DimaMazin Posted November 21, 2014 Share Posted November 21, 2014 Same as anything else: [math]\displaystyle m_o c^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - 1 \right)[/math] Do you think scientists measure neutrino KE by Joules? Link to comment Share on other sites More sharing options...
Strange Posted November 21, 2014 Share Posted November 21, 2014 (edited) Do you think scientists measure neutrino KE by Joules? I don't expect they measure the mass in kg, either. But you can multiply by the appropriate factor to convert to whatever units you want; it doesn't change the equation. Edited November 21, 2014 by Strange 1 Link to comment Share on other sites More sharing options...
swansont Posted November 21, 2014 Share Posted November 21, 2014 Do you think scientists measure neutrino KE by Joules? eV is more likely because it's the appropriate scale for atomic/nuclear/particle systems, but ultimately it doesn't matter since the conversion is trivial. 2 Link to comment Share on other sites More sharing options...
DimaMazin Posted November 22, 2014 Share Posted November 22, 2014 I don't expect they measure the mass in kg, either. But you can multiply by the appropriate factor to convert to whatever units you want; it doesn't change the equation. I only have wrongly called energy of rest mass by gravitational energy . Therefore you think I use another equation. Link to comment Share on other sites More sharing options...
Strange Posted November 22, 2014 Share Posted November 22, 2014 I only have wrongly called energy of rest mass by gravitational energy . Therefore you think I use another equation. No, you were just talking nonsense. Not particularly funny. Link to comment Share on other sites More sharing options...
DimaMazin Posted November 23, 2014 Share Posted November 23, 2014 No, you were just talking nonsense. Not particularly funny. EK(eV units)=(gamma-1)E0(eV units) * 6.24150934*1018 E0 - rest energy Link to comment Share on other sites More sharing options...
swansont Posted November 23, 2014 Share Posted November 23, 2014 EK(eV units)=(gamma-1)E0(eV units) * 6.24150934*1018 E0 - rest energy That won't get you energy in Joules, it will get you an answer in eV2/J. However, this doesn't affect the physics of the problem. So what's the point? Link to comment Share on other sites More sharing options...
DimaMazin Posted November 23, 2014 Share Posted November 23, 2014 That won't get you energy in Joules, it will get you an answer in eV2/J. However, this doesn't affect the physics of the problem. So what's the point? Yes. E=gamma*E0 EK=E-E0 EK(eV units)=gamma*E0(eV units) - E0(eV units)=(gamma-1)E0(eV units) Link to comment Share on other sites More sharing options...
Strange Posted November 23, 2014 Share Posted November 23, 2014 Yes. E=gamma*E0 EK=E-E0 EK(eV units)=gamma*E0(eV units) - E0(eV units)=(gamma-1)E0(eV units) Is there a point to these rather mangled equations? Link to comment Share on other sites More sharing options...
swansont Posted November 23, 2014 Share Posted November 23, 2014 Yes. E=gamma*E0 EK=E-E0 EK(eV units)=gamma*E0(eV units) - E0(eV units)=(gamma-1)E0(eV units) What is the point? While unit consistency is important (e.g. everything here has units of energy) the actual unit system used is a secondary consideration — it does not need to be specified in the equation. 1 Link to comment Share on other sites More sharing options...
DimaMazin Posted November 23, 2014 Share Posted November 23, 2014 Is there a point to these rather mangled equations? Let's check. gamma=2 E0=1eV then Etotal=2eV and EK=1eV E0=EK it's correct when gamma=2 Link to comment Share on other sites More sharing options...
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