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Electromagnetic absorption by water

Infrared Multispectral
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#1 Stefano

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Posted 19 June 2020 - 19:12

This reference is still work in progress. There can be errors, unnecessary repetitions, and so on. Anyway, even after I fixed everything (hopefully!), any advice or correction (as stated just below) is appreciated.

Note: if you are viewing this reference on a smartphone, especially in a vertical position, it may not be rendered properly. I suggest you to rotate your phone in a horizontal position or use a PC.

I wanted to write this topic as a reference for everything regarding the electromagnetic absorption spectrum by water. Any advice/correction is appreciated.

Definitions:
- attenuation length: length or thickness of a material required to bring down transmission to 1/e, or 36.788%. To avoid confusion with the symbol for the wavelength (λ), I will use "µ" instead (Note: this is a correction I made after posting the original post). Its reciprocal is the absorption coefficient, indicated by the symbol "α".
- SWIR: Short Wave InfraRed (about 1100-3000 nm, the definition of this band in particular can vary a lot between sources).
- MWIR: Medium Wave InfraRed (3000-8000 nm).
- LWIR: Long Wave InfraRed (8000-14000 nm).

Everything below refers to liquid water, unless otherwise specified.

Water is an almost colorless liquid. It is very transparent in visible light, even at reasonably large “thicknesses”. But anything except vacuum has a color. Water is actually slightly blue, it isn’t an illusion. The reflection of the sky above it contributes to the color, but even with a white background and enough water (one meter should be enough) the blue color can be seen.

In the image below, one can see that water is very transparent in the visible spectrum, as stated above, but strongly absorbs deep UV, infrared, microwaves and radio waves. Although not visible here, water is quite transparent at the extremes of the EM spectrum, that is hard X-rays, gamma rays and very low frequency radio waves. Soft X-rays are strongly absorbed by water, and that's one of the reasons only hard X-rays are used in radiography. According to Wikipedia, µ for water at 2 nm is less than a micrometer (https://en.wikipedia...y#Energy_ranges).

Water has a peak transmission at 485 nm (data taken from the Schott filter calculator 2017 program). One meter of water transmits about 98.164% at that wavelength (removed some decimals). This means that the attenuation length is almost 54 meters, (absorption coefficient 0.0185 m-1).
How to read the graph below: the numbers on the Y axis represent the number of attenuation lengths in one meter, or simply the absorption coefficient.
Attached Image: Absorption_spectrum_of_liquid_water.png
Credits: https://commons.wiki...iquid_water.png. Original image resized for the forum.

Water has a series of absorption peaks in the infrared region, that are stronger and stronger as the wavelength increases. In the NIR-SWIR region, they are found at 970 nm, 1200 nm, 1450 nm, 1950 nm and at around 2900 nm (actually, according to Wikipedia, there are two peaks at 2898 and 2766 nm). In the MWIR region, the strongest peak is at 6097 nm (still Wikipedia) (data may vary a little between different sources). In the LWIR region and above (THz, microwaves and radio waves) there are no longer strong peaks, and the absorption stays very high, slowly going down in the long waves. The absorption peak at 970 nm is within the sensitivity range of a silicon sensor, and thus it can be seen with a conventional full-spectrum camera and a suitable narrow bandpass filter. There's also a weak peak at 755 nm (µ = 0.38 m, data from the 2017 Schott filter calculator program). Note that this is too weak to be clearly seen in a glass of water (assuming a path of 7 cm, the transmission would be around 83%), but more than enough to be seen in a swimming pool (assumig a path of 1.2 m, the transmission would only be 4.25%, which is basically black).

Andy Perrin calculated the values of some peaks, and I will report also his data. For his source (and the others) see at the end of the reference.

