Talk:Atomic units

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Atomic units related to magnetic quantities

" In atomic units, the Bohr magneton ${\displaystyle \mu _{B}=1/2}$, ..."

However, as described in Bohr magneton page, ${\displaystyle \mu _{B}}$ is expressed in S.I. units as

${\displaystyle \mu _{B}={{e\hbar } \over {2m_{e}}}}$

and in Gaussian centimeter-gram-second units as

${\displaystyle \mu _{B}={{e\hbar } \over {2m_{e}c}}}$.

If ${\displaystyle e}$, ${\displaystyle \hbar }$, and ${\displaystyle m_{e}}$ are set equal to unity in each of these expressions, the Bohr magneton is

${\displaystyle \mu _{B}=1/2}$

in, S.I.-a.u., say, and

${\displaystyle \mu _{B}=1/(2c)}$

in, Gaussian-a.u., say, and they have different values.

By virtue of the equality

${\displaystyle 4\pi \epsilon _{0}=1}$,

electric quantities have the same values in both S.I.-a.u. and Gauss-a.u.. For the magnetic quantities like ${\displaystyle \mu _{B}}$, however, this does not seem to be the case, and there seems to be a freedom or ambiguity which of S.I. or Gauss to be chosen.

Now, my question is, when a.u. is referred to, does it imply what I wrote as S.I.-a.u., as can be read from the part of article cited above? Is there a consensus, or a rule?

NorioTakemoto 15:14, 15 February 2006 (UTC)

A nice little problem in undergraduate physics! Your (and now my) perplexity purely concerns the cgs formula for μB, and I gather that cgs units are only of historical interest nowadays. That pesky little c... tsk-tsk. I will think about this.202.36.179.65 19:09, 9 April 2006 (UTC)
NIST defines the magnetic dipole atomic unit as 2 ${\displaystyle \mu _{B}}$. See http://physics.nist.gov/cgi-bin/cuu/Value?aumdm%7Csearch_for=atomic+unit However, then one would have to introduce another factor of c in the Maxwell equations if I am right - otherwise atomic dipoles would not have the same field in a.u. and in the SI system ( prefactor is ${\displaystyle {\frac {\mu _{0}}{4\pi }}{\frac {m}{r^{3}}}}$ for SI and ${\displaystyle {\frac {1}{c}}{\frac {m}{r^{3}}}}$ according to the formulation in the article. But if you actually calculate some numbers (with the a.u. for m as defined by NIST) this is off by a factor of c. If one calculates the field of a infinite straight wire (to get rid of the amiguity in the definition of the magnetic moment), we get ${\displaystyle B={\frac {\mu _{0}I}{2\pi r}}}$ in SI and would get ${\displaystyle B={\frac {2I}{rc}}}$ with the formulation as in the article. If we run some numbers (with the a.u. of current again defined the same way as NIST http://physics.nist.gov/cgi-bin/cuu/Value?aucur%7Csearch_for=atomic+unit ), it is also off by a factor of 1/c. So either one needs to define current and magnetic dipole moment with another factor of c or write Maxwell's equation with >math>c^{-2}[/itex] See also:http://en.wikipedia.org/wiki/Cgs_units#Various_extensions_of_the_CGS_system_to_electromagnetism http://en.wikipedia.org/wiki/Cgs_units#Electromagnetic_units_in_various_CGS_systems The article (and NIST) as of the unit definitions uses the a.u. equivalent to ESU-CGS, and therefore the prefactors to Maxwell's equations need to be ${\displaystyle 4\pi }$ (correct in article), ${\displaystyle -1}$ (incorrect in the article) and ${\displaystyle 4\pi c^{-2}=4\pi \alpha ^{2}}$ (also incorrect in the article). I will edit the article in a minute. 128.200.93.197 (talk) 18:47, 24 August 2009 (UTC)

Ultimately the problem is that under Gaussian conventions you get a magnetic field from the combination

${\displaystyle {\frac {e}{a_{0}^{2}}}=1.72\times 10^{7}}$ gauss

and under SI conventions you get a magnetic field from the combination

${\displaystyle {\frac {\hbar }{ea_{0}^{2}}}=2.35\times 10^{9}}$ gauss

So there's an ambiguity when you say "the magnetic field is 1 in atomic units", it could be 1.72E7 gauss or 2.35E9 gauss. I conclude that there are definitely two different "atomic units"s based on the definition in the article. So the question is, are both actually in use? Or only one? Can we find any sources that use one or the other, or that say explicitly that it's ambiguous? --Steve (talk) 15:53, 18 December 2009 (UTC) UPDATE: I checked, both are in use. Added this to the article. --Steve (talk) 23:34, 25 December 2009 (UTC)

