Classical field theory
Field theory is theory with continous degrees of freedom
# Hamiltonians and Lagrangians
Hamiltonian and Lagrangian defined as integrals over densities $$H = \int d^3x \mathcal{H}, L = \int d^3x\mathcal{L}.$$ The Hamiltonian depends on the fields $\phi$ and their conjugate momenta $\pi$.
The Lagrangian is the Legendre transformation of the Hamiltonian $$\mathcal{L}[\phi,\dot{\phi}] = \pi[\phi,\dot{\phi}]\phi - \mathcal{H}[\phi,\pi[\phi,\dot{\phi}]],$$ where $\pi[\phi,\dot{\phi}]$ is defined by $\frac{\partial\mathcal{H}[\phi,\pi]}{\partial\pi}=\dot{\phi}$. The inverse transformation is defined accordintly
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 one derivative to be dynamical
- Interaction terms have three or more fields
- We also write interaction terms as $$\mathcal{L}\text{int} = -\mathcal{V} = -\mathcal{H}\text{int}.$$
The hamiltonian is not lorentz invariant, as the energy is not a lorentz scalar.
# Euler-lagrange equations
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 $$\delta S = 0,$$ from which we can derive the Euler-lagrange equations using partial integration and the assumption that the fields are zero at infinities $$\delta S = \int d^4x \left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right]\delta\phi.$$
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
# Noether’s theorem
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 is given by $$J_\mu = \sum_n \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_n)}\frac{\delta\phi_n}{\delta\alpha},$$ where we sum over all fields the symmetry applies to. $J_\mu$ is known as Noether current.
Corresponding to the conserved current, there is a conserved charge $Q = \int d^3x J_0$ with $\partial_tQ=0$.
# Energy momentum Tensor
Noethers theorem applied to translational invariance leads to the conserved current the Energy Momentum tensor $\mathcal{T}_{\mu\nu}$. 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 $\xi^\mu$ $$\phi(x+\xi) = \pi(x) + \xi^\nu \partial_\nu \phi(x) + …$$ 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 $$\partial^\mu \mathcal{T}{\mu\nu} = 0,$$ with the energy momentum tensor $$\mathcal{T}{\mu\nu} = \sum_n \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_n)}\partial_\nu \phi_n - g_{\mu\nu}\mathcal{L},$$ where $n$ sums over all fields. This corresponds to the conservation of four noether currents, namely energy and momentum. Integrating over spacial dimensions gives the conservation of total energy and momentum over time. The component $\mathcal{T}_{00}$ is the energy density which becomes the legendre transformation of the lagrangian $\mathcal{L}$ and therefore the hamiltonian.
# Coulombs law
Example: Maxwell equations
As an example we derive the inhomogenous Maxwells equations for a source current $J_\mu$ from the lagrangian $$\mathcal{L} = - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - A^\mu J_\mu.$$ The Euler Lagrange equations $$\frac{\partial \mathcal{L}}{\partial A_\alpha} - \partial_\beta \frac{\partial \mathcal{L}}{\partial (\partial_\beta A_\alpha)}= 0$$ 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 equation can be solved with Fourier transformation in position space with $j_0 = \delta^3, j_i = 0$ $$\square A(x) = \delta^3(x) \Rightarrow \triangle A_0(x) = \delta^3(x)$$ $$\Rightarrow \tilde{[\square A(x)]} = -|\vec{k}|^2\tilde{A}(k)$$ $$\Rightarrow \tilde{A}(k) = -\frac{1}{|\vec{k}|^2} \Rightarrow A(x) = \int \frac{d^3k}{(2\pi)^3} \frac{e^{i\vec{k}\vec{x}}}{|\vec{k}|^2}$$
After switching to spherical coordinates ($r$ the radius in position space $k$ in momentum space) this leads to the solution $$A_0(x) = -\int \frac{d^3k}{(2\pi)^3}\frac{1}{|\vec{k}|^2}e^{i\vec{k}\vec{x}}=-\frac{1}{8\pi^2 i r}\int^\infty_{-\infty}dk \frac{e^{ikr}-e^{-ikr}}{k}.$$
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 $$\int^\infty_{-\infty}dk \frac{-e^{-ikr}}{k+i\epsilon} = -(2\pi i)(-e^{-\epsilon r}).$$
Finally we obtain the result $$A_0(x) = -\frac{1}{4\pi r}.$$