This question came from studying Rutherford scattering.

As you can see in the diagram below, there is an alpha particle moving past a heavy nuclei and being deflected.

In the following analysis of the situation, it's said that since the force acts radially, there is no net torque due to the coulomb interaction on the alpha particle about the heavy nuclei and so its angular momentum is conserved.

This means $mvb=mv'b'$.

My question is, what if we took our origin to be at some point other than the heavy nuclei. Thus resulting on a net torque on the alpha particle about that point. I cant make the same claim about the product of $vb$ being equal to $v'b'$. But I feel like this should be true regardless of the point I choose to measure my torque around. What is special about setting your origin to be the heavy nuclei and why does any result that follows from doing so represent the reality of the situation? enter image description here

  • 2
    $\begingroup$ The point is that $b$ and $b'$ are defined as the distance of the asymptotic lines of the particle to the origin. If you move the origin, you change what $b$ and $b'$ are. (Or, if you choose to define $b$ and $b'$ so that they are the same in all coordinate systems, then if you move the origin, $L = m v b$ won't be true anymore.) $\endgroup$ – knzhou Mar 20 at 16:29

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