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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

    dextercioby

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    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.
     
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