Radiation pressure and the Stefan-Boltzmann law (2/4)

By 0x7df, Fri 25 May 2018, in category Uncategorized

physics, thermodynamics

In a previous post on Kirchhoff's law (1859) and black bodies , we saw that the energy density of thermal radiation is a function of temperature only. The first measurements of thermal radiation (from hot platinum wire) were made by Tyndall, and from his results Stefan concluded, in 1879, that the energy radiated went as the fourth power of the absolute temperature. This empirical relationship was later theoretically determined, for black bodies, by Boltzmann in 1884. The law that bears both their names is:

$$ R_B = \sigma T^4 $$

and \( \sigma\) is known as the Stefan-Boltzmann constant, and \( R_B\) is the emissive power, the radiant power emitted per unit area.

Radiation pressure

Before we begin, it's necessary to understand that radiation exerts a pressure. The easiest way to calculate the radiation pressure is to assume that photons are particles of mass \( m\), where \(m\) is given by:

$$ e = mc^2 $$

\(e\) being the photon energy. If we do this, then we can treat the radiation field as a photon gas and find its pressure in the same way as we would for a normal gas, from kinetic arguments; we assume the pressure is the sum of all the impulses delivered to a unit area of the wall in unit time by particles colliding with it.

The collision rate of a single particle is:

$$ \frac{|u_x|}{2 L_x}$$

where \( u_x\) is the component of the particle's velocity normal to the wall (assumed conserved), and \( L_x\) is the dimension of the container along the normal to the wall. The impulse delivered in each collision is equal to the momentum change of the particle, i.e.:

$$ 2m|u_x|$$

So the impulse per unit time is:

$$ \frac{m u_x^2}{L_x} $$

for a single particle. This leads to:

$$ \frac{N \bar m c^2}{3 L_x}$$

for \(N\) particles, where \( c^2/3\) is the mean value of \( \bar{u_x^2}\) and \( \bar m\) is the mean mass. This force corresponds to a pressure of:

$$ p = \frac{N \bar m c^2}{3 V} $$

where we have assumed the volume \( V = L_x A\); or:

$$ p = \frac{n \bar m c^2}{ 3}$$

where \( n = N/V\) is the number density of particles. Replacing \( \bar m c^2\) with \(/bar e\) gives:

$$ p = \frac{1}{3} n\bar e$$

which is the same as:

$$ p = \frac{1}{3} \frac{E}{V} $$

Derivation of the Stefan-Boltzmann law - energy density

Let's begin with the general thermodynamic expression:

$$ dE = T dS - p dV $$

which applies when volume is the only constraint to be be varied. If we divide by \( dV\) at fixed \( T\) we obtain:

$$ \left( \partial E/\partial V\right)_T = T \left(\partial S/\partial V\right)_T - p $$

We can see that for an enclosure at constant temperature:

$$ \left(\partial E/\partial V\right)_T = E/V $$

(i.e. the energy density in the cavity), since increasing the size of the cavity just adds more radiation at the same density (we proved in the previous post that the energy density is constant throughout and depends only on the temperature of the walls, not on the size or shape or material). Hence:

$$ E/V = T \left(\partial S/\partial V\right)_T - p $$

Next we can also use the Maxwell relation:

$$ \left(\partial S/\partial V\right)_T = \left(\partial p/\partial T\right)_V $$

which comes from the fact that both:

$$ dF = -S dT - p dV $$
$$ dF = \left(\partial F/\partial T\right)_V dT + \left(\partial F/\partial V\right)_T dV $$

are true, so that we can identify:

$$ S = -\left(\partial F/\partial T\right)_V $$
$$ p = - \left(\partial F/\partial V\right)_T $$

and, bearing in mind that \( \left(\partial/\partial V\right)_T \left(\partial F/\partial T\right)_V = \left(\partial/\partial T\right)_V \left(\partial F/\partial V\right)_T\), arrive at:

$$ \left(\partial S/\partial V\right)_T = \left(\partial p/\partial T\right)_V$$


Inserting this result into:

$$ E/V = T \left(\partial S/\partial V\right)_T - p $$


$$ E/V = T \left(\partial p/\partial T\right)_V - p $$

Now, the radiation pressure is \( p = (1/3) E/V\), so:

$$ 4p = T \left(\partial p/\partial T\right)_V $$


$$ dp/p = 4 dT/T $$

at fixed \( T\). It follows that \( p \propto T^4\), and since \( p \propto E/V\), then:

$$ E/V \propto T^4 $$

That is, the radiation energy density is proportional to the fourth power of the temperature. (We will call the constant of proportionality here \(a\) to distinguish it from \( \sigma\), since here we are considering energy density, and the version of the Stefan-Boltzmann law we gave at the beginning was for the radiated power per unit area.) Hence, and finally:

$$ E/V = a T^4 $$

Stefan-Boltzmann law - radiated power

If the energy density in the interior of our constant-temperature enclosure is \( E/V = a T^4\), then the flux onto unit area of the cavity wall would be \( ac T^4\) if the radiation were all incident normally on the wall. If it were incident at an angle \( \theta\) to the normal, the flux would be \( ac T^4 \cos \theta\). In reality the radiation is isotropic, so the flux is given by:

$$ R_B = \int{ac T^4 \cos \theta}df $$

where we have equated the radiated power per unit area, \(R_B\), with the flux of radiation falling on the wall, which must be true if the wall temperature stays constant; and where \( df\) is the area element of the unit sphere centred at the point of interest on the wall. This is given by:

$$ df = \frac{2\pi r^2 \sin \theta d\theta}{ 4\pi r^2} $$

so the flux is:

$$R_B = \frac{1}{2} \int_0^{\pi /2} {ac T^4 \cos\theta\sin\theta} d\theta  = \frac{1}{4} acT^4 $$

If we define \( \sigma = (1/4) ac\) then:

$$ R_B = \sigma T^4 $$

which is where we began.

Note that, since we know from Kirchhoff's law that a non-black body radiator has emissive power \( R = a R_B\), where \( a\) is the absorptivity of the non-black body at the temperature in question, then in general:

$$ R = a\sigma T^4 $$

Care is needed using this formula because \( a\) often has a strong dependence on temperature.

In practice, it is usually the difference between emitted and absorbed radiation that is measured. A black body at temperature \( T_1\) emits radiation at a rate of \( \sigma T_1^4\) per unit area, but if it is placed in a sealed enclosure with wall temperature \( T_2\), so that its surroundings are at temperature \( T_2\), it is subject to incident radiation at a rate of \(\sigma T_2^4\). (This latter must be true since, if the body were in equilibrium with the radiation field of the enclosure, it would be radiating at a rate of\(\sigma T_2^4\), and since it's in equilibrium, must absorb radiation at the same rate. The rate of absorption is no different when the body is at a different temperature  (Assume the body is small enough compared to the enclosure that its presence doesn't disturb the radiation field in the enclosure - \( T_2\) is constant.))


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