Will CO2 absorb photon in all directions?

There are some big steps taken in the above post ones that I have never taken before. Being comfortable talking the language of blueshift and redshift, and where the energy and momentum components come from and ultimately end up will need a bit of brushing up.

I want to use the Doppler shift maths to show its proof beyond dispute. That is my goal to complete over the next couple of weeks.

If parts of it are wrong hopefully the maths will show it up.
Where do we start with the maths?

 

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Quick comments (as busy with complex taxes)

Doppler shift is when the SOURCE of photon is moving.

Gravity shift is when the photon is climbing out of gravity field (red shift) or falling into one (blue shift) - much like a brick loses energy if thrown away from gravity source of gains energy if falling towards gravity source.

Motion of the absorber of photon towards source makes a blue shift. I.e. for that absorber the photon has more energy. This energy is not well thought of as coming from the slowing down of the absorber, as perhaps the absorber is massive, like the earth. In the frame of reference in which the absorber is at rest (Ie, an inertial frame moving with the absorber) that photon has more more energy, is blue shifted EVEN IF IT IS NOT ABSORBED - GET NO ENERGY FROM THE FRAME IT IS JUST MOVING THRU.

like wise for red shift with frame moving away from the soures. Frame ned not have high speeds - this is not based on relativity. Just that when headded at the approaching photon with say c/100 while 100 cycle of photon go by a stationary frame, your 0.01c frame will move thru one cycle more - I.e. instead of seeing photon 100 cycle in that time you will see 101 cycle pass or come at you if moving towards light source at 0.01c - not a speed with significant relativistic effect but a frame with photons having 1% more energy due to the blue shift.

Sorry no time to fix typos etc.

 

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It was good of you to take a break from the bookwork. When I read what you say I see we are very close to agreement on 75% of the issues.

The other 25% obviously I feel I am right and I will try and explain my reasoning from a different angle so you see where I am coming from.

Now above you are taking a situation where the difference in speed on the two sides is 0.02c (0.01 faster on one side and 0.01 slower on the opposite side.) So a photon that normally produces 100 cycles shows up as 101 on the blue shift side and presumably 99 on the red shift side.

Now you are thinking the higher frequency photon is going to have more momentum to impart but in fact it only looks like it has more frequency because the absorber is moving toward the photon. The photon has still only got the 100 momentum quality. The extra impact comes from the absorber moving. The momentum cancellation may be 101 quality but the extra 1 comes from the molecule.

The other thing to remember is that when there is a specific frequency (say 100) that can be absorbed it is the 99 photon blue shifting to the 100 and on the other side (red shift side) the 101 red shifting to 100.
So from the planets perspective it is a weaker photon being effective on the blue shift side whereas the heavier duty photon is effective on the red shift side.

We should be able to demonstrate that in a lab to see who is right.

 

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No, part of your post I have made bold is false. The blue shifted photon HAS more energy and momentum in the reference frame where the molecule that absorbs it is at rest. Everyone, not accelerating, molecules included, is in some rest inertial frame. No one of them is more “correct” than any other. In that rest frame, the frequency, energy and momentum is larger than in the frame where the source (the sun) is at rest.

You are not even consistent in your thoughts. You have understood my blue shift with 0.01c making there be 101 cycles instead of the 100 an observer in the sun´s rest frame would see. You also, I think, know that the energy & momentum of a photon is directly proportional to the photon frequency. And that if there are 101 cycle in the same time that there are only 100 in the sun´s rest frame, then in the blue shift frame the frequency is 1% higher. Yet you want to inconsistently claim the 1% higher frequency photon has no more momentum, still only 100 units. That it transfers more when absorbed by "borrowing" 1 unit from the absorber. - You don´t stop to think that it many not be absorbed and yet still in the frame that potential absorber is at rest you agree it has higher frequency, i.e. is "blue shifted." When you accept the POV I am trying to explain to you (which is correct), this inconsistent - self contradictory problem in your {wrong} POV goes away.

