Any central force implies conservation of angular momentum

Question

It is well known that not only a central force of 1/r^2, but any central force directed to one central point implies kepler's second law or the conservation of angular momentum. But how exactly can this be seen?

Answer 1

Define the angular momentum $\mathbf{L}$ by the cross product $$\mathbf{L} = \mathbf{r} \times \mathbf{p}.$$ By the product rule: $$ \dot{\mathbf{L}} = \dot{\mathbf{r}} \times \mathbf{p} + \mathbf{r} \times \dot{\mathbf{p}}. $$

The first term vanishes. This is because $\dot{\mathbf{r}} = \mathbf{v}$ and $\mathbf{p} = m\,\mathbf{v}$. Thus, in the first term we have the cross the cross product of two parallel vectors (i.e. $\dot{\mathbf{r}}\times \dot{\mathbf{r}}$), which is zero.

The second term also vanishes. From Newton's second law we know that $\dot{\mathbf{p}} = \mathbf{F}$. Since $\mathbf{F}$ is central, $\mathbf{F}$ is parallel to $\mathbf{r}$. Once again we have the cross product of two parallel vectors (i.e., $ {\mathbf{r}}\times {\mathbf{r}}$), which is again zero.

We have found that under the influence of a central force, $\dot{\mathbf{L}} = 0$. Thus, we conclude that under the influence of a central force the angular momentum, $\mathbf{L}$, is conserved.

Answer 2

Wizact has a nice concise answer. This result can also be seen as a result of symmetry of the Lagrangian under rotations, due to Noether's Theorem, which states

Theorem 1 (Noether's Theorem, 3 Dimensions): Suppose we have a Lagrangian $\mathcal{L}(x_1, x_2, x_3)$. If for some transformation of coordinates $$T: x_1 \to x_1 + f_1(x_1, x_2, x_3) \epsilon, ~ x_2 \to x_2 + f_2(x_1, x_2, x_3) \epsilon, ~ x_3 \to x_3 + f_3(x_1, x_2, x_3)\epsilon$$ leaves the Lagrangian invariant to linear order in $\epsilon$, that is, $\mathcal{L}(T(x_1), T(x_2), T(x_3)) = \mathcal{L}(x_1, x_2, x_3) + O(\epsilon^2)$, then there is a conserved quantity. The conserved quantity is $$P = \sum_{i = 1}^3 \left[ f_i(x_1, x_2, x_3) \cdot \frac{\partial \mathcal{L}}{\partial \dot x_i} \right]$$ Proving this is not terribly difficult, we just have to show that $dP/dt = 0$. I'll leave that as an exercise to you.

Example: That's quite a mouthful, but here's a simple example that might clear things up. Consider a free particle, whose Lagrangian is $\mathcal{L} = \frac{1}{2}m {\dot r}^2 = \frac12 m (\dot x_1^2 + \dot x_2^2 + \dot x_3^2)$. Under the transformation $$T: x_1 \to x_1 + \epsilon, ~ x_2 \to x_2 + \epsilon, ~ x_3 \to x_3 + \epsilon$$ which is just a shift of coordinates by the constant vector $\epsilon \cdot (1, 1, 1)$, it's clear that the Lagrangian is left the same, since $\dot x_i = \dot T(x_i)$. For example, $T(x_1) = x_1 + \epsilon$, whose time derivative is just $\dot x_1$. We've essentially taken $f_1 = f_2 = f_3 = 1$ in Noether's Theorem above. By Noether's Theorem, this would imply that $$P = \sum_{i = 1}^3 \left[ f_i(x_1, x_2, x_3) \cdot \frac{\partial \mathcal{L}}{\partial \dot x_i} \right] = \sum_{i=1}^3 \frac{\partial \mathcal{L}}{\partial \dot x_i} = p_x + p_y + p_z$$ And we expected this! The total momentum is conserved because of the symmetry under translations. In fact, we know more: we know each of $p_x$, $p_y$, and $p_z$ is conserved. Can you find a transformation and then use Noether's Theorem to show this?

Conservation of Angular Momentum: The conservation of angular momentum in a central potential can be seen as a result of symmetry due to infinitesimal rotations in each of the coordinate axes. Our Lagrangian has the form $$\mathcal{L} = \frac12 m (\dot x_1^2 + \dot x_2^2 + \dot x_3^2) + V(x_1^2 + x_2^2 + x_3^2)$$ I claim that this lagrangian is invariant to linear order under the following three continuous transformations: \begin{align} T_1: x_1 \to x_1, ~ x_2 \to x_2 - \epsilon x_3, ~ x_3 \to x_3 + \epsilon x_2 \\ T_2: x_1 \to x_1 + \epsilon x_3, ~ x_2 \to x_2, ~ x_3 \to x_3 - \epsilon x_1 \\ T_3: x_1 \to x_1 - \epsilon x_2, ~ x_2 \to x_2 + \epsilon x_1, ~ x_3 \to x_3 \end{align} These can be seen as infinitesimal rotations in the $x_1$, $x_2$, and $x_3$ axes, respectively. Let's just consider the effect of the first transformation $T_1$. Then \begin{align} \mathcal{L}(T(x_1), T(x_2), T(x_3)) &= \frac12 m (\dot x_1^2 + (\dot x_2 - \epsilon \dot x_3)^2 + (\dot x_3 + \epsilon \dot x_2)^2) + V(x_1^2 + (x_2 - \epsilon x_3)^2 + (x_3 + \epsilon x_2)^2) \\ &= \frac12 m (\dot x_1^2 + \dot x_2^2 + \dot x_3^2 + \epsilon^2 (\dot x_2^2 + \dot x_3^2)) + V(x_1^2 + x_2^2 + x_3^2 + \epsilon^2 (x_2^2 + x_3^2)) \end{align} And thus we can see that the Lagrangian is invariant to linear order: the largest correction in $\epsilon$ of $\mathcal{L}$ is proportional to $\epsilon^2$. Thus, there is a conserved quantity $$P = \sum_{i = 1}^3 \left[ f_i(x_1, x_2, x_3) \cdot \frac{\partial \mathcal{L}}{\partial \dot x_i} \right] = x_2 \frac{\partial \mathcal{L}}{\partial \dot x_3} - x_3 \frac{\partial \mathcal{L}}{\partial \dot x_2} = x_2 p_{x_3} - x_3 p_{x_2}$$ which is exactly the component of angular momentum in the $x_1$ direction! The exact same technique can be applied to transformations $T_2$ and $T_3$ to show that angular momentum in the $x_2$ and $x_3$ direction is conserved.

Conclusion: This explanation is a little long-winded, but this is just a part of a much deeper physics result (Noether's theorem), which roughly states that for every continuous symmetry (transformation of coordinates that leaves the Lagrangian invariant to leading degree), there exists a corresponding conserved quantity. You will definitely learn about this in your Classical Mechanics class, and this fact will remain important in many higher level physics classes.