0
$\begingroup$

I've started working through Analytical Mechanics for Relativity and Quantum Mechanics by Oliver Johns and I'm stuck on deriving a formula.

In the section titled "Change of Angular Momentum", Johns states that the rate of change for spin angular momentum, $ S $, for a collection of $ N $ many particles is equal to the spin external torque, so:

$$ \frac{\text{d}S}{\text{d}t} = \tau_s^{\text{(ext)}} $$

where spin external torque is defined as:

$$ \tau_s^{\text{(ext)}} = \sum_{\text{n = 1}}^N\rho_n \times{f_n^\text{(ext)}} $$

where $ \rho_n $ is the relative position vector of the nth particle.

I've been able to get this far in my derivation:

$$ 1)\space S = \sum_\text{n = 1}^N\rho_n\times (m_n\dot{\rho_n}) $$ which is given as the definition for spin angular momentum

$$ 2)\space \frac{\text{d}S}{\text{d}t} = \frac{\text{d}}{\text{d}t}\text(\sum_\text{n = 1}^N\rho_n\times\text(m_n\dot{\rho_n})) $$

$$ 3)\space \frac{\text{d}}{\text{d}t}\text(\sum_\text{n = 1}^N\rho_n\times\text(m_n\dot{\rho_n})) = \sum_\text{n = 1}^N\text[(\dot{\rho_n}\times m_n \dot{\rho_n}) \space+\space \text(\rho_n \times m_n\ddot{\rho_n})] $$

where $ \dot{\rho_n}\times m_n \dot{\rho_n} = 0 $ by the properties of cross products, so I've gotten to this point:

$$ 4)\space \frac{\text{d}S}{\text{d}t} = \sum_\text{n = 1}^N(\rho_n \times m_n \ddot{\rho_n}) $$

I know that $ m_n \ddot{\rho_n} = m_n (a_n - A) = f_n - m_n A $ where $ A $ is the acceleration of the center of mass.

So if my derivation is correct so far, then we must have the following:

$$ m_n \ddot{\rho_n} = m_n (a_n - A) = f_n - m_n A = f_n^\text{(ext)} $$

This is where I'm stuck. How can I show this last equivalence? My intuition tells me that if $ m_n A $ can be shown to be the internal force of the nth particle then the derivation is complete as $$ f_n - f_n^\text{(int)} = f_n^\text{(ext)} $$

So is it possible to show that this is true? Or is there another method that I'm not seeing?

Thanks

$\endgroup$
  • $\begingroup$ $\rho_n$ is the position vector relative to what? $\endgroup$ – FGSUZ Mar 4 at 21:09
  • $\begingroup$ The way that Johns defines $ \rho_n $ is as the difference between the position $ r_n $ of the nth particle and the center of mass of the collection of particles, $ R $, so he gives the definition as $ \rho_n = r_n - R $ $\endgroup$ – Z_z_Z Mar 4 at 21:14
  • $\begingroup$ Ok. Then there's nothing wrong. Torque is the sum of torque of CM plus torque wrt the CoM $\endgroup$ – FGSUZ Mar 4 at 21:30
  • $\begingroup$ That makes sense. The only thing tripping me up is the equivalence $ m_n\ddot{\rho_n} = f_n^\text{(ext)} $. I can't see how internal force on the nth particle is eliminated here. I suspect that $ m_n A $ is equivalent to the internal force on the nth particle, but I can't see how to derive that mathematically. Or is there some concept I'm forgetting entirely? $\endgroup$ – Z_z_Z Mar 4 at 21:51
0
$\begingroup$

There are two things here:

  1. The torque can be written as the torque on the center of mass plus the torque with respect to the center of mass: $\tau=\tau_{CM}+\tau'$
  2. Internal forces act by pairs: for each force on particle $i$ due to $j$, there is an opposite force from $j$ to $i$ of the same magnitude.

Consequently, $m_n \ddot{\rho}_m=f^{(ext)}$.


Full developent:

$$\frac{d}{dt}\vec{L}=\sum_i m_i \dot{\vec{r}_i}\times \dot{\vec{r}_i} + \sum_i m_i \vec{r}_i \times \ddot{\vec{r}_i}$$

The first term vanishes and

$$\frac{d}{dt}\vec{L}=\sum_i m_i \vec{r}_i \times \ddot{\vec{r}_i}$$

But $\vec{r}$ can eb split in $\vec{R}+\vec{\rho}$ where $\vec{R}$ is the position of the center of mass.

$$\frac{d}{dt}\vec{L}=\sum_i m_i (\vec{R}+\vec{\rho}_i) \times (\ddot{\vec{R}}+\ddot{\vec{\rho}_i}) $$

Appliying the distributive property, 4 terms appear. Only two of them survive due to properties of cross product. The remaining two terms are

$$\frac{d}{dt}\vec{L}=\sum_i m_i \vec{R} \times \ddot{\vec{R}}+ \sum_i m_i \vec{\rho}_i \times \ddot{\vec{\rho}_i} $$

Since the $R$'s are constant, you get $\sum_im_i =M $ and then

$$\frac{d}{dt}\vec{L}= M\vec{R} \times \ddot{\vec{R}}+ \sum_i \vec{\rho}_i \times (m_i\ddot{\vec{\rho}_i}) $$

The first term is the torque of the center of mass.

The second tem contains $(m_i\ddot{\vec{\rho}_i}) $ which is the force on particle $i$.

$$ [...] + \sum_i \vec{\rho}_i \times f_i $$

The force on particle $i$ will be the sum of internal + external. Since you're summing for all aprticles, the internal of one will cancel out with the internal of another.

Only external forces survive.

$\endgroup$
  • $\begingroup$ Thank you, this is much clearer now. I really appreciate the help. $\endgroup$ – Z_z_Z Mar 5 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.