Kirchhoff’s Law and Pressure of Diffusion

According to Maxwell, radiation is transformation of heat energy without significant heat of intervening medium. Some properties of radiation are:

1. Thermal radiation moves with the velocity of light.
2. It moves in a straight line as light.
3. It shows reflection and refraction phenomenon.
4. It obeys the inverse square law.
5. It also shows the diffraction, interference and polarization phenomenon.
6. It lies in infrared region.

Total Energy Density

It is defined as energy per unit volume at a point due to all possible wave lengths.

Spectral Energy Density

Energy per unit volume at a point due to specified wavelength range
$$ V = \int _0^b V_\lambda. d\lambda $$

Total Emissive Power

The amount of energy emitted per unit area per second by a body due to all possible wavelengths is called total emissive power.

Spectral Emissive Power (\(E_\lambda \))

The amount of energy emitted per unit area per second by a body due to specified wavelength.

Absorptive Power (\(a_\lambda \))

The ratio of amount of energy absorb per unit area per second to the amount of energy incident per unit area per second of the body.

Kirchhoff’s Law

It states that the ratio of the emissive power to the absorptive power for the radiations of a particular wavelength at a particular temperature is constant for all bodies. Moreover the ratio is equal to the emissive power of a perfectly black body.

Proof of Kirchhoff’s Law

Consider an object in an isothermal enclosure, let \(e_{\lambda }\) and \(a_{\lambda}\) be the emissive and absorptive power of that object. If ‘dQ’ amount of per unit area per unit time be incident on the object, the amount of energy absorbed per unit area per time.
$$= a_{\lambda }dQ\dots (i) $$
Here \(e_{\lambda} \) be the spectral emissive power in the range of \(\lambda \) to \(\lambda + d\lambda \). The amount of energy emitted per unit area per time is,
$$=e_{\lambda }d_{\lambda } \dots (ii) $$

Since object is isothermally enclosed the amount of energy emitted must be equal to amount of energy absorbed,

\begin{align*} e_{\lambda }d_{\lambda } &= a_{\lambda }dQ \\ \frac {e_{\lambda }}{a_{\lambda }} &= \frac {dQ}{d_{\lambda }} \\ \end{align*}

The quantity \(\frac {dQ}{d_{\lambda }}\) depends only on temperature therefore \(\frac {e_{\lambda}}{a_{\lambda}}\) is constant for all object at a fixed temperature.

Pressure of Radiation

The radiation posses the properties of light, so like light, it exerts a small but definite pressure on the surfaces on which it is incident, called pressure of radiation. Maxwell proved on the basis of electromagnetic theory that pressure is equal to the energy density for normal incidence on the surface. on a surface.

Let a photon of energy \(h\nu \) move with the velocity of light c. According to the theory of relativity,

\begin{align*} h\nu &= mc^2 \\ \text {or,}\: m &= \frac {h\nu}{c^2} \\ \text {Then momentum of photon} &= \text {mass} \times \text {velocity} \\ &= \frac {h\nu }{c^2} = \frac {h\nu }{c} = \frac ec \\ \end{align*}

where e being energy of photon.

Now the momentum incident on the surface PQ per unit per second is given by

$$ P = \frac {\sum e}{c} \dots (2) $$

where \(\sum e =\) total energy incident on the surface per unit area per second = E
$$ \therefore P = \frac EC \dots (3) $$

let u is the energy density, energy per unit volume, then the total energy passing through any area s os the surface normal to the radiation per second \(= u\times s\)

Energy flux is energy ration per unit area per second
\begin{align*} E &= \frac {usc}{s} = uc \\ \text {Substituting value in}\: (3) \\ P &= \frac {uc}{c}= u \dots (5) \end{align*}

So, for normal incidence on the surface, the pressure of radiation is equal to the energy density.

$$ P = u $$

Pressure of Diffuse Radiation

Consider a case in which radiation is diffuse so that the beam of radiation makes an angle \(\theta \) with the normal to surface QR. The energy incident on the surface PQ per unit area per sec = uc.

\(\therefore \) Energy incident per unit area per second = ucIf S is the surface area of the plane PQ, the energy incident on the surface PQ, the energy incident on the surface PQ per second = ucS

But the surface area of the plane \( PQ = (\text {surface area of PR) \times \cos \theta = S^1\cos \theta \)

Total energy crossing the plane PQ per sec is equal to the energy incident on the plane QR per sec.

\(\therefore \) The energy incident on QR per sec. \(= ucS’\cos \theta \)

So the energy incident on the plane QR per unit area per se
$$= \frac {uS’ \cos \theta }{S’} =us\cos \theta $$

\(\therefore \) the momentum incident on the surface QR per unit area per second

$$QR = (u\cos \theta )\cos \theta = u\cos ^2\theta \)$$

The rate of change of momentum per second per unit area is called pressure radiation. The pressure radiation on QR is \(u\cos ^2\theta \). In case of diffused radiations, the radiation is incident from all possible directions with equal probability.

\(\text {i.e.}\: P = u(\vec {\cos ^2\theta}\), where \(\vec {\cos^2\theta } represents \(\cos ^\theta \) over all directions whose value is \(\frac 13 \)

\begin{align*} \therefore P &= \frac u3 \\ \therefore \text {Pressure of diffused radiation} &=\frac 13 \times \text {energy density} \\ \end{align*}

Bibliography

S.S. Singhal, J.P. Agarwal, Satya Prakash. heat and thermodynamics and statistical physics. pragati prakashan, 2010.

—. Heat and Thermodynamics and Statistical Physics. Pragati Prakashan, 2010.

Vatsyayan, Dr. Rakesh Ranjan. Refresher Course in Physics. Kathmandu: Surya Book Traders, 2015.