Proof related to angular momentum

1. Oct 12, 2013

Saitama

1. The problem statement, all variables and given/known data
Show that if the total linear momentum of system of particles is zero, the angular momentum of the system is same about all origins.

2. Relevant equations

3. The attempt at a solution
I really don't know where to begin with this. I am not good at these kind of proofs. I need a few hints on how to approach such problems.

Any help is appreciated. Thanks!

2. Oct 12, 2013

TSny

You want to prove a general result. So, start with the most general definition of the total angular momentum of a system of particles. See what happens to this expression if you shift the origin.

3. Oct 12, 2013

Filip Larsen

Start by writing up what you know, namely the expression for the total linear momentum (which you know is zero) and the total angular momentum for some arbitrary points a and b. You then may want to look at how each relative vector from point b to a particle relates to the relative vector from point a to same particle?

4. Oct 12, 2013

Saitama

Angular momentum for a particle is defined as
$$\vec{L}=\vec{r}\times \vec{p}$$
where $\vec{r}$ is the radius vector of the particle from some fixed point (origin) and $\vec{p}$ is the momentum of particle.

For a system of particle,
$$\vec{L}=\sum_i \vec{r_i} \times \vec{p_i}$$
where i denotes the ith particle.

Shifting the origin, the new position vector for the ith particle is
$$\vec{r_i'}=\vec{r_i}-\vec{R}$$
Hence,
$$\vec{L'}=\sum_i \vec{r_i'}\times \vec{p_i}=\sum_i(\vec{r_i}-\vec{R})\times p_i$$
$$\Rightarrow \vec{L'}=\vec{L}-\sum_i \vec{R}\times \vec{p_i}$$
The second term is zero, hence proved.

Is this the correct way? Did I use the right words?

Thank you TSny! :)

5. Oct 12, 2013

TSny

Looks right to me. The final result is more obvious if you write out explicitly that $$\sum_i \vec{R}\times \vec{p_i} = \vec{R}\times \sum_i \vec{p_i}$$

6. Oct 12, 2013

Saitama

Got it, thanks a lot TSny!