# Why is $F=-\nabla V$?

I came across this equation $$F=-\nabla V$$ where $V$ is potential energy. I do understand that $$F(r)=-\frac{dV}{dr}.$$ Hence does this mean the nabla operator in this case means derivative? Because I checked up the definition of $\nabla$ and none of them suits this definition.

• If you checked the definition of $\nabla$, but none of them was the derivative, you checked the wrong definition. – ACuriousMind Jul 5 '15 at 23:19
• When the nabla $\nabla$ is written right next to its argument (without dot, cross, etc), it is usually known as the "grad" operator. Gradient... – Floris Jul 6 '15 at 0:02

On spherical coordinates, the gradient of a general function $V$ is: $$\nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi$$

If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the gravitational potential, then the partial derivatives with respect to $\theta$ and $\phi$ are zero, and therefore the $\nabla$ resumes to: $$\nabla V = \frac{\partial V}{\partial r}\mathbf e_r = \frac{dV}{dr}\mathbf e_r$$

Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time.

# Multivariate functions

When we transition from a function $f(x)$ to a field, which is a function of many variables $f(x, y, z)$, we suddenly have many ways to take the derivative. Instead of our iconic equation valid for small differences in x $\delta x$,$$f(x + \delta x) \approx f(x) + f'(x)\cdot\delta x$$we now have to write$$f(x + \delta x, y + \delta y, z + \delta z) \approx f(x, y, z) + f_x(x, y, z)\cdot\delta x + f_y(x, y, z)\cdot\delta y + f_z(x, y, z)\cdot\delta z.$$The principle is the same, but each variation that we're making must be made independently. Instead of us saying "when you zoom in far enough, the graph of $f(x)$ looks like a straight line," we're saying, "when you zoom in far enough, $f(x, y, z)$ looks like the expression $a x + b y + c z + d$." In this case the full name of $a = f_x(x,y,z)$ is "the partial derivative of $f$ with respect to its first variable, holding its second and third variables constant, evaluated at the point $(x, y, z)$."

All of that technicality is needed because you could hold other things constant, too. For example, you could ask how $f$ varies with $x$ when $y - x$ is held constant. Clearly this would correspond to the above case where $\delta y = \delta x$ while $\delta z = 0$. To succinctly express this kind of stuff, it's useful sometimes to write $(\partial f/\partial u)_{P,Q}$ as "the partial derivative of $f$ with respect to $u$ holding the expressions $P$ and $Q$ constant." So for example our observation about $y - x$ being held constant would be phrased as$$\left(\frac{\partial f}{\partial x}\right)_{y-x,z} = \left(\frac{\partial f}{\partial x}\right)_{y,z} + \left(\frac{\partial f}{\partial y}\right)_{x,z}.$$But there are some very common operators which we might leave out the extra detail. You will variously see all of these conventions: $$\partial_x f = \frac{\partial f}{\partial x} = f_x = \left(\frac{\partial f}{\partial x}\right)_{y,z}.$$Similarly, when we unqualifiedly talk about $\partial_r$ we're talking about keeping either, in cylindrical coordinates, $\theta$ and $z$ fixed, or else, in spherical coordinates, $\theta$ and $\phi$ fixed.

Work out some examples for yourself, like determining that the partial derivatives for $f(x, y, z) = x^3 + y^2 z^2 + x y z - x^2 z$ are $f_x = 3x^2 + y z - 2 x z$, $f_y = 2 y z^2 + x z$, $f_z = 2 y^2 z + x y - x^2$.

# Vectors

If you're doing physics with derivatives you've probably already seen vectors. The technical definition of vector in physics is, "a vector anything which transforms like a vector does under coordinate transformations." Lest that sound circular, it comes with an example: any arrow between two points in space. With that example, this definition extends that definition to anything which essentially "does the same thing" as a displacement vector does in space. One particular vector, the position vector, is usually denoted $\vec r = \left[x; y; z\right]$, and it points outward from some "origin" (0, 0, 0) to some point $(x, y, z)$.

One nice thing about vectors is that we have worked out which expressions are "coordinate independent" with them. Supposing that $\vec a$ is given by the coordinates $[a_x; a_y; a_z]$ and $\vec b$ is, in those same coordinates, $[b_x; b_y; b_z]$. Assume that the coordinate directions are all orthogonal. Then the number $\vec a \cdot \vec b = a_x b_x + a_y b_y + a_z b_z$ is known to be coordinate independent. It is 0 if and only if two vectors are perpendicular, having an alternate expression as $|\vec a| |\vec b| \cos \psi$ where $\psi$ is the angle between the two vectors in the plane that they both occupy. Furthermore, the "length" or "magnitude" of a vector, $|\vec a| = +\sqrt{\vec a \cdot \vec a}$ is also coordinate-independent as a special case. An almost-coordinate-independent "pseudovector" in 3D is given by $\vec a \times \vec b = \left[a_y b_z - a_z b_y;~a_z b_x - a_x b_z;~ a_x b_y - a_y b_x\right]$. It is "pseudo" because it changes sign when the coordinates change their "orientation" as in $(x,y,z)\rightarrow(x, z, y)$ or $(x, y, z)\rightarrow(-x, y, z)$.

