10
$\begingroup$

I was thinking about the definition of the conservation of momentum, which says that momentum is conserved unless outside forces are acting on the system, and I was wondering that if the system is the universe, if momentum must be conserved.

To put it more clearly, is momentum in the entire universe conserved?

(I know about the second law of thermodynamics, but I do not think that it applies if we are considering individual particles since there are no "heat losses" at that level (elastic collisions in Kinetic Molecular theory, etc).)

$\endgroup$
4
  • 2
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/2838/2451 and links therein. $\endgroup$ – Qmechanic Dec 6 '14 at 1:44
  • $\begingroup$ I'm not entirely sure how it relates; the forum is about energy, not momentum. Unless your point is that conservation of both momentum and energy are outdated Newtonian concepts altogether. $\endgroup$ – k_g Dec 6 '14 at 1:54
  • $\begingroup$ Wouldn't it also depend if our universe was part of a greater universe? Then, the universe's energy could change... $\endgroup$ – TanMath Dec 11 '14 at 23:19
  • $\begingroup$ Momentum is not conserved globally, it is required to change because the energy changes and the energy-momentum tensor must be conserved $\endgroup$ – Jim Dec 12 '14 at 20:22
7
+50
$\begingroup$

If classical mechanics were valid at cosmological levels, the answer would be yes. But general relativity is what describes the dynamics at this larger scale, and it is not generically possible in GR (in an arbitrary spacetime) to define conservation of momentum or conservation of mass-energy.

$\endgroup$
2
  • $\begingroup$ What about the conservation of the energy-momentum tensor? Or if we are only looking at the metric -the Einstein tensor? It is true that this involves covariant derivatives so in general you won't get something like the time derivative (in some coordinates) of a total momentum is zero, but it seems this is a good enough generalization of the conservation principle. $\endgroup$ – octonion Dec 12 '14 at 2:35
  • $\begingroup$ I didnt say it was not, but still is not satisfactory enough, that is, in the end mass-energy is not necessarily conserved $\endgroup$ – Wolphram jonny Dec 12 '14 at 2:42
2
$\begingroup$

In as far as I know, the universe is actually gaining energy.

And as far as we can see, the universe is expanding as a product of pressure, in the direction of the difference between opposing pressures, blah blah....

So would momentum be conserved under this rule? I suppose, if the exact same amount of energy were being "destroyed" as "created" within the system (universe) to the surroundings (extra-universe).. I might hastily speculate the exchange routes inward are called stars, and the 'way out' called dark matter, but that's unfair without measure, all of this is, completely immeasurable. (at this moment)

$\endgroup$
2
$\begingroup$

Short answer: No. Momentum-energy conservation arises by dint of Noether's theorem, which says that if a system's Lagrangian is invariant with respect to a continuous transformation, there is one conserved quantity, called the "Noether Charge" for each such transformation (technically: for each linearly independent tangent vector in the Lie algebra of the group of such transformation).

Momentum-energy is the Noether charge corresponding to the continuous "symmetry" of spacetime translation. If the physics of the physical system you're dealing with doesn't change if you continuously shift the origin of your spacetime co-ordinates, then we get momentum-energy conservation. (there is one Noether charge for each part of this shift: $p_x,\,p_y,\,p_z$ for shifts in the $X,\,Y,\,Z$ direction and energy for time-translation).

General solutions of the Einstein field equations do not have these symmetries. No global time or space co-ordinates are generally definable. Spacetime over here has, in general, a different curvature from spacetime over there, so the physics does care where we put our origin.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.