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Linear momentum as state

Linear momentum combines an object's inertia and motion into one vector. For a particle of constant mass moving much slower than light,

\[\mathbf p=m\mathbf v\]

The mass \(m\) is a nonnegative scalar, velocity \(\mathbf v\) is a vector, and momentum \(\mathbf p\) points in the same direction as velocity when mass is positive. Its SI unit is kilogram metre per second, equivalently newton second.

Why momentum belongs in the state

Position alone does not determine how motion continues. Two equal objects at the same place can have different futures if their velocities differ. Momentum records that motion while also including mass, which makes it especially convenient when objects of different masses interact.

For constant mass, velocity and momentum contain equivalent information because

\[\mathbf v=\frac{\mathbf p}{m}\]

Momentum is often the better system variable because total momentum is a direct vector sum. It is also the natural quantity changed by force and impulse. A state representation should not store velocity and momentum as independent authoritative fields unless it also enforces their consistency; otherwise they can silently disagree.

import numpy as np

state = {"mass_kg": 2.0, "position_m": 1.0, "momentum_kg_m_per_s": 6.0}
velocity = state["momentum_kg_m_per_s"] / state["mass_kg"]
print(f"Velocity: {velocity:.1f} m/s")

The division requires positive nonzero mass in this classical particle model. Massless particles are outside the formula's domain, and relativistic particles use a different momentum-velocity relationship.

Momentum is a vector

In one dimension, a sign can represent direction. Choosing rightward as positive makes a left-moving object's momentum negative. In three dimensions, calculate components:

\[\mathbf p=(mv_x,mv_y,mv_z)\]

Opposite momenta can cancel even though neither object is at rest. For two equal masses moving with equal speeds in opposite directions, total momentum is zero while total kinetic energy is positive. Momentum conservation therefore does not mean that every object keeps the same momentum, nor does zero total momentum mean no motion.

Never add vector magnitudes when the directions differ. Add components first, then take the magnitude only if the question asks for it:

\[\mathbf P=\sum_{i=1}^{N}\mathbf p_i,\qquad P=|\mathbf P|\]

Here \(\mathbf P\) is the system total and \(P\) is a nonnegative scalar. These are not interchangeable.

Force is the rate of momentum change

Newton's second law in its general momentum form is

\[\mathbf F_{\mathrm{net}}=\frac{d\mathbf p}{dt}\]

For a single constant-mass particle, substituting

\(\mathbf p=m\mathbf v\)

gives the familiar

\[\mathbf F_{\mathrm{net}}=m\mathbf a\]

The momentum form should not be expanded naively as

\(m\mathbf a+\mathbf v\,dm/dt\)

for a rocket without accounting for the momentum carried across the changing system boundary. Variable-mass problems require an explicit control volume or momentum-flux analysis.

Integrating force over a time interval produces impulse:

\[\mathbf J=\int_{t_1}^{t_2}\mathbf F_{\mathrm{net}}\,dt=\Delta\mathbf p\]

This relationship is valuable in collisions because the contact force may be large and brief while the total impulse remains measurable from the before-and-after momenta.

System momentum and internal exchange

For a system of particles,

\[\mathbf P=\sum_i m_i\mathbf v_i\]

Internal forces redistribute momentum between parts. Under the usual classical assumptions that internal interaction forces cancel in pairs, their contributions vanish from the system total, leaving

\[\frac{d\mathbf P}{dt}=\mathbf F_{\mathrm{external}}\]

If the net external impulse is zero over an interval, total momentum at the end equals total momentum at the beginning. The system does not have to be completely free of external forces; several external impulses can cancel. Conversely, a short interaction with a wall, track, person, expelled gas, or Earth can transfer momentum even if that external body barely moves visibly.

masses = np.array([2.0, 3.0])
velocities = np.array([[3.0, 0.0], [-1.0, 2.0]])
total_momentum = np.sum(masses[:, None] * velocities, axis=0)
print("Total momentum (kg m/s):", total_momentum)

The list of bodies defines the boundary. Record it with the result so another reader can tell whether a force is internal or external.

Reference-frame dependence

Classical momentum depends on the inertial frame. If a frame moves at constant velocity \(\mathbf V\) relative to the original, then Galilean transformation gives

\[\mathbf v_i'=\mathbf v_i-\mathbf V\]

and for a fixed-mass system,

\[\mathbf P'=\mathbf P-M\mathbf V\]

where \(M=\sum_i m_i\). Numerical momentum values change between frames, but conservation holds in every inertial frame when the assumptions hold. Before comparing two results, confirm that their axes, origin motion, and time interval match.

