Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gradient and Hessian of the Coulomb/Electrostatic Energy

  1. Feb 15, 2016 #1
    I have a function

    $$\displaystyle V(x)=\frac{1}{2}\sum_i \sum_{j \neq i} q_i q_j \frac{1}{\left|r_i - r_j\right|}$$ where ##r_i=\sqrt{x_i^2+y_i^2+z_i^2}## which is the coulomb potential energy of a system of charges.

    I need to calculate ##\frac{\partial V}{\partial x_k}## and ##\frac{\partial^2 V}{\partial x_k \partial x_l}## for an optimization routine.

    I guess I want to treat the set ##(x_i , y_i, z_i)## where i runs from 1 to N as independent variables ##(x_j)## where j runs from 1 to 3N (the direct sum of the position vectors).

    Would ##r_i = \sqrt{x_{3i-2}+x_{3i-1}+x_{3i}}## then?

    I ask because but I am not sure how to calculate the ##\frac{\partial r_i}{\partial x_k}## term

    I was also thinking maybe I can just calculate ##\frac{\partial r_i}{\partial x_k}, \frac{\partial r_i}{\partial y_k}, \frac{\partial r_i}{\partial z_k}## separately and then just relabel them as independent variables but I am not sure if this will work.

    Any help would be appreciated
     
    Last edited by a moderator: Feb 15, 2016
  2. jcsd
  3. Feb 15, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi,

    ##r_i## is not ##\sqrt{x_i^2+y_i^2+z_i^2}## !
    ## r_i ## is simply a vector ##\vec r_i = (x_i, y_i,z_i)## !
    You need ##| \vec r_i - \vec r_j | ## which is the square root of ## (\vec r_i - \vec r_j) \cdot (\vec r_i - \vec r_j) ##

    In your notation (somewhat awkward) this would be $$
    | \vec r_i - \vec r_j | = \sqrt{ ( x_{3i-2} - x_{3j-2} )^2 + ( x_{3i-1} - x_{3j-1} )^2 + ( x_{3i} - x_{3j} )^2 }
    $$
     
  4. Feb 15, 2016 #3
    Thanks for that correction, bit of an oversight!

    How would you suggest I calculate the gradient and the hessian analytically without that notation?
     
  5. Feb 15, 2016 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your ##V## looks an awful lot like the ##\ \ W\quad (1.51)\ \ ## in my 1974 Jackson 2nd edition, except that ##W## is a scalar and you write ##V(x)##. What does this ##x## stand for ? And how do you intend to optimize if ##V## diverges when ##\vec r_i \rightarrow \vec r_j ## ? Are there any boundary conditions ? In short: could you tell us a little more of your plans :smile: ?
     
  6. Feb 16, 2016 #5
    Sorry V is the coulomb potential for a system of charges each with position ##r_i=(x_i, y_i, z_i)##, it is a scalar function as you say of these positions. I want to create a gradient ##\frac{\partial V}{\partial x_k}## and a hessian ##\frac{\partial^2 V}{\partial x_k \partial x_l}## where the coordinates ##x_k## are the direct sum of the ##(x_i, y_i, z_i)##

    So for a system of two charges I would have ##r_1=(x_1, y_1, z_1) \ , \ r_2=(x_2, y_2, z_2)## so ##\textbf{x}=(x_1, y_1, z_1, x_2, y_2, z_2)## and my gradient would look like ##(\frac{\partial V}{\partial x_1}, \frac{\partial V}{\partial y_1}, \frac{\partial V}{\partial z_1}, \frac{\partial V}{\partial x_2}, \frac{\partial V}{\partial y_2}, \frac{\partial V}{\partial z_2})##

    I originally thought it would be a good idea to rewrite ##\textbf{x}=(x_1, y_1, z_1, x_2, y_2, z_2)## as ##(x_1, x_2, x_3, x_4, x_5, z_6)## using the coordinate transformations ##x_i \to x_{3i-2}, \ y_i \to x_{3i-1}, \ z_i \to x_{3i}## but as you said the notation is clunky.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gradient and Hessian of the Coulomb/Electrostatic Energy
Loading...