SI and Gaussian unit conversion in electrodynamics

jaobyccdee
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Homework Statement


I am reading about electrodynamics, and in different formula, there is a SI form and a Gaussian form.


Homework Equations


What is the relationship in the different unit system for c or B or E and other relative variables?



The Attempt at a Solution


sometimes there's a difference of 4(pie)/c, sometimes there's only 1/c, but i don't see when is which and which.
 
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hi jaobyccdee! :smile:

there's a good overview (but unfortunately no table) in http://en.wikipedia.org/wiki/Gaussian_units#Major_differences_between_Gaussian_and_SI_units

in particular …
One difference between Gaussian and SI units is in the factors of 4π in various formulas. SI is called "rationalized", because Maxwell's equations have no explicit factors of 4π in the formulas. On the other hand, the force laws, Coulomb's law and the Biot–Savart law, do have factors of 4π in them.

In Gaussian units, which are not "rationalized", the situation is reversed: Two of Maxwell's equations have factors of 4π in the formulas, while both of the force laws, Coulomb's law and the Biot–Savart law, have no factors of 4π.​
and
In Gaussian units, unlike SI units, the electric field E and the magnetic field B have the same dimension. This amounts to a factor of c difference between how B is defined in the two unit systems, on top of the other differences​
 
thank you!:)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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