Perhaps it will help you to consider a bullet instead of a photon. Lets say it is approaching you at 1000mph. I.e. in your rest frame it has closing speed of 1000mph. Bob, however is in a rest frame moving the same direction as the bullet but wrt to you 999mph. The very same bullet has lethal speed in your rest frame and Bob will hardly feel it if it hits him. Point is that the energy of a bullet OR a photon is a strong function of what frame it is measured in.

You falsely think the wave length, the energy and the momentum of a photon does not depend upon the frame it is measured (or absorbed) in. You falsely think the photon has a constant amount of energy and momentum for all frames. I.e. you think the photon (but not the bullet, I assume) has 100 units of momentum in all frames. In the inertial frame where the photon is blue shifted it has more energy, momentum and a shorter wavelength than in a frame where it is red shifted – This is true whether or not it is absorbed, just like it is for the bullet, whether or not it hits someone.

SUMMARY: Your POV is wrong, about both bullets and photons, and you know it is wrong about bullets. The energy and momentum of a photon or a bullet strongly depends upon the frame they are measured (or absorbed) in (or just are passing thru with no interactions).

Yes if the absorber in a frame where the photon is red shifted makes the photon energy IN THAT FRAME, match the absorption frequency, then it, not the blue shifted photon can be absorbed.

What you are failing to consider in the planet limb absorption problem is how many photons can transfer energy and momentum to CO2 and how many to other molecules (like ionize O2, N2 or disassociate them to O + O and N + N, etc.). I.e. you are not considering the solar distribution of photon at the Earth, nor the fact that there are few IR absorbers in the earth´s atmosphere compared to those that UV can ionize or disassociate. Solar spectrum as low level is like:

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This is the ground level distribution. There is much more UV at top of the atmosphere.
Now it is true that the limb going away from the sun will have more of visible and near IR (right side of graph) photons shifted to match CO2 (and H2O, CH4, etc.) molecule absorption bands than the limb moving towards the sun. The blue shifted limb has only the lesser number of IR photons to the right in the graph with less energy than the absorption bands, energized by the blue shift to have the required IR energy to be absorbed.

For example, consider the first deep absorption notch of the graph at about 0.94 micrometers. On the left of it there are many more photons that can be red shifted into that absorption notch than there are photons of lower energy (on the right side of the notch) that can be blue shifted into the notch. Thus, you would be correct if the main transfer of momentum were by IR absorption of solar photons. These absorbers make up tiny part of the atmosphere. It is 98 or so percent of molecules that do not absorb IR. Momentum is transferred to them by absorption of solar UV (to ionize, making the ionosphere or to disassociate making O, N, or even O2 from ozone (O3) etc.) Now these UV photons are on the left side of the solar peak (in the visible) so the argument reverses.

I.e. in the UV there are more photons just lacking the required energy (too close to visible wavelengths) to ionize or disassociate the molecules making up about 98% of the atmosphere. I.e. a little blue shift will give these greater in number near visible UV photons the energy they need to be absorbed or to ionize the dominate gases of the atmosphere. That is the limb with blue shift interacts more with the atmosphere than the limb with red shift. Also these more energetic UV photons do NOT need to match some narrow absorption band.

Note how steeply the number of UV photons decreases as you move to the left in the graph – i.e. as you move to higher energy UV. For interactions of disassociation (O2 goes to: O + O etc.) is a reasonably well defined energy required but any UV photon with the ionization energy OR MORE can ionize. Thus almost all of the UV is absorbed in the air and does not reach the ground. Because of the steep decrease in number of photons with increase of UV energy there are very few with too much energy to disassociate molecules that a little red shift would give them the correct disassociation energy compared to the larger number (nearer to the visible) that lack just a little energy to disassociate which the blue shift of the limb moving towards the sun can give.

Graph from: http://solarcellcentral.com/solar_page.html which is mainly concerned with solar cells. I did not read text, but hope they discussed there is an "energy matching" problem for them too. Only the photons with exactly the band gap energy can be converted 100% into electrical power. Those with less energy can not lift a valence electron up across the band gap to the conduction band. Those with more than the band gap energy will initially lift an electron up to an higher energy level of the conduction band states than bottom energy level, but as most of the states in the insulating material of the solar cell in the conduction band (it is not a metal where many states of the conduction band are occupied) the extra energy of the electron higher up than the bottom of the conduction band is quickly converted into heat as the electron falls down (in energy) to an unoccupied state near or at the bottom of the conduction band.

