Lagrangian for a relativistic free particle can be written as

$$L=-m_0c^2\sqrt{1-\frac{v^2}{c^2}} .\tag{1}$$

It gives correct expression of Hamiltonian which is

$$H=\sqrt{p^2 c^2+m_0^2c^4}.\tag{2}$$

During a lecture today, my professor told me that Lagrangian for a free particle does not have a specific form and can be written in many ways. Then, he wrote another expression of Lagrangian that can also represent a relativistic free particle

$$L=p^{\mu}p_{\mu}+ m^2.\tag{3}$$

My Questions are: What are the other forms that We can write our Lagrangian in? Does It have something to do with Lorentz invariance of the Lagrangian? How does above expression for Lagrangian can give us the correct equation of motion and Hamiltonian?

  • 1
    $\begingroup$ Comment to eq. (3) in the post (v2): Lagrangian depends on momenta? $\endgroup$
    – Qmechanic
    Oct 14 '17 at 16:12
  • $\begingroup$ This is confusing to me also. But that's what he wrote. $\endgroup$
    – physics101
    Oct 14 '17 at 16:17

It is impossible to give an exhaustive list$^1$ of Lagrangians for a relativistic point particle, but most of the physically relevant ones can be reduced from the following "master" Lagrangian$^2$

$$\begin{align}L~:=~&\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}, \cr \dot{x}^2~:=~&g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0, \cr \dot{x}^{\mu}~:=~& \frac{dx^{\mu}}{d\tau}, \end{align}\tag{A}$$

where $e=e(\tau)$ is an einbein field, and $\tau$ is a world-line parameter (not necessarily proper time).

There is a corresponding Hamiltonian formulation as well, cf. e.g. this Phys.SE post.

The point is that this Lagrangian (A) is world-line reparametrization covariant, cf. e.g. this Phys.SE post. By fixing this gauge symmetry in various gauges, and perhaps integrating out some of the fields, the Lagrangian and Hamiltonian can take on many different appearances, cf. e.g. this Phys.SE post.


  1. If we integrating out the einbein $e$ in the Lagrangian (A), and go to the static gauge $x^0=\tau$, we get OP's square root Lagrangian (1).

  2. We could get a non-square root Lagrangian $\frac{\dot{x}^2}{2}-\frac{ m^2}{2}$ by going to the gauge $e=1$, cf. e.g. this Phys.SE post. And so forth.


$^1$ This non-relativistic example should serve as a warning against trying to make a complete classification.

$^2$ We use the Minkowski sign convention $(−,+,+,+)$ and set the speed of light $c=1$.


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