Is there a reason why

$$\dfrac{h}{ec}\sim 1.38\cdot 10^{-23} V/(m\cdot s^{-2})$$


Dimensionally, the Unruh effect temperature reads

$$T_U=\dfrac{\hbar a}{2\pi k_B c}$$

from which the acceleration prefactor is

$$\left[\dfrac{ET}{E/t\times LT^{-1}}\right]=t\times \dfrac{a}{LT^{-2}}$$ Is an accident the above quantity mimic the value of almost the Boltzmann constant?

Take now the equation:


and $$q=\dfrac{h}{k_Bc}$$


$$\dfrac{ET}{(E/t)\times L/T}=\dfrac{t}{LT^{-2}} $$

  • 2
    $\begingroup$ Which values mimic? You've got one number and one dimensional analysis. $\endgroup$ – Brick Aug 6 '19 at 15:11
  • $\begingroup$ Notice that the units in the denominator of your first line look like miliseconds (if we ignore that fact that you have set them in italics)? Use a thing space or a dot to set them apart: m \, s gets $m \, s$ or m \cdot s gets $m \cdot s$ And while you're fixing that set them in an unright font (I like to use \mathrm for that): $\mathrm{m \cdot s}$. $\endgroup$ – dmckee --- ex-moderator kitten Aug 6 '19 at 15:28
  • $\begingroup$ Does the dimensional analysis of the Unruh effect have something to do with the numerical coincidence? $\endgroup$ – G. Smith Aug 6 '19 at 15:58

Yes, it's a coincidence. The units on the two expressions don't match, which means the "numerical agreement" between the two numbers is definitely artificial.

Boltzmann's constant is given in SI units by

$$k_B = 1.3806 \times 10^{-23} \frac{\text{J}}{\text{K}}.$$

The ratio $h / ec$ is given in SI units by

$$\frac{h}{ec} = 1.3795 \times 10^{-23} \frac{\text{kg}\, \text{m}}{\text{C}}.$$

The second expression has a unit of charge in it, while the first does not. We are free to rescale our definition of a "Coulomb" by any number we like, and would obtain a drastically different numerical value for $h / ec$ in our chosen units without changing the numerical value of $k_B.$ In general, if two quantities don't have the same units, then there is no meaningful way of comparing their numerical values, since those values are highly dependent on the choice of units.


It may be premature to call this a numerical coincidence.

In the near future, it may be possible to define the independent units which are in question in terms of each other, thus showing that this may indeed be a valid expression.

  • 2
    $\begingroup$ I have never heard of this system of units. I am highly skeptical about this entire post. $\endgroup$ – Dale Sep 9 '19 at 2:35

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