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This tag is for questions relating to physical constants which are any of a set of fundamental invariant quantities observed in nature and appearing in the basic theoretical equations of physics. Accurate evaluation of these constants is essential in order to check the correctness of the theories and to allow useful applications to be made on the basis of those theories.

1 vote

Why are the anthropometric units (which are about as big as we are) as large as they are rel...

Every dimensionful constant is a comparison to some standard. The oldest standards were based on the sizes of a typical person, because most full-grown people are more or less the same size and you al …
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2 votes
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How can vacuum have a permittivity?

The constant $\epsilon_0$ almost never appears in an expression where it's added to another term; it's multiplied instead. Contrast with electric charges, which combine by addition, and for which the …
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5 votes
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Is Relative Permeability $\mu_0$ still $4\pi \times 10^{-7} $H/m?

The organization which defines the SI is the Bureau Internationale de Poids et Mesures (BIPM). The current version of the SI brochure contains the definitive answer to your question (page 132): The p …
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0 votes

Is there a unit system, where all physical laws looks nice?

When you're doing Fermi estimation, it's convenient to use a "system of units" where $$ \sqrt{10} = \pi = 3 = 1 = -1 = c $$ though this is probably more tongue-in-cheek than you were hoping for. For s …
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3 votes
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Running Constant Values At Very Low Temperatures

The fine structure constant $\alpha\approx\frac1{137}$ appears in the Coulomb force between fundamental charges: $$ \alpha\hbar c = e^2/4\pi\epsilon_0, \quad\text{so}\quad |E_\text{Coulomb}| = \frac{e …
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4 votes

Redefinition of everything on May 20th, 2019

The new value of $\hbar$ will be consistent with the current CODATA value; it's at the BIPM website. Same for $e$. From here: the vacuum magnetic permeability $\mu_0$ is equal to $4\pi × 10^{–7}\ …
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0 votes

How to calculate the Avogadro constant from the Loschmidt constant and ideal gas law?

The constant $R$ is the molar gas constant. In your formula, $n$ is the amount of gas measured in moles, and $n/V$ is the molar density. To convert between the number of moles and the number of part …
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6 votes
Accepted

What exactly is Planck's constant? how did they calculate it?

The equations that govern quantum mechanics predict that the angular momentum (that is, spinning or orbiting) in a system can't take on any value, but instead come in lumps. The "reduced Planck const …
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2 votes

Measurement of the gravitational constant $G$ in free fall

Two platinum spheres in free space might orbit each other in six hours, but near the Earth they would orbit Earth every 1.5 hours. In practice you would be measuring their orbits around the Earth, wh …
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0 votes

How accurate are constants in cgs units?

benrg makes an excellent point that frequently combinations of constants are known better precision than individual constants. We also have the fact that both the cgs and SI units for electromagnetism …
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5 votes
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Why does the ladder operator contain the $\hbar ω$ part?

Your product $a^\dagger a = \frac{H}{\hbar\omega} + \frac12$ is dimensionless, because your $a$ and $a^\dagger$ are dimensionless. If the $ \newcommand{\ket}[1]{\left|#1\right>} \ket{\psi_n} $ are the …
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5 votes
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$G$ expressed as a function of $c$?

You cannot. The constants $G$ and $c$ have different units, so $G$ cannot be equal to $c$ times some dimensionless constant as in your question. From a metrology perspective, $c$ defines how we relat …
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0 votes

Angular momentum dimensions

The classical angular momentum is $$\vec L = \vec r \times \vec p$$ which has dimension $$ \rm [m] \times [kg\ m\ s^{-1}] = [J\,s] $$ The quantum-mechanical angular momentum operator has the same unit …
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67 votes
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Physicists adding 3 decimals to the fine structure constant is a big accomplishment. Why?

The fine structure constant tells us the strength of the electromagnetic interaction. There are some misleading statements in the news story. The big one is how to read the result, \begin{align} \alp …
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36 votes
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Why isn't it $E \approx 27.642 \times mc^2$?

It's a side effect of the unreasonable effectiveness of mathematics. You are in good company thinking it is a little strange. Many quantities in physics can be related to each other by a few lines o …
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