Referance Frame Calculating Newtonian Friedmann Equation

[Quarlep]

In friedmann equation we start to make a model
kinetic energy-potantial energy=U
kinetic energy depends observer and referance frame so what's the referance frame and observer in this calculation.
Thanks

 

[marcus]

The class of observers which are at CMB rest.
(Older term was observers "comoving with the Hubble flow", or isotropic ones----who see the universe as on average the same in all directions.)

 

[Chalnoth]

In friedmann equation we start to make a model
kinetic energy-potantial energy=U
kinetic energy depends observer and referance frame so what's the referance frame and observer in this calculation.
Thanks

The Friedmann equations don't make use of potential energy. Where did you get this from?

 

[Quarlep]

MmG/R

 

[Quarlep]

Is CMB is the edge of the universe ? We cannot see beyond it.So I will think a universe (infinite or finite). Inside the universe we have a sphere which shpere shell is CMB and the observer in the CMB looks the universe.Then he makes this Newtonian equation.

 

[PeterDonis]

MmG/R

This equation is not valid for the universe as a whole; it's only valid for an isolated gravitating body. As Chalnoth said, there is no potential energy in the Friedmann equations.

Is CMB is the edge of the universe ?

No. The CMB is everywhere in the universe. If you mean the "surface of last scattering", the spacelike surface at which the CMB was emitted, that is in the past; it's not an "edge" of anything.

 

[Chalnoth]

In fact, General Relativity, as it's usually formulated, doesn't include potential energy at all. This is because it's not actually possible to define a total energy in General Relativity that works in all cases.

There are, to be fair, some ways of writing down General Relativity that make use of a total energy, but those only work for special cases.

No. The CMB is everywhere in the universe. If you mean the "surface of last scattering", the spacelike surface at which the CMB was emitted, that is in the past; it's not an "edge" of anything.

To elaborate on this a little bit, the surface of last scattering is a surface in time (this is what PeterDonis means by spacelike). Before that time, the universe was opaque, so we can't see anything that happened before (directly, anyway: we can see the imprint of previous events on the CMB itself). This only represents a distance because the further away we look, the further in the past we look. The amount of distance to the currently-visible portion of the surface of last scattering changes over time.

 

[Quarlep]

This equation is not valid for the universe as a whole; it's only valid for an isolated gravitating body. As Chalnoth said, there is no potential energy in the Friedmann equations.

But it the simple friedmann equation starts with that

1/2mv²-MmG/R=U
v²-8ρπGR²/3=-kc²
H²R²-8ρπGR²/3=-kc²
H²-8ρπG/3=-kc²/R²
Friedmann equation

To elaborate on this a little bit, the surface of last scattering is a surface in time (this is what PeterDonis means by spacelike). Before that time, the universe was opaque, so we can't see anything that happened before (directly, anyway: we can see the imprint of previous events on the CMB itself). This only represents a distance because the further away we look, the further in the past we look. The amount of distance to the currently-visible portion of the surface of last scattering changes over time.

I understand this idea.
I asked about referance frame. Can we change referance frame ? I mean CMB is the only solution ?

 

[Chalnoth]

But it the simple friedmann equation starts with that

1/2mv²-MmG/R=U
v²-8ρπGR²/3=-kc²
H²R²-8ρπGR²/3=-kc²
H²-8ρπG/3=-kc²/R²
Friedmann equation

Where did you get this from? It doesn't conform to any derivation of the Friedmann equations I'm aware of.

Also, you can make equations look a lot better on these forums by using LaTeX equations:
https://www.physicsforums.com/help/latexhelp/

I understand this idea.
I asked about referance frame. Can we change referance frame ? I mean CMB is the only solution ?

It's always possible to change the reference frame. The CMB is just one choice. It's convenient because it makes the equations simple when talking about the behavior of our universe on large scales.

 

[Quarlep]

Where did you get this from? It doesn't conform to any derivation of the Friedmann equations I'm aware of.
.

Start watching 54:40 and in 55:30 you will see my equation

It's always possible to change the reference frame. The CMB is just one choice. It's convenient because it makes the equations simple when talking about the behavior of our universe on large scales.

Thank for this also

 

[Chalnoth]

I see. He's using a Newtonian approximation to derive the first Friedmann equation assuming nothing but matter makes up the universe. It is interesting that the answer in this case is identical to the answer given by General Relativity.

The full derivation using General Relativity doesn't make use of any potential energy in the equation, and it provides the right answer for a universe with matter, radiation, and dark energy. The way he went through this Newtonian approximation can be useful for getting a basic idea of how this works, just bear in mind is only a partial answer, and not at all how these equations were derived originally.

