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Let $f:\mathbb{R}^{m+n}\to \mathbb{R}^m,\ \mathbf{t}=[x_1,\cdots,x_m,y_1,\cdots,y_n]\to [f_1(\mathbf{t}),\cdots,f_m(\mathbf{t})]$ be a differentiable function. Here $[a_1,\cdots,a_m]$ means a column vector with a height of $m.$ I think that the symbol $\partial f\over {\partial \mathbf{y}}$ represents the matrix \begin{equation*}     \left(                \begin{array}{cccc}   \displaystyle{{\partial f_1}\over {\partial y_1}}&\displaystyle{{\partial f_1}\over {\partial y_2}}&\cdots&\displaystyle{{\partial f_1}\over {\partial y_n}}\\    \displaystyle{{\partial f_2}\over {\partial y_1}}&\displaystyle{{\partial f_2}\over {\partial y_2}}&\cdots&\displaystyle{{\partial f_2}\over {\partial y_n}}\\  \vdots&\vdots&\ddots&\vdots\\ \displaystyle {{\partial f_m}\over {\partial y_1}}&\displaystyle{{\partial f_m}\over {\partial y_2}}&\cdots&\displaystyle{{\partial f_m}\over {\partial y_n}} \end{array} \right) ,     \end{equation*} so when $m=1,$ we have ${\partial f\over {\partial \mathbf{y}}}=({\partial f\over {\partial y_1}},\cdots,{\partial f\over {\partial y_n}}).$ Thus $\partial f\over{\partial\mathbf{y}}$ is a row vector.

But in mechanics the following equation can be seen everywhere: $$m{d\mathbf{v}\over{dt}}=-{\partial U\over \partial{\mathbf{r}}}. $$ But if we view radius vector $\mathbf{r}$ as a column vector, then $\displaystyle{d\mathbf{v}\over{dt}}$ is a column vector, so it seems that $$ {\partial U\over \partial{\mathbf{r}}}=[{\partial U\over {\partial x}},{\partial U\over {\partial y}},{\partial U\over {\partial z}}],$$ which contradicts my definition.

Question: Should the equation be rewritten as $$m{d\mathbf{v}\over{dt}}=-\Big({\partial U\over \partial{\mathbf{r}}}\Big)^T ?$$

painday
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  • Nothing stops you from writing the acceleration vector as a column vector in your equation which makes it all correct, there are actually two layouts for the gradient https://en.m.wikipedia.org/wiki/Matrix_calculus – Triatticus Nov 09 '17 at 11:08

1 Answers1

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The correct way to write the equation you are considering is $m\frac{d\mathbf{v}}{dt} = -\nabla U$, where $\nabla U$ is the gradient, which spits out a vector. As you observe, the Jacobian $\frac{d U}{d\mathbf{x}}$ is not the same as the gradient $\nabla U$ (strictly speaking, the entries are the same, but the Jacobian has a preferred matrix dimension, thus we can use Jacobian notation, but it's a bit of an abuse of notation).

Whether you choose to write $\frac{d\mathbf{v}}{dt}$ and $\nabla U$ as row or column vectors in this equation is really your personal preference.

See https://en.wikipedia.org/wiki/Conservative_vector_field and https://en.wikipedia.org/wiki/Conservative_force

AnonymousCoward
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  • Ummm, but when the potential energy is determined by more than one objects, we have $U=U(\mathbf{r_1},\cdots,\mathbf{r_s}),$ and it seems that many physicists use the symbol ${\partial U\over{\partial \mathbf{r_i}}}$ to represent a column vector... – painday Nov 09 '17 at 09:05
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    physicists like to abuse all kinds of symbols. What they mean is gradient of a real valued function. Note that if each $\mathbf{r_i}$ is the distance from your particle to some fixed object of a given mass, then it is in fact just a function of the position of your particle (i.e. the coordinate $\mathbf{x}$). In general, if multiple particles are moving then you have a bigger configuration space, but the $\mathbf{r_i}$ are still all functions of the position coordinates, so really $U = U(\mathbf{r}_1, \ldots \mathbf{r}_s) = U(\mathbf{r}_1(\mathbf{x}), \ldots \mathbf{r}_s(\mathbf{x}))$. – AnonymousCoward Nov 09 '17 at 09:23
  • $\frac{\partial U}{\partial \vec{x}}$ is a gradient though, or it's transpose depending on your layout choice, it satisfies all the conditions of being the vector of partial derivatives of the scalar function. https://en.m.wikipedia.org/wiki/Matrix_calculus – Triatticus Nov 09 '17 at 11:14
  • Yes, the entries are the same. But the Jacobian has a fixed choice of matrix dimensions, which is the source of OP's confusion. The definition as a gradient and not a Jacobian is the reason that taking transpose doesn't matter. – AnonymousCoward Nov 09 '17 at 11:56