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Vacuum energy density

  1. Feb 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that, in natural units [itex]h=c=1[/itex], an energy density may be expressed as the fourth power of a mass. If the vacuum energy contributed by a cosmological constant is now of order of the critical density, what is the mass to which this density corresponds?


    2. The attempt at a solution
    For the first part I think that

    [tex]\rho_E \propto \frac{m_Pc^2}{l_P^3} = \frac{m_P^4c^5}{h^3}[/tex]

    where the index P is for the Planck units.
    Then I'm stuck. I'm not sure about how to relate the vacuum energy, the cosmological constant and the critical density. Anyone?
     
  2. jcsd
  3. Feb 6, 2007 #2

    Dick

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    Write the problem this way:

    [tex]\rho_C = \frac{m_C^4c^5}{h^3}[/tex]

    where the index C is for critical. You know a number for the critical density. Put it in and find m_C.
     
  4. Feb 6, 2007 #3
    I know a number for the critical density?
     
  5. Feb 6, 2007 #4

    Dick

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    You used to. Isn't it pretty closely related to H^2, H=70 km/sec/mpc? I won't put the error bars in there.
     
  6. Feb 6, 2007 #5
    Is it the same critical density for the Planck scale then?
    So the energy density and the matter density is the same?
     
  7. Feb 6, 2007 #6

    Dick

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    I think the current universe is pretty far from the planck scale. Your job to figure out how far. The problem says: vacuum energy is of the same order as the critical density NOW.
     
  8. Feb 6, 2007 #7

    Dick

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    BTW are you sure the powers of h and c are correct in the expressions? I'm having unit problems...
     
  9. Feb 6, 2007 #8
    The problem before this one is to express energy, length and time in terms of the Planck mass. I then got

    [tex]E_P=m_Pc^2[/tex]
    [tex]l_P=\frac{h}{m_Pc}[/tex]
    [tex]t_P=\frac{h}{m_Pc^2}[/tex]

    and to express energy density in terms of the Planck mass i used energy per unit volume

    [tex]\frac{E_P}{V} \propto \frac{E_P}{l_P^3} = \frac{m_P^4c^5}{h^3}[/tex]

    I'm not sure if this is correct.
     
  10. Feb 6, 2007 #9

    Dick

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    It's correct. But we want the critical density as a matter density rather than an energy density to do this problem. Try m_p/l_p^3. Otherwise there's an extra c^2 around.
     
  11. Feb 6, 2007 #10

    Dick

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    Or write the Freidmann eqn with rho_c/c^2. Your choice.
     
  12. Feb 6, 2007 #11
    Then I get

    [tex]\rho_c \propto \frac{m_P^4c^3}{h^3}[/tex]

    and by using

    [tex]\rho_c = \frac{3H_0^2}{8 \pi G}[/tex]

    I get

    [tex]m_P^4 \approx 1.379 \cdot 10^{-153}[/tex]

    which cannot be correct. Maybe my powers are wrong somewhere but I don't think so..
     
  13. Feb 6, 2007 #12

    Dick

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    You are not computing the planck mass here. You already know what that is. You are computing the mass scale of the 'vacuum energy' aka 'dark energy'. I get something more like m=10^(-27) kg. rho=m/meter^3. Might want to check it again.
     
    Last edited: Feb 6, 2007
  14. Feb 6, 2007 #13
    Ah ok, but my expression is ok then?
     
  15. Feb 6, 2007 #14

    Dick

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    Change this to [tex]m_c^4 \approx 1.379 \cdot 10^{-153} kg^4[/tex] and I'll say I pretty much believe you.
     
  16. Feb 6, 2007 #15
    Damn it! The Hubble constant is

    [tex]65 \cdot km \cdot s^{-1} \cdot Mpc^{-1} = 65 \cdot 10^3 \cdot m \cdot s^{-1} \cdot Mpc^{-1} = 65 \cdot 10^3 \cdot 10^{-6} \cdot m \cdot s^{-1} \cdot pc^{-1} =[/tex]
    [tex]= 65 \cdot (3.0857)^{-1} \cdot 10^3 \cdot 10^{-6} \cdot 10^{-16} \cdot m \cdot s^{-1} \cdot m^{-1} = 65 \cdot (3.0857)^{-1} \cdot 10^3 \cdot 10^{-6} \cdot 10^{-16} \cdot s^{-1}[/tex]

    right?

    Then I get, using [itex]\bar{h} = h/2\pi[/itex],

    [tex]m_P^4 = \frac{3H_0^2 \bar{h}}{8 \pi Gc^3} \approx \frac{3 \cdot (65 \cdot 10^3 \cdot 10^{-6} \cdot 10^{-16})^2 \cdot (6,626 \cdot 10^{-34})^3}{64 \pi^4 \cdot (3.0857)^2 \cdot 6.6726 \cdot 10^{-11} \cdot 27 \cdot 10^{24}} =[/tex]

    [tex]= \frac{3 \cdot 290.907 \cdot 4225 \cdot 10^{-102} \cdot 10^6 \cdot 10^{-12} \cdot 10^{-32}}{64 \pi^4 \cdot 9.522 \cdot 6.6726 \cdot 27 \cdot 10^{-11} \cdot 10^{24}} =[/tex]

    [tex]= 0.0345 \cdot 10^{-152}[/tex]

    [tex]\Rightarrow m_P \approx 0.431 \cdot 10^{-38}[/tex]

    I'll look for something wrong with this.
     
    Last edited: Feb 6, 2007
  17. Feb 6, 2007 #16

    Dick

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    You are STILL in the right ballpark providing the units are kg. But stop calling this the planck mass! And start putting units on your masses! And the statement of the problem defined h=1, not hbar=1.
     
  18. Feb 6, 2007 #17
    Ok, the statement of the problem should be h-bar, I was just lazy. ;)
     
  19. Feb 6, 2007 #18

    Dick

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    Well. That explains why I wasn't getting exactly the same numbers as you...
     
  20. Feb 6, 2007 #19
    Hehe sorry, and thanks again for your help.
     
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