0
$\begingroup$

I am trying to derive the equation for gravitational potential energy $E=-GMm/r$. On the Wikipedia page for energy: https://en.wikipedia.org/wiki/Work_(physics)#Work-energy_principle it says that $W=-\Delta PE$, and I can successfully derive the equation for gravitational potential energy using $W=-\Delta PE$ as a basis, however I do not understand why $W=-\Delta PE$, which means I'm not fully satisfied by my answer.

The Wikipedia page mentions the work-energy principle $W=\Delta KE$, which I do understand the origins of, and it somehow justifies $W=-\Delta PE$ using $W=\Delta KE$, but I get lost in the transition.

The link between the work energy principle and potential energy

How can the equation $W=-\Delta PE$ be justified/derived using the work-energy principle or otherwise?

$\endgroup$
11
  • $\begingroup$ Note: I presumed that W was the work done by the gravitational force, not an external agency. $\endgroup$ Dec 2, 2017 at 15:57
  • $\begingroup$ I believe it is in error (Wiki). Say we lift something from $h=0$ to $h$. The change in potential energy $U$ is $\Delta U=mgh$. This is clearly the work done, i.e. $W=\Delta U$. In more general terms we write: $\Delta W=\Delta U+\Delta K$ ($K$ is kinetic energy). $\endgroup$
    – Gert
    Dec 2, 2017 at 16:40
  • $\begingroup$ I have just edited my question, notice that I am using the equation $E=-GMm/r$ rather than $E=mgh$, so would $W=-\Delta U$ be correct for that equation due to the minus sign in $E=-GMm/r$? @Gert $\endgroup$ Dec 2, 2017 at 17:19
  • $\begingroup$ Also, I have seen a number of other sources which state or imply that the work done by gravity is minus the change in potential energy: $W=-\Delta PE$. @Gert $\endgroup$ Dec 2, 2017 at 17:27
  • $\begingroup$ $U(r)=-\frac{GMm}{r}$ is correct but it is the gravitational potential energy in a central gravitational field. It IS NOT equal to $\Delta U$. Hint: calculate $\Delta U$ for $r_2>r_1$. It is positive, just like in $\Delta U=mg\Delta h$ (for smallish, positive values of $h$). $\endgroup$
    – Gert
    Dec 2, 2017 at 17:32

1 Answer 1

3
$\begingroup$

Short answer

There is no need to use the Work-Energy Theorem to justify or derive the equation $W = -\Delta PE$ for the force of gravity because the equation $W = -\Delta PE$ for the force of gravity is true by the definition of potential energy.

Long answer

The definition for a conservative vector field is as follows:

A vector field $\mathbf{v}: D \to \mathbb{R}^n$, where $D$ is an open subset of $\mathbb{R}^n$, is said to be conservative if and only if there exists a $C^1$ scalar field $\varphi$ on $D$ such that $$\mathbf{v} = \nabla \varphi \tag{1}$$ where $\nabla \varphi$ denotes the gradient of $\varphi$. When the equation above holds, $\varphi$ is called a scalar potential for $\mathbf{v}$.

(Note: I have used $D$ instead of the $U$ that Wikipedia uses to denote the open subset so that I may use $U$ to denote the potential energy function later in this answer)

Since you have stated that you have been able to derive the correct expression for the gravitational potential energy $U_g = -GMm/r$ using the fact that $W_g = -\Delta U_g$, I am going to assume that you at least have a decent understanding of single-variable calculus and vectors. If so, you will be able to recognize that the gradient of a scalar function $\varphi$, defined in $\mathbb{R^3}$ as

$$\nabla \varphi (x,y,z) = \frac{\partial \varphi}{\partial x} \hat{\mathbf{i}} + \frac{\partial \varphi}{\partial y} \hat{\mathbf{j}} + \frac{\partial \varphi}{\partial z} \hat{\mathbf{k}} \tag{2}$$

where $\varphi$ is a function of the Cartesian coordinates $x$, $y$, and $z$, and $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$ are respectively the unit vectors pointing along the $x$, $y$, and $z$ axes, is similar to taking the derivative (i.e. "finding the slope") of a function $f(x)$ with respect to $x$, the difference being that the resulting function $\nabla \varphi$ is a vector function instead of a scalar function.

