# Energy of an Expanding Universe Decreases?

1. Jul 24, 2014

### yogi

The energy 'U' of a uniform spherical volume (e.g., a Hubble sphere) is (3/5)(M^2)G/R

Conventional wisdom has it that G and M are constant - at least in the preferred models that we are permitted to discuss on these forums. So when the Hubble sphere expands, we really don't expect a loss of energy. I would assume a universe expanding at constant velocity 'c' where the recessional rate and the Hubble rate are equal, would neither lose nor gain energy.

2. Jul 24, 2014

### Staff: Mentor

According to what physical theory?

First you need to show why the concept of "the energy of the universe" even has any meaning. Normally when you think of some object as having a certain energy, that tells you something about the interactions it can undergo. But the universe as a whole has nothing else to interact with.

Also, even if the concept of "the energy of the universe" makes sense, you would only expect it to be constant if the universe was static, because conservation of energy is linked to time translation symmetry. In an expanding universe like ours, there is no time translation symmetry, so there's no reason to expect the total energy to be constant.

3. Jul 25, 2014

### yogi

This is simply the standard derivation for a sphere of uniform mass distribution - consider the universe a series of shells (rho)4(pi)r^2 and integrate over the volume to get the positive energy - see Feynman's lectures on gravity.

While it is true the net energy in the universe may be zero - due to negative g fields equal to the pressure of positive matter, there is a positive component represented by all the mass and the distribution of the mass in terms of the gravitational attraction between the masses.

I do not see why a coasting universe (q = 0) cannot have the same energy as a static universe. All velocities remain the same, only the distance between the nebula change - there is no energy difference between a nebula at rest and one moving at constant velocity when judged from its own perspective. However there would be a gravitational deceleration that would tend to convert kinetic energy to potential - but by the definition of the constant c universe, there is no deceleration - so it would seem no energy need be supplied.

So back to my original question, it would appear that the dependence of energy U upon 1/R would mean the universe should lose positive energy during expansion if M and G are invariant - so where does it go?

4. Jul 25, 2014

### Ich

That is the gravitational binding energy only.
If G and M are constant, and R increases, at least I would really expect (3/5)(M^2)G/R to decrease. That's difficult to interpret as a loss of energy, though.
What does "recssional rate" mean, as opposed to "Hubble rate"? The latter meaning "Hubble parameter"? What is a velocity of expansion "c"?

5. Jul 25, 2014

### Staff: Mentor

You mean the part where he shows that, at least to a first approximation, the (negative) gravitational binding energy (which is the number with the 3/5 factor in front, as Ich pointed out) exactly cancels the (positive) rest energy? I assume you do, since you go on to say:

Yes, and Feynman's number for that is just $M$; he doesn't calculate it as a function of $R$ or anything else. If we use your suggestion of integrating a constant density (i.e., $\rho$ independent of $r$) over spherical shells, then we get (in units where $G = c = 1$)

$$M = \int_{0}^{R} 4 \pi \rho r^2 dr = \frac{4}{3} \pi \rho R^3$$

As $R$ increases, we would expect $\rho$ to decrease, so it's at least possible that $M$ remains constant. It's also possible that $M$ changes in such a way as to maintain the cancellation between the positive rest energy (which is just $M$ itself) and the negative gravitational binding energy $U \approx M^2 / R$. The latter, I believe, is what most cosmologists would say if you insisted on an interpretation in these terms of what happens as the universe expands (though I think many of them would also say that an interpretation in these terms is not really a good way of looking at it).

6. Jul 25, 2014

### yogi

Hi ICH - long time no chat

If the Universe is assembled by building it from thin spherical layers of thickness dr the differential work at each stage is:

dU = GMr(dM)/r

since Mr = ρ(4/3)(pi)(r^3) then dM = ρ(4pi)r^2 )dr;
the work in bringing-up the universe is as per my first post.

Are you saying this only represents gravitational binding energy? I was always under the impression that this formulation includes all forms of energy (appropriately converted to mass and then distributed over the volume against the work required to overcome the gravitational field.

The recessional rate for an accelerating universe is greater than the Hubble rate c. The two are only always equal in a q = 0 universe

7. Jul 25, 2014

### yogi

On further reflection - it would seem that these two ideas are one in the same -

For example, imagine a universe composed of two equal masses M, one at the Hubble center, and the other a shell of matter at the Hubble sphere with the same mass uniformly spread of the surface. The G force between the two surfaces is

GM(M)/R^2

There is no net force since the each mass is acted upon by an isotropic field. So now consider the energy represented by the totality of the isotropic field. The inward attraction at each point of the Hubble sphere and the outward pull of the Hubble sphere upon the central sphere of mass M, creates potential.

