# Angular velocity and the speed of light

So, accelerate a particle to just under the speed of light, then send it round a bend, technically it accelerates due to the change of direction, will its speed exceed or match $c$?

Keep it simple, I'm only a high school teacher!

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Yes it would accelerate, but the particle's instantaneous speed will never exceed $c$ if it is massive. Recall that acceleration does not imply a change in speed. Any acceleration that simply changes the direction of travel of the particle (namely an acceleration perpendicular to the direction of travel) will not affect the speed of the particle.
Let $\mathbf a_\perp(t)$ denote the component of the acceleration perpendicular to the particle's velocity at time $t$ and let $\mathbf a_\parallel(t)$ denote its component parallel to the velocity, then we have $$\dot{\mathbf v}(t) = \mathbf a_\perp(t) + \mathbf a_\parallel(t)$$ Using this, notice that $$2v(t)\dot v(t) = \frac{d}{dt}v(t)^2 = \frac{d}{dt}\mathbf v(t)\cdot\mathbf v(t) = 2\mathbf v(t)\cdot\dot{\mathbf v}(t)$$ where $v(t)$ is the speed at time $t$, but using the decomposition of the acceleration above gives $$\mathbf v(t)\cdot\dot{\mathbf v}(t) = \mathbf v(t)\cdot\mathbf a_\perp(t) +\mathbf v(t)\cdot\mathbf a_\parallel(t) = \mathbf v(t)\cdot\mathbf a_\parallel(t)$$ so combining these results we have $$v(t)\dot v(t) = \mathbf v(t)\cdot\mathbf a_\parallel(t)$$ If the parallel component of acceleration is zero, and if the particles speed is nonzero, then we get $$\dot v(t) = 0$$ as desired!