In a certain two-dimensional field of force the potential energy of a particle has the form U = αx2 + βy2, where α and β are positive constants whose magnitudes are different. Find out: (a) whether this field is central; (b) what is the shape of the equipotential surfaces and also of the surfaces for which the magnitude of the vector of force F = const.
(a) The field is not central, because the torque x·Fy − y·Fx = 2(α − β)xy is not zero when α ≠ β. (b) The surfaces of equal U and the surfaces of equal force size are both ellipses: x²/(U₀/α) + y²/(U₀/β) = 1 and x²/(F₀/2α)² + y²/(F₀/2β)² = 1.
Problem statement
Problem 1.93. In a certain two-dimensional field of force the potential energy of a particle has the form U = αx2 + βy2, where α and β are positive constants whose magnitudes are different. Find out: (a) whether this field is central; (b) what is the shape of the equipotential surfaces and also of the surfaces for which the magnitude of the vector of force F = const.
We assume the problem gives α and β only as symbols, so the page uses α = J/m² and β = J/m² as starting values. Both are sliders, and the answers hold for any two different values.
We assume a chosen value U₀ = J for the equipotential and a chosen force size F₀ = N for the constant-force curve. Both are sliders. They change the size of an ellipse and never its shape.
We assume the plane is the x-y plane, x and y are in metres, and the centre of the field is the origin. The probe on the drawing moves round the equipotential so the force can be seen at every point.
Symbols: x and y are the coordinates of the particle, Fx and Fy the force components, ψ the angle between the force and the line pointing to the centre.
The big idea in plain words
A central force always points straight at the centre, or straight away from it. To test this, draw the line from the centre to the particle and see if the force lies along it. Here the potential bends differently in the x direction and in the y direction, so the force is pulled more strongly toward one axis than the other. It only points at the centre on the two axes. Elsewhere it tilts. The curves of equal U and of equal force size follow from the same two squared terms, so both are ellipses.
The solution in each frame
Only one frame is needed. The centre of the field is at rest, and U depends only on x and y.
The solution step by step
Step 1, the force. Each component is minus the slope of U in that direction: Fx = −∂U/∂x = −2αx and Fy = −∂U/∂y = −2βy. So F = −2αx i − 2βy j.
Step 2, the central test. A central force is parallel to the position vector r = x i + y j, so r × F = 0. The z part of the product is xFy − yFx = x(−2βy) − y(−2αx) = 2(α − β)xy.
Step 3, the verdict. Since α ≠ β, this is not zero at any point off the axes. At the probe on the drawing it is and in the formula, and the live torque is shown on the page. So the field is not central. The tilt of the force away from the line to the centre is at most degrees for these values.
Step 4, equipotential surfaces. On such a surface U is constant: αx² + βy² = U₀. Divide by U₀: x²/(U₀/α) + y²/(U₀/β) = 1. This is an ellipse with semi-axes √(U₀/α) = m along x and √(U₀/β) = m along y.
Step 5, surfaces of constant force size. The size is |F| = √(Fx² + Fy²) = 2√(α²x² + β²y²). Setting it equal to F₀ and squaring gives 4α²x² + 4β²y² = F₀². Divide by F₀²: x²/(F₀/2α)² + y²/(F₀/2β)² = 1. This is also an ellipse, with semi-axes F₀/(2α) = m and F₀/(2β) = m.
Step 6, how the two compare. Both ellipses have their long side along the same axis. The equipotential has the axis ratio √(β/α) = . The constant-force ellipse has the ratio β/α = , so it is more stretched. They would both become circles only if α = β, and then the field would be central. This is why the problem says α and β differ.
A separate calculation does not use the force formulas. It finds the force from the slope of U by small differences at 720 points round the ellipse, and the largest difference from F = (−2αx, −2βy) is of the scale. U stays equal to U₀ round the ellipse to . The torque agrees to and the tilt to . Points on the constant-force ellipse give |F| = F₀ to . A bisection search along the x-axis gives the semi-axis m, against m from the formula. A scan of the tilt finds degrees at most, against degrees from arcsin(|β − α|/(β + α)). The largest relative difference is .
Five methods
1. Force method
Find F from U, then test whether it points along r. This is steps 1 to 3.
2. Energy method
Equal U is the definition of an equipotential, so steps 4 follows at once from U = U₀. A central field has equipotentials that are circles around the centre, so an ellipse already shows the field is not central.
3. Graph method
On the drawing the force arrow and the dashed line to the centre agree only on the two axes. The torque graph is zero there and nowhere else, with the sign changing in each quadrant.
4. Calculus method
Take the partial slopes, form the cross product, and square the force size. These are steps 1, 2 and 5.
5. Check method
Units: αx² is energy, so α is J/m². Then 2αx is J/m, which is newtons. Limits: with α = β the formulas give circles and zero torque, which is the central case.
Fastest: the energy method for the circle test, then the calculus method. Most reliable: the force method with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| α | Yes | The force, both ellipses, the tilt and the torque |
| β | Yes | The force, both ellipses, the tilt and the torque |
| Chosen U₀ | Size only | The size of the equipotential ellipse and where the probe sits |
| Chosen F₀ | Size only | The size of the constant-force ellipse |
| Given-answer mode | Sets β | The page finds β for a chosen largest tilt, and warns above 3 J/m² |
| Play and time bar | No | Play moves the probe once round, holds 1.2 s, then repeats |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Coefficient of x² | α | J/m² |
| Coefficient of y² | β | J/m² |
| Chosen potential energy | U₀ | J |
| Chosen force size | F₀ | N |
| Equipotential semi-axes | √(U₀/α), √(U₀/β) | m, m |
| Constant-force semi-axes | F₀/2α, F₀/2β | m, m |
| Largest tilt of the force | arcsin(|β − α|/(β + α)) | degrees |
Exam traps
- Saying the field is central because U depends on position only. It must depend on the distance r only.
- Calling the equipotentials circles. That needs α = β.
- Dropping the factor 2 in the force: Fx = −2αx, not −αx.
- Writing the constant-force ellipse with the semi-axes of the equipotential. Its semi-axes are F₀/(2α) and F₀/(2β).
- Mixing up the force direction: it points toward lower U, so it has the sign opposite to x and y.