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Maximum entropy implies linear expansion? 
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#1
Mar612, 03:03 PM

P: 373

Imagine we are at the center of a sphere of radius R containing a mass M.
The Beckenstein bound states that the entropy inside that sphere, S, must be given by the inequality: [itex] \large S \leq \frac{2 \pi k c R M}{\hbar}. [/itex] In order to maximise the entropy we need to fill the sphere of radius R with as much mass M as possible. The limit is reached when we have created a black hole. This occurs when the following relationship holds: [itex] \large \frac{G M}{R} = \frac{c^2}{2}. \ \ \ \ \ \ \ \ \ \ (1)[/itex] Now in the case of a black hole the event horizon is at a constant radius R and we are sitting on the singularity at the center. But instead let us assume that the sphere is expanding. Let us also assume that Equation (1) always holds so that the expanding sphere always has a maximum entropy. This of course means that the mass M must increase with radius R. Thus matter is being continuously created. If we assume flat space then Equation (1) implies that the mass density, [itex]\rho[/itex], is given by [itex] \large \rho = \frac{3 c^2}{8 \pi G R^2}. \ \ \ \ \ \ \ \ (2) [/itex] If we assume that the radius [itex]R(t)[/itex] is expanding with the Universal scale factor [itex]a(t)[/itex] then we can say: [itex] R(t) = R_0 a(t) \ \ \ \ \ \ \ \ (3)[/itex] where t is the cosmological time, [itex]R_0[/itex] is the radius at the present time [itex]t_0[/itex] and [itex] a(t_0) = 1 [/itex]. Now let us consider the Friedmann equation for flat space with no cosmological constant: [itex] \large (\frac{\dot{a}}{a})^2 = \frac{8 \pi G}{3} \rho \ \ \ \ \ (4) [/itex] Substituting Equations (2) and (3) into Equation (4) we obtain [itex] \large (\frac{\dot{a}}{a})^2 = \frac{c^2}{R_0^2 a^2}. [/itex] As the Hubble parameter at the present time, [itex]H_0[/itex], is given by [itex] H_0 = c / R_0,[/itex] we finally arrive at [itex] \large (\frac{\dot{a}}{a})^2 = \frac{H_0^2}{a^2}, [/itex] which has the simple linear solution [itex] a = H_0 t. [/itex] This solution to the Friedmann equation is thus the maximum entropy solution. It is very close to what is observed. 


#2
Mar612, 04:28 PM

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P: 11,878




#3
Mar612, 04:47 PM

P: 95

If it's at the edge of our Observable Universe, it may mean it's not created but simply enters our Observable Universe as its edge is moving further away.
Just a thought, I might be wrong. 


#4
Mar612, 05:03 PM

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P: 11,878

Maximum entropy implies linear expansion?



#5
Mar712, 02:59 AM

P: 373

[itex] \large S = \frac{k A}{4} [/itex] where A is the surface area of the sphere in units of Planck area [itex]\hbar G / c^3[/itex]. Thus, following the Holographic principle, we can think of all the entropy of the sphere as residing on its surface. Now we also have the following thermodynamic relationship between the change in the mass/energy of the system and the change in its entropy [itex] \large dM = \frac{T}{c^2} dS, [/itex] where T is the Unruh temperature at the surface of the sphere due to the gravity of the mass inside it. This shows that the mass change of the expanding sphere occurs due to the increase of the entropy in its expanding surface area. 


#6
Mar712, 06:51 AM

P: 939

The horizon radiates Hawking radiation, but I guess the universe is way too big to be kept in balance by this  it might be a fun exercise to check if there exists some radius where this is true. If I remember correctly, the radiation power depends on M like ~M^2, so there might be a "sweet spot".



#7
Mar712, 07:49 AM

P: 373

[itex] R = c t. [/itex] I presume that the expanding sphere will not radiate Hawking radiation outside itself as the radiation could not "outrun" the receeding horizon. 


#8
Mar812, 09:09 AM

P: 939




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