SCIENCE · INVESTIGATE AND EXPLAIN

Check every direction before calling an equilibrium stable

For U=(kx x²+ky y²)/2, the negative gradient gives force(−kx x,−ky y). The origin has zero force for every coefficient pair, but that alone does not classify its stability. Inspect how potential changes under displacements in both independent directions.

Learning outcome: Compare all sign combinations and one zero coefficient. Derive forces from the potential and classify using both slices, naming exactly what the quadratic model omits.

Starting recap

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For kx=−1,ky=−1, both slices fall and the origin is a maximum. For kx=1,ky=0, there is a restoring x force but no y force anywhere in the exact model.

Open the experiment

Predict

With kx1 and ky−1, the origin is:

Your prediction and reason

Experiment

Compare all sign combinations and one zero coefficient. Derive forces from the potential and classify using both slices, naming exactly what the quadratic model omits.

Classify stability in both independent directions

Worked teaching model · synthetic values · Exact planar quadratic U=(kx x²+ky y²)/2 with diagonal coefficients. Classify the origin using both directions; zero stiffness is exactly flat in this stipulated model, not permission to ignore unknown higher-order terms in a real potential.

East stiffness: 1 N/m North stiffness: -1 N/m

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Independent coordinate slices through the origin

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: east-axis slice y0 (solid).2: north-axis slice x0 (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Independent coordinate slices through the origin

east-axis slice y0

selected coordinate (m)	potential energy (J)
-2	2
-1.95	1.90125
-1.9	1.805
-1.85	1.71125
-1.8	1.62
-1.75	1.53125
-1.7	1.445
-1.65	1.36125
-1.6	1.28
-1.55	1.20125
-1.5	1.125
-1.45	1.05125
-1.4	0.98
-1.35	0.91125
-1.3	0.845
-1.25	0.78125
-1.2	0.72
-1.15	0.66125
-1.1	0.605
-1.05	0.55125
-1	0.5
-0.95	0.45125
-0.9	0.405
-0.85	0.36125
-0.8	0.32
-0.75	0.28125
-0.7	0.245
-0.65	0.21125
-0.6	0.18
-0.55	0.15125
-0.5	0.125
-0.45	0.10125
-0.4	0.08
-0.35	0.06125
-0.3	0.045
-0.25	0.03125
-0.2	0.02
-0.15	0.01125
-0.1	0.005
-0.05	0.00125
0	0
0.05	0.00125
0.1	0.005
0.15	0.01125
0.2	0.02
0.25	0.03125
0.3	0.045
0.35	0.06125
0.4	0.08
0.45	0.10125
0.5	0.125
0.55	0.15125
0.6	0.18
0.65	0.21125
0.7	0.245
0.75	0.28125
0.8	0.32
0.85	0.36125
0.9	0.405
0.95	0.45125
1	0.5
1.05	0.55125
1.1	0.605
1.15	0.66125
1.2	0.72
1.25	0.78125
1.3	0.845
1.35	0.91125
1.4	0.98
1.45	1.05125
1.5	1.125
1.55	1.20125
1.6	1.28
1.65	1.36125
1.7	1.445
1.75	1.53125
1.8	1.62
1.85	1.71125
1.9	1.805
1.95	1.90125
2	2

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

north-axis slice x0

selected coordinate (m)	potential energy (J)
-2	-2
-1.95	-1.90125
-1.9	-1.805
-1.85	-1.71125
-1.8	-1.62
-1.75	-1.53125
-1.7	-1.445
-1.65	-1.36125
-1.6	-1.28
-1.55	-1.20125
-1.5	-1.125
-1.45	-1.05125
-1.4	-0.98
-1.35	-0.91125
-1.3	-0.845
-1.25	-0.78125
-1.2	-0.72
-1.15	-0.66125
-1.1	-0.605
-1.05	-0.55125
-1	-0.5
-0.95	-0.45125
-0.9	-0.405
-0.85	-0.36125
-0.8	-0.32
-0.75	-0.28125
-0.7	-0.245
-0.65	-0.21125
-0.6	-0.18
-0.55	-0.15125
-0.5	-0.125
-0.45	-0.10125
-0.4	-0.08
-0.35	-0.06125
-0.3	-0.045
-0.25	-0.03125
-0.2	-0.02
-0.15	-0.01125
-0.1	-0.005
-0.05	-0.00125
0	0
0.05	-0.00125
0.1	-0.005
0.15	-0.01125
0.2	-0.02
0.25	-0.03125
0.3	-0.045
0.35	-0.06125
0.4	-0.08
0.45	-0.10125
0.5	-0.125
0.55	-0.15125
0.6	-0.18
0.65	-0.21125
0.7	-0.245
0.75	-0.28125
0.8	-0.32
0.85	-0.36125
0.9	-0.405
0.95	-0.45125
1	-0.5
1.05	-0.55125
1.1	-0.605
1.15	-0.66125
1.2	-0.72
1.25	-0.78125
1.3	-0.845
1.35	-0.91125
1.4	-0.98
1.45	-1.05125
1.5	-1.125
1.55	-1.20125
1.6	-1.28
1.65	-1.36125
1.7	-1.445
1.75	-1.53125
1.8	-1.62
1.85	-1.71125
1.9	-1.805
1.95	-1.90125
2	-2

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
East potential curvature	1	N/m
North potential curvature	-1	N/m
East force at(1,1)	-1	N
North force at(1,1)	1	N
Strict potential minimum	0	flag
Saddle	1	flag
Exactly flat coordinate directions	0	count

One direction rises and one falls: the origin is a saddle, not stable in every direction.

The gradient gives F=(−kx x,−ky y). Independent second derivatives are kx and ky; stability cannot be certified from a favorable one-dimensional slice alone. A zero coefficient is a perfectly flat direction only because this exact quadratic has no higher-order terms. A real zero-curvature measurement would require checking those omitted terms.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

One direction rises and one falls: the origin is a saddle, not stable in every direction.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Check every direction before calling an equilibrium stable

If both stiffnesses are positive, both coordinate slices rise and the origin is a strict minimum. If both are negative, it is a strict maximum. Opposite signs produce a saddle: one direction restores and the other drives the point away. A reassuring plot along just the restoring direction would miss the instability.

With kx1 and ky−1 N/m, the force at(1,1) m is(−1,+1) N. Along the x axis potential increases as x²/2, while along the y axis it decreases as−y²/2. The positive x curvature cannot certify the full planar equilibrium.

A zero coefficient creates an exactly flat coordinate direction only because this stipulated potential has no higher-order terms. It is not a strict isolated minimum even if the other direction rises. For a real potential with only measured zero quadratic curvature, higher-order terms might create a minimum, maximum or other structure, as the previous quartic example shows.

Changed-case check

Predict before reading the comparison. For kx=−1,ky=−1, both slices fall and the origin is a maximum. For kx=1,ky=0, there is a restoring x force but no y force anywhere in the exact model.

Retained investigation

Compare all sign combinations and one zero coefficient. Derive forces from the potential and classify using both slices, naming exactly what the quadratic model omits. Why is a zero measured curvature not enough to infer an exactly flat real potential?

Source and boundary

Primary conceptual reference checked30 October2026: OpenStax University Physics Volume1. All functions, numerical packets and teaching prose are original. Bounded supplied models are not measured experiments, apparatus instructions, independent practical skill or institutional credit. Numerical algorithm properties are derived for the disclosed model; do not treat them as universal solver guarantees.

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Why is a zero measured curvature not enough to infer an exactly flat real potential?

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