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This question already has an answer here:

the formula for potential energy by a conservative force is given by:

$$ F = -\nabla U(r), $$

which in one dimension may be simplified to:

$$ F = -\frac{dU}{dx} .$$

My question is how is it derived and why do we use a negative sign in the formula?. Is this by definition or is there some other reason?

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marked as duplicate by John Rennie, Bernhard, Ali, Kyle Kanos, Chris White Sep 3 '14 at 13:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

possible duplicate of Force as gradient of scalar potential energy – John Rennie Sep 3 '14 at 7:54
up vote 5 down vote accepted

If the particle moves from the point $x$ to $x+dx$, and assume $dx\gt 0$ for simplicity, then its potential energy increases by $$ dU = \frac{dU}{dx}dx $$ Well, it increases if $dU$ is positive and decreases if $dU$ is negative. So far I have only used the definition of the derivative – pure mathematics.

However, the total energy is conserved. The sum of the kinetic energy and the potential energy $$ E = T + U = {\rm const} $$ is constant. It means that if the potential energy increases, the kinetic energy decreases, and vice versa. However, an increasing kinetic energy is exactly the situation when the force $F$ is positive (directed in the same direction as the speed or $dx$).

In other words, the equation $$ dU = \frac{dU}{dx} dx $$ may be rewritten as $$ dT = -\frac{dU}{dx} dx $$ because $T$ is effectively $-U$, up to the constant whose differential is zero, but because the kinetic energy increases if the force and $dx$ have the same sign i.e. $$ dT = F\cdot dx $$ (pushing a right-moving particle by a right-directed force accelerates the particle; the expression above is the infinitesimal work), we may compare the two equations and see that $$ F = -\frac{dU}{dx} .$$ So the sign effectively arises from the "anticorrelation" of the kinetic and potential energy (along with the convention that all the terms are included in the total energy with the plus sign; the convention that the kinetic energy is positive, and so on).

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Perhaps a good thing to look at for someone not yet familiar is this: The work-energy theorem. It is this that allows us to use $dT=F\cdot dx$. – Danu Sep 3 '14 at 7:57
"However, an increasing kinetic energy is exactly the situation when the force F is positive (directed in the same direction as the speed or dx)." I don't think this is correct; force can be in the arbitrarily-chosen negative $x$-direction and speed can still increase (imagine the canonical gravitational free fall situation). – BMS Sep 3 '14 at 8:11
No, BMS, the kinetic energy increases exactly when the force and the velocity are "equally oriented" (or with angle less than 90 degrees). It's because $\vec F \cdot \vec v$ is the power which is the change of the kinetic energy per unit time. If the angle between $F,v$ is below 90 degrees, the inner product is positive, the power is positive, and the kinetic energy goes up. – Luboš Motl Sep 3 '14 at 8:13
(It's possible we are using different conventions.. here's my understanding.) I agree that when force and velocity are in the same direction, speed tends to increase, which is consistent with $dK/dt=\vec F\cdot \vec v$. But a positive dot product does not mean a positive force. Directions (signs) of forces obey the imposed coordinate systems (e.g., the gravitational force is negative if upward is positive, and positive if downward is positive, regardless of the direction of motion). The sign that is associated with vector components are coordinate system dependent, not velocity dependent. – BMS Sep 3 '14 at 8:18
In general I allow the velocity to be any vector, too. But due to the rotational (or parity) symmetry, this question about the sign, and many others, may be answered without a loss of generality by assuming a particular direction/sign. – Luboš Motl Sep 3 '14 at 8:56

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