dot x$, and then work out the Euler-Lagrange equations, the only way to get a term of the form $m\ddot x$ at the end is to start with a term of the form ${1\over 2}m\dot x^2$ in the Lagrangian.

That's the classical-mechanics answer. If you were looking for a field-theory answer instead, the details are different, but the idea is the same. Think about how you expect mass to show up in the field equation, and work out what term in the original Lagrangian will give rise to the appropriate expression when you use the Lagrangian to derive the field equation.

Answer 2

Take the Lagrangian with only the kinetic and quadratic terms. Find the wave equation and propagator and you'll see that they correspond to those of a free relativistic particle with mass squared equal to the quadratic coefficient. So the interpretation is correct in free field theory.

Adding higher order terms doesn't change this interpretation, since the higher order terms correspond to interactions.

Answer 3

The mass is a quadratic term in a Lagrangian because it is the Lagrangian density, the Lagrangian per unit of volume. The Lagrangian itself is obtained by integrating over the volume which contains the field. The "classical" relation is.

$H ~-~pv~=~-L$

These integrated quantities, the Hamiltonian $H$, the Lagrangian $L$ and the term $pv$ transform like.

$ \begin{array}{lll} H & \mbox{transforms as} & \gamma \\ pv & \mbox{transforms as} & \beta^2\gamma \\ L & \mbox{transforms as} & 1/\gamma \\ \end{array} $

The volume of a wave-functions transforms like $1/\gamma$ due to Lorentz contraction. So, the densities become higher by a factor $\gamma$, hence the Hamiltonian density ${\cal H}$, the Lagrangian density ${\cal L}$ and the density of the $pv$ term transform like.

$ \begin{array}{lll} \mathcal{H} & \mbox{transforms as} & \gamma^2 \\ \mathcal{pv} & \mbox{transforms as} & \beta^2\gamma^2 \\ \mathcal{L} & \mbox{transforms as} & "1" \\ \end{array}$

The corresponding relation of the densities transforms like the basic energy/momentum relation.

$ {\cal H}\ -\ {\cal pv}\ =\ -{\cal L} \qquad \mbox{transforms as} \qquad E^2-p^2\ =\ m^2 $

We see that the Lagrangian density is the same in all reference frames. It is a Lorentz scalar. This makes the Lagrangian density a fundamental quantity in quantum field theory. The Standard Model of physics is based on the Lagrangian density which in quantum physics is generally called just the Lagrangian, without the density.

The "equations of motion" are based on the derivatives of the Lagrangian density which is a Lorentz scalar. This assures that the whole Standard Model of physics transforms in the right way, that is, the laws of physics are the same in every reference frame.

Regards, Hans

Answer 4

For a relativistic particle (free or in an external filed), the Lagrangian contains mass differently, not in a "quadratic" way.

Answer 5

Well, this is because $p^2=m^2$, well known relativistic relation, where p is 4-momentum of a particle, and m is mass. Take free Lagrangian for some classical field and then derive classical equation of motion out of it. Any such equation will be consistent with $p^2=m^2$ (you can see this by doing Fourier transform into p-space), and to keep this condition, you interpret m from term $m^2\phi^2$ as mass of particle.

Answer 6

For the free quantum field, the quadratic term determines that the energy-momentum 4-vector observable of a single-particle state in a single fourier mode would be on the forward mass-shell instead of on the forward light-cone. That is, $E^2+\mathbf{p}^2=m^2$ is a constant, which is algebraically of the same form as the relativistic mass-energy-momentum relationship. [Note, however, that fourier modes are strictly speaking an improper limit in quantum field theory.]

Again for the free quantum field, but in terms of QFT understood as a stochastic formalism for field observables instead of in terms of a particle interpretation, the two-point correlation function $C(x-y)=\left<0\right|\hat\phi(x)\hat\phi(y)\left|0\right>$ satisfies the classical Klein-Gordon field equation, where we can discuss energy-momentum density.

More abstractly, one could take mass, or inertia, to denote a generalized "resistance" of a given simple evolution to a given deformation of the background or to the introduction of a new force law. In these terms, every constant term in a Lagrangian is part of a multi-dimensional description of what the response of an initial system would be to any given additional term in the Lagrangian. The mass term is singled out as the lowest order component of the inertia.

There seem to me to be a surprising number of ways in which this Question can be taken. I decided only after a while that the Question is useful enough to obviate the lack of clarity. For me it's only because of the QFT tag that the Question becomes productive.