Do lasers suffer R^2 propagation loss

[apchar]

The intensity (W/m^2) of an electromagnetic wave from an ordinary antenna decreases with the square of the distance from the emitter (in the far field.) Is the same true for a laser beam?

 

[cepheid]

No, the inverse square law only applies for an isotropic emitter, not for beamed emission. That having been said, intensity does decrease with distance, since a laser beam is always somewhat divergent.

 

[Born2bwire]

No, the inverse square law only applies for an isotropic emitter, not for beamed emission. That having been said, intensity does decrease with distance, since a laser beam is always somewhat divergent.

Not just isotropic sources, but any finite source will follow the inverse square law, including lasers. The intensity of the beam is inversely proportional to the square of the beam width. The beam width is proportional to the distance along the beam's axis. So it still works out to be inverse square. However, the beam width also has a constant term and it scales the distance along the axis by another reference value. So it takes an appreciable distance until the beam width behaves asymptotically as proportional to the distance.

This can be thought of in the same way as we treat the problem with antennas. Antennas follow the inverse square law when we are in the far field. At such distances the antenna looks like a directed point source. The far field region is thus dependent upon the physical size of the antenna and the wavelength of the radiation which indicates at which point the radiator electrically looks like a point source (typically the distance is something like 2D^2/\lambda where D is the largest dimension of the radiator). If we apply the same idea to a laser pointer we see that the far field only occurs at astronomical (figuratively) distances. A handheld laser pointer is what, a 1-2 cm in diameter and emits red light? So that means that the far field is on the order of 1 km away.

So we can see that while a laser beam will eventually fall prey to the inverse square law, the typical distances that the beam must first travel to do so are far greater than the distances over which they operate. As such, we do not consider this to be the over riding behavior of loss as we do with antennas which are generally designed to operate in the far field.

 

[apchar]

This can be thought of in the same way as we treat the problem with antennas. Antennas follow the inverse square law when we are in the far field. At such distances the antenna looks like a directed point source. The far field region is thus dependent upon the physical size of the antenna and the wavelength of the radiation which indicates at which point the radiator electrically looks like a point source (typically the distance is something like 2D^2/\lambda where D is the largest dimension of the radiator). If we apply the same idea to a laser pointer we see that the far field only occurs at astronomical (figuratively) distances. A handheld laser pointer is what, a 1-2 cm in diameter and emits red light? So that means that the far field is on the order of 1 km away.

But the dimensions of an antenna have to be comparable to the wavelength to be effective (D ~ lambda), making the point at which the far field begins a wavelength or two away. Is the same not true for lasers?

What about scattered laser light?

 

[Born2bwire]

But the dimensions of an antenna have to be comparable to the wavelength to be effective (D ~ lambda), making the point at which the far field begins a wavelength or two away. Is the same not true for lasers?

What about scattered laser light?

In general, the size of a single radiator of an antenna is on the order of a wavelength for efficiency purposes. But we can construct an array of antennas built out of such components that the overall size of the antenna is many wavelengths. The advantage being that an array allows us control over the beam width and angle. Just look at the Very Large Array. Wikipedia gives dimensions of 21 km and a wavelength of 7 mm. That means that the far field of the VLA is around 10^{11} km. That's only 842 Au so that's perfectly suitable for deep space observation. We can also note that the size of the dish is much much larger than the wavelength which is practical because the physics of a dish antenna is different from that of a wire antenna.

Lasers work by a different principle. Antennas are based off of classical electrodynamics and rely on the acceleration of charges. Lasers work off of quantum mechanics and make use of the energy states of atoms and molecules to emit light. We restrict the size of an individual radiator (generally a wire antenna) in a classical antenna to that of the wavelength because for the radiator to be efficient we need the excited currents to be coherent. If we have a simple wire antenna that is multiple wavelengths long, then there are sections of currents on the antenna which will be 180 degrees out of phase from each other which causes cancellation. So in the far field these cancellations will reduce the propagated power. In a laser, we can generate coherent light by virtue of how we excite and stimulate the emission from the atoms of the active media. We do not experience the same limits on the size of the active media because of this. As such, the electrical size of the radiator of a laser can be very very large compared to the wavelength of the radiated light.

