derivation of Rayleigh-Jeans law from classical physics PDF

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derivation of Rayleigh-Jeans law from classical physics
Calculating the total number of modes then differentiating the result to get an incremental number of modes...

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The waves satisfy the wave equation: ∇2 ฀฀ −

1 ฀฀ 2 ฀฀=0 ฀฀ 2฀฀฀฀ 2

For example, an electromagnetic wave will have electric field as well as magnetic field. Then each component of the electric and magnetic field will satisfy the above equation i.e. ∇2 ฀฀ −

1 ฀฀ 2 ฀฀=0 ฀฀ 2฀฀฀฀ 2

� , i.e. ฀ ฀ = �฀฀฀฀ , ฀฀฀฀ , ฀฀฀฀ , ฀฀฀ ฀ , ฀฀฀฀ , ฀฀฀฀ �. where ฀฀ represents any of the ฀฀, ฀฀, or ฀฀ –component of any of�,฀฀฀฀ There are different possible solutions of the above equation including the propagating wave solution and standing wave solution. A propagating wave solution may be written as: �. ฀฀ ± ฀฀฀฀�� ฀฀(฀฀, ฀฀, ฀฀, ฀฀) = ฀ ฀ exp�฀฀�฀฀ For a standing wave solution, we must specify the boundary conditions where the nodes must occur. For a standing wave solution with nodes at the walls of the cavity, we must have solution that must vanish at the walls at ฀ ฀ = 0, ฀฀, ฀ ฀ = 0, ฀฀ or ฀ ฀ = 0, ฀฀. � = ฀฀1 ฀฀� + ฀฀2 ฀฀� + ฀฀3 ฀฀ is: One possible standing wave solution, standing in a general direction given by ฀฀ ฀฀(฀฀, ฀฀, ฀฀, ฀฀) = ฀ ฀ sin(฀฀1 ฀฀) sin(฀฀2 ฀฀) sin(฀฀3 ฀฀) sin(฀฀฀฀) For standing waves, the components of the wave vector must be written as: ฀฀1 =

฀฀1 ฀฀ ฀ ฀

, ฀฀2 =฀

฀฀2 ฀฀ , ฀฀3 ฀

=฀

฀฀3 ฀฀ ฀

where ฀฀1 , ฀฀2 and ฀฀3 must be positive integers. The angular frequency ฀฀ is related with the wavelength as: ฀฀=

2฀฀฀฀ ฀ ฀

= ฀฀฀฀

� � = �฀฀� . ฀฀ Hence, we get: ฀ ฀ = ��฀฀  = ฀฀21 + ฀฀22 + ฀฀23 = ⇒ ฀฀ 2 = �฀฀. �฀฀

฀฀2 ฀฀ 2

฀฀1 ฀฀ 2 2฀฀ 2 ฀฀3 ฀฀ 2 ฀฀2 ฀฀ 2 ⇒� � =� � � +� � +� ฀ ฀ ฀ ฀ ฀ ฀ ฀฀ 2 4฀฀ ⇒ ฀฀21 + ฀฀22 + ฀฀23 = 2 ฀฀ Now the above is true for each standing modes. We must now calculate ALL POSSIBLE standing wave modes. Calculation of Modes � −space: Consider a coordinate system in the ฀฀ −space or ฀฀ � = (฀฀1 , ฀฀2 , ฀฀3 ) = ฀฀(฀฀1 , ฀฀2 , ฀฀3 ) ฀฀ ฀฀

In this coordinates, each set (triplet) of integer values of (฀฀1 , ฀฀2 , ฀฀3 ) corresponds to one standing wave. Now, we must remember that whatever the mode may be it must satisfy the condition: ฀฀21+ ฀฀22 + ฀฀23 =

4฀฀2 ฀฀2

The above condition may be satisfied by different choices of (฀฀1 , ฀฀2 , ฀฀3 ) and ฀฀. Remember, ฀฀ is fixed here. The calculation mode implies calculation of total number of all possible combinations of (฀฀1 , ฀฀2 , ฀฀3 ) and ฀฀ that satisfy the above condition. For example, take ฀ ฀ = ฀฀/2. Then the constraint becomes: ฀฀21 + ฀฀22 + ฀฀23 =

