# Describing Force using potentials

randybryan
Just started revision for Quantum Mechanics after a very long break so everything is a bit rusty. Don't have anyone nearby at hand to ask, so hoping someone here will help.

Been given Newton's second law of motion as

F= m d^2 x/ dt^2

which is a second order equation hence requiring two initial conditions, often given as x(t=0) and dx/dt (t=0). Then, to quote from my lecture notes, ''In quatum mechanics, we will work with potentials, rather than forces, so we will be restricted to conservative systems where energy is constant. The aboe Newton's law equation can be written as

AND THIS IS WHERE I CAN'T DEDUCE WHAT'S GOING ON

F= md^2x/dt^2 = -dV/dx

Where does the -dV/dx come from? I know this is a potential, but I don't know how it was derived

Homework Helper
hi randybryan!

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''In quatum mechanics, we will work with potentials, rather than forces, so we will be restricted to conservative systems where energy is constant. The aboe Newton's law equation can be written as

AND THIS IS WHERE I CAN'T DEDUCE WHAT'S GOING ON

F= md^2x/dt^2 = -dV/dx

Where does the -dV/dx come from? I know this is a potential, but I don't know how it was derived

"conservative system" means that the force F is conservative, ie that it is the gradient of a potential, V

in other words: by definition F.i = dV/dx (i'm not sure about the "minus" )

netheril96
hi randybryan!

(try using the X2 icon just above the Reply box )

"conservative system" means that the force F is conservative, ie that it is the gradient of a potential, V

in other words: by definition F.i = dV/dx (i'm not sure about the "minus" )

Force is the minus gradient of potential. And this is a classical physics question, not quantum.

randybryan
Thanks guys, it all makes sense now!

The reason it is negative is because it tends to zero at infinity as it moves away from potential well.

The only reason I put it in Quantum physics section is because it is used to rationalise Schrodingers equation