# Fine structure constant

## Homework Statement

e is charge
epsilon is permittivity

## Homework Equations

$$e^2 = 4\pi \epsilon GM^2$$

## The Attempt at a Solution

This is just a question really because I am a bit confused about something. Was reading up on the fine structure constant and apparently the fine structure constant is

$$\alpha = \frac{e^2}{GM^2}$$

I know it is supposed to be dimensionless, but where did epsilon go? Shouldn't it be

$$\frac{e^2}{\epsilon GM^2}$$

Or has epsilon been set to 1?

You're completely confused. The fine structure constant does not involve G. Try Google or Wikipedia.

actually it does.

I have been doing reading on it, and apparently

$$\frac{\hbar c}{GM^2}$$ is called the gravitational fine structure.

$$\frac{e^2}{\hbar c}$$ is the electromagnetic fine structure

http://en.wikipedia.org/wiki/Gravitational_coupling_constant

''αG is to gravitation what the fine-structure constant is to electromagnetism and quantum electrodynamics. The physics literature seldom mentions αG. This may be due to the arbitrariness of the choice among particles to use (whereas α is a function of the elementary charge e, about which there is no debate), and the relatively low precision with which αG can be measured. Unless stated otherwise, αG is here defined in terms of a pair of electrons.''

From the same link, Barrow and Tipler defined

$$e^2 / GM^2$$ as the a ratio of the fine structure.

So, can anyone please answer my question why epsilon doesn't appear in ratio's?

after all, the charge relationship is

$$e^2 = 4 \pi \epsilon GM^2$$

where apparently $$GM^2 = \hbar c$$ since the ratio $$\frac{\hbar c}{GM^2}$$ gives the fine structure constant, but so does

$$\frac{e^2}{GM^2}$$

but the epsilon seems to disappear... why?

Perhaps you're confusing the different units for electric charge? When using SI units, the charge q on an electron is measured in coulombs, and the force between two electrons is given by:
$$F = \frac{q^2}{4\pi \epsilon_0 r^2}$$
But when using CGS units, which is what you are using when you write the fine structure constant:
$$\alpha = \frac{e^2}{\hbar c}$$
, the charge e on an electron is measured in statcoulombs, and the force between two electrons is given by:
$$F = \frac{e^2}{r^2}$$
The 4 pi and the epsilon0 have basically been absorbed into redefining the unit of charge. Is this your question?

$$F = \frac{q^2}{4\pi \epsilon_0 r^2}$$
$$\alpha = \frac{e^2}{\hbar c}$$
$$F = \frac{e^2}{r^2}$$