We have seen how cosmic redshift can be understood as photons conserving angular momentum by 'shedding energy' in terms of frequency as the cosmic balloon expands. Is it reasonable to think that particles simply shed speed in order to also maintain a constant angular momentum? So it seems.
Intro
Fig. 1 (right) shows a single massive particle (pink arrow) being shot along the balloon surface at a time when R=25 units. As the balloon expands, the particle maintains its angular momentum relative to the balloon center by traveling more slowly along the balloon surface. This is visible as the shrinking of the angle between the arrow and the radial lines, representing static observers on the surface. Instead of continuously spiraling out, as a photon would have done, the particle's angle to a local radial line eventually approaches zero. This means that over time the particle's surface velocity (momentum) decays to zero.
This effect is qualitatively independent from the expansion profile, be it decelerating, constant or accelerating expansion, as long as there is expansion and not contraction. This is a rather surprising result of cosmic expansion, because it looks like a 'drag' or 'friction' that the cosmic balloon surface (space) exerts on particle movement. However, when there is no expansion (a static universe), the effect disappears, ruling out that it is a friction. What is more, should the cosmic balloon shrink, the particle will gain speed relative to balloon surface, as it has to in order to maintain a constant angular momentum.
Analytics
In Galilean dynamics, it is the kinetic energy (½mv^{2}) and the momentum (mv) of a particle that decay due to an expanding universe. The velocity v is also known as the 'peculiar velocity' in order to distinguish it from the 'Hubble velocity', which is the apparent recession velocity (relative to us) of a distant object that is static relative to the balloon surface.
As the radius R of the balloon increases, the v of the particle decreases in order to keep angular momentum L=mRv constant  at least in Galilean (low speed) dynamics. In relativistic (high speed) dynamics, things are not much different, just with the special relativistic Lorentz factor gamma (γ) entering the equation.^{[1]}
L = m R v γ = constant  (1)
where γ = (1v^{2}/c^{2})^{}^{0.5}. It is relatively easy to plot the curve of Fig. 1 from this equation, together with the expansion equations of Friedman, of course. Depending on the speed of the particle and the expansion profile, the particle path may be virtually straight, as in Fig. 1, or it may be curved. In the end, it always ends up going more and more radially, meaning with less and less momentum along the surface.
Figure 3.5 above is from a doctoral thesis by Tamara M. Davis, under supervision of prof. Charles Lineweaver, titled "Fundamental Aspects of the Expansion of the Universe and Cosmic Horizons".^{[2]} It shows the decay of the velocity of particles for various initial particle speeds, with light (v=c) the top curve. It is clear that the larger the kinetic (momentum) part of the total energy of a particle (see Eq. 2 below), the more it behaves like a photon.
Particle Redshift?
It is possible to analyze this case in terms of redshift of de Broglie waves.^{[3]} Following the thoughts in the previous Blog entry ("The linear momentum of a photon is given by p = h/λ, where h is the Planck constant") and Roger's suggestions on de Broglie waves, the following interpretation is very interesting.
The relativistic energy of a moving particle is given by
E = √[p^{2} + m^{2}] c^{2} = √[(h/λ)^{2} + m^{2}] c^{2}  (2)
where p is the momentum, m is the rest mass and λ the de Broglie wavelength of the particle. It seems like a "de Broglie momentum" (h/λ) is added vectorwise to the rest mass (m), giving the particle its waveparticle duality. Particle mass is presumed constant, so in order to keep angular momentum relative to the balloon center constant, the de Broglie wavelength of the moving particle must redshift more or less like for a photon.
Angular momentum of a particle moving on the balloon is given in special relativity by eq. (1) above as: L = m R v γ = constant.
But, according to ref. note [3]:
γ = h / (λ m v)  (3)
so we can also write:
L = R h / λ = constant  (4)
exactly as for a photon. If R is increasing, λ must be increasing by the same ratio in order to keep L constant. This is same as a velocity decrease in order to keep the relativistic angular momentum of the particle constant on the expanding balloon. The velocity equivalent to a specific λ can be extracted from Eq. 3, taking into account that γ is also a function of v.
Caution
One must always remember that the balloon analogy is just that, an analogy. It may help to get 'handles' on some of the puzzles of the cosmos, but it does not represent the real thing. As an example: the cosmos might be precisely flat, with no curvature, yet the behavior discussed here is apparently still present. How and why, I don't know.
Jorrie
Notes
[1] This is essentially the special relativistic form of Kepler's second law of planetary motion: "a planetary orbit sweeps out equal areas around the Sun in equal time intervals",
where the equal time intervals are now measured in proper time of the
moving particle. We do not need the general relativistic form here, because particles on
the balloon feel zero net gravitational acceleration (the whole surface
is at the same gravitational potential).
[2] http://arxiv.org/abs/astroph/0402278, a doctoral thesis by Tamara Davis, under supervision of Charles Lineweaver, page 50.
[3] http://en.wikipedia.org/wiki/De_Broglie_Wave
J

Re: Cosmic Balloon Application III: Particle Momentum Decay