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Newton’s laws as update rules

Newton's laws connect interactions to changes in motion. They do not say that force causes velocity; in an inertial frame, net force causes momentum to change. For a body of constant mass this becomes acceleration. A numerical update approximates those continuous laws over finite time, so the force model and integrator must remain visibly separate.

First law: define the inertial baseline

Newton's first law says that a body with zero net external force remains at rest or moves with constant velocity in an inertial reference frame. In symbols,

\[\sum\mathbf F_{\mathrm{ext}}=\mathbf 0\quad\Longrightarrow\quad\frac{d\mathbf p}{dt}=\mathbf 0\]

For constant nonzero mass, momentum

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

is constant exactly when velocity is constant. Zero net force does not require zero velocity. An object coasting at a steady velocity needs no continuing net force in this ideal description.

The first law also identifies the frames in which the other Newtonian laws take their standard form. An accelerating or rotating frame needs additional inertial-force terms if equations are written using that frame's measured acceleration.

Second law: momentum form first

Newton's second law is most generally written

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

For one body with constant mass,

\[\sum\mathbf F=m\mathbf a\]

The force and acceleration point in the same direction when mass is a positive scalar. The equation concerns the vector sum of all forces acting on the chosen system, not one arbitrarily selected force. Units provide an immediate check:

\[1\,\mathrm N=1\,\mathrm{kg\,m/s^2}\]

The shortcut

\(\mathbf a=\sum\mathbf F/m\)

requires constant mass and \(m>0\). Variable-mass systems require careful system boundaries and momentum flux; blindly differentiating \(m\mathbf v\) for a rocket without accounting for expelled material gives an incomplete model.

Third law: interactions come in pairs

If body A exerts a force on body B, then B exerts an equal-magnitude, opposite-direction force on A:

\[\mathbf F_{A\to B}=-\mathbf F_{B\to A}\]

The two forces act on different bodies. They therefore do not cancel on either body's free-body diagram. They cancel in the net internal-force sum only when A and B are both inside the chosen system boundary.

A force record should name source and target. A generic array called forces can lose which object receives the vector and can tempt code to apply a pair twice or to the same body.

import numpy as np

positions_m = np.array([0.0, 1.2])
masses_kg = np.array([1.0, 2.0])
stiffness, rest_length = 10.0, 1.0
force_on_first = stiffness * (positions_m[1] - positions_m[0] - rest_length)
forces_n = np.array([force_on_first, -force_on_first])
print("Pair forces (N):", forces_n)
assert np.isclose(forces_n.sum(), 0.0)

This exact opposite assignment preserves total momentum up to integration and floating-point error when the interaction law and closed-system assumptions support it.

Separate force evaluation from state advancement

At time \(t_n\), evaluate forces from a coherent snapshot of all bodies. For body \(i\),

\[\mathbf a_i(t_n)=\frac{1}{m_i}\sum_j\mathbf F_{j\to i}(t_n)\]

Only after every body's acceleration is known should states be advanced. Updating one body immediately and then using its new position to calculate another body's force mixes times and makes the result depend on iteration order.

def accelerations(positions, masses):
    if np.any(masses <= 0):
        raise ValueError("Masses must be positive.")
    pair_force = 10.0 * (positions[1] - positions[0] - 1.0)
    return np.array([pair_force, -pair_force]) / masses


print("Accelerations (m/s^2):", accelerations(positions_m, masses_kg))

The snapshot includes position, velocity, orientation if relevant, and any environmental fields at one time. Velocity-dependent forces such as drag must read the correct frame- relative velocity.

From differential law to a finite update

One simple velocity-first semi-implicit Euler step is

\[\mathbf v_{n+1}=\mathbf v_n+h\mathbf a_n,\qquad \mathbf r_{n+1}=\mathbf r_n+h\mathbf v_{n+1}\] 
velocities_m_per_s = np.zeros(2)
dt_s = 0.01
old_positions = positions_m.copy()
new_velocities = velocities_m_per_s + dt_s * accelerations(old_positions, masses_kg)
new_positions = old_positions + dt_s * new_velocities
print("New positions (m):", new_positions)
assert np.isclose(np.dot(masses_kg, new_velocities), 0.0)

This is a numerical approximation, not Newton's second law itself. Forward Euler, velocity Verlet, Runge-Kutta methods, and specialized constrained integrators use different finite updates while approximating the same continuous equations. The method and time step affect numerical error and stability.