4฀฀2 4฀฀2 = = 16 ฀฀2 ฀฀ 2 � � 2

This condition may be satisfied by different combinations of (฀฀1 , ฀฀2 , ฀฀3 ), all positive integers. Now consider the ฀฀ −space picture. In ฀฀ −space, each point with integer (฀฀1 , ฀฀2 , ฀฀3 ) will correspond to a volume of unity (with no unit or dimension). This implies the volume of a region in ฀฀ −space, will be equal to (numerically) the number of discrete lattice points contained within the region (since each lattice point has an associated unit volume with it). Now, ฀฀21 + ฀฀22 + ฀฀23 = constant implies a number of possible lattice points, having different 2 2 2 combinations (฀฀1 , ฀฀2 , ฀฀3 ) such that ฀฀ 1+ ฀฀2 + ฀฀3 = constant. These are lattice points on the sphere of radius �฀฀12 + ฀฀22 + ฀฀23 in –space. Of course we have to consider only 1/8 of the number times 2 for having two independent modes of polarization. This can be calculated in two ways:

A. Method-1 Calculating the total number of modes with 0 < |฀฀�| ≤ �(4฀฀2 )/฀฀2 = 2฀฀/฀฀ and then differentiating the result to get an incremental number of modes (the way we did it in the lecture): B. Method-2 Calculating directly the incremental number of modes within a spherical shell of radius from ฀฀ to ฀ ฀ + ฀฀฀฀, where ฀ ฀ =�|.|฀฀ � | ≤ (4฀฀2 )/฀฀2 is, with �|฀฀ | = Method-1: The number of standing waves ฀฀, in ฀฀ −space, for 0 < |฀฀

�฀฀12 + ฀฀22 + ฀฀23 is

4 1 ฀ ฀ = � � (2) ฀฀(฀฀21 + ฀฀22 + ฀฀23)3/2 8 3 In terms of wavelength ฀฀, since ฀฀21+ ฀฀22 + ฀฀23 =

4฀฀2 , ฀฀2

the number of modes ฀฀ is:

8฀฀฀฀3 ฀฀ ฀ ฀ = (฀฀21 + ฀฀22 + ฀฀23)3/2 = 3 3฀฀3 The number of standing waves (i.e. modes) per unit wavelength is found from differentiation of the above with respect to ฀฀: ฀฀฀฀ 8 ฀฀฀฀3 =− 4 ฀฀ ฀฀฀฀ The negative sign indicates that the number of modes decreases with increase of wavelength. The number of modes per unit wavelength per unit volume of the cavity is: −�

1 ฀฀฀฀ 8฀฀ � = ฀฀3 ฀฀฀฀ ฀฀4

The number of standing waves per unit volume with wavelengths between ฀฀ and ฀฀ + ฀฀฀฀ is (ignoring the negative sign) : 8฀฀ ฀฀฀฀ = ฀฀(฀฀)฀฀฀฀ = 4 ฀฀3 ฀฀฀฀ ฀฀ Method-2: The number of standing waves ฀฀฀฀, in ฀฀ −space, within a spherical shell of radius from ฀฀ to

= �฀฀12 + ฀฀22 + ฀฀23 is: ฀ ฀ + ฀฀฀฀, where ฀ ฀ =� ||฀฀

1 ฀฀฀฀ = ฀฀(฀฀)฀฀฀฀ = �� 2 (4 ฀฀฀฀2 )฀฀฀฀ 8 But ฀ ฀ = �(4฀฀2 )/฀฀2 = 2฀฀/฀฀ gives ฀฀฀฀ = (2฀฀) (−1)฀฀฀฀/฀฀2 . Hence, ignoring the negative sign, we get ฀฀฀฀2 2฀฀฀฀฀฀ 4฀฀22฀฀ 8฀฀ 1 2 )฀฀฀฀ = (4 � 2 ฀฀฀฀ = ฀ ฀ 2 2 ฀฀฀฀ = 4 ฀฀3 ฀฀฀฀ ฀฀฀฀ = ฀฀(฀฀)฀฀฀฀ = 2 ฀฀ ฀฀ 8 ฀฀ ฀฀...