= Angular momentum under an origin shift
{title2=$\mathbf L_{O'}=\mathbf L_O-\mathbf a\times\mathbf P$}
Move the fixed origin from $O$ to $O'=O+\mathbf a$, keeping the axes parallel and the <inertial frame>'s <velocity> measurements unchanged. The new <positions> are $\mathbf r'_i=\mathbf r_i-\mathbf a$, so distributivity of the <cross product> gives
$$
\mathbf L_{O'}=\sum_i(\mathbf r_i-\mathbf a)\times\mathbf p_i
=\mathbf L_O-\mathbf a\times\mathbf P,\qquad
\mathbf P=\sum_i\mathbf p_i.
$$
Here $\mathbf P$ is total <linear momentum>. For the <angular momentum> example with $\mathbf a=(1,0,0)\,\mathrm m$, the new <position> is $(1,0,0)\,\mathrm m$ and $L_{O',z}=6-3=3\,\mathrm{kg\,m^2/s}$. If total <momentum> is zero, <angular momentum> is independent of this origin shift. A moving origin additionally contributes $-\dot{\mathbf a}\times\mathbf P$ when this expression is differentiated while retaining the original <momenta>; see <angular momentum about the centre of mass> for an important case where that term vanishes.
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