Note that the scalar potential $\varphi$ does not have to be unique, since a different potential $\bar \varphi$ that satisfies $\varphi = k\bar \varphi + C$ where $k$ and $C$ are arbitrary constants also satisfies the criteria given by (1).

Visualizing the Gradient

Consider a hiker who wants to plan his ascent up Mauna Loa in Hawaii. After making some assumptions, such as the assumption that the mountain's surface -- a 2-dimensional surface embedded in a 3-dimensional space -- is relatively smooth and features like vertical cliffs or overhangs can be neglected, he can construct a continuous scalar function $h(x,y): D \subseteq \mathbb{R}^2 \to \mathbb{R}$ that takes in a coordinate pair $(x,y)$ and outputs the height of the mountain $h$ at that coordinate. Once he has this function, he can determine a set of coordinate pairs that satisfy the condition $h(x,y) = k$ for some constant $k$ and depict that set of points in $\mathbb{R}^2$. After doing this for several different values of $k$, the picture he ends up with looks something like this:

Topographic Map of Mauna Loa

This is a topographic map of Mauna Loa, and the curves seen on the map, each of which are constructed from the set of coordinate pairs that satisfy the condition $h(x,y) = k$ for some constant $k$ (the $k$ values are written on the curves), are known as contour curves. The gradient $\nabla h(x,y)$ is defined so that at any given point $(x_0,y_0)$, $\nabla h(x_0,y_0)$ will point both in the direction locally perpendicular to the contour curve that $(x_0,y_0)$ falls on and also in the direction of locally increasing altitude. With this information, he can plan his ascent up the mountain.

While $\nabla h(x,y)$ is a vector field in $\mathbb{R}^2$, forces in Newtonian mechanics are in general described by vector fields in $\mathbb{R}^3$. The drawback of a conservative vector field $\mathbf{v}$ in $\mathbb{R}^3$ is that the scalar potential $\varphi$ associated with $\mathbf{v}$ is more fully depicted as a 3-dimensional surface embedded in a 4-dimensional space -- a geometric depiction we cannot easily visualize. However, just as the hiker constructed contour curves in $\mathbb{R}^2$ to visualize the mountain, we can similarly construct equipotential surfaces for a scalar potential $\varphi(x,y,z)$ to depict it in $\mathbb{R}^3$ using the same method.

Line Integrals

The dot product between two vectors $\mathbf{a} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}} + a_z \hat{\mathbf{k}}$ and $\mathbf{b} = b_x \hat{\mathbf{i}} + b_y \hat{\mathbf{j}} + b_z \hat{\mathbf{k}}$ in $\mathbb{R}^3$ is defined as,

$$\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z \tag{3}$$

The dot product can not only be easily generalized to $\mathbb{R}^n$ but also used in $\mathbb{R}^n$ to define the concept of a line integral of a vector field:

For a vector field $\mathbf{F} : D \subseteq \mathbb{R}^n → \mathbb{R}^n$, the line integral along a piecewise smooth curve $C \subset D$, in the direction of $\mathbf{r}$, is defined as $$\int _{C}\mathbf {F} (\mathbf {r} )\cdot \,d\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)\,dt \tag{4}$$ where $\cdot$ is the dot product and $\mathbf{r}: [a, b] → C$ is a bijective parametrization of the curve $C$ such that $\mathbf{r}(a)$ and $\mathbf{r}(b)$ give the endpoints of C.