Below are the features of the most important absorption peaks:

The 970 nm peak is strong enough that only a few centimeters of water are required to see it. It is the last peak visible with a common silicon-based sensor.
Andy covered it very well, using a narrow 980 nm bandpass filter to isolate that band (transmission rises outside it, on both sides). He calculated an α value of 48.64 m-1 at 976.6 nm (I converted the units). The peak according to the 2017 Schott filter calculator is at 970 nm, with α = 45.34 m-1 (calculated by me).
https://www.ultravio...absorption-peak
https://www.ultravio...0nm-bp10-filter
https://www.ultravio...-980nm-bandpass
https://www.ultravio...loucester-980nm
https://www.ultravio...32-hello-mother

The 1200 nm peak is even stronger, but outside the sensitivity of silicon sensors, which are blind over 1100-1150 nm (see here: https://www.ultravio...nsitivity-limit). It is broader than the other peaks, and its beginning can be seen with a silicon-based camera, although only at the extreme limits of the sensor sensitivity. Andy's values are 1196.2 nm and α = 128.16 m-1.

The next peak, at 1450 nm is very strong. It is one of the very special things visible only with a SWIR camera, and probably one of the reasons to buy such a camera. Water is basically black at this wavelength, and the strong absorption by it is the main responsible for the darkness of human skin in SWIR. Andy's values are 1453.5 nm and α = 3276.68 m-1.

The 2900 nm peak is extremely strong. It is the band with the strongest absorption in the infrared/microwave/radio waves region. According to this, the peak wavelength is at around 2936 nm, with α = 1226200 m-1 (µ is less than a micrometer).

Water blocks radio signals, so a radio won't work under water, but radio waves in the lowest frequency bands (longest wavelengths) can actually penetrate water deep enough to be useful for communication purposes. Radio waves in the VLF band (3-30 KHz) can penetrate seawater up to approximately 20 meters, while radio waves in the ELF band (3-300 Hz) can penetrate seawater for several hundred meters (source: https://en.wikipedia...y_low_frequency).

On the other side of the spectrum, water starts heavily absorbing in the UVC region, below 200 nm. Imaging below 200 nm is very hard, it requires fused silica lenses (or other materials, such as CaF2), special light sources, phosphors, filters and so on. So far, as I am aware of, only David (dabateman) broke the 200 nm barrier, if we don't consider X-rays. He managed to image using the mercury 185 nm line, which also syntetizes ozone, a gas you don't want to breathe (at least, you can smell it).
This is his work:
https://www.ultravio...l-185nm-attempt
Was there a prevoius attempt? I can't find it.

Consequences:
The first obvious consequence is that water, especially in big amounts (lakes, rivers, and even swimming pools) appears very dark, usually black, seen in the infrared spectrum, as the sky does (the reason the sky is dark in the infrared is completely different, mainly due to very little Rayleigh scattering, but also partially due to absorption by water vapour).

Water can't be used as a polar solvent in the infrared spectrum if transparency is important.

As said above, skin becomes dark in the SWIR region and above. People have black skin and white hair up to LWIR and beyond. After around 3 μm, blackbody radiation emitted by people begins to be visible, but that's not reflected light (what you would expect from a light body), it is emitted light, which doesn't change the reflected color of something. As a comparison, it can be said that the Sun is black, because it is a good approximation of an ideal blackbody. the Sun doesn't reflect a lot of light, but does emit a ton of it.

Water glows in thermal infrared, because it absorbs light almost completely and thus it behaves as an almost perfect blackbody, with an emissivity approaching 1. Note that not all liquids are opaque in thermal infrared: liquid nitrogen is transparent, for example. This is another occasion to remind everyone a thing (members and readers): thermal cameras do NOT see heat, they see light emitted from hot objects, which is usually the same but not always. Liquid nitrogen doesn't appear black because it is cold, it appears transparent.

To sum up...

Formulas:
Attached Image: formula5.png This will give you the transmission (as a dimentionless number, between 0 and 1), given a depth x and µ.

Attached Image: formula6.png This will give you the attenuation length µ for a given transmission at a depth x.

Attached Image: formula7.png This will give you the depth x in function of everything else.

Also remember that "α" and "µ" are the reciprocal of each other. To convert the attenuation length into the corresponding absorption coefficient and viceversa, you just have to calculate its reciprocal. α = 1/µ and µ = 1/α.

