Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Page 580, Chapter 12 in Jackson's 3rd edition text carries the statement:

From the first postulate of special relativity the action integral must be a Lorentz scalar because the equations of motion are determined by the extemum condition, $\delta A = 0$

Certainly the extremeum condition must be an invariant for the equation of motion between $t_1$ and $t_2$, whereas I don't see how the action integral must be a Lorentz scalar. Using basic classical mechanics as a guide, the action for a free particle isn't a Galilean scalar but still gives the correct equations of motion.

share|cite|improve this question
You may find this helpful: – Ben Crowell Aug 14 '11 at 20:06
up vote 3 down vote accepted

First, observe that although the non-relativistic Lagrangian is not invariant. It changes by a total derivative, thus the equations of motions remain invariant. The reason of the difference between the Lorentzian and the Galilean cases is that the group action of the Lorentz group on the classical variables (positions and momenta) is a by means of a true representation, while in the case of the Galilean group the representation is projective. In the Language of geometric quantization, $exp(i \frac{S}{\hbar})$, where $S$ is the action is a section in $L \otimes \bar{L}$, where $L$ is the prequantization line bundle and $\bar{L}$ its dual. In other words, the action needs not be a scalar, only an exprssion of the form: $\bar{\psi}(t_2)exp(i \frac{S(t_1, t_2)}{\hbar})\psi(t_1)$, where $\psi(t)$ is the wavefunction at time $t$ and $S(t_1, t_2)$ is the classical action between $t_1$ and $t_2$. The reason that the representation in the Galilean case is projective is related to the nontriviality of the cohomology group $H^2(G, U(1))$ in the Galilean case in contrast to the Lorentz case. I have given a more detailed answer on a very similar subject in my answer to Anirbit: Poincare group vs Galilean group and in the comments therein.

share|cite|improve this answer

Yest, it must. It does not guarantee that the equations have physical exact solutions but at least everything looks relativistic.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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