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".

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 $(5)$.

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    $\begingroup$ It would be very easy to express with the Einstein summation convention... But I'm guessing that's not what you're after? $\endgroup$ – Marius Ladegård Meyer May 17 '20 at 6:15
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    $\begingroup$ You can use this equation $ \overrightarrow{a}=\left[ \dfrac{\partial \overrightarrow{v}}{\partial\overrightarrow{x}}\right] \overrightarrow{\dot x}$ a is matrix times vector $\endgroup$ – Eli May 17 '20 at 6:16
  • $\begingroup$ @Eli How can you differentiate a vector with respect to a vector? $\endgroup$ – user258881 May 17 '20 at 6:19
  • $\begingroup$ There is a reference here. But the quantity in brackets is normally associated with the Jacobian, which is also what the link seems to mean. $\endgroup$ – Marius Ladegård Meyer May 17 '20 at 6:20
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    $\begingroup$ @Eli Hmm... I see. That's quite close to what I wanted. Maybe you could turn it into an answer :) I would love to accept it. $\endgroup$ – user258881 May 17 '20 at 6:26

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

  • $\begingroup$ But isn't $\frac{\partial \mathbf v}{\partial x}$ different from $\frac{\mathrm d v_x}{\mathrm dx}$? $\endgroup$ – user258881 May 17 '20 at 6:30
  • $\begingroup$ Yes, it’s different. The former has three components and the latter is a single number. $\endgroup$ – G. Smith May 17 '20 at 6:32
  • $\begingroup$ So how come your expression and my equation $(5)$ are the same? $\endgroup$ – user258881 May 17 '20 at 6:35
  • $\begingroup$ They aren’t the same. Your equation is wrong. $\endgroup$ – G. Smith May 17 '20 at 6:35
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    $\begingroup$ You applied a one-dimensional result to three dimensions. To me, your derivation doesn’t make sense. My derivation is just the 3D chain rule. It’s not the same as three 1D chain rules. For example, $v_x$ can vary in the $y$ and $z$ directions. $\endgroup$ – G. Smith May 17 '20 at 6:43

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 ) :

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


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

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

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

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


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

  • $\begingroup$ +1! Good answer :) I did not accept your answer because I found G.Smith's answer more closer to what I expected. I didn't know that it could be solely expressed using vector calculus, though your answer provides another POV on the question. Thanks! $\endgroup$ – user258881 May 17 '20 at 7:06

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