Physics 301 07-Oct-1999 7-1 More on Blackbody Radiation Before Moving

Physics 301 07-Oct-1999 7-1

More on Blackbody Radiation

Before moving on to other topics, we’ll clean up a few loose ends having to do with blackbody radiation.

In the homework you are asked to show that the pressure is given by

π2τ 4 p = , 45¯h3c3 from which one obtains 1 pV = U, 3 for a photon gas. This is to be compared with

2 pV = U, 3 appropriate for a monatomic ideal gas.

In an adiabatic (isentropic—constant entropy) expansion, an ideal gas obeys the relation pV γ = Constant ,

where γ = Cp/CV is the ratio of heat capacity at constant pressure to heat capacity at constant volume. For a monatomic ideal gas, γ =5/3. For complicated gas molecules with many internal degrees of freedom, γ 1. A monatomic gas is “stiﬀer” than a polyatomic gas in the sense that the pressure in→ a monatomic gas rises faster for a given amount of compression. What are the heat capacities of a photon gas? Since

π2 U = Vτ4 , 15¯h3c3

2 ∂U 4π 3 CV = = Vτ . ∂τ 15¯h3c3 V How about the heat capacity at constant pressure. We can’t do that! The pressure depends only on the temperature, so we can’t change the temperature without changing the pressure. We can imagine adding some heat energy to a photon gas. In order to keep the pressure constant, we must let the gas expand while we add the energy. So, we can certainly add heat at constant pressure, it just means the temperature is constant as well, so I suppose the heat capacity at constant pressure is formally inﬁnite!

If one recalls the derivation of the adiabatic law for an ideal gas, it’s more or less an accident that the exponent turns out to be the ratio of heat capacities. This, plus the fact that we can’t calculate a constant pressure heat capacity is probably a good sign

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that we should calculate the adiabatic relation for photon gas directly. We already know σ Vτ3 Vp3/4, so for an adiabatic process with a photon gas, ∝ ∝ pV 4/3 = Constant , and a photon gas is “softer” than an ideal monatomic gas, but “stiﬀer” than polyatomic gases. Note that γ =4/3 mainly depends on the fact that photons are massless. Consider a gas composed of particles of energy E and momentum P = E/c,wherec is the speed of light. Suppose that particles travel at speed c and that their directions of motion are isotropically distributed. Then if the energy density is u = nE,wheren is the number of particles per unit volume, the pressure is u/3. This can be found by the same kind of argument suggested in the homework problem. The same result holds if the particles have a distribution in energy provided they satisfy P = E/c and v = c. This will be the case for ordinary matter particles if they are moving at relativistic speeds. A relativistic gas is “softer” than a similar non-relativistic gas!

On problem 3 of the homework you are asked to determine the power per unit area radiated by the surface of a blackbody or, equivalently, a small hole in a cavity. The result is (c/4)u where the speed of light accounts for the speed at which energy is transported by the photons and the factor of 1/4 accounts for the eﬃciency with which the energy gets through the hole. The ﬂux is

2 4 2 4 π τ π k 4 4 J = = T = σB T 60¯h3c2 60¯h3c2 where the Stefan-Boltzmann constant is 2 4 π k 5 erg σB = =5.6687 10− . 60¯h3c2 × cm2 sK4

Last lecture, we saw that the Planck curve involved the function x3/(exp(x) 1) with x =¯hω/τ. Let’s ﬁnd the value of x for which this curve is a maximum. We have− d x3 0= , dx ex 1 − 3x2 x3ex = , ex 1 − (ex 1)2 − − (3x2 x3)ex 3x2 = − − (ex 1)2 or − 0=(x 3)ex +3. − This transcendental equation must be solved numerically. The result is x =2.82144. At maximum, ¯hω h ν 2.82 = max = max , τ k T

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or T h Kelvin = =0.017 . νmax 2.82k Gigahertz So the above establishes a relation between the temperature and the frequency of the maximum energy density per unit frequency.

You will often see uν which is the energy density per unit Hertz rather than the energy density per unit radians per second. This is related to uω by the appropriate number of 2π’s. You will also see uλ which is the energy density per unit wavelength. This is found from u dλ = u dω . λ| | ω| | This says that the energy density within a range of wavelengths should be same as the energy density within the corresponding range of frequencies. The absolute value signs are there because we only care about the widths of the ranges, not the signs of the ranges. We use ω =2πc/λ and dω =(2πc/λ2) dλ , | | dω u = u , λ ω dλ

¯h (2πc/λ)3 2πc = , π2c 3 (exp(2 π¯hc/λτ) 1) λ2 − 8πhc = , λ5 (exp(hc/λτ) 1) − 8πτ 5 x5 = , h4c4 ex 1 − where x = hc/λτ. At long wavelengths, u 8πτ/λ4, and at short wavelengths u is λ → λ exponentially cut oﬀ. The maximum of uλ occurs at a wavelength given by the solution of

(x 5)ex +5=0. − The solution is x =4.96511 .... From this, we have hc λ T = =0.290 cm K . max 4.97k This is known as Wien’s displacement law. It simply says that the wavelength of the maximum in the spectrum and the temperature are inversely related. In this form, the constant is easy to remember. It’s just 3 mm Kelvin. (Note that the wavelength of the maximum in the frequency spectrum and the wavelength of the maximum in the wavelength spectrum diﬀer by a factor of about 1.6. This is just a reﬂection of the fact that wavelength and frequency are inversely related.)

