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Why is it so easy to imagine a universe with translational symmetry but non-conserved momentum?

Physics Asked by Ryder Rude on November 12, 2020

Suppose, in a universe of $n$ particles, the initial position vectors of the particles are $vec{r_1},vec{r_2}…..vec{r_n}$, and their initial velocities are $vec{v_1}, vec{v_2}…..vec{v_n}$

The evolution of these particles over time is given by this law:

$frac{dvec{v_i(t)}}{dt}=sum_{j=1}^n d_{ij}(t) vec{u}$

where, $d_{ij}(t)$ is the distance between particle $i$ and particle $j$ at time $t$, and $vec{u}$ is the unit vector along the direction of $vec{v}$

Clearly, this is a deterministic rule for the evolution of particles. This universe has translation symmetry because I can perform this experiment with $n$ particles anywhere in the universe, and as long as the initial conditions are same, the rule $frac{dvec{v_i(t)}}{dt}=sum_{j=1}^n d_{ij}(t)vec{u}$ will give me the same evolution for the particles, and hence the same result of the experiment.

But clearly, momentum is not being conserved in this universe. So what am I missing?

One Answer

A special result from Noether's theorem is that, if

(a) a physical system, in a reference frame, is described by a lagrangian of the form $L = T-U$ where $T= frac{1}{2}sum_{j=1}^n m_i dot{vec{r}}_j^2 $ and $U(vec{r}_1,ldots, vec{r}_n)$ are respectively the kinetic energy and the potential energy and

(b) the Lagrangian (i.e., $U$) is translationally invariant in the rest space of the reference frame,

then the momentum vector of the system (in the said reference frame) is constant along the solution of the equation of motion.

Evidently, your physical system either cannot be described by a Lagrangian with the said form, or, it exists but it is not invariant under rigid translations of the system.

From a much more basic viewpoint, using only the Newtonian formulation, the forces your wrote (I assume that every point has the same mass $m_j=1$) do not satisfy the 3rd law, so that internal forces do not cancel each other and the total momentum is not conserved. The reason is that, above, $$d_{ij}= d_{ji}:.$$ If instead $$d_{ij} vec{u}$$ were replaced for $$vec{d}_{ij}:=vec{r}_j-vec{r}_i:,$$ we would have $$vec{d}_{ij}= -vec{d}_{ji}:,$$ which easily would imply that the internal forces chancel each other producing a conserved total momentum.

Answered by Valter Moretti on November 12, 2020

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