# Non-inertial frames in Lagrangian mechanics?

Building on this Phys.SE post I am interested in how non-inertial frames can be considered in Lagrangian mechanics. My understanding is that changing the reference frame causes a transformation of the Lagrangian that will transform in such a way as to account for the apparent fictitious forces if the resultant frame is non-inertial. Colloquially we might say that it has all the information in it from the start to describe a system. Therefore a Lagrangian is not unique and transforms like a scalar. Can we describe the Euler-Lagrange equation (for discrete particles) as covariant? (normally this is the topic of field equations, which are covariant).

Bringing this back to more familiar topics in Lagrangian mechanics, should we view this as a point or a gauge transformation?

On the one hand a point transformation is the changing of the coordinate system, say $(q,\dot q)\rightarrow (Q,\dot Q)$, this will in the most general sense change the Lagrangian and leave equations of motion invariant.

A gauge transformation however has the same coordinates and the same general form of the Lagrangian except that there is a total time derivative added on the end.

Therefore how should one, if indeed one can, consider inertial frames in analytical mechanics in terms of point/gauge transformations?

Let the equations of motion be expressed in a frame with coordinates $q$. We now want to switch over to another (arbitrarily moving) frame, whose corresponding coordinates are $Q$, given by: $$Q = f(q, t)$$ For example, if the frame itself is moving with position $x(t)$, we will have: $$Q = q - x(t)$$ (where $x$ is not dynamic, but is completely specified in advance).
This does not necessarily leave the equations of motion invariant. It's true that Euler-Lagrange equations (note that the Lagrangian must now be allowed to be time dependent) $$\frac{d}{dt}\frac{\partial L(q, \dot{q}, t)}{\partial \dot{q}} = \frac{\partial L(q, \dot{q}, t)}{\partial q}$$ continue to hold, but the change in the form of the Lagrangian effected by the change of frame means that the equations of motion can 'look' different.
In the case that this point transformation is also a gauge transformation, we have a special situation. Consider the following/relevant example. In classical mechanics, from an inertial frame, the Lagrangian is: $$L(q, \dot{q}) = \frac{1}{2}m\dot{q}^2 - V(q)$$ The general transformation to an arbitrary moving (non-rotating, for simplicity) frame is given by $q = Q + x(t)$, so that $\dot{q} = \dot{Q} + \dot{x}(t)$, and the Lagrangian becomes: $$L(Q, \dot{Q},t) = \frac{1}{2}m\dot{Q}^2 + m\dot{Q}\dot{x}(t) + \frac{1}{2}m\dot{x}(t)^2 - V(Q+x(t))$$ The term quadratic in $\dot{x}$ produces only a pure-boundary term in the action $S = \int L dt$, and is irrelevant. The main term of interest is the second one (responsible for the fictitious forces, and part of the generalized potential, as mentioned in this answer to the question you linked), and it's contribution to the action: $$S_2 = \int m\dot{Q}\dot{x}(t)\ dt$$ Integrating by parts and neglecting the boundary term, we get $$S_2' = -\int mQ\ddot{x}(t) \ dt$$ This readily gives the answer - for the "time-dependent point transformation", which corresponds to a change of frame, to be a gauge transformation, we must have $\ddot{x}(t) = 0$ for all time. In this case, we get the important part of the action as: $$S' = \int \left[ \frac{1}{2}m\dot{Q}^2 - V(Q, t) \right] dt$$ This is hardly any different from the one we started with (the time dependence in $V$ is not an issue, and is just a reflection of the fact that the "field" would also appear to move in a moving frame; the important thing is that at a given time $t$, the particle sees the same force $-\nabla V$ at its location in both frames).
This is indeed, what makes inertial frames ($\ddot{x} = 0$ as seen from another inertial frame) special - the general point-transformation due to switching frames reduces to a gauge transformation, and the equation of motion 'looks' the same i.e. 'Galilean invariance'. That this doesn't occur in non-inertial frames leads to the fictitious forces seen in such frames.