Let’s apply some of these formulae to the sun. First, the peak of the spectrum is in about the middle of the visible band (do you think this is a coincidence or do you suppose

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there’s a reason for it?), at about λ 5000 A=5˚ 10 5 cm. Using the displacement max ≈ × − law (and assuming the sun radiates as a blackbody), we find Tsun 5800 K. The luminosity 33 1 ≈ 10 of the sun is L =3.8 10 erg s− . The radius of the sun is r =7.0 10 cm. The flux × 2 10 2 1 × emitted by the sun is J = L/4πr =6.2 10 erg cm− s− .Thisisabout60Megawatts ×4 per square meter! We equate this to σB T and find Tsun 5700 K, very close to what we estimated from the displacement law. ≈

Problem 17 in chapter 4 of K&K points out that the entropy of a single mode of thermal radiation depends only on the average number of photons in the mode. Let’s see if we can work this out. We will use

∂τlog Z σ = , ∂τ V where Z is the partition function and the requirement of constant volume is satisﬁed by holding the frequency of the mode constant. We’ve already worked out the partition function for a single mode 1 Z = . 1 e ¯hω/τ − − The average occupancy (number of photons) in the mode is

1 n = , e¯hω/τ 1 − from which we ﬁnd n +1 ¯hω n +1 = e¯hω/τ , or =log . n τ n Now let’s do the derivatives to get the entropy

∂ σ = (τ log Z) , ∂τ ∂ ¯hω/τ =logZ τ log 1 e− , − ∂τ − 1 1 ¯hω 1 =log τ e ¯hω/τ , ¯hω/τ ¯hω/τ − 1 e− − 1 e− − − τ −τ − − 1 ¯hω e ¯hω/τ =log + − , 1 n/(n +1) τ 1 e ¯hω/τ − − − n +1 n/(n +1) = log(n +1)+log , n 1 n/(n +1) − n +1 = log(n +1)+n log , n =(n + 1) log(n +1) n log n, −

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Which is the form given in K&K. This is another way of making the point that the expansion of the universe does not change the entropy of the background radiation. The expansion redshifts each photon—stretches out its wavelength—in proportion to the expansion factor, but it does not change the number of photons that have the redshifted wavelength—the number of photons in the mode. So, the entropy doesn’t change. (This assumes that the photons don’t interact with each other or with the matter. Once the universe is cool enough ( 4000 K) that the hydrogen is no longer ionized, then the interactions are very small.) ≤

Johnson Noise

This is another application of the thermal equilibrium of electromagnetic modes. Con- sider an ideal transmission line, like a long piece of lossless coaxial cable. Suppose it’s length is L and suppose it is shorted out at each end. Then any wave that travels along the line is reflected at each end and to have an appreciable amplitude, the length of the line must contain an integral number of half wavelengths. In other words this is just a one-dimensional cavity of length L. There are modes of the electromagnetic fields, most conveniently represented by the potential difference between the inner and outer conduc- tors. Vn = Vn,0 sin(ωt)sin(nπx/L), where n is any positive integer. Vn and Vn,0 represent the potential difference and the amplitude of the potential difference. The fields must satisfy Maxwell’s equations, so ω = nπc/L. Actually, if the coax is filled with a dielectric, the speed of propagation can be different from c, let’s assume it’s filled with vacuum. If this line is in thermal equilibrium at temperature τ, each mode acts like an oscillator and has average energy ¯hω/(e¯hω/τ 1). Let’s consider the low frequency limit so the average energy in each mode is just τ.− The number of modes per integer n is just 1. Then the number of modes per unit frequency is

L n(ω) dω = dω . πc The energy per unit length per unit frequency is then τ u = , ω πc at low frequencies.

