Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am just curious that if constants in cgs units change the answer of an equation. For example, Coulomb's constant, in SI units it equals to $8.98...\times 10^9 \,\mathrm{N\,m^2\,C^{-2}}$. However in cgs units it equals to 1. I think the difference between an answer calculated with Coulomb's constant in SI units and Coulomb's constant in cgs units would be a lot. I don't know if my logic is right.

share|cite|improve this question
$k_e$ is absorbed into the units for $q$. – Kyle Kanos Aug 14 '14 at 20:27
Coulomb's constant is an exact defined value in both SI and Gaussian-cgs units. – Stan Liou Aug 14 '14 at 20:49
@StanLiou I mean let's use Coulomb's law. Let's say there are two point charges which is 1 C and 1 cm apart from each other. If we calculate the force in cgs units $k_e\bullet \frac {1 C\times 1 C} {1^2 cm^2} = 1 N$ When we use SI units $k_e\bullet \frac {1 C\times 1 C} {0.01^2 m^2} = 8.98\times 10^{15} N$ There is a huge difference between these answers right? Can you tell me where I am wrong? – FarStar Aug 14 '14 at 21:02
Neither the Newton nor the Coulomb are units in the version of the cgs system where k= 1... – DJohnM Aug 14 '14 at 21:05
@User58220 Sorry I couldn't find the transformed units for N and C. Could you tell me what the correct ones are? – FarStar Aug 14 '14 at 21:08

I think there's a genuine and interesting physical point to be made here.

Taking a slightly different example, the gravitational acceleration of a massive body on a test particle is $a = GM/r^2$. If you can measure $a$ and $r$ accurately then you can find $GM$ to equal accuracy. But to find $M$ you also need to know $G$, and $G$ is rather difficult to measure. So it's entirely possible in principle to know $GM$ for an astronomical body with better accuracy than $M$, which would make $GM$ a more useful description of the object's mass than $M$, and might make the mass unit in units with $G=1$ more useful than the SI or cgs mass unit. I don't know whether there was any historical era where this was actually the case for any astronomical body, though.

More generally, the measurability/reproducibility of the base quantities of a unit system affects the maximum accuracy of other quantities stated in those units, so some unit systems are actually better than others.

(Edit: according to Wikipedia, "For several objects in the solar system, the value of $\mu$ [= $GM$] is known to greater accuracy than either $G$ or $M$.")

share|cite|improve this answer
The Gaussian gravitational constant is $$k\equiv\sqrt{G} = 0.01720209895\,\mathrm{AU}^{3/2}\,\mathrm{day}^{-1}\,\mathrm{M}_\odot^{-1/2}$$ which has been introduced with this value in 1809 and has a much higher precision than either $G$ or the solar mass alone. Also, the quantity $\mu = GM$ of a body is usually called the standard gravitational parameter. – Stan Liou Aug 15 '14 at 0:02
I would think that for most major solar system bodies, $GM$ is known far better than $G$ or $M$, and this isn't going to change any time soon. The only non-$GM$ ways of getting mass I can think of are compositional inference and shoving the thing with a known force. – Chris White Aug 15 '14 at 0:15

Take a specific example: two charges of 1 coulomb separated by 1 meter.

In MKSA, (now better known as SI) the force between them is given in Newtons, by:$$F=\frac{k q_1q_2}{r^2}=8.98\times10^9\text{ Newtons}$$since all the variables are 1.

So now you want to do the same problem in cgs-electrostatic units. $k=1$, $r=100$, and most importantly, $q_1=q_2=2.997925\times 10^9 \text{ stat-coulombs}$, the value of one coulomb in cgs-esu units.

So, the force equation becomes::$$F=\frac{k q_1q_2}{r^2}=\frac{1\times (2.997925\times 10^9)^2}{100^2}=8.98\times10^{14}\text{ dynes}$$which is the same as the previous result in Newtons...

share|cite|improve this answer

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 are, to some extent, historical accidents that predate the modern understanding of the theory. The "natural" units for electromagnetism take advantage of the dimensionless fine-structure constant $\alpha$, defined by $$ \frac{e^2}{4\pi\epsilon_0} = \alpha\hbar c $$ Consider that

  • $c$ and $\epsilon_0$ are defined constants, with zero uncertainty
  • charge is quantized, in integer multiples of $e$
  • angular momentum is quantized, in units of $\hbar$

If you're interested in precision, then, instead of cgs or mks units you should use $e\hbar c$ units where many of your quantities of interest are exact integers. Note that the error on $\hbar$ doubles if you insist on using joule-seconds rather than MeV-seconds. If you can choose not to infect your problem with macroscopic units, the uncertainty on the dimensionless $\alpha$ is nearly a hundred times smaller than the uncertainty on the joule-second value for $\hbar$.

However between CGS and MKS there is no difference in precision, nor in the accuracy of the predicted dynamics, since the two systems also use different units for charge and for force.

share|cite|improve this answer

The set of constants currently maintained by CODATA goes back as far as 1929 in varying degrees. These in general have been improving in precision, because this is something of metrological importance.

Measures like $\hbar$ and $e$ are generally derived, there are an assortment of other points that are generally more exact than these. What the CODATA table reflects, is that the errors are actually pre-linked, so something like $\hbar / m$ is more exact than $\hbar$ or $m$. It's the other way, really: $\hbar = m \cdot \hbar/m$.

One can deconstruct the CODATA tables, especially for the electron, into something like C.L.M.T.Q.Þ, where C is a hundred-like number 137.036, and Þ the temperature unit. Units derived from these base units (rydberg length, electron mass, speed of light, electron charge), are more exact than values like $\hbar$, and even though where the rydberg constant is $4pi$ L, the bohr orbit is L/C, and the classical electron radius is L/C³, is more exact than using $e$ and $m$ to derive these.

The older tables are in CGS, then the current system. The transition to SI occured after 1947, but the bulk of the conversions in 1960s or so. So data expressed in SI units are from fresher data, and that is what makes them seem more accurate than the CGS.

CGS and SI use different formulae. You can construct a common theory by supposing that where $S=U=1$ in SI, and $S=4\pi$, $U=c$ in CGS. Note that c has dimensions of velocity, and appears when both electric and magnetic quantities appear in the same equation.

$S$ does not appear in the CODATA tables, but it does appear when converting electric flux from cgs to SI, since the correct dimensions here is $QS$.

$U$ makes some appearence, since one sees in older tables $e/c$ as a magnetic quantity, comes as $e/U$. The equation $\epsilon\mu c^2 U^2 = 1$ is the correct form here. Since $U=c$ in cgs, it is a constant which has units, dimensions, experimental values and error, so a value known exactly in esu is not exact in emu.

A set of tables from ancient times in CGS units, might be updated using the most exact values of the 2010 CODATA, the values for the missing constants calculated, is as every bit as exact as the current SI data.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.