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energy density of photons and matter

 
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Apr27-10, 02:39 AM   #1
 

energy density of photons and matter


hi

im doing a question at the moment and am having issues understanding the question, and i cant 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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Apr27-10, 03:14 AM   #2
 
Quote by indie452 View Post
hi

im doing a question at the moment and am having issues understanding the question, and i cant 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 [tex] \rho_{m,0} [/tex] means:
The present density of baryonic matter in the universe. This is the measured mass of baryons in our universe per cubic meter.
Apr27-10, 03:20 AM   #3
 
Hint: Where does the matter get its energy from??
Apr27-10, 03:29 AM   #4
 

energy density of photons and matter


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
Apr27-10, 03:30 AM   #5
 
But why? I know the units work and that is what I got as well. Where does energy from mass come form?
Apr27-10, 03:40 AM   #6
 
oh from the E=mc^2 relation of mass to energy.
Apr27-10, 03:43 AM   #7
 
Quote by indie452 View Post
oh from the E=mc^2 relation of mass to energy.
Yep and since energy density [tex] \epsilon = \frac{E}{V} [/tex]


[tex] \epsilon = \frac{mc^2}{\frac{m}{\rho_0}} = \rho_0c^2 [/tex]
Apr27-10, 03:56 AM   #8
 
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
Apr27-10, 04:15 AM   #9
 
ok i just tried

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

i know this doesnt take into account 3 types of neutrinos
do i times 1.8x10-4 by 1.68 to take them into account?
Apr27-10, 04:17 AM   #10
 
Quote by indie452 View Post
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
[tex] \epsilon_m = \epsilon_{m,0}(1+z)^3 [/tex] and

[tex] \epsilon_{rad} = \epsilon_{rad, 0}(1+z)^4 [/tex]
Apr27-10, 04:37 AM   #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 doesnt seem right. surely temp should be larger than now?
Apr27-10, 04:48 AM   #12
 
Quote by indie452 View Post
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 doesnt seem right. surely temp should be larger than now?
No, [tex]

\frac{\epsilon_{rad}}{\epsilon_{m}} = \frac{\epsilon_{rad,0}}{\epsilon_{m,0}}(1+z) [/tex]
Apr27-10, 05:17 AM   #13
 
but surely
erad/em = 1 [cause im 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
Apr27-10, 05:25 AM   #14
 
O right, my bad.
Apr27-10, 05:31 AM   #15
 
thats ok, its just that it doesnt 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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