# Landau/Lifshitz "Mechanics" Mechanical Similarity

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.

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

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".