Gradient of the potential function

  • Thread starter Belginator
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  • #1
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Hi guys,

I'm trying to take the gradient of the potential function, and know the answer, but am not sure how to go about it. Can someone help me step by step as to how to do this.

So the potential function is:

\begin{equation}
U = \frac{1}{2} G \sum^{N}_{i=1} \sum^{N}_{j=1,j \neq i} \frac{m_i m_j}{\| \mathbf{r}_{ji} \|}
\end{equation}

Now I'm trying to take the gradient or partial with respect to $$\mathbf{r}_i$$

\begin{equation}
\frac{\partial U}{\partial \mathbf{r}_i} = -G \sum^{N}_{j=1,j \neq i} \frac{m_i m_j}{\| \mathbf{r}_{ji} \|^3} \mathbf{r}_{ji}
\end{equation}

So my question is, how do you go from the first equation to the answer (the second equation). If you could explain step by step with math that'd be appreciated. Thanks!
 

Answers and Replies

  • #2
AlephZero
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Sometimes it helps to go back to basics: start with one term of the sum and write out what it means.
$$U = \frac{1}{2}G\frac{m_1m_2}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}$$
Then find the three components of the vector ##\displaystyle\frac{\partial U}{\partial \mathbf{r}_1}##, i.e. ##\displaystyle\frac{\partial U}{\partial x_1}, \frac{\partial U}{\partial y_1}, \frac{\partial U}{\partial z_1}##.
 
Last edited:
  • #3
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I see what you're saying, but I think that's making it unnecessarily complicated. I know there's a way to do it by simply keeping it in the vector form. I was originally breaking it down using the following equation:

\begin{equation}
\mathbf{r}_{ji} = \mathbf{r}_{i} - \mathbf{r}_{j}
\end{equation}

But still couldn't quite get it to work out.
 
  • #4
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do you understand that?

\begin{equation}
\frac{\partial}{\partial \mathbf{r}_i} \frac{1}{\| \mathbf{r}_{i} \| } = - \frac{1}{\| \mathbf{r}_{i} \|^3} \mathbf{r}_{i}
\end{equation}
 
  • #5
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Hi dauto,

I'm not sure where the ##\displaystyle \mathbf{\hat{r}_i} ## comes from, I would have said:

\begin{equation}
\frac{\partial}{\partial \mathbf{r}_i} \frac{1}{\| \mathbf{r}_i \|} = - \frac{1}{\| \mathbf{r}_i \|^2}
\end{equation}
 
  • #6
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Actually I think I may understand where the ##\displaystyle \mathbf{\hat{r}}_i## comes from, it's because it's a gradient, and you're taking the partials with respect to the vector components of ##\displaystyle \mathbf{r_i}##. But I'm still missing something in the original problem, I don't know if it's the sums are throwing me off or the indices of ##\displaystyle \mathbf{r}##.
 
  • #7
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Note that its not r-hat, its r vector in dauto's post. I think your expression should have an r-hat in it, then it would be equivalent to dauto's expression, because r hat is r vector divided by the magnitude of r.
 
  • #8
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@ModusPwnd right I know. By simple taking the derivative you get my expression, but then there is the r-hat which is also added, but it's usually written as dauto wrote it, broken up as r-vector and mag. of r-vector. So I believe my #6 post is correct, I understand where the r-hat comes from because that accounts for his r-vector and the cubed term on the bottom instead of the squared. And it is due to essentially taking the gradient. But still I can't seem to solve the original problem step by step and see where everything comes from.
 
  • #9
AlephZero
Science Advisor
Homework Helper
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I see what you're saying, but I think that's making it unnecessarily complicated. I know there's a way to do it by simply keeping it in the vector form....
But still couldn't quite get it to work out.
Hi dauto,

I'm not sure where the ##\displaystyle \mathbf{\hat{r}_i} ## comes from...
...But I'm still missing something in the original problem, I don't know if it's the sums are throwing me off or the indices of ##\displaystyle \mathbf{r}##.
I rest my case about going back to basics. Sure, you can do this neatly using vectors - provided you understand what you are doing.

But thrashing around with notation you (apparently) don't quite understand isn't a good way to learn, IMO.
 

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