Friedmann Equation

1. Dec 27, 2008

pi.rootpi

Hey! I'd like to see an easy way to arrive to Friedmann equation by using Newton's Mechanics. I've seen many ways but they skip many steps so I don't understand the whole.

THANKS!

2. Dec 27, 2008

Orion1

Friedmann equations...

As stated by Wikipedia, the Friedmann equations are derived from General Relativity, not from Newtonian mechanics. If you are interested in how these equations were derived, I recommend reading Friedmann's actual papers themselves, they are listed in reference 1 below, at the end of page in Reference as well as most of the important equations.

Reference:
"http://en.wikipedia.org/wiki/Friedmann_equations" [Broken]

Last edited by a moderator: Apr 24, 2017 at 9:44 AM
3. Dec 27, 2008

pi.rootpi

Re: Friedmann equations...

Yes, but for people who doesn't have the level:tongue2: there is another way to get them, you can arrive by using Newtonian Mechanics, and when finally you have the constant then you have to compare with the real one and then you substitute your constant by the one that appears in the original one. That's the way I was asking for.

Thanks!

Last edited by a moderator: Apr 24, 2017 at 9:44 AM
4. Dec 27, 2008

Kurdt

Staff Emeritus
5. Dec 28, 2008

Orion1

Friedmann Equation derived from Newton mechanics...

The Friedmann_equation derived from Newton mechanics.

This model describes a open finite Universe relativistic harmonic oscillator with a positive non-zero cosmological constant:
$$\boxed{\Lambda = \frac{1}{dt^2} = 10^{-35} \; \text{s}^{-2}}$$

Harmonic oscillator force:
$$F = -kx$$

Newtonian cosmological constant force:
$$F_{\Lambda} = m \frac{d^2 r}{dt^2} = \frac{\Lambda m r}{3}$$

Harmonic oscillator potential energy:
$$U = \pm \frac{dF \cdot dL}{2} = - \frac{kx^2}{2} = \frac{\Lambda m r^2}{6}$$

Co-moving coordinates:
$$\boxed{r = a(t) x}$$

Kinetic energy:
$$E_k = \frac{m \dot{r}^2}{2} = \frac{m \dot{a}^2 x^2}{2}$$

$$\boxed{E_k = \frac{m \dot{a}^2 x^2}{2}}$$

cosmological constant potential energy:
$$U_{\Lambda} = \frac{\Lambda m r^2}{6}} = \frac{\Lambda m a^2 x^2}{6}}$$

$$\boxed{U_{\Lambda} = \frac{\Lambda m a^2 x^2}{6}}$$

Particle of mass m and radius r from a uniform expanding medium of density ρ.
The total mass within radius r:
$$m = \frac{4 \pi \rho r^3}{3}$$

Newton's universal law of gravitation:
$$F = - \frac{G m^2}{r^2} = - \frac{4 \pi G \rho m r}{3}$$

Particle gravitational potential energy:
$$V = - dF \cdot dL = - \frac{4 \pi G \rho m r^2}{3} = - \frac{4 \pi G \rho m a^2 x^2}{3}$$

$$\boxed{V = - \frac{4 \pi G \rho m a^2 x^2}{3}}$$

Relativistic harmonic oscillator potential energy:
$$\boxed{U = - mc^2 \left( \frac{kx^2}{2} - \frac{\Lambda a^2 x^2}{6} \right)}$$

Particle energy conservation:
$$U = E_k + V$$

Integration by substitution:
$$- mc^2 \left( \frac{kx^2}{2} - \frac{\Lambda a^2 x^2}{6} \right) = \frac{m \dot{a}^2 x^2}{2} - \frac{4 \pi G \rho m a^2 x^2}{3}$$

Factor out m:
$$- c^2 \left( \frac{kx^2}{2} - \frac{\Lambda a^2 x^2}{6} \right) = \frac{\dot{a}^2 x^2}{2} - \frac{4 \pi G \rho a^2 x^2}{3}$$

Multiply completely through by: $$\frac{2}{a^2 x^2}$$
$$- \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3} = \left( \frac{\dot{a}}{a} \right)^2 - \frac{8 \pi G \rho}{3}$$

Solve for Friedmann equation:
$$\boxed{H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi G \rho}{3} - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}}$$

Reference:
"http://en.wikipedia.org/wiki/Harmonic_oscillator#Simple_harmonic_oscillator" [Broken]
"http://zebu.uoregon.edu/lambda2.html" [Broken]
"http://en.wikipedia.org/wiki/Cosmological_constant" [Broken]
"http://en.wikipedia.org/wiki/Scale_factor_(universe)" [Broken]
"http://en.wikipedia.org/wiki/Friedmann_equations" [Broken]
"http://en.wikipedia.org/wiki/Hubble_parameter#Derivation_of_the_Hubble_parameter" [Broken]

Last edited by a moderator: Apr 24, 2017 at 9:45 AM
6. Jan 1, 2009

Orion1

Compton wavelength and general relativity...

