In a two-body problem, each body accelerates because of the other. In an N-body problem, every body responds to the combined field of all the others. The rule is simple, but its coupling matters: the trajectories must be solved together from one simultaneous state.
For bodies at positions \(\mathbf r_1\) and \(\mathbf r_2\), define
\[\mathbf d=\mathbf r_2-\mathbf r_1\]The force on body 1 due to body 2 is
\[\mathbf F_{1\leftarrow2}=Gm_1m_2\frac{\mathbf d}{|\mathbf d|^3}\]and Newton's third law gives
\[\mathbf F_{2\leftarrow1}=-\mathbf F_{1\leftarrow2}\]The forces are equal and opposite, but accelerations are generally not. Dividing by each body's mass gives acceleration inversely proportional to that body's inertial mass. A lighter body accelerates more, while the pair's total momentum remains constant in the absence of external force.
import numpy as np
masses = np.array([1.0, 2.0])
positions = np.array([[-1.0, 0.0], [1.0, 0.0]])
separation = positions[1] - positions[0]
force = masses[0] * masses[1] * separation / np.linalg.norm(separation) ** 3
print("Pair forces with G=1:", force, -force)
Computing both ordered pairs and adding each force twice is a common error. Visit only indices with \(i<j\).
Center-of-mass position is
\[\mathbf R=\frac{m_1\mathbf r_1+m_2\mathbf r_2}{m_1+m_2}\]With no external force, it moves at constant velocity. Relative position
\[\mathbf r=\mathbf r_1-\mathbf r_2\]obeys
\[\ddot{\mathbf r}=-G(m_1+m_2)\frac{\mathbf r}{|\mathbf r|^3}\]This turns the coupled relative motion into an equivalent one-body central-force problem. The reduced mass
\[m_{\mathrm r}=\frac{m_1m_2}{m_1+m_2}\]is useful for writing the relative kinetic energy and angular momentum. After solving relative motion, recover each orbit about the center of mass. Both bodies trace similar conics scaled according to the mass ratio; the heavier body has the smaller orbit.
When \(m_1\ll m_2\), the center of mass lies close to body 2 and treating body 2 as fixed can be an excellent approximation. It is still an approximation, not a different law of gravity.
For body \(i\), acceleration is
\[\mathbf a_i=G\sum_{j\ne i}m_j\frac{\mathbf r_j-\mathbf r_i}{|\mathbf r_j-\mathbf r_i|^3}\]Direct evaluation requires roughly one pair calculation for every pair of bodies, so its cost grows as the square of body count. The exact pair count is \(N(N-1)/2\).
def all_forces(positions, masses):
forces = np.zeros_like(positions)
for i in range(len(masses) - 1):
for j in range(i + 1, len(masses)):
offset = positions[j] - positions[i]
radius = np.linalg.norm(offset)
if radius == 0:
raise ValueError("Point bodies overlap.")
force = masses[i] * masses[j] * offset / radius**3
forces[i] += force
forces[j] -= force
return forces
forces = all_forces(positions, masses)
assert np.allclose(forces.sum(axis=0), 0.0)
print(forces)
The state used by every pair must correspond to the same time. Advancing body 1 before calculating body 2's interactions makes the answer depend on array order and no longer implements the intended simultaneous equations.
Total linear momentum is
\[\mathbf P=\sum_i m_i\mathbf v_i\]and pair-force cancellation gives a constant \(\mathbf P\). Therefore the center of mass
\[\mathbf R_{\mathrm{cm}}=\frac{\sum_i m_i\mathbf r_i}{\sum_i m_i}\]moves uniformly. A drifting center-of-mass velocity in an isolated calculation often reveals asymmetric force accumulation or numerical error.
Total angular momentum about a fixed inertial origin is
\[\mathbf L=\sum_i m_i\mathbf r_i\times\mathbf v_i\]and Newtonian pair forces conserve it because each force lies along its pair separation. Total mechanical energy is
\[E=\sum_i\frac12m_i|\mathbf v_i|^2-G\sum_{i<j}\frac{m_im_j}{r_{ij}}\]The potential sum also visits each unordered pair once. These quantities are exact invariants of the ideal isolated continuous equations. A numerical method only approximates them.
At each step, first assemble accelerations or forces. A symplectic method such as velocity Verlet often gives useful long-term energy behavior for conservative gravity. Adaptive high-order methods can resolve encounters and event times efficiently. The best choice depends on whether the goal is long statistical evolution, precise short arcs, or close-encounter timing.
Whatever method is chosen, compare results at fixed physical times under step or tolerance refinement. Monitor each invariant separately. Small total energy error does not guarantee that individual trajectories or encounter times are accurate in a chaotic system.
Point-mass acceleration diverges when separation goes to zero. A real body may collide, deform, merge, fragment, or pass through only if the model deliberately permits that. Define radii and event handling when collisions matter.
A Plummer-style softened acceleration is
\[\mathbf a_i=G\sum_{j\ne i}m_j\frac{\mathbf r_j-\mathbf r_i}{(r_{ij}^2+\epsilon^2)^{3/2}}\]Softening length \(\epsilon\) removes the point singularity but changes small-scale force and potential. It can represent limited spatial resolution in a large collisionless system, but it is not an exact point-body solution. Report it in physical units and repeat important conclusions with other appropriate values.
Adaptive time steps introduce another issue: if each body uses a different time without a consistent algorithm, pair symmetry and synchronization can be lost. Encounter methods must preserve a well-defined common evolution policy.
If the model says colliding bodies merge perfectly inelastically, conserve total mass and linear momentum:
\[m_{\mathrm new}=m_1+m_2\qquad\mathbf v_{\mathrm new}=\frac{m_1\mathbf v_1+m_2\mathbf v_2}{m_1+m_2}\]Place the new body at the pair center of mass. Mechanical energy is not conserved by an inelastic merge; the missing kinetic energy represents deformation, heat, radiation, or other unresolved internal processes. Angular momentum associated with internal spin may also need storage if the merged object is not treated as a point.
velocities = np.array([[0.0, 0.2], [0.0, -0.1]])
merged_mass = masses.sum()
merged_position = np.sum(masses[:, None] * positions, axis=0) / merged_mass
merged_velocity = np.sum(masses[:, None] * velocities, axis=0) / merged_mass
print(merged_mass, merged_position, merged_velocity)
Direct summation is clear and useful for validation, but large systems need other approaches. Tree methods group distant masses, particle-mesh methods solve a field on a grid, and hybrid methods combine near direct forces with distant approximations. Each introduces parameters such as opening angle, grid spacing, or assignment kernel.
These methods should be compared against direct summation on smaller fixtures. Parallel reductions may change floating-point addition order, producing tiny differences that later grow in chaotic systems. Record deterministic ordering or acceptable statistical comparisons rather than expecting bit-identical trajectories automatically.
Start with two unequal masses in their center-of-mass frame. Confirm equal-and-opposite pair force, unequal acceleration, stationary center of mass, and agreement with a known circular or elliptic orbit. Then add a distant third body and predict the direction of its initial perturbation.
Permute body storage order and confirm that physical output agrees within the declared floating-point policy. Reduce time step, examine minimum separations, and compare energy, momentum, angular momentum, and center-of-mass velocity. For softened or approximate forces, vary the approximation parameter independently of time step.
An N-body trajectory is credible only with the initial state, unit system, force law, collision or softening policy, integration method, and error evidence. Determinism of the equations does not imply that a long chaotic trajectory is predictable to arbitrary precision.
Calculate the mutual force between point masses 1 and 2 in units with G equal to one. The question for this calculation is precise: how does attractive force magnitude change when we change pair separation, 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 same force magnitude acts on each mass in opposite directions. 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 0.5 to 3 scaled length units. 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.
In this small example, q stands for pair separation, expressed in scaled length units. The output is
attractive force magnitude, expressed in scaled force units. 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.
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 1 scaled length units, the expression gives 2 scaled force units. 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**2", locals={"q": q})
reference_input = 1.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "scaled length units")
print("Predicted attractive force magnitude:", reference_output, "scaled force units")
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.
Before drawing hundreds of points, calculate three. The lower endpoint gives 8 scaled force units. The midpoint, at 1.75 scaled length units, gives 0.653061 scaled force units. The upper endpoint gives 0.222222 scaled force units. Read these together with the assumptions: they describe different choices of pair separation 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([0.5, 1.75, 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} scaled length units -> "
f"{output_value:.6g} scaled force units"
)
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.
The horizontal axis shows pair separation; the vertical axis shows attractive force magnitude. 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(0.5, 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("Pair separation (scaled length units)")
ax.set_ylabel("Attractive force magnitude (scaled force units)")
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.
A derivative asks a more specific question than the whole graph: how quickly does attractive force magnitude change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 1 scaled length units is -4 in those ratio units. Locally, the output decreases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.
\[f'(q) = - \frac{4}{q^{3}}\]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 = 1.0125
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, "scaled force units")
print("Full expression:", full_prediction, "scaled force units")
print(
"Approximation difference:",
full_prediction - linear_prediction,
"scaled force units",
)
Here the changed input is 1.0125 scaled length units. The tangent estimate is 1.95 scaled force units, while the full expression gives 1.95092 scaled force units. 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.
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 = 2.5
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.
Make one copy of the reference calculation before editing it. Choose a second value of pair separation inside the stated interval, predict whether attractive force magnitude 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 same force magnitude acts on each mass in opposite directions. 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.
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 scaled force units: 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(0.5))
for points in [101, 201, 401]:
grid = np.linspace(0.5, 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.