What causes Free Space Path Loss?
Last Post: April 5:

We know that the RF signal will attenuate as radiated from the transmitter antenna even if no obstacles exist on the path. This is called the Free Space Path Loss (FSPL), or Free Path Loss.
Someone may consider the distance itself as the reason that causes the FSPL, but this is incorrect. According to a Cisco document, two reasons actually cause the loss:
(1) The sender is one point, and the signal is sent around itself. The energy has to be distributed over a larger area (a larger circle) , but the amount of energy originally sent does not change. Therefore, the amount of energy available on each point of the circle is higher if the circle is small (less points) than large (more points among which the energy has to be divided).
(2) The receiver is not exactly one point, and will receive an amount of energy depending on its size: a large antenna will collect more points of the circle than a small one. But the antenna will never be able to pick up more than a portion of the signal originally sent; the rest of the energy sent will be lost.
Below is my assumption:
If RF energy could be radiated towards an exactly single direction instead of omni direction, and if the receiver could fully catch the signal sent, there would be no any FSPL at any distance.
Not quite sure whether what I think is correct. 
Hi cwnpchina
A lot of what you have said is basically correct. Free Space Path Loss comes from one thing and one thing only  the physical spreading of the RF energy as the signal moves through space. That is, free space with no obstacles.
Let's imagine a perfect omnidirectional antenna with one watt of physical, electrical power pumped into the antenna. Let's forget about EIRP for just now and just concentrate on that one watt of power. Let's also imagine that we have a perfectly efficient antenna [ again, no such thing exists in real life ].
Now let's imagine that we take the whole assembly up into space well away from any planets [ we actually get a very small amount of attenuation and reflection from spacedust, but we'll ignore all of that.] We now put on special glasses that allow us to see the radiated signal every 10 meters.
Firstly, we see a beautiful sphere of energy at 10 meters from the antenna. Smooth and round [ in actual fact we would see wavelets as defined in Huygen's principles, but we'll ignore that ].
Now, the watt is the unit which measures the RATE at which energy is transferred. So, in one second, one joule of energy is transferred. This sphere contains one joule of energy.
We now, wait a little time and look at the sphere at 20 meters from the source. The surface area has gone up by four times, since the surface area of a sphere is given by 4pi [ R squared ]. The ratio of area 2 to area 1 is 4pi [ 20 squared ] to 4pi [ 10 squared ] = [ 20 squared ] / [ 10 squared ] = 4. Now 10 log 4 = 6 [ approximately ]. If you've ever wondered on the CWNA books why it says that when we double the distance, the loss is 6 dB, there is the reason.
Now this sphere of energy [ where the energy is over the surface of the sphere not inside in the volume ] will just keep expanding forever. [ In fact most TV signals from earth thirty years ago are still expanding across the universe !!...albeit at tiny levels. Some of the signal, as well as going to TV receivers, ends up passing through the atmosphere. Some military satellites operate at VHF and UHF. It's the HF signals [ 330 Mhz] that mostly get 'bent'? or refracted by the ionosphere ].
So if you developed a special massive spherical antenna to wrap around everything you could capture all of that expanding signal. Assuming that the receive system had no losses.
So, that's the wonderful world of theory. What happens in real life ? Omnidirectional antennas do not radiate equally in all direction, but some do a pretty good job. So, if you had a small antenna to receive with, you could capture a portion of that energy.
Now, what about your system at the end of your post ?
If it were possible to make a radio signal almost like a laser beam, and focus it exactly so that you could capture the exact amount of energy that was launched, you would indeed have a system with zero path loss. Congratulations on working that out bye the way. Very, very few people ever get to the stage of understanding that. They get too focused on the formula. But the formula says.... But the formula assumes that there is spreading of energy, and unfortunately and fortunately at the same time, real systems do spread energy. Even a highly directional antenna like a parabolic spreads energy in it's main beam, so it does indeed have free space path loss. With every mathematical formula, we have to look at it's physical origins whenever we can to get a 'feel'? for what it is telling us.
There are very real plans for power stations in space, where solar energy would be converted into microwave energy and beamed down to earth in very narrow beams [ the larger the antenna, the narrower the beamwidth ]. The safety issues however could be really horrific.
http://www.youtube.com/watch?v=rXlEqhCkxbM
http://www.youtube.com/watch?v=iIXI_plAkk&feature=related
http://www.youtube.com/watch?v=SxZKqhDQujk&feature=fvwhttp://cleantech.com/news/4361/solarensplanouterspace
http://en.wikipedia.org/wiki/Spacebased_solar_power
http://photos.state.gov/libraries/usinfophoto/39/week_4_0807/082007suntower500.jpg
http://spaceset.org/g.micropower.jpgLook at the power values in the last one !!
In satellite stations and large radio systems, we use a waveguide, where we physically prevent the signal from expanding and "bounce"? it down a narrow, special pipe. Losses [ from reflection on the copper surface etc ] can be quite low, and you can get a large percentage of power transfer from one end to the other for moderate distances.
http://images.google.com/imgres?imgurl=http://www.vias.org/wirelessnetw/img/wndwprint_img_38.png&imgrefurl=http://www.vias.org/wirelessnetw/wndw_06_03.html&usg=__eIzFZsN3UUjVV5FM9JXq77mH9zo=&h=256&w=443&sz=16&hl=en&start=6&um=1&itbs=1&tbnid=jPBC1ls_fg3qFM:&tbnh=73&tbnw=127&prev=/images%3Fq%3Dwaveguides%26hl%3Den%26rlz%3D1T4ADBS_enUS249US249%26sa%3DN%26um%3D1
So, to summarize, if we could indeed transmit a beam that never spread and had a perfect receiving system, we would indeed have a system with zero free space loss.
Alas, it lives next door to the salesman's comment of "600 Mbps of .11n 24/7". Look, it's in the spec!!
Dave

