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Is antigravity a bunch of hot air?
22 years 3 months ago #2588
by Jim
Replied by Jim on topic Reply from
I have no complaint with any of the speed/temperature models you are describing. And this is good info for a scale many magnetudes higher than a molecule on nanometer scale. What I am trying to learn is how one molecule gets hotter. What do you add to one molecule make this happen? If you look at this as one molecule receiving energy how can it be that a photon(E=hf) is being added? And if not a photon then what is the cause? Sunlight is many photons warming many molecules true enough. One molecule-that is not as easy to explain.
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22 years 3 months ago #2589
by Jim
Replied by Jim on topic Reply from
After rereading the original question that started this chain I see my issue is not related in any direct why to the issue of hot air rising in cold air masses. Maybe it would be better to drop this because it is irrelevant.
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22 years 3 months ago #2686
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
> [Jim]: What I am trying to learn is how one molecule gets hotter. What do you add to one molecule make this happen? If you look at this as one molecule receiving energy how can it be that a photon(E=hf) is being added?
When a molecule gets hotter, it travels faster. The speed *is* the temperature, not something inside the molecule.
The mechanics of the absorption of a photon are that an electron absorbs it and is elevated to a higher orbital level. Later, the electron may re-emit the photon and drop back to its original level. Both absorption and re-emission can change the speed of the molecule, altering its temperature. The molecule is otherwise unchanged after this cycle is complete. -|Tom|-
When a molecule gets hotter, it travels faster. The speed *is* the temperature, not something inside the molecule.
The mechanics of the absorption of a photon are that an electron absorbs it and is elevated to a higher orbital level. Later, the electron may re-emit the photon and drop back to its original level. Both absorption and re-emission can change the speed of the molecule, altering its temperature. The molecule is otherwise unchanged after this cycle is complete. -|Tom|-
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22 years 3 months ago #2590
by Jim
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The QM result you describe is a model that works well too, but, in this case it fails in my opinion because the photon in QM is E=fh and the energy of this photon is way to much to be dispersed into the vast number of molecules that much energy would heat. I hope this makes a little sense so it is not misunderstood. The photon is indivisible according to QM and the energy in heating a molecule is many times less than one photon would give the electron in the molecule model you offer. This is a puzzle not even investigated by QM as are other items the QM method avoids in total.
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22 years 3 months ago #2824
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
> [Jim]: The QM result you describe is a model that works well too, but, in this case it fails in my opinion because the photon in QM is E=fh and the energy of this photon is way too much to be dispersed into the vast number of molecules that much energy would heat.
You lost me here. As I explained, the molecule is left unchanged by the cycle that includes absorption and re-emission of a photon. So it doesn't matter how much energy is involved. The only aspect of the molecule changed by the process is its speed, and that is the manifestation of its "heat". I agree QM has problems, but don't see that this is one of them. -|Tom|-
You lost me here. As I explained, the molecule is left unchanged by the cycle that includes absorption and re-emission of a photon. So it doesn't matter how much energy is involved. The only aspect of the molecule changed by the process is its speed, and that is the manifestation of its "heat". I agree QM has problems, but don't see that this is one of them. -|Tom|-
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22 years 3 months ago #2956
by Jim
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The in and out of the photon leaves the molecule as it was; no heating at all. The heating is quite a bit different.
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