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In one dimension, the acceleration of a particle can be written as:

$$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$$

Does this equation imply that if:

$$v = 0$$

Then,

$$\Rightarrow a = 0$$

I can think of several situations where a particle has a non-zero acceleration despite being at instantaneous rest. What's going on here?

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  • $\begingroup$ Throw a ball up into the air. Once it leaves your hand it will continuously (ignoring air resistance) accelerate downward at 32f/s/s. At some point the ball will reach the top of the arc and have zero velocity, but it will still be accelerating downward at that point. (If it did not accelerate downward it would get "stuck" in the air.) $\endgroup$ – Hot Licks Mar 3 '15 at 17:05
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    $\begingroup$ I'm having some difficulty following your mathematics here because it seems you are defining v as both a function of time and a function of position. Also when you say "v = 0" do you mean "there exists a time when the velocity is zero" or "there exists a position when the velocity is zero", or "the velocity function is the function which is zero everywhere"? I think if you more carefully define what v is then you'll clear up your confusion. $\endgroup$ – Eric Lippert Mar 3 '15 at 18:39
  • $\begingroup$ @HotLicks: the original question is not "what is a situation where the velocity is zero but the acceleration is not?" -- the OP notes that there are many such situations. The question is "what is wrong with my math?", and what is wrong with the math is the conflation of v as a function of time with v as a function of position. However, your example is very useful because it illustrates the fundamental problem: for one-dimensional motion of a ball that is thrown up and falls back down velocity is not a function of position because there are positions that have two different velocities! $\endgroup$ – Eric Lippert Mar 3 '15 at 22:54
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The correct thing to say would be that "if v=0 and dv/dx is finite then a=0".

A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero.

In this case $$ x=-\frac{gt^2}{2}+v_0t $$ $$ v=v_0-gt $$ and $$ a=-g $$ and, clearly, "a" can never be zero, but "v" can be zero... so what gives? Well... solving for t(x) gives $$ t(x)=\frac{1}{g}(v_0\pm\sqrt{v_0^2-2gx}) $$ and $$ v(x)\equiv v(t(x))=\pm\sqrt{v_0^2-2gx} $$ and so $$ \frac{dv}{dx}=\frac{\pm g}{\sqrt{v_0^2-2gx}} $$ ...which is conveniently infinite whenever v is zero.


Furthermore, I think that a more natural way to think about this issue can be found by considering what we really mean by $$ v(x) $$ and how we go about taking the derivative w.r.t. x.

What we really mean is that, given some functional form for "v" as a function of "t" called "v(t)", and given some functional form for "x" as a function of "t" called "x(t)", and given that "x(t)" can be inverted to find "t(x)", then, as mentioned above $$ v(x)=v(t(x))\;, $$ which is silly physicist notation. It is clearly silly notation because the "v()" on the left-hand side cannot actually have the same form as the "v()" on the right-hand side. That's clear, right? So really let's call it $\tilde v$. I.e., $$ \tilde v(x)=v(t(x)) $$ This function $\tilde v$ is a function of x and the derivative with respect to x is $$ \frac{d\tilde v}{dx}(x)=\frac{dv}{dt}(t(x))\frac{dt}{dx}=\frac{\frac{dv}{dt}(t(x))}{\frac{dx}{dt}}=\frac{a(t(x))}{v(t(x))} $$ I.e., (switching back to silly notation and not writing arguments of functions) $$ \frac{dv}{dx}=\frac{a}{v}\;, $$ so, clearly, for constant a, dv/dx is infinite whenever v=0.

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    $\begingroup$ "Infinite"? Are you sure? You are apparently conflating undefined values (at limits) with infinity, especially in the case of square rooting a negative value (presumably you're disallowing complex numbers): the result is non-existent, not infinite. $\endgroup$ – Lightness Races in Orbit Mar 3 '15 at 2:52
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    $\begingroup$ Colloquially speaking. $\endgroup$ – hft Mar 3 '15 at 6:33
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    $\begingroup$ That's not colloquial, that's wrong... I don't think it would be less clear $\endgroup$ – mattecapu Mar 3 '15 at 16:00
  • $\begingroup$ Consider that my ball thrown into the air may have a velocity of exactly zero at time T, but any delta-T away from that time (at least in Newtonian mechanics) it will necessarily have a non-zero velocity. I think the math is wrong. $\endgroup$ – Hot Licks Mar 3 '15 at 17:12
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    $\begingroup$ You think what math is wrong? Clearly the velocity is zero in the example at the apex of the trajectory and is non-zero before and after... so what? $\endgroup$ – hft Mar 3 '15 at 17:35
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No, it doesn't imply that $a = 0$.

