# What is the full QED Lagrangian with physics units written out?

I wonder what the QED Lagrangian would look like if you carefully write out all units of the terms and make sure they are consistent. I think there is something missing about Coulomb.

Can you write down the QED Lagrangian such that the physical units of all terms agree and state the units of all variables and terms?

• what is tripping you up here? This is a simple dimensional analysis problem. Mar 18 at 12:58
• I haven't found any reference on the internet and I cannot get the units right myself (for $\psi$, $A$, $F$ etc). Mainly the electric charge is unclear.
– Gere
Mar 18 at 12:59
• the electrical charge would need to be included in more terms I’ve never seen $e$ omitted; it’s $\hbar$ and $c$ that get set to 1. Mar 18 at 16:36
• The Minkowski metric is dimensionless. It’s the diagonal matrix (-1,1,1,1). Mar 18 at 18:50
• People downvoting not because it's a bad question, but because they think you are too lazy to solve it by yourself or look for Griffiths' book on particle physics. Mar 18 at 19:24

The Wikipedia article you linked to gives the action, not the Lagrangian or Lagrangian density, so I’ll do the same.

Note that since $$x^0 \equiv ct$$, we have $$d^4x = c \, dt \, d^3x$$ and therefore

$$S \equiv \int dt \, L \equiv \int dt \int d^3x \, \mathscr L = \int \frac{d^4x}{c} \mathscr L$$

where $$L$$ is the Lagrangian and $$\mathscr L$$ is the Lagrangian density. Thus $$\mathscr L$$ is the quantity inside the square brackets below, and $$L$$ is the integral of this density over the spatial coordinates.

$$S$$ must have units of action, which is the same as energy times time. The Lagrangian $$L$$ therefore must have units of energy, and the Lagrangian density $$\mathscr L$$ must have units of energy density.

The units of electromagnetic quantities such as $$e$$, $$\phi$$, $$\mathbf A$$, $$\mathbf E$$, $$\mathbf B$$, $$A^\mu$$, and $$F^{\mu\nu}$$ are the same in QED as in classical electromagnetism. Both the classical theory and the quantum theory can be done in various unit systems. I discuss four of them below.

The Minkowski metric $$\eta_{\mu\nu}$$ is dimensionless in Cartesian coordinates, and few people do QED calculations in anything other than Cartesian coordinates. In any case, since $$\mathscr L$$ must be a Lorentz scalar, if $$\mathscr L$$ has the right dimensions in Cartesian coordinates, it will have the right dimensions in any coordinates.

# SI units

$$S = \int\frac{d^4x}{c}\left[ -\frac{1}{4\mu_0}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \hbar c \gamma^\mu D_\mu - m c^2) \psi \right]$$

$$D_\mu \equiv \partial_\mu + \frac{ie}{\hbar}A_\mu$$

$$x^\mu \equiv (ct, \mathbf x)$$

$$A^\mu \equiv \left( \frac1c \phi, \mathbf A \right)$$

Base units: kilograms, meters, seconds, amperes

Table of $$\text{[quantity]} = \text{kg}^a \text{m}^b \text{s}^c \text{A}^d$$

$$\begin{array}{c|c|c|c|} \text{quantity} & \text{kg} & \text{m} & \text{s} & \text{A} \\ \hline S & 1 & 2 & -1 & 0 \\ \hline d^4x & 0 & 4 & 0 & 0 \\ \hline c & 0 & 1 & -1 & 0 \\ \hline \mu_0 & 1 & 1 & -2 & -2 \\ \hline F^{\mu\nu} \text{ or } F_{\mu\nu} & 1 & 0 & -2 & -1 \\ \hline\ \psi \text{ or } \bar\psi & 0 & -3/2 & 0 & 0 \\ \hline \hbar & 1 & 2 & -1 & 0 \\ \hline \gamma^\mu \text{ or } \gamma_\mu & 0 & 0 & 0 & 0 \\ \hline m & 1 & 0 & 0 & 0 \\ \hline \partial_\mu \text{ or } \partial^\mu & 0 & -1 & 0 & 0 \\ \hline e & 0 & 0 & 1 & 1 \\ \hline A^\mu \text{ or } A_\mu & 1 & 1 & -2 & -1 \\ \hline g_{\mu\nu} \text{ or } g^{\mu\nu} & 0 & 0 & 0 & 0 \\ \hline \end{array}$$

# Gaussian units

$$S = \int\frac{d^4x}{c}\left[ -\frac{1}{16\pi}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \hbar c \gamma^\mu D_\mu - m c^2) \psi \right]$$

$$D_\mu \equiv \partial_\mu + \frac{ie}{\hbar c}A_\mu$$

$$x^\mu \equiv (ct, \mathbf x)$$

$$A^\mu \equiv \left(\phi, \mathbf A \right)$$

Note: These are sometimes called "unrationalized" Gaussian units. The $$F^2$$ term has an extra factor of $$1/4\pi$$; this introduces factors of $$4\pi$$ into the field equations but removes them from solutions. For example, in these units the classical potential of a point charge is simply $$q/r$$.

