0

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 $t$.

Community
  • 1
PhyEnthusiast
  • 2,908
  • Can't find $r_i\times m_i dv/dt$, have you made simplifications somewhere or is it missing? – Emil Mar 12 '18 at 17:35
  • I haven't looked at this in detail, but note that Newton's Laws are applicable only for constant mass systems, suggesting that $\mathrm{d}m_i/\mathrm{d}t = 0$. I suppose care must be taken if you are allowing particles to break up but stay within the system. – garyp Mar 12 '18 at 17:40
  • Are rotations not considered here? Are all the particles moving with the same velocity or only the same direction or what? – John Alexiou Mar 12 '18 at 19:40
  • Something is way wrong. The units are inconsistent in the expressions and ${\rm d}m_i /{\rm d}t=0$ by definition. – John Alexiou Mar 12 '18 at 20:41
  • @Emil $dv/dt = 0$ since $v$ is the relative speed between inertial frames – PhyEnthusiast Mar 13 '18 at 13:40
  • @ja72 Which expression is wrong? And why are all particles moving with same velocity? I wrote $\mathbf{\vec{v}}_i$, ryt? – PhyEnthusiast Mar 13 '18 at 13:41
  • @garyp Why can't Newton's laws be applied? We just use $\mathbf{\vec{F}}_i = \dfrac{d(m_i \mathbf{\vec{v}}_i)}{dt}$? – PhyEnthusiast Mar 13 '18 at 13:44
  • Newton's laws are valid only for closed systems, which means the total mass of the system does not change. See this Wikipedia entry, and this SE answer where I elaborate on the issue a little in a different context. Your situation is a little different, but as I understand what you are doing, you ought to set $dm_i/dt=0$ – garyp Mar 13 '18 at 14:07
  • I did get $\sum_i \dfrac{dm_i}{dt} = 0$, but this doesn't necessarily mean $\dfrac{dm_i}{dt} = 0$ or does it? – PhyEnthusiast Mar 13 '18 at 14:14
  • You are right, it does not, but if one of your particles gains energy, another must lose energy in order to satisfy your first condition. The analysis would be more involved in that case. You'd need to add that constraint to the analysis. I'm not sure how to proceed in that case. – garyp Mar 13 '18 at 14:21
  • Please show how you got to $ \frac{{\rm d}}{{\rm d}t} \sum m_i = 0$. – John Alexiou Mar 13 '18 at 17:47
  • If a particle changes mass the force field generated by gravity should change? But if we look at the mass of individual molecules we probably need to change to a model with quantum effects? Or what scale are the $i$ particles on? Or are they volume integrals in a (fixed? comoving?) region? – Emil Mar 13 '18 at 18:21
  • Does the third law of Newton say something useful perhaps? Is the sum of forces 0 everywhere? – Emil Mar 13 '18 at 18:30
  • @ja72 I have added that. – PhyEnthusiast Mar 13 '18 at 23:59

1 Answers1

0

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 $\frac{{\rm d}\boldsymbol{L}}{{\rm d}t}=0$)

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

John Alexiou
  • 38,341
  • But why assume rigid body? I tried for a system of particles that may or may not together form a rigid body – PhyEnthusiast Mar 13 '18 at 13:48
  • If you find something wrong with what I have done, please point them out; so that I may rectify them – PhyEnthusiast Mar 13 '18 at 14:18
  • You need to show your initial work, where you define momentum in each frame. Work out linear momentum first, before tackling angular momentum. – John Alexiou Mar 13 '18 at 18:05
  • I edited my post to reflect more of your notation. I hope you can compare your work with my answer to figure out what went wrong. – John Alexiou Mar 13 '18 at 18:50
  • But this matches exactly with what I have done if you put $\sum_i \mathbf{\vec{F}}_i = 0$ and $\dfrac{dm_i}{dt} = 0$ and thus shows what I did is correct? – PhyEnthusiast Mar 13 '18 at 23:48
  • All I want to ask is: does $\dfrac{d}{dt} \sum_i m_i = 0$ and $\sum_i \dfrac{dm_i}{dt} \mathbf{\vec{r}}_i = \mathbf{0}$ imply $\dfrac{dm_i}{dt} = 0$? Plz, if you had the time :) – PhyEnthusiast Mar 14 '18 at 00:11
  • $m_i$ is constant unless you are working with a fluid domain. $\frac{{\rm d}m_i}{{\rm d}t}=0$ is an input (an assumption) not an output of the calculation. – John Alexiou Mar 14 '18 at 13:11
  • So $m_i$ is not constant for fluids? But each fluid particle has constant mass, ryt? – PhyEnthusiast Mar 15 '18 at 14:15
  • In a fluid, you consider a fixed volume but varying density in general (unless incompressible flow is considered). $${\rm d}m_i = \rho(\boldsymbol{r}_i), {\rm d}V_i$$ – John Alexiou Mar 15 '18 at 15:01
  • But even then the mass of each classical particle that makes up the fluid doesn't change with time, ryt? It's only that more particles have entered into the "compressed" region? – PhyEnthusiast Mar 16 '18 at 04:28
  • From a kinetic theory of gases point of view, you are correct. – John Alexiou Mar 16 '18 at 13:52
  • So which is correct? But I suppose we approximate it as continuous in continuum mechanics? – PhyEnthusiast Mar 17 '18 at 07:27
  • Also, can you please tell me whether it is necessary in the domain of application of classical mechanics, that $\dfrac{dm_i}{dt} = \mathbf{0}$ – PhyEnthusiast Mar 17 '18 at 17:57
  • I would say it is an assumption rather than a result. Things work out for us when it is so. – John Alexiou Mar 18 '18 at 01:09