Potential energy is the integral of the force GM^2/R over the volume of the universe which is crudely

GM^2/2R which is the expression for the energy of a two sphere (1/2)(M^2)G/R.

However, the total energy of the universe is M so we are off by a factor of 2. We simply remove the central mass M and conclude that the energy of matter in a two sphere universe can be expressed as
(1/2)(M^2)G/R. More particularly, all the energy of any form when converted to its M(c^2) equivalent can be considered as gravitational field energy.

This all sounds good after a couple of drinks - I wonder if I will like it as well in the morning.

Last edited: Jul 25, 2014
8. Jul 26, 2014

### Ich

Yes, it's been a few years. Good to hear from you again.

Of course. It's the energy you get when you assemble the sphere. Look it up in Wikipedia.
"c" was a velocity in the OP, so now it's a rate?
If Hubble rate means Hubble parameter, it is defined as recession velocity divided by distance.
I'm not aware of the term "recessional rate", except as a rarely used synonym for the Hubble parameter. What is its definition?
No. The difference of potential between two points is the line integral of force on a connecting curve. And whatever you wanted to calculate, the volume integral of something will never have the same units as said something.

I have no idea what you're up to. What you started with looked like Newtonian cosmology, which is a good thing to think about. But now I'm lost.

9. Jul 27, 2014

### yogi

When I use rate in connection with the Hubble sphere - generally - I have in my own mind rate of expansion 'c' at the Hubble scale. When I use rate in connection with recessional velocity, I have in my own mind the velocity of space at the Hubble limit which may be greater than c, equal to c or less than c depending upon the model and the epic. After many years of writing Patents, I tend to be my own lexicographer - sorry for the confusion.

Also, I left the square off the R in the denominator when I wrote the Force equation GM^2/R^2
when corrected this leads to the energy expression G(M^2)/2R for the 2-sphere construction.

Your point regarding the above expression as assembly energy only is conceded:

Substituting the Black whole radius 2GM/c^2 for R in the above expression for the two sphere leads to a mass energy (1/4)Mo(c^2) rather than Mo, so there is not sufficient energy to account for the rest mass energy of the Hubble contents in either a 2 sphere or 3 sphere expansion.

10. Jul 27, 2014

### Ich

A rate is usually something of units % per year or so. "c" is usually the speed of light. The Hubble scale is usually c/H0. How is this intended to fit together?
No, the Hubble limit = Hubble scale = Hubble sphere is defined as the distance where the recessional velocity = the "velocity of space" equals c. Further, as said before, a rate has not the right dimensions to be a velocity.
There's a literature out there, and I know that you are reading on the subject for more than 10 years now. So please try to use the words in the usual way.
Ok. In a "dust" cosmology, you have the kinetic energy of the outward moving particles to compare with their binding energy. The comparison yields those well-known results:
kinetic energy < binding energy: closed, positively curved universe with Big Crunch.
kinetic energy = binding energy: open, flat universe, coming to rest in the infinite future.
kinetic energy > binding energy: open, negatively curved universe with eternal expansion.

The mass energy of an expanding region is not changing in this case.

11. Jul 27, 2014

### yogi

Exact quote: Webster dictionary for rate: Quantity or degree of a thing measured per unit of something else as rate of speed.

AT the edge of the Hubble sphere the recessional velocity is transluminal in an accelerating universe.

12. Jul 27, 2014

### Ich

Couldn't find your quote on their web page. I found:
FWIW, I don't really care. But please provide an exact definition if you use words in a nonstandard way. If it's a velocity and has units m/s, please call it a velocity. If it's a Hubble parameter and has units of 1/s, call it the Hubble parameter or the expansion rate (which would be in my understanding something like km/s/MPc or %/s or %/billion years (<- @ marcus )

AT the edge of the Hubble sphere the recessional velocity is c. Any other definition, please provide citations.

The more important part of my post was the one with the dust cosmology, IMHO. At least I thought so.

13. Jul 28, 2014

### yogi

Interesting -I wasn't aware that most cosmologists would embrace a Machian type of mass enhancement during expansion - or perhaps I am misreading your post.

14. Jul 28, 2014

### Staff: Mentor

I'm not sure what "Machian type of mass enhancement" means, so I can't say for sure whether you're misreading or not.

