Let’s say we have a number of particles (charged, massive or anything that can create potential energy). The total potential energy of any particle can be given by $$U_i (\vec r_1, \vec r_2, ... \vec r_N) = \sum_{j\gt i}^{N} U_{ij}(|r_i -r_j|)$$ and hence total potential of the system is $$ U_{total}(\vec r_1, \vec r_2, ..., \vec r_N)= \sum_{i=1}^{N} U_i$$

($U_{ij}$ means potential energy of the particle $i$ due to the particle $j$)

Now, to get the force on the particle $i$ we write $$ \mathbf F_i = -\nabla_{r_i} U_{total}(\vec r_1, \vec r_2, ..., \vec r_n) $$

Now, that subscript after $\nabla$ is posing some problems for me.

  1. What is meant by “gradient with respect to $r_i$? Isn’t $r_i$ a fixed position of $i$ th particle? It’s same as saying “take the derivative of $f$ with respect to number 2”.

  2. Why not to take the simple gradient, that is only $\nabla$, of $U_{total} (\vec r_1, \vec r_2, ... \vec r_N)$ and evaluate the gradient at $r_i$ ?

Can you please resolve my doubts?

Here is a Stackexchange post which address the similar problem, but I’m quite unsatisfied with the answers over there.


The total potential energy is a function $U: \left(\Bbb{R}^3\right)^N \to \Bbb{R}$. Where the physical idea is that the $i^{th}$ copy of $\Bbb{R}^3$ tells us the position $(x_i, y_i, z_i)$ of the $i^{th}$ particle. So, writing something like $\mathbf{F}_i = - \nabla_{\mathbf{r}_i}U$ means: $\mathbf{F}_i : \left(\Bbb{R}^3\right)^N \to \Bbb{R}^3$ is the vector valued function defined as \begin{align} \mathbf{F}_i &= - \left( \dfrac{\partial U}{\partial x_i}\, \mathbf{e}_1 + \dfrac{\partial U}{\partial y_i}\, \mathbf{e}_2 + \dfrac{\partial U}{\partial z_i}\, \mathbf{e}_3\right) \\ &\equiv - \left(\dfrac{\partial U}{\partial x_i}, \dfrac{\partial U}{\partial y_i}, \dfrac{\partial U}{\partial z_i} \right), \end{align} where I use $\equiv$ to mean they're the same thing, expressed in different notation.

In other words, the force $\mathbf{F}_i$ on the $i^{th}$ particle is obtained by differentiating the total potential energy with respect to the $3$ cartesian coordinates of that particle.


The short answer to the question in the comments is that "no, $\mathbf{F}_a = - \nabla_{\mathbf{r}_a}(U_{\text{total}})$" is a mathematical statement which needs to be proven and is not a "law". Here, we crucially make use of the fact that the potentials $U_{ij}$ depend only on $|\mathbf{r}_i - \mathbf{r}_j|$.

By definition, the total force on $a^{th}$ particle, due to all other particles is \begin{align} \mathbf{F}_a = \sum_{j \neq a} \mathbf{F}_{\text{$j$ on $a$}} \end{align} What is $\mathbf{F}_{\text{$j$ on $a$}}$? Well, you simply take the potential energy $U_{aj}$ and differentiate with respect to $\mathbf{r}_a$, i.e $\mathbf{F}_{\text{$j$ on $a$ }} = - \nabla_{a}(U_{aj})$ (for ease of typing, I use $\nabla_a$ to mean $\nabla_{\mathbf{r_a}}$, which I defined above to mean differentiation with respect to $x_a, y_a, z_a$). Hence, \begin{align} \mathbf{F}_a = \sum_{j \neq a} \mathbf{F}_{\text{$j$ on $a$}} = - \sum_{j \neq a} \nabla_{a}(U_{aj}) \tag{$*$} \end{align}

I now claim that this is also equal to $-\nabla_a U_{\text{total}}$. To prove this, note first of all that $U_{ij} = U_{ji}$ and that $\nabla_{a}(U_{ij}) = 0$ if $a \notin \{i,j\}$ (because the potential energy depends only on the distance between the $i$ and $j$ particles, so clearly it doesn't depend on some other $\mathbf{r}_a$).

