# Question about $a = v\ \mathrm dv/\mathrm dx$

Consider $\vec{v}$ Now differentiating this w.r.t time, $$\vec{a} = d/dt( \vec{v}) = \vec{v}(d\vec{v}/dx)$$ Now this multiplication of vectors obviously makes no sense. This along with the fact that on integrating we get $v^2$ leads me to believe that this 'formula' is only applicable when $v$ is really $|v|$ (speed) and $a$ is really the rate of change of speed.

But while studying one-dimensional motion, I remember countless times getting an acceleration function like $a(x) =\text{something }\hat{i}$ and applying this same result and integrating which gave me the velocity. Why is this?

• The $v^2$ is absolutely OK. Is it possible that you mix up integration over time and position here? – mikuszefski Feb 6 '17 at 12:03
• I'm confused by your first equation there. $\vec{a} = \frac {d}{dt}( \vec{v})$ is not the same as "$\vec {v} \frac {d \vec{v}}{dx}$" . – JMac Feb 6 '17 at 12:06
• Yes and no. In 3D you have to do it by component as in vbrasic's answer. – mikuszefski Feb 6 '17 at 12:12
• @mikuszefski sorry but I don't see anywhere in vbrasic's answer something like $d\vec v/dx$. In fact the expression $\vec v (d\vec v/dx)$ makes no sense since it involves some sort of "double-vector": either this is some kind of tensor or dyadic, or else it's a scalar product, neither of which could equal a vector $\vec a$. If you have additional insight can you briefly expand? – ZeroTheHero Feb 6 '17 at 13:32
• $\dfrac{d^2 x}{dt^2}=v\dfrac{dv}{dx}$ is usual mathemtical tricks using frequently when I was studying AL Applied Mathematics. – Ng Chung Tak Feb 6 '17 at 13:45

Well, let me present my version as well: By definition we have

$$\vec a =\frac{\mathrm d}{\mathrm d t }\vec v = \frac{\mathrm d}{\mathrm d t }\left(v_x,v_y,v_z\right)^\mathrm{T} =\left(\frac{\mathrm d}{\mathrm d t} v_x,\frac{\mathrm d}{\mathrm d t} v_y,\frac{\mathrm d}{\mathrm d t} v_z\right)^\mathrm{T}$$ as in 1D $$\frac{\mathrm d}{\mathrm d t}v=\frac{\mathrm d x}{\mathrm d x}\frac{\mathrm d}{\mathrm d t}v=\frac{\mathrm d x}{\mathrm d t}\frac{\mathrm d}{\mathrm d x}v=v\frac{\mathrm d v}{\mathrm d x}$$ that is

$$\vec a =\left(v_x\frac{\mathrm d v_x}{\mathrm d x},v_y \frac{\mathrm d v_y}{\mathrm d y},v_z\frac{\mathrm d v_z}{\mathrm d z}\right)^\mathrm{T}$$

and we easily can just look at one component. Now concerning the integration: From $$a=v \frac{\mathrm d v}{\mathrm dx}$$ follows $$a \mathrm d x=v \mathrm d v$$ i.e. $$\int_{x_1}^{x_2} a \mathrm d x=\int_{v_1}^{v_2}v \mathrm d v =\frac 1 2 (v_2^2-v_1^2)$$ and if $a=\mathrm{const}$

$$a(x_2-x_1)=\frac 1 2 (v_2^2-v_1^2)$$ which is one of the "SUVAT" equations. This should work fine if generalized to 3D.

EDIT Due to some misleading concepts and some comments, let me give a sort of interpretation and an extreme example.

