Classical Mechanics – Why Omit the $V^2$ Term in Non-Inertial Frame?

classical-mechanicslagrangian-formalismreference frames

I'm trying to find trying to find the Lagrangian and Hamiltonian for a particle in a non-inertial frame, but when I try to do so, I always get a quadratic term, which textbooks like Landau & Lifshitz vol. 1 ignore for a reason I will specify later. Here's what I have:

The Lagrangian in the inertial frame is
$$L = \frac12 m \mathbf{\dot{r}}^2 – U(\mathbf{r})$$
This means, that in the tranformation to a non-inertial frame, where $\mathbf{v}=\mathbf{v'}+\mathbf{V}(t)$,
then we have
\begin{align}
L &=\frac12 m (\mathbf{\dot{r'}}+\mathbf{V})^2 – U(\mathbf{r'})\\
&=\frac12 m \mathbf{\dot{r'}}^2 + \frac12 m\mathbf{V}^2 + m\mathbf{\dot{r'}}\cdot\mathbf{V} – U(\mathbf{r'})
\end{align}
Landau & Lifshitz says the following though:

"$\mathbf{V}^2(t)$ is a given function of time, and can be written as the total derivative with respect to $t$ of some other function; the [$\frac12 m \mathbf{V}^2$] term can therefore be omitted."

Could someone explain this logic to me? I don't see the connection.

Best Answer

  1. It is important that the velocity ${\rm V}(t)$ of the reference frame is a given/prescribed external parameter (as opposed to an active dynamical variable) of the theory. We can then consistently view $\frac{m}{2}{\bf V}^2(t)$ as a total time-derivative. See also e.g. this related Phys.SE post.

  2. L&L are next relying on the fact that the Euler-Lagrange (EL) equations are not changed if the Lagrangian is changed by a total time derivative, cf. e.g. this Phys.SE post.

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