Gravitational effect and pressure of photons in Friedmann equation?

[icarus2]

Gravitational effect and pressure of photons in Friedmann equation?

 

[math]

\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) =  - \frac{{4\pi G}}{3}({\rho _m} + {\rho _r} + {\rho _\Lambda } + \frac{{3({P_m} + {P_r} + {P_\Lambda })}}{{{c^2}}})

[/math]

ρ is the unit of mass density.

Pressure of matter : P=0
Pressure of radiation : P=(1/3)ρc^2

Regarding the source of pressure, there are the following expressions in major books:

Quote

 

Note that the effect of the pressure P is to slow down the expansion (assuming P>0). If this seems counterintuitive, recall that because the pressure is the same everywhere in the universe, both inside and outside the shell, there is no pressure gradient to exert a net force on the expanding sphere. The answer lies in the motion of the particles that creates the fluid’s pressure. The equivalent mass of the particle’s kinetic energy creates a gravitational attraction that slows down the expansion just as their actual mass does.

- An Introduction to Modern Astrophysics 1161P

 

 

 

Quote

 

 Pressure in the kinetic theory of gases

3P=(Nmv^2)/Volume

 

 

In special relativity,

[math]{E^2} = {({m_0}{c^2})^2} + {(pc)^2}[/math]

For matter, [math]{E^2} = {({m_0}{c^2})^2}[/math]

For, relativistic particle, [math]{E^2} = {({m_0}{c^2})^2} + {(pc)^2}[/math]

For, radiation or photon, [math]{E^2} = {(pc)^2}[/math]

 

So, here's the question.
For a relativistic particle with rest mass it would be:

[math]{E^2} = {({m_0}{c^2})^2} + {(pc)^2}[/math]

In the case of relativistic particles, there is a pressure component due to kinetic energy or momentum, and this value cannot be ignored compared to mass energy.

[math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) =  - \frac{{4\pi G}}{3}({\rho _{{\rm{relativistic particle}}}} + \frac{{3{P_{{\rm{relativistic particle}}}}}}{{{c^2}}}) =  - \frac{{4\pi G}}{3}({\rho _{{\rm{rel - par}}}} + \frac{{3(\frac{1}{3}{\rho _{{\rm{rel - par}}}}{c^2})}}{{{c^2}}}) =  - \frac{{4\pi G}}{3}(2{\rho _{{\rm{rel - par}}}})[/math]

[math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) =  - \frac{{4\pi G}}{3}(2{\rho _{{\rm{rel - par}}}}) =  - \frac{{4\pi G}}{3}(\frac{{2{\rho _{{\rm{rel - par}}}}{c^2}}}{{{c^2}}})[/math]

When the energy density of matter is ρ, in the Friedmann equation, matter generates the gravitational effect of ρ.
When the energy density of a relativistic particle is ρ, in the Friedmann equation, the relativistic particle generates a gravitational effect of 2ρ.

But what about photons, which are not relativistic particles? Which expression is correct for photons that have no rest mass?

1)

[math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) =  - \frac{{4\pi G}}{3}({\rho _{rad}} + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}}) =  - \frac{{4\pi G}}{3}(2{\rho _{{\rm{rad}}}}) =  - \frac{{4\pi G}}{3}(\frac{{2{\rho _{{\rm{rad}}}}{c^2}}}{{{c^2}}})[/math]

Considering radiation(When limited to photons) with energy density ρ, it generates a gravitational effect equal to twice the energy density : 2ρ ?

 

2)

[math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) =  - \frac{{4\pi G}}{3}({\rho _{{\rm{radiation}}}} + \frac{{3{P_{{\rm{radition}}}}}}{{{c^2}}}) =  - \frac{{4\pi G}}{3}(0 + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}}) =  - \frac{{4\pi G}}{3}({\rho _{{\rm{rad}}}}) =  - \frac{{4\pi G}}{3}(\frac{{{\rho _{{\rm{rad}}}}{c^2}}}{{{c^2}}})[/math]

Considering radiation(When limited to photons) with energy density ρ, it generates a gravitational effect equal to 1 times the energy density : 1ρ ?

 

1) Is the expression correct? 2) Is the expression correct?

