1.3. Conservation of momentum and energy?
Are total momentum and energy conserved in cosmology? This is a nontrivial question because the canonical momentum and Hamiltonian differ from the proper momentum and energy.
Consider first the momentum of a particle in an unperturbed Robertson-Walker universe. With no perturbations, = 0 so that Hamilton's equation for p becomes dp / d = -am = 0, implying that the canonical momentum p is conserved. But, the proper momentum mv = a-1p measured by a comoving observer decreases as a increases. What happened to momentum conservation?
The key point is that v = dx / d is measured using a non-inertial (expanding) coordinate system. Suppose, instead, that we choose v to be a proper velocity measured relative to some fixed origin. Momentum conservation then implies v = constant (if = 0, as we assumed above). At = 1 and 2, the particle is at x1 and x2, respectively. Because dx / d gives the proper velocity relative to a comoving observer at the particle's position, at 1 we have dx / d = v - (/a)1 x1, while at 2, dx / d = v - ( / a)2 x2. (The proper velocity relative to the fixed origin is v in both cases, but the Hubble velocity at the particle's position - the velocity of a comoving observer - changes because the particle's position has changed.) Combining these, we find [(2) - (1)] / (2 - 1) - ( / a)[x(2) - x(1)] / (2 - 1) + O(2 - 1) or, in the limit 2 - 1 0, d2x / d2 = - (/a) dx / d. This is precisely our comoving equation of motion in the case = 0. Thus, the "Hubble drag" term ( / a)dx / d is merely a "fictitious force" arising from the use of non-inertial coordinates. Stated more physically, the particle appears to slow down because it is continually overtaking faster moving observers.
Energy conservation is more interesting. Let us check whether the Hamiltonian H(x, p, ) is conserved. Using Hamilton's equations for a single particle, we get
Using H = p2 / (2am) + am, we obtain dH / d = - ( / a)(p2 / 2am) + md (a) / d which is nonzero even if d / d = 0. Is this lack of energy conservation due to the use of non-inertial coordinates? While the appearance of a Hubble-drag term may suggest this is the case, if we wish to obtain the total Hamiltonian (or energy) for a system of particles filling all of space, we have no choice but to use comoving coordinates.
Perhaps the Hamiltonian is not conserved because it is not the proper energy. To examine this possibility, we use the Hamiltonian for a system of particles in comoving coordinates, with H = a(T + W). The proper kinetic energy (with momenta measured relative to comoving observers) is
while the gravitational energy W is given in eq. (1.11). Holding fixed the momenta, we see that a2T is a constant, implying (aT) / = - T. Similarly, holding fixed the particle positions, we find that a is a constant, implying (aW) / = 0. We thus obtain the Layzer-Irvine equation (Layzer 1963, Irvine 1965)
Total energy (expressed in comoving coordinates) is not conserved in Newtonian cosmology. (This is also the case in GR - indeed, there is generally no unique scalar for the total energy in GR.) However, if almost all of the mass is in virialized systems obeying the classical virial theorem 2T + W 0, we recover approximate total energy conservation.