15. Jul 29, 2014

### yogi

Some interpretations of Mach's principle are based upon bootstrapping - inertial mass is hypothesized to increase during expansion - as for example in a zero energy universe the increase in the negative G field due to expansion would require compensation in the form of positive matter energy. From your post I assumed this could be what you had in mind - the only other theory I can think of would be that of William McCrea (expanding negative pressure creating positive energy) later adapted by Guth and others as a basis for inflationary eras.

16. Jul 29, 2014

### Staff: Mentor

Hm, ok. This wasn't really what I had in mind, and I don't think it's really what most cosmologists would have in mind (since it seems to imply "continuous creation" of matter, which doesn't happen in our current cosmological models--the stress-energy tensor is locally conserved, as it must be in GR).

I also see that I made a misstatement in my earlier post: I said that as $R$ increases, we would expect $\rho$ to decrease, but that's only really true if we treat $R$ as increasing because of the expansion of the universe. Mathematically, there's nothing requiring us to do that: we could just as easily be comparing the "mass" of two different spherical regions with different $R$ at the same instant of cosmological time, so $\rho$ would be the same for both. In that case, of course, $M$ would increase with increasing $R$ simply because a larger spherical region includes more matter.

Furthermore, even if we treat increasing $R$ as due to the expansion of the universe, we still have to decide *how* $R$ increases as the universe expands, i.e., we have to decide what we want $R$ to describe, physically. I'm not sure what Feynman had in mind for that, but it seems to me that we would like to distinguish two effects: the effect of a larger $R$ "enclosing more matter" (similar to the comparison of two spherical regions at the same instant of time, as above), and the effect of the expansion of the universe on the value of $R$ that encloses a "constant amount of matter". (This would correspond to $R$ being the radius of a region enclosing a constant number of "comoving" worldlines in an FRW spacetime.)

I think the second type of effect is really what you are interested in: does the "total mass" of a region containing a constant number of "comoving" worldlines change as the universe expands, and if so, how? Intuitively, it shouldn't change, because (at least to the level of approximation of the FRW model) no stress-energy should be passing through the boundary of such a region (since all stress-energy in this model travels on "comoving" worldlines). But also, intuitively, the "gravitational potential energy" of such a region *should* change with expansion, because the distance between the masses changes. (Of course, one way of responding to this is simply to say that the concept of "gravitational potential energy" doesn't make sense in a non-stationary spacetime.)

I don't have time right now to work out how all this would be expressed mathematically, but at least it's food for thought.

17. Jul 30, 2014

### yogi

That is correct - the picture I envision would be much like a Weyl universe with divergent world lines so the potential would be expected to increase - but the 1/R relationship would thus come into play - so conceptually, something is missing

18. Jul 30, 2014

### yogi

I am assuming you are saying, in the negatively curved universe, the pressure equals [(rho)c^2]/3 so the universe is de sitter-like and undergoing exponential expansion ---ergo there is no energy change (I suppose because negative pressure is somehow balanced to equal positive density so the universe reverts to a low density state where it can expand as though it were an empty universe.

19. Aug 6, 2014

### Ich

Sorry, I missed that post.
Not at all, for two reasons:
1) In a de Sitter universe, pressure is -(rho)c^2, not [(rho)c^2]/3.
2) The "dust"cosmology I was referring has vanishing pressure. Therefore also no Dark Energy.

You asked for Newtonian equations. You can find them for example in this link, pp. 949-953. The gist of it is: kinetic energy and potential energy can be converted into each other, but the sum is always conserved.
Note that the same equations also describe the dust models in relativistic cosmolgy, so this ís not only a toy model, but a useful way of thinking about the dynamics of expansion.

20. Aug 8, 2014

### yogi

I meant to say negative pressure = ρ(c^2)/3. De Sitter originally took ρ = P = k = 0. The interesting aspect of equating negative pressure to ρ(c^2)/3 is that you get the same equation for a matter universe where density is always balanced by negative gravitational pressure (a real matter universe exponentially expanding with net energy always zero)

R(t) = exp[λ/3)^1/2]t

The P = - ρ(c^2) universe expands and contracts at constant energy density (e.g. inflation like).

Lambda = - 8πGρ - 3q(H^2)

This seems to be a necessary artifact for all expanding universes - either initially or as an ongoing process. Interestingly, when Lambda = 0, you get Hoyle's steady state universe