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- I want to know how the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## can be derived, because it is a key part in understanding how hamiltons principle can be derived.
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term
##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the action-integral can be derived.
If needed I can add the steps from how to get from (1) to (2).
The paper from where I took (1) is the following: https://www.researchgate.net/public...ple_for_the_Derivation_of_Equations_of_Motion
Thanks for your help
##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the action-integral can be derived.
If needed I can add the steps from how to get from (1) to (2).
The paper from where I took (1) is the following: https://www.researchgate.net/public...ple_for_the_Derivation_of_Equations_of_Motion
Thanks for your help
