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If both velocities would be the same, then the relative position of the E-field would remain static when compared to the envelope. With a group velocity different than that of phase, it just means that the constructive interference will slip when compared to the speed of the phase. Only when you arrive at the second term do you get a spreading of the interference effect of the group's components, as now the phase relationship is no longer linear.

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If you ever wish to compensate for the dispersion caused by the variation in $k(\omega)$ you need to know what $k_1=\frac{dk}{d\omega}|_{\omega_0}$ is. By itself $k_1$ gets you a good linear approximation at one frequency point but if you wish to be more precise then you will need other derivatives, such as $k_2=\frac{d^2k}{d\omega^2}|_{\omega_0}$. This is standard consideration in the so-called group delay compensating filter that is to be combined with a bandpass filter to reduce the dispersion caused by the latter. You may also find it used in the receiver of a chirp radar whose LC pulse compression filter needs group delay compensation or just to estimate the size and number of ghost targets that such dispersion may cause.

You are right that indeed there is a difference between the velocities of the group's constituent frequencies, when there is a group velocity that is different from that of the phase. But due to the linear relationship of the phase difference as the pulse propagates the only effect you will have is a phase slippage between the maximum of the E-field and the envelope, as I illustrate in the .gif I uploaded below:

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To word this question another way, is the inclusion of the second order term in the taylor series necessary to observe the "chirp" effect that is shown in the first image I linked. In my head it seems like the first derivative should already contain the fact that different wavelengths move at different speeds and that even with only the first order approximation we should still see the same effect. Can someone explain or point me to understanding what I am thinking about wrong?

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A wave packet can be described as a sum of sinusoidal with frequencies distributed around a central frequency $\omega_0$. As a result, one can show that the wave packet in the time-position domain may be written in the form

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If now the dispersion is weak, a first order Taylor development on $k(\omega)$ around $\omega_0$ is justified. In this second case, the group and phase velocities aren't equal anymore but the enveloppe remains unchanged during propagation (well represented on the gif provided by José Andrade).

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I am a bit confused on the topic of GVD. Most information I read on the topic tells me something like this: "GVD means that the group velocity will be different for different wavelengths in the pulse. Thus, in ultrashort pulses with broad bandwidth, pulse spreading can cause red (larger wavelength) to lead blue (smaller wavelength) as the pulse spreads." I attached two images below of what I mean.

This is the product of one function oscillating very quickly at frequency $\omega_0$, propagating at the phase velocity $v_{\phi}=\omega_0/k_0$; and a function enveloppe oscillating at a lower frequency, propagating at the group velocity $v_g=\frac{d\omega}{dk}\big|_{\omega=\omega_0}$. If the medium isn't dispersive, the phase and group velocity are equal.

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Now, my confusion lies in why the second derivative is necessary to consider here in order to see the "effect of GVD". In the case where we only count up to the first order term in the Taylor Series shown (aka only consider k(w0) & k'(w0)) aren't we already accounting for the fact that group velocity exists and hence that different wavelengths will travel at different speeds within the medium?

In the case of highly dispersive medium, a first order development isn't sufficient. You have to consider second (or higher) orders; it is typically the case for pulses with large bandwidth. In this case, the group velocity depends on $\omega$, leading to a deformation of the wave packet.

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