When we see these properties we can immediately rewrite the above expression as$$f(\vec r + \delta\vec r) \approx f(\vec r) + \left[\begin{array}{c} f_x(\vec r) \\ f_y(\vec r) \\ f_z(\vec r)\end{array}\right] \cdot \delta r$$When we see this we see that we can write the del operator $\nabla = [\partial_x; \partial_y; \partial_z]$. Taking a scalar field $f(\vec r)$, the operation which converts it to the vector field $\nabla f = [f_x, f_y, f_z]$ is called the gradient operation. At each point this gives you a vector, so it is a vector field.

By its nature, the gradient points, at any point it's evaluated, in the local direction of greatest increase of its scalar function, and perpendicular to a hyperplane $\nabla f \cdot \vec r = C$ wherein that function is approximately constant for small deviations from the point where the gradient was taken.

# Conservative fields

Combining our "almost vector" $\nabla$ and our "almost vector product" $\times$, it turns out in 3D that if a vector field $\vec v(\vec r)$ obeys $\nabla \times v = 0$ then it can be written as $\vec v = \nabla \phi$ for some function $\phi(\vec r)$. Usually for forces we instead choose $\vec F = -\nabla U$, with a negative sign. We do this because it fits our intuition that your potential energy U gets larger as you travel higher, fighting "against" the force of gravity. (Basically it makes a potential energy function for a hill look like the profile of that hill.) There is a similar law which is a little less useful, saying that if $\vec v$ obeys $\nabla\cdot v = 0$ then $\vec v = \nabla\times u$ for some $u$.

It is in general a little harder to find expressions for $\nabla$ in the different coordinates; this article is pretty useful to see what it looks like in non-Cartesian coordinates. In the special case for a spherically symmetric potential $U(r, \theta, \phi) = U(r)$ we do indeed discover that $\vec F(\vec r) = -\nabla U = - U'(r) ~\hat r$, where $\hat r = \vec r / |\vec r|$ is the unit vector in the positive $r$-direction. This is really helpful for a lot of different force laws.

• Gotta laugh and cry at this: "a vector anything which transforms like a vector does under coordinate transformations." I prefer the definition as a quantity with a direction and magnitude that can be calculated from a set of numbers called its components. And the components transforming under rotations such that its direction and magnitude remains invariant. – Larry Harson Jul 6 '15 at 3:00
• That line is shamelessly stolen from Prof. Bruce Kusse at Cornell. He said, "A tensor, of course, is anything that transforms like a tensor," sparking some laughs from the audience. He gave them a look and said, "don't laugh, this is science," getting more laughs, then explained that it's not really a joke, here's how we apply the coordinate-change matrix to these tensors and those tensors... and anything which transforms under coordinate changes like those entities is, of course, a tensor. – CR Drost Jul 6 '15 at 3:41
• I thought that quote came from Zee – Larry Harson Jul 6 '15 at 21:57
• Kusse might have stolen it too; I don't know. Or, intriguingly, maybe Kusse is the scientist mentioned by Zee. – CR Drost Jul 6 '15 at 22:35
• Nobody stole it, I've heard it used many times... and it's not wrong either, but it does require some elaboration, of course. – Gwynbleidd Aug 23 '15 at 4:02

The work-energy theorem leads us to the following result; $$\oint \vec F\cdot d\vec s=0$$

$$\oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n$$ Using the rules of vector calculus there must exist some scalar function such that;

$$\vec F=-\nabla U(r)$$ Since, $$\nabla \times \nabla U(r)=0$$ The negative sign is just for convention it ensures that a given force accelerates the particle to a lower potential energy, so that the work is done against the force. Note that any $\nabla \times \nabla \phi$ is equal to zero!

Let $F$ be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero $$\oint_{O} F \cdot dr = 0$$ The del operator $\nabla$ is defined in 3 dimensions as $$\nabla \equiv \frac{\partial}{\partial x}\hat{e_x}+\frac{\partial}{\partial y}\hat{e_y}+ \frac{\partial}{\partial z}\hat{e_z}$$ Furthermore from Stokes theorem we have that $$\oint_{O} F \cdot dr = \iint_{B}\nabla \times F \cdot dB$$ Therefore for a conservative vector field the curl of the force field is also zero $$\nabla \times F = 0$$ Let $V$ be a function called the potential such that $$V = -\oint_{O} F\cdot dr$$ Then it follows that $F = -\nabla{V}$, which in the 1 dimensional case reduces to $$F = -\frac{\partial{V}}{\partial{x}}\hat{i} = -\frac{dV}{dx}$$