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!


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.

Mathematical proof in case you want it (or need it for smart alecs ;))

I claim that

If a particle experiences no acceleration parallel to its direction of travel, and if its speed is nonzero, then its speed will not increase.

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!

| cite | improve this answer | |
  • $\begingroup$ okay, thanks! so slightly different question: Earth is accelerating but not speeding up as it journeys around the sun, is there any evidence to suggest Earths speed does change due to gravity of other planets/moons etc? - sorry ive got a ton of these from the kids today! $\endgroup$ – steve Mar 8 '13 at 19:57
  • $\begingroup$ The earth does changes its speed relative to the sun since the orbit it takes is not circular. Due to conservation of momentum, the speed follows Kepler's law. We can also compute how the other planets affect the motion of the earth. $\endgroup$ – user18764 Mar 8 '13 at 20:08
  • 1
    $\begingroup$ @steve I second what user18764 says. There are also small perturbations due to the other planets, although I'm not sure offhand what the size of these perturbations are. You might find en.wikipedia.org/wiki/Earth's_orbit helpful. By the way, I love how your students ask these questions, and I'm even happier that they have a teacher that cares enough to find the answers! Yay you! Respect. $\endgroup$ – joshphysics Mar 8 '13 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.