Interpretation

Usually we have acceleration a change of speed over a certain time. There is no reason why you could not "measure" it as a change of speed over a certain distance. The problem here is: if the accelerated object is already very fast, the distance required to see an according change needs to be very large. The other way around it works like this: If over the same distance two objects change there speed by the same amount but one was much faster than the other in the beginning, its acceleration must have been larger as well. That is the multiplication with $v$ of $\frac{\mathrm d v}{\mathrm dx}$

Example

Lets take a charged particle in a magnetic field (suggested by a comment) Without losing generality we may look only at 2 dimensions. In principle we must be careful as there is no one-to-one mapping neither for $x$ and $t$ nor for $y$ and $t$, but it turns out that this is no problem here. Otherwise use the implicit function theorem. We can simplify this to a circular motion:

$$\vec x = R \binom {\sin \omega t}{\cos \omega t}, \vec v = R \omega \binom {\cos \omega t}{-\sin \omega t},\vec a = R\omega^2 \binom {-\sin \omega t}{-\cos \omega t}$$ Hence $$v_x=\omega\sqrt{R^2 -R^2 \sin \omega t}= \omega\sqrt{R^2-x^2}$$ Therefore $$\frac{\mathrm d v_x}{\mathrm d x}=-\frac{x\omega}{\sqrt(R^2-x^2)}=-\frac{x \omega}{\frac{v_x}{\omega}}= -\frac{x \omega^2}{v_x}$$ So $$v_x \frac{\mathrm d v_x}{\mathrm d x}= -x \omega^2$$ And it is easy to see that $a_x=-\omega^2 x$ as well. The same way you do it for $y$. So if you treat the vectors correctly, $a=v \mathrm d v/\mathrm d x$ is true even for a motion where the acceleration in $x$ depends on the velocity in $y$ and vice-verse, and where there is no one-to-one global mapping from time to coordinate.

I'd like to point out here that the expression «$\vec{a}=d\vec{v}/dt=\vec{v} d\vec{v}/dx$» is not entirely senseless.

In the definition

$$\vec{a} = \frac{d\vec{v}}{dt}$$

velocity is considered to be the function of time $\vec{v} = \vec{v}(t)$ because we follow an individual particle along its trajectory. So in this case the expression is wrong.

But if you look at a continuous system of particles e.g. flowing water, you can define the velocity to be a function of time and coordinates $\vec{v} = \vec{v}(t, x, y, z)$ - it would be equal to the velocity of a small droplet of water that happened to be in point $(x,y,z)$ at time $t$. As time goes by, this droplet would move to a different point in space $(x',y',z')$ , while another droplet would take its place at $(x,y,z)$ and so on. The $\vec{v}(t,x,y,z)$ is called "velocity field".

To calculate the acceleration of an individual droplet in this case, you should compare droplet's velocity at time $t$

$$\vec{v}_d(t) = \vec{v}(t,x,y,z)$$

with the velocity it will have at time $t+dt$ having moved in space by $\vec{v}_d(t)dt$:

$$\vec{v}_d(t+dt) = \vec{v}\left(t+dt, x+v_x(t,x,y,z)dt, y+v_y(t,x,y,z)dt, z+v_z(t,x,y,z)dt\right)$$

Expanding this you get

$$\vec{v}_d(t+dt) = \vec{v}_d(t) + \vec{a}dt = \vec{v} + \left(\frac{\partial\vec{v}}{\partial t} + v_x \frac{\partial\vec{v}}{\partial x} + v_y \frac{\partial\vec{v}}{\partial y} + v_z \frac{\partial\vec{v}}{\partial z}\right)dt$$

so finally the acceleration is

$$\vec{a} = \frac{\partial\vec{v}}{\partial t} + v_x \frac{\partial\vec{v}}{\partial x} + v_y \frac{\partial\vec{v}}{\partial y} + v_z \frac{\partial\vec{v}}{\partial z}$$ which is close to your original expression, but rather be written as

$$\vec{a} = \frac{\partial\vec{v}}{\partial t} + (\vec{v} \cdot \frac{\partial}{\partial \vec{r}})\vec{v}$$

