How Are Group Velocity and Phase Velocity Related in Wave Packets?

[Hyperreality]

Wave packets has a group velocity of

v_{group}=\frac{d\omega}{dk}

and its phase velocity is

v_{phase}=\frac{\omega}{k}

Show that the group velocity and the phase velocity are related by:

v_{group} = v_{phase} - \lambda\frac{dv_{phase}}{d\lambda}

Can someone please tell me where to start on this problem?

 

[dextercioby]

All u need to know besides the rules of differentiation is
k=:\frac{2\pi}{\lambda}

Daniel.

PS.Compute the RHS of your equality and show it matches the LHS.

 

[wais]

To start, we can rewrite the group velocity equation as:

v_{group}=\frac{1}{\frac{dk}{d\omega}}

Next, we can use the chain rule to rewrite the phase velocity equation:

v_{phase} = \frac{d\omega}{dk} = \frac{d\omega}{d\lambda} \cdot \frac{d\lambda}{dk}

Plugging this into the group velocity equation, we get:

v_{group} = \frac{1}{\frac{dk}{d\lambda} \cdot \frac{d\lambda}{d\omega}}

Now, we can use the chain rule again to rewrite the denominator:

\frac{dk}{d\lambda} = \frac{dk}{d\omega} \cdot \frac{d\omega}{d\lambda}

Substituting this into the previous equation, we get:

v_{group} = \frac{1}{\frac{dk}{d\omega} \cdot \frac{dk}{d\omega} \cdot \frac{d\omega}{d\lambda}}

Simplifying, we get:

v_{group} = \frac{1}{\left(\frac{dk}{d\omega}\right)^2} \cdot \frac{d\omega}{d\lambda}

Finally, we can substitute the phase velocity equation for \frac{d\omega}{dk} into this equation:

v_{group} = \frac{1}{v_{phase}^2} \cdot \frac{d\omega}{d\lambda}

And since \frac{d\omega}{d\lambda} = \lambda \frac{dv_{phase}}{d\lambda}, we get:

v_{group} = v_{phase} - \lambda\frac{dv_{phase}}{d\lambda}

This shows that the group velocity and the phase velocity are related by the given equation.

 

[wais]

Sure, I can help you with this problem!

To start, we need to understand what the variables in the equations represent. The group velocity (v_{group}) and the phase velocity (v_{phase}) both refer to the speed at which a wave packet moves. The difference between the two is that the group velocity is the speed at which the overall envelope or shape of the wave packet moves, while the phase velocity is the speed at which the individual waves within the packet move.

Now, let's look at the first equation given: v_{group}=\frac{d\omega}{dk}. This equation tells us that the group velocity is equal to the change in angular frequency (d\omega) with respect to the change in wavenumber (dk). This makes sense because as the wavenumber increases, the frequency of the waves within the packet also increases, and therefore the packet must move faster.

Next, let's look at the second equation given: v_{phase}=\frac{\omega}{k}. This equation tells us that the phase velocity is equal to the angular frequency (ω) divided by the wavenumber (k). This means that the phase velocity is determined by the frequency and wavelength of the waves within the packet.

Now, to show the relationship between the two velocities, we can start by taking the derivative of the second equation with respect to the wavelength (λ). This gives us:

\frac{dv_{phase}}{d\lambda} = \frac{1}{k}\frac{d\omega}{d\lambda} - \frac{\omega}{k^2}\frac{dk}{d\lambda}

Next, we can substitute the first equation into the second term of this derivative, giving us:

\frac{dv_{phase}}{d\lambda} = \frac{1}{k}\frac{d\omega}{d\lambda} - \frac{v_{phase}}{k}\frac{dk}{d\lambda}

Finally, we can substitute the second equation into the first term of this derivative, giving us:

\frac{dv_{phase}}{d\lambda} = \frac{1}{k}v_{group} - \frac{v_{phase}}{k}\frac{dk}{d\lambda}

Now, if we rearrange this equation, we get:

v_{group} = v_{phase} - \lambda\frac{dv_{phase}}{d\lambda