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#55
Feb1212, 01:32 PM

P: 152

Right? If you use the concept of "relativistic mass" the relation still holds when there are moving masses. Right? 


#56
Feb1212, 01:39 PM

P: 3,188




#57
Feb1212, 03:50 PM

P: 480

Assuming we know that energy is linear correlated with mass (as used in Newton's definition of momentum) but we do not know the proportionality factor k: [itex]E = k \cdot m[/itex] Then the change of mechanic energy is [itex]dE = F \cdot ds = k \cdot dm[/itex] According to Newton's second law the force is [itex]F = \frac{{dp}}{{dt}} = \frac{{d\left( {m \cdot v} \right)}}{{dt}} = m \cdot \frac{{dv}}{{dt}} + v \cdot \frac{{dm}}{{dt}}[/itex] Integration of the resulting differential equation [itex]\frac{{dm}}{m} = \frac{{v \cdot dv}}{{k  v^2 }}[/itex] results in [itex]m = \frac{{m_0 }}{{\sqrt {1  \frac{{v^2 }}{k}} }}[/itex] The constant of integration m_{0} is the mass of the body at rest and as this equation gives rational results for k<v^{2} only the unknown proportionality factor k must be the square of a maximum velocity that no body can reach or exceed. In relativity there is such a velocity: the speed of light in vacuum. Therefore there is only one possibility for a linear correlation of energy and mass: [itex]E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1  \frac{{v^2 }}{c^2}} }}[/itex] 


#58
Feb1212, 04:23 PM

P: 3,188




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