Center of mass and REduced mass

  • Thread starter PeteGt
  • Start date
  • #1
51
0
I forget how these two relate both conceptually and mathematically...any help?
 

Answers and Replies

  • #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
 
  • #3
51
0
So what about more than one mass? So would the reduced mass formula really be 1/u=1/m1+1/m2+1/m3+...
 
  • #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:
  • #5
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
134
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.
 
  • #6
krab
Science Advisor
896
2
cookiemonster said:
[tex]m_{cm} = \frac{\sum x_n m_n}{\sum m_n}[/tex]

cookiemonster

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

Related Threads on Center of mass and REduced mass

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
881
Replies
2
Views
619
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
1K
Top