Can conservation of angular momentum be proven?

Question

It's been a while I was thinking about conservation of angular momentum. The fact which makes me uncomfortable is why does uniform angular velocity implies, $$\vec{\tau}^{\text{EXT}}=0.$$

I was trying to prove it and found it in an attempt but suspect its validity.

My proof:

Suppose an object is moving with a constant angular velocity, say $\omega$. The linear velocity is related to it by the equation $$v=r\omega$$

If $\omega$ is constant then for body rotating about an axis (i.e. $r$ is constant) then $v$ is also constant (We will use this at last).

$$\vec{L}=\vec{r}×\vec{p}$$ Differentiating on both sides, $$\frac{d\vec{L}}{dt}=\vec{r}×\frac{d\vec{p}}{dt}+\frac{d\vec{r}}{dt}×\vec{p}$$ $$\frac{d\vec{L}}{dt}=\vec{r}×m\frac{d\vec{v}}{dt}+\vec{v}×\vec{p}$$

Second term goes to $0$.

The first term is $0$ too as our linear velocity $v$ was constant according to our first claim. Since $d\vec L/dt=0$ implies that $\vec{L}$ is constant vector.

Answer 1

I am answering the title

Can conservation of angular momentum be proven?

Conservation laws are called laws because they are axiomatic, experiments and observations of the last centuries have given the three conservation laws of energy, momentum and angular momentum. Any physics mathematical theory to be valid has to include these conservation laws.

The proof you are asking for is related with "how the mathematics leads to consistency with the conservation law" . Here is a review Symmetries and conservation laws: Consequences of Noether’s theorem

Answer 2

You write

our linear velocity $v$ was constant according to our first claim

where the “first claim” was

The linear velocity is related to it by the equation $=$

But $v=r\omega$ is a scalar statement about the magnitude of the linear velocity. The direction of the linear velocity changes as your object moves in its circular motion: at times it goes “to” and at other times it goes “fro.” So your expression for $\vec L$ becomes

\begin{align} \vec L &= \vec r \times \frac{\mathrm d\vec p}{\mathrm dt} + \frac{\mathrm d\vec r}{\mathrm dt}\times \vec p \\ &= \vec r \times \vec F + \vec v \times \vec p \\ &= \vec r \times \vec F \end{align}

where $\vec F = \mathrm d\vec p/\mathrm dt$ is Newton’s second law.

In uniform circular motion, there is only a center-pointing force with $\vec r \times \vec F = 0$. Otherwise $\vec r \times \vec F_\text{external}$ is the usual definition of an external force.

Note that a free particle moving in a straight line, with $\vec F=0$ and $\vec p=\text{constant}$, has constant angular momentum for any choice of origin. A free particle moving in a straight line does not have a well-defined frequency.

Answer 3

While both answers already given by rob and Anna v are correct, let me add my answer, this answer will provide a proof. As it has been pointed out to you, angular momentum is well defined without any angular frequency.

Suppose that the only force acting on a particle is a central force, that is, a force that has the form $\mathbf{F}=f(r)\hat{r}$, where $r=|\mathbf{r}|$, the magnitude of the position vector, and $\hat{r}=\frac{\mathbf{r}}{r}$, the unitary vector in the direction of $\mathbf{r}$. Newton's second law implies that $$\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}=f(r)\hat{r}, \tag{1}$$ where $\mathbf{p}=m\mathbf{v}$ is the momentum. The derivative of the angular momentum, $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ is $$\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}=\mathbf{r}\times\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}+\mathbf{v}\times\mathbf{p}. \tag{2}$$ Both terms in the RHS vanish, due to the properties of the cross product. The first one vanishes due to the equation of motion, labeled equation (1) in this answer, which implies that the net force (equal to the rate of change of momentum) is perpendicular to $\mathbf{r}$. The second term vanishes because $\mathbf{p}$ and $\mathbf{v}$ are in the same direction. We are then left with $$\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}=\mathbf{0}, \tag{3}$$ which is conservation of momentum. This is a vector theorem, it is telling you that the time derivative of each of the three components of angular momentum is zero. The reason why this is so useful is that many commonly encountered forces are central, such as Newtonian gravity, or the spring force given by Hooke's law.