What Does -∂V/∂x Represent in Newton's Law?

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The discussion clarifies that the term -∂V/∂x represents the negative partial derivative of potential energy (V) with respect to position (x), indicating how potential energy changes as position changes while holding other variables constant. This concept is integral to Newton's second law, where the force (F) is expressed as F = ma = -∂V/∂x. The negative sign indicates that an increase in potential energy corresponds to a decrease in force acting in the direction of increasing x. The terms -∂V and ∂x do not have meaning in isolation; they must be understood together within the context of derivatives. Understanding these derivatives is crucial for analyzing systems in quantum mechanics and classical physics.
rb120134
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Homework Statement
What does -∂V/∂x mean?
Relevant Equations
F= ma
In one of my textbooks about quantum mechanics, they mention a vehicle moving in a straight line along the x axis. With Newtons first law they take the second derivative from a which is
d^2x/dt^2 and that should be equal to
-∂V/∂x. What exactly does -∂V indicate?
The complete equation becomes
F=ma md^2x/dt^2= -∂V/∂x. It is a partly derivative. What does " -∂V/∂x" indicate, which units belong to -∂V and
∂x? They were talking about potential energy in the textbook.
 
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rb120134 said:
Homework Statement: What does -∂V/∂x mean?
Homework Equations: F= ma

In one of my textbooks about quantum mechanics, they mention a vehicle moving in a straight line along the x axis. With Newtons first law they take the second derivative from a which is
d^2x/dt^2 and that should be equal to
-∂V/∂x. What exactly does -∂V indicate?
It doesn't indicate anything! The "-∂V" alone means nothing without the accompanying "/∂x".

The complete equation becomes
F=ma md^2x/dt^2= -∂V/∂x. It is a partly derivative
"partial derivative", not "partly derivative"
What does " -∂V/∂x" indicate, which units belong to -∂V and
∂x? They were talking about potential energy in the textbook.
As I said above, "-∂V" and "∂x" have no meaning separately. "-∂V/∂x" is the "negative of the partial derivative with respect to x. You titled this "second derivative of Newton's Law" and mentioned "partial derivative". Can we assume that you know what "derivatives" and "partial derivatives" are? The derivative of a function is instantaneous rate of change of the function. If a function, like V, depends on more that one variable (here V is defined in "three space" so depends on the variables x, y, and z and might depend on time, t) then the partial derivative of V with respect to x is the instantaneous rate of change of V as x changes, all other variables held constant.
 
Hi! sorry for sending this so late, I hope it eventually gets to you but for the term "delV/delx" "V" would be the function of for the potential energy of the system with its partial derivative taken with respect to "x" representing it's position.
 
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