# Energy Density & Pressure

1. Aug 29, 2008

### mysearch

Part-1:Mass
In the context of cosmology, the Friedmann-Acceleration equation highlights the need to understand the components that contribute to both the energy density $$[\rho]$$ and pressure [P]:

[1] $$\left(\frac {\ddot a}{a} \right) = - \frac {4 \pi G}{3} \left( \rho + \frac {3P}{c^2} \right)$$

However, I was hoping somebody might be able to answer some questions, in part-2, about the relationship between $$[\rho]$$ and [P] specifically related to radiation. First some initial references with respect to mass [m]:

[2] Energy Density $$\rho = Energy/Volume$$
[3] Pressure $$P = Force/Area$$
[4] $$P = \omega \rho c^2$$

On this basis, I can see how the units between $$[\rho]$$ and [P] are resolved, albeit that there is no explanation of $$[\omega]$$ other than it is a ‘weighted’ number with no units. If I expand [2] for matter as follows:

[5] $$\rho = mc^2/Volume$$ or
[6] $$m =\rho V/c^2$$

As such, I have a relationship between $$[\rho]$$ and [m], but while [4] alludes to a corresponding relationship between [m] and [P], I must first define $$[\omega]$$. This seems to be done on the basis that matter in the universe is so low its density can be treated on par with dust and, as such, has no pressure and so $$[\omega=0]$$.

2. Aug 29, 2008

### mysearch

On the basis of [1], in post #1, mass only has a gravitational decelerating effect?

This seems a reasonable enough conclusion, so I will now try and repeat the same process for radiation, i.e. E=hf, but extending this definition to $$E = hf = mc^2$$:

[7] $$\rho = E/V = hf/Volume = mc^2/V$$ or
[8] $$m =\rho V/c^2$$; where $$m=hf/c^2$$

Clearly [6] and [8] are essentially identical, which seems to imply that radiation energy density will also have an effective mass that will cause a deceleration with respect to [1]. Again, this seems a reasonable conclusion.

Many sources define $$[\omega=1/3]$$ for radiation, but I am not sure how they derived this result plus it leads to the conclusion that radiation not only causes deceleration due to its effective gravitational mass, but an additional deceleration due to its pressure. This doesn't seem so intuitive and I would really like to be able to review a more detailed derivation that explains this result.

Can anybody either explain the result or provide a link?

Many Thanks.

3. Aug 29, 2008

### George Jones

Staff Emeritus

Here's one way:

http://www.sfu.ca/~boal/390lecs/390lec23.pdf.

4. Aug 30, 2008

### mysearch

Response to #3

George, thanks for the link, it was a useful article that helps explain why $$[\omega=1/3]$$ for radiation. It also explains why most papers just quote the result. However, I am still confused as to why the Acceleration equation below implies that radiation pressure slows expansion in the same way as gravitational mass.

[1] $$\left(\frac {\ddot a}{a} \right) = - \frac {4 \pi G}{3} \left( \rho + \frac {3P}{c^2} \right)$$

This statement is based on the fact that $$[\omega_{radiation}=+1/3]$$, such that the radiation pressure [P] seems to combine with the gravitational energy density to decelerate expansion. As such, it would seem only a negative $$[\omega]$$ value would result in a pressure that counters the effects of gravity and drive the expansion of the universe, as associated with dark energy. This said, the following extract, relates to the lifecycle of star, and would seem to suggest that radiation pressure counters the pull of gravitational mass, at least, within a star.
Therefore, if I were asked to breakdown this process in terms of radiation pressure and gravitational energy density, it would go along the following lines:

- Imagine a spherical surface within the star corresponding to a surface of particles.

- Within this surface, hydrogen fusion produces photons that travel outwards towards our spherical surface.

- These photons collide with particles at this conceptual surface and energy is transferred from photons to particles.

- This can be interpreted as an outward pressure, i.e. force per unit area, resulting from the energy density of photons, i.e. energy per unit volume, within the spherical surface.

- However, the energy density of the photons within this surface also has an effective mass and therefore an inward gravitational effect on the particles at our surface.

- At the physical surface of a star, the outward pressure of radiation is matched by the net inward gravitational effect of the mass of the particles and the effective mass of the radiation.

I assume I missing some key point, but the net result is still a degree of confusion over the sign of $$[\omega_{radiation}=+1/3]$$.

5. Aug 31, 2008

### mysearch

Rationalising the Friedmann Model?

While I understand the historical significance of $$[\Lambda]$$ in respect of Einstein’s initial reluctance to accept the idea of an expanding universe, I am not sure I understand the need to perpetuate its use today. In part, my earlier postings are an attempt to understand whether a basic model of the universe can be described in terms of an energy density plus an associated pressure without recourse to any other symbolic references, e.g. $$[\Lambda]$$?

