Question on wave equation of plane wave.

  • Thread starter yungman
  • Start date
  • #1
yungman
5,614
225
For plane wave travel in +ve z direction in a charge free medium, the wave equation is:

[tex]\frac{\partial^2 \widetilde{E}}{\partial z^2} -\gamma^2 \widetilde E = 0[/tex]

Where [itex]\gamma^2 = - k_c^2 ,\;\; k_c= \omega \sqrt {\mu \epsilon_c} \hbox { and } \epsilon_c = \epsilon_0 \epsilon_r -j\frac{\sigma}{\omega}[/itex] .


[tex]\widetilde E = E_0^+ e^{-\gamma z} \;+\; E_0^- e^{\gamma z} \;=\; E_0^+ e^{-\alpha z}e^{-j \beta z} \;+\; E_0^- e^{\alpha z}e^{j \beta z} [/tex]

Notice the second term is the reflected wave AND is growing in magnitude as it move in -ve z direction!!!! That cannot be true. It should still decay at rate of [itex] e^{-\alpha z } [/itex].

What am I missing?
 
Last edited:

Answers and Replies

  • #2
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,472
erm as we move towards -ve z direction z gets smaller and smaller hence [tex]e^{az}[/tex] gets smaller too for a>0.
 
  • #3
yungman
5,614
225
Why didn't I think of that!!!:surprised

Thanks
 
  • #4
Delta2
Homework Helper
Insights Author
Gold Member
5,696
2,472
Happens to me too, when i focus my mind on something i miss some relatively simple things.
 

Suggested for: Question on wave equation of plane wave.

Replies
2
Views
4K
Replies
22
Views
9K
  • Last Post
  • Optics
Replies
2
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
5
Views
7K
Replies
1
Views
4K
Replies
1
Views
3K
  • Last Post
Replies
2
Views
3K
Replies
2
Views
8K
Top