How can I write the vector form of this equation, $a = vdv/dx$?

Question

My Physics teacher was deriving the 'Work-Energy Theorem' for a single particle in the class; where after doing the vector addition of all the forces acting on the particle, he put the resultant of all those forces equal to the product of mass and acceleration of the particle, which was totally correct for me.
After that, he put the acceleration as $v(dv/dx)$, so that he could derive the theorem after integrating all the forces with respect to $x$ (which is displacement) and integrating velocity of the object with respect to $v$ (which is velocity itself); and thus equating the sum of work done by all the forces on the object with change in the kinetic energy of the object.

This was also correct in my sight during the class when he was explaining, but later on while studying the class-notes by myself, I got confused in the step in which he put acceleration as $v(dv/dx)$. I was not able to understand how the acceleration which had been written with the vector sign (I mean, the arrow above any vector quantity), could be substituted with v(dv/dx), which do not have any vector sign.
To solve this problem, I thought to add the vector sign (the arrow, again) in the latter, but there, the main problem had arisen, for which I am writing this question, which was, "which quantity/quantities is/are vector quantity/quantities in $v(dv/dx)$?".

In simpler words, how can I write $a = v(dv/dx)$ vectorially?

A good example of what I am asking could be $F = ma$, which can also be written as $F$ (with vector sign) = mass times acceleration (with vector sign). In the same way, how can I write $a = v(dv/dx)$, with the vector sign?

Answer 1

For the $x$-component of acceleration we have$^\dagger$ \begin{align} a_{\color{blue}{x}}&=\frac{\mathrm d v_{\color{blue}{x}}}{\mathrm dx}\frac{\mathrm d x}{\mathrm dt}+\frac{\mathrm d v_{\color{blue}{x}}}{\mathrm dy}\frac{\mathrm d y}{\mathrm dt}+\frac{\mathrm d v_{\color{blue}{x}}}{\mathrm dz}\frac{\mathrm d z}{\mathrm dt}\\ &=v_x\frac{\mathrm d v_{\color{blue}{x}}}{\mathrm dx}+v_y \frac{\mathrm d v_{\color{blue}{x}}}{\mathrm dy}+v_z\frac{\mathrm d v_{\color{blue}{x}}}{\mathrm dz} \end{align} which we can write as \begin{align} a_x=\vec v\cdot\nabla v_x \end{align} where $\nabla$ is defined as the 'gradient' of a function, given by $$\nabla f=\begin{pmatrix}\frac{\partial f}{\partial x}\\\frac{\partial f}{\partial y}\\\frac{\partial f}{\partial z}\end{pmatrix}$$ For the entire vector $\vec a$ we now get \begin{align} \vec a=(\vec v\cdot\nabla)\vec v \end{align} where $(\vec v\cdot\nabla)=\left(v_x\frac{\partial }{\partial x}+v_y\frac{\partial }{\partial y}+v_z\frac{\partial }{\partial z}\right)$ is a differential operator.

$\dagger$ Note: technically we would have to use partial derivatives for $\frac{\mathrm d v_x}{\mathrm dx}$. We can write the velocity as only depending time, as $v(t)$, or as depending on position which in turn depends on time, i.e. as $v(x(t),y(t),z(t))$. When you differentiate a function depending on multiple arguments with respect to just one of its arguments you have to use partial derivatives.

Answer 2

Approach. You only need to be able to perform (partial) derivatives of composite function, and handel them carefully.

In vector form, you may think the position of point $P$ as $\mathbf{r}_P(t)$, its velocity $\mathbf{v}_P(t)$, and its acceleration $\mathbf{a}_P(t)$.

Method 1. If you define the velocity as a function of the position of the particle $\mathbf{v}_P(t) = \mathbf{v}(\mathbf{r}_P(t))$, you can evaluate the time derivative of this function as

$\dfrac{d}{dt}\mathbf{v}_P(t) = \dfrac{d}{dt} \mathbf{v}(\mathbf{r}_P(t)) = \dfrac{d v_{P,i}}{dt} = \dfrac{\partial v_i}{\partial x_j} \dfrac{dx_j}{dt} = v_j \dfrac{\partial v_i}{\partial x_j} = \mathbf{v}_P(t) \cdot \nabla\mathbf{v}(\mathbf{r}_P(t))$,

being $\mathbf{v}_P \cdot \nabla$ the directional derivative in the direction of $\mathbf{v}_P$ multiplied by the magnitude of the velocity $v_P = |\mathbf{v}_P|$,

$\dfrac{d}{dt}\mathbf{v}_P(t) = v_P(t) \mathbf{\hat{t}}(\mathbf{r}_P(t)) \cdot \nabla\mathbf{v}(\mathbf{r}_P(t))$,

being $\mathbf{\hat{t}}$ the unit vector, tangent to the trajectory.

