How to Fix 'Covariance Could Not Be Estimated' in SciPy curve_fit for a Boltzmann-Type Model: Units, Scaling, and Parentheses

Step-by-step fix for SciPy curve_fit on a Boltzmann-type model: stop covariance errors, add amplitude, use correct units (eV/K), and get stable T estimates.

Fitting non-linear models to experimental curves is routine, but a few subtle pitfalls can make optimizers stall or return meaningless parameters. Here is a compact walkthrough of a real-world case: a one-parameter fit based on a Boltzmann-type expression that triggers “covariance could not be estimated”, converges to T=1, and produces a misleading plot. We will unpack what goes wrong and how to fix it without changing the intended physics.

Problem setup

The target model follows an expression proportional to B(2J+1)/(kB·T)·exp(−B·J(J+1)/(kB·T)). The independent variable is J, the unknown parameter is T, and kB and B are constants. A direct attempt to fit with SciPy might look like this:

# Constants
k_b_const = 1.380649e-23
beta_coeff = 0.3808

# Fit function
def fit_fun(j_var, t_param):
    return (beta_coeff * (2 * j_var + 1) / k_b_const * t_param *
            np.exp(-beta_coeff * j_var * (j_var + 1) / k_b_const * t_param))

p_opt, p_cov = optimize.curve_fit(fit_fun, x_vals, y_vals)
print(p_opt)

plt.plot(x_vals, y_vals, "o")
plt.plot(x_vals, fit_fun(x_vals, p_opt[0]))

The optimizer responds with an error about the parameter covariance and the single parameter becomes 1:

OptimizeWarning: Covariance of the parameters could not be estimated

Here is the dataset used for the fit:

x_vals = [2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48]
y_vals = [0.185,0.375,0.548,0.695,0.849,0.931,0.996,0.992,1.0,0.977,0.927,0.834,0.691,0.68,0.575,0.479,0.421,0.351,0.259,0.208,0.162,0.135,0.093,0.066]

What actually goes wrong

The first issue is operator precedence. Writing a/kb*T is not the same as a/(kb*T). In the original form, both the prefactor and the exponential effectively multiply by T rather than divide by (kB·T). In this particular structure that inversion makes the optimizer learn 1/T instead of T, which alone doesn’t explain the covariance failure but does distort the parameterization and aggravates instability.

The second issue is the normalization of the target vector. The provided y-values form a distribution normalized to unity area. Fitting a model with a hard-coded amplitude equal to one prevents the curve from adjusting to the observed scale and blocks convergence. Introducing an explicit scale parameter N resolves this by letting the model match the overall level while T shapes the distribution.

The third issue is unit consistency and scaling. The Boltzmann constant is given in J/K, while the units of B are not stated. The exponent must be dimensionless. Using kB in eV/K instead of J/K provides a more appropriate numerical scale in this context and yields stable, credible results for T.

There are also two practical gotchas. Always parenthesize denominators such as (kB*T) to avoid precedence pitfalls. And when plotting the fitted curve, pass the same x-array you used for fitting; referencing an undefined or different j variable leads to mismatched or misleading plots.

Fix, step by step, and a working example

The remedy is minimal: add parentheses around kB*T, include a free amplitude N, and switch kB to eV/K. The rest of the pipeline remains the same. Below is a complete, self-contained example that fits the provided data and validates the result.

import numpy as np
import matplotlib.pyplot as plt
from scipy import optimize, integrate
from sklearn import metrics

# kB in eV/K for consistent scaling
kB_ev = 8.617333262e-5
B_const = 0.3808

def shape_fn(j_idx, temp, amp):
    return amp * (
        B_const * (2 * j_idx + 1) / (kB_ev * temp) *
        np.exp(- B_const * j_idx * (j_idx + 1) / (kB_ev * temp))
    )

j_data = np.array([2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48])
y_data = np.array([0.185,0.375,0.548,0.695,0.849,0.931,0.996,0.992,1.0,0.977,0.927,0.834,0.691,0.68,0.575,0.479,0.421,0.351,0.259,0.208,0.162,0.135,0.093,0.066])

# Initial guess [T, N]
params_hat, cov_hat = optimize.curve_fit(shape_fn, j_data, y_data, p0=[1e5, 1])

# Uncertainties from covariance
stderr = np.sqrt(np.diag(cov_hat))

# Predictions and R^2
pred = shape_fn(j_data, *params_hat)
r2 = metrics.r2_score(y_data, pred)

# Smooth curve for visualization
j_grid = np.linspace(j_data.min(), j_data.max(), 200)
y_grid = shape_fn(j_grid, *params_hat)

fig, ax = plt.subplots()
ax.scatter(j_data, y_data)
ax.plot(j_grid, y_grid)
ax.set_xlabel("J'')")
ax.set_ylabel("N(J'')")
ax.grid()

print("Estimated [T, N]:", params_hat)
print("Std errors        :", stderr)
print("R2                :", r2)

With this setup, the regression converges cleanly. Reported values are approximately T = 2.51e6 ± 0.03e6 and N = 2.76e1 ± 0.02e1, and the fit quality is strong with R² ≈ 0.994.

If you instead use kB in J/K, the inferred temperature becomes on the order of 1.6e+25, which is numerically unwieldy here. The reason is strictly about units and scaling; the exponent must stay dimensionless.

Since the data represent a normalized distribution, you can also verify that the fitted curve integrates to unity after removing the amplitude factor. This check can be done numerically:

j_check = np.linspace(0, 100, 2000)
y_check = shape_fn(j_check, *params_hat)
area = integrate.trapezoid(y_check / params_hat[-1], j_check)
print("Area under curve (normalized):", area)

The area is effectively one, which is consistent with the interpretation of y as a unit-area distribution.

Why this matters

Non-linear least squares is sensitive to formulation. A missing pair of parentheses silently changes how the parameter enters the model. A distribution normalized to unit area forces an additional free scale if you want the optimizer to adjust both height and shape. And ignoring units can inflate or deflate parameters by many orders of magnitude, confusing both the solver and anyone reading your results.

Practical takeaways

Parenthesize denominators like (kB*T) in both the prefactor and the exponent. Include a free amplitude N when fitting distributions whose area is fixed by construction. Keep the exponent dimensionless by aligning units; using kB in eV/K is a pragmatic choice in this case and leads to stable estimates. Finally, be consistent when plotting: evaluate the fitted function on the same x-array used for fitting to avoid mismatches.

Adhering to these small but crucial details turns an intractable fit with a singular covariance into a well-behaved regression with interpretable parameters and a high-quality match to the data.