Proving that $\vec{v}\times\sum_i\dfrac{dm_i}{dt}\vec{r}_i=t(\vec{v}\times\sum_i\vec{F}_i)$ when no external torque

Question

There is this idea of relativity in Classical Mechanics:

The laws of mechanics valid in an inertial frame must also be valid in any frame moving uniformly with respect to it.

I was just trying to apply these to the case of the law of conservation of momentum and the law of conservation of angular momentum.

Let there be an inertial frame S and another frame S' moving with velocity $\mathbf{\vec{v}}$ w.r.t to S with:

$$\mathbf{\vec{r}}'_i = \mathbf{\vec{r}}_i - \mathbf{\vec{v}}t$$

$$\mathbf{\vec{v}}'_i = \mathbf{\vec{v}}_i - \mathbf{\vec{v}}$$

For momentum conservation: In frame S', putting $\dfrac{d}{dt} \sum_i \mathbf{\vec{p}}'_i = \mathbf{0}$ and substituting $\dfrac{d}{dt} \sum_i \mathbf{\vec{p}}_i = \mathbf{0}$ of frame S in it:

$$\dfrac{d}{dt} \sum_i \mathbf{\vec{p}}'_i = \dfrac{d}{dt} \sum_i \mathbf{\vec{p}}_i - \dfrac{d}{dt} \sum_i m_i \mathbf{\vec{v}} = \mathbf{0} - \mathbf{\vec{v}} \dfrac{d}{dt} \sum_i m_i$$

If this has to be $\mathbf{0}$, then $\sum_i m_i = 0$

Now, on to angular momentum. In frame S:

$$\dfrac{d}{dt} \sum_i \mathbf{\vec{L}}_i = \dfrac{d}{dt} \sum_i (\mathbf{\vec{r}}_i \times m_i\mathbf{\vec{v}}_i) = \mathbf{0}$$

Am trying to prove the law in frame S' from the law in S:

$$\dfrac{d}{dt} \sum_i \mathbf{\vec{L}}'_i = \dfrac{d}{dt} \sum_i \mathbf{\vec{L}}_i - \dfrac{d}{dt} \sum_i (\mathbf{\vec{r}}_i \times m_i\mathbf{\vec{v}}) - \dfrac{d}{dt} \sum_i (\mathbf{\vec{v}}t \times m_i \mathbf{\vec{v}}_i)$$

$$= \mathbf{0} - \dfrac{d}{dt} \sum_i (\mathbf{\vec{r}}_i \times m_i\mathbf{\vec{v}}) - \dfrac{d}{dt} \sum_i (\mathbf{\vec{v}}t \times m_i \mathbf{\vec{v}}_i)$$

$$= - \sum_i m_i (\mathbf{\vec{v}}_i \times \mathbf{\vec{v}}) - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{r}}_i \times \mathbf{\vec{v}}) + \sum_i m_i (\mathbf{\vec{v}}_i \times \mathbf{\vec{v}}) - \sum_i m_i (\mathbf{\vec{v}}t \times \mathbf{\vec{a}}_i) - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{v}}t \times \mathbf{\vec{v}}_i)$$

$$= - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{r}}_i \times \mathbf{\vec{v}}) - \sum_i m_i (\mathbf{\vec{v}}t \times \mathbf{\vec{a}}_i) - \sum_i \dfrac{dm_i}{dt} (\mathbf{\vec{v}}t \times \mathbf{\vec{v}}_i)$$

$$= \mathbf{\vec{v}} \times \sum_i \dfrac{dm_i}{dt} \mathbf{\vec{r}}_i - \mathbf{\vec{v}}t \times \sum_i \mathbf{\vec{F}}_i$$

But this is what I wanted to prove to be $\mathbf{0}$. I stil have to prove the following:

For a system of particles at $\mathbf{\vec{r}}_i$ with mass $m_i$, which have forces $\mathbf{\vec{F}}_i$ acting on them such that $\sum_i \mathbf{\vec{r}}_i \times \mathbf{\vec{F}}_i = \mathbf{0}$, given $\sum_i \dfrac{dm_i}{dt} = 0$; how do I prove:

$$\mathbf{\vec{v}} \times \sum_i \dfrac{dm_i}{dt} \mathbf{\vec{r}}_i = \mathbf{\vec{v}}t \times \sum_i \mathbf{\vec{F}}_i$$

for any arbitrary $\mathbf{\vec{v}}$ and for all time $t$.