Table with the peaks in the infrared region. Note: I excluded some sources as the "resolution" was too low (the distance between two consecutive measurements was too wide), and could have impacted the precision of the data. This table (but, really, the entire reference) is still work in progress. You will see two tables temporairly, I am re-writing it to be more ordered. I may also add a second (permanent) table with the transmission peaks (the opposite of the absorption peaks), but that will be later, after I finished working on this one first.

Source 1Source 2Source 3Source 4Source 5Source 6Source 7Source 8Source 9Average
peak 1λ = 746 nm
µ = 367.24 mm
α = 2.723 m-1
λ = 750 nm
µ = 384.62 mm
α = 2.6 m-1
λ = 752 nm
µ = 355.87 mm
α = 2.81 m-1
λ = 750 nm
µ = 382.83 mm
α = 2.6121 m-1
λ = 750 nm
µ = 381.68 mm
α = 2.62 m-1
λ = 755.3 nm
µ = 347.43 mm
α = 2.8783028 m-1
λ = 764 nm
µ = 360.46 mm
α = 2.7742 m-1
λ = 755 nm
µ = 380.31 mm
α = 2.6294 m-1
λ = 752.7875 nm
µ = 369.57 mm
α = 2.7059 m-1
peak 2λ = 970/975 nm
µ = 22.222 mm
α = 45 m-1
λ = 973 nm
µ = 19.455 mm
α = 51.4 m-1
λ = 975 nm
µ = 22.295 mm
α = 44.8523 m-1
λ = 976.6 nm
µ = 20.56 mm
α = 48.6390343 m-1
λ = 970 nm
µ = 22.054 mm
α = 45.344 m-1
λ = 973.42 nm
µ = 21.255 mm
α = 47.047 m-1
peak 3λ = 1180/1200 nm
µ = 9.6156 mm
α = 104 m-1
λ = 1190/1205 nm
µ = 8 mm
α = 125 m-1
λ = 1199 nm
µ = 7.9577 mm
α = 125.6637 m-1
λ = 1196.2 nm
µ = 7.8025 mm
α = 128.1639538 m-1
λ = 1195.675 nm
µ = 8.2845 mm
α = 120.71 m-1
peak 4λ = 1440 nm
µ = 357.14 µm
α = 2.8 mm-1
λ = 1443 nm
µ = 315.46 µm
α = 3.17 mm-1
λ = 1489 nm
µ = 325.79 µm
α = 3.069435 mm-1
λ = 1447 nm
µ = 327.21 µm
α = 3.0561 mm-1
λ = 1453.5 nm
µ = 305.19 µm
α = 3.276680057 mm-1
λ = 1454.5 nm
µ = 325.26 µm
α = 3.0744 mm-1
peak 5λ = 1940 nm
µ = 83.452 µm
α = 11.983 mm-1
λ = 1927 nm
µ = 80.645 µm
α = 12.4 mm-1
λ = 1949 nm
µ = 83.927 µm
α = 11.9151625 mm-1
λ = 1927.9 nm
µ = 76.558 µm
α = 13.062 mm-1
λ = 1930.5 nm
µ = 73.154 µm
α = 13.66971162 mm-1
λ = 1934.88 nm
µ = 79.327 µm
α = 12.606 mm-1
peak 6λ = 2950 nm
µ = 787.77 nm
α = 1.2694 µm-1
λ = 2937 nm
µ = 833.52 nm
α = 1.199731068 µm-1
λ = 2935.1/2936 nm
µ = 815.528 nm
α = 1.2262 µm-1
λ = 2940.85 nm
µ = 811.84 nm
α = 1.2318 µm-1
peak 7λ = 4700 nm
µ = 23.81 µm
α = 42 mm-1
λ = 4699 nm
µ = 23.817 µm
α = 41.9859564 mm-1
λ = 4697 nm
µ = 24.084 µm
α = 41.522 mm-1
λ = 4698.666 nm
µ = 23.903 µm
α = 41.836 mm-1
peak 8λ = 6100 nm
µ = 3.7051 µm
α = 269.9 mm-1
λ = 5955 nm*
µ = 5.481 µm*
α = 182.4497649 mm-1*
λ = 6086.4 nm
µ = 3.6888 µm
α = 271.09 mm-1
λ = 6093.2 nm
µ = 3.6969 µm
α = 270.5 mm-1
peak 9λ = 15000 nm
µ = 2.9691 µm
α = 336.8 mm-1
λ = 14588/14689 nm
µ = 2.9645 µm
α = 337.33 mm-1
λ = 14819.25 nm
µ = 2.9668 µm
α = 337.065 mm-1
peak 10λ = 50000 nm
µ = 7.7399 µm
α = 129.2 mm-1
λ = 50000 nm
µ = 7.7399 µm
α = 129.2 mm-1


