Solving Wave Equation with Paraxial Approximation

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Homework Help Overview

The discussion revolves around the application of the paraxial approximation to the wave equation, specifically focusing on the transition between different forms of the equation and the implications of substituting variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the terms in the wave equation after making substitutions, particularly questioning how derivatives change with the substitution of U_0. There is also a discussion about the presence of the e^{ikz} term in the final equation and its implications.

Discussion Status

The conversation is ongoing, with participants actively working through the mathematical details and questioning the reasoning behind certain steps. Some guidance has been offered regarding the derivatives, but there is no explicit consensus on the interpretation of the terms involved.

Contextual Notes

Participants express uncertainty about the treatment of the e^{ikz} term and the overall structure of the equations, indicating a need for clarity on the assumptions made in the problem setup.

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If U_0(r,z)=V(r,z)e^{ikz}, then:

\frac{\partial U_0}{\partial z}=\frac{\partial V}{\partial z}e^{ikz}+ikVe^{ikz}

due to the product rule.

so... \frac{\partial^2 U_0}{\partial z^2}=\ldots? :wink:
 
Oops, I see now I was looking past the obvious. I'll do the calculation in the morning. Is there a reason why they leave out the e^{ikz} from the final equation?

Thanks.

edit: V = V(r,z)e^{ikz}?
 
Last edited:
Confundo said:
Oops, I see now I was looking past the obvious. I'll do the calculation in the morning. Is there a reason why they leave out the e^{ikz} from the final equation?

They haven't "left it out"...To see what becomes of that term, work out the derivatives in equation (4) using the assumed form of U_0.


edit: V = V(r,z)e^{ikz}?

Huh?! :confused:...What are you trying to ask here?
 
Worked that out now, must learn not to try and do derivatives in my head while tired.
 

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