 

Edited March 31 by icarus2

[Mordred]

\[\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) =  - \frac{{4\pi G}}{3}({\rho _m} + {\rho _r} + {\rho _\Lambda } + \frac{{3({P_m} + {P_r} + {P_\Lambda })}}{{{c^2}}})\]

that's the math you posted that didn't latex properly as you forgot \ in your opening [math] statement. Massless particles are set up as ultra relativistic in terms of the equations of state.

https://en.wikipedia.org/wiki/Equation_of_state_(cosmology)

\[\rho=-4\] for ultra relativistic you can use the \(\omega\) for the ratio between pressure and energy density for ultra relativistic particles.

That book also doesn't show the cosmological constant portion. At least not what you posted. You don't have the author to see if I have the same textbook . 

Edited March 31 by Mordred

[icarus2]

I think I didn't phrase the question appropriately.
 

[math]\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) =  - \frac{{4\pi G}}{3}({\rho _m} + {\rho _r} + {\rho _\Lambda } + \frac{{3({P_m} + {P_r} + {P_\Lambda })}}{{{c^2}}})[/math]

[math]\rho  + \frac{{3P}}{{{c^2}}} = ({\rho _m} + \frac{{3P{}_m}}{{{c^2}}}) + ({\rho _r} + \frac{{3P{}_r}}{{{c^2}}}) + ({\rho _\Lambda } + \frac{{3P{}_\Lambda }}{{{c^2}}})[/math]

In this question the first and third terms are not of interest.
What happens if you summarize(Energy density of photon ρ) the second term?
When the energy density of a photon is ρ, I am curious as to whether its gravitational effect in the acceleration equation is 1ρ or 2ρ.
*When considering the dimension, it must be divided by c^2, but this is not an important issue.

[math]{\rho _r} + \frac{{3P{}_r}}{{{c^2}}} = [/math]

Edited April 1 by icarus2

[Mordred]

ok first off what those relations are designed to show you is the critical density relations used to determine the geometry. As well as expansion rates of the geometry. in essence of the following statement

\[\Omega_{total}=\Omega_m+\Omega_r+\Omega_\Lambda\] so in your equation they are showing the pressure relations on the portion of the LHS of the equal sign where I have the total density.  You can normalize the LHS.  The RHS should be the percentages of each except they have the wrong values in each category. Matter for example has 0 pressure, Its energy density has zero influence on pressure. Lambda gets funky in that currently its a negative pressure relation w=-1. You would need to use the scalar field equation of state in the link I gave for the kinetic vs potential energy terms. 

the statement would be properly presented as this

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

that details the following. 

\[\frac{d}{dt^2}(\rho a^3)+P\frac{d}{dt}a^3=0\]

(notice your 1 to 3 ratio )

where \(\rho_m=0, \ P_r=\rho/3, \)

matter exerts no pressure so the term in the bracket is the radiation energy density to pressure relation.

I'm not sure why the relations look wrong in what you have if that's from the book you used. However hopefully the format I used is far more clear.

 it may help to know that the critical density formula uses matter only with  no pressure term p=0

 

\[\rho_{crit} = \frac{3c^2H^2}{8\pi G}\] 

that may help to understand the ratio between pressure and density

Edited April 1 by Mordred

[MigL]

Mordred has already explained, but Wiki does also .

"The scale factor � is a parameter of the FLRW metric, and its time evolution is governed by the Friedmann equations. The second Friedmann equation,

�¨�=−4��3(�+3��2)+Λ�23,

shows how the contents of the universe influence its expansion rate. Here, � is the gravitational constant, � is the energy density within the universe, � is the pressure, � is the speed of light, and Λ is the cosmological constant. A positive energy density leads to deceleration of the expansion, �¨<0, and a positive pressure further decelerates expansion. On the other hand, sufficiently negative pressure with �<−��2/3 leads to accelerated expansion, and the cosmological constant also accelerates expansion. Nonrelativistic matter is essentially pressureless, with |�|≪��2, while a gas of ultrarelativistic particles (such as a photon gas) has positive pressure �=��2/3. Negative-pressure fluids, like dark energy, are not experimentally confirmed, but the existence of dark energy is inferred from astronomical observations."

