Momentum in a Scalar Field Theory

Given a scalar field theory described by the Lagrangian

$$\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}\partial_\nu\phi\partial_\mu\phi – \frac{1}{2}m^2\phi^2,$$

we can consider the effect of an infinitesimal transformation

$$x^\mu \to x^\mu – \epsilon^\mu.$$

A quick computation shows that the change in the Lagrangian is

$$\delta\mathcal{L} = -\epsilon^\rho\partial_\rho\mathcal{L},$$

which is a total derivative. Hence, Noether’s theorem tells us that there are conserved currents for the four choices of $\epsilon_\rho$, given by

$$T^{\rho\mu} := \left(j^\rho\right)^\mu = \frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi\right)}\partial_\rho\phi – \delta^{\mu\rho}\mathcal{L}.$$

There are then four conserved charges:

$$Q^0 = \int\mathrm{d}^3x\; T^{00} = \int\mathrm{d}^3x\left[\frac{1}{2}\dot{\phi}^2+\frac{1}{2}\left(\nabla\phi\right)^2+\frac{1}{2}m^2\phi^2\right]$$


$$Q^i = \int\mathrm{d}^3x\; T^{i0} = -\int\mathrm{d}^3x\;\dot{\phi}\partial^i\phi.$$

The first conserved charge is easily interpreted as the total energy: $E=Q^0$. The other three charges correspond to the total momentum in the field: $P^i=Q^i$. But does this make sense? What does the quantity $\dot{\phi}\partial^i\phi$ signify?

Well, using the symmetry of the Lagrangian under Lorentz boosts, it is possible to derive another conserved charge:

$$Q^{0i} = \int\mathrm{d}^3x\left(x^0T^{0i}-x^iT^{00}\right).$$

Since this is conserved, we find that

$$0 = \frac{\mathrm{d}Q^{0i}}{\mathrm{d}t} = \int\mathrm{d}^3x\;T^{0i}+t \int\mathrm{d}^3x\frac{\partial T^{0i}}{\partial t} – \int\mathrm{d}^3x\;x^iT^{00}\\
= P^i + t\frac{\mathrm{d}P^i}{\mathrm{d}t} – \frac{\mathrm{d}}{\mathrm{d}t} \int\mathrm{d}^3x\;x^i T^{00}.$$

Since $P^i$ is conserved, this reduces to

$$P^i = \frac{\mathrm{d}}{\mathrm{d}t}\int\mathrm{d}^3x\;x^iT^{00}.$$

This makes a lot more sense. Since $T^{00}$ is the energy density (and $E=m$), we immediately recognise the familiar rule from classical mechanics:

$$p = \frac{\mathrm{d}}{\mathrm{d}t}(mx).$$

This also suggests an interpretation for $\dot{\phi}\partial^i\phi$. Finding the expression for $T^{00}$ and taking the time derivative, we find

$$\frac{\partial}{\partial t}T^{00} = \frac{\partial}{\partial t}\left(\frac{1}{2}\dot{\phi}^2+\frac{1}{2}\left(\nabla\phi\right)^2+\frac{1}{2}m^2\phi^2\right)\\
= \dot{\phi}\ddot{\phi}+\nabla\phi(\nabla\dot{\phi})+m^2\phi\dot{\phi}.$$

We now use the equation of motion

$$\partial_\mu\partial^\mu\phi + m^2\phi = 0 \Longrightarrow \ddot{\phi} + m^2\phi = \nabla^2\phi,$$

to write this as

$$\frac{\partial}{\partial t}T^{00}  = \left(\dot{\phi}\nabla^2\phi + \nabla\phi(\nabla\dot{\phi})\right) = \nabla\left(\dot{\phi}\nabla\phi\right).$$

This is a continuity equation of the form

$$\frac{\partial\rho}{\partial t} + \mathbf{\nabla}\cdot\mathbf{j} = 0,$$

where $\rho=T^{00}$ is the energy density and $\mathbf{j} = -\dot{\phi}\mathbf{\nabla}\phi$ is the energy density flux. In other words $j^i=-\dot{\phi}\partial^i\phi$ is the flow of energy in the $x^i$-direction.

Let us see how this relates to our earlier definition of $P^i$. We have

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(x^iT^{00}\right)  = x^i\nabla\left(\dot{\phi}\nabla\phi\right).$$

Integrating this over space, we find

$$P^i = \int\mathrm{d}^3x\;x^i\nabla\left(\dot{\phi}\nabla\phi\right).$$

Using the divergence theorem, we have in general for scalar functions $u,v,w$ that

$$\int\mathrm{d}^3x\;u\nabla\cdot\left(v\nabla w\right) = -\int\mathrm{d}^3x\;\nabla u\cdot\left(v\nabla w\right) + \text{boundary term}.$$

Hence, in our case, we find

$$P^i = -\int\mathrm{d}^3x\;\dot{\phi}\partial^i\phi,$$

which is what we had originally. So indeed, the total momentum of the scalar field in the $x^i$-direction is simply the energy flux in the $x^i$-direction integrated over all space.