Explain electroweak unification me

Factual Questions
1

Correct me if I’m wrong with any of my prior knowledge.
In electromagnetism, the force is carried by the photon and the the weak force is carried by the W+, W- and Z bosons. I get that. AIUI, At suitable energies the electromagnetic force and weak force become unified.

What does that mean? EM and weak forces act so differently what would an electroweak force look like as opposed to when they are broken?
Why do the forces break at low energies?
What is the carrier for the EMW force?
What other questions do I want answered but I don’t know enough to ask?

2

I’ll start with a question that isn’t directly about unification. Why is the weak force weak? And I’ll bring in only the Z boson part of the weak force, not the W parts.

A simple particle interaction might be two electrons “bouncing” off each other:

e^- + e^- \rightarrow e^- + e^-

This interaction can be mediated by a photon or a Z boson. Both bosons are electrically neutral, and the electrons in the interaction are charged under both the EM and the weak forces, so the electrons can talk to both the photon and the Z boson. In reality, both possibilities are always present; you don’t get to pick and choose, and the two paths quantum mechanically combine to give the full dynamics. But for everyday particle energies, the Z boson contribution is negligible. It’s weak. Why?

The contribution of each piece ultimately has a factor something like

\frac{1}{|E^2-M^2|}

in it, where E is the energy scale of the process and M is the mass of the mediating boson. For the EM (photon) case, M=0 since the photon is massless. For the weak (Z boson) case, M=M_Z=91~\mathrm{GeV}.

At energies well below M_Z, the factor above is:

EM: \frac{1}{E^2} (since M=0)
Weak: \frac{1}{M_Z^2} (approximately, since M_Z is large compared to E)

That big honkin’ M_Z in the denominator is why the weak force is weak at lower energies. As a quantitative case, for energies around 1 GeV (which is already hundreds of times higher than typical nuclear energies), you’re looking at a fractional contribution of 10-4 for the weak force.

If you go to energies much higher than M_Z, however, the factor above is:

EM: \frac{1}{E^2} (since M=0)
Weak: \frac{1}{E^2} (approximately, since M_Z is now small compared to E)

EM and weak look the same.

So, the vast difference in strength at low energies is due to the mass of the Z boson, and this aspect falls away at high energy. Two questions immediately come about:

  • Why does the Z boson have a mass while the photon does not?
  • What does this have to do with unification? The forces can still be separate forces even if they might have comparable strengths as high energy.

I’ll come back to those in a next post. In the meantime, feel free to dial up or down the desired technical level of the discussion.

3

So does this mean that at high enough energies, one particle does the interaction but that below a certain energy it turns into a photon and/or boson?

4

No, there are always the same number of bosons. (The photon is also a boson.) But your question is a good segue into the two hanging questions I left above.

Their answers stem from same underlying ideas, but it will be easier initially to pretend they don’t. But know that a layer of elegance is coming in the end.

Basic unification
While one often says that the EM and weak forces unify at high energies, a different framing is to say they are unified in the theory regardless of the energies involved. The fact that different practical aspects of the unified force manifest at low energy doesn’t remove the unification. In this framing, unification was a step in the historical understanding when it was realized that the EM and weak forces were two sides of one coin.

In the “broken” (un-unified) description, you have the following fundamental values:

  • electromagnetic coupling constant*, g_{e}
  • charged-current (W-mediated) coupling constant, g_W
  • neutral-current (Z-mediated) coupling constant, g_Z
  • W boson mass, M_W
  • Z boson mass, M_Z

These are all independent choices, as it were. In the unified picture, these will become interrelated. There’s also stuff related to chirality (handedness) of the interactions, but we will be able to skirt around those.

Forces from symmetries
The fundamental forces come about through symmetries. The relevant symmetries have to do with how physics needs to stay unchanged under certain “rotations” of the quantum fields or aspects of them. (Note: particles are described mathematically using “fields”, which take on values across all of spacetime.)

