# Wave energy dissipated due to geometry

• A
miraboreasu
Hey

Condition 1:
A 2D infinite plane and there is a circular hole in the middle. When t=0, an impulsive loading, P=f(t), is applied to the boundary of the circle(outward), so the wave will start at the boundary of the circle and propagate in the plane

Condition 2:
A 3D infinite plane and there is a circular hole in the middle. When t=0, an impulsive loading, P=f(t), is applied to the boundary of the sphere(outward), so the wave will start at the boundary of the sphere and propagate in space

During the wave propagation in the plane, assuming there is dissipation (actually I don't know what caused it, thermal maybe, but I think in the world, waves cannot travel endlessly, and I only need to know from the geometry perspective). How to model the energy dissipated regarding the geometry? Basically, the modeling for the energy dissipated during 2D propagation in a plane and 3D spherical propagation?

Thanks

Gold Member
2022 Award
but I think in the world, waves cannot travel endlessly
uh, no, EM waves (light for example) can travel endlessly (assuming they don't run into anything). I think you probably mean that MECHANICAL waves can't travel endlessly.

In 2D the energy density falls in proportion to the inverse of the radius.
In 3D the energy density falls in proportion to the inverse of the radius squared.

• tech99
Gold Member
Further to post #3, the material has a finite thickness, so the initial intensity can be found. For 2D the energy is then distributed over the surface of a cylinder of increasing radius, giving a 1/2 pi H R dependency for the intensity. For the 3D case, the energy is distributed over the surface of a sphere of increasing radius, giving a 1/4 pi R^2 dependency. For the case of dissipated energy, this falls exponentially with distance, so in the case of microwaves passing through water vapour, for instance, we would simply apply a factor in decibels per metre.

miraboreasu
In 2D the energy density falls in proportion to the inverse of the radius.
In 3D the energy density falls in proportion to the inverse of the radius squared.
Thanks, can you please provide more materials, so I can take look at the derivation? I think this is coming from area and volume, but I want to see the details,

miraboreasu
Further to post #3, the material has a finite thickness, so the initial intensity can be found. For 2D the energy is then distributed over the surface of a cylinder of increasing radius, giving a 1/2 pi H R dependency for the intensity. For the 3D case, the energy is distributed over the surface of a sphere of increasing radius, giving a 1/4 pi R^2 dependency. For the case of dissipated energy, this falls exponentially with distance, so in the case of microwaves passing through water vapour, for instance, we would simply apply a factor in decibels per

Thanks, can you please provide more materials, so I can take look at the derivation? I think this is coming from area and volume, but I want to see the details,
As the wave energy radiates, it is distributed over an ever-increasing line in 2D, or area in 3D.

As the radius of a circle increases, so does the circumference.
C = 2π · r

As the radius of a sphere increases, the surface area increases as the square of the radius.
A = 4π · r²

miraboreasu
uh, no, EM waves (light for example) can travel endlessly (assuming they don't run into anything). I think you probably mean that MECHANICAL waves can't travel endlessly.
Yes, sorry I miss this part, assume the mechanical wave in a 2D and 3D elasticity pane and space

miraboreasu
As the wave energy radiates, it is distributed over an ever-increasing line in 2D, or area in 3D.

As the radius of a circle increases, so does the circumference.
C = 2π · r

As the radius of a sphere increases, the surface area increases as the square of the radius.
A = 4π · r²
Thanks, is there any equation or material showing how C and A in the energy dissipation term? Like the whole form.

Homework Helper
Gold Member
2022 Award
Hi @miraboreasu. There may be a language problem causing some confusion. Maybe this will help...

Dissipation = the wave's energy being converted to other forms of energy (usually heat).

For example, for an EM wave traveling through free (empty) space, there is no dissipation; the total energy in the wave remains constant. When the wave spreads out as it travels, its energy gets ‘spread out’ - but the total wave energy remains the same.

In simple geometry, a 2D plane is a flat surface (e.g. a table top) of zero thickness. (There are other meanings at more advanced levels.)

In simple geometry 3D, means ordinary space. (There are other meanings at more advanced levels.) Therefore we don’t refer to a ‘3D plane’ (unless maybe you are a mathematician working in 4 or more dimensions!).

A wave in 2D would be confined to a surface. The nearest real-life example I can think of is dropping a stone in a pond giving waves which spread outwards on the water’s surface.

A wave in 3D occupies space – e.g. sound waves when you speak, with the energy spreading out in all possible directions.

For a brief intuition about the inverse square law, you could read ‘The fable of the butter gun’ here: https://www.khanacademy.org/science...ce-and-electric-field/a/ee-inverse-square-law.

Edit: you might find it useful to do a little research of your own on the 'intensity' of a wave.

• sophiecentaur, miraboreasu and vanhees71