lecture 8

Lecture 8
Wave Velocity - Phase and Group Velocities
PHASE VELOCITY eq. 31 describes an oscillatory function that has a sinusoidal variation with either z or t. If we set t equal to a constant (we choose t=0) then we have a sinusoidally varying function of the form e^(-ik_z z). If we set z equal to a constant (we choose z=0) then we have a similar function of the form e^i?t. In either case, it is understood that we take the real part of the function to obtain the value of the amplitude.
The argument of eq. 31 is a phase factor ? such that:
?=k_z z-?t ………….. (35)
By considering the value of the wave for ? = constant we can track how the wave moves with time, since the amplitude of the wave function remains constant when the argument of the wave function is constant. The velocity at which the wave form propagates is called the phase velocity. We obtain this velocity by taking the derivative of ? and equating it to zero, which denotes the maximum value of the wave form.
For d?=0
k_z dz=?dt ……….. (36)
Therefore, the phase velocity is
V_phase=dz/dt=?/k_z =c …………… (37)
This proves that the constant c in eq. 22 is the speed at which the wave moves in time. By comparing eqs. 22 and 21 one gets that:
c=1/(?_o ?_o )=299,792,457.4 ± 1.1 m/s. …………… (38)
GROUP VELOCITY
An infinite wave train of a single frequency has no starting or stopping point; it can t be modulated. Any attempt at modulation - to start and stop a wave train - requires a wave involving more than one frequency. We will first consider two plane waves with electric fields of E_1 and E_2 described as:
E_1=cos?(k_1 z-?_1 t) and E_2=cos?(k_2 z-?_2 t)
Adding these two wave results in:
E=E_1+E_2=cos?(k_1 z-?_1 t)+cos?(k_2 z-?_2 t)
E=2cos 1/2?[?(k_1+k?_2)z-(?_1+?_2)t]×cos 1/2?[?(k_1-k?_2)z-(?_1-?_2)t] …(39)
If ?_1 and ?_2 can be considered nearly the same, one can write:
k ?=(k_1+k_2)/2 …………….. (40)
? ?=(?_1+?_2)/2 …………….. (41)
k_m=(k_1-k_2)/2 …………….. (42)
?_m=(?_1-?_2)/2 …………….. (43)
For optical frequencies
? ???_m ……………(44)
Hence, we can determine the velocity V_m by considering the phase factor ?_m associated with the modulation frequency?_m by taking the derivative of ?_m and equating it to zero.
?d??_m=k_m dz-?_m dt=0 ……………. (44)
Solving for V_m=dz/dt results in:
V_group=V_m=dz/dt=?_m/k_m =(1/2(?_1-?_2))/(1/2(k_1-k_2))=??/?k=d?/dk ……..(45)
where, V_group is the speed of the two combined waves E_1 and E_2.

In general the group velocity is derived for a wave of centre frequency ?_o but now with a finite width ??. In effect, the finite width allows for the wave to consist of more than one frequency and hence modulation could occur. Assume a propagation constant k_o at frequency ?_o but now allow it to vary with frequency ? in the region of ?_o such that:
k=k_o+?(dk/d?) ?_(?=?_o ) (?-?_o) ……………. (46)
Let us take a Fourier expansion of the electric field of the wave as follows:
E(z,t)=???E(?)e^(-i(kz-?t)) d?? ……….. (47)
Inserting eq. (46) into eq. (47) yields:
E(z,t)=e^(-i(k_o z-?_o t)) ?_(-??_o/2)^(+??_o/2)?E(?)Exp[-i((dk/d?)_(?_o ) z-t)??] d?? .. (48)

where ??=?-?_o.
Note that the plane wave is now modulated by a wave involving the function[(dk/d?)z-t]. We can now obtain the velocity of this modulated wave in the same way as we did for a plane wave by evaluating a constant phase factor or [(dk/d?)z-t]= constant. Solving for the velocity of this "group" or bundle of waves, we find that
dz/dt=V_group=d?/dk …………… (49)
A finite group velocity implies a change in velocity with a change in wavelength. A medium with such characteristics is referred to as a dispersive medium. In a non-dispersive medium the group velocity and the phase velocity are identical. In a dispersive medium, the group velocity is different from the phase velocity. The group velocity is the velocity at which information can be transmitted, because sending information in involves modulation.

Transverse Electromagnetic Waves and Polarized Light
E(z,t)=E_o e^(-i(k_z z-?t)) ………… (50)
The Equation above represents the electric component of an electromagnetic wave whose amplitude is a vector quantity. A similar relationship can be expressed for the magnetic component H. Since ??E=0, and since ??E=ik?E, we can conclude that ik?E=0 and thus E is perpendicular to k, the direction of propagation. A similar argument can be made for the magnetic portion of the wave, since ??H=0. Therefore, both the amplitudes E_o and H_o are perpendicular to the direction of the propagation vector k and thus to the propagation of the wave. (They are also perpendicular to each other.) Hence these electromagnetic waves are referred to as transverse waves.
Because the amplitude factor in an electromagnetic wave is a vector quantity that lies in a plane perpendicular to the direction of propagation, that amplitude can be resolved into independent orthogonal components within that plane. For example, the wave expressed in eq. (50), which is traveling in the z direction, can be resolved into its two orthogonal components or polarizations as:
E(z,t)=E_ox ?i ?e?^(-i(k_z z-?t))+E_oy ?j ?e?^(-i(k_z z-?t+?)) ………… (51)
where i ? and j ? are unit vectors in the x and y directions, respectively. The relative phase factor ? suggests that the two independent polarizations need not be in exact phase. If ?=0, or integral multiples of 2?, then the combined waves represent a single wave with the transverse amplitude vector always oriented in the same direction. Such a wave is referred to as plane or linearly polarized light, as shown in Figure 4. Other values of ? cause the amplitude vector of the wave to spiral in either an elliptical or a circular manner, as indicated in the figure. Such waves are referred to as either circularly or elliptically polarized light as shown in figure 5.