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Hunus Group TitleBest ResponseYou've already chosen the best response.0
In a book I'm reading, the author keeps moving quantities to the other side of the differential element. \[ m\vec G \cdot d \vec r =  dU\] He then says to divide by m \[\vec G \cdot d \vec r = \frac{dU}{m}\]he then says \[\phi = \frac{U}{m}\] so that \[\vec G \cdot \vec r = d \phi\]
 2 years ago

Hunus Group TitleBest ResponseYou've already chosen the best response.0
Can someone explain this to me?
 2 years ago

hamza_b23 Group TitleBest ResponseYou've already chosen the best response.1
alright do you get the first step atleast where he brought the m over?
 2 years ago

hamza_b23 Group TitleBest ResponseYou've already chosen the best response.1
i am going to assume you know that because it is a constant and you can divide it if you like the author chooses too thats not important , ok so now to the second line we got G.dr = dU/m right after what the author did was he took the integral of both sides lets do the left hand side first G.dr the integral of that in respect to r is G.r because dr says we have a function of r. Now lets do the right sidedU/m the integral of that is U/m, we dont do anything to 1/m because that is just a constant next then he says phi = U/m where they let G.r=dphi
 2 years ago
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