# Angular momentum in different points

I have a question about angular momentum: Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?

Consider central-force motion, such as a planet moving around a (very massive) star. The angular momentum of such a planet is constant if we take the origin as the center of the star. It is not constant if we take the origin to be any other point.

• But "constant" is different from "conserved." In the new system, there is a torque on the planet about any other point, so the torque is providing the change of the angular momentum. That torque-time integral is part of the conservation law. Dec 27 '18 at 23:10
• @BillN: If you're defining angular momentum to be conserved whenever $\Delta \mathbf{L} = \int \pmb{\tau} \, dt$, then I'm pretty sure that the statement "angular momentum is conserved" is tautological. Dec 27 '18 at 23:39
• @BillN How do you define "conserved"? Dec 28 '18 at 0:07
• @BillN Okay, I see your point, but I'm sorry to disagree, or not disagree, but just not joining your view. For me, they'll keep being the same, but I'll be very careful to clarify this point from now on. Let me explain my view: I only say "conserved" when talking about "total" amounts. So total momentum is conserved because it is constant. I never use "conserved" with partial systems, because saying "momentum 1 is conserved but not constant" is like "okay, it's not a random number, but it's still unspecified", so it's not useful for me. Hope you understand. Dec 28 '18 at 21:29
• @BillN We are just using different definitions of what it means for something to be conserved within a system, but we agree on a deeper level. I agree with you within your definition, which I don't think is necessarily wrong. Thanks for the discussion. Dec 29 '18 at 2:20

Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?

Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point.

Angular momentum relative to an origin $${\mathcal O_1}$$

$$\mathbf{L_{\mathcal O_1}} = \mathbf{r_{\mathcal O_1} \times p_{\mathcal O_1}}$$

where $$\mathbf r_{\mathcal O_1}$$ is the position vector to the particle relative to some origin $${\mathcal O_1}$$. Now suppose that angular momentum is conserved in $${\mathcal O_1}$$. Then

$$\frac{d \mathbf L_1}{dt} = \mathbf{\dot{r_1} \times p_1} + \mathbf{r_1 \times \dot{p_1}} = \frac{1}{m} \mathbf{p_1 \times p_1} + \mathbf{r_1 \times \dot{p_1}} =0$$

but since the direction of momentum is frame-independent, the first term vanishes (that is, $$\mathbf{p_1} = \mathbf{p}$$). It then follows that

$$\mathbf{r_1 \times F_1} =0 .$$

Now, let's look at some other origin $$\mathcal{O}_2$$, given that $$L$$ is conserved in $$\mathcal O_1$$. Well the first term much vanish again, that's fine but what about the second term? Does

$$\mathbf{r_2 \times F_2} \stackrel{?}{=}0.$$

Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector.