Solution:

The relation between the wavelength $\lambda$ and the frequency $\nu$ is given by:

$\frac{c}{\lambda} = \sqrt{\nu^2 - \nu_0^2}.$

This can be rewritten as:

$k = \frac{2\pi}{\lambda} \quad \Rightarrow \quad k = \frac{2\pi}{c} \sqrt{\nu^2 - \nu_0^2}.$

The dispersion relation is therefore:

$k = \frac{2\pi}{c} \sqrt{\nu^2 - \nu_0^2}.$

Now, we analyze the group velocity and phase velocity:

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1. Group Velocity ($v_g$):

The group velocity is defined as:

$v_g = \frac{\partial \nu}{\partial k}.$

From the dispersion relation $k = \frac{2\pi}{c} \sqrt{\nu^2 - \nu_0^2}$, we solve for $\nu$ as a function of $k$:

$\nu = \sqrt{\left( \frac{kc}{2\pi} \right)^2 + \nu_0^2}.$

Taking the derivative of $\nu$ with respect to $k$:

$v_g = \frac{\partial \nu}{\partial k} = \frac{kc}{2\pi \nu}.$

Substitute $\nu = \sqrt{\left( \frac{kc}{2\pi} \right)^2 + \nu_0^2}$:

$v_g = \frac{\frac{kc}{2\pi}}{\sqrt{\left( \frac{kc}{2\pi} \right)^2 + \nu_0^2}} \cdot c.$

Simplify:

$v_g = \frac{c^2 k}{2\pi \nu}.$

Comparison of $v_g$ to $c$:

Since $\nu^2 = \left( \frac{kc}{2\pi} \right)^2 + \nu_0^2$, the group velocity is always less than $c$ due to the contribution of $\nu_0^2$ in the denominator.

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2. Phase Velocity ($v_p$):

The phase velocity is defined as:

$v_p = \frac{\nu}{k}.$

Using the dispersion relation $k = \frac{2\pi}{c} \sqrt{\nu^2 - \nu_0^2}$:

$v_p = \frac{\nu}{\frac{2\pi}{c} \sqrt{\nu^2 - \nu_0^2}} = \frac{\nu c}{\sqrt{\nu^2 - \nu_0^2}}.$

Comparison of $v_p$ to $c$:

The phase velocity is always greater than $c$ because the numerator $\nu c$ is greater than the denominator $\sqrt{\nu^2 - \nu_0^2}$.

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Summary:

1. Group Velocity ($v_g$):
   $v_g = \frac{\frac{kc}{2\pi}}{\sqrt{\left( \frac{kc}{2\pi} \right)^2 + \nu_0^2}} \cdot c,$
   and $v_g < c$.

2. Phase Velocity ($v_p$):
   $v_p = \frac{\nu c}{\sqrt{\nu^2 - \nu_0^2}},$
   and $v_p > c$.

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Conceptual Explanation:

• Group velocity represents the speed at which the energy or information propagates through the waveguide. It is slower than the speed of light $c$.
• Phase velocity represents the speed of the wave's phase, which can exceed $c$ due to the dispersive nature of the waveguide. This does not violate special relativity because no information travels at the phase velocity.

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Do you have any further questions or need a detailed breakdown of any step?

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Related Questions:

1. What is the significance of the constant $\nu_0$ in this context?
2. Can the group velocity ever exceed $c$ in any physical system?
3. How does the dispersion relation affect wave propagation in a waveguide?
4. What are some real-world applications of group velocity and phase velocity in waveguides?
5. How does the relationship between $\nu_0$ and $\nu$ determine whether a wave is propagating?

Tip:

Always distinguish between group velocity and phase velocity in wave mechanics. Group velocity relates to energy transfer, while phase velocity is a mathematical property of wave propagation.