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michael154
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Hey,
In a derivation of relativistic energy (in Physics for Scientists and Engineers, 5th edition, Serway and Beichner) they use a method of integration by substitution:
Given that
[itex]F=\frac{dp}{dt}[/itex]
and relativistic momentum is given by
[itex]p=\frac{mv}{\sqrt(1-(v^2/c^2))}[/itex]
[itex]W=∫F dx=∫\frac{dp}{dt} dx[/itex]
which is fine then they say
[itex]\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt(1−(\frac{(v^2)}{c^2}))}=\frac{m\frac{dv}{dt}}{(1−(\frac{v^2}{c^2}))^(\frac{3}{2})}[/itex]
(This is about halfway through the derivation.)
its that last step that i don't understand. why does the denominator term now have a 3/2 index? and why is du/dt given like that? Any help would be appreciated.
In a derivation of relativistic energy (in Physics for Scientists and Engineers, 5th edition, Serway and Beichner) they use a method of integration by substitution:
Given that
[itex]F=\frac{dp}{dt}[/itex]
and relativistic momentum is given by
[itex]p=\frac{mv}{\sqrt(1-(v^2/c^2))}[/itex]
[itex]W=∫F dx=∫\frac{dp}{dt} dx[/itex]
which is fine then they say
[itex]\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt(1−(\frac{(v^2)}{c^2}))}=\frac{m\frac{dv}{dt}}{(1−(\frac{v^2}{c^2}))^(\frac{3}{2})}[/itex]
(This is about halfway through the derivation.)
its that last step that i don't understand. why does the denominator term now have a 3/2 index? and why is du/dt given like that? Any help would be appreciated.
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