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

[Niles]

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?

 

[Niles]

No suggestions?

 

[astrorob]

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.

 

[Niles]

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?

 

[astrorob]

It'll give you a number with units: {Js^{-1}m^{-2}}

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.

 

[Niles]

Hmm, in my book the constant alfa = 7.56*10^-16 J * m^-3 * K^-4?

 

[astrorob]

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

 

[Shooting Star]

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.

 

[Niles]

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?

 

[Shooting Star]

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 h\nu.

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.

 

[Niles]

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?

 

[Shooting Star]

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.

 

[Niles]

I did.

Thanks for helping! :-)