Angular momentum conservation under Galileo transformation

Question

I was trying to see when angular momentum is independent of choice of origin, but then it seems angular momentum no longer conserved under Galileo transformation to me :

Given a point mass is doing circular orbital motion in an inertial frame: $$\vec L = \vec r \times \vec p $$

In a new relatively stationary frame with displacement $\vec R$:$^{\dagger}$

$$\vec {L'}=\vec {r'} \times \vec {p'}$$ $$\vec {L'}=({\vec R +\vec {r}} )\times \vec {p}$$

$$\vec {L'}=({\vec R +\vec {r}} )\times \vec {p}$$

Take time derivative: $$\dot {\vec {L'}}=({\dot{\vec R} +\dot{\vec {r}}} )\times \vec {p} +({\vec R +\vec {r}} )\times \dot{\vec {p}}$$ $$\dot {\vec {L'}}=0 +({\vec R +\vec {r}} )\times \dot{\vec {p}}$$ Given angular momentum is conserved in an orbital motion in the old frame ($\vec {r} \times \dot{\vec {p}} = 0$): $$\dot {\vec {L'}}=\vec R \times \dot{\vec {p}}$$ However, this term is not always zero - it is absurd, since we will not have a new torque by picking up a new stationary inertial frame.

What's wrong with my reasoning?

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$\dagger$: both $\vec p$ and $\vec {p'}$ should be same in both inertial frames.

Answer 1

The thing is that you are mixing two things. On the one hand, if you move your origin, all positions obviously suffer the transformation

$x_{new}=x_{old}+x_{OO'}$ (and so on).

But, on the other hand, the radius angular momentum $L$ is NOT the position of the particle, but its position with respect to the point you chose.

In other words, $\vec{L}_Q=\vec{r}_{PQ}\times \vec{p}_{PQ}$

You don't take the radius to the origin (unless the origin is the point you chose). You take the radious from the particle's position to the point you chose.

You obviously have $\vec{r}_{PQ}\equiv \vec{PQ}=\vec{OQ}-\vec{OP}$

and changing $O$ doesn't matter because that cancels out. Of course you were right: changing the reference frame doesn't create a new torque... provided that the point you calculate $L$ respect with remains the same one.

If you change origin AND the point, then you have a new value, but that's logical because $L$ depends on the point you chose to compute it.