Answer 1

Using your notation of

$$ \boldsymbol{r}_{i}' =\boldsymbol{r}_{i}-\boldsymbol{v}\,t \\ \boldsymbol{v}_{i}' =\boldsymbol{v}_{i}-\boldsymbol{v} $$

and with the assumption that $\dot{\boldsymbol{v}}=0$ (uniform motion of frame S') form the linear and angular momentum expressions on the S frame.

$$ \boldsymbol{p} =\sum_{i}m_{i}\boldsymbol{v}_{i} \\ \boldsymbol{L} =\sum_{i}\left(\boldsymbol{r}_{i}\times m_{i}\boldsymbol{v}_{i}\right) $$

Now look at linear and angular momentum in the S' frame and relate them to the ones from S.

$$\require{cancel} \begin{aligned} \boldsymbol{p}'&=\sum_{i}m_{i}\boldsymbol{v}_{i}'=\sum_{i}m_{i}\left(\boldsymbol{v}_{i}-\boldsymbol{v}\right)=\boldsymbol{p}-\left(\sum_{i}m_{i}\right)\boldsymbol{v}=\boldsymbol{p}-m\,\boldsymbol{v}\\\boldsymbol{L}'&=\sum_{i}\left(\boldsymbol{r}_{i}'\times m_{i}\boldsymbol{v}_{i}'\right)=\sum_{i}\left(\boldsymbol{r}_{i}-\boldsymbol{v}\,t\right)\times m_{i}\left(\boldsymbol{v}_{i}-\boldsymbol{v}\right)\\&=\sum_{i}\left(\boldsymbol{r}_{i}\times m_{i}\boldsymbol{v}_{i}\right)-\sum_{i}(\boldsymbol{r}_{i}\times m_{i}\boldsymbol{v})-\boldsymbol{v}\,t\times\left(\sum_{i}m_{i}\boldsymbol{v}_{i}\right)+\cancel{\boldsymbol{v}\,t\times\left(\sum_{i}m_{i}\right)\boldsymbol{v}}\\&=\boldsymbol{L}+\boldsymbol{v}\times\sum_{i}\left(m_{i}\boldsymbol{r}_{i}\right)+\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)\times\boldsymbol{v}\,t \end{aligned}$$

To show that these quantities are conserved, take the derivative (assuming that $\frac{{\rm d}\boldsymbol{p}}{{\rm d}t}=0$ and that $\frac{{\rm d}\boldsymbol{L}}{{\rm d}t}=0$)

$$\begin{aligned} \frac{{\rm d}}{{\rm d}t}\boldsymbol{p}'&=\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{p}}-m\,\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{v}}=0\\\frac{{\rm d}}{{\rm d}t}\boldsymbol{L}'&=\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{L}}+\frac{{\rm d}}{{\rm d}t}\left[\boldsymbol{v}\times\sum_{i}\left(m_{i}\boldsymbol{r}_{i}\right)\right]+\frac{{\rm d}}{{\rm d}t}\left[\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)\times\boldsymbol{v}\,t\right]\\&=\boldsymbol{v}\times\sum_{i}\left(m_{i}\frac{{\rm d}}{{\rm d}t}\boldsymbol{r}_{i}\right)+\cancel{\frac{{\rm d}}{{\rm d}t}\boldsymbol{p}}\times\boldsymbol{v}\,t+\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)\times\boldsymbol{v}\\&=\boldsymbol{v}\times\sum_{i}\left(m_{i}\boldsymbol{v}_{i}\right)+\boldsymbol{p}\times\boldsymbol{v}\\&=\boldsymbol{v}\times\boldsymbol{p}+\boldsymbol{p}\times\boldsymbol{v}=0 \end{aligned}$$