Irodov 1.132: Unequal Stiffness Makes the Field Elliptical

Problem Irodov 1.132

A particle moves in a two-dimensional field whose potential energy is $U(x,y) = \alpha x^2+\beta y^2$ , where $\alpha$ and $\beta$ are positive constants with $\alpha\neq\beta$ . Determine:

whether this force field is central
the shape of the curves of constant $U$ , and separately the shape of the curves along which the force magnitude $F$ is constant

Given

$U = \alpha x^2+\beta y^2$
$\alpha\neq\beta$ , both positive

Find

(a) is the field central; (b) shape of

U=\mathrm{const}

and of

F=\mathrm{const}

Solution

Step 1The force field from the potential

The force is minus the gradient of the potential energy, so its two Cartesian components follow directly from partial derivatives of $U$ .

\vect F = -\nabla U = -\frac{\partial U}{\partial x}\,\vect i-\frac{\partial U}{\partial y}\,\vect j = -2\alpha x\,\vect i-2\beta y\,\vect j \tag{1}

Step 2Testing whether the field is central

A central force obeys $\vect r\times\vect F=0$ everywhere, since two parallel vectors have a vanishing cross product. Form that product from the position vector and the components in (1); the cross product of a unit vector with itself vanishes, and the two mixed products are $\pm\vect k$ with opposite signs.

\begin{align*}\vect r\times\vect F &= (x\vect i+y\vect j)\times(-2\alpha x\,\vect i-2\beta y\,\vect j) \\ &= -2\beta xy\,(\vect i\times\vect j)-2\alpha xy\,(\vect j\times\vect i) \\ &= -2\beta xy\,\vect k+2\alpha xy\,\vect k \\ &= 2(\alpha-\beta)\,xy\,\vect k\end{align*}

Since $\alpha\neq\beta$ , this is nonzero at every point off the two axes, so $\vect F$ is not parallel to $\vect r$ there.

(a) The field is not central.

Step 3Shape of the equipotential curves

Setting $U$ to a fixed value $C$ in the given potential is the equation of the family of equipotential curves.

\alpha x^2+\beta y^2 = C

Dividing through by $C$ casts this as the standard form of an ellipse with semi-axes $a=\sqrt{C/\alpha}$ along $x$ and $b=\sqrt{C/\beta}$ along $y$ .

\frac{x^2}{C/\alpha}+\frac{y^2}{C/\beta} = 1

\dfrac{a}{b} = \sqrt{\dfrac{\beta}{\alpha}}

(2)

The equipotential curves

U=\mathrm{const}

are ellipses, semi-axis ratio

a/b=\sqrt{\beta/\alpha}

Step 4Shape of the constant force curves

The force magnitude from (1), set to a fixed value $F$ , gives the second family of curves.

F^2 = F_x^2+F_y^2 = 4\alpha^2x^2+4\beta^2y^2

Dividing through by $F^2$ casts this too as an ellipse, with semi-axes $a'=F/(2\alpha)$ along $x$ and $b'=F/(2\beta)$ along $y$ .

\frac{x^2}{\bigl(F/2\alpha\bigr)^2}+\frac{y^2}{\bigl(F/2\beta\bigr)^2} = 1

\dfrac{a'}{b'} = \dfrac{\beta}{\alpha}

(3)

The constant-force curves

F=\mathrm{const}

are also ellipses, semi-axis ratio

a'/b'=\beta/\alpha

Both families for α=1, β=2: the equipotential U=2 (gold), with semi-axes 2 along x and 1 along y, and the constant-force curve F=4 (royal), with semi-axes 2 and 1. They meet at (0,1). — Both families for $\alpha=1$ , $\beta=2$ : the equipotential $U=2$ (gold), with semi-axes $\sqrt2$ along $x$ and $1$ along $y$ , and the constant-force curve $F=4$ (royal), with semi-axes $2$ and $1$ . They meet at $(0,\pm1)$ .

Check

Both ratios reduce to $1$ when $\alpha=\beta$ , recovering circular curves, the isotropic case the problem excludes by requiring $\alpha\neq\beta$ . Since $\beta/\alpha\neq(\beta/\alpha)^2$ unless $\beta=\alpha$ , the two ellipse families, (2) and (3), are genuinely different shapes: $U=\mathrm{const}$ and $F=\mathrm{const}$ are not the same set of curves here.

Remark

This potential is two independent one-dimensional harmonic oscillators glued into one particle, with spring-like constants $2\alpha$ along $x$ and $2\beta$ along $y$ ; a central field would need those two constants equal, which is exactly the condition part (a) shows fails. More generally, a separable potential $U(x,y)=U_1(x)+U_2(y)$ gives a central field only when $U_1'(x)/x$ and $U_2'(y)/y$ are the same constant for every $x$ and $y$ , which forces $U_1$ and $U_2$ to be the same quadratic, the isotropic harmonic potential $U=k(x^2+y^2)/2$ : a separable potential built from any other matching power, $U_1(x)=U_2(y)=x^4$ say, fails to be central for the same reason this one does, even though its two halves are identical functions.