In this system, $$\mu_0$$ and $$\epsilon_0$$ do not exist. This is because Gaussian units don't have a electromagnetic base unit; charge is defined in terms of the mechanical base units.

Base units: grams, centimeters, seconds

Table of $$\text{[quantity]} = \text{g}^a \text{cm}^b \text{s}^c$$

$$\begin{array}{c|c|c|c|} \text{quantity} & \text{g} & \text{cm} & \text{s} \\ \hline S & 1 & 2 & -1 \\ \hline d^4x & 0 & 4 & 0 \\ \hline c & 0 & 1 & -1 \\ \hline F^{\mu\nu} \text{ or } F_{\mu\nu} & 1/2 & -1/2 & -1 \\ \hline\ \psi \text{ or } \bar\psi & 0 & -3/2 & 0 \\ \hline \hbar & 1 & 2 & -1 \\ \hline \gamma^\mu \text{ or } \gamma_\mu & 0 & 0 & 0 \\ \hline m & 1 & 0 & 0 \\ \hline \partial_\mu \text{ or } \partial^\mu & 0 & -1 & 0 \\ \hline e & 1/2 & 3/2 & -1 \\ \hline A^\mu \text{ or } A_\mu & 1/2 & 1/2 & -1 \\ \hline g_{\mu\nu} \text{ or } g^{\mu\nu} & 0 & 0 & 0 \\ \hline \end{array}$$

# Heaviside-Lorentz units

$$S = \int\frac{d^4x}{c}\left[ -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \hbar c \gamma^\mu D_\mu - m c^2) \psi \right]$$

The rest is the same as for Gaussian units.

These are sometimes called "rationalized" Gaussian units. The $$F^2$$ term has a prefactor of $$-1/4$$ instead of $$-1/16\pi$$. This removes factors of $$4\pi$$ from the field equations.

# Particle physics units

$$S = \int d^4x \left[ -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \gamma^\mu D_\mu - m) \psi \right]$$

$$D_\mu \equiv \partial_\mu + ieA_\mu$$

$$x^\mu \equiv (t, \mathbf x)$$

$$A^\mu \equiv \left(\phi, \mathbf A \right)$$

Base units: GeV

Table of $$\text{[quantity]} = \text{GeV}^a$$

$$\begin{array}{c|c|} \text{quantity} & \text{GeV} \\ \hline c\equiv 1 & 0 \\ \hline \hbar\equiv 1 & 0 \\ \hline S & 0 \\ \hline d^4x & -4\\ \hline F^{\mu\nu} \text{ or } F_{\mu\nu} & 2 \\ \hline\ \psi \text{ or } \bar\psi & 3/2 \\ \hline \gamma^\mu \text{ or } \gamma_\mu & 0\\ \hline m & 1 \\ \hline \partial_\mu \text{ or } \partial^\mu & 1 \\ \hline e & 0 \\ \hline A^\mu \text{ or } A_\mu & 1 \\ \hline g_{\mu\nu} \text{ or } g^{\mu\nu} & 0\\ \hline \end{array}$$

• Wow, thanks! That's more extensive than I wished for. My main stumbling block was $\mu_0$ which did not appear in the references I found. I don't have all books at hand. Do you know which book would mention $\mu_0$?
– Gere
Mar 19 at 8:03
• I don't understand how $\mu_0$ can get dropped in other units though. For that you would need to fix units such that $\varepsilon_0=1$? And then $e$ isn't 1 anymore? Maybe I'm lost. Do you know a reference discussing how to make $\mu_0=1$?
– Gere
Mar 19 at 8:07
• Do you know which book would mention $\mu_0$? Wikipedia’s article on the covariant formulation of classical electromagnetism has it in $\mathscr L$. Mar 19 at 17:16
• This article explains how charge can be defined in terms of mass, length, and time. A unit of charge defined this way is such that if you put it on each of two point particles separated by one unit of distance you get a repulsion of one unit of force. Mar 19 at 17:24
• This makes Coulomb’s Law be $F=q_1q_2/r^2$, so there is no silly $1/4\pi\epsilon_0$ as in SI. Similarly, there is no need for a $\mu_0$ in the force between two currents. Mar 19 at 17:26