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Center of inertia?

  1. Sep 22, 2007 #1
    Center of momentum?

    Hi there,i'm giving exams this tuesday on modern physics and I stumpled upon a problem which i couldn't find an answer in my book and internet.The problem in a few words asks you to find the speed of the center of momentum when 2 particles collide.Can someone plz give a definition of the "center of momentum speed" and formulas or links plz?
     
    Last edited: Sep 22, 2007
  2. jcsd
  3. Sep 22, 2007 #2
    I don't know what is "center of momentum". They probably had in mind "center of mass" [itex] \mathbf{R} [/itex]. In relativistic quantum mechanics it is also known as the Newton-Wigner position operator. For a classical system of non-interacting particles with positions [itex] \mathbf{r}_i [/itex] and energies [itex] e_i [/itex] the definition is

    [tex] \mathbf{R} = E^{-1} \sum_i e_i \mathbf{r}_i [/tex]..............(1)

    where

    [tex] E= \sum_i e_i [/tex]

    is the total energy of the system. In the non-relativistic case, when the rest energy dominates, one gets [itex] E \approx Mc^2 [/itex], [itex] e_i \approx m_ic^2 [/itex], and

    [tex] \mathbf{R} = M^{-1} \sum_i m_i \mathbf{r}_i [/tex]

    the familiar definition of the "center of mass".

    The center of mass of any isolated system always moves along straight line with constant velocity. This is true also for interacting systems of particles, whose center-of-mass is given by formulas more complicated than (1).

    Eugene.
     
  4. Sep 22, 2007 #3

    pervect

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  5. Sep 22, 2007 #4

    robphy

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    Here is a geometric interpretation.

    Consider a system of particles. Add up their 4-momenta [using either the parallelogram rule or the tail-to-tip method] and obtain the resultant 4-momentum vector. The center of momentum is the reference frame with unit vector tangent to that resultant 4-vector.

    In the current frame, this vector has components of the form
    [tex]E \hat t + P \hat s = M(\cosh\theta \hat t + \sinh\theta \hat s)=M\cosh\theta(\hat t+\tanh\theta \hat s) = M\cosh\theta(\hat t+V \hat s)[/tex],
    where [itex]V\hat s[/itex] is the velocity of that center-of-momentum-frame.
    Writing
    [tex]E \hat t + P \hat s = E (\hat t +\frac{P}{E}\hat s),[/tex]
    we can express that velocity as the ratio of relativistic-spatial-momentum to relativistic-energy: [tex]V=\frac{P}{E}[/tex].

    In that center of momentum frame, the relativistic-spatial-momentum [itex]P'[/itex] in that frame is zero.

    In short, the center-of-momentum-frame is along the resultant,
    and the velocity of that frame [with respect to our frame] is "the slope of the resultant" in our coordinates.
     
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