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A tricky question. Here is the famous graph of the running of the three coupling constants in the standard model: Running of the gauge couplings

The graph shows, in its top curve, the running of the coupling constant $\alpha_1$. This is the coupling of the weak hypercharge coupling constant for the weak hypercharge group $\mathrm{U}(1)_{Y}$, which is one of the three gauge groups of the Standard Model of particle physics.

But there is a tricky detail. In that curve, $\alpha_1$ is multiplied with $5/3$. This factor $5/3$ comes from the assumption that GUTs are valid. The factor ensures that the various group traces of $\mathrm{U}(1)_Y$, $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$ are normalized in the correct way when they form the $\mathrm{SU}(5)$, $\mathrm{SO}(10)$ or any other grand unification gauge group.

In the case that grand unification is wrong, the factor $5/3$ cannot be deduced. Which factor would be natural in this case?

Clarification added, after remarks by Lubos Motl: it is assumed in the question that the usual definition of the weak hypercharge is used, $Y_W = 2 (Q - T_3)$, in which left-handed quarks have hypercharge $1/3$.

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There is of course no unique answer because if the normal GUT embedding and theory is wrong, there may still be another valid theory that naturally involves the same or (almost) any other rescaling of the hypercharge fine-structure constant.

If you say that GUT is wrong, it is far from specifying what is actually right. When it comes to coupling constants and other things, GUT is more predictive than theories without any unification - that's a characteristic feature of unifying theories. So if you say that GUT is wrong, you just reduce the ability to predict and explain patterns in Nature, so you can't expect that by this negative assumption, you will obtain positive predictions. One shouldn't expect that by throwing out theories that are meant to explain something, one automatically finds different answers to the questions.

Of course, if you assume a normalization with no relationship between any groups where the hypercharge $U(1)$ is normalized just like any other $U(1)$ would be, and your goal is to make the formulae as simple as possible on the paper (which is not really a physical criterion), then $5/3$ is replaced by $1$. You just omit the $5/3$ factor. But this is a kind of vacuous statement because one may only compare the fine-structure constants of the different group factors if there is some relationship between them which is either grand unification or plays the same role.

One more preemptive comment: at low energies, it is not true that the hypercharge fine-structure constant renormalized by another simple factor such as $5/3$ yields the electromagnetic fine-structure constant. At the GUT scale, similar relations exist but the electromagnetic fine-structure constant is not well-defined there.

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  • $\begingroup$ Thank you for your remarks. You can see that I do not speak about the fine structure constant at all in this question, but only, as you kindly explained to me, about the hypercharge coupling constant. $\endgroup$
    – Claude
    Commented May 13, 2011 at 9:50
  • $\begingroup$ The issue I have is that I am not sure at all that 1 is more natural than 5/3; it could equally be another value. In fact, a value of 1 would mean that U(1) and SU(2) have equal coupling constants, and that the weak mixing angle has sin^2 = 1/2. It is not obvious that this is the most natural choice. $\endgroup$
    – Claude
    Commented May 13, 2011 at 9:59
  • $\begingroup$ I was only thinking what happens if GUT is wrong. Taking your remark to the extreme, it would mean that normalization of the weak hypercharge is not fixed at all if GUTs are wrong. $\endgroup$
    – Claude
    Commented May 13, 2011 at 10:01
  • $\begingroup$ But the normalization must be fixed somehow, because the normalization fixes/enters the weak mixing angle. $\endgroup$
    – Claude
    Commented May 13, 2011 at 10:02
  • $\begingroup$ After all, coupling constants can be measured, so it must be possible to define them uniquely. $\endgroup$
    – Claude
    Commented May 13, 2011 at 10:13

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