From definition of Lagrangian: $L = T - U$. As I understand for free particle ($U = 0$) one should write $L = T$.

In special relativity we want Lorentz-invariant action thus we define free-particle Lagrangian as follows:

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

At the same point, we have that definition of 4-momentum implies the kinetic energy is: $$T = (\gamma - 1) m c^2.$$

As you might guess, 1) question is how to relate all these formulas?

2) I do not understand why there is no $1/\gamma$ near $U$ in relativistic Lagrangian?

3) What is meaning of the first term in $L$ for relativistic case?


2 Answers 2


It helps to write the full action: $$S = \int \frac{-mc^2}{\gamma}dt - \int U dt $$

The first term can be put in a much better form by noting that $d\tau = \frac{dt}{\gamma}$ represents the proper time for the particle. The action is then: $$S = -mc^2\int d\tau - \int U dt$$ The first term is Lorentz invariant, being only the distance between two points given by the Minkowski metric, and is good in relativity. The second term however, isn't (assuming that $U$ is a scalar); there is no way it can be a relativistic action.

There are two easy ways out:

  1. The first is simply to change the term to $\frac{U}{\gamma}$. This gives the action: $$S = -\int (mc^2+U)d\tau$$
  2. The second is to "promote" the term (a terminology used in Zee's Einstein Gravity in a Nutshell) to a relativistic dot product, giving the action: $$S = -mc^2\int d\tau - \int U_\mu dx^{\mu}$$

The former has no real world classical analog (that I know of), and the latter is more or less the interaction of a particle with a static electromagnetic field. But the original form is recovered from the latter when the spatial components of $U_\mu$ vanish, leaving only $U_0$.

The kinetic energy is obtained by transforming the Lagrangian to the Hamiltonian (see here).


A very informal approach would be to understand how the mathematics develops: since 'Action' in the Lagrangian sense is never a vector, it must be a scalar. It is in this case the energy. From special relativity we have the postulate that the laws of physics are the same for all observers in all inertial reference frames. Hence: $$ L\gamma = -mc^2 $$ $$or, L=-mc^2/\gamma $$

We understand that for any coordinate the following quantity represents the generalised momentum: $$p_i=\delta L/\delta\dot q_i=-mc^2\delta\sqrt{1-\beta^2}/\delta v =v\gamma mc^2/c^2 =\gamma mv$$(Consistent with relativistic momentum)

Therefore the final expression for the Lagrangian can be expected to take the form: $${\mathcal{L}}=-mc^2/\gamma -V(x)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.