Astrophysics - How much CMB-photons do you absorb per second

In summary, in order to calculate the number of Cosmic Microwave Background-photons absorbed per second in space, the formula n = beta * T^3 can be used, where beta = 2.03*10^7 m^{-3} * K^{-3}. This formula gives the number of photons per cubic meter, which is also the amount that would pass through a spherical body per second. To convert this to photons/s, one must multiply by the surface area of the body. The energy density of photons can be obtained by integrating E = alfa*T^4, where alfa = 7.56*10^-16 J * m^-3 * K^-4. To obtain the number density of photons, one must
  • #1
Niles
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Homework Statement


1) Calculate approx. how many Cosmic Microwave Background-photons that you would absorb per second if you were in space.

The Attempt at a Solution


Ok, I know that the CMB-photons fit with a blackbody spectrum with T = 2.725 K. So the total amount of photons I find by using

n = beta * T^3, where beta = 2.03*10^7 m^{-3} * K^{-3}.

This is the amount of photons per cubicmeter, which is also the amount that would pass through my body (assumed spherical) per second.

Am I correct?
 
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  • #2
No suggestions?
 
  • #3
Are you not given any more details?

Edit:

I misread,

If you use the Stefan-Boltzmann law to find the corresponding energy flux density, you can then find the energy/s incident on the surface. It's then simple to convert this energy to photons/s.
 
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  • #4
The Stefan-Boltzmann law

\epsilon_{gamma} = alfa * T^4.

The temperature I use is T = 2.750 K. Then I get a number with units J/m^3. How do I get from this to a time-unit involving s?
 
  • #5
It'll give you a number with units: [tex]{Js^{-1}m^{-2}}[/tex]

So you'll need to multiply by the surface area of body absorbing the energy to give you the total amount of energy absorbed per second. Dividing by the energy of the photon will then give you the number of photons/s absorbed.
 
  • #6
Hmm, in my book the constant alfa = 7.56*10^-16 J * m^-3 * K^-4?
 
  • #8
Niles said:

The Attempt at a Solution


Ok, I know that the CMB-photons fit with a blackbody spectrum with T = 2.725 K. So the total amount of photons I find by using

n = beta * T^3, where beta = 2.03*10^7 m^{-3} * K^{-3}.

This is the amount of photons per cubicmeter, which is also the amount that would pass through my body (assumed spherical) per second.

Am I correct?

Where did you get this formula? Show it with h, c etc. so that we can understand. (At least, I don't have this ready made value.)

Why do you have to be spherical? Are you very fat?

EDIT: After I submitted, I see that astrorob has given a link. But you need the number density, not the energy density. You have to divide E(ν)dν by hν before integrating.
 
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  • #9
Ehh, now I'm confused.

In my book I have the energy density of photons, which when integrated gives E = alfa*T^4. But this is not correct, since we want the number density of the photons. So I divide with the energy of one photon, and then I integrate?

EDIT: The equation in my book is the one on the bottom of this page: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.html#c1

And the thing about me possibly being fat (which I'm actually not.. I'm quite skinny) - I guess that doesn't matter. All I need is my area, right?
 
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  • #10
Niles said:
In my book I have the energy density of photons, which when integrated gives E = alfa*T^4. But this is not correct, since we want the number density of the photons. So I divide with the energy of one photon, and then I integrate?

That sounds reasonable, doesn't it? In fact, the energy density per unit frequency width was obtained by multiplying the number density per unit frequency width by [itex]h\nu[/itex].

And the thing about me possibly being fat (which I'm actually not.. I'm quite skinny) - I guess that doesn't matter. All I need is my area, right?

Yes, then you'll have to find the number of photons incident on your body per m² per sec.
 
  • #11
Ok, I found an expression for the number density of photons in blackbody radiation:

n = beta * T^3, where

beta = 2.03 *10^7 m^-3 * K^-3.

But again, I don't get seconds from anywhere?
 
  • #12
Sorry for the delay in replying. Since you have solved it in the mean time, I hope you got the value as 3*10^16/m²/s.
 
  • #13
I did.

Thanks for helping! :-)
 

1. How is the number of CMB-photons absorbed per second calculated?

The number of CMB-photons absorbed per second is calculated by multiplying the area of the object by the energy density of the CMB and the speed of light. This gives the total number of CMB-photons that pass through the object each second, which is then multiplied by the absorption coefficient of the material to determine the number of photons absorbed.

2. What is the significance of CMB-photons in astrophysics?

CMB-photons, or cosmic microwave background photons, are significant in astrophysics because they provide a window into the early universe. These photons were emitted shortly after the Big Bang and have been traveling through space for billions of years, carrying information about the conditions of the early universe. Studying CMB-photons can provide insights into the structure and evolution of the universe.

3. How does the amount of CMB-photons absorbed per second vary between different objects?

The amount of CMB-photons absorbed per second varies between different objects depending on their size, temperature, and composition. Objects with a larger surface area and higher absorption coefficient will absorb more CMB-photons, while cooler objects will emit more CMB-photons than they absorb.

4. Can CMB-photons be used to study objects in space?

Yes, CMB-photons can be used to study objects in space through a process called imaging. By measuring the intensity and direction of CMB-photons, scientists can create images of celestial objects such as galaxies, stars, and planets. This allows for the study of distant objects that may be difficult to observe using other methods.

5. How does the absorption of CMB-photons affect the temperature of an object?

The absorption of CMB-photons can affect the temperature of an object by transferring energy to the object. When a CMB-photon is absorbed, its energy is converted into heat, causing the object to warm up. This is why objects with a higher absorption coefficient will have a higher temperature than those with a lower absorption coefficient, as they are absorbing more CMB-photons per second.

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