As you may know, all transmission lines have a characteristic impedance, R.Ifa resistor R is connected across the end of the line, then a wave traveling down the line is completely absorbed by the resistor. So, let’s take a resistor, in equilibrium at temperature τ, and connect it to the end of the line. Since the resistor and the line are at the same temperature, they are already in thermal equilibrium and no net energy transfer takes place. Each mode in the line is a standing wave composed equally of traveling waves headed in both directions. The waves traveling towards the resistor will be completely

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absorbed by the resistor. This means that the resistor must emit waves with equal power in order that there be no net transfer of energy. The energy in the frequency band dω per unit length headed towards the resistor is τdω/2πc. This is traveling at speed c,sothe power incident on the resistor is τdω/2π which is also the power emitted by the resistor,

What we’ve established so far is that the line feeds power τdω/2π into the resistor and vice-versa. This means that a voltage must appear across the resistor. This will be a ﬂuctuating voltage with mean 0 since it’s a random thermal voltage. However, its mean square value will not be zero. Let’s see if we can calculate this. As an equivalent circuit, we have a resistor R, a voltage generator (the thermally induced voltage source), a ﬁlter (to limit the frequencies to dω), and another resistor of resistance R representing the transmission line. Then the current is I = V/2R. The average power delivered to the resistor is then I2 R = V 2 /4R = τdω/2π. In the lab, one measures frequencies in Hertz rather than radiansh i perh second,i ν = ω/2π. Finally

V 2 =4Rτ dν . h i This relates the mean square noise voltage which appears across a resistor to the temperature, resistance, and bandwidth (dν). Of course, this voltage results from ﬂuctuations in the motions of electrons inside the resistor, but we calculated it by considering electromagnetic modes in a one-dimensional cavity, a much simpler system! This thermal noise voltage is called Johnson noise.

Debye Theory of Lattice Vibrations

A little thought will show that sound waves in a solid are not all that diﬀerent from electromagnetic waves in a cavity. Further thought will show that there are some important diﬀerences that we must take into account.

The theory of lattice vibrations that we’ll discuss below applies to the ion lattice in a conductor. In addition, one needs to account for the thermal eﬀects of the conduction electrons which behave in many respects like a gas. We’ll consider the electron gas later in the term. For now, we imagine that we’re dealing with an insulator.

We will treat crystalline solids. This is mainly for conceptual convenience, but also because we want reasonably well deﬁned vibrational modes.

As a model, suppose the atoms in a solid are arranged in a regular cubic lattice. Each atom vibrates around its equilibrium position. The equilibrium and the characteristics of the vibrations are determined by interactions with the neighboring atoms. We can imagine that each atom is connected to its six nearest neighbors by springs. At ﬁrst sight, this seems silly. But, the equilibrium position is determined by a minimum in the potential energy, and the potential energy almost surely increases quadratically with

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displacement from equilibrium. This gives a linear restoring force which is exactly what happens with a spring. So our solid is a large number of coupled oscillators. In general, the motion of a system of coupled oscillators is very complex. You probably know from your classical mechanics course, that the motion of a system of coupled oscillators can be resolved into a superposition of normal modes with the motion of each mode being simple harmonic in time. So, we can describe the motion with the N vectors ri which represent the displacement of each atom from its equilibrium position, or we can describe the motion with 3N normal mode amplitudes. For those of you that know about Fourier transforms, the normal modes are just the Fourier transforms of the position coordinates.

These normal modes represent elastic vibrations of our solid. They are standing elastic waves, or standing sound waves. In this respect, they are similar to the cavity modes we discussed earlier. There are two main diﬀerences. First, there are three polarizations: there are two transversely polarized waves (as we had in the electromagnetic case) and one longitudinally polarized wave (absent in the electromagnetic case). Second, there is a limit to the number of modes. If our solid contains N atoms, there are 3N modes. In the electromagnetic case, there is no upper limit to the frequency of a mode. High frequency modes are with ¯hω τ are not excited, but they are there. In the elastic case, frequencies which are high enough that the wavelength is shorter than twice the distance between atoms do not exist.

For simplicity, we are going to assume that the velocity of sound is isotropic and is the same for both transverse and longitudinal waves. Also, we’ll assume that the elastic properties of the solid are independent of the amplitude of the vibrations (at least for the amplitudes we’ll be dealing with).

We’ll carry over as much stuﬀ from the electromagnetic case as we can. A typical mode will look something like n πx n πy n πz displacement component = A sin ωt sin x sin y sin z , L L L where the sine factors might be cosines depending on the mode, A represents an amplitude, and for convenience, the solid is a cube of side L. The frequency and mode numbers are related by the speed of sound, v, π2v2 ω2 = n2 + n2 + n2 . L2 x y z If the solid contains N atoms, the distance between atoms is L/N 1/3. The wavelength must be longer than twice this distance. More precisely 2L 2L > 1/3 , nx N

with similar relations for ny and nz. In other words, the mode numbers nx, ny,andnz are integers within the cube N 1/3 N 1/3 N 1/3. This lower limit on the wavelength (upper limit on the frequency) is an example× of× the Nyquist limit discussed later in these notes.

