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?

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

  • $\begingroup$ If the angular momentum in this circular orbital motion is conserved, then the particle is moving under the action of a central force $\:\mathbf{F}\:$ with center let the point $\:\mathrm{O}\:$. But if you apply the law "moment of force around a new point $\:\mathrm{O'}\:$=rate of angular momentum change with respect to $\:\mathrm{O'}\:$" then the angular momentum is changing with rate equal to the moment of the force $\:\mathbf{F}\:$ around this new point $\:\mathrm{O'}\:$. $\endgroup$
    – Frobenius
    Commented Sep 15, 2017 at 12:10
  • $\begingroup$ @Frobenius thanks, I am not sure if I understand what you mean, anyway, I mean a Galileo translation, nothing extra had been done in that thought experiment. $\endgroup$
    – Shing
    Commented Sep 15, 2017 at 14:42

1 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.

  • $\begingroup$ isn't $\vec r$ simply position vector? $\endgroup$
    – Shing
    Commented Sep 15, 2017 at 14:44
  • $\begingroup$ No, if you're calculating the angular momentum with respect to a point $Q$, then $r$ is $QP$. Normally, we place our origin of coordinates in the same point, just for simplicity. If you change the origin but you maintain the point $Q$, nothing changes. If you change the point $Q$, it changes, because torque and $L$ depend on the point respect which they're taken. That's what my answer was about. I don't know if I was clear... do you see it? $\endgroup$
    – FGSUZ
    Commented Sep 15, 2017 at 17:43
  • $\begingroup$ @Shing Quite a few introductory treatments set the origin at the place around which they want to compute $\vec{L}$ to reduce the notational complexity of the first discussion, but $\vec{L}$ can be computed around arbitrary points. And if you are computing it around, say, the axle of a machine it should be clear that the position of the axle is also affected by the change of coordinate system as FGSUZ says. $\endgroup$ Commented Sep 16, 2017 at 1:34

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