Phase velocity in oblique Angle propagation (Plane wave)

baby_1
Messages
159
Reaction score
16
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 )})
 

Attachments

Last edited:
Physics news on Phys.org
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##.
 
  • Like
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.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top