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PHYS 445 Lecture 26 - Black-body radiation I
26 - 1

Lecture 26 - Black-body radiation I

What's Important:

··

number density energy density

Text: Reif

In our previous discussion of photons, we established that the mean number of photons with energy is

n =

(26.1)

e^ß- 1

Here, the questions that we want to address about photons are:

···

what is their number density what is their energy density what is their pressure?

Number density n

Eq. (26.1) is written in the language of discrete states. The first thing we need to do is replace the sum by an integral over photon momentum states:

S_i® òd ³p

Of course, it isn't quite this simple because the density-of-states issue that we introduced before: for every spin state there is one phase space state for every h ³, so the proper replacement more like

S_i® (1/h ³) òd ³p òd ³r

The units now work: the left hand side is a number, and so is the right hand side. But this expression ignores spin - it just deals with the states in phase space. For every photon momentum, there are two ways of arranging its polarization (i.e., the orientation of its electric or magnetic field vectors):

where the photon momentum vector is perpendicular to the plane. Thus, we have

S_i® (2/h ³) òd ³p òd ³r.

(26.2)
Assuming that our system is spatially uniform, the position integral can be replaced by the volume

òd ³r = V.

PHYS 445 Lecture 26 - Black-body radiation I The total number of photons is then
26 - 2

V
d³p

N =2

.
(26.3)

h³e - 1

This integral is written in terms of energy
and momentum p; for photons, these

quantities are related through Einstein's relation (m = 0)

= pc.

Dividing N by V gives the integrated number density of photons n_g, which can be written in an integral form as
2

h³d ³p n =

(26.4)

ßpc

- 1

or as with

[number per unit volume between p and p+dp] = n_g(p) d ³p
2 d³p n (p)d³p =

.
(26.5)

h³e^ßpc- 1

Let's integrate Eq. (26.4) for an isotropic system, where ò_anglesd ³p = 4pp ²dp:

8p p²dp n =

ßpc

- 1

Changing variables to x = ßpc gives

x²dx n =

(ßhc)³₀e - 1

The integral has the form of a Reimann Zeta function, and is

x²dx

=2 (3) =2 · 1.202

₀e - 1

Thus,

æk T hc

öø

n =1.202· 16p

(26.6)

Example

The universe is filled with relic radiation left over from the Big Bang with a temperature of 2.7 K. What is n_gof this radiation?

ö³

1.38 ´ 10^-23· 2.7

æç

n =1.202· 16p

6.63 ´ 10^-34· 3.0 ´ 10⁸

= 4.0´ 10⁸m^{- 3}=400cm^{- 3}

PHYS 445 Lecture 26 - Black-body radiation I
26 - 3

Energy density u

We saw that the number of photons per unit volume between p and p+dp is

2 d³p h³e^ßpc- 1 n (p)d³p =

These photons have an energy of pc, giving an energy density u_g(p)d p in this momentum range of

2 d ³p h³e^ßpc- 1 u (p)d³p =pc

.
Taking the integral of this over momentum gives the total energy density u_gfor an isotropic system,

p²dp
8pc p³dp

u = pc

h³e^ßpc- 1 h³e

- 1

ßpc

where ò_anglesd ³p = 4pp ²dp as before. Once again changing variables to x = ßpc

4¥

4 ¥

8pc ækT x³dx ækT x³dx

^ö_øò
_øò

u =
=8phc

₀e^x- 1

hc ₀e - 1

The integral is equal to

x³dx

p
=

₀e - 1 15

Thus,

8p (pk_BT)⁴u =

.
(26.7)

15 (h c)³
Example u =

Find the energy density of photons at room temperature T = 293 K.
8p (p· 1.38 ´ 10^{- 23}· 293)⁴
=5.5 ´ 10^{- 6}J/m³.
15 (6.63 ´ 10^-34· 3.0 ´ 10⁸)³

Mean energy per photon

From Eqs. (26.6) and (26.7) we can determine the mean energy per photon

8p (k_BT)⁴

15 (h c)³u n

k_BT =2.7k_BT

(26.8)

(k_BT )³

1.202· 30
1.202· 16p

(hc)³

Notice once again how k_BT sets the energy scale.

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