
#55
Jan813, 07:09 PM

P: 18





#56
Jan813, 07:20 PM

P: 784

In relativistic physics momentum isn't defined as ##p=mv## but rather as ##p= \gamma m v## where ##\gamma \equiv \frac{1}{\sqrt{1  v^2/c^2}}##. So your derivation doesn't quite work. It's been a while since I've had to derive E=mc^2 (I had to do it for an extra credit problem on a test) so I'm not sure whether or not the derivation on that site can be simply expanded to be correct or not, but in it's presented form it does not work.
As I said though I approve of it as a teaching method  it's quite simple  as long as readers are aware that a more thorough proof is required. 



#57
Jan813, 07:39 PM

P: 18





#58
Jan813, 08:08 PM

P: 77

Landau's derivation of the equations of relativistic mechanics in his book Classical Field Theory might help you better understand relativity.




#59
Jan813, 08:11 PM

P: 784





#60
Jan813, 08:35 PM

P: 18





#61
Jan813, 08:42 PM

P: 18





#62
Jan813, 11:54 PM

P: 492





#63
Jan913, 01:48 AM

P: 18





#64
Jan913, 02:20 AM

C. Spirit
Sci Advisor
Thanks
P: 4,925

The Lagrangian [itex]L = mc^{2}\sqrt{1  \frac{v^{2}}{c^{2}}}[/itex], [itex]p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1  \frac{v^{2}}{c^{2}}}}[/itex], [itex]E = p\cdot v  L = \frac{mv^{2}}{{\sqrt{1  \frac{v^{2}}{c^{2}}}}} + mc^{2}\sqrt{1  \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{{\sqrt{1  \frac{v^{2}}{c^{2}}}}}[/itex] so in the rest frame we have that [itex]E = mc^{2}[/itex]. The speed of light squared quantity is just a unit conversion. If you work in natural units then you won't even see it explicitly.




#65
Jan913, 02:36 AM

P: 18

And I dont know what is "ρ" in the equation . Could you explain? 



#67
Jan1013, 10:02 AM

P: 18




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