Source 1Source 2Source 3Source 4Source 5Source 6Source 7Source 8Source 9Average
peak 1λ = 746 nm
µ = 367.24 mm
α = 2.723 m-1
λ = 750 nm
µ = 384.62 mm
α = 2.6 m-1
λ = 752 nm
µ = 355.87 mm
α = 2.81 m-1
λ = 750 nm
µ = 382.83 mm
α = 2.6121 m-1
λ = 750 nm
µ = 381.68 mm
α = 2.62 m-1
λ = 755.3 nm
µ = 347.43 mm
α = 2.8783028 m-1
λ = 764 nm
µ = 360.46 mm
α = 2.7742 m-1
λ = 755 nm
µ = 380.31 mm
α = 2.6294 m-1
λ = 752.7875 nm
µ = 369.57 mm
α = 2.7059 m-1
peak 2λ = 970/975 nm
µ = 22.222 mm
α = 45 m-1
λ = 973 nm
µ = 19.455 mm
α = 51.4 m-1
λ = 975 nm
µ = 22.295 mm
α = 44.8523 m-1
λ = 976.6 nm
µ = 20.56 mm
α = 48.6390343 m-1
λ = 970 nm
µ = 22.054 mm
α = 45.344 m-1
λ = 973.42 nm
µ = 21.255 mm
α = 47.047 m-1
peak 3λ = 1180/1200 nm
µ = 9.6156 mm
α = 104 m-1
λ = 1190/1205 nm
µ = 8 mm
α = 125 m-1
λ = 1199 nm
µ = 7.9577 mm
α = 125.6637 m-1
λ = 1196.2 nm
µ = 7.8025 mm
α = 128.1639538 m-1
λ = 1195.675 nm
µ = 8.2845 mm
α = 120.71 m-1
peak 4λ = 1440 nm
µ = 357.14 µm
α = 2.8 mm-1
λ = 1443 nm
µ = 315.46 µm
α = 3.17 mm-1
λ = 1489 nm
µ = 325.79 µm
α = 3.069435 mm-1
λ = 1447 nm
µ = 327.21 µm
α = 3.0561 mm-1
λ = 1453.5 nm
µ = 305.19 µm
α = 3.276680057 mm-1
λ = 1454.5 nm
µ = 325.26 µm
α = 3.0744 mm-1
peak 5λ = 1940 nm
µ = 83.452 µm
α = 11.983 mm-1
λ = 1927 nm
µ = 80.645 µm
α = 12.4 mm-1
λ = 1949 nm
µ = 83.927 µm
α = 11.9151625 mm-1
λ = 1927.9 nm
µ = 76.558 µm
α = 13.062 mm-1
λ = 1930.5 nm
µ = 73.154 µm
α = 13.66971162 mm-1
λ = 1934.88 nm
µ = 79.327 µm
α = 12.606 mm-1
peak 6λ = 2950 nm
µ = 787.77 nm
α = 1.2694 µm-1
λ = 2937 nm
µ = 833.52 nm
α = 1.199731068 µm-1
λ = 2935.1/2936 nm
µ = 815.528 nm
α = 1.2262 µm-1
λ = 2940.85 nm
µ = 811.84 nm
α = 1.2318 µm-1
peak 7λ = 4700 nm
µ = 23.81 µm
α = 42 mm-1
λ = 4699 nm
µ = 23.817 µm
α = 41.9859564 mm-1
λ = 4697 nm
µ = 24.084 µm
α = 41.522 mm-1
λ = 4698.666 nm
µ = 23.903 µm
α = 41.836 mm-1
peak 8λ = 6100 nm
µ = 3.7051 µm
α = 269.9 mm-1
λ = 5955 nm*
µ = 5.481 µm*
α = 182.4497649 mm-1*
λ = 6086.4 nm
µ = 3.6888 µm
α = 271.09 mm-1
λ = 6093.2 nm
µ = 3.6969 µm
α = 270.5 mm-1
peak 9λ = 15000 nm
µ = 2.9691 µm
α = 336.8 mm-1
λ = 14588/14689 nm
µ = 2.9645 µm
α = 337.33 mm-1
λ = 14819.25 nm
µ = 2.9668 µm
α = 337.065 mm-1
peak 10λ = 50000 nm
µ = 7.7399 µm
α = 129.2 mm-1
λ = 50000 nm
µ = 7.7399 µm
α = 129.2 mm-1




