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Numerical integration using Weber force

  1. Jun 18, 2014 #1
    I need to compute numericaly n-body sys. interacting acording to the Weber force:

    and I have a problem with the acceleration on rhs: r'', because the acceleration is unknown, due to the Newton law: F = ma, and we need just 'a' to do next step of integration.

    For the standard Newton/Coulomb force there is no problem:
    F = k/r^2 r^0; so, we can directly compute: a = F/m = ...

    but if a force F depends on r'', this is impossible, because:
    r'' = |a| + v_t^2/r, where: v_t is a tangential speed (velocity vector: r' = v = v_r + v_t).

    Is it solvable the Weber force - usable in some way?
  2. jcsd
  3. Jun 19, 2014 #2
    Maybe I would ask to move this issue to the section of physics... mathematicians probably don't know too much about numerical problems of solving equations of motion.

    i correct a mistake in my recent post:
    |r|'' = a_r + v_t^2/r
  4. Jun 19, 2014 #3


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    I don't really understand the conceptual problem here. If you set up a finite difference scheme to solve the equations, you know an approximation to r'', in exactly the same way as you know approximations to r' and r. If you are doing the timestep from t to t+h, and you need an approximation for r'' that depends on the (unknown) solution at time t+h, use an iterative solution algorithm.

    On the other hand if you are asking us to invent an efficient and accurate algorithm to solve that particular equation, that's a different question, and considering that your Wiki link says
    you are unlikely to find the answer in a handbook of numerical methods.
  5. Jun 19, 2014 #4


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    Especially those pesky chaos theorists. :rolleyes:
  6. Jun 19, 2014 #5
    No, i don't konow the r'', because the force is defined as a function of the r'' also.

    With a standard electric force:
    F = ke^2/r^2 r^0
    and: F = ma, then we can solve: a = F/m = ke^2/mr^2 r^0, simply.

    But with the Weber force we get:
    a = F/m = ke^2/ma (1 - r'^2/2c^2 + r''/c^2) r^0;

    and the r'' = |r|'', ie. a radial acceleration, so it depends obviously on tha 'a' in some way, which we just want to compute.

    Numerical methods have nothing to the equation, which we want to integrate.
    These methods work for any equation - computable of couse.
  7. Jun 19, 2014 #6


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    That makes no sense at all to me.

    Anyway, if you have already calculated r at times t-2h, t-h, and t, you can easily estimate the value of r'' at time t (fit a parabola through the 3 points) and use that to calculate F. (You may need to tweak that basic idea - I haven't done any error and stability analysis on it)

    Or, you know the potential energy is a function of r and r', so you can set up a solution algorithm based on energy conservation (i.e. a weak formulation of the original differential equation).
  8. Jun 20, 2014 #7
    You can't compute in this way, because the r'', which you suggest to compute just geometricaly, will be incorrect - some static version, ie. without any forces (you assume here: v = const, at a position r(t)).

    This is probably the same, bacause you need to konow acceleration to modify a velocity v, and a position r. So, we must compute the gradient of these potential, thus we get the Weber force.
  9. Jun 20, 2014 #8
    Maybe I underappreciated too much these 'estimation' method of r''.
    This will be an extrapolation.
    Very little precision... perhaps 2-nd order with three points, but locally, thus globally 1-st only.
    I plan to compute with 8 order... at last. :)

    This is probably just implicit ode problem, formally:
    y'' = f(t,y,t',y''); or F(t,y, y,',y'') = 0

    in every step we must to solve some system of equations, to get a force, ie. y''.

    For the simplest case: two body only, we can compute this:

    a = -k/r^2 (1 - r'^2/2c^2 + r.r''/c^2)

    and in this case: r'' = a + v_t^2/r, thus the explicit form is:
    [tex]a = -\frac{k}{r^2}\frac{1 - r'^2/2c^2 + v_t^2/c^2}{1 + k/r}[/tex]

    and now we can directly compute the acceleration a.
    Esspecialy for a cilcular motion we have: v_r = 0, v = v_t, so:

    [tex]a = -\frac{k}{r^2} \frac{1 + v^2/c^2}{1 + k/c^2r}[/tex]; but now k/r is just v^2 thus:
    [tex]a = -\frac{k}{r^2} \frac{1 + v^2/c^2}{1 + v^2/c^2} = -\frac{k}{r^2}[/tex]

    This means for a circular motion the Weber force is identical with the Coulomb force. :)
    Last edited: Jun 20, 2014
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