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 →v w.r.t to S with:
→r′i=→ri−→vt
→v′i=→vi−→v
For momentum conservation: In frame S', putting ddt∑i→p′i=0 and substituting ddt∑i→pi=0 of frame S in it:
ddt∑i→p′i=ddt∑i→pi−ddt∑imi→v=0−→vddt∑imi
If this has to be 0, then ∑imi=0
Now, on to angular momentum. In frame S:
ddt∑i→Li=ddt∑i(→ri×mi→vi)=0
Am trying to prove the law in frame S' from the law in S:
ddt∑i→L′i=ddt∑i→Li−ddt∑i(→ri×mi→v)−ddt∑i(→vt×mi→vi)
=0−ddt∑i(→ri×mi→v)−ddt∑i(→vt×mi→vi)
=−∑imi(→vi×→v)−∑idmidt(→ri×→v)+∑imi(→vi×→v)−∑imi(→vt×→ai)−∑idmidt(→vt×→vi)
=−∑idmidt(→ri×→v)−∑imi(→vt×→ai)−∑idmidt(→vt×→vi)
=→v×∑idmidt→ri−→vt×∑i→Fi
But this is what I wanted to prove to be 0. I stil have to prove the following:
For a system of particles at →ri with mass mi, which have forces →Fi acting on them such that ∑i→ri×→Fi=0, given ∑idmidt=0; how do I prove:
→v×∑idmidt→ri=→vt×∑i→Fi
for any arbitrary →v and for all time t.
Answer
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}
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