Source 1

Source 2

Source 3

Source 4

Source 5

Source 6

Source 7

Source 8

Source 9

Average

Peak 1

λ = 746 nm

λ = 750 nm

λ = 752 nm

λ = 750 nm

λ = 750 nm



λ = 755.3 nm

λ = 764 nm

λ = 755 nm

λ = 752.7875 nm

µ = 367.24 mm

µ = 384.62 mm

µ = 355.87 mm

µ = 382.83 mm

µ = 381.68 mm



µ = 347.43 mm

µ = 360.46 mm

µ = 380.31 mm

µ = 369.57 mm

α = 2.723 m-1

α = 2.6 m-1

α = 2.81 m-1

α = 2.6121 m-1

α = 2.62 m-1



α = 2.8783028 m-1

α = 2.7742 m-1

α = 2.6294 m-1

α = 2.7059 m-1

Peak 2



λ = 970/975 nm

λ = 973 nm

λ = 975 nm





λ = 976.6 nm



λ = 970 nm

λ = 973.42 nm



µ = 22.222 mm

µ = 19.455 mm

µ = 22.295 mm





µ = 20.56 mm



µ = 22.054 mm

µ = 21.255 mm



α = 45 m-1

α = 51.4 m-1

α = 44.8523 m-1





α = 48.6390343 m-1



α = 45.344 m-1

α = 47.047 m-1

Peak 3



λ = 1180/1200 nm

λ = 1190/1205 nm

λ = 1199 nm





λ = 1196.2 nm





λ = 1195.675 nm



µ = 9.6154 mm

µ = 8 mm

µ = 7.9577 mm





µ = 7.8025 mm





µ = 8.2845 mm



α = 104 m-1

α = 125 m-1

α = 125.6637 m-1





α = 128.1639538 m-1





α = 120.71 m-1

Peak 4



λ = 1440 nm

λ = 1443 nm

λ = 1489 nm



λ = 1447 nm

λ = 1453.5 nm





λ = 1454.5 nm



µ = 347.22 µm

µ = 315.46 µm

µ = 325.79 µm



µ = 327.21 µm

µ = 305.19 µm





µ = 325.26 µm



α = 2.88 mm-1

α = 3.17 mm-1

α = 3.069435 mm-1



α = 3.0561 mm-1

α = 3.276680057 mm-1





α = 3.0744 mm-1

Peak 5



λ = 1940 nm

λ = 1927 nm

λ = 1949 nm



λ = 1927.9 nm

λ = 1930.5 nm





λ = 1934.88 nm



µ = 83.452 µm

µ = 80.645 µm

µ = 83.927 µm



µ = 76.558 µm

µ = 73.154 µm





µ = 79.327 µm



α = 11.983 mm-1

α = 12.4 mm-1

α = 11.9151625 mm-1



α = 13.062 mm-1

α = 13.66971162 mm-1





α = 12.606 mm-1

Peak 6



λ = 2950 nm



λ = 2937 nm



λ = 2935.1/2936 nm







λ= 2940.85 nm



µ= 787.77 nm



µ = 833.52 nm



µ = 815.528 nm







µ = 811.84 nm



α = 1.2694 µm-1



α = 1.199731068 µm-1



α = 1.2262 µm-1







α = 1.2318 µm-1

Peak 7



λ = 4700 nm



λ = 4699 nm



λ = 4697 nm







λ = 4698.666 nm



µ = 23.81 µm



µ = 23.817 µm



µ = 24.084 µm







µ = 23.903 µm



α = 42 mm-1



α = 41.9859564 mm-1



α = 41.522 mm-1







α = 41.836 mm-1

Peak 8



λ = 6100 nm



λ = 5955 nm*



λ = 6086.4 nm







λ = 6093.2 nm



µ = 3.7051 µm



µ = 5.481 µm*



µ = 3.6888 µm







µ = 3.6969 µm



α = 269.9 mm-1



α = 182.4497649 mm-1*



α = 271.09 mm-1







α = 270.5 mm-1

Peak 9



λ = 15000 nm







λ = 14588/14689 nm







λ = 14819.25 nm



µ = 2.9691 µm







µ = 2.9645 µm







µ = 2.9668 µm



α = 336.8 mm-1







α = 337.33 mm-1







α = 337.065 mm-1

Peak 10



λ = 50000 nm















