Phase velocity in oblique Angle propagation (Plane wave)

baby_1
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Homework Statement
Maxwell equation
Relevant Equations
V=xt
Hello,
Regarding the wave oblique angle propagation and based on Balanis "Advanced engineering Electromagnetic" book on page 136 ( it has been attached) I need to know why the phase velocity in x direction is not important to keep in step with a constant phase plane( Just equation 4-23).
question.JPG
I derived the electric filed:
\vec{E}=E_{x}(z)e^{-j\beta (xSin(\theta )+zCos(\theta ))}
For calculating the phase velocity we need to obtain the time-domain form:
\vec{E(t)}=Re\{E_{x}(z)e^{-j\beta (xSin(\theta )+zCos(\theta ))}e^{jwt}\}=E_{x}(z)Cos(\omega t-\beta (xSin(\theta )+zCos(\theta ))
Calculating the Z- phase velocity (X=0):
\frac{d}{dt}(\omega t-\beta zCos(\theta ))=0=>\omega -\beta \frac{dz}{dt}Cos(\theta )=0=>v_{z}=\frac{\omega}{\beta Cos(\theta )}
which is the same as the book. However,

1)First: I need to know why the x velocity (on x-axis) is not important.
x phase velocity:(z=0)
\frac{d}{dt}(\omega t-\beta xSin(\theta ))=0=>\omega -\beta \frac{dx}{dt}Sin(\theta )=0=>v_{x}=\frac{\omega}{\beta Sin(\theta )}
2)Second: Why the resultant phase velocity is not the same as the light velocity(v_{c}=\frac{\omega}{\beta})
\sqrt{v_{z}^2+v_{x}^2}=\sqrt{({\frac{\omega}{\beta Cos(\theta )})}^2+({\frac{\omega}{\beta Sin(\theta )})}^2}=(\frac{\omega}{\beta Sin(2*\theta )})
 

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baby_1 said:
1)First: I need to know why the x velocity (on x-axis) is not important.
The "x velocity" is just as important as the "z velocity". I guess that the textbook just decided not to mention the x velocity since it can be found in a similar manner to the z velocity. Below, I denote these velocities as ##u_x## and ##u_z##.

baby_1 said:
2)Second: Why the resultant phase velocity is not the same as the light velocity(v_{c}=\frac{\omega}{\beta})
\sqrt{v_{z}^2+v_{x}^2}=\sqrt{({\frac{\omega}{\beta Cos(\theta )})}^2+({\frac{\omega}{\beta Sin(\theta )})}^2}=(\frac{\omega}{\beta Sin(2*\theta )})
Imagine a stick moving with translational velocity ##\vec v## as shown below. The stick is the black line and ##\vec v## is perpendicular to the stick. The z and x components of ##\vec v## would be ##v_z= v \cos \theta## and ##v_x = v \sin \theta##.
1627072157553.png


The points of intersection of the stick with the z and x axes are shown in green. The speed at which the intersection with the z-axis moves along the z-axis is ##u_z = \frac{v}{\cos \theta}##. Similarly, for the speed of the intersection along the x-axis, ##u_x = \frac{v}{\sin \theta}##.

##u_z## and ##u_x## are not the z and x components of the velocity of some single point. In particular, they are not the z and x components of the velocity of some point on the stick. So, we do not expect ##v = \sqrt {u_z^2+u_x^2}##.

The textbook uses the notation ##v_{pz}## and ##v_{px}## for ##u_z## and ##u_x##. These should not be confused with the ##z## and ##x## components of the vector ##\vec v##: ##v_z## and ##v_x##.
 
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Likes Delta2, baby_1 and hutchphd
I see the Like is nothing to thank you. I really appreciate your explanations and the time you spent to draw the shape and explain more and more.
 
You are welcome.
 
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