Deriving the equality between the external torque and the rate of change of angular momentum, for a system of particles

Question

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

Answer 1

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

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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.