λ = 50000 nm



µ = 7.7399 µm















µ = 7.7399 µm



α = 129.2 mm-1















α = 129.2 mm-1


Note:
- Data marked with "*" is data that I don't consider accurate enough, and I didn't use it to calculate the average.
- Where there are two wavelengths in the same cell (ex. 970/975 nm), I first averaged them (in this case, 972.5 nm), and then used that result to calculate the average. (The commutative property doesn't hold for the average operation.)
- Since all μ values have been calculated (not copied, unlike λ and α), they have been rounded to 5 digits. The same thing has been done in the "Average" column, where I first calculated the average α value, and then used it to calculate the average μ value. (Calculating the average directly for μ is NOT the same as calculating it for α and then inverting it.)

Sources in the table:

Source 1: "Experimental study of the absorption in distilled water, artificial sea water, and heavy water in the visible region of the spectrum", 1963. Data: https://omlc.org/spe.../sullivan63.txt
Source 2: "Optical constants of water in the 200nm to 200µm wavelength region", 1973. Data: https://omlc.org/spe...data/hale73.txt
Source 3:"Optical properties of water in the near infrared", 1974. Data: https://omlc.org/spe...ta/palmer74.txt
Source 4: "Split-pulse laser method for measuring attenuation coefficients of transparent liquids: application to deionized filtered water in the visible region", 1978. Data: https://omlc.org/spe...ta/querry78.txt
Source 5: "Physical Optics of Ocean Water", 1988. Data: https://omlc.org/spe...a/shifrin88.txt
Source 6: "Wedge shaped cell for highly absorbent liquids: infrared optical constants of water", 1989. Note: this is the source cited in the post several times. Data: https://omlc.org/spe...wieliczka89.txt
Source 7: Andy Perrin's calculations (source below);
Source 8: "The optical properties of pure water", 1994. Data: https://omlc.org/spe...buiteveld94.txt
Source 9: Schott filter calculator program (2017);

Andy's source for his calculations: "Refractive indices of water and ice in the 0.65- to 2.5-μm spectral range"
Applied Optics Vol. 32, Issue 19, pp. 3531-3540 (1993)

Some of the sources listed above can be found here: http://omlc.ogi.edu/.../abs/index.html. This is a collection of the sources from which data was taken to draw the graph posted at the beginning, by its author (read the licence in the credits). I suggest checking the data yourself, if you want more. I had to keep things
simple, so I didn't take data from all this sources (in fact, I only used one of them, the linked one).