The center-of-momentum frame is the inertial frame in which total momentum is zero. For nonrelativistic fixed masses, it moves with the center-of-mass velocity

\[\mathbf V_{\mathrm{cm}}=\frac{\mathbf P}{M}\]

Individual bodies can still move rapidly in this frame.

Worked two-body calculation

Take a \(2\,\mathrm{kg}\) cart moving right at \(3\,\mathrm{m/s}\) and a \(1\,\mathrm{kg}\) cart moving left at \(4\,\mathrm{m/s}\). With right positive,

\[P=(2)(3)+(1)(-4)=2\ \mathrm{kg\,m/s}\]

The total points right even though one cart moves left. If the carts interact on an effectively isolated track, any acceptable outgoing velocities must produce the same total. Momentum alone may not determine both outgoing velocities; collision type or a coefficient of restitution supplies additional information.

Numerical updates and residuals

For a computed system, update each body's momentum from forces and then evaluate the total. A simple step has the structure

\[\mathbf p_i^{n+1}=\mathbf p_i^n+\mathbf F_i^n\Delta t\]

This is first-order time integration, not an exact formula for changing force. Internal force pairs should be generated consistently so equal-and-opposite impulses cancel to roundoff. Computing the two sides through unrelated approximations can create artificial system momentum.

momenta = masses[:, None] * velocities
before = momenta.sum(axis=0)
forces = np.array([[4.0, 0.0], [-4.0, 0.0]])
dt = 0.1
momenta += forces * dt
external_impulse = np.zeros(2)
residual = momenta.sum(axis=0) - before - external_impulse
assert np.allclose(residual, 0.0)
print("Momentum residual:", residual)

Compare the residual with a physically scaled absolute-plus-relative tolerance. Reduce the time step and verify convergence. A conserved total can still coexist with incorrect individual trajectories, so also test analytic cases, energy when appropriate, and contact conditions.

Domain and model limits

The expression \(\mathbf p=m\mathbf v\) is a nonrelativistic approximation. At speeds approaching the speed of light, momentum is

\[\mathbf p=\gamma m\mathbf v,\qquad \gamma=\frac{1}{\sqrt{1-v^2/c^2}}\]

Momentum conservation remains fundamental, but the relationship to velocity changes. At ordinary mechanical speeds, the classical expression is accurate enough.

Reliability checks

Momentum is useful state because it connects microscopic exchanges to a simple system balance. It can change dramatically for one object while remaining exactly accounted for across a well-defined isolated system.

A worked Python notebook

Store the momentum of a 2 kilogram body and inspect its relation to velocity. The question for this calculation is precise: how does momentum change when we change velocity, 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 mass is fixed; changing mass requires an explicit open-system momentum balance. 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 -5 to 5 m/s. 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 velocity, expressed in m/s. The output is momentum, expressed in kg m/s. 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.

\[f(q) = 2 q\]

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 3 m/s, the expression gives 6 kg m/s. 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("2*q", locals={"q": q})
reference_input = 3.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m/s")
print("Predicted momentum:", reference_output, "kg m/s")

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 -10 kg m/s. The midpoint, at 0 m/s, gives 0 kg m/s. The upper endpoint gives 10 kg m/s. Read these together with the assumptions: they describe different choices of velocity 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([-5.0, 0.0, 5.0])
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/s -> {output_value:.6g} kg m/s")
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 velocity; the vertical axis shows momentum. 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(-5.0, 5.0, 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("Velocity (m/s)")
ax.set_ylabel("Momentum (kg m/s)")
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 momentum change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 3 m/s is 2 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.

\[f'(q) = 2\]

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 = 3.05
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, "kg m/s")
print("Full expression:", full_prediction, "kg m/s")
print("Approximation difference:", full_prediction - linear_prediction, "kg m/s")

Here the changed input is 3.05 m/s. The tangent estimate is 6.1 kg m/s, while the full expression gives 6.1 kg m/s. 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 = 10.0
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 velocity inside the stated interval, predict whether momentum 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 mass is fixed; changing mass requires an explicit open-system momentum balance. 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 kg m/s: 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(5.0) - evaluate(-5.0))
for points in [101, 201, 401]:
    grid = np.linspace(-5.0, 5.0, 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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