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Center of mass and REduced mass

  1. Apr 2, 2004 #1
    I forget how these two relate both conceptually and mathematically...any help?
     
  2. jcsd
  3. Apr 2, 2004 #2
    [tex]\mu = \frac{m_1m_2}{m_1 + m_2}[/tex]

    [tex]m_{cm} = \frac{\sum x_n m_n}{\sum m_n}[/tex]

    cookiemonster
     
  4. Apr 2, 2004 #3
    So what about more than one mass? So would the reduced mass formula really be 1/u=1/m1+1/m2+1/m3+...
     
  5. Apr 2, 2004 #4
    Honestly, I've never seen it for more than two masses, but I suppose the most logical extension would be

    [tex]\frac{1}{\mu} = \sum \frac{1}{m_n}[/tex]

    That being said, the above expression may have absolutely no physical meaning. I don't know.

    Anyway, for the uses of the center of mass and reduced mass together is a 2-body problem. The math simplifies in the center of mass coordinate system and when you use the center of mass coordinate system, the math involved makes it convenient to define reduced mass as such.

    cookiemonster

    Edit: I say "the most logical extension" because reduced mass is also defined as

    [tex]\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}[/tex]

    You can use a bit of algebra to demonstrate that this equation is simply a rearrangement of the first one I gave.

    Edit2: Which I would have known you already knew had I read your post more carefully...
     
    Last edited: Apr 2, 2004
  6. Apr 5, 2004 #5

    arildno

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    I think the reduced-mass concept should be generalized to a matrix rather than to a scalar. If we consider a N-particle system where the particle positions x(i) are measured relative to the C.M., we have, for a conservative system, that the energy equation (f.ex.) may be written as:
    1/2*MV^(2)+1/2*(v^(T)*Q*v)+U=const.
    Here, M is the system's total mass, V is the speed of C.M, U is the potential energy; whereas v is the N-vector of particle velocities, v^(T)=(v(1),..,v(N))
    (T is for transpose)
    (v(i)=dx(i)/dt)).
    Q is the N*N diagonal mass matrix, Q(j,j)=m(j), where m(j) is the mass of the j-th particle.
    (The resulting product of velocities, f.ex. v(j)^(2) is the dot product if v(j) is a vector.)

    We have, by definition of particle positions relative to C.M, m(i)*x(i)=0, where summing over i=1,..N is implied.
    Hence, we may eliminate a particle (the N'th, f.ex.), from our system, and
    we represent the other particles by their distances x(i,N) (i=1,..N-1):
    x(i)=x(N)+x(i,N), i=1,..N-1, v(i,N)=dx(i,N)/dt, vrel^(T)=(v(1,N),...v(N-1,N))

    The energy equation may now be rewritten as:
    1/2*MV^(2)+1/2*(vrel^(T)*R*vrel)+U=const.

    Here, R is the (N-1)*(N-1) reduced mass matrix with respect to particle N:

    R(j,j)=r(j,j)=m(j)(M-m(j))/M
    R(i,j)=R(j,i)=-r(i,j), r(i,j)=m(i)*m(j)/M (i not equal to j)

    We see that r(i,j) is less than both m(i) and m(j).
    Again, products of velocities should be regarded as inner products if the velocities v(i,N) are vectors.
     
  7. Apr 5, 2004 #6

    krab

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    The centre of mass is a position, not a mass.
     
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