Thus some solar photons lack the energy to make any power and others (most) have too much and make heat as well as electric power. For silicon and its band gap, this means the theoretical max efficiency (with our sun) is only 22%. If the sun had photon distribution more like a red giant star has, the max efficient possible with silicon would be higher. On earth to get greater efficiency, two different layers of different materials are used. The thin one nearer the sun has the greater band gap energy difference, so most photons just pass thru it but the UV and "blue" photons have enough energy to pump electrons up into its conduction band with less heat and more power made; but of course there are not so many of these more energetic photons.

Many of those that pass thru the outer layer do have enough energy to lift electrons up across the smaller band gap of the inter material (perhaps germanium) so greater than 22% efficiency can be achieved, with more expensive solar cells - not worth the extra cost on Earth where space is available to just buy more simple one-layer cells to get more power per dollar spent. On a space craft however, where collection area costs money and weight to make, the dual (or more layer cells) do pay off - are more economic per kWh generated.

PS I hate doing taxes and love to teach but must get back to the IRS. At least Schedule D and new confusing form 8949 of it is done now.

 

You gave me a good idea. Let's just work with bullets only for a while. The shooters are standing on the Sun firing bullets toward the planets and the planet is spinning fast enough so that the .22 calbre are non-injurious on the Red shift (RS) side but injurious on the Blue Shift side (BS).
When a bullet is effective it lodges and transfers its moment through the victim and hence to the planets spin rate.
Non-injurious bullets are not absorbed.
The shooters have a range of guns available, and they find that .303 are injurious on RS side.
The number of momentum changes are limited not by the bullets or therate of firing but the number of combatants available which are evenly spread and never really dies but carry on round and round the planet in their daily lives. Suffering wounds and recovering.
Now as even I can see the ones on the BS side are getting the same impact Injury from the bullet for they are running into it, as on the RS where they are running with the bullet.

It must be possible to see that since only high momentum bullets are being captured on one side and low momentum on the other side, there is a net momentum imbalance which is going into the angular momentum of the planet.

OK as I picture it is not being transferred entirely at the moment of impact but it is still being transferred somewhere somehow for the high momentum bullets are being lodged.

But it has me a bit stumped just at the moment why that situation would result in the planet spinning faster and faster but because it is such a basic mechanical analogy once I'm awake again I'll lay out the problem on an Excel sheet and let the computer do the maths.

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Fine, but you must tell the relative number of bullets stopped in each ot the two limbs.

I.e. give data on the source (how many are 22s, how many are 45s, etc) and what fraction get stopped by the atmosphere.

I gave graph of the sun´s different type of photon bullets (at ground level. For top of atmosphere distribution, fill in all the absorption notches on the right of peak and greatly increase the UV on left of peak.

Note point I made that all of the "UV bullets" transfer momentum because (1) not only a critical wavelength or two that can be absorbed and (2) ~98% of the atmosphere cannot absorbe IR.

Thus the UV monentum transfer is dominate and on the blue shift limb there is more of it as the near visible UV is made more energetic by the blue shift and there is very little "too energetic" at left edge of graph to be red shifted to match the disassociation energy reqirements.

 

Well at first I’m going to see if I can do the maths for just one lump of inelastic plasticine on each side of a spinning disc into which bullets of the differing mass and different speeds will do if they strike the plasticine with the same momentum and energy.

Did you know I have no idea where to begin!

But I'm going to look at the energy momentum relationship for mass.
Can you have different masses and speeds yielding the same momentum and energy as does a photon where the momentum and energy are proportional?
With matter there is the V^2 relationship so it might not be linear.

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Could this be right?

http://physics.stackexchange.com/qu...ationship-between-kinetic-energy-and-momentum

So if the above quote is correct we can't get two different bullets to behave, like in the analogy, to have the same momentum and energy. I didn't know that.

