In Landau/Lifshitz "Mechanics", 3e, subsection 10, a question asks to "find the ratio of the times in the same path for particles having the same mass but potential energies differing by a constant factor". The solution provided is

$$ \frac{t'}{t}=\sqrt{\frac{U}{U'}} $$

I do not understand this result. If $U'=U+U_0$ for a constant $U_0$, then $U_0$ is a total time derivative which can be added to the Lagrangian without changing the equations of motion. How then could it alter a measurable quantity like $t$? Or by "differing", does Landau imagine a rescaling of the potential energy; i.e. $U'=\alpha U$?

EDIT: If $U\rightarrow\alpha U$ and $t\rightarrow\beta t$ then

$$ L\rightarrow \sum_a\frac12m_a\beta^{-2}\vec{v}_a^2-\alpha U =\beta^{-2}\left[\sum_a\frac12m_a\vec{v}_a^2-\beta^2\alpha U\right] $$

and the equations of motion remain unchanged if

$$ 1=\beta^2\alpha =\left(\frac{t'}{t}\right)^{\!2}\frac{U'}{U} $$

which gives the desired result. I assume $U'=\alpha U$ is what Landau had in mind.

  • 1
    $\begingroup$ Differing by a constant factor means $U'=kU$. The relation $U' = U+U_0$ is described as "differ by a constant term". $\endgroup$ Commented Aug 29, 2023 at 16:44
  • $\begingroup$ @JánLalinský Thank you - if you make your comment into an answer, I'll be happy to accept it. $\endgroup$
    – CW279
    Commented Aug 29, 2023 at 16:50

1 Answer 1


Differing by a constant factor means U′=kU ($k$ is the factor). The relation $U′=U+U_0$ is described as "$U,U'$ differ by a constant term".


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