E 0 k i ( r k r r t ) r = r = c | r E E 0 e k | B 0 i ( r k r r - - PowerPoint PPT Presentation

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Thermal Radiation Radiation in thermal equilibrium with its surroundings E 0 k i ( r k r r t ) r = r = c | r E E 0 e k | B 0 i ( r k r r t ) r r r r r B = B 0 e B 0 = 1 k E 0 /c 8.044 L 15 B1 - 0 | 2 1 Time

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SLIDE 1

Thermal Radiation Radiation in thermal equilibrium with its surroundings E k E r = r

i(r k·r r−t)

= c|r k| E0 e B0 r r

i(r k·r r−t)

r r r B = B0 e B0 = 1k × E0 /c

8.044 L15B1

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SLIDE 2

1

Time average energy density u = E0|E

  • 0|2

2

Time average energy flux

  • jE = (cu)-

1k Time average pressure (⇒ to - k) P = u Thermal radiation has a continuous distribution of frequencies.

u(ν,T)

Peaks near hν = 3kBT (h/kB 5 × 10−11 K-sec)

ν

8.044 L15B2

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SLIDE 3

Spectral Region ν (Hz) T (K) Thermal Rad. Radio Microwave Infrared Visible Ultraviolet X ray γ ray 106 1010 1013

1 2 × 1015

1016 1018 1021 1.7 × 10−5 0.17 1.7 × 102 8.5 × 103 1.7 × 105 1.7 × 107 1.7 × 1010 cosmic background room temp. sun’s surface black holes

8.044 L15B3

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SLIDE 4

ENERGY ABSORBED ABSORPTIVITY α( ν,T) ENERGY INCIDENT

ISOTROPIC

ENERGY EMITTED EMISSIVE POWER e( ν,T) AREA

ISOTROPIC

8.044 L15B4

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SLIDE 5

THERMAL RADIATION: PROPERTIES

2 ENERGY FLUXES, IN AND OUT OF CAVITY B FILTER: FREQUENCY OR POLARIZATION CAVITY A CAVITY B

TA TB

ASSUME TA = TB AND THERMAL EQUILIBRIUM

8.044 L15B5

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SLIDE 6

CONCLUSIONS:

  • u(ν, T ) is independent of shape and wall material
  • u(ν, T ) is isotropic
  • u(ν, T ) is unpolarized

8.044 L15B6

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SLIDE 7

CONSIDER AN OBJECT IN THE CAVITY, IN THERMAL EQUILIBRIUM

dA T

COMPUTE THE ENERGY FLUX n

∆A

c∆t

θ

8.044 L15B7

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SLIDE 8

∆E = (E in cylinder) p(θ, φ) dθ dφ sin θ 1 = (u ∆A cos θ c∆t) dθ dφ 2 2π

π/2 cos θ sin θ 2π 1

= c u ∆A ∆t dθ dφ 2 2π

1/4 1

1

⇒ energy flux onto dA = 4c u(ν, T )

8.044 L15B8

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SLIDE 9

Momentum Flux Plane wave momentum density p r = u r 1k

c

|∆p| = 2|p⇒| since r −r p⇒ in = p⇒ out

8.044 L15B9

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SLIDE 10

2 cos θ |∆p|ν = (E in cylinder) p(θ, φ) dθ dφ c

π/2 2π 1

= u(ν, T ) ∆A ∆t cos2 θ sin θ dθ dφ 2π

1/3 1

1

=

3u(ν, T ) ∆A ∆t

⇒ P (T ) = 1

u(ν, T ) dν

3

8.044 L15B10

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SLIDE 11

Apply detailed balance to the object in the cavity. Eout = Ein e dA = α (4

1c u(ν, T )) dA

e(ν, T )

1

⇒ =

4c u(ν, T )

α(ν, T ) This ratio has a universal form for all materials. The result is known as KIRCHOFF’S LAW.

