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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.

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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 $\vec v$, $E\rightarrow \frac{1}{2} mv^2 + mc^2$.

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The frame is called a momentarily co-moving inertial frame. The frame is a standard inertial frame, which means that it covers all of spacetime. The “momentarily” does not refer to its existence but rather that it is only co-moving with the particle momentarily.

At that moment indeed $dE/dt=0$ provided the particle is not changing mass.

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    $\begingroup$ perhaps it is even better to put it the other way round: the particle is only co-moving with the frame momentarily $\endgroup$ – Andrew Steane May 7 at 22:45

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