Ask your own question, for FREE!
Physics 15 Online
OpenStudy (anonymous):

A laconic least action question.

OpenStudy (anonymous):

S, action always 'tries' to be at a minimum. \[\S=\int\limits_{}^{}(T-V)dt\] ,so: \[\int\limits_{}^{}dS=\int\limits_{}^{}(T-V)dt\] \[\int\limits_{}^{}(dS/dt)dt=\int\limits_{}^{}(T-V)dt\] so \[dS/dt=T-V\] where S is always at a minimum. 1.)Is the mathematics correct so far? 2.) So this means T is always at a minimum, V at a maximum, yes?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!