# Conservation of momentum

References: Anthony Zee, Einstein Gravity in a Nutshell, (Princeton University Press, 2013) – Chapter I.2, problem 1.

The nature of the dependence of a force or potential on the underlying position coordinates can determine certain conservation laws. In his chapter I.2, Zee shows that a central force (a force that is always directed towards its source, such as the Earth’s gravity or a point charge’s electrostatic field) conserves angular momentum. Actually his derivation is a generalization to any number ${D\ge2}$ dimensions of the more familiar proof in 3-d, which goes like this:

The angular momentum of a mass ${m}$ is defined as

$\displaystyle \mathbf{L}=\mathbf{r}\times\mathbf{p} \ \ \ \ \ (1)$

where ${\mathbf{p}=m\mathbf{v}=m\dot{\mathbf{r}}}$ is the linear momentum. Taking the time derivative we get

 $\displaystyle \dot{\mathbf{L}}$ $\displaystyle =$ $\displaystyle \dot{\mathbf{r}}\times\mathbf{p}+\mathbf{r}\times\dot{\mathbf{p}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m\dot{\mathbf{r}}\times\dot{\mathbf{r}}+\mathbf{r}\times\mathbf{F}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+0\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (5)$

where the third line uses the fact that ${\mathbf{r}\parallel\mathbf{F}}$ for a central force, so their cross product is zero. Thus ${\mathbf{L}}$ doesn’t change with time.

Now suppose that a force ${F}$ is the negative gradient of a potential function ${V}$ so that by Newton’s law:

$\displaystyle m_{a}\frac{d^{2}x_{a}^{i}}{dt^{2}}=-\frac{\partial V\left(x\right)}{\partial x_{a}^{i}} \ \ \ \ \ (6)$

where the index ${a}$ refers to particle ${a}$ in a collection of ${N}$ interacting particles, and ${i}$ is the component of the coordinate ${x}$. Note that the ${x}$ in ${V\left(x\right)}$ represents all ${D}$ components of ${x}$ (if we’re doing the calculation in ${D}$-dimensional space) and not just the magnitude of the distance.

Now suppose that ${V}$ is a function only of the coordinate differences ${x_{a}^{i}-x_{b}^{i}}$ between particles ${a}$ and ${b}$, where ${a,b=1,\ldots,N}$ with ${a\ne b}$. In this case, the total linear momentum is given by

$\displaystyle p^{i}=\sum_{a}m_{a}\frac{dx_{a}^{i}}{dt} \ \ \ \ \ (7)$

The time derivative is

 $\displaystyle \dot{p}^{i}$ $\displaystyle =$ $\displaystyle \sum_{a}m_{a}\frac{d^{2}x_{a}^{i}}{dt^{2}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sum_{a}\frac{\partial V\left(x\right)}{\partial x_{a}^{i}} \ \ \ \ \ (9)$

Since ${V}$ is a function of the set of all possible differences ${x_{a}^{i}-x_{b}^{i}}$, the terms in the sum 9 cancel in pairs. For example, for 3 particles if ${V=f\left(\left(x_{a}^{i}-x_{b}^{i}\right),\left(x_{a}^{i}-x_{c}^{i}\right),\left(x_{b}^{i}-x_{c}^{i}\right)\right)}$, then ${\frac{\partial V}{\partial x_{a}^{i}}}$ will contain a term equivalent to ${\frac{\partial V}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}}$ (plus another term resulting from ${\frac{\partial V}{\partial\left(x_{a}^{i}-x_{c}^{i}\right)}}$). However, ${\frac{\partial V}{\partial x_{b}^{i}}}$ will contain a term equivalent to ${-\frac{\partial V}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}}$ which cancels the first term. That is

 $\displaystyle \frac{\partial V}{\partial x_{a}^{i}}$ $\displaystyle =$ $\displaystyle \frac{\partial f}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}+\frac{\partial f}{\partial\left(x_{a}^{i}-x_{c}^{i}\right)}\ \ \ \ \ (10)$ $\displaystyle \frac{\partial V}{\partial x_{b}^{i}}$ $\displaystyle =$ $\displaystyle -\frac{\partial f}{\partial\left(x_{a}^{i}-x_{b}^{i}\right)}+\frac{\partial f}{\partial\left(x_{b}^{i}-x_{c}^{i}\right)}\ \ \ \ \ (11)$ $\displaystyle \frac{\partial V}{\partial x_{c}^{i}}$ $\displaystyle =$ $\displaystyle -\frac{\partial f}{\partial\left(x_{a}^{i}-x_{c}^{i}\right)}-\frac{\partial f}{\partial\left(x_{b}^{i}-x_{c}^{i}\right)} \ \ \ \ \ (12)$

so adding up all the derivatives causes the terms to cancel in pairs. The argument is fairly easily extended to ${N}$ particles.

Thus ${\dot{p}^{i}=0}$ and linear momentum is conserved. [I realize this isn’t a very elegant or mathematical way of proving it; there is probably a better way of writing it down, but hopefully you get the idea.]