• @xaxa: yeah that must be it: it's a sensible way to bridge the what the OP wrote with anything close to an acceleration. – ZeroTheHero Feb 6 '17 at 13:45
• I am sorry, but the initial part is not true. In case of a single particle the expression is not wrong. As $v$ is a function of $t$ so is the position $x$ and in principle, for some region in space and time, one can invert $x(t)$ to $t(x)$ and therefore $v(x)$ is absolutely ok and is just a "coordinate" transformation. – mikuszefski Feb 6 '17 at 14:10
• @mikuszefski Strictly speaking the expression is wrong in any case because it has scalar on the right and vector on the left - this can happen only in one dimensional case - while OP is clearly asking about 3D – xaxa Feb 6 '17 at 14:25
• True; yet, my point is that it is true for a single particle (in 1D .... in 3D you have to take care how to write the vector) – mikuszefski Feb 6 '17 at 14:28
• Even a single particle can be moving in a complex 3D trajectory e.g. charged particle in magnetic field. – xaxa Feb 6 '17 at 14:31

If you knew differential geometry, velocity and acceleration can be written as:

\begin{align*} \mathbf{v} &= \dot{s} \, \mathbf{T} \\ &= v \, \mathbf{T} \\ \mathbf{a} &= \ddot{s} \, \mathbf{T}+ \kappa \, \dot{s}^2\mathbf{N} \\ &= \frac{d^2s}{dt^2} \mathbf{T} + \frac{v^2}{\rho} \mathbf{N} \\ &= v\frac{dv}{ds} \mathbf{T} + \frac{v^2}{\rho} \mathbf{N} \\ \end{align*}

where $\kappa$ is the curvature and $\rho$ is the radius of curvature.

Note that $v\dfrac{dv}{ds}$ is the tangential component of acceleration whereas $\dfrac{v^2}{\rho}$ is the centripetal acceleration.

Also \begin{align*} \int \mathbf{F} \cdot d\mathbf{r} &= \int m\mathbf{a} \cdot \mathbf{v} \, dt \\ &= m\int v^2 \, \frac{dv}{ds} dt \\ &= m\int v^2 \frac{dt}{ds} \, dv \\ &= m\int v \, dv \\ &= \frac{m(v^2-u^2)}{2} \end{align*} which is work done by the force $\mathbf{F}$ and the centripetal component does zero work.

Some facts from differential geometry \begin{align*} s &= \int |\mathbf{v}| \, dt \tag{arclength} \\ \dot{s} &= |\mathbf{v}| \tag{speed} \\ &= v \\ \mathbf{T} &= \frac{\mathbf{v}}{v} \tag{tangent vector}\\ \mathbf{B} &= \frac{\mathbf{v} \times \mathbf{a}}{|\mathbf{v} \times \mathbf{a}|} \tag{binormal vector} \\ \mathbf{N} &= \mathbf{B} \times \mathbf{T} \tag{normal vector} \\ \kappa &= \frac{|\mathbf{v} \times \mathbf{a}|}{v^3} \tag{curvature} \\ \rho &= \frac{1}{\kappa} \tag{radius of curvature} \end{align*}

• I'm in 8th grade. I don't know differential geometry – xasthor Feb 6 '17 at 12:37
• All you need to know is $$mv \, dv=m\mathbf{a} \cdot d\mathbf{r}$$ that relates kinetic energy with work. So $v\dfrac{dv}{ds}$ really has physical meaning. – Ng Chung Tak Feb 6 '17 at 12:57
• You should introduce all letters that you use, especially as you virtually already assume that the OP is not familiar with the topic. – mikuszefski Feb 6 '17 at 14:26

When we do $\frac{d}{dt}\vec{v}$, when $\vec{v}$ has components along multiple axes, we simply bring the derivative inside the vector, such that for a vector $\vec{v(t)}=(x(t),\,y(t),\,z(t))$, $$\frac{d}{dt}\vec{v(t)}=(\frac{d}{dt}x(t),\,\frac{d}{dt}y(t),\,\frac{d}{dt}z(t)),$$ which you can verify will be something of the form, $$\frac{d}{dt}x(t)\hat{i}+\frac{d}{dt}y(t)\hat{j}+\frac{d}{dt}z(t)\hat{k}.$$ Similarly, we integrate to respect to each component to get velocity from acceleration. The total rate of change in velocity (i.e. acceleration) is the rate of change in all components, and it does indeed have directionality. Think of "multiplication" by $\frac{d}{dt}$ as scalar multiplication.