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Work as path integrals

Mechanical work measures energy transferred by a force through displacement. Because force and displacement are vectors, only the force component tangent to the motion contributes. For a varying force or curved path, the contributions must be accumulated along the actual trajectory.

Constant force over one displacement

For constant force \(\mathbf F\) and displacement \(\Delta\mathbf r\), work done by that force is

\[W=\mathbf F\cdot\Delta\mathbf r=F\,\Delta r\cos\theta\]

The angle \(\theta\) is between force and displacement. Work is positive when force has a component along the displacement, negative when it opposes displacement, and zero when it is perpendicular.

In SI units,

\[1\,\mathrm J=1\,\mathrm{N\,m}\]

Torque can also use newton-metres, but torque is an axial vector and energy or work is a scalar. Equal unit dimensions do not make the quantities interchangeable.

From small steps to a line integral

Along a small displacement \(d\mathbf r\), the differential work is

\[dW=\mathbf F\cdot d\mathbf r\]

Adding these contributions along an oriented path \(C\) gives

\[W_C=\int_C\mathbf F(\mathbf r,t)\cdot d\mathbf r\]

If the path is parameterized by time, \(\mathbf r(t)\), then

\(d\mathbf r=\mathbf v(t)dt\)

and

\[W=\int_{t_1}^{t_2}\mathbf F(\mathbf r(t),\mathbf v(t),t)\cdot\mathbf v(t)\,dt\]

This form also handles velocity- and time-dependent forces. Reversing the path reverses work only when the same position-dependent force field is traversed in reverse. With time dependence, velocity dependence, hysteresis, or changing constraints, the reversed experiment may not sample the same force.

Cartesian component form

In Cartesian coordinates,

\[W=\int_C(F_x\,dx+F_y\,dy+F_z\,dz)\]

This is not three unrelated integrals unless the path and force permit that separation. The coordinate changes occur together along one curve.

For a parameter \(u\) with endpoints \(u_a\) and \(u_b\),

\[W=\int_{u_a}^{u_b}\mathbf F(\mathbf r(u))\cdot\frac{d\mathbf r}{du}\,du\]

The tangent derivative and parameter interval encode path orientation. Reparameterizing the same oriented curve does not change exact work.

Worked curved-path example

Consider the force field \(\mathbf F=(ky,0)\) and the quarter-circle path

\[\mathbf r(\phi)=(R\cos\phi,R\sin\phi),\qquad 0\le\phi\le\frac{\pi}{2}\]

The path derivative is

\[\frac{d\mathbf r}{d\phi}=(-R\sin\phi,R\cos\phi)\]

and the field on the path is \(\mathbf F=(kR\sin\phi,0)\). Therefore

\[W=\int_0^{\pi/2}-kR^2\sin^2\phi\,d\phi=-\frac{\pi kR^2}{4}\]

The negative sign follows because the force points right while the path's horizontal motion points left over the quarter circle. A straight path between the same endpoints gives a different result, demonstrating that this field is not conservative over the region.

Work-energy theorem

For a particle of constant mass in an inertial frame, net work equals change in kinetic energy:

\[W_{\mathrm{net}}=\int_C\mathbf F_{\mathrm{net}}\cdot d\mathbf r=\Delta K\]

with

\[K=\frac12mv^2\]

This follows by using

\(\mathbf F=m\,d\mathbf v/dt\)

and \(d\mathbf r=\mathbf vdt\). It is net work by all forces on the particle, not automatically the work by one selected force.

The theorem is an independent check: calculate the force integral and kinetic-energy difference separately. Agreement tests force signs, path order, and numerical quadrature. It does not identify how energy is divided among thermal, elastic, or other stores in a larger system.

Conservative forces and endpoints

A conservative force can be written as

\[\mathbf F=-\nabla U\]

Then its work depends only on endpoints:

\[W_{\mathrm{cons}}=U(\mathbf r_1)-U(\mathbf r_2)=-\Delta U\]

For a closed path,

\[\oint_C\mathbf F\cdot d\mathbf r=0\]

These statements need an appropriate domain. A locally curl-free field on a region with holes or singularities may require additional global analysis before a single-valued potential is assumed.

