What is the value of the fine structure constant at Planck energy?

[heinz]

At low energy, the value is around 1/137.036. What is the value at the Planck energy? Is it listed somewhere?

After all, the renormalization of the charge (as it is also called) is mentioned everywhere. But numbers are rare ...

Hz

 

[Naty1]

No specific answer but this is interesting:

According to the theory of the renormalization group, the value of the fine structure constant (the strength of the electromagnetic interaction) grows logarithmically as the energy scale is increased. The observed value of α is associated with the energy scale of the electron mass; the electron is a lower bound for this energy scale because it (and the positron) is the lightest charged object whose quantum loops can contribute to the running. Therefore 1/137.036 is the value of the fine structure constant at zero energy.

via wikipedia at http://en.wikipedia.org/wiki/Fine-s...the_fine_structure_constant_truly_constant.3F

 

[Avodyne]

At low energy, the value is around 1/137.036. What is the value at the Planck energy?

It depends on whether there are any presently-undiscovered charged particles with masses below the Planck scale. Also, above the W mass (~80 GeV), it makes more sense to talk about the SU(2) X U(1) coupling constants g2 and g1, rather than the electromagnetic coupling e, though we can still formally define it via 1/e^2 = 1/g2^2 + 1/g1^2. Because of all this, no one bothers to compute the formal value of alpha at the Planck scale.

 

[Civilized]

At low energy, the value is around 1/137.036. What is the value at the Planck energy? Is it listed somewhere

The answer is definitely that no one knows because it would involve summing all the diagrams to all orders of perturbation theory. I can, however, give you the second order correction:

\alpha(q^2) = \alpha(0)(1 + \frac{\alpha(0)}{3\pi}f(\frac{-q^2}{m^2c^2}))

where:

<br /> q^2 = -4|p|^2 sin(\theta / 2)<br />
<br /> f(x) = 6 \int_0^1 z(1-z) ln(1 + x z(1 - z)) dz<br />

It's irritating that we have to leave f(x) in the form of an integral that cannot be reduced to elementary functions, but the asymptotic behaviors are f(x) = x/5 when x << 1 and f(x) = ln(x) when x >> 1. As you can see, 1/137 is still pretty much as valid as ever, although the corrections would be measurable if we could get up to the Planck energy.

 

[heinz]

The answer is definitely that no one knows because it would involve summing all the diagrams to all orders of perturbation theory. I can, however, give you the second order correction:

\alpha(q^2) = \alpha(0)(1 + \frac{\alpha(0)}{3\pi}f(\frac{-q^2}{m^2c^2}))

where:

<br /> q^2 = -4|p|^2 sin(\theta / 2)<br />
<br /> f(x) = 6 \int_0^1 z(1-z) ln(1 + x z(1 - z)) dz<br />

It's irritating that we have to leave f(x) in the form of an integral that cannot be reduced to elementary functions, but the asymptotic behaviors are f(x) = x/5 when x << 1 and f(x) = ln(x) when x >> 1. As you can see, 1/137 is still pretty much as valid as ever, although the corrections would be measurable if we could get up to the Planck energy.

Thank you! And what are p and theta?

Hz