Resonance condition

Resonance condition

Thanks again! The resonance condition is the most important feature of the m theory of LENR, so the comparison with the Woods Saxon (WS) potential can be carried out in various ways, giving expressions for m(r) and dm(r)/dr in terms of the parameters of the WS potential. This goes beyond the models used so far for m(r) and dm(r) / dr. It is also necessary to be careful about the differentiation of the WS potential, Eq. (1) of Note 431(5), to give the WS force Eq. (3), because the parameter a0 must appear in the force in order to keep dimensional units right. If r =0 in the WS potential it becomes a constant so the WS force vanishes. Clearly, this is incompatible with an infinite m force at r = 0. So a more general solution of 2m(r) = rdm(r) / dr is needed, because the solution r = 0 means no WS force. The comparison of Eqs. (3) and (4) can be carried out in various ways with computer algebra, so m(r) and dm(r) / dr can be expressed in tems of R, U sub 0, a sub N and a0. Firstly my hand calculation of Eq. (3) should be run through Maxima to check it. Maxima will give a result without a sub 0 being present, but that is dimensionally incorrect. So a sub 0 has to be devised to keep the dimensions correct.

Resonance condition

The condition 2m(r) –> r*dm(r)/dr is only reached for r–>0 in the
models we have investigated so far. I am not sure if this has an impact
on the comparison with the Woods-Saxon potential.

Horst

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