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Understanding the Sommerfeld radiation condition?

Physics Asked on November 28, 2021

I assumed the Sommerfeld radiation condition which is satisfied by radiating solutions to the Helmholtz equation ensured that a wave decays to zero as the wave becomes infinitely far from the source. However, a colleague says that the Sommerfeld radiation condition doesn’t imply that the wave decays to zero at infinity, instead he says it actually means that a radiating wave decays to a plane wave at infinity.

$$biggvertfrac{partial u}{partial vert x vert} – ik ubiggvert le C frac{1}{vert xvert^{d+1/2}}$$

  1. So what is the true meaning of the Sommerfeld radiation condition, does it imply a wave decays to zero, or a radiating wave decays to a plane wave, or something else?
  2. What is the LHS of the equation saying? It doesn’t very intuitive what with the multiplication of $u$ by $ik$?

2 Answers

Both you and your friend are partly right. (I was thinking as your friend just 20 min ago)

Let me rewrite Sommerfeld's radiating condition for the 2D case (just as an example, it works the same way in 3D).

$$lim_{|mathbf{x}|toinfty}sqrt{|mathbf{x}|}left(frac{partial u}{partial|mathbf{x}|}-ikuright)=0quadforalltheta$$

that is, the limit has to be $0$ when going to infinity in every direction.

As you say, if the function $u$ is smooth and goes to zero then the limit is fulfilled.

On the other hand your friend may have thought that the term between parentheses is null in the case of a plane wave. This is not completely true.

In 2D a plane wave takes the form: $$uleft(mathbf{x}right)=Ae^{ikmathbf{d}cdotmathbf{x}}$$

being $A$ the amplitude of the wave, $k$ the wave number (the same as in the Helmholtz equation), $i$ the unit imaginary number, and $mathbf{d}$ the unit vector in the direction of propagation of the wave, that is, perpendicular to the wavefront.

If we meassure our angles with respect to $mathbf{d}$ then the plane wave can also be represented as:

$$uleft(mathbf{x}right)=Ae^{ik|mathbf{x}|cos(theta)}$$

where we only evaluated the scalar product.

The derivative with respect to $|mathbf{x}|$ is then:

$$frac{partial u}{partial|mathbf{x}|}=Aikcos(theta)e^{ik|mathbf{x}|cos(theta)}=ikcos(theta)uleft(mathbf{x}right)$$

and we see that this only satisfies Sommerfeld's radiation condition when $theta=0$.

So... to sumarize:

Does a plane wave satisfy Sommerfeld's radiation condition?

No, it must be satisfied in all directions and it only satisfies it in the direction of propagation of the wave.

What does de $iku$ term means?

It is what you get when you differentiate a plane wave with respecto to $|mathbf{x}|$ along its propagation direction.

What is the meaning of Sommerfield's radiation condition?

Sommerfeld's radiation conditions only allows waves to radiate energy towards infinity (outgoing waves) but not the infinity to radiate back. In the case of a plane wave the propagation of energy goes in only one direction in the whole plane, and with no attenuation (this is not very physical). It means that depending which direction are you looking at the energy is going towards infinity or coming from infinity.

The condition says that in every direction in space, the wave has to tend to a plane wave propagating in that direction, and the difference between the actual wave and a plane wave propagatin in that direction has to decrease faster than $sqrt{|mathbf{x}|}$.

To correct you, if a function $u$ decreases faster than $sqrt{|mathbf{x}|}$ in all directions, then this function will satisfy Sommerfeld's radiation condition, but this condition allow for more general cases.

Hope everything is clearer now.

Answered by Manuel Pena on November 28, 2021

So what is the true meaning of the Sommerfeld radiation condition, does it imply a wave decays to zero, or a radiating wave decays to a plane wave, or something else.

From the quote by Sommerfeld below, it seems to me that "scatter to infinity" can be taken as a wave decaying to zero.

From Wikipedia Sommerfeld Radiation,

Arnold Sommerfeld defined the condition of radiation for a scalar field satisfying the Helmholtz equation as

"the sources must be sources, not sinks of energy. The energy which is radiated from the sources must scatter to infinity; no energy may be radiated from infinity into ... the field."

Mathematically, consider the inhomogeneous Helmholtz equation${displaystyle (nabla ^{2}+k^{2})u=-f{mbox{ in }}mathbb {R} ^{n}}$where ${displaystyle n=2,3}$ is the dimension of the space, ${displaystyle f}$ is a given function with compact support representing a bounded source of energy, and ${displaystyle k>0}$ is a constant, called the wavenumber.

A solution ${displaystyle u}$  to this equation is called radiating if it satisfies theSommerfeld radiation condition$${displaystyle lim _{|x|to infty }|x|^{frac {n-1}{2}}left({frac {partial }{partial |x|}}-ikright)u(x)=0}{displaystyle {hat {x}}={frac {x}{|x|}}}$$(above, ${displaystyle i}$ is the imaginary unit and ${displaystyle |cdot |}$ is the Euclidean norm). Here, it is assumed that the time-harmonic field is ${displaystyle e^{-iomega t}u.}$ If the time-harmonic field is instead ${displaystyle e^{iomega t}u,}$ one should replace ${displaystyle -i}$ with ${displaystyle +i}$ in the Sommerfeld radiation condition.

For your second point:

What is the LHS of the equation saying? It doesn't very intuitive what with the multiplication of u by ik?

From Eighty Years of Sommerfeld's Radiation Condition - CiteSeerX (There is a PDF associated with this file, which, for some reason I cannot link directly to). My apologies for this, but the PDF does provide a possible rational for the formulation of the equation, i. e. in order to remove unwanted unphysical solutions.

One of the difficulties with formulating a wave propagation problem in this way is that the solution may not be unique. Besides the expected outgoing waves which result when the incident wave is scattered by the object, the mathematical solution also provides incoming waves which originate at infinity and move towards the object. These incoming waves are physically meaningless and must be rejected by some criterion built into the mathematical formulation of the problem. Sommerfeld was the first to state a mathematically precise and easily applicable condition which, when added to exterior boundary value problems for the Helmholtz equa- tion, ensures a unique solution. This condition is applied at infinity and for three- dimensional problems requires that the solution u satisfy $Y= Vx’ + y= + 22, i=CT$, uniformly with respect to all directions in which the limit is approached.

In the (very likely) case that there are subtle aspects to this that are well beyond me, I would ask that you google the above reference and download the PDF.

The Sommerfeld radiation condition is used to solve uniquely the Helmholtz equation. For example, consider the problem of radiation due to a point source ${displaystyle x_{0}}$ in three dimensions, so the function ${displaystyle f}$ in the Helmholtz equation is ${displaystyle f(x)=delta (x-x_{0}),}$ where ${displaystyle delta }$ is the Dirac delta function. This problem has an infinite number of solutions, for example, any function of the form

${displaystyle u=cu_{+}+(1-c)u_{-},}$w here ${displaystyle c}$ is a constant, and ${displaystyle u_{pm }(x)={frac {e^{pm ik|x-x_{0}|}}{4pi |x-x_{0}|}}.}$

Of all these solutions, only ${displaystyle u_{+}}$ satisfies the Sommerfeld radiation condition and corresponds to a field radiating from ${displaystyle x_{0}.}$. The other solutions are unphysical. For example, ${displaystyle u_{-}}$ can be interpreted as energy coming from infinity and sinking at ${displaystyle x_{0}.}$

Answered by user108787 on November 28, 2021

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