For scalar fields, there is a generalization of the fundamental theorem of calculus, known as the fundamental theorem of calculus for line integrals:

Let $\varphi :D\subseteq \mathbb {R} ^{n}\to \mathbb {R}$ and $\gamma$ is any curve from $\mathbf{p}$ to $\mathbf{q}$. Then $$\varphi \left(\mathbf {q} \right)-\varphi \left(\mathbf {p} \right)=\int _{\gamma [\mathbf {p} ,\,\mathbf {q} ]}\nabla \varphi (\mathbf {r} )\cdot d\mathbf {r} \tag{5}$$

This means that for any conservative vector field $\mathbf{v}$ with an associated scalar potential $\varphi$ given by (1),

$$\varphi \left(\mathbf {r}_2 \right)-\varphi \left(\mathbf {r}_1 \right)=\int _{\gamma [\mathbf {r}_1 ,\,\mathbf {r}_2 ]}\mathbf{v} \cdot d\mathbf {r} \tag{6}$$

and since the path $\gamma$ is arbitrary, the line integral is said to be path-independent.

The Work-Energy Theorem

All this being said, conservative vector fields do play an important role in the Work-Energy Theorem, which more specifically states that the work $W_{net}$ performed by the net force acting on a particle moving along a path $\gamma$, defined as,

$$W_{net} = \int _{\gamma [\mathbf {r}_1 ,\,\mathbf {r}_2 ]}\mathbf{F}_{net} \cdot d\mathbf {r} \tag{7}$$

will be equal to that particle's change in kinetic energy $\Delta K$; a proof, which involves a simple application of Newton's Second Law, can be found in the link to the Wikipedia article that you provided in your original question statement.

$W_{net}$ can be divided into two contributions: the work $W_{con}$ performed by the sum of the conservative forces acting on the particle,

$$W_{con} = \int _{\gamma [\mathbf {r}_1 ,\,\mathbf {r}_2 ]}\mathbf{F}_{con} \cdot d\mathbf {r} \tag{8}$$

and the work $W_{ncon}$ performed by the sum of the non-conservative forces acting on the particle,

$$W_{ncon} = \int _{\gamma [\mathbf {r}_1 ,\,\mathbf {r}_2 ]}\mathbf{F}_{ncon} \cdot d\mathbf {r} \tag{9}$$

While $W_{con}$ is always path-indepedent, $W_{ncon}$ is generally path-dependent and does not have an associated scalar potential $\varphi$ that satisfies (1).

The Negative Sign

So why do we use the convention $W_{con} = -\Delta U$ and not $W_{con} = \Delta U$?

Consider a particle moving along a path $\gamma$ in space and, to simplify the scenario, assume that only conservative forces are acting on the particle. The change in kinetic energy of the particle along the path is given by the Work-Energy Theorem,

$$\int _{\gamma [\mathbf {r}_1 ,\,\mathbf {r}_2 ]}\mathbf{F}_{con} \cdot d\mathbf {r} = \Delta K \tag{10}$$

But we also know that since $\mathbf{F}_{con}$ is a conservative force field, there must exist a scalar energy function $U$ whose gradient is proportional by some constant $c$ to $\mathbf{F}_{con}$. Therefore,

$$\int _{\gamma [\mathbf {r}_1 ,\,\mathbf {r}_2 ]}\mathbf{F}_{con} \cdot d\mathbf {r} = c\Delta U \tag{11}$$

Since we are dealing with Newtonian mechanics, we can apply the Law of Conservation of Energy, which states that energy in all of its forms can neither be created nor destroyed -- it can only be converted from one form to another. In other words, for this system, the change in kinetic energy of the particle $\Delta K$ cannot be created out of nothing; it must be extracted directly from the energy field $U$ itself -- no more, no less -- and so $\Delta K = -\Delta U$, which means $c = -1$ and therefore $W_{con} = -\Delta U$.

From this we can define the total mechanical energy of a given particle $E_{mech} = K + U$, and so in general, the change in the mechanical energy of the system $\Delta E_{mech}$ is simply the work performed by non-conservative forces in transferring energy to and from the system's surroundings,

$$\begin{align} \Delta E_{mech} & = \Delta K + \Delta U \\ & = W_{net} - W_{con} \\ & = W_{ncon} + W_{con} - W_{con} \\ & = W_{ncon}\end{align} \tag{12}$$

$\endgroup$
1
  • $\begingroup$ Brilliant and comprehensive answer, this gave me a completely new perspective on the topic, thank you so much! $\endgroup$ Dec 16, 2017 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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