Stefan-Boltzmann Law & Blackbody Curve Integration

Mathematical Proof & Spectral Breakdown

• 1. Mathematical Link Between Planck & Stefan–Boltzmann: Integrating Planck's equation over all wavelengths from $0$ to $\infty$ mathematically evaluates to: $$\int_0^\infty E_\lambda(\lambda, T) \, d\lambda = \int_0^\infty \frac{2\pi h c^2}{\lambda^5 \left(e^{\frac{hc}{\lambda k_B T}} - 1\right)} d\lambda = \frac{2\pi^5 k_B^4}{15 c^2 h^3} T^4 = \sigma T^4$$ The shaded area under the Planck curve is physically equal to the total power emitted by the Stefan–Boltzmann law!
• 2. Why the Bulb (2800 K) Curve Looked "Flat" Without Auto-Zoom: Because power scales with $T^4$, the peak emission at $6000\text{ K}$ is over $21\times$ taller than at $2800\text{ K}$[span_0](start_span)[span_0](end_span). On a fixed global axis, $2800\text{ K}$ is compressed near the bottom[span_1](start_span)[span_1](end_span). Clicking Auto-Zoom automatically rescales the vertical axis to the exact peak at any temperature so you can inspect its shape, peak location ($\lambda_{\text{max}} = 1035\text{ nm}$), and the rainbow overlap[span_2](start_span)[span_2](end_span)!
• 3. Where the Light Actually Goes at 2800 K: As calculated in real time above, only ~9% of the total area falls inside the visible rainbow window ($380\text{ nm} - 750\text{ nm}$)[span_3](start_span)[span_3](end_span). Over 90% is radiated into the invisible infrared as heat[span_4](start_span)[span_4](end_span).