Talk:Optical depth

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There is a disparity here... I think an optical depth is something like transmission=exp(-depth). But I'm not sure. William M. Connolley 19:54, 1 January 2006 (UTC).[reply]

corrected definition

The optical depth tau is related to the fraction of light scattered. The following equation expresses this relationship:

I/I_0= exp{-tau},

where I_0 is the incident light and I is the light that passes through the medium without being scattered. Equivalently, for a homogenous medium, tau is the ratio of the path length to the mean free path.

Corrected definition inserted in the article, some inexactitudes corrected, some concepts in atmospheric science added. I hope it is appreciated.Marenco 26 September 2006 (UTC)

Great! that's cleared up the definition finally. Deuar 20:13, 27 September 2006 (UTC)[reply]

Cognitive problem with explanation

I think there is a cognitive problem here. As I understand it, optical depth is a property of the material you are looking through, not of you the observer or the object you are observing. However, the article talks about it in analogy to taking an object and moving it backward, which makes it sound like the medium is not what is being described by optical depth.

It might be better to speak of optical depth increasing as the fog gets thicker, and the value of optical depth being the farthest thing you can see through that fog.

Raddick 20:10, 4 November 2007 (UTC)[reply]

I think the explanation can be reformulated to avoid misunderstandings, but is important to state that the optical depth change when there is more medium between the observer and the object (more light get scattered or absorved). So the optical depth is not just a property of the material, but it also depends on the distance between the object and the observer. The farther the object is, the less you see of it (because there is more medium on the way). Hsxavier 23:58, 5 November 2007 (UTC)[reply]

I'm not convinced that the first sentence is correct :

Optical depth is a measure of transparency, and is defined as the fraction of radiation (or light) that is scattered or absorbed on a path.

The equation is correct, but if tau is the fraction of the radiation that is removed then tau=0.5 should mean that half of the radiation has been absorbed or scattered. However e^(-0.5) ~ 0.6 and e^(-1) is certainly not equal to zero. --Maddoug (talk) 15:59, 22 January 2008 (UTC)[reply]

The current (Apr. 25, 2009) first sentence "Optical depth, or optical thickness is a measure of transparency, and is defined as the negative logarithm of the fraction of radiation (or light) that is scattered or absorbed on a path." is not correct. Given I/I0=e^-tau, tau=-ln(I/I0). Thus the first sentence needs to be "Optical depth, or optical thickness is a measure of transparency, and is defined as the negative logarithm of the fraction of radiation (or light) that IS NOT scattered or absorbed (i.e., IS transmitted) on a path." —Preceding unsigned comment added by Random25 (talk • contribs) 18:47, 25 April 2009 (UTC)[reply]

Nuclear Trivia

For what it's worth: optical thickness (depth, whatever) is an important concept in understanding the design of thermonuclear weapons.

Basesurge (talk) 07:41, 22 March 2010 (UTC)[reply]

Optical Depth/Optical thickness

Is there no difference between the 2 concepts in English stellar physics ? I can't be sure this is a mistake, I am neither an English native-speaker nor a stellar physicist, so... Anyway, French physicists would see a big difference between optical depth (profondeur optique) and optical thickness (épaisseur optique), although the terms are often subject to confusion and related mistakes. In stellar physics :

Optical depth deals with intensity of light generally emitted by a star,
Optical thickness deals with transparency of a sheet of material.

Same in English ? Should it be clearly mentioned both here (the article I have been partially translating) and in Optical depth ?
Mianreg (talk) 02:39, 25 August 2010 (UTC)[reply]

Clarification of definition

There is one thing I do not understand with this concept. According to the scattering properties of light, it can be influenced even in a far distance from a small scatterer (but this influence is very small). So if we say that light which is scattered (even a little bit) has been 'influenced' and is taken out from the original light beam, than optical depth would be infinite everywhere (in our universe). So what is the minimum scattering angle or minimum momentum transfer until which scattering has not to be taken into account for optical depth? Or is optical depth dependent on observer properties like a camera resolution (and this would define the minimum angle not to care about)? 134.76.234.36 (talk) 10:08, 14 October 2010 (UTC)[reply]

An optical depth of infinity would mean that 100% of the light is absorbed or scattered at zero distance. A mirror has this property as do most solid, opaque objects. In order to photograph distant galaxies, the optical depth of space must be near zero. I know that this is counterintuitive, but that is the way it is defined. Q Science (talk) 20:33, 14 October 2010 (UTC)[reply]

Thank you for your answer, but you didn't actually come to the core point of my understanding problem.

Consider for example the models of Elementary Particle physics for scattering: Scattering is caused by forces. Forces have an infinite expansion of influence (but their influence decreases by - sometimes high - power laws). So every particle everywhere at (almost) every time is scattered 'a little bit', changing the particles properties like direction only very slightly. Where is the limit when I say here it's scattered, but there it's not? Scattering happens much, much more often than just that scattering what is easily observed. That is why I think the definiton of optical depth must somehow depend on the observer (what kind of scattering the observer is able to distinguish, e.g. what number of photons the observer looses compared to a theoretical unaffected beam).

