View Full Version : "Casi.Voy equation"
In Summer 2010 I was travelling through southern Poland and Slovakia.
In the attic of some old house, I found a pile of papers filled with
notes and dated 1945. They are hardly readable - the Author used
almost only abbreviations in Latin, Cyrillic and Hebrew alphabets.
There is a signature: "Casi.Voy" (with cyrillic). There are a few
equations, and this was the only one I was able to fully decypher:
1/(4πε_0 ) e^2/r^2 =G (((2cℏG^(-1))^1/2 )/(4/3 π (√(2.5))^3 ))^2/r^2
would anyone try to interpret it?
FLUndergrad
Jan11-12, 10:34 AM
Well I'm not a Phd or anything but it looks like its got too many arbitrary constants in it to be anything earth-shattering. Just my 2 cents though Im still an undergrad physics major.
BishopUser
Jan11-12, 02:35 PM
Maybe I am wrong, but it looks similar to coulomb's law http://en.wikipedia.org/wiki/Coulomb%27s_law
1. The way you formatted the equation it is completely unreadable, please try to use LaTeX (as described here (http://www.physicsforums.com/showthread.php?t=546968)).
2. There are not many arbitrary constants in the equation - G, c, e, \hbar, \epsilon_0 and \pi have a rather obvious meaning. Only 2.5 looks strange (and apparently is a part of a formula for sphere volume).
jhae2.718
Jan12-12, 06:00 PM
Here is an attempt to TeXify the equation:
\frac{1}{4 \pi \varepsilon_0 } \frac{e^2}{r^2} = G\frac{\left(\frac{\sqrt{\frac{2 c \hbar}{G}}} {\frac{4}{3}\pi (2.5)^{\frac{3}{2}}} \right)^2 } {r^2}
r2 cancels out.
I can't read pdf version, I get some error message.
Here's the correct version as in the original. Thanks for TeXifying jhae2.718!
\frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2} = G \frac{ (\frac{\sqrt{2 c \hbar G^{-1}}} {\frac{4}{3} \pi (\sqrt{2.5})^3 })^2}{r^2}
The meaning of left hand side is quite obvious as BishopUser noted - it's Coulomb's interaction between two electrons. The right hand side is rather strange - it seems a bit like gravitational interaction between masses of some kind.
At least looks like units are consistent.
Note that r2 and G cancel out.
collinsmark
Jan13-12, 09:02 PM
\frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2} = G \frac{ \left(\frac{\sqrt{2 c \hbar G^{-1}}} {\frac{4}{3} \pi \left(\sqrt{2.5} \right)^3 } \right)^2}{r^2}
Hmmm. Yes, as Borek points out it is dimensionally correct.
The \sqrt{2 c \hbar G^{-1}} has units of mass, as expected. But mass of what? Notice there are no variables in the term. Everything is a constant. And the term contains both the gravitational constant and Planck's constant? weird.
The \frac{4}{3} \pi \left(\sqrt{2.5} \right)^3 seems to be a ratio of volumes of some kind -- Like the ratio of the volume of a sphere of radius r per volume of a cube with a length of \frac{r}{\sqrt{2.5}} . Or instead of a cube, maybe the 2.5 has something to do with the shape of some sort of pyramid or something -- something with flat faces so things like pi don't come in.
Or perhaps, the \frac{4}{3} \pi \left(\sqrt{2.5} \right)^3 might be a ratio of densities, such as the density of some shape with straight faces per density of a sphere containing the same mass.
I can only speculate.
The form of this equation suggests, that author was thinking about connections between electromagnetism (or at least electrostatics) and gravitation. As long as I understand, we don't have any "standard and official" theory combining those kinds of interactions, but we have a number of more or less interesting (or even crazy) attempts to create such - many of them "homemade".
I find this particular equation so interesting, not only because I found it written among old, notes of unknown origin, but because it connects electromagnetism with gravitation in a quantitative manner (gives a "mass equivalent" of a unit charge).
I really wonder if there's something to it.
Vanadium 50
Jan22-12, 03:04 PM
Mulkay, it sounds like you are trying to convince us that they're is something to it.
There is not.
First, this is just numerology - attempting to relate different numbers together. There is no "something" yet for there to be "to it". That would require some sort of explanation as to why these numbers are related.
Second, the person who wrote this doesn't know the most elementary physics, or they would have noticed that many of the terms they wrote down cancelled.
Finally, when you complete the path of cancellations, and replace the ε and c with μ (which has the numeric value of 4π x 10), you will be left with an hbar, and a pile of algebra involving numbers. I can always take a single number and represent it as a pile of algebra.
Sorry.
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