Newtonian Mechanics – Action of Free Particle Invariance Under Galilean Transformation

Question

I want to show that the action of a free particle is invariant under a Galilean transformation
$$
(t,\vec{x})\rightarrow (t+a,R\vec{x}+\vec{v}t+b)=(t^\prime, \vec{x}^\prime) \quad\text{where}\quad R\in O(3).
$$
Transforming the action yields
$$
S=\frac{m}{2}\int\mathrm{d}t\:\left(\frac{\mathrm{d} \vec{x}}{\mathrm{d}t}\right)^2\rightarrow S^\prime=\frac{m}{2} \int\mathrm{d}t^\prime\:\ \left(\frac{\mathrm{d} \vec{x} ^\prime}{\mathrm{d}t}\right)^2 =\frac{m}{2} \int\mathrm{d}t\:\ \left(R \frac{\mathrm{d} \vec{x}}{\mathrm{d}t}+\vec{v}\right)^2
$$
which can be rewritten as a total derivative
$$
S^\prime=\frac{m}{2}\int\mathrm{d}t\left[ \:\left(\frac{\mathrm{d} \vec{x}}{\mathrm{d}t}\right)^2 +2\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\vec{v}\cdot R\vec{x}+\frac{1}{2}\vec{v}^2t\right)\right].$$
However, I‘m not sure if this is correct since I assumed that the time derivative doesn‘t transform as it has a fixed definition.

Answer 1

Transformation of the time derivative

From the Galilean transformation you defined, i.e. $t' = t + a$, you can evaluate how the time derivative is transformed: $$ \frac{d}{d t} = \frac{dt'}{dt} \frac{d}{dt'} = \frac{d(t + a)}{dt} \frac{d}{dt'} = \frac{d}{dt'} \quad \,.$$

Invariance of classical action of free particle

The usage of the term 'invariance' here is a bit context-dependent. In this question, invariance is defined as a transformation behaviour of an object. An invariant action would transform as a scalar meaning that $S' = S$. In the case of Galilean transformation however (to continue your calculation)

$$ S' = \frac{m}{2} \int_{t_1}^{t_2} dt \, \bigg[\Big(\frac{dx}{dt}\Big)^2 + 2 \frac{d}{dt} \Big(\vec{v} \, \cdot \, R\vec{x}(t) \Big) + v^2 \bigg] \equiv \, S \, + \int_{t_1}^{t_2} dt \bigg[ \frac{d}{dt} H(\vec{x}(t)) \bigg] \, + \, C $$

the action receives an additional integral over a total time derivative as well as a constant, ergo not invariant. But whenever we physicists talk about an action being invariant, we usually refer to the equations of motions, the physics, we obtain from it. Now as discussed here and here, the total time derivative will, after integration, just give you another constant term $\Delta H = H(\vec{x}(t_2)) - H(\vec{x}(t_1))$ that is added to your action which will not have any influence on the physical quantities that are described by your action since $\delta S' = \delta S$.