It's important to understand the meaning and to be able to handle the directional derivative to avoid the evaluation of the gradient of the velocity $\mathbf{v}$, since the velocity $\mathbf{v}_P$ of the particle is only defined along the trajectory.

Thus, with the trajectory paramterized with its arc-length parameter $s$, $\tilde{\mathbf{r}}_P(s) = \mathbf{r}_P(t)$ the last equation reads

$\dfrac{d\mathbf{v}_P}{dt} = v_P \dfrac{d}{ds} \mathbf{v}(\tilde{\mathbf{r}}_P(s))$.

Method 2. We could have found the very same equation introducing the parametrization of the trajectory with its arc-length parameter $s$, $\mathbf{r}_P(t) = \tilde{\mathbf{r}}_P(s) = \tilde{\mathbf{r}}_P(s(t))$ and performing the derivative of the composite function $\mathbf{v}_P(t) = \mathbf{v}_P(s(t))$, as

$\dfrac{d \mathbf{v}_P}{dt} = \dfrac{d s}{dt} \dfrac{d \mathbf{v}_P}{ds} = v_P \dfrac{d \mathbf{v}_P}{ds}$,

being the magnitude of the velocity $v_P = \frac{ds}{dt}$ by definition of the arc-length parameter.

Answer 3

The equation $a=\frac{dv}{dx}v$ is non-trivial to generalise to multiple dimensions, because $dx$ becomes a vector in multiple dimensions and you can't divide by a vector. The other answers have mentioned an approach with the Jacobian matrix (which is a way to kinda differentiate vector w.r.t. to a vector)

A simpler (but less general) generalisation can be:

$$\vec{a}\cdot\vec{dx}=\vec{v}\cdot \vec{dv}$$

$\vec{dx}$ and $\vec{dv}$ are small changes in position and velocity

To see why this is correct, just expand the dot product component wise:

$$\sum a_i dx_i=\sum v_idv_i$$

This equation is true because $a_i=v_i\frac{dv_i}{dx_i}$ (and hence, $a_idx_i=v_idv_i$) is separately true for each of the three individual axes $i=1,2,3$.

Furthermore, you can use $\vec{a}\cdot\vec{dx}=\vec{v}\cdot \vec{dv}$ to get the multiple dimension version of the work energy theorem. Just integrate both sides and multiply by $m$ to get:

$$\int m\vec{a}\cdot \vec{dx}=\sum \int mv_i dv_i$$

$$=\sum m\frac{v_i^2}{2}+C$$

Answer 4

I have problem with your equation

the general case is:

that Position vector $~\mathbf{R}~$ is a function of the generalized coordinates $~\mathbf{q(t)}$ \begin{align*} &\mathbf{R}= \mathbf{R}(\mathbf{q(t)})\quad\Rightarrow\text{the velocity }\\ &\mathbf{v}=\mathbf{\dot{R}}=\frac{\partial \mathbf{R}}{\partial\mathbf{q}}\,\mathbf{\dot{q}}=\frac{\partial {R_i}}{\partial{q_j}}\,{\dot{q}_j}\quad\Rightarrow\text{the acceleration }\\ &\mathbf{a}=\mathbf{\dot{v}}=\frac{d}{dt}\left(\frac{\partial {R_i}}{\partial{q_j}}\,{\dot{q}_j}\right)= \frac{\partial {R_i}}{\partial{q_j}}\,{\ddot{q}_j}+ \frac{\partial}{\partial q_k}\left(\frac{\partial {R_i}}{\partial{q_j}}\,{\dot{q}_j}\right)\dot{q_k}\\ &\text{Newton}\\ &m\,\mathbf{a}=\mathbf{F} \end{align*} your case is: \begin{align*} &\mathbf{R}(t)=\begin{bmatrix} x_1(t) \\ x_2(t) \\ x_3(t) \\ \end{bmatrix}\quad, \mathbf{q}=\mathbf{R}\quad\Rightarrow\\ &\mathbf{v}=\frac{\partial {R_i}}{\partial{q_j}}\,{\dot{q}_j}\quad, v_i=\dot{x}_i \end{align*} thus the velocity is not a function of the x coordinates

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if $~ x_i~$ are generalized coordinates and the velocity vector is a function of the generalized coordinates then :

\begin{align*} &\mathbf v=\mathbf{v}(\mathbf{x}(t))\quad\Rightarrow &a_i=\frac{d}{dt}\,v_i=\frac{\partial v_i}{\partial x_j}\dot{x}_j \end{align*}

but this just „algebra“