1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hessian matrix of the Newtonian potential is zero?

  1. Aug 29, 2014 #1
    So I'm looking at the hessian of the Newtonian potential:

    [itex] \partial^2\phi / \partial x_i \partial x_j [/itex]

    Using the fact that (assuming the mass is constant):

    [itex] F = m \cdot d^2 x / d t^2 = - \nabla \phi [/itex]

    This implies:

    [itex] \partial^2\phi / \partial x_i \partial x_j = -m \cdot \frac{\partial}{\partial x_j} (d^2 x_i / d t^2) = -m \cdot \frac{\partial}{\partial x_j} (\partial^2 x_i / \partial t^2)[/itex]

    As we can swap the total derivatives for partial derivatives since for Cartesian coordinates:

    [itex] \partial x_i / \partial x_j = \delta_{ij} [/itex]

    Using the fact that we can swap the order of differentiation for mixed partials (assuming continuity of the partial derivatives) we obtain:

    [itex] \partial^2\phi / \partial x_i \partial x_j = -m \cdot \partial^3 x_i / \partial x_j \partial t^2 = -m \cdot \frac{\partial}{\partial t^2} \partial x_i / \partial x_j = -m \cdot 0 = 0 [/itex]

    Hence I obtain the result that the hessian of the Newtonian potential is zero which can't possibly be correct but I can't find the error in my calculation.

    Any help would be much appreciated :)
  2. jcsd
  3. Aug 29, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper

    What you wrote doesn't make too much sense and the mathematical manipulations are illegal. Acceleration depends on time, coordinate depends on time: a(x) = a(t(x)). Good luck reverting x(t) into t(x).
  4. Aug 29, 2014 #3
    So the problem is in the last step where I swap the order of differentiation because it is not possible to find time as a function of position?

    I guess the proper expression for the differential of acceleration with respect to a spatial coordinate is:

    [itex] \partial a(t(x)) / \partial x = \frac{\partial a(t)}{\partial t} \cdot \frac{\partial t}{\partial x} = \frac{\partial a(t)}{\partial t} \cdot (\frac{\partial x}{\partial t})^{-1} [/itex]

    Which is clearly non-zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook