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Referring to unidimensional motion, it is obvious that it doesn't always make sense to write the speed as a function of position. Seems to me that this is a necessary condition to derive formulas like:

$$v^2=v_0 ^2 +2\int_{x_0}^{x}a\cdot dx$$

In fact, in the first step of the demonstration (the one I saw, but I think that this step is crucial) it's required to write $a=dv/dt=(dv/dx)(dx/dt)$, that doesn't make sense if $v$ isn't a function of $x$.

When can one rigorously write $v=v(x)$?

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If you have $x(t)$, you can obtain $t(x)$ and inject it in $v(t)=v(t(x))$. –  Vladimir Kalitvianski Feb 21 '13 at 20:03

3 Answers 3

up vote 2 down vote accepted

This is going to be essentially the same in content as Jerry Schirmer's response, but I thought you might like to hear it in more mathematical terms. The velocity function $v$ is defined as $$ v(t) = \dot{ x}(t) $$ Let's take the domain of the position function to be the open interval $(t_1, t_2)$ and suppose that it has the property that given any point $x_0$ in the range of $x$, there is a unique point $t_0$ in its domain $(t_1, t_2)$ such that $x(t_0) = x_0$. Then there exists a function $x^{-1}$ (the inverse of $x$) defined on the range of $x$ satisfying $$ x^{-1}(x(t)) = t $$ Now we define a function $\bar v$ on the range of $x$ by $$ \bar v(x) = v(x^{-1}(x)) $$ It is common to abuse notation here and use $v$ in place of $\bar v$ for this function, but let's keep things notationaly rigorous. Then on one hand the chain rule gives $$ \frac{d}{dt}\bar v(x(t)) = \frac{d\bar v}{dx}(x(t))\,\dot x(t) = \frac{d\bar v}{dx}(x(t))\,v(t) $$ While on the other hand we use the definition of $\bar v$ to write $$ \frac{d}{dt}\bar v(x(t)) = \frac{d}{dt} v(x^{-1}(x(t))) = \frac{dv}{dt}(t) = a(t) $$ and combining these observations gives the identity you wanted $$ a(t) = \frac{d\bar v}{dx}(x(t))\,v(t) $$ Notice that if we indulge in the usual abuse of notation, then we can simply write this as $$ a = v \frac{dv}{dx} $$

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Generally, the total derivative can be broken into a sum of partial derivatives. If the acceleration $a$ is taken to be a function only of $x$ and $t$, then the total derivative is \begin{equation} a=\frac{\mathrm{d} v}{\mathrm{d} t} = \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x}\frac{\partial x}{\partial t} \end{equation}

One can safely write $\mathbf{a}=\mathbf{v}\cdot\nabla \mathbf v$, then, when $\partial v /\partial t=0$. This is true when you are considering a single particle or object, as the velocity of the particle at a point where the particle doesn't exist is not changing. However, for distributions of particles, the distinction is meaningful.

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Thank you, but I don't understand the explaination of why $\dfrac {\partial v}{\partial t} = 0 $ . Why is $ \partial v / \partial t $ evaluated at a point where the particle does not exist? –  pppqqq Feb 21 '13 at 21:01
    
This idea comes from distributions of particles, where you might ask "what happens at this point in space?" Particles may move in and out, so you could talk about the velocity of particles at that point in space changing without changing the point in space that you are talking about. However, when you are talking about a single object, then you are asking about the velocity of the object at the point in space where the object is. The notion only really applies if you adopt a "lab frame" that is stationary, with respect to which the particles can move. –  KDN Feb 22 '13 at 0:45
    
I realize I stated that badly... the notion only really applies in the lab frame if you are interested in what is at a particular point in space, rather than following individual objects through their trajectories in space. –  KDN Feb 22 '13 at 2:19
    
Unfortunately I haven't got enough mathematical notions to fully understand the matter , however writing down the partial derivative of the function $v$ (seen as a function of "points of my lab frame at a precise instant") I have an intuition about what you mean by $\partial v / \partial t = 0 $ if we have just one particle. In fact (if the particle is not still, after the little $\partial t $ time there is no more particle in that point in space and so no speed. Sorry if i'm simplifying too much. However thank you for the answer, that makes a lot of sense out of that. –  pppqqq Feb 22 '13 at 10:29
    
Nope, that's about the gist of it. –  KDN Feb 22 '13 at 14:09

It can be done in any case where the velocity can be written as a function of the position. This can be done if the velocity is not constant, and if there are no turnaround points in the motion. For instance, consider $x = r\sin(\omega t)$, $v = \omega r \cos(\omega t)$.

Then, we have:

$$\begin{align} x &= r \sin(\omega t)\\ t &= \frac{1}{\omega}\sin^{-1}(x/r)\\ v &= \omega r \cos (sin^{-1}(x/r))\\ &= \omega \sqrt{r^{2} - x^{2}} \end{align}$$

Which is a valid transformation so long as $\sin^{-1}(x/r)$ is defined, which means that you only cover the right half of the unit circle.

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