From Landau, Lifshitz Mechanics p.127

$$ \renewcommand{\vec}[1]{\mathbf{#1}} L'=\frac{1}{2}m(\vec{v}'^2+\vec{v'}\cdot\vec{V}+\vec{V}^2)-U $$

He states that "$\vec{V}^2(t)$ can be written as the total derivative with respect to $t$ of some other function.". I understood that but the point that I could not understand is why we can't write the terms $\vec{v'}\cdot\vec{V}$ and $\vec{v}'^2$ as the total derivative with respect to $t$ of some other function?

$\vec{v}'$ is the velocity of particle measured in non-inertial frame.

$\vec{V}$ is translational velocity of non-inertial frame with respect to inertial frame.


1 Answer 1


The main point is that:

  1. On one hand, ${\bf r}^{\prime}(t)$ is an dynamical active variable of the system. Its time evolution is determined by the equations of motion of the system, i.e. Newton's 2nd law. Here ${\bf v}^{\prime}:=\dot{\bf r}^{\prime}$.

  2. On the other hand, ${\bf R}(t)$ is a passive (background) variable with pre-defined time-dependence. Here ${\bf V}:=\dot{\bf R}$.

We could in principle define e.g.

$$\tag{1} F_1~:=~\frac{m}{2}\int_0^t \! dt^{\prime}~ \dot{\bf r}^{\prime}(t^{\prime})^2,\qquad \dot{F}_1~=~\frac{m}{2}\dot{\bf r}^{\prime 2}, $$

$$\tag{2} F_2~:=~\frac{m}{2}\int_0^t\! dt^{\prime}~ \dot{\bf r}^{\prime}(t^{\prime})\cdot \dot{\bf R}(t^{\prime}),\qquad \dot{F}_2~=~\frac{m}{2}\dot{\bf r}^{\prime}\cdot\dot{\bf R} ,$$

$$\tag{3} F_3~:=~\frac{m}{2}\int_0^t \! dt^{\prime}~ \dot{\bf R}(t^{\prime})^2,\qquad \dot{F}_3~=~\frac{m}{2}\dot{\bf R}^2. $$

The problem with e.g. the definition (1) is that $F_1$ would be a non-local quantity. It would depend on the dynamical active variable ${\bf r}(t^{\prime})$ in the past $t'<t$. See also this Phys.SE post.

On the other hand, there is no problem with locality in the definition (3) of $F_3$ because the passive variable ${\bf R}(t)$ is pre-defined for all times $t$. In other words, $F_3$ is just another passive background variable.


  1. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, (1976), p. 127.

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