LOG#227. Cosmic energy.

Short post number two! Surprise!

Have you ever wondered what is the cosmic energy of the Universe? Well, giving up certain General Relativity issues related to the notion of energy in local sense, there is indeed a global notion of energy for the Universe as a whole. I am not considering the Multiverse as an option today. Let me begin with the High School notion of mass and density, particularized for the Universe:

    \[M_U=\rho_UV_U\]

We are considering a closed spherical Universe with 3-d geometry, and then its volume reads

    \[V_U=\dfrac{4}{3}\pi R_U^3\]

What is the radius of the Universe? Well, we could take it as the Hubble radius of the observable Universe, i.e.,

    \[R_U=\dfrac{c}{H}\]

where H\approx 70km/s/Mpc. The density of the Universe can be written as the cosmological value of the vacuum/Hubble scale

    \[\rho_U=\dfrac{\Lambda c^4}{8\pi G}=\dfrac{3c^2H^2}{8\pi G}\]

so \Lambda c^4=3c^2H^2. Therefore, the formula for the mass of the Universe in terms of fundamental constants is

    \[\boxed{M_U=\dfrac{c^3}{2GH}=\dfrac{c^2}{2G}\sqrt{\dfrac{3}{\Lambda}}}\]

and the expression for the cosmic energy follows up from special relativity greatest formula E_U=M_Uc^2 as

    \[\boxed{E_U=\dfrac{c^5}{2GH}=\dfrac{c^4}{2G}\sqrt{\dfrac{3}{\Lambda}}}\]

Also, defining L_\Lambda=\sqrt{3/\Lambda} and L_P ^2=G\hbar/c^3, you get

    \[\boxed{E_U=\dfrac{\hbar}{2}\left(\dfrac{cL_\Lambda}{L_P^2}\right)}\]

Remark: the cosmological constant fixes not only the biggest mass as the Universal mass of the Universe (I am sorry for pedantic expression), but also fixes the smallest possible mass (the so-called Garidi mass in de Sitter group or de Sitter relativity):

    \[M_\Lambda=\dfrac{\hbar}{c}\sqrt{\dfrac{\Lambda}{3}}\]

And you can thus prove that

    \[\dfrac{M_U}{M_\Lambda}=\dfrac{1}{2}\dfrac{c^3}{G_N\hbar}\dfrac{3}{\Lambda}\sim \dfrac{L_\Lambda^2}{L_P^2}\sim 10^{122}\]

Note, that

    \[M_P=\dfrac{\hbar}{c}\dfrac{1}{L_P}\]

    \[M_\Lambda=\dfrac{\hbar}{c}\dfrac{1}{L_\Lambda}\]

    \[M_U=\dfrac{\hbar}{c}\dfrac{L_\Lambda}{L_P^2}\]

    \[V_P\sim L_P^3\]

    \[V_\Lambda=V_U\sim L_\Lambda^3 \]

    \[\dfrac{M_P}{M_\Lambda}\sim\dfrac{L_\Lambda}{L_P}\]

    \[\dfrac{M_U}{M_P}\sim\dfrac{ L_\Lambda}{L_P}\]

    \[\dfrac{M_U}{M_\Lambda}\sim\dfrac{ L_\Lambda^2}{L_P^2}\]

and thus

    \[\dfrac{\rho_P}{\rho_\Lambda}=\dfrac{L_\Lambda^4}{L_P^4}\]

    \[\dfrac{\rho_P}{\rho_U}=\dfrac{L_\Lambda^2}{L_P^2}\]

    \[\dfrac{\rho_U}{\rho_\Lambda}=\dfrac{L_\Lambda^2}{L_P^2}\]

See you in other blog post!!!!!

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