In post #3, I tried to outline my understanding of the relationship between energy density and pressure in connection with the stability of a star, which might be summarised for reference by way of 2 numbered bullets:

1. Any volume of space that contains energy must be considered to have an associated mass by virtue of $$E=mc^2=hf$$ and, as such, this effective mass must have a gravitational effect.

2. Equally, any volume of space that contains energy must also be considered to have a pressure by virtue of the relationship $$P=\omega \rho c^2$$.

While possibly overly simplistic, it might be said that the cosmology model is based on just 3 equations, i.e. Friedmann, Fluid and Acceleration, which are often presented with a $$[\Lambda]$$ term. However, there are also references that seem able to discuss this model in terms of just energy density and pressure without any direct reference to $$[\Lambda]$$, e.g.

http://arxiv.org/abs/astro-ph/0309756
http://relativity.livingreviews.org/...es/lrr-2001-1/

As such, these papers discuss a cosmological model in terms of the following forms of energy density to which there may or may not be an associated pressure:

- Matter = $$\rho \propto a^{-3};\ P =\omega \rho c^2;\ \omega =0$$

- Radiation = $$\rho \propto a^{-4};\ P =\omega \rho c^2;\ \omega =+1/3$$

- Cold Dark Matter = $$\rho \propto a^{-3};\ P =\omega \rho c^2;\ \omega =0$$

- Dark Energy = $$\rho \propto a^{-0};\ P =\omega \rho c^2;\ \omega =-1$$

- Space Curvature = $$\rho \propto a^{-2};\ P =\omega \rho c^2;\ \omega =-1/3$$

Note: Not sure whether CDM follows the same profile as matter?

If we try to reconcile the description of bullets (1) and (2) within an expanding universe, as described by the Friedmann equations, it seemed logical to initially consider the associated pressure [P] as a source of expansion and energy density as a source of gravitational mass resulting in a slow down of expansion. This is why the value of $$[\omega_{radiation}=+1/3]$$ for radiation was questioned in post #4, because it seems to imply that the pressure of radiation actually contributed to the slow down of the earlier universe, not its expansion.

Last edited: Aug 31, 2008
6. Sep 1, 2008

### jonmtkisco

Hi mysearch,

1. It has positive pressure, causing it to push outward with force against any "surface" or manifold at which there is a pressure gradient (i.e., lower pressure on the other side of the surface.)

2. It has a rest mass which gravitates.

3. Its positive pressure additionally gravitates. The equation of state for radiation is w = 1/3, so this pressure causes an effective doubling of the gravitational density.

In the case of a star, there is a pressure gradient at the surface of the star, so the positive pressure creates an outward force. This outward force counterbalances the gravity of the star's baryons, as well as the tiny contribution of gravity from the free photons inside the star.

In the case of the universe, the standard assumption is that there is no outer "edge" of the universe. Therefore the positive pressure of free radiation equally permeates the entire universe, and there is no pressure gradient anywhere. The pressure is homogeneous. So the outward force of positive radiation pressure has nothing to push against, and therefore it has NO EFFECT on the expansion rate of the universe.

The only effect of the positive radiation pressure on the Friedmann equation is to double the deceleration caused by radiation's gravity. Observed radiation density also decreases more quickly than the volume of the universe expands, because it is redshifted in the rest frames of observers in the galaxies it passes, which themselves are receding faster and faster away from the radiation's source as a function of distance and time.

Matter on the other hand has the equation of state w = 0, meaning that it has no pressure effect at all in the Friedmann equations. Only its rest mass contributes to gravitational deceleration.

Lambda can have whatever pressure is assigned to it by the particular model. The cosmological constant model assigns it a pressure of w = -1. Thus the cosmological constant contributes a net acceleration effect equivalent to 2X its rest mass, with the first 1X of the negative pressure going to offset the gravitational deceleration of the cosmological constant's own rest mass. As with positive pressure, there is nothing for the negative pressure of Lambda to pull against, no pressure gradient, so it exerts no collapse force per se.

If Lambda with negative pressure could theoretically be confined temporarily to a certain volume surrounded by space with no Lambda, the negative pressure would cause an inward "pulling" force or tension towards the center of the Lambda sphere. But no one has figured out how to create such an experiment, and the negative pressure force of lambda is insignificant except at astronomically large volumes.