Other works by other members that involve water absorption (tell me if I missed some):

Andy Perrin:
https://www.ultravio...p-and-some-pics

Bernard Foot:
https://www.ultravio...attach_id=19348 (complete topic: https://www.ultravio...__fromsearch__1);

https://www.ultravio...dpost__p__32824

David (dabateman):
https://www.ultravio...r-test-lp1100nm

Eka (ins13):
https://www.ultravio...ng-apples-950nm

Mine (Stefano):
https://www.ultravio...h-a-solar-panel

About the other states of water...

Solid water (aka ice) behaves in a similar way to liquid water. The absorption spectrum is similar, but the peaks are not exactly the same, and the difference can be seen using a narrow bandpass filter (see here).

Water vapour has a very different absorption spectrum than those of liquid and solid water. This spectrum is analised in the Wikipedia article I already linked several times above (https://en.wikipedia...rption_by_water). This absorption is partially the responsible for the atmospheric windows in the infrared spectrum.

Maybe this needs to be refined, any improvement is welcome. Again, tell me any errors/omissions or needed modifications.

Edit: changed and fixed a few things:
- Fixed some broken links;
- Changed the symbol used for the attenuation length from "λ" (lambda) to "µ" (mu), to avoid confusion with the wavelength symbol;
- Changed the formulas shown above (thanks Andy);
- Included "Eka" as the real name for the member "ins13" (tell me if this is ok or if I need to change it);
- Differentiated the symbols for the attenuation length and the absorption coefficient ("µ" and "α" respectively);

- Added a brief description of the absorption spectrums by ice and water vapour;

- Added an image made with the "full color IR" tecnique by Bernard Foot showing blue water;
- Added some details about the penetration of very low frequency radio waves in (sea)water (idea from UlfW);
- Added a table with the absorption peaks in the infrared range (still work in progress);
- Added another blue water image by Bernard Foot;

Edited by Stefano, Today, 17:57.


#2 Andy Perrin

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Posted 19 June 2020 - 19:24

Stefano please don’t use lambda for both attenuation length and wavelength in the same document. Also log base e should be ln().

#3 bobfriedman

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Posted 19 June 2020 - 19:42

perhaps use "Mu" for linear attenuation??

Edited by bobfriedman, 19 June 2020 - 19:42.


#4 Stefano

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Posted 19 June 2020 - 19:46

ex (in the second formula) is the entire base. Maybe my formulas can be simplified, if someone can tell me. Also in the third one the base is e with an exponent.

#5 Stefano

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Posted 19 June 2020 - 19:48

The problem is that symbols are recycled in physics. In this case we have two lambdas, one for the wavelength and the other one for the attenuation length. Should I change the symbols?

#6 Stefano

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Posted 19 June 2020 - 19:56

View Postbobfriedman, on 19 June 2020 - 19:42, said:

perhaps use "Mu" for linear attenuation??
The lowercase version may generate confusion with "µm" (maybe not). Should I use another symbol not present in the document?

#7 Stefano

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Posted 19 June 2020 - 20:42

Also, tell me if there are any broken links.

#8 bobfriedman

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Posted 19 June 2020 - 22:09

View PostStefano, on 19 June 2020 - 19:56, said:

The lowercase version may generate confusion with "µm" (maybe not).

maybe not.

Edited by bobfriedman, 19 June 2020 - 22:09.


#9 Andy Perrin

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Posted 19 June 2020 - 22:12

µ won't create any confusion with micrometers. The base of a logarithm shouldn't have any variables like x in it. If you do, you should be rewriting it differently, but in this case I think you have simply done it wrong. Remember, e is a number like π. e^x is a function.

ln(x) = loge(x).

My suggestions would be:

Attached Image: Screen Shot 2020-06-19 at 6.16.34 PM.png

Edited by Andy Perrin, 19 June 2020 - 22:17.


