Energy density of photons and matter

AI Thread Summary
The discussion revolves around calculating the energy density ratio of Cosmic Microwave Background Radiation (CMBR) photons to baryonic matter. The current baryonic matter density is given as 2.56x10^-27 kg/m³, and the CMBR temperature is 2.725 K, leading to an energy density calculation of 0.417x10^-13 J/m³ for CMBR. To convert baryonic matter density into energy density, the mass density is multiplied by c², resulting in a value of 1.44 GeV/m³. The participants also explore the redshift at which energy densities of matter and radiation equalize, concluding that the redshift calculation yields approximately 5526, although there are concerns about the accuracy of temperature derived from this value.
indie452
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hi

im doing a question at the moment and am having issues understanding the question, and i can't ask my lecturer as he is stuck abroad with no internet.

the question asks to calculate the ratio of current energy density of CMBR photons to that of baryonic matter.
the present density of baryonic matter in the universe is pm,0=2.56x10-27kg/m3
CMBR Temp = 2.725K

ok so i calculated energy density of the CMBR = 4/c * [sigma]T4 = 0.417x10-13J/m3
sigma=stefans constant
i'm not quite sure how to turn the pm,0 of baryonic matter into an energy density. but i can see it needs to be multiplied by dimensions (m/s)^2

thanks for any hints

also what does pm,0 mean?
 
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indie452 said:
hi

im doing a question at the moment and am having issues understanding the question, and i can't ask my lecturer as he is stuck abroad with no internet.

the question asks to calculate the ratio of current energy density of CMBR photons to that of baryonic matter.
the present density of baryonic matter in the universe is pm,0=2.56x10-27kg/m3
CMBR Temp = 2.725K

ok so i calculated energy density of the CMBR = 4/c * [sigma]T4 = 0.417x10-13J/m3
sigma=stefans constant
i'm not quite sure how to turn the pm,0 of baryonic matter into an energy density. but i can see it needs to be multiplied by dimensions (m/s)^2

thanks for any hints

also what does pm,0 mean?

The problem states what \rho_{m,0} means:
The present density of baryonic matter in the universe. This is the measured mass of baryons in our universe per cubic meter.
 
Hint: Where does the matter get its energy from??
 
i am thinking that i times the baryonic mass density by c^2
this would give 1.44Gev/m^3
so the ratio would be 0.261Mev/m3 / 1.44Gev/m^3 = 1.8125x10-4
 
But why? I know the units work and that is what I got as well. Where does energy from mass come form?
 
oh from the E=mc^2 relation of mass to energy.
 
indie452 said:
oh from the E=mc^2 relation of mass to energy.

Yep and since energy density \epsilon = \frac{E}{V}\epsilon = \frac{mc^2}{\frac{m}{\rho_0}} = \rho_0c^2
 
ok thanks for that the reasoning behind that makes more sense now

btw the next part says to calculate th redshift at which the energy density of matter = that of radiation (i.e. the CMBR photons)

i tried:

e = energy density

e[rad] / e[matter] = 1 when equal and this is also proportional to a-4/a-3 where a is cosmological scale factor. and i am told a ~ (z+1)

this would mean however that 1= 1/a = 1/(z+1) and so z=0
i know this is wrong

i think it should be ~3600 this value i found many times in my reading but didnt understand how it was found
 
ok i just tried

z+1 ~ (1.8x10-4)-1
so z is ~5526

i know this doesn't take into account 3 types of neutrinos
do i times 1.8x10-4 by 1.68 to take them into account?
 
  • #10
indie452 said:
ok thanks for that the reasoning behind that makes more sense now

btw the next part says to calculate th redshift at which the energy density of matter = that of radiation (i.e. the CMBR photons)

i tried:

e = energy density

e[rad] / e[matter] = 1 when equal and this is also proportional to a-4/a-3 where a is cosmological scale factor. and i am told a ~ (z+1)

this would mean however that 1= 1/a = 1/(z+1) and so z=0
i know this is wrong

i think it should be ~3600 this value i found many times in my reading but didnt understand how it was found

\epsilon_m = \epsilon_{m,0}(1+z)^3 and

\epsilon_{rad} = \epsilon_{rad, 0}(1+z)^4
 
  • #11
ok so em0/erad0 = 1+z

z= (1.8x10-4)^-1 - 1 = 5526.0066027

this gives a temp of the CMBR => T=1/(z+1) = 1.809x10-4 K
this doesn't seem right. surely temp should be larger than now?
 
  • #12
indie452 said:
ok so em0/erad0 = 1+z

z= (1.8x10-4)^-1 - 1 = 5526.0066027

this gives a temp of the CMBR => T=1/(z+1) = 1.809x10-4 K
this doesn't seem right. surely temp should be larger than now?

No, <br /> <br /> \frac{\epsilon_{rad}}{\epsilon_{m}} = \frac{\epsilon_{rad,0}}{\epsilon_{m,0}}(1+z)
 
  • #13
but surely
erad/em = 1 [cause I am looking for equality]
and so
erad/em = 1 = erad0(z+1)/em0
so
erad0/em0 = 1/(z+1)
therefore
z+1 = em0/erad0
which is what i put
 
  • #14
O right, my bad.
 
  • #15
thats ok, its just that it doesn't seem right.
is it correct to get the temp by saying T~1/(z+1)?
cause this just gives values i would have thought as being too small. or is this the overall temp of radiation and matter? and i only want the CMBR temp
 
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