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The number of modes per unit frequency is just as it was for the electromagnetic case except that we must multiply by 3/2 to account for the three polarizations instead of two,

3Vω2 n(ω) dω = dω . 2π2v3 This works for frequencies low enough that the corresponding n’s are within the cube. It’s messy to deal with this cubical boundary to the mode number space. Instead, let’s approximate the upper boundary as the surface of a sphere which gives the same number of modes. In other words, there will be an upper limit to the frequency, called ωD,such that ωD 3V ωD V 3N = n(ω) dω = ω2 dω = ω3 , 2π2v3 2π2v3 D Z0 Z0 which gives 6π2N 1/3 ω = v. D V

Each mode acts like a harmonic oscillator and its energy is an integer timeshω ¯ .The quanta of sound are called phonons. A solid contains a thermally excited phonon gas. The average energies of these oscillators are just as they were in the electromagnetic case. We ﬁnd the total energy by adding up the energies in all the modes,

3V ωD ¯hω U = ω2 dω , 2π2v3 exp(¯hω/τ) 1 Z0 − 3V xD x3 dx = τ 4 , 2π2¯h3v3 ex 1 Z0 − where ¯hω 6π2N 1/3 ¯hv kθ θ x = D = = = , D τ V τ kT T where θ is called the Debye temperature and is given by

6π2N 1/3 ¯hv θ = . V k

The Debye temperature is not a temperature you can change by adding or removing heat from a solid! Instead, it’s a characteristic of a given solid. The way to think of it is that a vibration with phonon energy equal to kθ is the highest frequency vibration that can exist within the solid. Otherwise the wavelength would be too short. (The weird factor of 6π2 occurs because we replaced a cube with a sphere!) Typical Debye temperatures are a few hundred Kelvin.

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The limit of integration depends on the temperature, so in general, we can’t look up the integral. Instead, we have to numerically integrate and produce a table for diﬀerent values of xD = θ/T. Such a table is given in K&K.

Therearetwolimitingcaseswherewecandotheintegral.Theﬁrstcaseisverylow temperature (T θ). In this case xD is very large and we can replace xD by .Then the integral is π4/15 and we have ∞

π2V 3π4N U = τ 4 = τ 4 . 10¯h3v3 5k3θ3 The heat capacity at constant volume is then

12 T 3 C = π4Nk . V 5 θ So a prediction of this theory is that at low temperatures, the heat capacities of solids should be proportional to T 3. This is borne out by experiment!

The other limit we can consider is very high temperature (T θ). In this case, we expect all modes are excited to an average energy τ, so the total should be U =3Nτ.Is this what we get? At very high temperatures, xD 1, so we can expand the exponential in the denominator of the integrand,

3V xD x3 dx U = τ 4 , 2π2¯h3v3 ex 1 Z0 − 3V xD = τ 4 x2 dx , 2π2¯h3v3 Z0 V = τ 4 x3 , 2π2¯h3v3 D V ¯h3v3 6π2N = τ 4 , 2π2¯h3v3 τ 3 V =3Nτ , as expected. Actually, we picked ωD so this result would occur “by construction.” The heat capacity goes to 3Nk in this limit.

In one of the homework problems you are asked to come up with a better approxima- tion in the limit of small xD.

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The Nyquist Frequency

We imagine that we have a function of time that we sample periodically, every T seconds. Then the Nyquist frequency is the frequency corresponding to a period of two samples. ωN =2π/2T = π/T. Consider a sine wave at some frequency ω,

y(t)=sin(ωt + φ) .

Since we are sampling, we don’t have a continuous function, but a discrete set of values:

ym =sin(ωmT + φ) .

Suppose the frequency is larger than the Nyquist frequency. Then we can write it as an even integer times the Nyquist frequency plus a frequency less than the the Nyquist frequency: ω =2nωN +Ω=2πn/T +Ω, where ω Ω +ω .Then − N ≤ ≤ N

ym =sin(2πnm +ΩmT + φ)=sin(ΩmT + φ) .

In other words, when we sample a sine wave periodically, waves with frequencies greater than the Nyquist frequency look exactly the same as waves with frequencies less than the Nyquist frequency. This is illustrated in the ﬁgure. The arrows along the bottom indicate

the times at which the signal is sampled. A signal at the Nyquist frequency would have one cycle every two sample intervals. The high frequency wave has 3.7 cycles every two

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sample intervals or a phase change of 3.7π =(4 0.3)π every sample. We can’t tell how many multiples of 2π go by between samples, so− the high frequency wave looks exactly like a low frequency wave with 0.3 cycles per two samples. The points show the value of the signal (either wave) at each− sampling interval. Of course the application to the Debye theory of lattice vibrations is that the Nyquist spatial frequency is the highest frequency a periodic lattice can support.