As for scattered light, it's behavior is no different. It's only a question of how far the light has traveled and how well it has preserved its Gaussian beam.

 

[Phrak]

The intensity (W/m^2) of an electromagnetic wave from an ordinary antenna decreases with the square of the distance from the emitter (in the far field.) Is the same true for a laser beam?

Ideally, a collimated Bessel light beam is non-diffractive, and therefore not divergent over distance in vacuum. This is also true of a phased array of antenna. The aperture is also of infinite diameter (the complete Bessel function), so it is only approximated. An antenna array is of discrete sources, even with smoothing techniques, is already an approximation to a Bessel function, and at a disadvantage already.

 

[Andy Resnick]

It's a little tricky even for the fundamental Gaussian mode, where:

[tex] I(r,z) = \frac{P}{\pi/2 \omega^{2}(z)} exp(-\frac{2r^{2}}{\omega^{2}(z)}) [/tex]

where I is the intensity, P the power, and [itex]\omega[/itex] the beam radius.

The on-axis intensity (r = 0) falls off as 1/z^2 for z>> [itex]z_{R}[/itex], where [itex]z_{R}[/itex] is the Rayleigh length. Asymptotically, the beam radius is a linear function of the propagation distance, and so the power/area falls off as 1/z^2.

I should mention that what I have been calling 'intensity' is in fact *irradiance*- power/area- intensity is power/solid angle, but the term 'intensity' is often used instead of irradiance.

 

[sophiecentaur]

Ideally, a collimated Bessel light beam is non-diffractive, and therefore not divergent over distance in vacuum. This is also true of a phased array of antenna. The aperture is also of infinite diameter (the complete Bessel function), so it is only approximated. An antenna array is of discrete sources, even with smoothing techniques, is already an approximation to a Bessel function, and at a disadvantage already.

In a phased array antenna (or a laser, surely), there will be a distance, beyond which the spreading loss is inverse square. How can there not be an effective sinx/x type diffraction pattern to the emerging beam?

 

[Phrak]

You're right, of course. No matter how careful you are, there is still a far field.
The sinx/x relation of obtained for a beam intensity constant in phase and amplitude across the aperture, as you know. In answer to the question, "what beam profile at the aperture is non-dispersive?" any of these curves is the solution for radial intensity of the source.

First order Bessel functions. Also see Wikipedia Bessel beam.

 

[Redbelly98]

Bessel beams require an infinite amount of power, and infinitely large aperture, and so cannot be produced in reality. I could say, tongue-in-cheekly, that the power and aperture size of actual lasers is considerably smaller than the quantities I mentioned in the previous sentence

As born2bwire said in Post #2, actual laser beams will diverge, due to diffraction, after a distance comparable to D²/λ (or less). Laser beam divergence is often measured in milliradians, so it is small but nonetheless the irradiance does diminish as 1/r².

 

[Phrak]

Yes, of course, Redbelly.

It's very disappointing that the Bessel functions don't decay faster so to better be approximated over a finite aperture. However, notice that J₁ and J₂ are nearly out of phase in the tails. Does this persist? I wounder if J₁+J₂ is not also a non-dispersive solution, and so far better approximated over an aperture of the same radius used for either J₁ or J₂, alone.

 

[sophiecentaur]

I suppose it should be pointed out that 1/r2 loss has nothing to do with the directivity pattern. The source needn't be isotropic. It's just that the d=0 point may not be within the source.

 

[Claude Bile]

Not just isotropic sources, but any finite source will follow the inverse square law, including lasers. The intensity of the beam is inversely proportional to the square of the beam width. The beam width is proportional to the distance along the beam's axis. So it still works out to be inverse square. However, the beam width also has a constant term and it scales the distance along the axis by another reference value. So it takes an appreciable distance until the beam width behaves asymptotically as proportional to the distance.

This is not true for an elliptical beam.

