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Tidal forces/fields

  1. Aug 25, 2005 #1
    Hey people,

    Im doing some analysis of some N-body simulation data. I'm trying to calculate the tidal forces exerted on the smaller groups of particles by the other mass. I have a model for the distribution of matter causing the tidal field so I can analytically calculate the gravitational potential and the directional second derivatives, but how do I translate that into the forces. Anybody have a reference for some good reading on the subject.
  2. jcsd
  3. Aug 25, 2005 #2


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    Science Advisor

    Well if you have a potential V(x,y,z) then the force in cartesian coordinates (x,y,z) is given by
    F_x = \frac{\partial{V}}{\partial{x}} \hspace{.25 in} F_y = \frac{\partial{V}}{\partial{y}} \hspace{.25 in} F_z = \frac{\partial{V}}{\partial{z}}

    and if you have a unit vector U, the tidal force T is another vector, the gradient of the force F in the direction of vector U, given by

    T_x = \frac{\partial^2{V}}{\partial x \partial x}} U_x +
    \frac{\partial^2{V}}{\partial x \partial y}} U_y +
    \frac{\partial^2{V}}{\partial x \partial z}} U_z
    T_y = \frac{\partial^2{V}}{\partial y\partial x}} U_x +
    \frac{\partial^2{V}}{\partial y\partial y}} U_y +
    \frac{\partial^2{V}}{\partial y\partial z}} U_z
    T_z = \frac{\partial^2{V}}{\partial z\partial x}} U_x +
    \frac{\partial^2{V}}{\partial z\partial y}} U_y +
    \frac{\partial^2{V}}{\partial z\partial z}} U_z

    You can write this in tensor notation

    T^i = K^i{}_j U^j

    where [tex] K^i{}_j = \frac{\partial^2{V}}{\partial x^i \partial x^j}[/tex]

    It gets more complicated if you want to use general (non-cartesian) coordinates

    But you can always say that the tidal forces at a point are given by a second rank tensor, one that takes in a vector (the displacement) and spits out a vector (the tidal force).

    I *think* that the partial derivates should normally alll commute, so [tex]
    \frac{\partial^2 V}{\partial x \partial y} = \frac{\partial^2 V}{\partial y \partial x} [/tex]

    Google finds "Clairaut's theorem"

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