Will CO2 absorb photon in all directions?

Discussion in 'Physics & Math' started by Robittybob1, Mar 16, 2012.

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  1. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    That is seems to be possible, but it is complex question of little concern to how ice ages ENDS.

    At end of an ice age, when the ice starts to melt a lot of 0 C water is flowing into the ocean and that will certainly have the presure on the hydrates increase. That stabalize the marginally stable hydrates. If they were not already covered by the 4C water (waters densest state and oceans should have a lot of it on the bottom, and perhaps nearly up to the top where even colder water may be far from the equator,) when the ice age starts to end, then some may have more cooling down to 4C which also stabalizes them.

    Thus I would think the opposite of what you are suggesting happens at the end of an ice age - I.e. some of hydrates that were on the verge of decomposition, become solidly stable if we are still speaking about the end of an ice age. I.e. CH4 does not accelerate the ending of and ice age, IMHO.

    At the start of an ice age with ocean level dropping, there would be CH4 release with the pressure on some marginal hydrates falling, but then the effect would be to REDUCE the rate of Earth´s cooling. I am not 100% sure of my facts, but think it is true that ice ages form / grow much more slowly than they end. Very consistent with the sudden volcanic release of GHGs and the reduction in rate of cooling some CH4 release at start would make.
     
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  3. Robittybob1 Banned Banned

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    I have the feeling the science in this thread has been left unproven either way. I want to come to a conclusion based on science. Can anyone help me with this please?

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  5. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Just because you can and have invented several unfounded alternatives to the accepted theories, does not mean the science is unproven, but of course no scientific theory is 100% "proven." - Science does not do that. It only builds a network of many inter-related, and always confrimed, concepts it accepts as true.
     
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  7. Robittybob1 Banned Banned

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    I am trying to resolve the momentum issues through 2 other threads but they must be too difficult or some other reason. I am at a loss how to proceed.
    http://www.physforum.com/index.php?showtopic=39361&st=0&#entry516238
     
  8. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    The problem you describe is really quite simple, but let me specifically state the there are constraints on the motion of the axis of the massless rotation table. (An equal an opposite impulse to that of the bullets so it does not tend to move away from the guns as it would if mounted on a cart with wheels.)

    I don´t like to work with specific numbers as that tends to destroy understanding, so your faster bullet has velocity V and the slower one v (both in the frame where the rotation axis is fixed). All the bullet mass are m and each of the block they stop in have mass M, initially. The pairs of bullets always arrive at their blocks at the same time and always tangential to the circular motion of the block. Initially for the first bullet pair (n =1) the table is not rotating until after the bullets have been stopped in the blocks.

    Thus the n =1 pair of bullets give the table (V-v)m units of rotational momentum (your radius was 1, so I will just say “momentum”) So the tangential speed, S, (at 1 meter from rotation axis) when the n =2 pair stikes the blocks is S(2M +2 m) = 2S(M+m) = (V-v)m., or

    S1 = (V-v)m / [2(M+m)] ( I added the blue color to help you see where the n =2 equation below for the speed S2 comes from.)

    Now when the n = 2 bullet pair has been stopped in the blocks, the momentum added to the table is:
    [(V-S1) – (v +S1)]m or [(V-v) -2(S1)]m so the rotation speed S2 is solved from:

    2(S2)(M + 2m) = [(V-v) -2(S1)]m. I have enclosed S1 and S2 in parends so you will know they are labels and not multiply by two, for example in the case of (S2) or by 13 in the case (S13). Etc.

    I.e. S2 = [(V-v) -2(S1)]m / [2(M + 2m)]. The two 2s in this demoninator are telling the total bullet mass is 4m.
    You should be able to continue for the n =3 final speed, S3 etc.

    I think* the speed Sn (after n pairs of bullets have been stopped in the blocks) is:

    Sn = [(V-v) -2(Sn´)]m / [2(M + nm)] where Sn´ = S(n-1) I.e. the rotational speed just before before the nth pair of bullets is stopped in the blocks.

    Note that 2(Sn´) is becoming larger as n increases and the difference [(V-v) - 2(Sn´)] is asymotically approaching zero. Thus, there is a Max rotation speed you can give the table, even if n becomes very large.

    *I have just guessed the form of the general nth term – you can check by doing the n =3 (or even the n =4) cases. Also I don´t guarantee I have not dropped a factor – I only wanted to show how the solution is easily achieved.
     
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  9. Robittybob1 Banned Banned

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    I'll put that equation into an Excel worksheet and increase "n" till we get the maximum "s".

    I liked the way you put the extra detail to the problem e.g.

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  10. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    You might want to add a column that computes the KE transfer efficiency.

    I.e. with every increment of n the constant KE of the bullets is used / contributed.

    I won´t do it but am quite sure the rotational KE of the table after n=2 is more than twice the rotational KE after only two bullets have been fired (due to KE going as the square) Thus, for some number of bullets (2n) the efficiency of KE transfer is increasing but then it will decrease.

    There will be a KE transfer efficiency max for some n, but even if you go to the bottom of your spread sheet, with very large n, you will find the "s" is still increasing ever so slightly - I.e. there is no "max s" only an asymptotic limit.
     
