Alternative expression of acceleration in vector form

Question

Let's imagine a one dimensional case, where a particle is moving with a velocity $v$ and an acceleration $a$. Thus

$$a=\frac{\mathrm dv}{\mathrm dt}\tag{1}$$

Applying the chain rule, equation $(1)$ can be rewritten as

$$a=\frac{\mathrm dv}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}\Longrightarrow \boxed{a=v\frac{\mathrm dv}{\mathrm dx}}\tag{2}$$

Now, if we were dealing with a 2D or a 3D case, then we would use vectors in the above expressions. Thus

\begin{alignat}{2} a&=\frac{\mathrm dv}{\mathrm dt}&&\Longrightarrow\mathbf a=\frac{\mathrm d \mathbf v}{\mathrm dt}\tag{3}\\ a&=v\frac{\mathrm dv}{\mathrm dx}&&\Longrightarrow \mathbf a=\:\:?\tag{4} \end{alignat}

As you can see, the vector form of equation $(1)$ (which is equation $(3)$) can be easily found, however I do not know of any way to express the equation $(2)$ in vector form.

The natural thought was to express the velocity into its components. For a 3D case, let $\mathbf v=v_x\mathbf{\hat i}+v_y\mathbf{\hat j}+v_z\mathbf{\hat k}$. Doing this, we have essentially converted the 3D case to three 1D cases. Thus using equation $(2)$:

$$\mathbf a =v_x\frac{\mathrm d v_x}{\mathrm dx}\mathbf{\hat i}+v_y\frac{\mathrm d v_y}{\mathrm dy}\mathbf{\hat j}+v_z\frac{\mathrm d v_z}{\mathrm dz}\mathbf{\hat k}\tag{5}$$

However, this expanded version doesn't seem particularly useful to me. Is there any way to express equation $(5)$ in a "closed form" (without explicitly writing out the components)? I feel that writing it in closed form might involve some common vector calculus operators (along with dot and cross products), though I am not exactly sure how to express it in a "closed form".

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Justification of equation $(5)$: We kow that $\mathbf a=a_x\mathbf{\hat i}+a_y\mathbf{\hat j}+a_z\mathbf{\hat k}$

Now since

$$a_x=\frac{\mathrm dv_x}{\mathrm dt}=v_x\frac{\mathrm d v_x}{\mathrm d x}$$

Thus subsitituting this for every component, we re-obtain equation $(5)$.

Answer 1

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{dx}{dt}\frac{\partial\mathbf{v}}{\partial x}+\frac{dy}{dt}\frac{\partial\mathbf{v}}{\partial y}+\frac{dz}{dt}\frac{\partial\mathbf{v}}{\partial z}=(\mathbf{v}\cdot\nabla)\mathbf{v}$$

Answer 2

I am a bit confused with your notation so I chose my notation.

Assume the components of the position vector $\vec{R}=[x_1,x_2,\ldots,x_{n_R}]^T $are function of the generalized coordinates $q_1,q_2,\ldots,q_{n_Q}$ thus: $x_j=x_j(q_i)$ where $j=1,(1),n_R$ and $i=1,(1),n_Q\quad ,n_Q \le n_R$

We want to obtain the velocity vector $\vec{v}=\frac{d\vec{R}}{dt}$

$$\dot{x}_1=\frac{\partial x_1}{\partial q_1}\,\dot{q}_1+\frac{\partial x_1}{\partial q_2}\,\dot{q}_2+\ldots$$

$$\dot{x}_2=\frac{\partial x_2}{\partial q_1}\,\dot{q}_1+\frac{\partial x_2}{\partial q_2}\,\dot{q}_2+\ldots$$

or $$\dot{x}_j=\sum_i^{nQ}\frac{\partial x_j}{\partial q_i}\,\dot{q}_i$$

or with vector notation (Engineer notation ) :

$$\vec{v}=\vec{\dot R}=\underbrace{\frac{\partial \vec{R}}{\partial \vec{q}}}_{n_R\times n_Q}\,\vec{\dot{q}}$$

Example:

$$\vec{R}=\left[ \begin {array}{c} x_{{1}} \left( q_{{1}},q_{{2}} \right) \\ x_{{2}} \left( q_{{1}},q_{{2}} \right) \\ x_{{3}} \left( q_{{1}},q_{{2}} \right) \end {array} \right] =\left[ \begin {array}{c} r\sin \left( q_{{1}} \right) \cos \left( q_{ {2}} \right) \\r\sin \left( q_{{1}} \right) \sin \left( q_{{2}} \right) \\ r\cos \left( q_{{1}} \right) \end {array} \right] $$

$$\vec{q}=\left[ \begin {array}{c} q_{{1}}\\ q_{{2}} \end {array} \right] $$

$$\underbrace{\frac{\partial \vec{R}}{\partial \vec{q}}}_{3\times 2}= \left[ \begin {array}{cc} {\frac {\partial }{\partial q_{{1}}}}x_{{1} } \left( q_{{1}},q_{{2}} \right) &{\frac {\partial }{\partial q_{{2}}} }x_{{1}} \left( q_{{1}},q_{{2}} \right) \\ {\frac { \partial }{\partial q_{{1}}}}x_{{2}} \left( q_{{1}},q_{{2}} \right) &{ \frac {\partial }{\partial q_{{2}}}}x_{{2}} \left( q_{{1}},q_{{2}} \right) \\{\frac {\partial }{\partial q_{{1}}}}x_{ {3}} \left( q_{{1}},q_{{2}} \right) &{\frac {\partial }{\partial q_{{2 }}}}x_{{3}} \left( q_{{1}},q_{{2}} \right) \end {array} \right] $$

Thus: $$\vec{v}=\left[ \begin {array}{cc} r\cos \left( q_{{1}} \right) \cos \left( q_ {{2}} \right) &-r\sin \left( q_{{1}} \right) \sin \left( q_{{2}} \right) \\ r\cos \left( q_{{1}} \right) \sin \left( q_{{2}} \right) &r\sin \left( q_{{1}} \right) \cos \left( q_{{ 2}} \right) \\ -r\sin \left( q_{{1}} \right) &0 \end {array} \right] \,\left[ \begin {array}{c} \dot{q}_{{1}}\\ \dot{q}_{{2}} \end {array} \right] $$

Remark:

The velocity vector $\vec{v}$ is a function of $\vec{q}$ and $\vec{\dot{q}}$. Your vector v is only a function of $\vec{q}$ this is not the general case