# How are constants derived? [closed]

With little to no background in mathematics and physics respectively, I hope that I might still receive a somewhat elaborate response, but understandable for a layman. I am curious how constants such as Planck’s constant are derived. I can’t imagine that it was measured before it was calculated... so how exactly do these numbers pop up? Is it a result of algebraic manipulations of equations?

Thank you!

## closed as too broad by stafusa, Aaron Stevens, David Z♦Oct 1 '18 at 9:14

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Hi and welcome to the Physics SE! These derivations are specific to each model/equation, which makes this question way too broad. Also, please note that you are expected to have thoroughly searched for an answer before asking your question. And it's important to detail where you're stuck and why, in order to attract good answers. You can consider checking the advice on writing good questions. – stafusa Oct 1 '18 at 0:36

$$E = \hbar \omega$$
$$p = \hbar k$$
Here, $$E$$ represents energy, $$p$$ represents momentum, $$\omega$$ represents angular frequency, and $$k$$ represents wavenumber. So, if you can take a photon (particle of light) and measure its angular frequency $$\omega$$ (i.e. - its color) and then also measure its energy $$E$$, you can compute Planck's reduced constant $$\hbar = E / \omega$$.
These deBrogile relationships were stated after scientists noticed that there was a linear relationship between $$E$$ and $$\omega$$ as well as $$p$$ and $$k$$ for elementary particles (and incidentally,their constant of proportionality $$\hbar$$ ends up being the same for both relationships--this is not obvious--only can be determined by experiment!). Once you have the equations, you can then measure the value of $$\hbar$$.