Does the Compton wavelength obey general relativity and special relativity?

$$\boxed{r_u = \overline{\lambda_u}}$$

$$m_u$$ - Universe mass

Universe Compton wavelength:
$$r_u = \overline{\lambda_u} = \frac{\hbar}{m_u c}$$
$$\boxed{r_u = \frac{\hbar}{m_u c}}$$

Universe quantum age:
$$t_u = \frac{r_u}{c} = \frac{\hbar}{m_u c^2}$$

$$\boxed{t_u = \frac{\hbar}{m_u c^2}}$$

Universe quantum sphere average density:
$$\rho_u(t_u) = \frac{m_u}{V_u} = \frac{3 m_u}{4 \pi r_u^3} = \frac{3 m_u}{4 \pi} \left( \frac{m_u c}{\hbar} \right)^3 = \frac{3 c^3 m_u^4}{4 \pi \hbar^3}$$

$$\boxed{\rho_u(t_u) = \frac{3 c^3 m_u^4}{4 \pi \hbar^3}}$$

Friedmann equation:
$$H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi G \rho_u(t_u)}{3} - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}$$

Integration by substitution:
$$H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi G}{3} \left( \frac{3 c^3 m_u^4}{4 \pi \hbar^3}\right) - \frac{kc^2}{a(t_u)^2} + \frac{\Lambda(t_u) c^2}{3} = \frac{2 G c^3 m_u^4}{\hbar^3} - \frac{kc^2}{a(t_u)^2} + \frac{\Lambda(t_u) c^2}{3}$$

Quantum Compton-Friedmann equation:
$$\boxed{H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{2 G c^3 m_u^4}{\hbar^3} - \frac{kc^2}{a(t_u)^2} + \frac{\Lambda(t_u) c^2}{3}}$$

Reference:
"http://en.wikipedia.org/wiki/Compton_wavelength" [Broken]
"http://en.wikipedia.org/wiki/Friedmann_equations" [Broken]
"http://en.wikipedia.org/wiki/General_relativity" [Broken]
"http://en.wikipedia.org/wiki/Special_relativity" [Broken]
"http://en.wikipedia.org/wiki/Friedmann-Lema%C3%AEtre-Robertson-Walker_metric#Solutions" [Broken]

Last edited by a moderator: Apr 24, 2017 at 9:54 AM
7. Jan 4, 2009

Orion1

Friedmann scale factor...

$$\boxed{r_u(t_0) = 4.4 \cdot 10^{26} \; \text{m}}$$

Friedmann scale factor:
$$\boxed{a(t) = \frac{r_u(t)}{r_u(t_0)}}$$

$$a(t_u) = \frac{r_u(t_u)}{r_u(t_0)} = \frac{\hbar}{m_u c r_u(t_0)}$$

$$\boxed{a(t_u) = \frac{\hbar}{m_u c r_u(t_0)}}$$

Quantum Compton-Friedmann equation:
$$H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{2 G c^3 m_u^4}{\hbar^3} - \frac{kc^2}{a(t_u)^2} + \frac{\Lambda(t_u) c^2}{3}$$

Integration by substitution:
$$H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{2 G c^3 m_u^4}{\hbar^3} - kc^2 \left( \frac{m_u c r_u(t_0)}{\hbar} \right)^2 + \frac{\Lambda(t_u) c^2}{3} = \frac{2 G c^3 m_u^4}{\hbar^3} - k \left( \frac{m_u c^2 r_u(t_0)}{\hbar} \right)^2 + \frac{\Lambda c^2}{3}$$

Quantum Compton-Friedmann equation:
$$\boxed{H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{2 G c^3 m_u^4}{\hbar^3} - k \left( \frac{m_u c^2 r_u(t_0)}{\hbar} \right)^2 + \frac{\Lambda c^2}{3}}$$

$$\Lambda = \frac{1}{dt^2} = 10^{-35} \; \text{s}^{-2}$$

Reference:
"http://en.wikipedia.org/wiki/Scale_factor_(universe)" [Broken]
"http://en.wikipedia.org/wiki/Universe" [Broken]

Last edited by a moderator: Apr 24, 2017 at 9:58 AM
8. Jan 5, 2009

mysearch

Orion1: Thought your presentation of the equations in posts #5, #6 & #7 was excellent, but wanted to try to better understand the rationale behind some of the steps by way of clarification.

