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

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

• $\rho_n$ is the position vector relative to what? Mar 4, 2019 at 21:09
• 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 $\rho_n = r_n - R$ Mar 4, 2019 at 21:14
• Ok. Then there's nothing wrong. Torque is the sum of torque of CM plus torque wrt the CoM Mar 4, 2019 at 21:30
• 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? Mar 4, 2019 at 21:51

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.

• Thank you, this is much clearer now. I really appreciate the help. Mar 5, 2019 at 9:14