In Goldstein’s Classical Mechanics (3rd edition), section 7.10 focuses on covariant lagrangians for point particles. Here, we begin by stating

$$L=-mc\sqrt{x’_{\nu} x’^{\nu}}, \tag{7.162}$$

with sign convention $(+,-,-,-)$. But later we note that other lagrangians work, such as

$$L=\frac{1}{2} m u_{\nu} u^{\nu}. \tag{7.164}$$

Or, more generally,

$$L=m f\left(u_{\nu} u^{\nu}\right).\tag{footnote}$$

For any function $f(y)$ whose derivative, evaluated at $y=c^2$, equals $1/2$. Under this generalization, 7.164 corresponds to $f(y)=y/2$ which indeed has the desired derivative. It is stated in the footnote at the bottom of page 322 that 7.162 corresponds to $f(y)=-c \sqrt{y}$, which it clearly must for $L=m f(u_{\nu} u^{\nu})$ to equal 7.162, but the derivative seems to be wrong. Explicitly,

$$\frac{d f(y)}{dy} =-\frac{1}{2}c ~ y^{-1/2}.$$

And thus the derivative, evaluated at $y=c^2$, is $-1/2$, not $+1/2$ as is stated in the footnote. Is there an error here? Regardless of what is written, what constraint must we apply on the derivative of such functions $f$ to describe the motion of relativistic point particles? Should it be $+1/2$ or should it be $-1/2$?

I have checked if this was listed errata but it did not seem to be so. Additionally, I checked the 2nd edition to see if this seeming error was there, and it was in fact not. There seems to be a significant change between the 2nd and 3rd editions [the former used $(-,+,+,+)$ convention while the latter uses $(+,-,-,-)$ convention] but, even in light of this change, the problem persists.


A related question: immediately following 7.164, Goldstein states that he’ll use 7.162 for the “kinetic energy” part of the lagrangian in following discussions, but he, like many other authors, seems to instead use 7.164, not 7.162. Is this likely a typo? I couldn’t find it in the errata.


1 Answer 1

  1. For what it's worth, one may argue that already the Lagrangian (7.164) has the wrong sign in the 3rd edition of Goldstein. Of course, an overall sign in the Lagrangian does not affect the Euler-Lagrange equations, but traditionally the sign is chosen to make the non-relativistic limit $\frac{1}{2}m{\bf v}^2$ appear with a plus sign. Note that this sign has further repercussions for the remainder of the section in Goldstein.

  2. It seems relevant to mention that the Lagrangian for a relativistic point particle is often taken to be $$ L~:=~\pm\frac{\dot{x}^2}{2e}-\frac{e (mc)^2}{2},\qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu},\tag{A}$$ where the Minkowski signature is $(\mp,\pm,\pm,\pm)$, respectively, cf. e.g. this Phys.SE post. Here $e$ is an einbein field. This Lagrangian (A) is world-line (WL) reparametrization covariant, cf. e.g. this Phys.SE post.

    • The Lagrangian (7.162) [which is also WL reparametrization covariant] follows from the Lagrangian (A) by integrating out $e$, cf. e.g. this Phys.SE post.

    • The corrected Lagrangian (7.164) is the Lagrangian (A) in the gauge $e=1/m$ up to an irrelevant constant term, cf. e.g. this Phys.SE post.

  • $\begingroup$ As always - your answers are invaluable! Some questions, as this material is somewhat new to me: 1. you call $e$ a gauge in the second bullet. Am I to interpret this as implying that I am free to choose $e$ to simplify my work? Like the gauges I know from EM? 2. If (1) is true, then why is there the whole process of integrating out $e$? Couldn't I just pick $e$ to be the final value that it 'integrates' to? 3. Is there any physical interpretation to $e$? $\endgroup$ Mar 14, 2022 at 15:25
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Mar 15, 2022 at 9:44

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