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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?

Qmechanic
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A wing on a fly
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2 Answers2

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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).

AV23
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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: $${\mathcal{L}}=-mc^2/\gamma -V(x)$$

Khandaker B. Ahmed
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