  11. Robittybob1 Banned Banned

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    That is interesting in that another person on another forum has suggested there will be a limit at 50 m/sec tangential speed.
    With an asymptote - that also means it is heading toward a limit too doesn't it?
    Thanks BillyT.

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  12. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Easy to show he/she has the asymptotic speed correct:

    Assume it rotating with 50m/sec, then the "fast" bullet slams into its block with relative velocity of 1000-50 = 950 AND so does the "slow" one hit its block at 900+50 = 950 so no more net impulsive torque is applied. Your spread sheet of formula I gave should almost reach 50m/ sec too, if neither of us made an error.
    Yes, but never gets to it exactly with finite n.
     
  13. Robittybob1 Banned Banned

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    Well it seems we are all in agreement on the asymptotic limit. I'll try and set up the sheet this weekend. Thanks Billy.

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  14. Believe Happy medium Valued Senior Member

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    The only limiting factor of how much IR radiation that molecule can take is how hot it can get before it simple breaks apart.

    There may be some particular configuration/orintation of the molucule that would not allow the radiation to be absorbed (think ramen spectra, di-pole moments) but, the molecule is moving very quickly and that orientation is not a major contributor over all (i.e. it may only be in that particular configuration 0.001% of the time)
     
  15. Believe Happy medium Valued Senior Member

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    I have to wonder what the point of this is. Is it simply a mental exercise?
     
  16. Robittybob1 Banned Banned

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    I was looking to see if the planets with greenhouse gas atmospheres, does the absorption of the photon energy cause the winds to blow pro-grade to the rotation of the planet and ultimately make the planet rotate faster.

    This is like saying the main factor when looking at a planet's rotation rate is its atmosphere. The energy is provided by the Sun.

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  17. Believe Happy medium Valued Senior Member

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    Oh shit dude!!!! Have never mind what I said before and good luck with that!!!!

    Pretty sure the answer is no, but I really don't know. It may factor into the speed in some way that I don't understand (but wouldn't it have to speed up then since the sun is always adding energy? I am sure that this has not been observed, but it may only work to some maximum that has already been reached then stop). I think that the energy comes pretty much straight on though so it would have to be indirect and the wind blows in every direction (though some more then others but I believe the direction depends on where you are) so idk.

    Good luck again and I hope you have fun doing it!
     
  18. Robittybob1 Banned Banned

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    The absorption of the infrared radiation is modified by the transfer of the absorbed energy to the kinetic movement of the other molecules around it.
    So the way I thought you suggested doesn't seem to be an an issue for the energy is always being transferred away from the CO2. Whether just heating will cause the gas molecules to disintegrate I not so sure, but I have heard of plasma which is extremely hot gas so it seems right at very high temperatures.

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  19. Believe Happy medium Valued Senior Member

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    Some will be transferred out in the form of re-radiation, and collisions but some will stick around. This is reason why emmisions always have a longer wavelength then the initial excitation, some energy is always lost as translational, vibrational, and rotational energy. If that motion gets strong enough (I.E. the visible/ir radiation source is much more intense then the sun, or a high energy particle like gamma, xray, UV in some cases) then the molecule can break the chemical bonds that hold it together. It needs a stronger source then the sun because the energy transfer due to collisions (i.e. the faster moving molecules hitting slower moving molecules and losing energy) is carrying it all away before it can get going fast enough.
     
  20. Robittybob1 Banned Banned

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    @ Believe - are you working in a science field? It is a long shot idea and my science was a bit rusty so that is why I have reduced the problem down to simple conservation of momentum problems. Once I get really confident what happens to the momentum and kinetic energy in the interactions I will go back to looking at the the photon situation.

    I imagine the problem will take a year to resolve at the rate I'm going.

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  21. Believe Happy medium Valued Senior Member

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    Not working, but I am on my last semester of my BS in chemistry

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    . The motion generated by the things I spoke of is random so it would generally be equal in every direction. (one goes up, one goes down, one goes east, one goes west, ext) As far as the sun shining on the earth is concerned the net motion is away from the side the sun is shining on (heat spreads out) but it would only favor one direction over another due to topography such as mountains as it will take the path of least resistance.

    I'm not an atmospheric scientist so I may be a bit off but the general idea should be correct.
     
  22. Believe Happy medium Valued Senior Member

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    The more I think about it the more sure I am that it would not effect it. Though it seems large the atmosphere is not a large part of the mass of the earth(I.E. the atmosphere is only about 11km thick and not very dense, the earth is around 12,000km thick and made of far denser stuff) so for it to effect the earths rotation I would think the wind would have to blow your house down and then some. It could still could have built up slowly to some critical value though but I'm about 99% sure now.
     
  23. Believe Happy medium Valued Senior Member

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    OK,(it's late and my brain no work good) now I'm 100% sure because the critical value thing would require the wind to blow in the same direction which we know it does not. In fact where I am at (not telling) the wind blows different directions in the same day and where I am is not unique. This could be possible on another planet with different topograhical features like a large mountain range that causes the wind to blow the same way all the time. The calcuations could still be fun!
     
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