Thanks, Dave. Pretty good explanation. I've learned a lot from your words every time.
A supplement to Dave's post.
To exactly calculate the Free Space Path Loss value, the formula below can be used:
FSPL = 32.44 + 20log(d) + 20long(f)
Where FSPL is the Free Space Path Loss (in dB), d is the distance from transmitter antenna (in km), and f is the frequency of RF signal (in MHz).
Assuming a 2.4 GHz AP is transmitting, using the formula above, we could determine the FSPL at the distance of 100 m and 200 m from the AP, respectively:
FSPL (100 m) = 32.44 + 20log(0.1) + 20log(2400) = 80.1 dB
FSPL (200 m) = 32.44 + 20log(0.2) + 20log(2400) = 86.1 dB
The 6 dB Rule also applies here.
Distance doubles > Attenuation increases by 6 dB > EIRP decreases by 6 dB
Distance halves > Attenuation decreases by 6 dB > EIRP increases by 6 dB 
A few notes about the "constants"? in the FSPL equation. This stuff is pretty horrible, and should be avoided at all costs unless you have insomnia, but it shows where the numbers come from. The equation gives us:
FSPL = 32.4 + 20log f + 20 log d
Where f = frequency in Ghz
d = distance in metersThis comes from the following derivation:
The power flux density [ power per square meter at the receiving antenna ] is given by:
PFD = EIRP/4pi d [ squared ]We now imagine a perfect receiving antenna whose gain is given by:
G = 4Pi A/lamda [ squared ]
A = area of the antenna [ parabolic assumed ]Rearranging:
A = G lamda [ squared ]/4Pi
Lamda = wavelength of the antenna [ meters ].
The received power = PFD x area of antenna
= EIRP/4Pi D [squared ] x G lamda [ squared ]/4Pi
= EIRP X G X lamda [ squared ]/4 Pi D [ all squared ]
= EIRP x G X [ lamda/4Pi D ] squared
Putting it all in logs, and using negative logs?:
= 10 log EIRP + 10 log G 10 log [4Pi D/ lamda] squared
They now say the receive power = EIRP + G Losses
Losses = 10 log [4Pi D/ lamda] squared
Lamda = c/f where c = speed of light = 3 X 10 to the 8 m/s
Substituting gives us:
Losses = 32.4 + 20 log d + 20 log f
When we transmit a fixed frequency, 32.4 + 20 log f will be a fixed constant. However, 20 log d will change as distance changes.
So the frequency factor really is related to the change in gain of an ideal antenna as frequency varies. This is not easy to visualize at all.
Mathematically, that is how the formula comes out: for a fixed frequency signal, a constant plus a varying component.
The constant stays fixed when the frequency does not vary, but the varying component changes with distance [ spreading ].
For two different frequency signals, they will both have different constants, but the 6dB per doubling of distance factor will remain the same for both.
Dave

