Let be $u=|\bar{u}|$ the speed of a free particle (at constant speed) of mass $m$ that is moving in relation to an inertial frame. Why we initially introduce the term $\epsilon$ to the free lagrangian

\begin{equation}\tag{1} L_\mathbf{free}=\epsilon\sqrt{1-\frac{u^{2}}{c^{2}}}=\epsilon \sqrt{1-\frac{\{[x'(t)]^{2}+[y'(t)]^{2}+[z'(t)]^{2}\}}{c^{2}}} \end{equation}

and then to prove that $\epsilon$ it is really $-mc^2$?

Probable references:

  1. Classical electromagnetic radiation 3rd, Marion (english version), pag. 516$\to$ 522.
  2. The classical theory of fields - 4th revised, English Edition, Landau-Lifshitz - Chapter 2 pag. 26 § 8 (*).
  3. Classical electrodynamics, Jackson, 3rd edition, pagg. 579-580-581-582 (english version), chapter 12, (formula (12.7) for the formula $(2)$).

  1. Classical electrodynamics, Jackson, pagg. 567-568 (italian version).
  2. Goldstein pag. 300 (italian version) chapter 7, eq.: 7.141.

1 Answer 1


This is partly a repetition of the argument presented in an earlier question of yours and in the chat therein.

Lagrangians are an exercise in repetition. You cannot derive a Lagrangian per se. You already have to know the answer.

  • step 1:
    You start with the equations of motion of a system (be electrodynamics, classical mechanics, QED etc.) that you know from other methods.

  • step 2:
    You know the Euler-Lagrange equation, that needs a Lagrangian $\mathcal{L}$ as input and that gives the equation of motion as output: $$ \frac{\partial\mathcal{L}}{\partial q} - \frac{\mathrm{d}}{\mathrm{d}t}\left (\frac{\partial \mathcal{L}}{\partial \dot{q}}\right ) = 0, $$where $q$ is your variable of choice (e.g. position $x$), and $p = \partial \mathcal{L}/\partial \dot{q}$ is the conjugate momentum.

  • step 3:
    You invent a Lagrangian $\mathcal{L}$ that, when put into E-L, gives you back your equation of motion.

  • step 4: So what's the point of re-writing known equations of motion in Lagrangian formalism? It's more compact, it's a scalar (does not require immediate choice of basis), it's more elegant, manifestly symmetric, easier to implement constraints.

In your specific case at hand, relativistic mechanics.

You already know that the momentum $\mathbf{p}$ is given by: $$ \mathbf{p} = \gamma_{\mathbf{u}} m_0 \mathbf{u}, $$ where $m_0$ is the rest mass and $\gamma_{\mathbf{u}} = 1/\sqrt{1-u^2/c^2}$, and the energy $E$ (or the Hamiltonian $H$, same thing) is given by: $$ E = H = \gamma_{\mathbf{u}}m_0 c^2. $$

Then you make stuff up until you notice that this specific choice: $$ \mathcal{L} = -\frac{m_0 c^2}{\gamma_{\mathbf{u}}}$$ gives you the correct answers.


  • momentum: $$ \mathbf{p} = \frac{\partial \mathcal{L}}{\partial \dot{\mathbf{r}}} = \frac{\partial \mathcal{L}}{\partial \mathbf{u}} = \gamma_{\mathbf{u}}m_0 \mathbf{u},$$


  • energy: $$ H = \mathbf{p}\cdot \dot{\mathbf{r}} - \mathcal{L} = \gamma_{\mathbf{u}}m_0 c^2.$$

In all of the above I assumed zero external potential(s) $V(\mathbf{r}, \dot{\mathbf{r}})$.


The OP seems to be wanting a "proof" for his factor of $\epsilon$.

So let's do that.

Let's "guess" a lagrangian of the form: $\mathcal{L} = \epsilon / \gamma_{\mathbf{u}} = \epsilon \sqrt{1-\frac{u^2}{c^2}}$ like in your question.

Let's calculate $\left (\frac{\partial \mathcal{L}}{\partial \dot{x}}\right ) = \left (\frac{\partial \mathcal{L}}{\partial u_x}\right ) = -\epsilon\frac{u_x/c^2}{\sqrt{1-u_x^2/c^2}} = -\epsilon \gamma u_x/c^2$. Where I'm doing it in 1D for clarity.

This has to be equal to the momentum $p_x = \gamma m_0 u_x$, because we know it from relativistic mechanics.

Equate the two: $$-\epsilon \gamma u_x/c^2 = \gamma m_0 u_x \quad \Rightarrow \quad \epsilon = -m_0 c^2 !$$

  • $\begingroup$ Kindest then my question not is very clear. :-( My question is: why do I have to introduce the term $\epsilon$ for the free lagrangian and then prove that $\epsilon=-mc^2$? $\endgroup$
    – Sebastiano
    Commented Sep 14, 2019 at 19:02
  • 1
    $\begingroup$ You don't have to do anything. You're just trying to guess what specific Lagrangian gives you the right answer. The way presented in your book is just another way of doing it, which happens to work (even though it is not guaranteed to). They're guessing it depends on the gamma factor $\gamma$ and on some proportionality constant $\epsilon$. You then have to figure out what $\epsilon$ is in order to give you the correct answers (like at the end of my answer). I'll edit the answer to try and address your concern now. $\endgroup$ Commented Sep 14, 2019 at 19:05
  • $\begingroup$ Now it is ok :-) is perfect! :-) My sincere grazieeeeeeeeeeeeeeeeeeeeee. $\endgroup$
    – Sebastiano
    Commented Sep 14, 2019 at 19:25
  • $\begingroup$ Out of curiosity, what was the reason behind the "-1" that someone gave this answer? $\endgroup$ Commented Sep 14, 2019 at 21:34
  • $\begingroup$ “Private” as in a chat room? Which is still public. $\endgroup$ Commented Sep 14, 2019 at 21:41

Your Answer

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

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