# Is the total energy of each particle constant in an expanding universe?

1. Apr 7, 2012

### johne1618

Assume that we have a particle $i$ at rest according to co-moving coordinates.

Then the total energy of the particle at rest, $E_i$, is its rest mass energy plus the sum of the (negative) gravitational potential energies between it and every other particle in the Universe. Thus we have

$\large E_i = m_i c^2 - \sum^{\infty}_{j=1} \frac{G m_i m_j}{r_{ij}}$ where $i \ne j$.

Do we expect this total energy to stay constant as the Universe expands?

If the gravitational potential terms become less negative as the Universe expands then the rate at which time passes at particle $i$ will increase causing its energy/frequency to increase. A clock at position $i$ would not tick at a constant rate and therefore could not register the passage of cosmological time properly which seems strange as it is in an inertial frame and therefore should be a "good" timekeeper.

In order that the total energy of each particle in the Universe should stay constant we would require that the number of particles in the Universe and hence its total mass should be proportional to the Universal scale factor. This puts a strong constraint on the scale factor forcing it to be linear in time.

2. Apr 8, 2012

### Chalnoth

Well, the first problem with this is that the resultant potential energy will be infinite. I'm not sure it makes sense to talk about the conservation of an infinite number.

The second thing to point out is that this is a result using only Newtonian gravity, and thus doesn't take into account the gravitation due to a cosmological constant or radiation.

That said, if you have a universe that has only normal matter in it, my understanding is that you get the exact same result for how the universe expands in Newtonian gravity as with General Relativity. So in principle, it must be possible to define gravitational potential energy for a matter-only universe in a sensible way. But you probably can't include the entire universe in the calculation. However, the gravitational attraction of everything outside of a spherical shell will always be identically zero, so you can truncate the calculation at a finite radius in comoving coordinates (comoving because you don't want your potential energy to track a different number of particles at different times).

Then what you would find, I believe, is that the sum of the kinetic and potential energies of the particles as estimated by an observer in the center would be a constant.

But this doesn't inform the truly interesting cases where you have radiation and dark energy.

3. Apr 8, 2012

### DrStupid

The gravitational potential energy does not belong to a single particle.