A conservative force is naturally the vector gradient of a given scalar field . I don't know why the convention to put the negative sign in front of the gradient operator. Or is this just a misconception some of my professors had?

The gradient of 1/r is automatically in the negative r direction, so saying the force field is negative gradient is straight up wrong, if it were true the Earth would explode..

Edit: It was brought to my attention that physicists like to define energies of classical systems to be negative. That answers the question of why they would take the negative gradient..

But really it just rises more questions than answers.. I thaught energy would almost always assumed to be non negative. Also energy is simply the hamiltonian of the lagrangian system.. so why even bother about some sign conventions in the definition of energy?


The intuition is that potential is like height. It takes work to climb up; forces push you down. Another benefit is that with this sign, the potential is the potential energy, rather than its negative.

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  • $\begingroup$ The gradient of 1/r is automatically in the negative r direction, so saying the force field is negative gradient is straight up wrong, if it were true the Earth would explode $\endgroup$ – Puppet master Apr 4 '19 at 20:02
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    $\begingroup$ @Schaurberger You've made a sign error. The potential of the Earth is like $-1/r$. The force, which is the negative gradient of that, indeed points to smaller $r$. $\endgroup$ – knzhou Apr 4 '19 at 20:05
  • $\begingroup$ @Schaurberger Alternatively, you could define the Schaurberger-potential of the Earth to be $1/r$, so that the force is $\nabla(1/r)$, which also points to smaller $r$. Obviously, both ways are going to give you the same force, because they differ by two minus signs. $\endgroup$ – knzhou Apr 4 '19 at 20:06
  • $\begingroup$ Ok so that is the convention. That the potential is actually negative. And it makes sense since for gravitational potential makes it so that the higher it is the bigger potential.. $\endgroup$ – Puppet master Apr 4 '19 at 21:56
  • $\begingroup$ It seems rather unnatural to assume a negative energy for a classical system. $\endgroup$ – Puppet master Apr 5 '19 at 5:05

For simplicity, take the Lagrangian of a one-dimensional point subject to some potential $V(x)$:

$$L = \frac12 m \dot x^2 - V(x).$$

Taking the variation of the action, we have,

$$\delta S = \int \mathrm dt \left( \frac{\partial L}{\partial x}\delta x + \frac{\partial L}{\partial \dot x}\delta \dot x\right)$$

and noting that $\delta \dot x = \partial_t(\delta x)$, we can apply integration by parts,

$$\delta S = \int \mathrm dt \left(\frac{\partial L}{\partial x}- \frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot x} \right)\delta x = 0$$

which we demand to be zero, and assume the boundary term vanishes. Plugging in our expression for the Lagrangian, we have,

$$\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot x} = m\ddot x = \frac{\partial L}{\partial x} = -\frac{\partial V}{\partial x}.$$

We identify $m\ddot x$ as the force, so the (one-dimensional) gradient of $V$ gives us the force, with a minus sign which is not arbitrary.

The minus sign in $L$ is by the definition - kinetic minus potential energies. The minus sign in the integral comes from the integration by parts rule, not forcefully put in.

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    $\begingroup$ "For simplicity, take the Lagrangian..." I think you're overestimating the OP's physics background, but maybe you know that and want to provide an answer for a different audience. $\endgroup$ – user113773 Apr 4 '19 at 22:50
  • $\begingroup$ @PiKindOfGuy The OP is doing vector calculus, so in an undergraduate physics program, he should have done Lagrangian mechanics or will be doing it in the very near future. $\endgroup$ – JamalS Apr 5 '19 at 9:30

So that the mechanical energy may be defined as kinetic plus potential and be conserved. Otherwise it would be kinetic minus potential that's conserved and that's silly because how is it that subtracting an energy from another gives you the total mechanical energy?

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  • $\begingroup$ how you ask? Well, the answer is just like that lol... $\endgroup$ – Puppet master Apr 5 '19 at 5:06
  • $\begingroup$ Energy fundamentally is a hamiltonian, a time symmetry of the lagrangian system. And yes of course it comes as a substraction of two terms.. $\endgroup$ – Puppet master Apr 5 '19 at 5:08
  • $\begingroup$ @Schaurberger I don't understand what you're trying to convey. $\endgroup$ – user113773 Apr 5 '19 at 11:38
  • $\begingroup$ It's just very confusing. First of all energy need not be conserved in general.. isnt giving an arbitrary constat to energy choosing a gauge? $\endgroup$ – Puppet master Apr 6 '19 at 12:59
  • $\begingroup$ You're right, it needn't be conserved in general, but when it is, defining $E = T + V$ makes more sense than $E = T - V$. $\endgroup$ – user113773 Apr 6 '19 at 19:49

Roughly speaking one can view it as wanting a potential function to look like a landscape where a ball will tend to roll “downhill,” if that helps. This means that the force should be in the direction of greatest decrease of potential energy, which is $-\nabla U$.

There is nothing wrong with the opposite sign convention; it is merely unintuitive because in the presence of drag things would tend towards potential maximums rather than minimums, and so we would call it something other than “potential” energy (indicating that it has the potential to be used for work): perhaps “wasted” or “spent” energy.

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  • $\begingroup$ Yea that makes sense. I just think that intuitive or not , conventions have no place in physics. Physics is about how nature works not about what pleases our intuition. Especially when introduction of negative energy is not very natural in the context of general relativity because it would produce negative curvature. $\endgroup$ – Puppet master Apr 6 '19 at 13:02

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