Mordred Posted Sunday at 03:36 PM Share Posted Sunday at 03:36 PM (edited) 38 minutes ago, MJ kihara said: That's not the actual volume if you have used it. 10^-15*10^-15*10^-15 cubic meter. Try again it's simple powers and division if you want to use inches 3.9374 ×10^-14 How many 10^{-15} meters fits 1 meter. Recall the author specified volume as well. Have you never learned exponent rules ? \[m^{-n}=\frac{1}{m^n}\] So take \(10^{-15}=\frac{1}{10^{15}}\) Edited Sunday at 03:59 PM by Mordred Link to comment Share on other sites More sharing options...
MJ kihara Posted Sunday at 04:20 PM Share Posted Sunday at 04:20 PM Then 1/10^15*1/10^15*1/10^15 it must have been a typo on the statement you made above. The author is using volume. 1 Link to comment Share on other sites More sharing options...
Mordred Posted Sunday at 04:27 PM Share Posted Sunday at 04:27 PM (edited) The authors unit is 10^-15 meters not meters^3 for range of force. Ie radius of each volume but just in case I will check them later after my meeting. It was 1 am when I did that set so will recheck Edited Sunday at 04:41 PM by Mordred Link to comment Share on other sites More sharing options...
Mordred Posted Sunday at 08:16 PM Share Posted Sunday at 08:16 PM 3 hours ago, MJ kihara said: Then 1/10^15*1/10^15*1/10^15 it must have been a typo on the statement you made above. The author is using volume. Yeah your right good catch must have entered something wrong on calculator and didn't spot it. Too early in am last night. Still incredibly large energy density Well above current cosmological constant in joules/meter^3 for simplicity we can ignore spherical 10^45 ev/m^3 gives 1.60*10^26 joules/m^3 Link to comment Share on other sites More sharing options...
MJ kihara Posted yesterday at 02:26 AM Share Posted yesterday at 02:26 AM 6 hours ago, Mordred said: Yeah your right good catch must have entered something wrong on calculator and didn't spot it. Too early in am last night. Still incredibly large energy density Well above current cosmological constant in joules/meter^3 for simplicity we can ignore spherical 10^45 ev/m^3 gives 1.60*10^26 joules/m^3 If instead of using 1 eV per his SU(3) unit and use what he has provided after U(1) symmetry has broken as the effective photon mass 10^-18 eV I suppose you can recover the appropriate energy density. Link to comment Share on other sites More sharing options...
Mordred Posted yesterday at 02:33 AM Share Posted yesterday at 02:33 AM (edited) Honestly I tried looking for extremely low order interactions, even the numerous graphs I looked through for any superconductor studies were well out of range. Consider this a single electron has greater energy than the value you just gave. The least massive SM particle with mass being neutrinos but the momentum term is still an issue In all honesty the best chance to get the idea working and not a terrible idea in and of itself is simply get rid of the 10^123 factor. Use the Bose-Einstein methodology would solve a great deal of the problem. It would be a very easy fix that way. Thst would go a long way back to feasible by just dropping that 10^123. Secondly use pions for the SU(3) atoms being the lightest meson and use the Maxwell boltzmann statistics to calculate number density. No confusion now you have a clear defined state to work from. That would also open up a large body of similar studies involving QCD Meissner. Epions actually require higher energy levels under SUSY so don't use a SUSY particle. If it were me those would be the route to address the issues. Edited yesterday at 02:52 AM by Mordred Link to comment Share on other sites More sharing options...
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