# relativistic Lagrangian

The relativistic kinetic energy $T_{\operatorname{rel}}$ for a free particle of (rest) mass $m$ is

$T_{\operatorname{rel}} = (\gamma - 1) m c^2 \quad \gamma = (1 - v^2 / c^2)^{- 1 / 2} \quad v = | \dot{\mathbf{r}} |$

That relation can be derived by considering the linear momentum being $\mathbf{p}= \gamma m \dot{\mathbf{r}}$. Then

$T = \int \dot{\mathbf{r}} d\mathbf{p}= \cdots = \gamma m c^2 - E_{0} \quad T (0) = 0 \Rightarrow E_{0} = m c^2$

For $\epsilon \ll 1$, it’s $(1 + \epsilon)^n \approx 1 + n \epsilon$. Utilizing this relation, for $v \ll c$, the $\gamma$ is

$\gamma = (1 - v^2 / c^2)^{- 1 / 2} \approx 1 + (1 / 2) (v^2 / c^2)$

Substituting in the first, the relativistic kinetic energy for small velocities is

$T_{\operatorname{rel}; u \ll c} \approx \frac{1}{2} m v^2 = T_{\operatorname{nonrel}}$

In the classical (nonrelativistic) framework the Lagrangian is formed

$L = T - V$

This isn’t the case in the relativistic framework. By integrating the generalized momenta

$p_{i} = \frac{\partial L}{\partial \dot{q}_{i}} = \gamma m \dot{q}_{i} \Rightarrow L = - \frac{m c^2}{\gamma} + \sum_{k} f_{k} (\dot{q}_{j \neq i})$

where $f_{k}$ are arbitrary functions of velocities. By assuming, without loss of generality, $f_{k} = 0$, the relativistic Lagrangian for a free particle is written

$L = - m c^2 \gamma^{- 1}$

For comparison, doing this in the classical framework will, as expected, recover the kinetic energy.

$p_{i} = \frac{\partial L}{\partial \dot{q}_{i}} = m \dot{q}_{i} \Rightarrow L = \frac{1}{2} m \dot{q}_{i}^2 + \sum_{k} f_{k} (\dot{q}_{j \neq i}) \Rightarrow{f_{k} = 0} L = T$

A rigorous derivation, also known as Lanczos’ parametric approach, can be done taking a Lorentz invariant action and a covariant Lagrangian (Greiner 2010; Cline 2019). Another approach, first shown in (Pars 1965), is utilizing d’Alembert’s principle (Nađđerđ et al 2014).

Expanding the relativistic relation, for small velocities, it takes the form

$L \approx - m c^2 + \frac{1}{2} m v^2 = - E_{0} + T_{\operatorname{nonrel}}$

For a particle in potential field, the relativistic Lagrangian is similar to the classical form. That is the potential energy is subtracted from a kinetic term that is the Lagrangian for a free particle.

$L = - m c^2 \gamma^{- 1} - V$

## Bibliography

• W. Greiner. Classical Mechanics: Systems of Particles and Hamiltonian Dynamics. Springer, 2010.

• D. Cline. Variational Principles in Classical Mechanics: Revised Second Edition. River Campus Libraries, 2019. [URL]

• L.A. Pars. A Treatise on Analytical Dynamics. Wiley, 1965. [URL]

• Laslo J Nađđerđ, Miloš D Davidović, and Dragomir M Davidović. American Journal of Physics, 82(11):1083–1086, 2014.