# Should 4 fundamental forces really be 3 because of electroweak unification?

Physicists concluded that, in fact, the weak and electromagnetic forces have essentially equal strengths. This is because the strength of the interaction depends strongly on both the mass of the force carrier and the distance of the interaction. The difference between their observed strengths is due to the huge difference in mass between the W and Z particles, which are very massive, and the photon, which has no mass as far as we know.

So, should all books teaching four fundamental forces change to three with third being Electroweak?

Or still there are four fundamental forces?

• I consider it as one step towards unification (or GUT). Maybe one day we might be able to explain all forces in terms of just one. – N.S.JOHN Feb 7 '16 at 10:01
• So, are we waiting for all to unify. – Anubhav Goel Feb 7 '16 at 10:03
• Their are two types of nuclear interaction, weak and strong, and two types of long range, macroscopic forces, gravity and electromagnetic. So observationally it is four. Unification shows interconnection between forces, so it is possible to explain particle behavior in a unified fashion. So its your choice, depending upon the application. – Peter Diehr Feb 7 '16 at 12:16

The standard model is written in the language of quantum field theory (QFT). In QFT one usually starts from a Lagrangian, which is a function of the fields that we suppose to be the elementary constituents of the world, or, to be even more precise, the elementary objects of our description of the world, and then one quantizes it by using the appropriate techniques. Given a Lagrangian, the whole theory can be derived from it. The Lagrangian of the electromagnetic field and a fermionic (charged) field, such as the electron field, can be written as $$\mathcal{L}=\bar{\psi}\gamma^{\mu}(i\partial_{\mu}-m-eA_{\mu})\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ Here $e$ is the electric charge of the fermion, $m$ its mass, $\psi$ its field, $A_{\mu}$ the photonic field and $F_{\mu\nu}F^{\mu\nu}$ is something like the product of the electric and magnetic fields. The two terms $$\mathcal{L}_{\psi}=-\bar{\psi}\gamma^{\mu}(i\partial_{\mu}-m)\psi$$ and $$\mathcal{L}_{A}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ tell you how the fields behave independently of each other. The term $$\mathcal{L}_{int}=e\bar{\psi}\gamma^{\mu}\psi A_{\mu}$$ contain both $\psi$ and $A_{\mu}$, and it tells you how the fields interact.  Now, the full Lagrangian of the electroweak theory is much more complex. It contains the Higgs field, a symmetry breaking potential and six generations of fermions, so I won't write it in its full form. Nevertheless, the interaction term between a single fermion and the "electroweak" field can be schematically written as
$$\mathcal{L}_{int}=-\bar{\psi}\gamma^{\mu}(g'B_{\mu}+g\tau^{a}W_{\mu}^{a})\psi$$ Here $a=1,2,3$, $B_{\mu}$ and $W_{\mu}^{a}$ are the four "electroweak" fields and $\tau^{a}$ are three appropriate matrices ($\psi$ is a column vector and $\bar{\psi}$ is a row vector, so they can have matrices in between). $g$ and $g'$ are the coupling constants of the theory, and they can be chosen different (and in fact they are). This is very important to the distinction between unification and mixing. Now, first of all, the $W$'s and $B$ cannot be directly identified with the photonic field and the $W^{\pm}$, $Z$ fields. As I told you, the full Lagrangian contains other terms. These terms, through the mechanism of spontaneous symmetry breaking, give mass to linear combinations of the $W$'s and the $B$. The combinations are as follows: $$W^{\pm}_{\mu}=\frac{1}{\sqrt{2}}(W^{1}_{\mu}\mp i W^{2}_{\mu})\\ Z_{\mu}=\frac{1}{\sqrt{g^{2}+g'^{2}}}\ (gW^{3}_{\mu}-g'B_{\mu})\\ A_{\mu}=\frac{1}{\sqrt{g^{2}+g'^{2}}}\ (g'W^{3}_{\mu}+gB_{\mu})$$ $W^{\pm}_{\mu}$ are the fields of the $W^{\pm}$ bosons. $Z_{\mu}$ is the field of the $Z$ boson. $A_{\mu}$ is the field of the photon. The former are given mass. $A_{\mu}$ turns out to be given no mass. When you rearrange the terms in the Lagrangian in terms of these combinations of fields, you get the Lagrangian of electrodynamics plus the Lagrangian of the weak theory. As you can see, the photonic field is a mixing between the $B$ field and the $W^{3}$ field. The $Z$ field is too. So one gets electrodynamics and the weak theory by mixing fields.  Now, as I told you, the coupling constants $g$ and $g'$ can be chosen to be different. This is because in the theory which contains $B$ and the $W$'s, symmetry principles (together with something called the renormalizability principle) don't constrain the $B$ field to interact directly with the $W$'s. So you can interpret the starting point of the electroweak theory to be a theory in which two different kind of interactions exist. But because the theory contains the symmetry breaking terms, at low energies the two mix to give the theories of electromagnetic and weak interactions. On the other hand, let's suppose that to begin with we had a theory of a truly single interaction. Then would $g$ and $g'$ be constrained to be chosen equal? The answer is still: no. The theory would have the very same form as the one with two different interactions. This has to do with the symmetry group of the theory, which in this case is called $U(1)\times SU(2)$ ($U(1)\equiv B, SU(2)\equiv W$'s). When you have a $U(1)$ multiplying some other group, it is difficult to give an interpretation to the multiplication itself. $U(1)$ doesn't change the structure of the second group in any case, i.e. it behaves as a somewhat distinct portion of the product group (and this is why the coupling constants can be chosen to be different).