If, at some value $t = t_0$, the acceleration is non-zero while the velocity is zero, the position function is either a minimum or maximum. That is, $x(t)$ is stationary there:

$$x(t_0 + dt) = x(t_0)$$

which means that at $t = t_0$

$$\frac{dx}{d\dot x} = \frac{dx}{dv} = 0$$

thus $\frac{dv}{dx}$ is undefined at $t = t_0$.

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You can apply chain rule if $v$ is differentiable wrt $x$ and $x$ is differentiable wrt $t$. I think there are no other conditions,as this post on MathSE seems to say, https://math.stackexchange.com/questions/688152/necessary-conditions-for-the-chain-rule-of-differentiation-to-be-valid#=

and this condition is not always available. When $v=0$,make sure $\frac{dv}{dx}$ exist.

Related post:When can one write $a=v \cdot dv/dx$?

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Note that when you apply chain rule , you assume dx not to be zero . This will clear it up for you .

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    $\begingroup$ No one multiplies or divides by $dx$. We are just using chain rule. $\endgroup$ – Ruslan Mar 3 '15 at 7:15
  • $\begingroup$ thats the same thing . Chain Rule's proof requires this criteria . I will edit the answer then . $\endgroup$ – stud Mar 3 '15 at 9:27
  • $\begingroup$ No it's not. Chain rule doesn't rely on differentials at all. $\endgroup$ – Ruslan Mar 3 '15 at 10:48
  • $\begingroup$ @Ruslan then what is its proof ? I thought chain rule was for differentiation so its proof would have that concept . $\endgroup$ – stud Mar 4 '15 at 10:37
  • $\begingroup$ Just read the wikipedia article. $\endgroup$ – Ruslan Mar 4 '15 at 11:54
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I want to take another tack than that of the other answers. This will be one big handwave rather than a rigorous mathematical argument, but I hope it gets the idea across intuitively.

First off, as I noted in a comment, and as hft notes, you are using "v" to mean both "velocity as a function of time" and "velocity as a function of position". That's confusing, but there's no fundamental problem there. Except...

Except that your mathematics depends on being able to differentiate velocity with respect to position. This requires that velocity actually be a function of position.

Under what circumstances can one-dimensional velocity actually be a function of position? There must be exactly one velocity for each position. What does this imply about our velocity? That it must never change sign! Because if it does change sign then our particle is sometimes going forwards and sometimes going backwards, and therefore there must be a position which is traversed both backwards and forwards, and therefore velocity would not be a function of position.

So without loss of generality, let us suppose that velocity is never negative. Let us also suppose that position, velocity and acceleration functions are continuous and differentiable and all that good stuff.

Now let's think about the physicality of this situation with respect to acceleration.

Suppose the velocity is positive and the acceleration is zero or positive.

The particle is speeding off to the right hand side, it's position is getting more and more positive, faster all the time if acceleration is positive, and not slower if it is zero. Plainly the velocity will never be zero if this continues.

So let's then suppose the velocity is positive and the acceleration is negative. Our particle is getting slower and slower. Always moving to the right, mind you, because by supposition velocity is a function of position. But getting slower and slower as it goes.

Now, suppose it gets slower and slower and slower but never reaches zero velocity at any time. No problem there. The acceleration has to get closer and closer to zero, but neither the acceleration nor the velocity get to zero.

OK, so we've eliminated a bunch of cases from consideration -- the case where the acceleration is zero and velocity never changes, the case where acceleration is positive and velocity never gets smaller, and the case where acceleration is negative and velocity gets closer and closer to zero but never gets there. We only care about situations where velocity gets to zero.

Now let's consider the case where velocity starts off as positive, acceleration is negative, and velocity gets to zero at some time as a result. What has to happen to acceleration at the point where the velocity becomes zero? The acceleration cannot be negative at that point, because if it were then the particle would start moving backwards and we know it does not do that. The acceleration has to be either zero at that point or positive.

Suppose the acceleration is positive at the moment that the velocity is zero. Plainly it was negative before the velocity became zero; we could not have slowed down to zero from a positive velocity if the acceleration was positive or zero. But this contradicts our supposition that the acceleration function was a nice smooth differentiable function! Acceleration went instantaneously from a negative value to a positive value without going through zero, and therefore was not a nice continuous function.

The only remaining possibility is that the acceleration is zero at the point where the velocity is zero. Which is precisely what you wanted to show.

I can think of several situations where a particle has a nonzero acceleration despite being at instantaneous rest. What's going on here?

What's going on is that in all of those situations, either the acceleration is discontinuous at that point, or velocity is not actually a function of position, as is required by your mathematics.

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Where do you get that substitution?

That is not a fraction as in numbers with rules of fractions

It is dv(t) / dt - you cannot substitute for dt like that
That is v with respect to t

Clearly derivatives have different product rules
d(uv)/dx = u ⋅ dv/dx + v ⋅ du /dx

There is no rule for the substitution or equivalent you made
There is no basis for that

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