I Could a Gaussian beam be described as parallel at a large distance?

AI Thread Summary
A Gaussian beam can be approximated as parallel at large distances, specifically when the propagation distance exceeds the Rayleigh distance, leading to negligible curvature of the wavefront. The discussion involves modeling the beam's electric field and simplifying it for analysis by treating the detected light as a beam with a waist radius corresponding to the detection area. The concept of phase error is mentioned, indicating that it becomes negligible in this context, although some participants seek clarification on its relevance to optics. The received power can be calculated based on the transmitter power and the areas involved, providing a way to relate the electric field strength to the beam's characteristics. Overall, the conversation emphasizes the need for further research and clarification on the approximation's validity.
Haorong Wu
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Hello, there. Suppose a Gaussian beam is sent and is received at a great large distance, i.e., the propagation distance ##z \gg z_R## the Rayleigh distance.

The Gaussian beam can be described by $$E_0 \frac {1}{w(z)} \exp \left ( \frac {-r^2}{w(z)^2}\right )\exp\left ( -i\left (kz+k\frac {r^2}{2R(z)}-\psi(z)\right )\right ) $$
where ##E_0## is the amplitude of the electric field, ##w_0## is the waist radius, ##w(z)## is the radius of the beam at ##z##, ##k## is the frequency, ##R(z)## is the radius of curvature of the beam's wavefronts at ##z##, and ##\psi(z)## is the Gouy phase.

When the distance ##z## is larger than the Rayleigh distance ##z_R##, the beam will diverge noticeably. In my scenario, the propagation distance ##z## is so large that the curvature of the wavefront will approach zero. In order to simplify my analysis, I would like to treat them as parallel light when received by a finite small detection area. And since the other part that is not detected is lost, I modeled the light to be as a beam with a waist radius ##w_d##, which is the radius of the detection area, such as $$E_d \frac {1}{w_d} \exp \left ( \frac {-r^2}{w_d^2}\right )\exp\left ( -ikz \right ) $$ where ## E_d## is the received amplitude, the curvature term vanishes and the Gouy phase can be harmlessly removed.

But my professor said that he is not sure whether this approximation is correct or not and asked me to find more related papers. However, I have searched in Google scholar for days without success. Could you help me with this analysis or share possible materials? Thanks in advance.
 
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In radio wave propagation beyond the Rayleigh Distance, we normally assume that phase error due to curvature is zero. Of course, all beams will diverge beyond the Rayleigh Distance.
 
Hi, @tech99. Thanks for your reply. I am not familiar with the concept of phase error. On some websites, it reads that phase err (phase error) is the phase difference between the I/Q reference signal and the I/Q measured signal, averaged over all symbol points. But this seems not to be related to optics. Could you refer me to some materials? Thanks!
 
I was just considering the phase error across a curved wavefront, which is negligible.
Not sure about the concept of using a waist diameter wd.
If you know the transmitter power, Pt, then received power Pr = Pt x detector area/beam area at distance z.
Received electric field strength = sqrt (377 x Pr)
If you don't know transmitter power but know Eo then you can find it from: Pt = waist area x Eo^2/377
 
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