(Bessel Beam) Wave vector question

[kire]

Hello, first post on these forums... I read up on the rules and such but please excuse me if i missed something.

I am making a simulation of a bessel beam (non-diffracting and self healing) in MatLab but I do not know how to define the wave vector. More specifically, the transverse (k_T) and the propagation constant (k_z). If i can find one of them, I can solve for the other.

The "governing" equation I am using is:

U(r)=J₀(k_T*rho)*exp(-j*k_z*z)

J₀ is the zero order Bessel function of the first kind, rho is just sqrt(x^2+y^2), j is imaginary number, z is distance along the z-axis.

The relationship between the wave vectors is:

k_T²+k_z²=k²

Now, a fellow labmate initially had told me that for a gaussian beam, you usually set z=0 and solve for k_z because at the z=0 plane, all the wave vectors point only in the z-direction and thus k_T=0. However this will not work because if z=0 and k_T=0 then the equation above yields a plane wave.




This is just a summary of my problem . If you need more information, I will be happy to provide it.

 

[AndyW]

Hello, thank you for your post and welcome to the forum! It seems like you are on the right track with your simulation of a bessel beam. The wave vector, also known as the propagation vector, is a crucial component in determining the behavior of a wave. In your equation, kT represents the transverse wave vector, which is perpendicular to the direction of propagation (z-axis). On the other hand, kz represents the longitudinal wave vector, which is parallel to the direction of propagation.

To define the wave vector, you can use the relationship you mentioned, kT2 + kz2 = k2, where k is the magnitude of the wave vector. This equation is derived from the Pythagorean theorem, where kT and kz represent the sides of a right triangle and k represents the hypotenuse.

In order to solve for the wave vector components, you will need to have some information about your system. For example, the wavelength of your bessel beam can help determine the magnitude of the wave vector. The angle of incidence and refractive indices of the materials involved can also provide valuable information in solving for the wave vector components.

It is important to note that in your equation, the value of kT can never be equal to 0, as this would result in a plane wave, as you mentioned. This is because a bessel beam is a non-diffracting wave, meaning it does not spread out as it propagates. Therefore, the transverse wave vector must have a non-zero value to maintain this behavior.

I hope this helps you in defining your wave vector and solving for the transverse and longitudinal components. If you need further assistance, please provide more information about your system and I will be happy to help. Good luck with your simulation!

 

[sir-pinski]

Hello, welcome to the forums!

Defining the wave vector for a Bessel beam can be a bit tricky, as it involves both transverse and longitudinal components. In general, the wave vector can be written as k = (kT, kz), where kT is the transverse component and kz is the longitudinal component. In your equation, J0(kT*rho) represents the transverse component and exp(-j*kz*z) represents the longitudinal component.

In order to find the wave vector, you can use the relationship you mentioned: kT^2 + kz^2 = k^2. This relationship is true for all types of waves, not just Gaussian beams. However, for a Bessel beam, the transverse component is given by kT = m/ρ, where m is the order of the Bessel function and ρ is the radial distance from the beam axis. So, in your equation, you can set kT = m/ρ and solve for kz. This will give you the longitudinal component of the wave vector.

I hope this helps with your simulation. If you have any further questions, please feel free to ask. Best of luck with your project!