I've been trying to understand this concept for hours without any success. I found similar questions on this forum (Derivative with respect to a vector is a gradient?) but I still don't understand.

Let's consider a multiple particle system with $n$ particles. I'm running multiple times into the notation $$\vec{F_k}=- \frac{\partial}{\partial \vec{x_k}} V(\vec{x_1},\vec{x_2},..., \vec{x_n})$$ where $F_k$ denotes the resulting force acting on the particle $k$.

I assume the right hand side is an $\mathbb{R}^3$ vector because the left hand side is.

I came up with one example: Let $f:\mathbb{R}^n \rightarrow \mathbb{R}, f(x_1, ... ,x_n)=3x_1+{x_2}^2x_4-5x_3+...+x_n$ Then (because the variables are real) I know: $$\frac{\partial}{\partial x_1}f(x_1,...x_n)=3 \quad \frac{\partial}{\partial x_2}f(x_1,...x_n)=2x_2x_4 \quad \frac{\partial}{\partial x_3}f(x_1,...x_n)=-5$$ Now I guess one could define $\vec{a}=(x_1,x_2,x_3)^T$, then $\frac{\partial}{\partial \vec{a}}f(x_1,...x_n)=(3,2x_2x_4,-5)^T$

Even if this is correct, I'm struggling to apply this idea to the potential $V$ above because the variables there aren't scalars. What if for example $V(\vec{x_1},\vec{x_2},..., \vec{x_n})=\vec{x_1}^3$ ? Raising a vector to the 3rd power doesn't make any sense to me. Should I think of V as a function $V(x_{11},x_{12},x_{13},...,x_{n1},x_{n2},x_{n3}): \mathbb{R}^{3n}\rightarrow \mathbb{R}$?

I'm sorry about the long question but I hope to have at least clarified my problems and possibly helped you to help me.


3 Answers 3


I think the notation $ \frac{\partial}{\partial{\vec{x_k}}} $ stand for the gradient $ \vec{\nabla} $. The gradient operator act on scalar funtion like $V$ and gives a vector : $$\vec{\nabla} V= \begin{pmatrix} \frac{\partial}{\partial_{x_1}}V\\ \dots \\ \frac{\partial}{\partial_{x_n}}V \end{pmatrix} $$ Here you describe $V$ as a function of multiple vectors but i think is like you use one vector that contains all the $\vec{x_i}$ coefficients $V(x_{11},x_{12},...,x_{n3})$. You give the example of $V=\vec{x_1}^3$ in this case $V$ depend only on $x_1$ and either $V=x_1^2 \vec{x_1}$ and $V$ is a vector or $V=|\vec{x_1}|^3$ and this is a scalar.

I hope this helps you.


If you're talking about newtonian mechanics, then $\mathbf{F} = - \nabla V$. I'm not sure why you're indexing $\mathbf{F}$, either it's a vector $\mathbf{F}$ or you're referring to the components $F_k$. The components are then $F_k = \left[-\nabla V \right]_k = -\frac{\partial V}{\partial x_k}$. In general if you have a scalar function $f: \mathbb{R}^n \to \mathbb{R}$ then you can define the directional derivative just like a normal derivative as \begin{align*} \partial_{\mathbf{v}} f\big|_{\mathbf{x_0}}& = \lim_{h \to 0} \frac{f(\mathbf{x}_0 + h \mathbf{v}) - f(\mathbf{x}_0)}{h} \\ &= \mathbf{v} \cdot \nabla f \end{align*} When you have multiple variables then the slope of the function can obviously be different if you're "walking" in different directions.

The notation $\mathbf{x}^3$ is usually meant as repeated scalar products, therefore $\mathbf{x}^2 = ||\mathbf{x}||^2 = x^2 + y^2 + z^2$ is actually just a scalar but $\mathbf{x}^3 = \mathbf{x}^2 \mathbf{x} = (x^2 + y^2 + z^2) \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ is then a vector. Taking the derivative along $\mathbf{v} = (1, 0, 1)$ would then be \begin{align*} \partial_{\mathbf{v}} \mathbf{x}^3 &= \mathbf{v} \cdot \nabla \mathbf{x}^3 \\ &= \mathbf{v} \cdot \begin{pmatrix} 3x^2 + y^2 + z^2 \\ x^2 + 3y^2 + z^2 \\ x^2 + y^2 + 3z^2 \end{pmatrix} \\ &= 1(3x^2 + y^2 + z^2) + 0 (x^2 + 3y^2 + z^2) + 1 (x^2 + y^2 + 3z^2) \end{align*}

To illustrate it with multiple particles, let's look at the Kepler problem and ignore the constants for a second, then $V(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} = \left((x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 + z_2)^2 \right)^{-1/2} = V(x_1, y_1, z_1, x_2, y_2, z_2)$. The force on the first particle is then \begin{align*} \mathbf{F}_1(\mathbf{r}_1, \mathbf{r}_2) &= - \frac{\partial V}{\partial \mathbf{r}_1} = -\nabla_{\mathbf{r}_1} V = (1, 1, 1, 0, 0, 0) \nabla V \\ &= - \begin{pmatrix} \partial_{x_1} V \\ \partial_{y_1} V \\ \partial_{z_1} V \end{pmatrix} \\ &= -\frac{1}{2}\frac{1}{((x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 )^{3/2}} \begin{pmatrix} 2(x_1 - x_2) \\ 2(y_1 - y_2) \\ 2(z_1 - z_2) \end{pmatrix} \\ &= - \frac{\mathbf{r}_1 - \mathbf{r}_2}{|\mathbf{r}_1 - \mathbf{r}_2|^3} \end{align*} Note that $(1, 1, 1, 0, 0, 0)$ points along $\mathbf{r}_1$. The last line is just Newton's law of gravity $\mathbf{F}_G = \frac{1}{r^2}\mathbf{e}_r$. So your comment is correct.

  • $\begingroup$ I think I'm getting there. I consider a multiple particle system in newtonian mechanics, $F_k$ denotes the resulting force on particle $k$. As 2piOmega suggested, is it correct to think of $V=V(x_{11},x_{12},x_{13},...,x_{n3})$ as a function $V:\mathbb{R}^{3n}\rightarrow \mathbb{R}$?. If yes, is it then correct to say $\frac{\partial}{\partial \vec{x_k}}V=\frac{\partial}{\partial (x_{k1},x_{k2},x_{k3})^T}V=(\frac{\partial}{\partial x_{k1}}V,\frac{\partial}{\partial x_{k2}}V,\frac{\partial}{\partial x_{k3}}V)^T $ ? I think here is where my main problem lies. $\endgroup$ Commented Apr 1, 2021 at 10:42
  • 1
    $\begingroup$ I edited my answer, I hope it helps. $\endgroup$
    – Wihtedeka
    Commented Apr 1, 2021 at 11:59
  • $\begingroup$ Your example was exactly what I needed. Thank you! $\endgroup$ Commented Apr 1, 2021 at 12:37

Your expression appears to be referring to the scalar potential, V, (of an object) the value of which depends on its position in a multi-space specified by coordinates: x, y, z, and perhaps more. Then the force on the object depends on the gradient of the potential which is found by summing the partial derivatives with respect to each dimension. Though I have not seen a formal definition of the operation of dividing a scalar by a vector, this expression seems to imply that would be done in terms of components.


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