Why not just answer questions here? Instead of having us go to that pop-up ridden tripod?
I looked at the page and there is a lot "given" without explanation. In the first page you calculat Fl, in the new page it is twice as big, because you use 2r. Okay, but then on the new page you take Fl as given (calculated on 1st page from m and 2r) and then take m and derive again 2r. Well, that's nice and circular.
Then the deBroglie wavelength. You say that v is some absurdly small value like 10
-34, where does that come from and what units are you using? I guess you chose that "velocity" to get the correct answer that 2r = lambda.
Now, answer please, and here on the board not on your page.
1. Why is this a force, when it does not have the units of force?
2. What units are you using anyway? Can you not just put it all in SI (i.e. kg m s). That would make all units much clearer (e.g. Fl = kg m instead of MeV fermi)
3. Why is the velocity v = 2.30078 x 10-34 and in what units?
4. All the other questions in previous posts.
Quote:
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Originally Posted by from new page
Therefore at the given speed λ = 2r. Demonstrating that the CLF model produces the same value for 2r as de Broglie produces for λ.
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But only for a very special value of the momentum of the particle that has not been explained. Any other momentum of the particle will show that lambda is not equal to 2r.
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Optimism does not change the laws of physics. (T'Pol)
A good scientist has freed himself of concepts and keeps his mind open to what is. (Dao De Jing 27)
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Martin (
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