Jon

7. Sep 2, 2008

### mysearch

Response to #6

Hi Jon,
Thanks for the bullets and subsequent rationale; collectively they provide a good argument for the accepted assumptions about radiation, which I was missing. However, could I also interpret your description in terms of a uniform expansion of each unit volume of space, which therefore does not give rise to any temperature gradient, i.e. net flow of photons, and so does not give rise to any force per unit area, i.e. pressure, with respect to any conceptual surface at radius [r]?

If so, does this suggest that radiation in the context of the universe has no associated pressure of expansion. Therefore, the value of $$\omega=+1/3$$ gives rise to a somewhat misleading concept of pressure, which doesn’t really act as pressure, but rather as an additional gravitational effect?

If possible, I would also like to try to clarify a few wider points associated with radiation that I have read recently:

- Baryogensis describes the process where for every 1 billion+1 particles annihilating with 1 billion antiparticles, the net result was 1 particle and 2 photons.

- 2 photons are required to maintain the conservation of momentum and this process accounts for there being ~2 billion photons per particle in the universe.

- The energy density of matter is a function of $$\frac {1}{a^3}$$.

- The energy density of radiation is a function of $$\frac {1}{a^4}$$.

- The additional power of [a] for radiation is associated with redshifting due to the expansion of the universe. As such, current estimates place the amount of energy in massive particles to be over 1000x than the energy in radiation.

Could I try to summary the state-of-play using a slightly non-standard form of Friedmann’s equation, which just uses energy density and pressure, which was one of the points raised in #5:

$$H^2=\frac{8}{3} \pi G \left[ \left(\rho_m + \rho_\lambda + \rho_\Lambda + \rho_k \right) + \left( \frac {P_m}{\omega_m c^2} + \frac {P_\lambda}{\omega_\lambda c^2} + \frac {P_\Lambda}{\omega_\Lambda c^2} + \frac {P_k}{\omega_k c^2} \right) \right]$$

- All components produce a gravitational effect, although [k=0] implies spatial curvature has no gravitational effect today and the $$a^{-4}$$ factor reduces radiation to near zero in the current era.

- While conceptually all components may be said to have a pressure effect, [k=0] again suggests this is now zero, $$\omega=0$$ for matter suggests its associated pressure is zero and $$\omega=+1/3$$ for radiation suggests this acts more like an additional gravitational mass rather than an expansive pressure.

Does CDM have the same attributes as matter?

Is dark energy the only effective agent of expansion?

8. Sep 2, 2008

### jonmtkisco

Re: Response to #6

Yes.

Correct, but it's not misleading if you understand how it works.
Sorry, all these equations make me weary, so I won't comment on them.
I'm not sure what the question is, but it is correct that the only role of pressure in the Friedmann equations is to add positive or negative gravitational density.

If you mean in terms of the pressure component of the Friedmann equations, the answer is yes.
There are two sources of expansion, the original one that is most commonly believed to have resulted from inflation shortly after the big bang, and the late-times acceleration believed to be occurring because the universe passed the balance point (at around 7 Gy) where it had more Lambda than matter (in terms of their effect on deceleration/acceleration).

Jon

9. Sep 5, 2008

### mysearch

Hi,
Could anybody help clarify the assumed nature of dark energy in terms of its energy density and pressure? In the standard form of Friedmann's equation the implication of the cosmological constant $$[\Lambda]$$ is that it overpowers any contraction due to gravity, as implied by $$\rho$$.

However, it seems possible to make the following assumptions about energy density associated with the cosmological constant based on the equivalence of units in the Friedmann equation, i.e.

[1] $$\rho_\Lambda \equiv \frac {\Lambda c^2}{8 \pi G}$$

As an aside, the units of $$[\Lambda=1/metres^2]$$ and the numbers suggest that this is equivalent to $$[1/R^2]$$, where [R] seems to be in the order of the radius of the visible universe?

[2] $$\rho = \frac {E}{V} = \frac {mc^2}{V}$$ such that $$m= \frac {\rho V}{c^2}$$

This suggests that dark energy also has an effective mass and therefore a gravitational effect that would slow down the expansion of universe in opposition to the expansion assumed by the following equation:

[3] $$P = \omega \rho c^2$$ where $$\omega_\Lambda=-1$$

Equation [1] suggests that the use of the cosmological constant in Friedmann’s equation can be replaced by an equivalent energy density. While I have shown a transposition into pressure, I am not sure how this ‘overpowers’ the effective gravitational mass given that $$P_\Lambda = -1 \rho_\Lambda c^2$$?

A similar problem seems to occur with this substitution into the Fluid equation, while the Acceleration equation is offset by the factor $$[\rho + 3P]$$ that allows [-P] to overcome $$[\rho]$$ and give a net positive acceleration.

Appreciate any insights. Thanks.

Last edited: Sep 5, 2008