If nature had chosen a very simple symmetry to impose, one related to adjusting a field’s complex phase (with “complex” as in “complex numbers”), then you would get an emergent EM-like interaction and a massless boson like the photon. But nature has chosen to complicate things. In the real situation, the imposed symmetry relates to rotations not only in phase but also in a vector-like aspect of the fields called “isospin”. The name stems from how the math of it looks like quantum mechanical “spin”. But it’s not related to angular momentum like spin is. It’s instead related to whether the particle is on the “up” or “down” side of the particle families (like the up / down quarks or the electron / electron-flavor neutrino).

With these symmetries imposed, you need two coupling constants or force “strengths”: one for the phase piece and one for the isospin piece. Further, the phase piece still yields a single boson, while the more complicated isospin piece leads to three bosons. But these 1+3 bosons are not the ones you know. Let’s call them B and C_1, C_2, and C_3. These are all massless bosons. In fact, you can’t write down a well-behaved theory like this with massive bosons. I’ll return to this later.

Hiding within the isospin symmetry operations is a linear combination that can convert particles between isospin down and isospin up cases. This combination of symmetry operations leads to a different combination of bosons being the physically relevant ones, namely:

W^{\pm} = \frac{1}{\sqrt{2}}(C_1\mp iC_2)

That is, the W bosons correspond to this particular quantum combination of two of the fundamental bosons that stem directly from the symmetries of nature.

The remaining two bosons (B and C_3) can also be combined into superpositions. We could just leave them alone, but later we’ll see that we have no choice.

Consider the 2D Cartesian axes x and y. If we rotate the axes by some angle \theta, we get new axes x' and y' related to the original ones as follows:

x'=x\cos\theta - y\sin\theta
y'=x\sin\theta + y\cos\theta

In the same way, treat the bosons B and C_3 as two axes and rotate these into new quantum mechanical superpositions through an angle \theta_w, forming two bosons that’s we’ll label as photon (\gamma) and Z:

\gamma=B\cos\theta_w + C_3\sin\theta_w
Z=-B\sin\theta_w + C_3\cos\theta_w
(with the minus sign in a different place than in the (x,y) example by choice of which way the rotation is going).

Why we do this and how masses get into the picture is next.

Summary of this piece:

  • Symmetries of nature lead to forces and their corresponding bosons.
  • The fundamental symmetries present lead to four massless bosons.
  • Two of these combine to give the W bosons, which cleanly allow conversion between “up”- and “down”-type particles.
  • The other two also combine by rotating by a so-far-arbitrary angle \theta_w.

This already starts to interrelate the fundamental constants at the top of the post, but I’ll save that part for when the full picture is laid out.

Footnote:
* g_{e} = \sqrt{4\pi\alpha}, where \alpha is the fine structure constant.

5

Aside, while the much-more-qualified @Pasta continues organizing and setting down his thoughts: It’s relevant that this number is less than 1, and the math would be much more difficult if it were greater than 1.

6

Some of our actual physical bosons are massive, so what gives?

Higgs mechanism
There is another field in the theory, the Higgs field, that has two important qualities: (1) it is “charged” under the symmetries we’ve been talking about, so it cares about isospin and all that, and (2) it has an unusual self-interaction potential that gives it a non-zero value everywhere in space even when there are no Higgs particles present.

The second piece means that setting the Higgs field to zero everywhere is not the lowest-energy configuration of the universe. That zero-Higgs version is the configuration that makes the overall symmetries of the universe clear. At that (unstable) zero value of the Higgs field, the rules about massless bosons and gauge symmetries hold. But once you build that universe, it is free to evolve to a lower energy state.

In that lower energy state, and because of point (1) above, some of the gauge bosons pick up mass as a consequence of the isospin-aware Higgs field having a non-zero value always. Also, in moving from the zero-Higgs-field case to the stable case, the B and C_3 rotation has to happen at an angle directly related to the coupling constants of the phase and isospin pieces of the symmetries.

