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
F=\frac{dp}{dt}
and relativistic momentum is given by
p=\frac{mv}{\sqrt(1-(v^2/c^2))}
W=∫F dx=∫\frac{dp}{dt} dx
which is fine then they say
\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})}
(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
F=\frac{dp}{dt}
and relativistic momentum is given by
p=\frac{mv}{\sqrt(1-(v^2/c^2))}
W=∫F dx=∫\frac{dp}{dt} dx
which is fine then they say
\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})}
(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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