Electromagnetic Wave Equation

In summary, the solution to the homework equation is E(y,z)=E(y,z)cos(\omega t-k_xx). The attempted solution was to determine \nabla^2\textbf{E}, but this was not necessary as the k_x^2 term was already determined by evaluating the ##\nabla^2## operator.
  • #1
rmjmu507
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0

Homework Statement


Show that the solution [itex]\textbf{E}=E(y,z)\textbf{n}\cos(\omega t-k_xx)[/itex] substituted into the wave equation yields

[itex]\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}=-k^2E(y,z)[/itex]

where [itex]k^2=\frac{\omega^2}{c^2}-k_x^2[/itex]

Homework Equations


See above.

The Attempt at a Solution


I plugged the given solution into [itex]\frac{\partial^2 \textbf{E}}{\partial y^2}+\frac{\partial^2 \textbf{E}}{\partial z^2}=\frac{1}{c^2}\frac{\partial^2 \textbf{E}}{\partial t^2}[/itex] and got:

[itex]\textbf{n}\cos(\omega t-k_xx)[\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}]=-\frac{\omega^2}{c^2}E(y,z)\textbf{n}\cos(\omega t-k_xx)[/itex]

Now, canceling like terms I get:

[itex]\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}=-\frac{\omega^2}{c^2}E(y,z)[/itex]

But I'm missing a [itex]k_x^2[/itex] term on the RHS, and cannot figure out where this could/would have come from...can someone please explain?
 
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  • #2
I was able to get the [itex]k_x^2[/itex] term by determining [itex]\nabla^2\textbf{E}[/itex] and rearranging, thus obtaining the desired relation.

However, I'm not entirely sure why it's necessary to determine [itex]\nabla^2[/itex]. Can someone please explain this to be?
 
  • #3
You had to evaluate the ##\nabla^2## operator because that is the definition of the wave function. ## \nabla^2 \vec{E} = \frac{\partial^2 \vec{E}}{\partial t^2}## Adding an ##x## dependence into your function for ##\vec{E}## meant you had to fully evaluate the Laplacian.
 
  • #4
I see...I was considering this equation as only a two-dimensional one...for some reason I was overlooking the x component in the cosine function. Not entirely sure why, perhaps because of the E(y,z) term, but I now realize this is simply a coefficient corresponding to the amplitude.

Thanks!
 
  • #5


The missing term on the right-hand side can be accounted for by using the relationship between wave number k and angular frequency ω, which is given by k = ω/c. Substituting this into the given expression for k^2, we get:

k^2 = (\frac{\omega}{c})^2 - k_x^2

= (\frac{\omega}{c})^2 - (\frac{\omega}{c})^2 (since k_x = \frac{\omega}{c})

= 0

Therefore, the missing k^2 term on the right-hand side can be neglected, and we can rewrite the equation as:

\frac{\partial^2 E(y,z)}{\partial y^2}+\frac{\partial^2 E(y,z)}{\partial z^2}=-k^2E(y,z)

where k^2 = 0. This is consistent with the wave equation for electromagnetic waves, as the transverse electric field component is independent of the propagation direction.
 

1. What is the electromagnetic wave equation?

The electromagnetic wave equation is a mathematical formula that describes the behavior of electromagnetic waves, which are a combination of electric and magnetic fields that travel through space at the speed of light. It was first developed by James Clerk Maxwell in the 19th century and is a fundamental equation in the study of electromagnetism.

2. What is the significance of the electromagnetic wave equation?

The electromagnetic wave equation is significant because it accurately predicts the behavior of electromagnetic waves, which play a crucial role in many areas of science and technology. It is used in fields such as telecommunications, optics, and electronics, and forms the basis for our understanding of light and other forms of electromagnetic radiation.

3. How is the electromagnetic wave equation derived?

The electromagnetic wave equation is derived from Maxwell's equations, which are a set of four fundamental equations that describe the behavior of electric and magnetic fields. The wave equation is a result of combining two of these equations, known as Gauss's law and Faraday's law, and applying the principles of vector calculus.

4. What are the variables in the electromagnetic wave equation?

The electromagnetic wave equation has two main variables: the electric field (E) and the magnetic field (B). These fields are related to each other by the speed of light (c) and the electromagnetic constant (ε0), which represents the permittivity of free space. The equation also includes the time derivative (d/dt) and the spatial derivative (∇).

5. How is the electromagnetic wave equation used in practical applications?

The electromagnetic wave equation is used in a variety of practical applications. For example, in telecommunications, it is used to design and optimize antennas that transmit and receive electromagnetic waves. In optics, it is used to understand the behavior of light and design optical devices such as lenses and mirrors. It is also used in the development of electronic devices and technologies such as radar, wireless communication, and MRI machines.

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