Is this angular momentum derivative always true?

Question

Recently, when asking for the derivative of angular momentum, I gave my own solution $$ \frac {d \vec L} {dt} = \frac {d \vec r_{(t1)}} {dt} \times \frac {d \vec r_{(t2)}} {dt} + \vec r_{(t1)} \times \frac {d^2 \vec r_{(t1)}} {dt} \tag 1$$ How to calculate the derivative of the angular momentum vector $ d\vec L = d(\hat I \vec \omega)?$

Even though for a few examples I have counted, the derivative gave the correct result the question remained: is it always true?

I already have an answer, but first I have to ask a question, these are the rules here

Answer 1

$$\forall m \in \Bbb R \land \forall \vec r_{(t1)}= (r_{1x},r_{1y},r_{1z}) \land \forall \vec r_{(t2)}= (r_{2x},r_{2y},r_{2z}) \land \forall \vec r_{(t3)}= (r_{3x},r_{3y},r_{3z})$$ $$(\frac {d \vec L} {dt}=\vec v_{(t1)}\times m \vec v_{(t2)} +\vec r_{(t1)} \times m \vec a_{(t1)} )$$

velocity and acceleration vectors

$$\frac {r_{2x} - r_{1x}}{dt} = \vec v_{x(t1)} ; \frac {r_{2y} - r_{1y}}{dt} = \vec v_{y(t1)}; \frac {r_{2z} - r_{1z}}{dt} = \vec v_{z(t1)} \tag 1 $$

$$\frac {r_{3x} - r_{2x}}{dt} = \vec v_{x(t2)} ; \frac {r_{3y} - r_{2y}}{dt} = \vec v_{y(t2)}; \frac {r_{3z} - r_{2z}}{dt} = \vec v_{z(t2)} \tag 2 $$

$$\frac {v_{2x} - v_{1x}}{dt} = \vec a_{x(t1)} ; \frac {v_{2y} - v_{1y}}{dt} = \vec a_{y(t1)}; \frac {v_{2z} - v_{1z}}{dt} = \vec a_{z(t1)} \tag 3 $$

angular momentum vectors

$$\vec L_{(t1)}= \vec r_{(t1)} \times m \vec v_{(t1)} $$ $$(r_{1y}v_{1z} - r_{1z}v_{1y}, r_{1z}v_{1x}-r_{1x}v_{1z},r_{1x}v_{1y} -r_{1y}v_{1x}) \tag 4 $$ $$\vec L_{(t2)}= \vec r_{(t2)} \times m \vec v_{(t2)} $$ $$(r_{2y}v_{2z} - r_{2z}v_{2y}, r_{2z}v_{2x}-r_{2x}v_{2z},r_{2x}v_{2y} -r_{2y}v_{2x}) \tag 5 $$ change of angular momentum vector with time $$\vec L_{(t2)}-\vec L_{(t1)} \tag 6$$

$$\frac {d l_{x}}{dt}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {6.a}$$ $$\frac {d l_{y}}{dt}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {6.b}$$ $$\frac {d l_{z}}{dt}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {6.c}$$

we calculate the components of the derivative

$$\vec v_{(t1)}\times \vec v_{(t2)}=(v_{1y}v_{2z} - v_{1z}v_{2y}, v_{1z}v_{2x}-v_{1x}v_{2z},v_{1x}v_{2y} -v_{1y}v_{2x}) \tag 7$$

$$v_{1y}v_{2z} - v_{1z}v_{2y}=r_{2y}v_{2z}-r_{1y}v_{2z} -r_{2z}v_{2y}+r_{1z}v_{2y}\tag {7.a}$$ $$v_{1z}v_{2x} - v_{1x}v_{2z}=r_{2z}v_{2x}-r_{1z}v_{2x} -r_{2x}v_{2z}+r_{1x}v_{2z}\tag {7.b}$$ $$v_{1x}v_{2y} - v_{1y}v_{2x}=r_{2x}v_{2y}-r_{1x}v_{2y} -r_{2y}v_{2x}+r_{1y}v_{2x}\tag {7.c}$$