Elliptical beams possesses two rates of divergence, one rate in the orientation of the major axis, and the other in the orientation of the minor axis. The area does not scale as r^2, but as a.b, where a and b are the major and minor axes of the ellipse, meaning the irradiance of an elliptical beam can scale anywhere from r to r^2, depending on the ellipticity of the beam.

Claude.

 

[Born2bwire]

This is not true for an elliptical beam.

Elliptical beams possesses two rates of divergence, one rate in the orientation of the major axis, and the other in the orientation of the minor axis. The area does not scale as r^2, but as a.b, where a and b are the major and minor axes of the ellipse, meaning the irradiance of an elliptical beam can scale anywhere from r to r^2, depending on the ellipticity of the beam.

Claude.

Are there any examples of laser beams that propagate as elliptical beams? I have only ever seen the beams be approximated as Gaussian.

 

[Andy Resnick]

Are there any examples of laser beams that propagate as elliptical beams? I have only ever seen the beams be approximated as Gaussian.

Diodes. Diode lasers- they have to be 'circularized' in order to couple to a fiber.

 

[sophiecentaur]

This is not true for an elliptical beam.

Elliptical beams possesses two rates of divergence, one rate in the orientation of the major axis, and the other in the orientation of the minor axis. The area does not scale as r^2, but as a.b, where a and b are the major and minor axes of the ellipse, meaning the irradiance of an elliptical beam can scale anywhere from r to r^2, depending on the ellipticity of the beam.

Claude.

If you think this then, perhaps, you could tell us how a particular part of the beam 'knows' which part it is and can know 'how to diverge'. What you say implies that putting an elliptical mask at some point to restrict an isotropic beam will suddenly affect parts of the beam that don't actually touch the sides.

It is true that the apparent source for some beams may not be a point but, at a sufficient distance, (far field) the finite extent of a source becomes a point and then you can't avoid the inverse square from kicking in. Even a laser or an array of phased dipoles can't avoid behaving that way although the apparent point of origin of the wave may be some way behind the laser.

 

[Andy Resnick]

If you think this then, perhaps, you could tell us how a particular part of the beam 'knows' which part it is and can know 'how to diverge'.

That's easy- pass it through a shaped aperture.

 

[sophiecentaur]

How is it that the distrubited mass of astronomical bodies doesn't promote the same 'inverse square failure of the gravitational field' worries? It is precisely the same geometrical situation. If people insist that lasers are, somehow, different then so are filament lightbulbs and stars. Get far enough away an we see the inverse square. Take the Solar System. Does that exhibit the inverse square law whilst you're using slingshot orbits? Likewise, one wouldn't expect the close behaviour of a laser to follow it.

 

[sophiecentaur]

That's easy- pass it through a shaped aperture.

If you introduce an aperture then that becomes a secondary source (it diffraction pattern modifies the original wave. If it's a large aperture (any shape you like) you won't see any difference. If it's a tiny aperture then its diffraction pattern will dominate and you then, effectively, have another radiating source and the ISL will start to apply from there onwards.

Another point: if you subscribe to reciprocity then you would agree that putting your aperture right next to your receiver should produce the same measurement of energy flux as putting the aperture right in front of the source. So, by your argument, as you vary the distance of a 'directive' measuring device from an isotropic source, you would also experience non-ISL behaviour. That is hard to credit and I don't think it agrees with anyone's experience. It would, for instance, imply that different telescopes would see a different brightness variation for stars at different distances.

Are you just considering near field in your model? That would, of course, make things different.

 

[Andy Resnick]

How is it that the distrubited mass of astronomical bodies doesn't promote the same 'inverse square failure of the gravitational field' worries? It is precisely the same geometrical situation. If people insist that lasers are, somehow, different then so are filament lightbulbs and stars. Get far enough away an we see the inverse square. Take the Solar System. Does that exhibit the inverse square law whilst you're using slingshot orbits? Likewise, one wouldn't expect the close behaviour of a laser to follow it.

Nobody (well, me at least) is disputing the limit z -> inf that a source behaves as a point source. I even showed this explicitly for a Gaussian beam in this thread. As a practical matter, if the Rayleigh length is many kilometers, making measurements in a lab will show deviations from the inverse-square law.

