Proving Magnitude of Centre of Mass Vector Equation

[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 r_ij², then multiply by -0.5m_im_j and then sum over the indices. My summation rules are a bit rusty.

 

[Reshma]

Work backwards? I will try that. My summation concepts are pretty outdated too. 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

 

[neutrino]

Very nicely done, Emanuel.

 

[Reshma]

Bravo, Emanuel!