Why 2pi

[Royston]

Sorry if this seems a stupid question, but why is 2pi used to complete a fine structure constant. I'm right near the end of my course, and working out Planck mass, and Planck energy which has 2pi just thrown in, with no explanation ?

[Royston]

Anyone ? If I want to get planck mass, it's the square root, of the planck constant x C, divided by '2pi' x gravitational constant. 2pi crops up in fine structure constants like Alpha Em as well, so why 2pi ?

[Severian]

It is just a convention. The energy of a wave is [math]E= h\nu[/math] where [math]\nu[/math] is the frequency. It is sometimes more conventient to express the frequency in terms of an angular speed [math]\omega = 2\pi \nu[/math] which is measuring the rate of change of the phase of the wave in radians per second.

 

Then the energy is [math]E= h \omega / 2 \pi[/math] which is a bit more messy, so we invented [math] \hbar = h/2\pi[/math] to give [math]E= \hbar \omega[/math] and keep the equations neat.

 

When you go deeper into the theory, you find that [math]\omega[/math] is really more natural than [math]\nu[/math] so it is more natural to use [math]\hbar[/math] rather than [math]h[/math] for things like the Planck energy.

 

 

 

For QED you calculate cross-sections as a perturbative series in the e, the (modulus of the) charge of the electron. This is because the QED electron photon vertex looks like:

 

[math] - i e \gamma^\mu [/math]

 

Since e is small, higher orders in e are smaller still and you can approximate the result by cutting off the expansion. So we would naturally expect a cross-section to look something like

 

[math]\sigma \sim A_4 e^4 +A_6e^6 +A_8e^8 + ...[/math]

 

where [math]A_i[/math] are of order 1. (It starts at [math]e^4[/math] because you need at least two interactions to conserve momentum and then need to square the matrix element.)

 

But when you do the calculations you find that the A's above are actually getting smaller as they go up, making your series converge faster. This is because you get extra factors of [math]\pi[/math] with each loop, which are actually coming from the [math]2 \pi[/math] in a Fourier series. In fact, you tend to get an extra [math]1/4 \pi[/math] for each order, so it makes more sense to rewrite the expansion as:

 

[math]\sigma \sim B_2 \alpha^2 +B_3\alpha^3 +B_4\alpha^4 + ...[/math]

 

where [math]\alpha = \frac{e^2}{4 \pi}[/math]

 

 

Of course, this definition is in natural units where [math]\hbar=c=\epsilon_0=1[/math]. In 'unnatural units' it would become [math]\alpha=\frac{e^2}{\hbar c 4 \pi \epsilon_0}[/math]. Notice that the [math]2 \pi[/math] in the [math]\hbar[/math] cancels with the [math]2 \pi[/math] in the denominator, so you could use this to say that [math]h[/math] was more 'natural' after all...

[Royston]

Right, thanks for the response Severian, though I must admit I havn't covered natural or unnatural units, and I'm clueless with the use of a matrix in QED, I've bookmarked this page for future reference though.

 

I can just about follow the first bit

 

EDIT: Ok, atleast I know what is meant by a natural unit now...it just wasn't mentioned in the book.