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I have just drawn up a pivot table for that formula and there are endless solutions to that formula that give the same energy by varying momentum and mass. Did I do it right?? Yes you can get the same energy in the two masses but at each time they will have a different level of momentum.

The Formula was OK but my worksheet is a bit suspect (found the error, and now working).

 

So when photons perform a task like vibrate a CO2 molecule it the result of the energy of the photon (frequency) and the momentum changes to the absorber is a secondary effect, so to make the analogy fit the situation we will look at bullets with the same energy rather than the same momentum for we can't get the energy and momentum to match when there is differing mass.
But how do we account for the motion of the absorber? OK when we look at the velocity we will look at the relative velocity. Is this right - can we add velocity like that?
Are 2 cars impacting at 20 mph each the same energy as a 40 mph car impacting a stationary car?

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1500 kg 8.9408 m/s (20 mph) 119906.857 Joules (59953.42848
Joules energy of one car X 2)
1500 kg 17.8816 m/s 239813.7139 Joules

So there is more energy in the relative motion frame. So look can't add the motion and treat it that way for energy but for momentum yet you can.
If a bullet is going at 1000 m/sec how much extra energy is added to the impact if the plasticine is moving toward the bullet? What about if the plasticine block is moving away?
So it definitely is an inelastic collision. There will be no bouncing off.
Looking at a few similar situations on the net - solution depends on the conservation of momentum that is guaranteed, and not the conservation of energy. A real catch .22 situation!

So what I'm thinking is that we'll need bullets of the required energy and once I can show they have sufficient energy then just look at the effects of the conservation of the momentum.

Question then is does it require a higher energy bullet to go through the thickness of the plasticine if it is already in motion in the same direction of motion. Gut feeling it is but we want the maths.

 

I will not go off subject and discuss in detail bullets impacting targets. I only mentioned bullets to help you understand that in general energy depends upon what frame you are measuring it in. Below quote of your post, I will do the same with cars. But I always try to answer direct questions, if interesting and I have time.

No. You are trying to compare, perhaps equate the energies in to different frames. The crushing of both cars will be identical as this is the same event seen or described in two different frames.

In frame A both cars move at 20mph. In frame B one is stationary (I´ll call that car, now and in next discussion, "car S") and other moves at it at 40 mph.

Consider frame C in which car S is moving to the right at 1000mph an the other car is also moving to the right, but at 960 mph. Same collision, Same crushing of the cars in the collision, but much more total KE in frame C than in either of the two frames you mentioned. I.e. again proving that energy is very frame dependent.*

Instead of going off on this tangent about bullets, do you now agree that the 25% difference (your post 263) between what I have been telling you and what you believe has been reduced to zero? I.e. that you now fully accept that the photon´s energy and momentum is NOT a constant for all frames, AND the idea that it is a constant (momentum always 100 units you said) is false? Also that during absorption by moving CO2 molecule at rest in frame where it is seen as a blue shifted photon, the CO2 is NOT supplying energy to permit the photon to have energy equal to the excitation energy? I.e. that in that CO2´s rest frame that the photon actually HAS more energy (and momentum) than it had in frame where the sun is at rest? And that is still true , even if there is no collision or absorption by a CO2 molecule?

If not quote the part of my post explaining (and I think proving) that the photon really does have more energy in the CO2 molecule´s rest frame so there is no "borrowing" from the KE or momentum from the CO2 molecule to get the required excitation energy as you suggested with your 25% disagreement with my POV. Your POV only makes sense IF there is something special about the frame you have chosen - the one where the CO2 is moving towards the sun at about 1000mph due to earth rotation.

Although there is no preferred frame in which to describe physical event (Frame C above is OK – just as good as the others) it is commonly easier to work / calculated in the center of mass frame or in case of photon absorbing in the frame where the absorber is at rest (as the photon has velocity C in all frames, only it energy and momentum is frame dependent.

