… {Consider a planet with 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.
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).
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. ....
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:
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.
