Re: E = 3/4 mc² or E = mc²? The forgotten Hassenohrl 1905 work.

SOME CORRECTIONS TO THE PREVIOUS POST:
On Mon, 2 Dec 2024 0:36:16 +0000, rhertz wrote:
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From their paper, this is the balance (as published in 1932):
>
>
Lithium7 amu 7.0104
Hydrogen amu 1.0072
        8.0176
>
Helium amu 4.0011
Helium amu 4.0011
        8.0022
>

Difference  Δamu = 0.0154 ± 0.003 amu = 14.3 MeV ± 2.8 MeV. To this, it
has to be added an extra energy of 2.7 MeV
Δamu varies between 0.0124 amu and 0.0187 amu.
The total change in energy ΔE varies between 14.3 MeV and 19.8 MeV.
It corresponds to equations:
ΔE = 2/3 Δmc²  for 14.3 MeV
ΔE = 7/5 Δmc²  for 19.8 MeV
Authors claimed that momentum was accounted and conserved.
These values are written, after hundred of experiments, following the
relationship:
7:3 Li + 1:1 H ---> 4:2 He + 4:2 He + energy
https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1932.0133
This corresponds to the equation:
Considering that momentum is conserved and the energy of the proton is
0.6 MeV, the final values are closer to Hassenohrl than to Einstein.
Hardly an experimental verification of ΔE = Δmc², as relativists have
claimed as this being the first experimental proof of such equation.
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