Why do physicists take the convention that a force field is the negative gradient of a scalar field?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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?

Answer 4

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.