So, we have \begin{align} \nabla_a(U_{\text{total}}) &= \nabla_a\left( \sum_{i=1}^n \sum_{j>i} U_{ij}\right) \end{align}

Now, we shall split the sum over $i$ into 3 pieces: $i=a$, $i<a$ and $i>a$. Then, we get: \begin{align} \nabla_a(U_{\text{total}}) &= \sum_{j>a} \nabla_a(U_{aj}) + \sum_{i<a} \sum_{j>i} \nabla_a(U_{ij}) + \sum_{i>a} \sum_{j>i} \nabla_a(U_{ij}) \end{align} Now, recall the above mentioned property that $\nabla_a(U_{ij}) = 0$ if $a \notin \{i,j\}$. So, the only non-zero terms in the above summation are: \begin{align} \nabla_a(U_{\text{total}}) &= \sum_{j>a} \nabla_a(U_{aj}) + \sum_{i<a} \nabla_a(U_{ia}) + 0 \\ &= \sum_{j \neq a} \nabla_a(U_{aj}), \tag{$**$} \end{align} where in the last line I used the fact that $U_{ia} = U_{ai}$, and I renamed summation indices and combined everything. So, if you combine $(*)$ and $(**)$ then you immediately see that \begin{align} \mathbf{F}_a &= - \nabla_a(U_{\text{total}}). \end{align}

  • $\begingroup$ obtained by differentiating the total potential energy with respect to the 3 cartesian coordinates of that particle. should I accept it as a law? $\endgroup$ May 24 '20 at 5:57
  • $\begingroup$ @Knight That sentence was merely meant to summarize my answer regarding your question about the meaning of the notation $\nabla_{\mathbf{r}_i}U$. What you're now asking is a completely different question. Anyway, in this context, it's not a fundamental postulate of any kind; this follows pretty much by the relationship between conservative forces and their associated potentials. $\endgroup$
    – peek-a-boo
    May 24 '20 at 6:08
  • $\begingroup$ Actually, I meant that should I accept that subscript thing as a law? That is to get the force on the $i$ th particle I should write $\nabla_{r_i} U_{total}$ (knowing that gradient of a potential is force) ? $\endgroup$ May 24 '20 at 6:38
  • $\begingroup$ The problem is arising because when we have just one particles then the force per unit mass/charge at any point is simply the negative gradient of potential, that is simply $$\nabla V = \left( \frac{\partial V}{\partial x} , \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right)$$ $\endgroup$ May 24 '20 at 6:40
  • $\begingroup$ @Knight take a look at the edit. $\endgroup$
    – peek-a-boo
    May 24 '20 at 7:47

What is meant by taking the gradient with respect to $\vec{r}_i$ is to calculate the quantity $$\vec{\gamma} = \left(\frac{\partial U_{total}}{\partial r_{i, 1}}, \frac{\partial U_{total}}{\partial r_{i, 2}}, \frac{\partial U_{total}}{\partial r_{i, 3}}\right)$$ where $r_{i, j}$ is the $j$th coordinate of the vector $\vec{r}_i$. What is crucial to note here is that neither the vector $\vec{r}_i$ nor its components are fixed quantities. They are variables, or degrees of freedom, that the system depends upon. That is why taking the derivative of a function with respect to them makes sense in the first place.

  • $\begingroup$ What is crucial to note here is that neither the vector 𝑟⃗𝑖 nor its components are fixed quantities I’m unable to understand why is it so? If $r_i$ is not the position of $i$ th particle then what it is? Is $r_1$ a fixed position of particle 1? $\endgroup$ May 24 '20 at 6:07
  • $\begingroup$ @Knight You have a nice answer above, but what I'm saying is that the system is dynamic - namely, the particles in the system will move. It's like in one dimension with a spring where $V = \frac12 kx^2$, and $x$ will be fixed at any specific moment in time, but it still makes sense to take the derivative $dV/dx$. In fact, Newton's laws state that $F = -dV/dx = ma$ and tell us that the evolution of the physical system in time depend on the displacement w.r.t. to the equilibrium position of the spring. $\endgroup$
    – paulinho
    May 24 '20 at 14:47
  • $\begingroup$ Yes. I agree with your comment. $\endgroup$ May 24 '20 at 15:18

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