References: Robert D. Klauber, *Student Friendly Quantum Field Theory*, (Sandtrove Press, 2013) – Chapter 2, Problem 2.6.

It’s important to understand the distinction between a *particle* theory and a *field* theory. To see how this works, we’ll start by looking at classical theories of particles and fields using the Lagrangian formalism.

** Classical particle theory **

First, let’s revisit the Euler-Lagrange equations for a system of classical particles. Suppose we have particles in 3-d space, for a total of degrees of freedom. If we define the Lagrangian as

where is the kinetic energy (that depends only on velocities ) and is the potential energy (that depends only on positions ). We want to find the path followed by the system between times and , that is, we want to find between those times, subject to the constraint that and are fixed at some known values. In general, there is an infinite number of paths the system *could* take between these two times, and each path is specified by choosing the functions (which in turn determines ). Each choice of path gives a different form for the Lagrangian.

The principle of least action states that the action , defined as a functional of the paths that can be followed, is an extremum (in practice, almost always a minimum, hence the principle of *least* action). The action is defined as

The condition that be an extremum is specified by requiring , which means that if the Lagrangian gives a minimum (we’ll assume the extremum is always a minimum from here on to avoid confusion), then any slight variation of the paths that make up increases . Thus the condition is just an extension of the usual condition that the first derivative of an ordinary function be zero in order for that function to have a minimum.

To calculate , we need to vary the paths slightly. Using the chain rule (actually, we should justify that the chain rule works when calculating variations in functions, but we’ll trust the mathematicians on this point) we get

where the repeated index is summed.

We can now integrate the second term by parts to get

The requirement that and are fixed means that at the limits of integration, so the middle term is zero. We’re then left with

This must be true for all possible variations so the quantity in brackets must be zero, which gives us the Euler-Lagrange equations:

It’s worth digressing at this point to explain why we can take and as independent variables. It would seem that they are *not* independent since once you’ve specified you can get by just taking the derivative. The point is that when we specify the Lagrangian , we don’t know what is; all we have is the function which depends on both and . The goal of minimizing the action is to find the curves such that these curves together with their derivatives minimize the integral of . The physics comes in specifying the Lagrangian; the mathematics then allows us to determine the paths followed by the particles. In principle, we can specify to be any old function of and , but once we’ve done this, the form of is fixed and we can then solve the Euler-Lagrange equations to find the particle paths. In other words, the Euler-Lagrange equations specify the so that the together with their derivatives minimize the action .

In an alternative universe, we could conceive of a Lagrangian that depended on , and , say. In that case all of , and would be independent variables in the derivation above, and we’d end up with a different form of the Euler-Lagrange equations. The fact that physical Lagrangians depend only on , and is a consequence of Newton’s second law , since this allows only the positions and velocities to be specified as independent variables.

All this is fine, except how do we know that these equations, when solved for , actually do give the path followed by the system? The key is to look back at the definition of in 1. Then

where is the momentum of degree of freedom . Therefore, the Euler-Lagrange equations are equivalent to

where is the force acting on degree of freedom . This is just Newton’s second law, so the Euler-Lagrange formulation is indeed equivalent to Newton’s laws.

** Classical field theory **

The main difference between particle theory and field theory is that the variables no longer describe the motion of anything, that is, they are no longer functions of time. Rather, they become fixed labels for points in space. The position variables become independent variables in the same way that the time is independent. Taken together, they label points in spacetime.

A *field* is some quantity that has a value for each point in spacetime, and it is this quantity that can change as we move from place to place or forward in time. For a scalar field such as temperature or density, the field consists of a single value attached to each point in spacetime, where we now use the notation to represent the space components together with time. (That is, is a four-vector in special relativity, with , and so on.) A vector field, such as the electric field , is actually composed of three separate fields, one for each spatial coordinate. Each of these fields again has a single value for each point .

To work out the Euler-Lagrange equations for classical field theory, we need to think about what is meant by a ‘path’ that the system follows. Because the spacetime coordinates are no longer dynamical variables, it doesn’t make sense to ask how changes with time. What *does* change is the value of the field (or if we have more than one field, as with the electric field, in which case the index ranges over all the fields), so it is the field that is the dynamical variable. As such, the path followed is determined by a function of the field values. By analogy with the Lagrangian in the particle case, we define the *Lagrangian density* . The notation is defined as

The Lagrangian density is the Lagrangian per unit volume, and each infinitesimal volume element follows a path through time, so the action element of this volume element between times and is

The total action of the entire system is the integral of this over some spacetime volume that encloses the entire system spatially during the time interval, so

The idea now is to apply the calculus of variations to this integral and require as in the particle case. Remember that we’re varying the *fields* at each spacetime point and not the coordinates . Therefore (I’ll drop the superscript to avoid confusion with the summation convention, so the following should be taken to apply to each field separately. A summation over is implied):

To work out the second term, we write out the derivative explicitly:

where the last line follows from the product rule. We therefore get

The last term is the integral of a 4-d divergence over a 4-d volume and (trusting the mathematicians again) we can use a 4-d analog of Gauss’s theorem to convert this to a surface integral over a 3-d surface that bounds . Making the usual assumption that this surface can be removed to infinity and that our system is finite so that at infinity, this integral goes to zero. We’re left with

The requirement that this is valid for all variations in the field gives us the field theory version of the Euler-Lagrange equations (where I’ve restored the index indicating which field we’re talking about; note that is still summed):

**Example** * Suppose we have a classical, non-relativistic field of dust particles. The dust is sparse enough that there is no appreciable inter-particle interaction and there is no external force (such as gravity). Thus the potential energy density is . Further, suppose that the particle mass density is and is constant in time.*

*
*
Suppose the dust particles move about their initial positions. We can describe this motion as a displacement field , where describes the direction of displacement. Note that still describes a fixed point in spacetime, while can vary with space and time. The fields have the dimensions of length, since they measure the displacement of a particle from its initial position.

The kinetic energy density can be described in terms of as

where here there *is* a sum over , since we’re adding up the kinetic energy contributions from the three spatial directions. A dot indicates a derivative with respect to time, as usual.

Taking

the Euler-Lagrange equations 24 give us

So

*
In other words, the acceleration of the field (and hence of the dust particles) is zero, so they move with constant velocity. This is just an expression of Newton’s law applied to a continuous medium.*