Mechanics queries Why the work done by internal forces is independent of frame of reference?
Since the work done by a body is the dot of force and displacement and displacement is a frame dependent quantity, then how come when that force is internal the work becomes independent of frame of reference?
 A: From action reaction law we have for n-particle system
$$dW = \sum_{i\neq j}\vec{F}_{ij}\cdot d\vec{s}_i=\sum_{j>i}\vec{F}_{ij}\cdot (d\vec{s}_i-d\vec{s}_j)$$
with $dW$ being infinitesimal work, $\vec{F}_{ij}$ being force acting on particle i by particle j and $d\vec{s}_i$ is infinitesimal displacement of particle i.
Now it turns out, that while $d\vec{s}_i$ depends on the frame, $d\vec{s}_i-d\vec{s}_j$ does not.
To see this, let us call the position vector of the particle i at the time t in the original frame $\vec{r}_i(t)$ and in the new system by $\vec{r}'_i(t)$. Rotation and translation has no influence on displacement, but boost does. If we assume our origins are aligned at $t_0$ and that the two frames are moving with velocity $\vec{V}$ relative to each other we get
$$\vec{r}_i(t)=\vec{V}(t-t_0)+\vec{r}'_i(t),$$
so under the boost we get that $d\vec{s}_i$ transforms according to the law
$$d\vec{s}_i=\vec{r}_i(t_0+dt)-\vec{r}_j(t_0)= \vec{r}'_i(t_0+dt)+\vec{V}dt-\vec{r}'_j(t_0)=d\vec{s}'_i + \vec{V}dt$$
and we can see that the term $\vec{V}dt$ cancels itself out in the subtraction of displacements, since it is same for all the particles