Just found this on the good old Wiki:
http://en.wikipedia.org/wiki/Freespace_path_loss
It also has the bizzare but true statement under Physical Explanation
The FSPL expression above often leads to the erroneous belief that free space attenuates an electromagnetic wave according to its frequency. This is not the case, as there is no physical mechanism that could cause this.?
Dave

Hi Dave, just a typo:
FSPL = 32.4 + 20log f + 20 log d
Where f = frequency in Ghz
d = distance in meters
In the formula above, the unit of distance d will be kilometer instead of meter.
But for the formula FSPL = (4*Pi*d / lamda) [squared], we need to use meter, not kilometer, to calculate the loss.
:) 
A little more mathematics about 6 dB Rule: Why is it always 6 dB?
As Dave states, if the signal frequency keeps constant, the FSPL value would only be related to the distance. Now let's demonstrate the following truth: The difference between FSPL(2d) and FSPL(d) is always 6 dB.
Since both 32.44 and the signal frequency are constants, only the elements of 20log(2d) and 20log(d) will be considered.
FSPL(2d)  FSPL(d)
= 20log(2d)  20log(d)
= 20[log(2) + log(d)]  20log(d)
= 20log(2) + 20log(d)  20log(d)
= 20log(2) < constant
= 6.02 dB 
Hi cwnpchina
Yes, that d should have been in km not meters. Typed that very late last night.
Dave 
As a summary:
If we have a perfect isotropic radiator in perfect space, we never have any loss !! The energy keeps spreading out forever. If you were to gather the energy over the surface of the sphere at any distance from the source, you would always get the exact same amount as was transmitted in the first instance [ conservation of energy due to no attenuation ]. Imagine you blow up a red balloon to six inches in diameter and you make a perfect little circle of diameter one inch on the surface. You try to see inside the balloon, but you cannot. The density of ballon material [ particles per square inch ] is too high to allow light to pass through. Now you blow it up to 24 inches in diameter. The little circle you marked has now grown bigger. We need to keep our measuring method constant, so we now make another one inch diameter circle within the original one. We can now start to see inside the balloon as the material has become stretched and there are now less balloon particles per sqaure inch than before. The total amount of ballon material has remained the same, however the surface area has now increased. Substitute balloon material? for power and balloon material per square inch? for power flux density and we can roughly make an analogy with the expanding energy from an omnidirectional antenna?.
One of the biggest sources of confusion comes from the term attenuation?. For a radio signal to be attenuated, there has to be a physical interaction of the radio wave with a physical material or object [ e.g. passing through air, passing through a wall etc ]. Free space path loss refers to [ as it says ] loss through a vacuum. Attenuation [ the small amount of signal reduction or large amount depending on frequency ] values may be added onto the free space path loss. The terms loss and attenuation are unfortunately used interchangeably in modern texts.
Now imagine that we have a magic parabolic antenna one square meter in diameter [ called a perfect unit area lossless antenna ] and we stand in front of an outdoor bridge link in front of the transmitting antenna. We gather all the energy that we can from our little dish and measure it. Now we keep in a straight line away from the antenna [ called the boresight? or right down the middle of the main lobe ] and walk several hundred feet away and gather more energy. This time we see a much reduced value. Has the energy been reduced by the air in between ? No [ well a tiny, tiny amount has, but we can ignore that ]. Like the balloon material, the energy that originally filled? the little dish at the first position has spread out and there is less per square meter [ the power flux density has reduced ].
The frequency component comes from the equations mentioned earlier to do with the ideal unit area antenna?.
Some RF concepts are a bit difficult to explain physically, so analogies are often used.
Hope this has shed a little light on this.
I need to lay down now and may never get up again....
Dave

What Dave says can be demonstrated using the Inverse Square Law, say, the power change is inversely proportional to the square of the distance change from the source:
P = 1 / (d*d)
A graph from Wiki illustrates the Law:
http://upload.wikimedia.org/wikipedia/commons/2/28/Inverse_square_law.svg
Both the FSPL and 6 dB Rule are based on the Inverse Square Law.