Waves on deep water with surface tension T and density ρ are governed by the dispersion relation $$ω^2 = gk + \frac{T}{\rho}k^3.$$

Calculate the phase and group velocities of the waves. Show that the phase velocity reaches a minimum at a wave number $k_c$ that you should calculate. What is the group velocity for this wave number?

the phase velocity is the speed of a point of constant phase which is $c_p^2=\frac{w^2}{k^2}$

the group velocity of a wave is the velocity with the overall shape of the waves' amplitudes $v_g =\frac{dω}{dk}$

But I am confused since the c can take either direction, does it mean that I need to consider both $c_p=\frac{ω}{k} or-\frac{ω}{k}$. And when I got $c_p=\sqrt{\frac{g}{k}+\frac{T}{\rho}k}$ then how should I get the minimun phase velocity.

what is the motion of the individual wave crests relative to the centre of a wave packet for $k<k_c$ and for $k>k_c$?

  • $\begingroup$ Firstly, you expression for phase speed is wrong - you should of divided by $k^2$. When I was dealing with dispersion relations we had postive and negative components. Were you trying to go from the phase speed to group speed? if so then yes it would be wrong. $\endgroup$ – Chinny84 May 16 '17 at 12:45
  • $\begingroup$ Sould it be like this? $\endgroup$ – stedmoaoa May 16 '17 at 12:50
  • $\begingroup$ no, you computed the initial dispersion re-arrangement wrong, if you had divided by $k^2$ correctly then you would of solved the problem (I did it below since I miss my days when I was playing with 5th order dispersion relations) $\endgroup$ – Chinny84 May 16 '17 at 13:32

The phase speed is given by $$ \frac{\omega^2}{k^2}=\frac{g}{k}+\frac{T}{\rho}k\implies c_p^2 = \frac{g}{k}+\frac{T}{\rho}k $$ so Phase speed is $$ c_p = \pm\sqrt{\frac{g}{k}+\frac{T}{\rho}k} $$ The group speed is given by $$ 2\omega\frac{\partial \omega}{\partial k} = g + \frac{3T}{\rho}k^2 $$ so we have $$ v_g = \frac{g + \frac{3T}{\rho}k^2}{2\omega} $$ To determine the minimum phase speed we are essentially compute $$ \frac{\partial c_p}{\partial k} = 0 $$ so we use the first relation $$ 2c_p\frac{\partial c_p}{\partial k} = -\frac{g}{k^2}+\frac{T}{\rho} $$ This is zero for a trial case, $c_p=0$, but in general we require $$ -\frac{g}{k^2}+\frac{T}{\rho} =0 \implies k_c^2 = \frac{g\rho}{T} $$ so the minimum phase speed occurs for $$ k_c = \sqrt{\frac{g\rho}{T}} $$

  • $\begingroup$ Thank you so much and it really makes sense! and how should I get the group velocity for this k_c? $\endgroup$ – stedmoaoa May 16 '17 at 13:40
  • $\begingroup$ Well, you have the equation for the $v_g$ in my post, so sub in $k_c$ or the r.h.s and then also compute $\omega(k_c)$ or $\omega(\sqrt{g\rho/T})$. $\endgroup$ – Chinny84 May 16 '17 at 13:41
  • $\begingroup$ It is what I think thank you! And one more question: what is the motion of the individual wave crests relative to the centre of a wave packet for k<k_c and for k>k_c? I am really new to this and can you please explain at little? $\endgroup$ – stedmoaoa May 16 '17 at 13:44
  • $\begingroup$ My idea is that the group velocity >phase velocity when k>k_c and vice versa. But how to relate it to the crest $\endgroup$ – stedmoaoa May 16 '17 at 13:47

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