Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

You know that for a potential function (conservative force/fields) that

$\nabla U = \pm \vec{F}$

In math, we don't have that minus sign, we have only the plus one.

What does it mean if you get rid of the plus sign? I remember the minus signs tells us that the force is always in the opposite direction of the increasing potential function. Does math tell me there exists a potential function such that it's force also increases as potential goes up?

share|improve this question
3  
I'm curious, where have you seen that equation with $\pm\vec{F}$? I have never seen it with a plus sign. –  David Z Dec 30 '11 at 1:40
    
pretty much in any calculus book? –  Hawk Dec 30 '11 at 2:37
1  
Not just any calculus book, or at least not the ones I've used as far as I remember. If you give a specific reference I can elaborate... but that should be done in Physics Chat. –  David Z Dec 30 '11 at 2:51
    
OP writes $\vec{\nabla}(\pm U) =\vec{F}$. The whole question seems to be just a sign flip $U\leftrightarrow -U$. –  Qmechanic Dec 30 '11 at 14:08

2 Answers 2

up vote 2 down vote accepted

Mathematically, it just gives a duality with vector fields and scalar fields in multivariable calculus, associated with conservative vector fields and line integrals. As such, the $\pm$ is irrelevant, because it can be absorbed into the force vector. For physics, we take the sign convention to be negative, so that it agrees with the fact that the force is restoring the object it acts on to a lower energy configuration. Note that we could alternatively absorb the negative sign into the potential! It is all a matter of sign convention, and when you define potential and force in physics (as stated above), the negative sign appears in your equation.

share|improve this answer

Mathematically, we can write $$\mathbf{F}\cdot\mathrm{d}\mathbf{l}=\pm\mathrm{d}E,\qquad\mathrm{if}\ \mathbf{F}\equiv\pm\mathbf{\nabla}E.$$ Here $\mathbf{F}$ is a vector field, $\mathbf{l}$ is a position vector, and $E$ is a scalar field.

In physics, the work done by a force $\mathbf{F}$ over the infinitesimal displacement $\mathrm{d}\mathbf{l}$ equals the left-hand side in the above relation. Thus, under the condition as given above, the right-hand side is also true in physics. The scalar field $E$ is called the energy corresponding to the given (conservative) force. When evaluating the above relationship and a minus sign appears, the corresponding energy is called potential energy (as for weight and the restoring force on a spring, for example).

For Newton's second law, $\mathbf{F}\cdot\mathrm{d}\mathbf{l}=+\mathrm{d}\left(\frac{1}{2}mv^2\right)$. This is an example of $\mathbf{F}=+\mathbf{\nabla}E$.

share|improve this answer

protected by Qmechanic Mar 16 '13 at 14:46

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.