Why are you allowed to omit the $V^2$ term in the non-inertial frame?

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.

• – Eli
Mar 15 at 15:55
• Earlier in the book they explain why you can add a total time derivative to a Lagrangian without changing the resulting dynamics. In my version it is on page 4 just before the paragraph about Galileo's relativity principle. Mar 15 at 15:58
• @AlmostClueless I'm now seeing this again, and I remember learning that. Thank you. Mar 15 at 16:07
• More generally, adding any fixed function of time or space to a Lagrange density (in the context of classical field theory) doesn't change the Euler-Lagrange equations, because when you take the functional derivatives for the Euler-Lagrange equations you differentiate with respect to the fields only. This argument still holds even though you can't write a function $f(\vec{x}, t)$ as a "total derivative" in the context of an integral over a multi-dimensional background. Mar 15 at 16:13
• The given answers are excellent, but if you haven't already done so, it would perhaps be worthwhile to forget about the theorems and general statements and simply compute the Euler-Lagrange equations. You will find that the $\frac{1}{2}mV^2$ term simply drops out and doesn't contribute, which can be understood to be a consequence of the nice theorems you have learned. Mar 15 at 16:55

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.

The Lagrangian in the inertial frame is $$L = \frac12 m \mathbf{\dot{r}}^2 - U(\mathbf{r})$$

... in the transformation to a non-inertial frame... we have \begin{align} L &=\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?

In general, if $$F=F(q,t)$$, we have $$\dot F = \frac{dF}{dt} = \frac{\partial F}{\partial q}\dot q + \frac{\partial F}{\partial t}\;,$$ which means that $$\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}\;.$$

If we have two Lagrangians that differ by the total time derivative of F: $$\bar L = L + \frac{dF}{dt}\;,$$ then the equations of motion derived from $$\bar L$$ and $$L$$ are the same, since $$\frac{\partial \bar L}{\partial \dot q} = \frac{\partial L}{\partial \dot q} + \frac{\partial \dot F}{\partial \dot q} = \frac{\partial L}{\partial \dot q} + \frac{\partial F}{\partial q}$$ and $$\frac{\partial \bar L}{\partial q} = \frac{\partial L}{\partial q} + \frac{\partial \dot F}{\partial q} = \frac{\partial L}{\partial q} + \frac{d}{dt}\frac{\partial F}{\partial q}\;.$$

Such that: $$\frac{\partial \bar L}{\partial q} - \frac{d}{dt}\frac{\partial \bar L}{\partial \dot q} = \frac{\partial L}{\partial q} + \frac{d}{dt}\frac{\partial F}{\partial q} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q} + \frac{\partial F}{\partial q}\right)$$ $$= \frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q} + \frac{d}{dt}\frac{\partial F}{\partial q} - \frac{d}{dt}\frac{\partial F}{\partial q}$$ $$= \frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q}$$

For the more specific case of interest here, $$F=F(t)$$, is only a function of time (not a function of $$q$$ and $$t$$): $$F(t) = \int^t dt' \frac{m}{2}V^2(t')$$

The relation $$\frac{\partial \dot F}{\partial \dot q}=\frac{\partial F}{\partial q}$$ still holds, but is trivial in this case, since both sides of the equation are zero. Nevertheless, since the equation still hold the equations of motion derived from $$\bar L$$ and $$L$$ are the same.

Even more specifically, for the case of interest here (note, I write q instead of OP's r', and I write $$\bar L$$ instead of OP's $$L$$): $$\bar L = L + \frac{dF}{dt} =\frac12 m \mathbf{\dot{q}}^2 + m\mathbf{\dot{q}}\cdot\mathbf{V} - U(\mathbf{q}) + \frac12 m\mathbf{V}^2$$ $$L = \frac12 m \mathbf{\dot{q}}^2 + m\mathbf{\dot{q}}\cdot\mathbf{V} - U(\mathbf{q})$$

$$\frac{\partial \bar L}{\partial \mathbf{q}} = -\nabla U$$ $$\frac{\partial L}{\partial \mathbf{q}} = -\nabla U$$

$$\frac{\partial \bar L}{\partial \mathbf{\dot q}} = m \mathbf{\dot q} + m\mathbf{V}$$ $$\frac{\partial L}{\partial \mathbf{\dot q}} = m \mathbf{\dot q} + m\mathbf{V}$$

So, regardless of whether you work with $$L$$ or $$\bar L$$ you end up with the equations of motion: $$m \mathbf{\ddot q} + m\mathbf{\dot V} = -\nabla U\;.$$