Minus sign in Force potential relation, a convention?

Question

The usual defination of force in terms of potential energy is $$\vec F=-\nabla U$$

This definition leads to

$$K_1+P_1=K_2+P_2\rightarrow \Delta K + \Delta P =0$$

Where $K$ and $P$ are kinetic and potential energies.

However the minus sign over here seems just a convention to me rather than something physically meaningful. That is, if we define

$$\vec F=\nabla U$$

Then the energy equation becomes

$$K_1-P_1=K_2-P_2\rightarrow \Delta K = \Delta P $$

Except for this change, nothing else really changes.

Now, some people like to argue that the flow of energy is from higher to lower potential, as an analogy to pressure, but again that is just a convention. One might as well say that energy flows from low to high potential.

Thus is minus sign really just a convention or something else that I am missing?

Answer 1

The usual defination of force in terms of potential energy is $$\vec F=-\nabla U$$

...However the minus sign over here seems just a convention

Yes. It is a convention. But it is a long-established convention and I assure you, you do not want to use the opposite convention.

Conservative forces are forces for which the work done depends only on the initial and final configuration of the system (rather than the entire path from the initial to the final configuration).

For such forces we can introduce a function of space and time called the "potential energy," which you have denoted by $U$.

By definition (by convention), the work done by these external conservative forces is equal to the decrease in $U$: $$ dW = -\frac{\partial U}{\partial q_i}dq_i\;. $$

If there is only one particle in the system then: $$ dW = -\vec \nabla U\cdot d\vec x\;. $$

This convention leads to all the usual expressions of Classical Mechanics, such as the Lagrangian being defined as: $$ L = T - U $$ and the total mechanical energy being defined as: $$ E = T + U $$

If you want to flip the sign of $U$, you can in principle do it, but it will flip a bunch of other long-held conventional signs and it will be likely end up being very confusing.

Answer 2

Consider this example of some field potential energy profile :

Field's potential energy gradient at particular point is a vector which points to zones with higher potential energy (by definition of gradient), while force in magnitude is same as field potential energy gradient vector but is opposite in direction, because force tries to lower body's potential energy.

If you would ask- why it tries to lower potential energy,- that would be an awesome question, which answer I don't know. Anyway, here's a correct version of potential energy gradient and force vectors relation:

$$ {\vec {F}}=-{\vec {\nabla }}U \tag 1$$

Btw, in this scheme quantized and non-equidistant potential energy levels depicted, but it doesn't matter for your type of question.

So, the final verdict is that minus sign has a deep meaning in Physics, - it shows that some target quantity direction is opposite compared to dependent quantity direction.

Answer 3

We want to uphold the convention that the potential energy is defined such that:

$$U(r)= -\int_{\infty}^{r} \vec{F} \cdot \vec{dr}$$

This convention implies that;

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

To see this, integrate the latter expression along a line integral,and notice that the second half is simply

$$ = -[U(r) - U(\infty)]$$ $$= -U(r)$$

Rearranging gives us:

$$U(r)= -\int_{\infty}^{r} \vec{F} \cdot \vec{dr}$$

This is why there is a negative sign, we artificially add the minus sign so that U is the same potential energy function that we have defined earlier. We simply could have not done this, and called this some other potential energy function and it would still make sense. However it is convenient to use the same potential energy function defined earlier.

Answer 4

Various times that I have tutored physics (way in the past) I use a math analogy to convince someone that the force MUST be the negative of the gradient of the potential function. No conventions need to be considered.

The equation $F = -\nabla{U}$ tells you that $F$ is the slope of the curve U. Usually when thinking about the slopes of a curve is that if it is slanted upwards to the right (say $+x$ direction) then it is positive. But, if you think of that slope as a ramp on which a ball can freely role in a gravitational field then it would accelerate down the ramp, not up the ramp. Therefore, acceleration is opposite to the increase of potential and therefore $F=-\nabla{U}$.

Of course, it is easy to say that choosing such directions is a convention but this example gives one a kind of feel for getting the signs correct.