# Is $\frac{dE}{dt}=0$ in an accelerating particle’s instantaneous rest frame?

My special relativity book uses an argument that involves $$\frac{dE}{dt}=0$$ in an accelerating particles rest frame (to show a force parallel to a particles velocity is parallel in all frames).

However I don’t understand how you can do the derivative $$\frac{dE}{dt}$$ in a frame which only exists for an instant so how can even an infinitesimal $$dt$$ be defined? To me this is like saying that $$\frac{dv}{dt}=0$$ in an accelerating particle’s instantaneous rest frame which is obviously wrong.

In non-relativistic mechanics, $$\frac{dE}{dt} = m\vec v \cdot \frac{d\vec v}{dt}$$, so $$\vec v =0 \implies \frac{dE}{dt} = 0$$. This is true e.g. for a vertically-launched projectile at the apex of its trajectory.
Essentially the same is true in relativistic physics, since $$\frac{dE}{dt} = mc^2 \frac{d\gamma}{dt}= \gamma^3 m \vec v \cdot \frac{d\vec v}{dt}$$. If you are in a frame where $$\vec v=0$$, even for a moment, then $$\frac{dE}{dt}=0$$ (at that moment).
Of course, we could have heuristically used the non-relativistic calculation too, since in the limit of infinitesimal $$\vec v$$, $$E\rightarrow \frac{1}{2} mv^2 + mc^2$$.
At that moment indeed $$dE/dt=0$$ provided the particle is not changing mass.