Question

The phase velocity of transverse waves in a crystal of atomic separation a is given byy = csin(ka/2) pka/2 1. What is the dispersion relation e(k)? 2. What is the group velocity as a function of k?

Answer 1

Answer:

a

e(k) =  \frac{2a}{c}  *  sin (\frac{k*a}{2} )

b

G_{v} =  \frac{d e(k ) }{dk }  =  \frac{a^2}{c}  *  cos (\frac{k* a}{2} )

Explanation:

From the question we are told that

    The  velocity  of transverse waves in a crystal of atomic separation is  

                  $b_y = c \frac{sin (\frac{k*a}{2} )}{\frac{k*a}{2} }$

Generally the dispersion relation is mathematically represented as

            $e(k) = b_y * k$

=>    $e(k) = c \frac{sin(\frac{k*a}{2} ) }{ \frac{k*a}{2} } * k$

=>    $e(k) = c * \frac{sin (\frac{k_a}{2} )}{ \frac{a}{2} }$

=>    $e(k) = \frac{2a}{c} * sin (\frac{k*a}{2} )$

Generally the group velocity is mathematically represented as

          $G_{v} = \frac{d e(k ) }{dk } = \frac{a^2}{c} * cos (\frac{k* a}{2} )$