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Probability of an IR photon colliding with a molecule of CO2?

Being curious yet skeptical is good. Here is a skeptical question.

Carbon dioxide is a trace gas with a concentration in the atmosphere of only 390 parts per million. Sure seems like not much to worry about; that's much less than 1%. I know that the sun warms the earth and sea with visible radiation, and that much of that is absorbed then re-radiated as infrared radiation. So trace a little IR photon in its path from the ground, and up through the air. If it hits a CO2 molecule, it is absorbed and then re-radiated in a random direction, trapping much of that photon's energy in the atmosphere. But there is so little of that CO2 gas! So I ask, what is the probability that a given IR photon will collide with a CO2 molecule as it travels through several miles of atmosphere in it's path to space? How do you calculate that probability?

9 Answers

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  • 10 years ago

    <<Being curious yet skeptical is good. >>

    No, it is not. "Skepticism" is a form of anti-science. When a theory is proposed you should examine it on its merits without a bias that you are trying to show it is wrong.

    <<So trace a little IR photon in its path from the ground, and up through the air. If it hits a CO2 molecule, it is absorbed and then re-radiated in a random direction, trapping much of that photon's energy in the atmosphere.>>

    This is not true. While you are "skeptical" of correct things you are not very skeptical of your beliefs that are incorrect physics. A photon is not localized. Technically every photon anywhere "near" a CO2 molecule has some overlap with it. Some have more overlap than others. But even ignoring that, if a photon strongly overlaps with a CO2 molecule it is not the case that there is a 100 percent certainty that it will interact.

    <<But there is so little of that CO2 gas!>>

    The density is small, but that does not mean that a photon travelling a long distance will not encounter one. For example, if there is only one dead squirrel on average per 100 miles on a road, if you travel 1,000 miles on that road you have a good chance of encountering one. Also, you seem to think that a high fraction of the photons need to be absorbed to have a significant effect. That is not correct.

    <<So I ask, what is the probability that a given IR photon will collide with a CO2 molecule as it travels through several miles of atmosphere in it's path to space? How do you calculate that probability?>>

    It is best to do experimentally, because the theoretical calculation would be too difficult. You would do an experiment where you pass the relevant infrared radiation thru a column of CO2 and see how much is blocked. The results could then be easily translated to the situation regarding the Earth's atmosphere.

  • 10 years ago

    A virtual certainty!

    An IR photon (of the correct wavelength) has a mean free path (the average distance it will travel before being absorbed) of only a few metres in the lower atmosphere. I'm not sure of the exact figure; I have heard 10 metres, also as little as 1 metre. (I assume this figure would increase at altitude; CO2 is well mixed, but with decreasing pressure, the molecules would be farther apart.)

    So, whilst there is no law that would prevent the photon from radiating directly into space, the probability of it happening is so low that it would be virtually impossible.

    If we made the simplifying assumption that it was the same photon, it would probably be absorbed and re-radiated millions of times (since, as you say, the direction after re-radiation is random) before eventually escaping to space.

    ==================

    Good answers from Linlyons and d/dx.

  • Vince
    Lv 7
    10 years ago

    After this sunlight is absorbed and re-radiated, it does not leave the atmosphere. Photons are light. I believe infrared radiation takes the form of a wave. When the photons enter the atmosphere, CO2 doesn't block them. But when the photons come in contact with an object and becomes infrared radiation waves, CO2 would keep these from leaving the atmosphere.

  • 10 years ago

    The math is NOT simple. I use an analogy in my head of cannonballs hitting spinning cannons, which spit the cannonballs back out after a delay. I finally found a wiki reference that at least begins to give the info you need to do the calculations.

    Edit to directly address the question. The number of interactions is large and the properties of the interaction are simple. This makes a Monte Carlo simulation the most accurate determinant of probabilities. You fill a mathematical space with randomly placed mathematical air molecules of the known interaction cross-section to the same density as the real atmosphere, shoot a known flux of mathematical photons at it and see what photons hit what molecules. It is an easier digital way of solving the same problem as neutrons hitting nuclei, which Feynman helped find ways for teams of people to solve during the Manhattan project.

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  • 10 years ago

    The probability of a photon interacting with a molecule is given by the absorption cross section. The absorption cross section varies with frequency, but is in the range of 1E-19 cm^2 for CO2 in the 700 cm-1 region. A detailed listing of the absorption cross sections for most of the gasses in the atmosphere is given in the HITRAN database.

    http://www.cfa.harvard.edu/hitran/

    The absorption cross section is empirically derived from the extinction coefficient (imaginary part of the refractive index. Theoretically, the absorption cross section can be derived from the transition dipole moment matrix, which is a quantum mechanical calculation. For anything beyond hydrogen, the mathematics gets quite involved. The absorption cross section of a molecule is understood in a classical sense as an antennae. The "reach" of the molecule as an antennae is much larger than the physical size of the molecule.

  • 10 years ago

    the greenhouse effect is based on the fact that CO2 is mostly transparent to visible light and mostly opaque to the lower frequency IR radiation from the surface. the question is not necessarily one of probability, but in the difference in probability from 50 years ago to now.

    if CO2 concentration was 300ppm 50 YAG, then the current probability is 30% higher now.

  • Anonymous
    7 years ago

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  • Anonymous
    10 years ago

    Dont know

  • 10 years ago

    FAmily guy

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