This question is about work done being dependent on the frame of reference, which is, obviously the net work done.

I ask what is the reason for the work done by internal forces to be independent of the frame of reference?

For example, some works by internal forces are due to friction, spring force, (I am not sure if I should include tension as well), mechanical work done by a person, etc. So are they independent of the frame of reference because they depend only on the "relative distance" where the force acts? If so, can someone answer this more elaborately and clearly?


1 Answer 1


So you've got a set of particles with trajectories $\vec{r}_i(t)$ and velocities $\vec{v}_i(t)$. Internal forces are the ones acting between the particles - lets denote $\vec{F}_{ij}$ the force acting on particle $i$ by particle $j$ . Crucially, they satisfy the 3rd law $\vec{F}_{ij} = - \vec{F}_{ji}$ as a result their sum over $i$ and $j$ should be zero:

$$\sum_{i,j}\vec{F}_{ij} = 0 $$

Computing the work done by the internal forces in time $dt$:

$$dA = \sum_{i,j} \vec{v}_i \cdot \vec{F}_{ij}\,dt$$

If we move to a moving reference frame, then we'll have to subtract the velocity $\vec{V}$ from all the particles' velocities. The work done in the new reference frame will be

$$dA' = \sum_{i,j} (\vec{v}_i -\vec{V}) \cdot \vec{F}_{ij}\,dt $$

Opening the brackets and using the equality for sum of the forces.

$$ dA' = dA - \vec{V}\cdot\sum_{i,j}F_{ij}\,dt = dA - \vec{V}\cdot 0\,dt= dA$$

  • $\begingroup$ @ Kostya also this $dA=dA‘=\overrightarrow{v‘}\cdot \overrightarrow{F‘}$ can be shown with the rotation matrix $\endgroup$
    – Eli
    Apr 25, 2021 at 6:16

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