# Does the position-time graph have to be a smooth function?

If at some time $t$ there were a discontinuity in the velocity-time graph, then the acceleration would be infinite at $t$. So intuitively, it seems that the velocity-time graph must be continuous. I was wondering if all derivatives of the position-time graph are continuous functions (i.e. if the position-time graph is smooth) and if there was a way to prove it.

• If it take a perfectly rigid ball (which is practically not possible), and bounce it off a perfectly hard surface, then it's velocity will be discontinuous in time. Are you asking whether such a situation is practically possible? – theindigamer Apr 5 '16 at 5:35
• I know that for velocity to be discontinuous there must be infinite rigidity, infinite force, or something else impractical. I am asking if all derivatives of position, i.e. acceleration, jerk, etc. also must be continuous. – Rogue Autodidact Apr 5 '16 at 6:11
• I think this is an interesting question. People are often very casual about how differentiable things need to be, and it is certainly useful to actually think about it carefully. My intuition is that everything is at least smooth but I don't know why I think that. One reason to ask for more than smooth is that, to do physics, you need to be able to approximate things in some nice way, by some power series say, and you need that series to converge. Well, if it's a power series then things need to be analytic. – tfb Apr 5 '16 at 6:45
• Related: physics.stackexchange.com/q/151399/2451 and links therein. – Qmechanic Apr 5 '16 at 6:48

## 2 Answers

As you said, the next derivative of the velocity with respect to time is the acceleration. And the acceleration could in principle have a step somewhere due to a force starting to act on the object.

• But doesn't the force need to increase continuously as opposed to having a step? For example, if I apply a force to an object by pushing it, wouldn't I have to continuously increase the force applied from zero to some nonzero value (as opposed to directly jumping from zero to some nonzero value)? – Rogue Autodidact Apr 5 '16 at 4:15
• @RogueAutodidact: if you think of all forces as deriving from some sort of field theory, sure. But no inconsistency is introduced into the Newtonian framework by allowing discontinuities in the acceleration. – Jerry Schirmer Apr 5 '16 at 4:39
• @JerrySchirmer if you allow acceleration to be discontinuous you lose determinism in classical mechanics: that's often regarded as bad, although not everyone would regard it as so. (See Norton's dome.) – tfb Apr 5 '16 at 7:04

If you plot the time versus position of 'the high point on a see-saw', there is an abrupt change when end A goes from high to low (while end B goes from low to high). This is not entirely trickery, there are lots of useful items that exploit some kind of discontinuity (a toggle switch or an electronic astable 'flip-flop').

What happens when light reflects from a mirror? How can we deny that the path of the light is sharply kinked, i.e. 'not smooth' or that the velocity abruptly reverses, which (if a point particle were involved) would imply infinite acceleration?

As Newtonian mechanics applies to an object, so center-of-mass motion of the object is always smooth because Newton's laws apply; a 'reflection' of a ball occurs by distortion of the ball's shape over a short timespan, and that distortion generates force just like compressing a spring, and the force accelerates the ball. This ought not to be generalized, however. Some things are beyond the scope of Newtonian mechanics.