In post #5:
You start with a definition of Lambda = 1/s^2, but in the final Friedmann equation, the units have switched to 1/m^2 by virtue of including c^2. I know a lot of people like using natural units, but it can sometimes be misleading when trying to correlate units in the derivation.

The 3rd equation was interesting and I followed the link that explained its premise. If you were to model a classical circular orbit, which accounted for the Lambda force, would the following form be correct?

[1] $$\frac {mv^2}{r} + \frac{\Lambda c^2 m r}{3} = \frac{GMm}{r^2}$$

I am assuming that in an expanding universe, Lambda would act in opposition to the force of gravity. I raise this point because presumably it is important when we switch from force to energy in the later equations.

As another point of clarification, would equating Lambda to dark energy imply that the effective mass (M), related to the energy density contained in the volume defined by [r], would increase with the expansion of the universe, if the density energy remains constant under expansion?

Finally, I am also trying to understand how the concept of a SHM oscillator is being introduced. Without damping, an oscillator would just maintain the conservation of energy between kinetic and potential energy, but I shall also speculatively introduce the Lambda energy terms such that we might write:

[2] $$dE_T = (dE_K + dE_\Lambda) + (-dE_P) = 0$$

I realise this is a dubious conservation of energy assumption in an expanding universe and I am using the differential form simply because the equation ignores the energy put into the system to start it oscillating in the first place. However, if I do a very quick and dirty derivation based on the premise of equation [2], but bearing in mind the energy associated with Lambda:

[3] $$E_K = E_P - E_\Lambda$$

In this form, the kinetic energy only relates to radial velocity and I am going to ignore that complexity of comoving coordinates by assuming [k=0] from the outset. I now substitute for Newtonian kinetic and gravitational energy plus your Lambda energy, but highlight the difference in sign:

[4] $$1/2 mv^2 = \frac{GMm}{r} - \frac{\Lambda c^2 m r^2}{6}$$

I can divide through by (m) and substitute for (M=density*volume):

[5] $$\frac{v^2}{2} = \frac{4 \pi G \rho r^2}{3} - \frac{\Lambda c^2 r^2}{6}$$

[6] $$H^2 = \frac{v^2}{r^2} = \frac{8 \pi G \rho}{3} - \frac{\Lambda c^2 }{3}$$

I know the sign is wrong, but wanted to discuss this point as you have clearly been looking at these equations in some detail and may be able to put me straight on a couple of issues that have been puzzling me. If I initially ignore Lamba, equation [6] would reduce to the form

[7] $$H^2 = \frac{v^2}{r^2} = \frac{8 \pi G \rho}{3}$$

Given that this equation seems to only relate the velocity [v] of our unit mass (m) to a gravitational term, why would [v] be described as an expansive velocity as normally associated with the Hubble constant (H)?

If we re-introduce Lamba as an expansive energy (negative pressure); wouldn’t it act in opposition to the gravitational potential in an expanding universe?

I will defer comment on #6 and #7 as this post is already a bit long, but would appreciate any clarification you might be able to offer on the points raised. Thanks.

9. Jan 5, 2009

Orion1

Classical Newtonian harmonic oscillator...

Affirmative, the units have changed in the relativistic harmonic oscillator potential energy well.

Non-relativistic classical Newtonian harmonic oscillator potential energy:
$$U = U_k + U_{\Lambda} = - \frac{kx^2}{2} + \frac{\Lambda m r^2}{6}} = - \frac{kx^2}{2} + \frac{\Lambda m a^2 x^2}{6} = - \left( \frac{kx^2}{2} - \frac{\Lambda m a^2 x^2}{6} \right)$$

$$\boxed{U = - \left( \frac{kx^2}{2} - \frac{\Lambda m a^2 x^2}{6} \right)} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}$$

Relativistic harmonic oscillator potential energy:
$$\boxed{U = - mc^2 \left( \frac{kx^2}{2} - \frac{\Lambda a^2 x^2}{6} \right)} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dL^2}}$$

When the harmonic oscillator switches from a classical newtonian harmonic oscillator to a relativistic harmonic oscillator, the Systeme International units change for the cosmological constant.