From   Expansion of the universe - Wikipedia

[Mordred]

There is another set of ratios that may be of interest resulting from the above equations that may prove useful in better visualizing the amount of influence each contributor has. First for all reader the critical density is the density that without the cosmological constant the universe is critically flat. 

one can use the above to get a ratio of \(\Omega_i=\rho_i/\rho_c\)

\[\Omega_M=\frac{8\pi G}{3H^2_0}\rho_M\]

\[\Omega_R=\frac{8\pi G}{3H^2_0}\rho_R\]

\[\Omega_{\Lambda}=\frac{\Lambda}{3H^2_0}\]

with curvature term

\[K=\frac{K}{3a_0^2H^2_0}\]

those are the ratios of each component

\[H^2(a)=H_o^2(\Omega_R\frac{a_0^4}{a^4}+\Omega_M\frac{a_0^3}{a^3}+\Omega_\Lambda+\Omega_K\frac{a_0^2}{a^2})\]

now using the same ratios of the density evolutions of matter, radiation and lambda setting k=0 and knowing that the  Hubble parameter changes over time. One can produce the following equation of how the universe evolves over time. This equation correlates the Hubble rate compared to the rate today at each redshift. 

\[H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}\]

as well as the density of each at a particular redshift

 

 

 

 

Edited April 1 by Mordred

[LaurieAG]

While we're on ratios, from the PLANCK and WMAP data.

[Mordred]

This is just an interesting tidbit as radiation equation of state is

\[p_R=\rho_R/3\equiv \omega=1/3 \]

when you apply Maxwell-Boltzmann statistics which is a mixed combination of the Bose-Einstein and Fermi-Dirac  statistics using the effective degrees of freedom of all bosons and fermions one can determine the temperature evolution. The result turns out to be the inverse of the scale factor. 

\[\rho_R=\frac{\pi^2}{30}{g_{*S}=\sum_{i=bosons}gi(\frac{T_i}{T})^3+\frac{7}{8}\sum_{i=fermions}gi(\frac{T_i}{T})}^3 \]

\[S=\frac{2\pi^2}{45}g_{*s}(at)^3=constant\]

temperature scales inversely to the scale factor giving

\[T=T_O(1+z)\]

as a function of redshift so now you can also follow the temperature evolution knowing this relation

Edited April 1 by Mordred

[icarus2]

Consider a situation where matter and radiation have the same energy density ρ_0. If we set c=1,

mass density = energy density

 

For matter,
[math]{\rho _m} + 3P{}_m = {\rho _m} + 3(0) = {\rho _m} = {\rho _0}[/math]

For radiation (photon),
[math]{\rho _r} + 3P{}_r = {\rho _r} + 3(\frac{1}{3}{\rho _r}) = 2{\rho _r} = 2{\rho _0}[/math]

 

Now if we think about the physical meaning of the two cases,

This means that if matter and radiation have the same energy density ρ_0, the gravitational effect of the radiation is twice that of the matter.

If photon and matter have the same energy density, does the photon exert twice the gravitational force (or effect) than the material exerts? Is this true?

In my opinion,
In the case of photon, it seems wrong to include both density in the (mass) density term and pressure term.

Edited April 1 by icarus2

[Mordred]

Why do you keep thinking pressure or energy/mass density is gravitational effect ? They do both effect gravity but by different ratios that those equations by themselves do not show if you place the pc^2 portion of radiation it would be the T^00 component of the stress energy momentum tensor.  The P in pc^2 is the momentum term not the density or pressure  in your equations. So you need further conversions 

However if both matter and radiation are uniformly distributed gravity at any point would be zero according to Newtons shell theorem. So any gravitational effect depends on its distribution.

\[E=pc^2\] where the p is momentum not pressure or density. 

Another indication of realizing thinking those equations equate to gravity effect is incorrect is that a critically dense universe being flat spacetime has no gravity as there is no curvature term.

Edited April 1 by Mordred

[joigus]

On 3/31/2024 at 6:12 AM, icarus2 said:

But what about photons, which are not relativistic particles? Which expression is correct for photons that have no rest mass?

What do you mean non-relativistic photons? Photons are always relativistic. Am I missing something? Everybody else seems to understand what you mean, but I don't.

Because they're always relativistic, the equation of state (pressure as a function of density) is what it is in the matter term of the Friedmann equation.

[icarus2]

23 hours ago, Mordred said:

Why do you keep thinking pressure or energy/mass density is gravitational effect ? They do both effect gravity but by different ratios that those equations by themselves do not show if you place the pc^2 portion of radiation it would be the T^00 component of the stress energy momentum tensor.  The P in pc^2 is the momentum term not the density or pressure  in your equations. So you need further conversions 

However if both matter and radiation are uniformly distributed gravity at any point would be zero according to Newtons shell theorem. So any gravitational effect depends on its distribution.