$$\vec r_{(t1)}\times \vec a_{(t1)}=(r_{1y}a_{1z} - r_{1z}a_{1y}, r_{1z}a_{1x}-r_{1x}a_{1z},r_{1x}a_{1y} -r_{1y}a_{1x}) \tag 8$$

$$r_{1y}a_{1z} - r_{1z}a_{1y}=r_{1y}v_{2z}-r_{1y}v_{1z} -r_{1z}v_{2y}+r_{1z}v_{1y}\tag {8.a}$$ $$r_{1z}a_{1x} - r_{1x}a_{1z}=r_{1z}v_{2x}-r_{1z}v_{1x} -r_{1x}v_{2z}+r_{1x}v_{1z}\tag {8.b}$$ $$r_{1x}a_{1y} - r_{1y}a_{1x}=r_{1x}v_{2y}-r_{1x}v_{1y} -r_{1y}v_{2x}+r_{1y}v_{1x}\tag {8.c}$$ We count the derivative $$\frac {d \vec L`} {dt}=\vec v_{(t1)}\times m \vec v_{(t2)} +\vec r_{(t1)} \times m \vec a_{(t1)} \tag 9$$

$$\frac {d l`_{x}}{dt}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {9.a}$$ $$\frac {d l`_{y}}{dt}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {9.b}$$ $$\frac {d l`_{z}}{dt}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {9.c}$$

$$(6)=(9)$$

C.K.D.

Answer 2

Here we are dealing with some rigid body $B$ with density $\rho$.

I will use $r,z$ as vectors, $I,A,X$ as matrices in 3-dimensional space.

Such a rigid body is usually normalized in that its center of mass is zero and its principal axes to the inertia tensor $I$ are the coordinate axes, that is, $I$ is diagonal, where $$ I=\int_B\rho(z) ((z^Tz)-zz^T)\,d{\rm vol}(z). $$

This body moves with some displacement and rotation $r(t)+A(t)z$, $z\in B$. The path in the orthogonal matrices has a derivative $\dot A=AX$ where $X$ is an anti-symmetric matrix whose operation on vectors can be given as $Xz= ω\times z$. Then $\ddot A=\dot A X+A\dot X=AX^2+A\dot X$, so that \begin{align} \ddot Az=A(ω×(ω×z))&=A((ω^Tz)ω-|ω|^2z)\\ z^TA^T\ddot Az&=ω^Tzz^Tω-|ω|^2|z|^2\\ \int_B\rho(z)z^TA^T\ddot Az\,d{\rm vol}(z)&=-ω^TIω \end{align}

The total angular momentum in an external or lab frame of the body is then \begin{align} L_e&=\int_B\rho(z)(r+Az)\times (\dot r+\dot A z)\,d{\rm vol}(z)\\ &=mr\times \dot r + \int_B\rho(z)(Az)\times (AX z)\,d{\rm vol}(z)\\ &=mr\times \dot r + A\int_B\rho(z) (z× (ω× z))\,d{\rm vol}(z)\\ &=mr\times \dot r + A\int_B\rho(z) ((z^Tz)ω- zz^Tω))\,d{\rm vol}(z)\\ &=mr\times \dot r + AIω.\\ \end{align} One could now name $L_b=Iω$ the internal moment of inertia, relative to the body frame.

Then the derivative of the external moment, which is what is driving the mechanics in reacting to external forces, is \begin{align} \text{Then }~\dot L_e&=mr\times \ddot r + \dot AIω + AI\dot ω\\ &=mr\times \ddot r + AXIω + AI\dot ω\\ &=mr\times \ddot r + A[(ω×(Iω)) + I\dot ω] \end{align} Again one can split off the second term and relate it to the time evolution of the internal moment. I would not call it the time derivative,...