Worked force sum

A cart of mass \(m=2.0\,\mathrm{kg}\) receives a horizontal push of \(10\,\mathrm N\) and kinetic friction of \(4\,\mathrm N\). Choosing the push direction as positive,

\[F_{\mathrm{net},x}=10\,\mathrm N-4\,\mathrm N=6\,\mathrm N\]

so

\[a_x=\frac{6\,\mathrm N}{2.0\,\mathrm{kg}}=3.0\,\mathrm{m/s^2}\]

If the horizontal forces stay constant for \(0.10\,\mathrm s\), the exact velocity change over that interval is \(0.30\,\mathrm{m/s}\). The normal force and weight cancel vertically only because vertical acceleration is zero under the stated contact conditions; they are still real forces on the cart.

Common force models and domains

Uniform near-surface weight is

\[\mathbf F_g=m\mathbf g\]

Hooke's law for an ideal spring along one coordinate is

\[F_s=-k(x-x_{\mathrm{eq}})\]

and a simple linear drag model is

\[\mathbf F_D=-b(\mathbf v-\mathbf v_{\mathrm{fluid}})\]

Each formula has a domain. Uniform weight is a local approximation. An ideal spring is linear only over a suitable deformation range. Linear drag applies in a different flow regime from quadratic drag. A force function should validate parameters and document its approximation rather than silently returning a vector everywhere.

Internal and external forces

Whether a force is internal depends on the system boundary. For a two-cart system, the spring force between carts is internal. For one cart alone, that same spring interaction is external. Summing the momentum equations for a closed multi-body system cancels third-law internal pairs and gives

\[\frac{d\mathbf P_{\mathrm{total}}}{dt}=\sum\mathbf F_{\mathrm{external}}\]

This system-level balance is a powerful check. If a closed pair interaction changes total momentum substantially, inspect sign, target assignment, update ordering, and boundary assumptions.

Forces are not impulses

A continuous force changes momentum through time integration:

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

An ideal instantaneous collision may be represented by an impulse that changes velocity discontinuously. Treating that event as an ordinary finite force evaluated once per frame makes the result depend arbitrarily on the time step. Collision response needs an event or contact model consistent with its time scale.

Validation tests

Newton's laws define continuous relationships between interactions and motion. A good computational translation preserves system boundaries, vector sums, pair ownership, units, and simultaneous state before choosing how to approximate the time evolution.

A worked Python notebook

Apply different net forces to a body whose inertial mass is 2 kilograms. The question for this calculation is precise: how does acceleration change when we change applied net force, 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.

This is the net external force in an inertial frame; internal forces cannot accelerate the whole isolated system. 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 N. 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 applied net force, expressed in N. The output is acceleration, expressed in m/s^2. 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.

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

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 3 N, the expression gives 1.5 m/s^2. 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("q/2", locals={"q": q})
reference_input = 3.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "N")
print("Predicted acceleration:", reference_output, "m/s^2")

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 -2.5 m/s^2. The midpoint, at 0 N, gives 0 m/s^2. The upper endpoint gives 2.5 m/s^2. Read these together with the assumptions: they describe different choices of applied net force 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} N -> {output_value:.6g} m/s^2")
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 applied net force; the vertical axis shows acceleration. 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("Applied net force (N)")
ax.set_ylabel("Acceleration (m/s^2)")
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 acceleration change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 3 N is 0.5 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.

\[f'(q) = \frac{1}{2}\]

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, "m/s^2")
print("Full expression:", full_prediction, "m/s^2")
print("Approximation difference:", full_prediction - linear_prediction, "m/s^2")

Here the changed input is 3.05 N. The tangent estimate is 1.525 m/s^2, while the full expression gives 1.525 m/s^2. 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 applied net force inside the stated interval, predict whether acceleration 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: This is the net external force in an inertial frame; internal forces cannot accelerate the whole isolated system. 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 m/s^2: 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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