How does Wien's Scaling Law unify experimental data in blackbody radiation?

  • Thread starter Thread starter Narcol2000
  • Start date Start date
  • Tags Tags
    Law Scaling
Narcol2000
Messages
25
Reaction score
0
How does Wien's scaling law

<br /> \frac{u(\lambda)}{T^5} = \frac{f(\lambda T)}{\lambda^5T^5}<br />

imply that if \frac{u(\lambda)}{T^5} is plotted as a function of \lambda T, all experimental data will lie on a single curve?
 
Physics news on Phys.org
I don't know anything about the physics here, but it seems obvious to me that if you plot y = u(\lambda)/T^5 versus x = \lambda T then Wien's scaling law tells you that y = f(x) / x^5. So if you take a data point (x', y') and Wiens' law is true, then y' should be (within error) f(x')/(x')^5.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top