Notes on Field Theory

18 Oct 2026 PDF

Lorentz Force

\[\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})\]

Q) Is this equation Lorentz invariant?

A) Proof by construction of Lorentz invariant Lagrangian: Assume a 4-vector $A_\mu(t, x)$ that encodes all the information about $\mathbf{E}$ and $\mathbf{B}$ in the following way.

\[\begin{align} F_{\mu \nu} &= \partial_\mu A_\nu -\partial_\nu A_\mu \\ &= \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & B_z & -B_y \\ E_y & -B_z & 0 & B_x \\ E_z & B_y & -B_x & 0 \\ \end{pmatrix} \end{align}\]

Convention: $x$ stands for all the spatial coordinates of $X^\mu$.

The following is a scalar including a product of the particle’s position with the field (hence represents the action of interaction):

\[S_{\mathrm{int}}=e\int dX^\mu A_\mu(t, x)\]

The following is the action for a free particle:

\[S_{\mathrm{part}} = \int -m\sqrt{1-\dot{x}^2}dt\]

We thus add the two together to obtain the total action $S = S_{\mathrm{int}} + S_{\mathrm{part}}$. To find the Lagrangian, we rewrite the integral to factor out $dt$, by which we obtain:

\[\mathcal{L} = -m\sqrt{1-\dot{x}^2} + eA_0(t, x) + e\dot{X^n}A_n(t, x)\]

Using Euler-Lagrange, we indeed obtain the equation for relativistic Lorentz force:

\[m \frac{d}{dt}\frac{\dot{X_p}}{\sqrt{1 - \dot{x}^2}} = e(\mathbf{E} + \mathbf{v} \times \mathbf{B})_p\]

Or more succintly,

\[m\frac{d^2 X_\mu}{d\tau^2} = e F_{\mu\nu} \frac{dX^\nu}{d\tau}\]

Homogeneous Maxwell Equations

Definition.

\[\begin{align} F_{\mu \nu} &= \partial_\mu A_\nu -\partial_\nu A_\mu \\ &= \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & B_z & -B_y \\ E_y & -B_z & 0 & B_x \\ E_z & B_y & -B_x & 0 \\ \end{pmatrix} \end{align}\]

The above definition yields the following mathematical identities.

\[\begin{gather} \nabla \cdot \mathbf{B} = 0 \\ \nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0 \end{gather}\]

Current density 4-vector

\[\rho = \frac{dQ}{dxdydz}, \, j_x = \frac{dQ}{dydzdt}\]

To write succintly,

\[J^\mu = \frac{dQ}{dxdydzdt}dX^\mu\]

We thus have the following local conservation equation:

\[\partial_\mu J^\mu = 0 \quad \left(\text{i.e. } \frac{\partial{\rho}}{\partial{t}} + \nabla \cdot\mathbf{J} = 0 \right)\]

This suggests that the following is a 4-vector.

\[J^\mu = \begin{pmatrix} \rho \\ J_x \\ J_y \\ J_z \end{pmatrix}\]

Indeed, a simple thought experiment proves that $J^\mu$ respects Lorentz transformation, see pg. 412-413 of The Theoretical Minimum: Special Relativity and Field Theory.

Deriving the non-homogeneous Maxwell’s equations

Simple algebra shows that the non-homogeneous equations are equivalent to:

\[\partial_\mu F^{\mu\nu} = -J^\nu\]

To derive this, we will first set up a Lagrangian that is

1. Lorentz invariant
2. Gauge-invariant
3. Local

First, we adopt the following notation:

\[\partial_\mu \phi = \phi_{,\mu}\]

By locality, the Lagrangian should depend only up to first derivative, i.e.

\[\mathcal{L}(\phi, \phi_{,\mu})\]

To respect gauge invariance, the Lagrangian must be invariant under the following transformation:

\[A_\mu \Rightarrow A_\mu + \frac{\partial S}{\partial X^\mu}\]

where $S$ is a scalar field. Since $F^{\mu\nu}$ is gauge invariant, we are free to use these terms in the Lagrangian, but we are prohibited from using such terms as $A_\mu A^\mu$ (although it is a perfectly fine scalar).

Lagrangian when $\mathbf{J} = 0$

The above considerations lead us to postulate the following Lagrangian:

\[\begin{align} \mathcal{L} &= -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \\ &= -\frac{1}{4}(A_{\nu,\mu}-A_{\mu,\nu})(A^{\nu,\mu}-A^{\mu,\nu}) \\ &= \frac{1}{2}(E^2 - B^2) \end{align}\]

To find the eqution of motion, use the Euler-Lagrange equations for field:

\[\frac{\partial}{\partial X^\mu} \frac{\partial \mathcal{L}}{\partial \phi_{,\mu}} = \frac{\partial \mathcal{L}}{\partial \phi}\]

Note that each $\phi$ is a field; since we are dealing with four fields $A_t, A_x, A_y, A_z$, we will obtain a system of four equations, each of them being:

\[\frac{\partial}{\partial X^\nu} \frac{\partial \mathcal{L}}{\partial A_{\mu,\nu}} = \frac{\partial \mathcal{L}}{\partial A_\mu}\]

Calculating the terms yield:

\[\frac{\partial F^{\mu\nu}}{\partial X^\nu} = 0\]

i.e.

\[\partial_\mu F^{\mu\nu} = 0 = J_\mu.\]

Lagrangian when $\mathbf{J} \neq 0$

Q) How should we add $\mathbf{J}$ to the Lagrangian?

A) Although $A^\mu A_\mu$ is not Gauge invariant, $J^\mu A_\mu$ is! Let’s check this fact.

Let the new action be:

\[\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \underbrace{- J^\mu(x)A_\mu(x)}_{\text{new term}}\]

The gauge transform is: \(\mathcal{L}' = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - J^\mu(x)\left( A_\mu(x) + \frac{\partial S}{\partial X^\mu} \right)\)

Thus,

\[\begin{align} \Delta S &= \int -J^\mu(x)\frac{\partial S}{\partial X^\mu} d^4x \\ &= \int -\frac{J^\mu(x)}{\partial X^\mu} S \,d^4x \quad(\because \text{partial integral}) \\ &= \int -(\partial_\mu J^\mu)S \,d^4x \\ &= 0 \quad (\because \text{continuity equation}) \end{align}\]

Thus, $\mathcal{L}$ is Gauge invariant. Let us now extract the equation of motion:

\[\frac{\partial}{\partial X^\nu} \frac{\partial \mathcal{L}}{\partial A_{\mu,\nu}} = \frac{\partial \mathcal{L}}{\partial A_\mu}\]

Since the new term only explicitly depends on $A_\mu$, its effect on LHS is nil. Calculating the RHS, we obtain:

\[\frac{\partial \mathcal{L}}{\partial A_\mu} = -J^\mu\]

Hence,

\[\partial_\mu F^{\mu\nu} = -J^\nu \qquad \blacksquare\]