Understanding the Electric Field of a Plane Wave: A Complex Vector Approach

In summary, the electric field for a ray(plane wave actually) at a distance s from the reference point is given by E(s) = E(0) * exp(-jks). The phase variation of the electric field along the ray is represented by exp(-jks).
  • #1
broli86
4
0
Hi I've working on a computer program where I need to calculate some
electric fields. I am referring to a thesis which provides these
formulas. For eg:

The electric field for a ray(plane wave actually) at a distance s from
the reference point:

E(s) = E(0) * exp(-jks)

where E(s) = electric field at a distance s from the reference point.
E(0) = electric field at reference point.
s = distance travelled.
k = wave number = 2 * PI / wavelength
exp(-jks) as per my thesis, represents the phase variation of the
electric field along the ray.

now, electric field is always a vector(3d in my case). so the user
will input a magnitude for the electric field at the reference point
and i can calculate the reference electric field vector[ E(0) ] easily
by multiplying it with the unit direction vector the wave.

exp(-jks) expands as

cos(ks) - j sin(ks)
^ ^ ^
if E(0) is a vector of the form a x + b y + c z,
^ ^ ^
Then E(s) = (a x + b y + c z) * ( cos(ks) - j sin(ks) )

I'm really confounded at this expression because how does one multiply
a complex number and a vector ? And even if it is possible then what
about E(s) ? If E(0) was defined to be a 3d vector then E(s) must also
be a 3d vector. What does it actually mean ? If there are some EE/
physics experts who can probably figure it out then it would be really
great.
 
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  • #2
Well really E(s) = Re{Eo * exp(-jks)} (ie the real part)... so in short you're supposed to ignore the sin term.
 
  • #3
nicksauce said:
Well really E(s) = Re{Eo * exp(-jks)} (ie the real part)... so in short you're supposed to ignore the sin term.

I think that equation was somewhat wrong. I think the EM field can be represented as a complex vector:

A complex vector as in a 3d vector whose components are complex
numbers :
(a1 + i * b1) xhat + (a2 + i * b2) yhat + (a3 + i * b3) zhat

where xhat, yhat, zhat are unit vectors along x, y, z respectively.

one can rearrange it to get real part as the electric field vector and
imaginary part as the magnetic field vector, both at right angles to
each other :
real + i * imag

electric field: (a1 * xhat + a2 * yhat + a3 * zhat)

magnetic field: (b1 * xhat + b2 * yhat + b3 * zhat)


instantaneous electric field: Re( EM(0) * exp( -jks ) )
instantaneous magnetic field: Imag( EM(0) * exp(-jks) )

where k (wave number ) = 2 * pi / wavelength
s = distance traveled in the direction of the plane wave
EM(0) is the EM field at reference point.

Here it seems the problem won't arise while multiplying EM(0) and exp(-jks) as both are complex numbers.

Is this correct ??
 
Last edited:

What is an electric field of a plane wave?

The electric field of a plane wave is a type of electromagnetic wave that is characterized by its electric field oscillating in a single plane perpendicular to the direction of wave propagation. It is a fundamental concept in electromagnetism and is used to describe various phenomena, such as light and radio waves.

How is the electric field of a plane wave calculated?

The electric field of a plane wave is calculated using the equation E = E0sin(kx - ωt), where E0 is the maximum amplitude of the electric field, k is the wave number, x is the distance from the source, ω is the angular frequency, and t is the time.

What is the relationship between the electric field of a plane wave and its frequency?

The electric field of a plane wave is directly proportional to its frequency. This means that as the frequency of the wave increases, the amplitude of the electric field also increases. This relationship is described by the equation E = hν, where h is Planck's constant and ν is the frequency.

How does the electric field of a plane wave behave at different angles of incidence?

The electric field of a plane wave behaves differently at different angles of incidence. When the wave is incident perpendicular to the surface, the electric field is parallel to the surface and has its maximum amplitude. As the angle of incidence increases, the amplitude of the electric field decreases and becomes perpendicular to the surface at an angle of 90 degrees.

What are some real-world applications of the electric field of a plane wave?

The electric field of a plane wave has many real-world applications, including in telecommunications, radar systems, and medical imaging. It is also used in various technologies, such as wireless charging and satellite communication. Understanding the behavior of the electric field of a plane wave is crucial in designing and optimizing these technologies.

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