Classical field theory

Last updated Dec 29, 2024

• Onepass

Field theory is a theory with continuos degrees of freedom

The Hamiltonian and Lagrangian are defined as integrals over densities:

Definition: Hamiltonian

The Hamiltonian depends on the field $\phi$ and the conjugate $\pi$ $$H = \int d^3x \mathcal{H}.$$ It is written as an integral over a Hamiltonian density.

The Lagrangian is always lorentz invariant!

Definition: Lagrangian

The Lagrangian is also written in terms of a density $$L=\int d^3x\mathcal{L}.$$

The Lagrangian density is given as the Legendre transformation of the Hamiltonian $$\mathcal{L}[\phi,\dot{\phi}]=\pi[\phi,\dot{\phi}]\dot\phi - \mathcal{H}[\phi,\pi[\phi,\dot{\phi}]].$$ The dynamical variables are now $\phi$ and $\dot\phi$.

The conjugate momentum is defined in terms of $\phi$ and $\dot\phi$ by $$\frac{\partial\mathcal{H}[\phi,\pi]}{\partial\pi}=\dot\phi.$$

The Hamiltonian corresponds to the energy and is therefore conserved, however it is not lorentz invariant! (Energy is a component of a four vector, so no lorentz scalar)

Usually The Hamiltonian is associated with the total energy as sum of kinetik $\mathcal{K}$ and potential $\mathcal{V}$ energy $$\mathcal{H} = \mathcal{K} + \mathcal{V}.$$ The Lagriangian is often $$\mathcal{L} = \mathcal{K} - \mathcal{V}.$$

• Kinetik term always has two fields and at least one derivative to be dynamical
• Interaction terms have three or more fields
• We also write interaction terms as $$\mathcal{L}_\text{int} = -\mathcal{V} = -\mathcal{H}_\text{int}.$$

Definition: Action

The Action is the integral over time of the Lagrangian $$S = \int dt L = \int d^4x\mathcal{L}(x)$$

The dynamics of a system follow from the principle of least action

Principle of least action

$$\delta S=0$$

from which we can derive equations of motion using $$\delta S = \int d^4x \delta\mathcal{L} = \int d^4x \left[\frac{\partial\mathcal{L}}{\partial\phi}\partial\phi - \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta(\partial_\mu\phi)\right]$$ Here we can use partial integration. Within the action integral all boundary terms of partial integration vanish, since we assume that the fields vanish at infinity. This partial integration essentially always lets us do $$A\partial_\mu B = - (\partial_\mu A)B$$ Then $$\delta S = \int d^4x\left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right]\delta\phi =0,$$ which is fulfilled for all variations $\delta\phi$ (you can vary the field in any way you want as long as the variation is small) so that we can conclude

Euler Lagrange Equations

The Euler Lagrange Equations for a scalar field are the equations of motion given in terms of its Lagrangian $$\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}=0$$

Example: Klein Gordon equation

For $\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi)-\frac{1}{2}m^2\phi^2$ the Euler-lagrange equations give the Klein Gordon equation $$(\Box + m^2)\phi = 0$$ This is the equation of motion for a free scalar field

The Noether theorem relates symmetries of a Lagrangian to conserved quantities

Noether’s theorem

If a Lagrangian has a continuous symmetry then there exists a curent associated with that symmetry that is conserved when the equations of motion are satisfied

If the symmetry is given by a parametrized transformations $f(\phi,\alpha)$ the current can be derived by requiring $\frac{\delta \mathcal{L}}{\delta \alpha}=0$ which leads to

$$J_\mu = \sum_n \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_n)}\frac{\delta\phi_n}{\delta\alpha},$$

where we sum over all fields the symmetry applies to. The variation $\frac{\delta\phi_n}{\delta\alpha}$ is the variation of the field under an infinitesimal variation of the symmery transformation. $J_\mu$ is known as Noether current. If the equations of motion are satisfied it fulfills

$$\partial_\mu J_\mu = 0$$

Corresponding to the conserved current, there is always a conserved charge

$$Q = \int d^3x J_0$$ with $$\partial_tQ = \int d^3x \partial_tJ_0=\int d^3x\vec{\nabla}\cdot\vec{J}=0.$$

The last integral is zero since we can perfom the integral of the derivative giving $\vec{J}$, which we assume to vanish at the boundaries (nothing enters or leaves out infinite box).

Example: Noether current for the scalar field

Say we have a complex scalar field $\phi$ (this amounts to two degrees of freedom e.g. $\phi$ and $\phi^*$) with the Lagrangian $$\mathcal{L}=|\partial_\mu\phi|^2-m^2|\phi|^2.$$ This Lagrangian is invariant under the continous symmetry $$\phi\to e^{-i\alpha}\phi.$$ The variation of the fields are $$\frac{\delta\phi}{\delta\alpha}=-i\phi\quad\frac{\delta\phi^* }{\delta\alpha}=i\phi^*.$$ The noether current therfore is $$J_\mu = -i\left(\phi\partial_\mu\phi^*-\phi^*\partial_\mu\phi\right).$$

Noethers theorem applied to translation invariance (Of the action, not only the Lagrangian!) leads to the conserved current the Energy Momentum tensor $\mathcal{T}_{\mu\nu}$.