 

[Quarlep]

Ok I will keep in my mind

 

[Quarlep]

Is Newtonian approach enough to make correct assuptions about universe ?

 

[Chalnoth]

Is Newtonian approach enough to make correct assuptions about universe ?

Sort of. The primary problem here is radiation: Newtonian gravity doesn't give the right answer for how radiation interacts with gravity (it underestimates how much radiation responds to gravity by a factor of two).

You can still get the right answer using purely-Newtonian arguments, but the problem is that there are also ways to get the wrong answer.

Straight Newtonian gravity also doesn't include a cosmological constant, but it's not too hard to modify Newtonian gravity to include a cosmological constant: the cosmological constant adds a repulsive acceleration between two masses that is proportional to their distance.

The way to get the right answer using Newtonian arguments, by the way, is to take the first Friedmann equation as derived in that video as correct, and consider $\rho$ to work for energy density as well as mass density (just with a difference of $c^2$ for a unit conversion factor). Then, to get how the energy density changes with time, use the pressure of the fluid under consideration:

Imagine an expanding box with walls, and the pressure of the stuff inside that box puts a force on those walls. As the box expands, the radiation inside the box is pushing in the direction of expansion, which does work. By energy conservation, we know that the work performed on the walls of this hypothetical box must come from the fluid, so the fluid loses energy as a result of this pressure. Working through the equations with radiation gives you that the energy density of radiation scales as $1/a^4$. Working through the equations with a cosmological constant (which has pressure equal to minus its energy density), and you correctly get that the energy density stays the same.

 

[Quarlep]

Later in lectures 3 and 4 he made that too.In Cosmology lecture 4 (2009) 1:10:00 times he wrotes that things and how he derives.Thing is radiation dominant matter dominat and later cosmological constant.I don't know GR so I don't know derive friedmann equatuon use GR.Thats why asked that question.I am working on cosmolgy.If I made $\rho$ to $\rho_m+\rho_r+\rho_d$
m matter ,r radiation,d dark energy.
And keep other things same I get same eq I guess isn't it ?

 

[Chalnoth]

Later in lectures 3 and 4 he made that too.In Cosmology lecture 4 (2009) 1:10:00 times he wrotes that things and how he derives.Thing is radiation dominant matter dominat and later cosmological constant.I don't know GR so I don't know derive friedmann equatuon use GR.Thats why asked that question.I am working on cosmolgy.If I made $\rho$ to $\rho_m+\rho_r+\rho_d$
m matter ,r radiation,d dark energy.
And keep other things same I get same eq I guess isn't it ?

I'm not entirely sure I understand you, but I think so. First, the first Friedmann equation:

$$H^2(t) = {8 \pi G \over 3}\rho(t) - {k \over a^2}$$

We can rewrite this as:

$$H^2(t) = {8 \pi G \over 3}\left(\rho_m(t) + \rho_r(t) + \rho_d(t)\right) - {k \over a^2(t)}$$

Then it's possible to make use of stress-energy conservation (in General Relativity) or energy conservation with the hypothetical expanding box (if using the Newtonian approximation) to see how each parameter changes over time.

$$\rho_m(t) = {\rho_{m0} \over a^3(t)}$$
$$\rho_r(t) = {\rho_{m0} \over a^4(t)}$$
$$\rho_e(t) = \rho_{e0}$$

The last equation assumes that the dark energy is a cosmological constant (the simplest possibility).

 

[Quarlep]

I'm not entirely sure I understand you, but I think so. First, the first Friedmann equation:

$$H^2(t) = {8 \pi G \over 3}\rho(t) - {k \over a^2}$$

We can rewrite this as:

$$H^2(t) = {8 \pi G \over 3}\left(\rho_m(t) + \rho_r(t) + \rho_d(t)\right) - {k \over a^2(t)}$$

Then it's possible to make use of stress-energy conservation (in General Relativity) or energy conservation with the hypothetical expanding box (if using the Newtonian approximation) to see how each parameter changes over time.

$$\rho_m(t) = {\rho_{m0} \over a^3(t)}$$
$$\rho_r(t) = {\rho_{m0} \over a^4(t)}$$
$$\rho_e(t) = \rho_{e0}$$

The last equation assumes that the dark energy is a cosmological constant (the simplest possibility).

I was talking about this idea.If I want a write a Friedmann equation without cosmlogical constant I will just remove $\rho_d$ isn't it.

 

[Chalnoth]

I was talking about this idea.If I want a write a Friedmann equation without cosmlogical constant I will just remove $\rho_d$ isn't it.

Correct.

 

[Quarlep]

Thanks