8.044 L15B11

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SLIDE 12

Black Body Radiation If α ≡ 1 ≡ “Black” Then e(ν, T ) = 1

4c u(ν, T )

OVEN CAVITY AT T

Measure e(ν, T ) and obtain u(ν, T )

8.044 L15B12

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SLIDE 13

Thermodynamic Approach

∞

u(T ) ≡ u(ν, T ) dν Then E(T, V ) = u(T )V

1

P (T, V ) = 3u(T )

8.044 L15B13

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SLIDE 14
  • This is enough to allow us to find u(T).

dE = TdS − PdV ∂E ∂S ∂P = T − P = T − P ∂V

T

∂V

T

∂T

V

1

= Tu(T) − 1

3u(T) 3

also = u(T)

  • u(T) = 4 u(T)

T u(T) = AT 4

8.044 L15B14

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SLIDE 15

Emissive Power of a Black (α = 1) Body e(ν, T ) = 1

4c u(ν, T ) ⇒ e(T ) = 1 4c u(T ) = 1AcT 4 4

e(T ) ∞ σT

4

This is known as the STEFAN-BOLTZMANN LAW. σ = 56.7 × 10−9 watts/m2K4

8.044 L15B15

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SLIDE 16

r Statistical Mechanical Approach H? Single normal mode (plane standing wave) in a rectangular conducting cavity.

Ex z L

E r (r r, t) = E(t) sin(nπz/L)r 1x

0,0,n,r 1x

B (r r, t) = (nπc2/L)−1E ˙ (t) cos(nπz/L)r 1y

0,0,n,r 1y

8.044 L15B16

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SLIDE 17

Energy density = 1E0E

  • · E
  • + 1

µ 1

0 B

  • · B
  • [no -

r or t average] 2 2

 

H = V 1E0E2(t) + 1 1 (nπc2/L)−2E ˙ 2(t)

2 2

2 µ0 V E0 = E2(t) + (nπc/L)−2 E ˙ 2(t) 2 2 Each mode corresponds to a harmonic oscillator.

8.044 L15B17

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SLIDE 18
  • Count the modes.

it

r Enx,ny,nz = |E|r j sin(nxπx/L) sin(nyπy/L) sin(nzπz/L)e The unit polarization vector E j has 2 possible orthog-

  • nal directions and ni = 1, 2, 3 · · ·.

2

2 E E

2 E

πc

2 2 2

− c 2E = 0

  • 2 =

(n + n + n )

x y z

t2 L

8.044 L15B18

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SLIDE 19
  • If the radian frequency < ω

nz R

L

2 2 2

R = n + n + n = ω

x y z

πc # modes (freq. < ω)

ny

1 4

GRID SPACING

= 2 × 8 × 3πR3

1 UNIT

3

π L

nx

= 3

πc

ω3

8.044 L15B19

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SLIDE 20
  • 3

d# L V D(ω) = = π ω2 = ω2 dω πc π2c3

D(ω)

ω

8.044 L15B20

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SLIDE 21

Classical Statistical Mechanics D() kBT < E() >= kBT u(, T) =< E() > =

3 2

V π2c

∞

u(T) = u(, T) d = ∞ u(ω, T)

8.044 L15B21

CLASSICAL MEASURED

ω

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SLIDE 22

Quantum Statistical Mechanics ¯ h < E() >= + ¯ h/2

¯ h/kT − 1

e D() ¯ h 3 u(, T ) =< E() > = + z. p. term π2c3

¯ h/kT − 1

V e du(, T ) To find the location of the maximum, set = 0. d The maximum occurs at ¯ h/kT ≈ 2.82.

8.044 L15B22

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SLIDE 23
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SLIDE 24
  • −¯

hω/2kT (1 − e −¯ hω/kT )−1

Z = Zi Zi = e

states i

The first factor in the expression for Zi comes from the zero-point energy. F (V, T ) = −kT ln Z = −kT ln Zi

states i

= −kT D(ω) [ln Zi] dω

8.044 L15B23

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SLIDE 25
  • −¯

hω/kT )

F (V, T ) = −kT D(ω) − ln(1 − e dω + · · ·

  • kT V

−¯ hω/kT ) dω

= ω2 ln(1 − e π2c3

  • V

= (kT )4 x

2 ln(1 − e −x) dx

π2c3¯ h3

v- _

−π4

45

1 π2 = − (kT )4 V 45 c3¯ h3

8.044 L15B24

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SLIDE 26
  • ∂F

1 π2 P = − = (kT)4 ∂V T 45 (c¯ h)3 ∂F 4 π2 S = − = k4T 3 V ∂T V 45 (c¯ h)3 1 4 1 π2 E = F + TS = − + (· · ·) = (kT)4 V 45 45 15 (c¯ h)3 Note: P = 1

3 E/V = 1 3 u(T) independent of V .

8.044 L15B25

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SLIDE 27

NOTE: THE ADIABATIC PATH IS T3V=CONSTANT

T

ADIABATIC T/T0 = (V/V0)-1 / 3

V

8.044 L15B26

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SLIDE 28

MIT OpenCourseWare http://ocw.mit.edu

8.044 Statistical Physics I

Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.