 

[Andy Resnick]

If you introduce an aperture then that becomes a secondary source (it diffraction pattern modifies the original wave. If it's a large aperture (any shape you like) you won't see any difference. If it's a tiny aperture then its diffraction pattern will dominate and you then, effectively, have another radiating source and the ISL will start to apply from there onwards.
.

I'm not sure what you are getting at- you asked how light 'knows' how much to diffract, given the existence of an elliptical beam. Non-axisymmetric apertures produce non-axisymmetric beams, even in the far field: Fraunhofer diffraction pattern.

 

[sophiecentaur]

That's easy- pass it through a shaped aperture.

Perhaps I didn't get what you meant here. I was just commenting that a large aperrture would make little difference and a small one would behave as a new source (pinhole with a sinx/x type beam).

 

[sophiecentaur]

I'm not sure what you are getting at- you asked how light 'knows' how much to diffract, given the existence of an elliptical beam. Non-axisymmetric apertures produce non-axisymmetric beams, even in the far field: Fraunhofer diffraction pattern.

I don't see how the asymmetry of a beam, formed as a result of some diffraction process (i.e. any optics you care to mention) can affect the spread at great distances. If each infinitesimal / omnidirectional contribution follows the ISL then the far field resultant will also follow the ISL (the patterns are multiplicative, aren't they?) Things could only depart from this if the light were to bend on its journey through free space.

I understand that lasers behave quantitatively differently from other radiators because of their relatively large extent (in wavelengths) but there's nothing, in principle, between a laser and an array of dipole radiators, fed from the same source. The spreading loss of such an array is always calculated on the grounds that the ISL applies. This is in no way affected by the actual radiation pattern of the array. The two factors just multiply to find the final signal level.

Or are you talking about something entirely different?

 

[Andy Resnick]

I don't see how the asymmetry of a beam, formed as a result of some diffraction process (i.e. any optics you care to mention) can affect the spread at great distances. If each infinitesimal / omnidirectional contribution follows the ISL then the far field resultant will also follow the ISL (the patterns are multiplicative, aren't they?) Things could only depart from this if the light were to bend on its journey through free space.

I understand that lasers behave quantitatively differently from other radiators because of their relatively large extent (in wavelengths) but there's nothing, in principle, between a laser and an array of dipole radiators, fed from the same source. The spreading loss of such an array is always calculated on the grounds that the ISL applies. This is in no way affected by the actual radiation pattern of the array. The two factors just multiply to find the final signal level.

Or are you talking about something entirely different?

Perhaps it's worthwhile clearly stating what the inverse-square law is: the power per unit area (the irradiance) scales as 1/r^2, where r is the distance from the source to the observation location.

I wonder if the confusion stems from the fact that the intensity (power per steradian) is not isotropic for non-point sources, or that intensity is not irradiance.

In any case, given a diffracting wavefront, the intensity is constant with propagation distance- the diffraction pattern is specified in terms of *angles* for this reason. The irradiance, however, varies as 1/r^2 because the diffraction pattern grows ever larger in spatial extent.

 

[Redbelly98]

Since I know from your prior posts that you're both knowledgeable about this stuff, I am wondering if there is some semantic issue underlying this discussion. Perhaps the intensity-vs.-irradiance definition Andy mentioned.

I think we all agree that, once you're a far enough distance from the source, the 1/r² behavior of the power-per-area holds. But #1: there are situations where it is not practical to be "far enough" away. But #2: the "far enough" distance can be different for different transverse dimensions of the beam.

 

[sophiecentaur]

Yes. I don't think we can really be disagreeing about anything this basic.
It's just that there is a fundamental difference between Energy Flux Density and Energy per Steradian (Which actually has the ISL built into it). If you're a comms Engineer then the Energy per Steradian is of no interest at all What counts is the Power getting to the 'receiving' area of your distant receiving antenna. That was where I was coming from and 'my' distances, angles and suchlike are not the same as for someone who is concerned with the shape of a focused laser beam.