* Note this is not the case for random thermal KE, which we call temperature. I.e. the temperature of glass with unchanging mass of ice floating in water, is 0C (or 32 F) in ALL frames. Showing this mathematically is not too hard. Has to do with fact only the random motion makes KE. If all molecules had in addition to their random motion a 1000 mph velocity componet in frame A, then yes their total KE is greater than in the frame where class is sitting on your table.

PS If I have been able to convert your nearly 100% false original POV to the correct POV, understanding, we both should be happy about this.

 

I've just woken up and reading your reply. Maybe I've got a migraine coming on or something but I am real difficulty to relating to it. Sorry about that.
But one thing I want to make clear is that I'm going to believe what comes out of my Excel sheet.
So that is why I am a bit mystified that you reckon 2 cars at 20 equals 1 at 0 and the other at 40. For using the energy equation the values were different.
For the momentum values maybe it was the same, but I hadn't put them in but off the top of my head M*V + m*v = M*V2. (1500 kg * 20 mph * 2 cars = 1500 kg * 40 mph * 1 car + 1500 kg * 0 mph * 1 car)
But even then the impact is going to result in a different type of net motion.
Head on they may lock together and have zero velocity but in the other the higher speed car will drive the other backward.
Look if we can't solve simple collision how are we going to tackle planets.

So if you are right or wrong I had better find out.

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Let the Myth Busters provide the answer ( but not quite the same one)
http://mythbustersresults.com/mythssion-control

 

Of course the computed KEs will be different for calculations in different frames - point I have been trying to get you to understand and accept (for cars or photons)

If there is garbage into spread sheet, what do you expect to come out?

There are an infinte number of frames in which this two car collision can be accurately described and all will have different KEs - Chosing one as correct is how you get garbage into your spread sheet, but if you must compute and use KE, do it in my frame C - Perhaps then you will more quickly understand that nothing valid comes from KE analysis you are planning (as I undersatnd it) - only garbage if switching to a different frame´s POV of the same event changes the results.

Note added later - I just read your post 273. It is totally irrelivent to your original question I answered as there are two entirely different events: one has two cars colliding and other has car run into brick wall. That is not just one event, described in two different frames.

 

But one of the replies had an interesting observation in it regarding capturing eggs, which could be analogous to catching photons.

He was right about the egg but wrong about some of the rest.

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New study shows "Ice age end was accelerated by CO2"
http://www.theregister.co.uk/2012/04/05/ice_age_ended_by_co2/

 

It is quite a mystery as to how the earth ever escapes from an ice age when during one, the earth´s albedo is much lower due to the snow and ice covering large parts of it reflecting sun light (IR included) – I.e. there is much less solar heating during an ice age.

I lean to the theory that Earth escapes the ice age´s grasp with a strong volcanic eruption that pumps large amounts of GHG into the atmosphere where it remains for decades. That is very consistent with the new observations your link reports as CO2 is relased with volcanic eruptions in great tonage along with many other GHGs.

The “earth wobble” theories would seem to make ice ages occur at much more regular intervals than they do. Large volcanic eruptions are less regular in their occurance, like ice age escapes are.

 

That could be right, but what they established (at this stage) was that the CO2 rose before the ice melted, rather that the CO2 rose after the ice melted.
So the end of the ice age is a global warming phenomenon. It certainly doesn't pay to bring the GHG levels down too low.

 

Yes, it is decade or so AFTER the volcano eruption before any significant amount of ice melts. For first year or so more ice forms as the sky is black with dust. Read about the volcano caused "little ice age" of couple 100 years ago - it even snowed in NYC in the summer time.

There is no danger the GHG level will get too low (dropping Earth´s average temperature to about freezing as I recall) as the most important GHG is H2O. There will a lot of that so long as Earth has oceans.

 

This is likely of little impact, but during an ice age two other things happen.

There is a reduced amount of vegetation and more decay of the stuff that died out as the ice advanced, releasing CO2. A very small impact.

What could be a potentially greater impact is that as ice increases, sea level falls. There is a great deal of methane hydrate which should sea level fall enough would be destabilized. Though methane has a shorter half life in the atmosphere it is a far more significant GHG than CO2 and degrades to CO2.