nstansbury Posted June 14, 2008 Share Posted June 14, 2008 (edited) Hi all, apologies for cross posting, but I really need some sanity checking done on these equations, especially in my proving [math]E=mc^2[/math] from the Planck constant. http://www.scienceforums.net/forum/showthread.php?p=414167 I think the SFN Mods would probably prefer comments in the originating thread. Cheers. Edited June 14, 2008 by nstansbury Link to comment Share on other sites More sharing options...
ydoaPs Posted June 14, 2008 Share Posted June 14, 2008 E2=(mc2)2+(pc)2 Link to comment Share on other sites More sharing options...
nstansbury Posted June 14, 2008 Author Share Posted June 14, 2008 Yep that's what I have too - I don't follow? Link to comment Share on other sites More sharing options...
Klaynos Posted June 14, 2008 Share Posted June 14, 2008 For stationary objects the pc term is 0. Link to comment Share on other sites More sharing options...
nstansbury Posted June 14, 2008 Author Share Posted June 14, 2008 For stationary objects the pc term is 0 Electromagnetic waves aren't stationary - they have momentum themselves. Link to comment Share on other sites More sharing options...
Klaynos Posted June 14, 2008 Share Posted June 14, 2008 Electromagnetic waves aren't stationary - they have momentum themselves. Yep and for EM waves the mc2 part is 0. For a moving massive object neither term are 0. Link to comment Share on other sites More sharing options...
nstansbury Posted June 14, 2008 Author Share Posted June 14, 2008 Yep and for EM waves the mc2 part is 0. For a moving massive object neither term are 0. I beg to differ: [math]p=\frac{h}{\lambda}=\frac{E}{c}[/math] http://en.wikipedia.org/wiki/Momentum#Momentum_in_relativistic_mechanics Link to comment Share on other sites More sharing options...
Klaynos Posted June 14, 2008 Share Posted June 14, 2008 I beg to differ: [math]p=\frac{h}{\lambda}=\frac{E}{c}[/math] http://en.wikipedia.org/wiki/Momentum#Momentum_in_relativistic_mechanics E2=(mc2)2+(pc)2 m=0 E2=0+(pc)2 =>E=pc Which is the same as you said. So we do infact agree... Link to comment Share on other sites More sharing options...
nstansbury Posted June 15, 2008 Author Share Posted June 15, 2008 Yep and for EM waves the mc2 part is 0. Apologies, I miss read - thought you were suggesting the pc value was 0. Link to comment Share on other sites More sharing options...
Radical Edward Posted June 15, 2008 Share Posted June 15, 2008 do bear in mind that being a relativistic equation you should be careful about what you mean by moving and stationary. Link to comment Share on other sites More sharing options...
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