Newton-Einstein relative unit conversion:
$$\boxed{\frac{\Lambda_N}{\Lambda_E} = \frac{dL^2}{dt^2} = c^2} \; \; \; \; \; \; \boxed{\frac{k_N}{k_E} = dF \cdot dL = \frac{mc^2}{a}} \; \; \; \; \; \; \boxed{\frac{\rho_N}{\rho_E} = 2}$$

Negative.

Newton's universal law of gravitation:
$$F_g = m \frac{d^2 r}{dt^2} = - \frac{G m^2}{r^2}$$

Newtonian centripetal force:
$$F_c = m \frac{d^2 r}{dt^2} = \frac{mv^2}{r}$$

Harmonic oscillator force:
$$F_k = m \frac{d^2 r}{dt^2} = - kx$$

Newtonian cosmological constant force:
$$F_{\Lambda} = m \frac{d^2 r}{dt^2} = \frac{\Lambda m r}{3} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}$$

Newtonian force theorem:
$$\boxed{F_g + F_c = F_k + F_{\Lambda}}$$

$$\boxed{F_g = -F_c + F_k + F_{\Lambda}}$$

$$F = m \frac{d^2 r}{dt^2} = - \frac{Gm^2}{r^2} = - \frac{mv^2}{r} - kx + \frac{\Lambda m r}{3}$$

$$\boxed{\frac{Gm^2}{r^2} = \frac{mv^2}{r} + kx - \frac{\Lambda m r}{3}}$$

Multiply completely through by $$\frac{r}{2}$$:
$$\frac{Gm^2}{2r} = \frac{mv^2}{2} + \frac{k r x}{2} - \frac{\Lambda m r^2}{6}$$

Kinetic energy:
$$\boxed{E_k = \frac{mv^2}{2} = \frac{Gm^2}{2r} - \frac{k r x}{2} + \frac{\Lambda m r^2}{6}}$$

Multiply completely through by $$\frac{2}{m}$$:

$$v^2 = \frac{G}{r} \left( \frac{4 \pi \rho r^3}{3} \right) - \frac{k r x}{m} + \frac{\Lambda r^2}{3} = \frac{4 \pi G \rho r^2}{3} - \frac{k r x}{m} + \frac{\Lambda r^2}{3}$$

$$\boxed{v^2 = \frac{4 \pi G \rho r^2}{3} - \frac{k r x}{m} + \frac{\Lambda r^2}{3}} \; \; \; \; \; \; \boxed{a = \frac{r}{x}}$$ - co-moving coordinates

Non-relativistic Newtonian Hubble parameter equation:
$$\boxed{H^2 = \frac{v^2}{r^2} = \frac{4 \pi G \rho}{3} - \frac{k}{ma} + \frac{\Lambda}{3}} \; \; \; \; \; \; \boxed{\Lambda = \frac{1}{dt^2}}$$

Last edited: Jan 6, 2009
10. Jan 9, 2009

Orion1

Friedmann Equation...

This 'Hubble-like' parameter is only a factor in the Friedmann equations relative only to the cosmological constant:

$$\boxed{\Lambda = 1.112 \cdot 10^{-52} \; \text{m}^{-2}}$$

Friedmann equation:
$$H^2 = \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi G \rho}{3} - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}$$

The Hubble parameter:
$$\boxed{H_0 = \sqrt{\frac{8 \pi G \rho}{3} - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}}}$$

$$\boxed{H_0 = 2.26 \cdot 10^{-}^{18} \; \text{s}^{-}^{1}}$$

$$H_{\Lambda}^2 = \frac{\Lambda c^2}{3}$$

$$H_{\Lambda} = c \sqrt{\frac{\Lambda}{3}}$$

$$\boxed{H_{\Lambda} = 1.825 \cdot 10^{-18} \; \text{s}^{-1}}$$

Reference:
"https://www.physicsforums.com/showpost.php?p=2015381&postcount=5" [Broken]
"http://en.wikipedia.org/wiki/Friedmann_equations" [Broken]

Last edited by a moderator: Apr 24, 2017 at 10:08 AM