For example, small scattering angles doesn't affect me if I'm looking at some object on Earth. If I'm looking into space, I might loose, say, a photon from a specific source due to one small scattering event near that source (not considering small Space Angle here) only if the mentioned source would be put at some very high distance (d_crit).

In other words: Particles are influenced quite often, but the majority of events don't affect them very much. 134.76.234.36 (talk) 08:55, 18 October 2010 (UTC)[reply]

Sorry, I think that I still don't understand your question.

Light deflected by gravity, or a lens, is neither absorbed nor scattered. As a result, simple deflection does not affect "optical depth". Random deflection (scattering) does. Think of "optical depth" as simply the natural logarithm of the percent transmitted times minus one (to make the value positive).

OD = -ln(percent transmitted)

You are right, this does depend on the observer.

Scattering is not just light lost from the distant object, but is also light from other objects scattered toward you. This is why we don't see stars during the day, the sky scatters sun light so that the sky is brighter than the stars. The stars are still there, and the atmosphere is still as transparent as it is at night, but so much light is scattered toward our eyes that the stars "disappear". Q Science (talk) 18:44, 18 October 2010 (UTC)[reply]

That is not true. Light deflected by gravity is (generally) scattered, just look up the definition of scattering or search papers about gravitational scattering.

What I want to point out is that the definiton of optical depth given in this article and also in other sources is somehow strange using "the fraction of radiation (e.g., light) that is not scattered" on its path. I regard that so, because practically all radiation was scattered on its path, but for some fraction the scattering was so small, that this fraction arrives at the observer almost (but never exactly) on a straight line. I strongly think, the dependence on observer properties (angle of view) must be included in the definition like ("the fraction of radiation (e.g., light) that is not scattered out of the observer's field of view or absorbed in the medium."). --134.76.234.36 (talk) 08:00, 19 October 2010 (UTC)[reply]

Gravitational scattering is always toward the gravitational source, and is not comparable to atmospheric scattering which is random (isotropic). However, you make an interesting point. In optics, scattering reduces the intensity of the radiation without changing the direction of the main beam.

Chandra discusses a "pencil of radiation traversing a medium", in other words, a narrow beam. The article discusses "radiation absorbed or scattered along a path", which is about the same thing. I don't see where adding the word "observer" clarifies that meaning. It does not matter if the definition uses "absorbed or scattered" or "transmitted" since the sum of those two values is one. When the medium is optically thick, then it absorbs or scatters more radiation and the intensity of a "pencil of radiation" is reduced. Q Science (talk) 18:31, 19 October 2010 (UTC)[reply]

First to gravitational scattering; it is the same principle than atmospheric scattering. Both scattering kinds are (in the borders of the uncertainty principle) deterministic; scattering on the electric or magnetic field of a molecule or aerosol particle and scattering on a gravitational field (of e.g. a molecule or aerosol particle; with that low masses it is weak, but nevertheless there is also scattering going on). Single scattering is in both cases deterministic, not random, and in both cases in the same sense of meaning isotropic (depending on the isotropy of the scattering field).

Only if you add multiple scattering events, it becomes more and more 'random' and also isotropic a posteriori, by adding a large number of different deterministic scattering events and trajectories. This is true for the situation of many scatterers in an aerosol, and also in a sufficiently large medium of gravitational scattering. The beam, however, becomes random after adding a large number of direction changes (which has to be larger the more little the changes are), which means a very large distance in space (if there are not enough Black Holes around), and a very small distance in an aerosol for visible photons. However, for neutrinos its different: Here the gravitational scattering is more important to come to a 'random isotropic state'.

So let me point out my critics again: No matter what particle flies there (photons,...) they are all the time scattered (incredibly much more than every nanosecond, because there are so many interaction particles like virtual photons around). The vast majority of these scattering events have a almost negligible influence. However, where you set the border of an apparent scattering influence, depends on the observer properties (How good observer distinguishes between a straight photon trajectory and a slant trajectory). If you just say every photon that undergoes any scattering event does not count anymore for you, than you can not count any photon.

I will point out this in an example: Imagine a beam of 5 absolutely perfect aligned photons (with a small time between each photon shot) which shall be observed by a camera after passing a gas medium (with a temperature T>0K). One photon hits almost directly a molecule, is scattered to 90 degrees, hits another molecule and another and comes never to the observer. Two other photons only one time scatter in the middle of the way at 1 molecule and then leave the medium at 30 degrees and 15 degrees off the original line. The fourth photon is scattered 1-time very near the observer 15 degrees off. And the last photon never passes a gas molecule close-by, but the far-influence of some molecules lets it come off the straight line by 0.0000000000001 degree ---> it is not propable in this medium that the photon comes off only by such a low angle; but "by chance" it happened (its deterministic, but we can only not really compute all the gas-beam interactions) . With a normal camera you may see photon 4 and 5 (because 4 is scattered directly in front of the camera and comes off only 0.1mm). With a narrow-angle camera you might only detect photon 5. With a camera 100000000000000 miles away through absolutely field-free vacuum (not more optical depth!) you will even not detect photon 5. But all the cameras would detect all 5 photons if there would be no medium!

The optical depth calculated by using the 3 cameras would be totally different. So which one should I choose? --134.76.234.36 (talk) 08:00, 20 October 2010 (UTC)[reply]