Gravity near Earth's surface gives \(W_g=-mg(y_2-y_1)\). Horizontal shape does not matter in this approximation. Kinetic friction usually gives path- dependent negative work because it acts opposite sliding over accumulated distance.

Numerical work from sampled data

Suppose measured or calculated positions are \(\mathbf r_0,\ldots,\mathbf r_N\). A left-endpoint approximation is

\[W\approx\sum_{i=0}^{N-1}\mathbf F(\mathbf r_i)\cdot(\mathbf r_{i+1}-\mathbf r_i)\]

A trapezoidal force average is usually more accurate for a smooth position-dependent field:

\[W\approx\sum_{i=0}^{N-1}\frac{\mathbf F_i+\mathbf F_{i+1}}{2}\cdot\Delta\mathbf r_i\] 
import numpy as np

positions_m = np.column_stack([np.linspace(0, 2, 101), np.zeros(101)])
forces_n = np.column_stack([3 * positions_m[:, 0], np.zeros(101)])
displacements = np.diff(positions_m, axis=0)
average_forces = (forces_n[:-1] + forces_n[1:]) / 2
work_j = np.sum(average_forces * displacements)
print(f"Work: {work_j:.3f} J")
assert np.isclose(work_j, 6.0)

Force and position samples must refer to compatible times and frames. If only force at one endpoint is measured, calling the result trapezoidal invents a missing sample.

Refine the path, not just the plotted curve

Numerical quadrature error depends on how force changes along each segment. Repeat the calculation with more samples or subdivided steps, compare the same full path, and look for convergence. Sharp force changes, collisions, and discontinuities should be split at known events rather than averaged across blindly.

For measured data, finer interpolation does not create new information. Measurement uncertainty and sampling bandwidth can dominate quadrature error. Report both when they matter.

Power is the rate of doing work

Instantaneous power delivered by a force is

\[P=\frac{dW}{dt}=\mathbf F\cdot\mathbf v\]

Power can be zero even when force is nonzero, as with an ideal centripetal force perpendicular to velocity. It can be negative when a force removes mechanical energy. Average power over a nonzero interval is

\(\Delta W/\Delta t\)

and should not be confused with the average of endpoint powers without an approximation rule.

System boundaries and sign conventions

Say “work done by force A on system B.” The same interaction transfers energy out of one system and into another. If the system includes both a block and spring, spring potential energy may be internal; for a block-only system, work by the spring crosses the boundary.

Friction illustrates why the boundary matters. For a sliding block alone, kinetic friction does negative work on the block. For a larger block-plus-surface system, the interaction can increase internal thermal energy while total energy is conserved.

Checks that expose mistakes

Work is not a property stored at one point on a path. It is an energy transfer accumulated from a force along an oriented displacement, and its correct calculation preserves both the geometry and the physical system boundary.

A worked Python notebook

Move through a one-dimensional force field F equals 3x and integrate the work to a chosen endpoint. The question for this calculation is precise: how does work from x=0 change when we change endpoint position, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

The path is along x and the coefficient has units of newtons per metre. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 3 m. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for endpoint position, expressed in m. The output is work from x=0, expressed in J. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = \frac{3 q^{2}}{2}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 2 m, the expression gives 6 J. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("3*q**2/2", locals={"q": q})
reference_input = 2.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m")
print("Predicted work from x=0:", reference_output, "J")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0 J. The midpoint, at 1.5 m, gives 3.375 J. The upper endpoint gives 13.5 J. Read these together with the assumptions: they describe different choices of endpoint position within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 1.5, 3.0])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} m -> {output_value:.6g} J")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows endpoint position; the vertical axis shows work from x=0. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 3.0, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Endpoint position (m)")
ax.set_ylabel("Work from x=0 (J)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does work from x=0 change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 2 m is 6 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = 3 q\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 2.015
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "J")
print("Full expression:", full_prediction, "J")
print("Approximation difference:", full_prediction - linear_prediction, "J")

Here the changed input is 2.015 m. The tangent estimate is 6.09 J, while the full expression gives 6.09034 J. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 3.0
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of endpoint position inside the stated interval, predict whether work from x=0 will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: The path is along x and the coefficient has units of newtons per metre. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of J: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(3.0) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 3.0, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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