Irodov 1.133: One Field Curls, One Does Not – Amit & Deepali Sinha Foundation

Problem Irodov 1.133

Two stationary force fields are defined over the $xy$ plane, $\vect F_1=ay\vect i$ and $\vect F_2=ax\vect i+by\vect j$ , where $\vect i$ and $\vect j$ are the unit vectors along the $x$ and $y$ axes and $a$ , $b$ are constants. Determine, for each of $\vect F_1$ and $\vect F_2$ separately, whether it is a potential (conservative) field.

Given

$\vect F_1=ay\vect i$
$\vect F_2=ax\vect i+by\vect j$

Find

Which, if either, of

\vect F_1

\vect F_2

is a potential (conservative) field

Solution

Step 1Test both fields with the same closed loop

A force is potential exactly when the work it does around every closed path is zero, since that is what lets the work between two points be assigned a path independent value, $-\Delta U$ . Take the same rectangular loop for both fields, $O(0,0)\to A(x_0,0)\to B(x_0,y_0)\to C(0,y_0)\to O$ , for an arbitrary $x_0,y_0>0$ : the loop only has to be tried once, symbolically, since a zero result for every $x_0,y_0$ rules nothing out by accident and a nonzero result is one counterexample, already enough to rule potential out.

F_1=ay i: horizontal everywhere, growing with height, not with x. Arrows are schematic (half scale, a=1) to fit inside the loop. — $\vect F_1=ay\vect i$ : horizontal everywhere, growing with height, not with $x$ . Arrows are schematic (half scale, $a=1$ ) to fit inside the loop.

F_2=ax i+by j: each component grows only with its own coordinate. Arrows are schematic (half scale, a=b=1). — $\vect F_2=ax\vect i+by\vect j$ : each component grows only with its own coordinate. Arrows are schematic (half scale, $a=b=1$ ).

Step 2Field

\vect F_1=ay\vect i

around the loop

$\vect F_1$ has no $y$ component, so only the horizontal legs $O\to A$ and $B\to C$ can contribute, and $O\to A$ sits at $y=0$ where $F_1=0$ :

\begin{align*}W_{OA}=0, \qquad W_{AB}=\int_0^{y_0} 0\,dy = 0, \qquad W_{BC}=\int_{x_0}^{0} a y_0\,dx = -ay_0x_0, \qquad W_{CO}=0.\end{align*}

Adding the four legs, only $W_{BC}$ survives:

W_1 = W_{OA}+W_{AB}+W_{BC}+W_{CO} = -ax_0y_0.

W_1=-ax_0y_0

(1)

Since $x_0,y_0>0$ , (1) is nonzero: the work around this loop does not vanish, so $\vect F_1$ is not potential.

Step 3Field

\vect F_2=ax\vect i+by\vect j

around the same loop

Both components now contribute on every leg:

\begin{align*}W_{OA}=\int_0^{x_0} ax\,dx = \frac{ax_0^2}{2}, \qquad W_{AB}=\int_0^{y_0} by\,dy = \frac{by_0^2}{2},\end{align*}

\begin{align*}W_{BC}=\int_{x_0}^{0} ax\,dx = -\frac{ax_0^2}{2}, \qquad W_{CO}=\int_{y_0}^{0} by\,dy = -\frac{by_0^2}{2}.\end{align*}

Adding all four:

\begin{align*}W_2 = \frac{ax_0^2}{2}+\frac{by_0^2}{2}-\frac{ax_0^2}{2}-\frac{by_0^2}{2} = 0.\end{align*}

W_2=0

(2)

This holds for every $x_0,y_0$ , not just one favourable rectangle, so $\vect F_2$ is potential.

Check

The same conclusion follows from the standard component test $\partial F_x/\partial y=\partial F_y/\partial x$ : for $\vect F_1$ , $\partial F_x/\partial y=a$ against $\partial F_y/\partial x=0$ , unequal, confirming not potential; for $\vect F_2$ , both partial derivatives are $0$ , confirming potential. This matches (1), (2) exactly, since (1) is $-x_0y_0$ times that same mismatched derivative $a$ .

\vect F=ax\vect i+by\vect j

is potential;

\vect F=ay\vect i

is not.

Remark

The potential energy for the conservative field follows from $-\partial U/\partial x=ax$ and $-\partial U/\partial y=by$ , giving $U(x,y)=-(ax^2+by^2)/2$ , up to an additive constant; no such $U$ exists for $\vect F_1$ .