# What does it mean to take the “ gradient with respect to the position $r_ i$”?

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.

Edit:

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: $$i=a$$, $$i and $$i>a$$. Then, we get: \begin{align} \nabla_a(U_{\text{total}}) &= \sum_{j>a} \nabla_a(U_{aj}) + \sum_{ii} \nabla_a(U_{ij}) + \sum_{i>a} \sum_{j>i} \nabla_a(U_{ij}) \end{align} Now, recall the above mentioned property that $$\nabla_a(U_{ij}) = 0$$ if $$a \notin \{i,j\}$$. So, the only non-zero terms in the above summation are: \begin{align} \nabla_a(U_{\text{total}}) &= \sum_{j>a} \nabla_a(U_{aj}) + \sum_{i where in the last line I used the fact that $$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}

• obtained by differentiating the total potential energy with respect to the 3 cartesian coordinates of that particle. should I accept it as a law? – Knight wants Loong back May 24 '20 at 5:57
• @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. – peek-a-boo May 24 '20 at 6:08
• 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) ? – Knight wants Loong back May 24 '20 at 6:38
• 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 $$\nabla V = \left( \frac{\partial V}{\partial x} , \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right)$$ – Knight wants Loong back May 24 '20 at 6:40
• @Knight take a look at the edit. – 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.

• 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? – Knight wants Loong back May 24 '20 at 6:07
• @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. – paulinho May 24 '20 at 14:47
• Yes. I agree with your comment. – Knight wants Loong back May 24 '20 at 15:18