Centre of mass problem

[Reshma]

Prove that the magnitude of R of the position vector for the centre of mass from an arbitrary origin is given by the equation:
M^2R^2 = M\sum m_ir_i^2 - {1\over 2}\sum m_i m_j r_{ij}^2

Well the centre of mass is given by:
\vec R = \frac{\sum m_i r_i}{M}

But squaring this doesn't seem to produce the result I require. I need more help.

[neutrino]

I'm not too sure about this but I think it should work if you start working backwards - from the defintion of rij2, then multiply by -0.5mimj and then sum over the indices. My summation rules are a bit rusty. :redface:

[Reshma]

Work backwards? I will try that. My summation concepts are pretty outdated too.:yuck: Got to relocate my high school number theory text.

[Emanuel84]

This is quite straightforward. Let's start with:

M\vec{R} = \sum_i m_i \vec{r}_i

Square it:

M^2 R^2 = \sum_{i,j} m_i m_j \vec{r}_i \cdot \vec{r}_j \qquad (1)

Then consider \vec{r}_{i j} \equiv \vec{r}_i - \vec{r}_j \Longleftrightarrow r^2_{ij} = (\vec{r}_i - \vec{r}_j)^2 = r^2_i - 2 \vec{r}_i \cdot \vec{r}_j + r^2_j \Longleftrightarrow \vec{r}_i \cdot \vec{r}_j = \frac{1}{2}(r^2_i+r^2_j-r^2_{ij})

Insert the last relationship in (1) and you'll obtain:

M^2 R^2 = \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j-r^2_{ij}) = \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) - \frac{1}{2} \sum_i m_i m_j r^2_{ij} \qquad (2)

Now, consider the first term of (2)'s rhs:

\frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) = \frac{1}{2}\sum_i m_i \left[\sum_j m_j (r^2_i + r^2_j) \right] = \frac{1}{2} \sum_i m_i \left[\sum_j m_j r^2_i + \sum_j m_j r^2_j \right] \qquad (3)

But \sum_j m_j r^2_i = M r^2_i so:

\frac{1}{2} \sum_i m_i \left[\sum_j m_j r^2_i + \sum_j m_j r^2_j \right] = \frac{1}{2} \sum_i m_i \left[ M r^2_i + \sum_j m_j r^2_j \right] = \frac{1}{2} \left[ M \sum_i m_i r^2_i + \sum_i m_i \sum_j m_j r^2_j \right] = \frac{1}{2} \left[ M \sum_i m_i r^2_i + M \sum_j m_j r^2_j \right] =

= \frac{1}{2} M \left[ \sum_i m_i r^2_i + \sum_j m_j r^2_j \right]

Because "j" is a summed index, you can call it "i" so (3) becomes:

\frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) = \frac{1}{2} M \left[ \sum_i m_i r^2_i + \sum_i m_i r^2_i \right] = M \sum_i m_i r^2_i \qquad (3)

Finally, just insert rhs of (3) in (2) and that's it :smile:

[neutrino]

Very nicely done, Emanuel. :approve:

[Reshma]

Bravo, Emanuel! :smile: