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Coordinate systems and basis changes

A coordinate system lets us describe locations and vector directions with numbers. It chooses an origin, axes, axis orientation, and units. These choices belong to the description, not to the physical object. Two observers can assign different components to the same displacement and still agree on its length and physical direction.

Coordinates need an origin and a basis

In a two-dimensional Cartesian system, the position of a point is

\[\mathbf r=x\hat{\mathbf i}+y\hat{\mathbf j}\]

The origin determines where \(x=0\) and \(y=0\) are located. The unit vectors \(\hat{\mathbf i}\) and \(\hat{\mathbf j}\) determine the positive directions. Cartesian axes are normally orthogonal and the basis vectors have unit magnitude.

A position coordinate changes if the origin is translated. A free vector such as a displacement or velocity does not change under a simple origin translation, provided the axes keep the same orientation and the compared frames have no relative motion. For two points, the translated origin cancels from their displacement:

\[\Delta\mathbf r=\mathbf r_2-\mathbf r_1\]

This distinction prevents a common data-model error: treating a position and a displacement as interchangeable arrays merely because each has the same number of components and the same units.

Polar and Cartesian coordinates

A point in a plane can also be described by radial distance \(r\) and angle \(\phi\). With the angle measured counterclockwise from the positive horizontal axis,

\[x=r\cos\phi,\qquad y=r\sin\phi\]

The inverse conversion is

\[r=\sqrt{x^2+y^2},\qquad \phi=\operatorname{atan2}(y,x)\]

The angle is not unique because adding a whole turn describes the same ray. At \(r=0\), the angle is undefined. Software must choose and document an output range such as

\(-\pi<\phi\le\pi\)

or \(0\le\phi<2\pi\). It must also state whether inputs are radians or degrees.

import numpy as np


def cartesian_to_polar(x, y):
    radius = np.hypot(x, y)
    angle = None if radius < 1e-12 else np.arctan2(y, x)
    return radius, angle


def polar_to_cartesian(radius, angle):
    if radius < 0:
        raise ValueError("Radius must not be negative.")
    return np.array([radius * np.cos(angle), radius * np.sin(angle)])


radius, angle = cartesian_to_polar(3.0, 4.0)
print(radius, angle)
assert np.allclose(polar_to_cartesian(radius, angle), [3.0, 4.0])

Polar coordinates are not just two interchangeable array slots. One component has length units and the other is an angle. Adding two polar coordinate pairs component by component does not perform vector addition. Convert to a common Cartesian basis, add the vectors, and convert back if a polar result is wanted.

Re-expressing a vector in rotated axes

Suppose the new basis is rotated counterclockwise by an angle \(\theta\) relative to the old basis. Its unit vectors are

\[\hat{\mathbf i}'=\cos\theta\,\hat{\mathbf i}+\sin\theta\,\hat{\mathbf j},\qquad \hat{\mathbf j}'=-\sin\theta\,\hat{\mathbf i}+\cos\theta\,\hat{\mathbf j}\]

The physical vector remains fixed. Only its component description changes. Taking dot products with the new unit vectors gives

\[\begin{bmatrix}v'_x\\v'_y\end{bmatrix}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}v_x\\v_y\end{bmatrix}\]

At a positive quarter-turn, a vector originally along the positive horizontal axis has new components \((0,-v)\). The negative sign is reasonable: the old positive horizontal direction lies clockwise from the new positive horizontal axis.

def components_in_rotated_basis(components, basis_angle):
    c, s = np.cos(basis_angle), np.sin(basis_angle)
    return np.array([[c, s], [-s, c]]) @ components


old = np.array([3.0, 4.0])
new = components_in_rotated_basis(old, np.pi / 2)
print("Components in rotated axes:", new)
assert np.allclose(new, [4.0, -3.0])
assert np.isclose(np.linalg.norm(old), np.linalg.norm(new))

The transformation preserves magnitude because the rotation matrix is orthogonal. If the old component column is \(\mathbf v\) and the new one is \(\mathbf v'\), then

\[|\mathbf v'|^2=(\mathbf v')^{T}\mathbf v'=\mathbf v^{T}\mathbf v=|\mathbf v|^2\]

This is both a mathematical result and a strong implementation test.

Passive basis changes and active rotations

There are two different questions:

  1. What components describe the same vector when the coordinate axes rotate?
  2. What components describe a vector that physically rotates while the axes stay fixed?

For the same positive angle, the matrices are transposes of each other. An active counterclockwise rotation in fixed axes is

\[\begin{bmatrix}v_x^{\mathrm{rot}}\\v_y^{\mathrm{rot}}\end{bmatrix}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}v_x\\v_y\end{bmatrix}\]

The passive basis-change matrix shown earlier is its inverse. Both formulas can be correct, so memorizing signs without naming the operation is unreliable. A function name such as rotateVectorInFixedBasis or expressInRotatedBasis is safer than a vague name such as rotate.

A worked basis-change example

Take a displacement with old components

\((4,2)\,\mathrm m\)

and rotate the axes counterclockwise by \(30^\circ\). The new components are

\[v'_x=4\cos30^\circ+2\sin30^\circ\approx4.46\,\mathrm m\]

and

\[v'_y=-4\sin30^\circ+2\cos30^\circ\approx-0.268\,\mathrm m\]

The old magnitude is \(\sqrt{4^2+2^2}=\sqrt{20}\,\mathrm m\). The new magnitude is the same apart from rounding. The negative second component does not mean the displacement physically changed direction; it says the fixed arrow points slightly below the new horizontal axis.

More than a rotation

A general coordinate transformation may include translation, rotation, reflection, scaling, or time-dependent motion. These should not all be called basis rotations. Reflection changes handedness, which reverses signs in cross-product orientation. Scaling an axis changes the metric used to compute lengths unless the representation is handled carefully. In a non-orthonormal basis, the sum of component squares is not generally the physical squared magnitude.

Cylindrical and spherical coordinates also have position-dependent unit vectors. The radial direction at one point is not parallel to the radial direction at another. When differentiating a vector expressed in such coordinates, derivatives of the basis vectors contribute. Simply differentiating the stored component numbers as though the basis were fixed can give a wrong velocity or acceleration.

A coordinate frame that moves relative to another introduces time-dependent effects. Galilean velocity transformations, rotating-frame fictitious forces, and Lorentz transformations are physical frame changes with additional assumptions; they are not covered by the simple static rotation matrix on this page.

Data and validation rules

A reliable coordinate value should identify:

Round-trip tests are especially effective. Convert a set of Cartesian points to polar coordinates and back, avoiding the undefined origin angle, then compare within a stated tolerance. Apply a basis rotation followed by its inverse and recover the original components. Check that rotations preserve vector magnitude and dot products. Test zero, quarter-turn, half-turn, and full-turn angles because their expected signs are easy to predict. Finally, verify that positions respond to an origin translation while free displacements do not.

Coordinates make physical reasoning computable, but they also introduce conventions. Keeping those conventions explicit lets different descriptions be translated without mistaking a change of numbers for a change in nature.

A worked Python notebook

Express the fixed velocity vector (3, 4) in axes rotated by a chosen angle. The question for this calculation is precise: how does first rotated component change when we change basis rotation angle, 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 vector is fixed while its coordinates change; this is a passive basis rotation. 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 -1 to 1 rad. 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 basis rotation angle, expressed in rad. The output is first rotated component, expressed in 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.

\[f(q) = 4 \sin{\left(q \right)} + 3 \cos{\left(q \right)}\]

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 0.3 rad, the expression gives 4.04809 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("3*cos(q)+4*sin(q)", locals={"q": q})
reference_input = 0.3
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "rad")
print("Predicted first rotated component:", reference_output, "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 -1.74498 m/s. The midpoint, at 0 rad, gives 3 m/s. The upper endpoint gives 4.98679 m/s. Read these together with the assumptions: they describe different choices of basis rotation angle 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([-1.0, 0.0, 1.0])
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} rad -> {output_value:.6g} 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 basis rotation angle; the vertical axis shows first rotated component. 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(-1.0, 1.0, 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("Basis rotation angle (rad)")
ax.set_ylabel("First rotated component (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 first rotated component change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.3 rad is 2.93479 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.

\[f'(q) = - 3 \sin{\left(q \right)} + 4 \cos{\left(q \right)}\]

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 = 0.31
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")
print("Full expression:", full_prediction, "m/s")
print("Approximation difference:", full_prediction - linear_prediction, "m/s")

Here the changed input is 0.31 rad. The tangent estimate is 4.07744 m/s, while the full expression gives 4.07724 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 = 2.0
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 basis rotation angle inside the stated interval, predict whether first rotated component 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 vector is fixed while its coordinates change; this is a passive basis rotation. 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: 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(1.0) - evaluate(-1.0))
for points in [101, 201, 401]:
    grid = np.linspace(-1.0, 1.0, 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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