Dear Jamil Kooli .

Both Planck constant h and Dirac constant ћ are values ​​with the dimension of angular moment and differ exactly by 2π. Therefore, the result of their division is also equal to 2π, and there is nothing to be surprised.

Dulin Mikhail.

Planck Constant and Reduced Planck Constant

The reduced Planck constant is defined as:

\hbar = \frac{h}{2\pi

So naturally, the ratio:

\frac{h}{\hbar} = 2\pi

is always true by definition. This isn’t a physical law, it’s a mathematical identity—it arises from the convenience of including 2\pi in circular/angular quantum mechanics (e.g., wave functions, spin, Fourier transforms). So the ratio itself doesn’t carry physical meaning; it just simplifies expressions in quantum field theory and related disciplines.

Relating h to Natural Constants G, c, and the Cosmological Constant \Lambda

This brings us into the realm of natural units and dimensional analysis, especially Planck units, which are derived from h, G, and c. For example:

• Planck length: \ell_p = \sqrt{\frac{\hbar G}{c^3}}

• Planck time: t_p = \sqrt{\frac{\hbar G}{c^5}}

• Planck mass: m_p = \sqrt{\frac{\hbar c}{G}}

All of these use \hbar, not h.

Now, when you ask:

“If \frac{h}{\hbar} = 2\pi always, how can we relate h to G and c without involving \pi?”

You’re right that \pi enters through the definition of \hbar, and it does appear in natural units. But that’s not necessarily embarrassing—it’s a reflection of the fact that our universe has rotational symmetry, and \pi is fundamental in the geometry of such systems. Circular or spherical geometries (as in wave functions, black hole horizons, and spacetime) inevitably bring in \pi.