I don’t expect that last paragraph will be super clear on a first read, so to reword in a summary fashion:

  • The fundamental bosons B, C_1, C_2, and C_3 arise from the gauge symmetries. They are massless.
  • The Higgs field interacts with itself in a way that makes it take on a non-zero value even in vacuum.
  • This non-zero value leads to certain linear combinations (quantum superpositions) of the fundamental bosons becoming physically relevant.
  • Three of these gain mass in the process, and two of these are charged. These are the Z, W+, and W-.
  • A fourth one is the massless photon.

All of this amounts to a unified description – the electroweak force. At very high energy, the Higgs field can get kicked out of its low-energy phase – melting, if you will. At that point, it’s a big stew that doesn’t lend itself to the language of perturbation theory (Feynman diagrams with virtual particles and bosons “mediating” things). The familiar four bosons lose their relevance, but the four fundamental bosons that were always actually there are still there.

Returning back to the fundamental constants, this unification leads to remarkable relationships between previously unrelated quantities. Now, you need only three things: a single mass or energy (e.g., the Higgs field value), a coupling constant g for the isospin symmetry, and a coupling constant g' for the “phase” symmetry. With this unification:

  • The rotation angle \theta_w is given by \tan \theta_w =\frac{g'}{g}.
  • The previously independent coupling constants are unified:

g_W=g
g_Z= g_W/\cos\theta_w
g_e = g_W\sin\theta_w

  • The masses of the Z and W bosons are not independent: M_W = M_Z \cos\theta_w.

Summarizing
The EM and weak forces are described by a unified model. This unification is a statement of our understanding, not a statement of energy scales, per se. At very high temperatures (of the sort needed to unbreak the broken Higgs symmetry), the whole stew is at play simultaneously, and interactions aren’t mappable onto the “such-and-such particles mediate the forces” language. (In technical jargon, one would say the theory is “non-perturbative” at such energies.)

At low energies, the behaviors of the Z and photon decouple due to the Z boson being the massive one. The behaviors of the W bosons also have this mass effect, making that piece of the weak force also weak at low energies. The W bosons also switch particles between “up” and “down” types, further separating that behavior from the photon (and Z boson).

At energies even well above the Z and W boson masses, nothing special really happens. You still have \gamma, Z, and W doing their usual thing. It’s just that the “weak” ones aren’t weak anymore since they’ve overcome the burden of their mass. The overall description of interactions is still the same, and just as unified at high energy as it is at low.

Even at obscenely large energies, a single collision doesn’t amount to a phase change of the universe, but you can start to look for the fingerprints of the transition between the broken and unbroken phases in interactions.

This is all a big core dump of info, at a randomly selected level of technical detail. Happy to adjust.

7

As one extra nugget of what it means to be “unified” –

The weak coupling strength is a little more complicated than just an overall coupling constant. There are two pieces. For the Z boson case, one of the pieces scales sort of simply with the g_Z from above. The other term, however, scales also by a numerical coefficient that looks like:

\pm(1 - 4q\sin^2\theta_w)

where q is the electrical charge of the interacting fermion involved, and the \pm out front is related to whether the fermion is the more positively charged or more negatively charged member of the isospin “doublet” (e.g.: up quark +, neutrino +, down quark -, electron -). This is to say, for an interaction mediated by the Z boson, the strength of that weak interaction very directly cares about the electrical properties of the particles, separately from the presence of any EM aspects of the interaction mediated by a photon.

For an electron, this weak coefficient is (-1+4\sin^2\theta_w) while for a charm quark it is (1-\frac{8}{3}\sin^2\theta_w). This is an window into the underlying unification. The forces are understood today in a unified way, rather than “becoming” unified at some energy.

8

One other thing I’ll mention here: It is possible to produce a more-or-less complete description of electromagnetism without the weak force. It’s a bit off because of the lack of the weak contribution, but at energy scales familiar to humans, that error is negligible. But there is no substantive description of the weak force, at all, without the electromagnetic unification.