#10 dabateman

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Posted 19 June 2020 - 22:29

View PostStefano, on 19 June 2020 - 19:48, said:

The problem is that symbols are recycled in physics. In this case we have two lambdas, one for the wavelength and the other one for the attenuation length. Should I change the symbols?

This reminds me of how to find a chem-physics person in a room. Ask what new (nu) and they or I will answer c/lambda.


#11 Stefano

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Posted 20 June 2020 - 01:42

View PostAndy Perrin, on 19 June 2020 - 22:12, said:

µ won't create any confusion with micrometers. The base of a logarithm shouldn't have any variables like x in it. If you do, you should be rewriting it differently, but in this case I think you have simply done it wrong. Remember, e is a number like π. e^x is a function.

ln(x) = loge(x).

My suggestions would be:

Attachment Screen Shot 2020-06-19 at 6.16.34 PM.png
Thanks Andy, they were so simple to obtain! I always do simple things in a too complicated way. I just had to do ln on both sides and than rearrange things. What I instead did was to separate the components of the exponent, and that led to my not-so-nice (but still working) formulas (I tested them). I don't know why sometimes I can't see a simple solution, even when I have it in front of my eyes.

#12 Stefano

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Posted 20 June 2020 - 01:43

Tomorrow I will fix a few things... thank you all for the suggestions.

Again, tell me if there is something else to fix, even minor things.

Edited by Stefano, 20 June 2020 - 12:24.


#13 StephanN

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Posted 20 June 2020 - 20:43

Maybe I am thick, but "Also remember that a m-1 simply means a times in one meter. The length in meter is the reciprocal of a (1/a). If µ = 100 m-1, it means that µ = 0.01 m, or 1 cm." makes no sense. How can a length have the reciprocal of a length as a unit? I think I get what you're intending to say (in a colloquial way), but "m-1" in science is the reciprocal of a metre and used for the wavenumber (https://en.wikipedia...wiki/Wavenumber), for example. IMHO, you'd better not use "100 m-1" but rather "10-2 m".
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#14 Andy Perrin

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Posted 20 June 2020 - 21:29

StephanN, mu is meters here. Not 1/m. It is the attenuation length. Alpha is the absorption coefficient which is in 1/m units. Alpha = 1/mu.

#15 Stefano

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Posted 20 June 2020 - 22:34

m-1 is not the length (although correlated), but something different. The length says how much units (meters, for example) are contained in a "length". The other thing says how much lengths are contained in a unit. (Yes, I know you know it). I wrote it to make things easier for people not used to this writing. It is another way to write a length, if you will.

I had another idea: should I specify the differences between water vapor, liquid water and ice?

Edited by Stefano, 20 June 2020 - 22:36.


#16 Andy Perrin

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Posted 20 June 2020 - 23:10

Stefano, you still have some problems up there. You are using µ for attenuation length (m) in some places but as absorption coefficient (typically called alpha) (m^-1) in other places. You can't use it for both.

Is there any way you could let me edit this?

Quote

If µ = 100 m-1, it means that µ = 0.01 m, or 1 cm.
That's an example where you used two units for the same quantity in one sentence.

Correct this (and all similar errors) to:

Quote

If a = 100 m-1, it means that µ = 0.01 m, or 1 cm.

Or you can send it to me and I'll fix it for you if that's easier.

Edited by Andy Perrin, 20 June 2020 - 23:20.


#17 Stefano

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Posted 20 June 2020 - 23:25

It is the same for me (I can edit myself). If you will modify it, do you mean to send you the text, and then you will send me back the edited version? Or, can Andrea give you the ability to modify it? I fear links will break if I copy-paste them.

#18 Stefano

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Posted 20 June 2020 - 23:40

Maybe it will be simpler if I modify them. Should I use the greek letter α?

#19 Andy Perrin

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Posted 21 June 2020 - 00:16

Yeah use the Greek. As long as you are clear feel free to edit yourself. Otherwise I meant to send me the text.

#20 Stefano

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Posted 21 June 2020 - 00:40

Keep in mind that I don't have a source for my text, I am writing it directly here. So what you see posted is the actual text.