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Forces that depend only on distance

  1. Aug 5, 2009 #1
    It is said that if a force depends only on distance (And not on time), the equations that involve those forces are invariant to a change of the frame of reference.....is this true?

    Coulomb's force law is only dependent on distance. At a certain space location the force has a value that does not change with time. But after the object is subjected to that force, it moves to a new spatial position and the force on it changes. So there is a change with time....

    A time changing source generates a time changing field and therefore a time changing force on an object.... what is special about these type of forces? Why don't they fit in the frame of reference transformations? after all, in classical mechanics time is absolute and distance differences are invariant...

    thanks,
    fisico30
     
  2. jcsd
  3. Aug 6, 2009 #2

    Jano L.

    User Avatar
    Gold Member

    Hi,
    in classical mechanics, motion equations are invariant to changes of inertial reference frame.
    Imagine the force acting on a particle given as a function of coordinates and time. Consider two mutually moving frames, with nocoinciding origins O, O'. The coordinates in these frames are related by Galilei transformation, which in the simplest case of two equally oriented coordinated systems and motion od S' along x with velocity v is

    [tex]
    x' = x - vt,
    y' = y,
    z' = z,
    t' = t.
    [/tex]

    (In general, x,y,z would enter in some linear combinatinons, but t' would be always t). Then in S and S' frames the equations of motion are

    [tex]
    \mathbf F(x,y,z, t) = m\mathbf a(t)~~,~~\mathbf F'(x',y',z', t) = m\mathbf a'(t).
    [/tex]

    The equations looks the same - every frame use its coordinates to describe the particle, but the form of law is the same. We say that Newton's law of motion is covariant with respect to Galilean transformations coordinates. Furthermore, because frames are inertial, forces are the same

    [tex]
    \mathbf F(x,y,z, t) = \mathbf F' (x',y',z', t).
    [/tex]

    There is no need to have the time independent force; equations are covariant in general.

    In relativistic description (which is closer to reality), the equations are again covariant in respect to, but Lorentz transformation of coordinates, which in our special case is

    [tex]
    x' = \gamma(x - vt),
    y' = y,
    z' = z,
    t' = \gamma(t-vx/c^2).
    [/tex]

    For example, the law of motion of a particle in EM field is in S

    [tex]
    qF^\mu_{~\nu}(x,y,z,t) u^\nu(t) = m\frac{d^2 x^\mu}{d\tau^2}(t)
    [/tex]

    and in S'

    [tex]
    qF'^\mu_{~\nu}(x',y',z',t') u^\nu(t') = m\frac{d^2 x'^\mu}{d\tau^2}(t').
    [/tex]

    So there is no need to have the time independent force - formulae are invariant with respect to spacetime coordinate transformations in general. Best,

    Jano
     
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