 

E=pc2

where the p is momentum not pressure or density. 

 

Another indication of realizing thinking those equations equate to gravity effect is incorrect is that a critically dense universe being flat spacetime has no gravity as there is no curvature term.

Pressure, energy, and mass are all sources of gravity. Therefore, these terms enter the Friedmann equation and play the role of -(4πG/3)p, which is the gravitational effect.

According to Birkhoff’s theorem,

Quote

As a corollary, the acceleration of an expanding shell in our fluid universe is determined solely by the fluid lying within the shell.

The first Friedmann equation without a cosmological constant is derived from general relativity, but there is also a form derived from the conservation of mechanical energy of Newtonian mechanics (dust model), and this is taught in major books. At this time, the terms entered are kinetic energy and gravitational potential energy.

[math]\frac{1}{2}m{v^2}(t) - G\frac{{{M_r}m}}{{r(t)}} = E[/math]

Therefore, the right-hand side term (8πG/3)ρ comes from the total mass M_r inside the shell.

Friedmann's second equation without a cosmological constant is also derived from general relativity, but can also be derived from F=ma=-GMm/R^2. This also uses gravitational force. Because this is intuitive, if we look at this derivation process

[math]ma =  - \frac{{GMm}}{{{R^2}}}[/math]

[math]m\frac{{{d^2}R(t)}}{{d{t^2}}} =  - \frac{{GM(t)m}}{{R{{(t)}^2}}}[/math]

[math]M(t) = \frac{{4\pi {R^3}(t)}}{3}\rho (t) = \frac{{4\pi {a^3}(t){R^3}}}{3}\rho (t)[/math]

[math]m\frac{{{d^2}a(t)R}}{{d{t^2}}} =  - Gm\frac{{\frac{{4\pi }}{3}{a^3}(t){R^3}\rho (t)}}{{{a^2}(t){R^2}}} =  - Gm\frac{{4\pi }}{3}a(t)R\rho (t)[/math]

[math]\frac{{\ddot a(t)}}{{a(t)}} =  - \frac{{4\pi G}}{3}\rho (t)[/math]

Adding pressure, we can create an acceleration equation.

[math]\frac{{\ddot a}}{a} =  - \frac{{4\pi G}}{3}(\rho  + \frac{{3P}}{{{c^2}}})[/math]

 

In other words, we can see that ρ(t) is the mass density inside the expanding shell, and the -(4πG/3)ρ term describes gravity or gravitational effects.

It can be seen that the total mass M inside the shell is included in both Friedmann equations 1 and 2. This means that the equivalent mass of total energy is included.
If matter and radiation have the same energy density ρ0, it is thought that they should be the same when entering the acceleration equation.

In my opinion,

We can put in the equivalent mass of the energy density of the radiation divided by c^2, and let the pressure of the radiation be 0, or since the rest mass of the radiation is 0, let the mass density be 0, and in the pressure term (1/3)ρ must be inserted.

18 hours ago, joigus said:

What do you mean non-relativistic photons? Photons are always relativistic. Am I missing something? Everybody else seems to understand what you mean, but I don't.

Because they're always relativistic, the equation of state (pressure as a function of density) is what it is in the matter term of the Friedmann equation.

Photons are also included as relativistic particles. However, in this explanation, I tried to distinguish between relativistic particles that have rest mass and photons that do not have rest mass.

For photons, which do you think is correct: 1ρ0 or 2ρ0 ?

Edited April 2 by icarus2

[Mordred]

No one is arguing that energy density can used to determine gravitational effects. 

That's an obvious what isn't obvious is amount of gravity one will experience due to that energy/mass will depend on its distribution.

You simply have to study and apply Newtons Shell theorem to understand that.

Let's put it this way in the Einstein field equations the stress energy momentum term it does not matter if the energy density term is from matter or from radiation. Its simply the energy mass density. 

How it is distributed in the tensor is what determines how the Stress tensor tells spacetime how to curve to produce gravitational effects.

Anyways you can easily convert units such as joules for energy into kg for mass. There are plenty of online calculators for that. 

It's trivial to take an equal amount of energy/density of matter to an equal amount energy/mass density of photons to see that if  doesn't matter if energy mass density is produced by photons or matter but simply the quantity and distribution

 

Edited April 2 by Mordred