Homogeneity of Space implies Conservation of Linear Momentum

This article needs to be tidied. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code.

This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.

Theorem

Because space is homogeneous in an inertial frame of reference, linear momentum is conserved if the system is closed.

Proof

In inertial frames of reference, space is homogeneous. Therefore, working exclusively with inertial frames of reference, in closed systems (those which do not interact in any ways with external agents and are therefore causally disconnected from anything external), we can make a translation of every particle in said system and nothing will change (broadly speaking, this means "no region of space is special or privileged"). In order to prove this, let us examine what effects it would have on the Lagrangian function describing a system of these characteristics if we were to make an inifinitesimally small translation of all the particles in it. For simplicity, the proof shall be carried out with a one-particle system, but it is trivial to extrapolate it to an arbitrary number of them.

Because the system is closed, the Lagrangian of the system (of the particle) does not depend on time, and it is a function exclusively of the particle's position and velocity:

$\LL = \map \LL {\vec r_i, \dot {\vec r_i} }$

Therefore, an infinitesimal change in the Lagrangian function can be written in cartesian coordinates as follows:

$\ds \delta \LL = \sum_i \frac {\partial \LL} {\partial x_i} \delta x_i + \sum_i \frac {\partial \LL} {\partial \dot x_i} \delta \dot x_i$

Now, taking into account we are just carrying out a (constant) translation of the position vector, this translation does not depend on time, and therefore:

$\delta \dot x_i = \dfrac \d {\d t} \delta x_i = 0$

Hence, we are only left with the first term.

Since our hypothesis is the Lagrangian remains unchanged under translations in space, we must then impose $\delta \LL = 0$ and see what conditions must be met in order to guarantee this equality.

Let us consider what we have so far:

$\ds \delta \LL = \sum_i \frac {\partial \LL} {\partial x_i} \delta x_i = 0$

Since each infinitesimal displacement is independent from the rest, the only way in which this summation can be null is if each partial derivative is null:

$\ds \therefore \sum_i \frac {\partial \LL} {\partial x_i} = 0$

Now, remembering Euler-Lagrange equations, we can state:

$\dfrac {\partial \LL} {\partial x_i} = - \dfrac \d {\d t} \dfrac {\partial \LL} {\partial \dot x_i}$.

We reach the conclusion that:

$\dfrac \d {\d t} \dfrac {\partial \LL} {\partial \dot x_i} = 0$

Therefore, $\dfrac {\partial \LL} {\partial \dot x_i} = \text{constant}$, but since $\dfrac {\partial \LL} {\partial \dot x_i} = \dfrac {\partial \paren {T + U} } {\partial \dot x_i}$ with $U = \map U {x_i}$, then we have:

$\ds T = \sum_i \frac 1 2 m \dot x_i^2$

It is clear that: $\dfrac {\partial \LL} {\partial \dot x_i} = p$, where $p$ is the particle's linear momentum.

Therefore, we have:

$p = \text{constant}$

$\blacksquare$