Missing step involving e for maxwell equ. solution

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In summary, the conversation is about a person having trouble understanding the steps in solving a differential equation in their book. The book presents two equations and gives the solutions as exponential functions. The person is confused about where the exponential part comes from and the author assumes the reader knows how to solve basic differential equations. The general solution is a linear combination of the two exponential functions.
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d3nat
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Homework Statement


I am following along in my book (Laser Fundamentals by Silfvast, pg 14), and I cannot understand how they got from one step to the next.



Homework Equations


Since it's quite lengthy, I'm going to dive into the middle of it. If you need more background information, let me know



The Attempt at a Solution



( d^2*A_z ) / dz^2 + (w^2*A_z) / v^2 = 0
and
( d^2*A_t ) / dt^2 + (w^2*A_t) = 0

and then the book says that these are familiar forms and have the following solutions

A_z = C_1 e^i(w/v)z + C_2 e^-i(w/v)z
A_t = D_1 e^iwt + D_2 e^-iwt

I have following everything up to that point. I don't understand where they are getting the exponential part. I can somewhat see how it relates since if you look at the A_z part, the w^2/v^2 looks similar to how it goes to e^i(w/v)z
But I don't understand where the author pulled all of this information from. What are the missing steps from the top two lines to get to the bottom two lines.

Any advice is greatly appreciated.
 
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  • #2
They are assuming you know how to solve basic differential equations. There are two independent solutions of A''+c^2*A=0 (where the prime is differentiation with respect to the independent variable, call it t), A(t)=exp(ict) and A(t)=exp(-ict). Substitute them into check. Given that, a general solution is a linear combination of those two. Does that make it look more familiar?
 

1. What is the "missing step" in Maxwell's equation solution involving e?

The missing step in Maxwell's equation solution involving e is the application of the electric displacement vector D.

2. Why is the electric displacement vector important in Maxwell's equation solution?

The electric displacement vector is important because it takes into account the effect of electric polarization on the electric field, which is necessary for a complete and accurate solution to Maxwell's equations.

3. How does the electric displacement vector relate to the permittivity of a material?

The electric displacement vector is directly proportional to the permittivity of a material. This means that the greater the permittivity of a material, the greater the electric displacement vector and the stronger the effect of electric polarization on the electric field.

4. Why is it important to include the electric displacement vector in the solution to Maxwell's equations?

Including the electric displacement vector in the solution to Maxwell's equations allows for a more complete and accurate representation of the behavior of electric fields in materials with different permittivities. This is important for understanding and predicting the behavior of electromagnetic waves and other electrical phenomena.

5. What are some real-life applications of the electric displacement vector in Maxwell's equation solution?

The electric displacement vector is important in many real-life applications, such as designing and optimizing electronic devices, understanding the behavior of electromagnetic waves in different materials, and calculating the capacitance of capacitors. It is also essential in the development of technologies such as radar, wireless communication, and medical imaging.

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