The derived au table needs some attention

i might get to it myself at a later time. it would be good if au, Planck, Stoney, etc and all "natural units" were tied together into articles of consistent format. r b-j 02:36, 16 May 2006 (UTC)

I spent some time cleaning up both tables in the article. It didn't make sense to have Boltzmann and gravitational constants in the table. They were defined very differently from the other dimensional scales. Wigie 14:18, 18 May 2006 (UTC)

The Fundamental atomic units table contradicts BIPM

" ... Similarly, in the a.u. system, any four of the five quantities charge, mass, action, length, and energy are taken as base quantities. The corresponding base units are the elementary charge e, electron mass me, action , Bohr radius (or bohr) a0, and Hartree energy (or hartree) Eh, respectively. In this system, time is again a derived quantity and the a.u. of time a derived unit, equal to the combination of units /Eh. Note that a0 = alpha/(4piRinfinity), where alpha is the fine-structure constant and Rinfinity is the Rydberg constant; and Eh = e2/(4piepsilon0a0) = 2Rhc0 = alpha2mec02, where epsilon0 is the electric constant and has an exact value in the SI. ..."

that is, energy is a base unit of the system. However for some reason, the article insists that the "electric constant" is a base unit, why is this? isn't it better to agree with BIPM. The article doesn't give any source for this choice.

--Alfredo.correa (talk) 06:57, 6 May 2010 (UTC)Alfredo

The units are all interrelated, because of equations like:
${\displaystyle {\frac {e^{2}}{4\pi \epsilon _{0}a_{0}}}=E_{h}}$
You define four of the units to be "1" and then prove that all the other units are "1". But it doesn't matter which ones are "fundamental" (or "base") and which ones are "derived", it just matters that they're all consistent. BIPM makes a different choice than this article does, but it's a pointless distinction anyway. The article doesn't make this very clear right now... --Steve (talk) 02:08, 11 May 2010 (UTC)

Comparison with Planck units

I made some changes here, but I'm still not happy with it. It seems that this section doesn't make its point very concisely, and in the process "hides" some interesting au values like the speed of light and the Bohr magneton. Wigie 14:23, 18 May 2006 (UTC)

Boltzmann's constant

The article says:

Finally, au normalize a unit of atomic energy to 1, while Planck units normalize to 1 Boltzmann's constant k, which relates energy and temperature.

However, if I understand the table of derived units correctly, atomic units also normalize Boltzmann's constant to 1. Am I missing something? Henning Makholm 15:01, 6 July 2006 (UTC)

Hi, same problem for me. Can someone give a reference to an external document where it is shown how the atomic unit for temperature is derived? As far as I can see, one has to define (not derive) ${\displaystyle {\frac {E_{h}}{k_{B}}}=1a.u.}$. I'm not convinced that this is a commonly used definition. Tovrstra 14:48, 19 March 2007 (UTC)

Electric constant

It might be worth mentioning that, in SI units, the electric constant is an exactly defined (not approximately) unit. Replacing the common definition ${\displaystyle {\frac {1}{4\pi \epsilon _{0}}}}$ with the equivalent ${\displaystyle {\frac {\mu _{0}c^{2}}{4\pi }}}$ simplifies to ${\displaystyle 10^{-}7c^{2}}$ exactly, or 8987551787.3681764 kg m^3 s^-2 C^-2.
--65.202.227.50 (talk) 15:46, 18 February 2010 (UTC)mjd

The exact definition of ε0 is important if you're talking about how SI works, but if you're talking about atomic units I don't think it matters too much. The expression ${\displaystyle {\frac {1}{4\pi \epsilon _{0}}}}$ is very common and familiar, it would be very confusing to people if we stopped using that expression and instead used the uncommon expression ${\displaystyle {\frac {\mu _{0}c^{2}}{4\pi }}}$. Since they're the same thing anyway we should use the more common and easy-to-recognize expression I think. :-) --Steve (talk) 00:10, 19 February 2010 (UTC)