Consider the transformation of translation: A global translation $x^\mu \to x^\mu + \xi^\mu$ is generated by infintesimal ones, which in turn can be described by an expansion in the translation $\xi^\mu$ $$\phi(x+\xi) = \phi(x) + \xi^\nu \partial_\nu \phi(x) + …$$ A small variation $\frac{\delta \phi}{\delta x^\nu}$ by $\xi$ is thus given by the derivative $\partial_\nu \phi$. If the action $S$ is invariant under such a variation we can derive the equations $$\frac{\delta S}{\delta \xi^\mu} = \int d^4x \frac{\delta \mathcal{L}}{\delta \xi^\mu} = \xi^\mu \int d^4x \partial_\mu \mathcal{L} = 0$$ Inserting the total derivative from the variation $\delta \mathcal{L}$ we can find

$$\partial^\mu \mathcal{T}_{\mu\nu} = 0,$$

where $\mathcal{T}_{\mu\nu}$ is the energy momentum tensor

Energy momentum tensor

The energy momentum tensor $\mathcal{T}{\mu\nu}$ is defined as $$\mathcal{T}_{\mu\nu} = \sum_n \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_n)}\partial_\nu \phi_n - g_{\mu\nu}\mathcal{L},$$ where $n$ sums over al fields in the Lagrangian. The energy momentum tensor consists of four conserved currents meaning $$\partial^\mu\mathcal{T}_{\mu\nu}=0.$$ The component $\mathcal{T}{00}$ can be identified with the Legendre transfomation of the Lagrangian and therefore corresponds to the Hamiltonian and the Energy density $\mathcal{E}$ respectively.

The conserved charges are $$Q_\nu = \int d^3x \mathcal{T}_{0\nu},$$ which correspond to total Energy (for $\nu=0$ as given above) and momentum repectively (for $\nu=1,2,3$).

As an example for a classical field theory we look at electrodynamics and derive Coulombs law directly from the Lagrangian.

Example: Coulombs law

We start with the Lagrangian for a source current $J_\mu$ $$\mathcal{L} = - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - A^\mu J_\mu.$$ The source is a single charge $e$ at the origin which would be $$J_0(x)=\rho(x)=e\delta^3(x)\qquad J_i(x)=0.$$ The Euler Lagrange equations (Remembering $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$) $$\frac{\partial \mathcal{L}}{\partial A_\alpha} - \partial_\beta \frac{\partial \mathcal{L}}{\partial (\partial_\beta A_\alpha)}= 0$$ lead to the inhomogenous maxwell equations $$\partial_\beta F^{\beta\alpha} = J^\alpha.$$ In Lorentz gaue $\partial_\mu A^\mu = 0$ we can simplify $$\square A_\nu(x) = J_\nu(x).$$ This is the equation of motion for the Field $A^\mu$. To solve it, we can use Fourier transformation in position space with $J_0 = \delta^3, J_i = 0$ (setting $e=1$) $$\square A(x) = \delta^3(x) \Rightarrow \triangle A_0(x) = \delta^3(x)$$ The Fourier transformation of both sides are given by $$\Rightarrow \tilde{\square A(x)} = -|\vec{k}|^2\tilde{A}(k), \quad\tilde{\delta(x)}=1.$$ Therefore we find: $$\Rightarrow \tilde{A}(k) = -\frac{1}{|\vec{k}|^2} \Rightarrow A(x) = \int \frac{d^3k}{(2\pi)^3} \frac{e^{i\vec{k}\vec{x}}}{|\vec{k}|^2}$$

This corresponds to the formal solution of the equation of motion $A_\mu = \frac{1}{\Box}J_\mu$. The quantitiy $\frac{1}{\Box}$ is formally the Propagator of the field.

After switching to spherical coordinates ($r$ the radius in position space $k$ in momentum space) this leads to the solution (using $\vec{k}\vec{x}=kr\cos{\Theta}$ and integrating the angular part) $$A_0(x) = -\int \frac{d^3k}{(2\pi)^3}\frac{1}{|\vec{k}|^2}e^{i\vec{k}\vec{x}}=-\frac{1}{8\pi^2 i r}\int^\infty_{-\infty}dk \frac{e^{ikr}-e^{-ikr}}{k}.$$

Since the expression in the integral is well behaved for $k\to 0$ we can use a trick to compute the integral $$\int^\infty_{-\infty}dk\frac{e^{ikr}-e^{-ikr}}{k} = \lim_{\epsilon \to 0}\left[\int^\infty_{-\infty}dk\frac{e^{ikr}-e^{-ikr}}{k + i\epsilon}\right].$$ This integral can be computed with Cauchy’s integral formula. We split the the numerator. For the $e^{ikr}$-term we choose the complex contour to be the semicircle in the upper plane, since there $k$ has a positive imaginary part which lets the boundary terms become zero for $|k|\to \infty$. Sinze there is no pole in the upper plane this part is zero. For the $e^{-ikr}$-term we choose the lower semicircle where we then have the $-i\epsilon$ pole. Cauchy’s integral formula leads to $$\int^\infty_{-\infty}dk \frac{-e^{-ikr}}{k+i\epsilon} = -(2\pi i)(-e^{-\epsilon r}).$$

Taking $\epsilon\to 0$, we finally obtain the result $$A_0(x) = -\frac{1}{4\pi r}.$$