Recently, I have been looking at the Lagrange equations that describe particles of different spin. I know lagrange takes the form $L=$kinetic-potential, and it seems these equations do take this form. But I dont see how they correspond to the kinetic or potential. How are these equations derived?

The Klein-Gordon equation spin 0 $$L=c^2\partial _\lambda \phi \partial^\lambda \phi ^* -\left(\frac{mc}{\hbar}\right)^2\phi \phi^*$$

Dirac Lagrangian spin 1/2 $$L=i\hbar c \tilde\psi \gamma ^\mu \partial _\mu \psi-mc^2\tilde\psi \psi $$

The Proca equation spin 1 $$L=-\frac{1}{16 \pi}F^{\mu \nu} F_{\mu \nu} +\frac{1}{8 \pi} \left(\frac{mc}{\hbar}\right)^2A^\mu A_\mu$$

I suppose I mean to say, is there one equation that you can get all three of these equations from?

  • $\begingroup$ Can you clarify what you want to derive them from? Often, we use a Lorentz-invariant lagrangian as the starting point for specifying a model, and then we deduce the rest of the model's properties (like the spins of its particles) from that. But sometimes we go the other way, too: we ask "what's the simplest Lorentz-invariant lagrangian I could start with if I want to end up with a particle of spin $j$?" Are you asking how we can find that simplest lagrangian without any guesswork? $\endgroup$ Commented May 6, 2021 at 0:59
  • 1
    $\begingroup$ I dont have a good enough background in this yet to even know where these equations come from. Im asking what is a method that we can use to derive these equations. Where do they come from? It sounds like the starting point is the Lorentz-invariant lagrangian. So can someone show me how to derive those three equations from it. $\endgroup$
    – David
    Commented May 6, 2021 at 1:48
  • $\begingroup$ Okay, I think that's sufficient clarification. The three quantities you wrote are all Lorentz-invariant lagrangians, so you're asking how to come up with a Lorentz-invariant lagrangian that correponds to a (non-interacting) particle with a given spin. It's a good question. $\endgroup$ Commented May 6, 2021 at 1:53
  • $\begingroup$ Your lagrangians are just kinetic parts, there is no porential here. $\endgroup$ Commented May 6, 2021 at 7:47
  • $\begingroup$ How do you know this? Also how are these Lagrangian just the kinetic part? $\endgroup$
    – David
    Commented May 6, 2021 at 8:02

1 Answer 1


As mentioned in the comment by @Chiral Anomaly, these are the Lagrangians for the particles of these spins. But it is seem like you need a bit more than that.

We require our theories to be invariant under Lorentz transformations. We know that from special relativity. Now the first question is: what particles are "compatible" with this symmetry. To answer that we need to look at the irreducible representations of the Lorentz group. These essentially describe "objects" that transform into one another under Lorentz transformations, i.e. they don't mix up with other such objects. These irreducible representations are characterised by something called spin. Spin 0 is a scalar field (it doesn't transform under a Lorentz transformation); spin 1 is a vector (it transforms as a vector). Now it turns out there are also irreducible representations with half-integer spin: spin 1/2 describes a fermion, spin 3/2 etc.

Now that we have these particles, let us ask what theory can we write for these. This means what Lorentz invariant Lagrangians can we write with these fields. The first step on this is to focus on what we call free Lagrangians. These are Lagrangians that are at most quadratic in the fields. Why do we first look at that? Because in the quantum theory we need to calculate the a path integral, which is essentially an integral of $e^{i\int L_{\text{free}}}$ where $L_{\text{free}}$ is the Lagrangian. Now if $L_{\text{free}}$ is quadratic these are just Gaussian integrals and we can solve this exactly. The Lagrangians you mention are the quadratic Lorentz invariant Lagrangians for spin 0,1/2 and 1.


  1. As they are quadratic you can view them as being only the kinetic part (in classical physics the kinetic part is quadratic too). But unfortunately free fields are not very interesting, as these particles don't interact with one another. To make the theory interesting you need terms that mix up these fields and are higher than quadratic. Just as a potential energy in classical physics. You can now write the full Lagrangian as $L_{\text{free}}+ L_{\text{interaction}}$ and you can consider the interaction as a correction to the free theory and perform perturbation theory.
  2. The spin one Lagrangian you mention is not the one we know describes spin one particles. Indeed we know from experiment that the spin one particles (photons) are massless so the $A^\mu A_\mu$ term is not there.
  3. I have simplified many things and in doing so I will probably attract many comments/corrections. But this is really the gist of where these equations come from.

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