I think that this topic is one that's likely to generate a lot of misconceptions when people answer it, since it's a complicated topic and analogies/explanations that are given by pop science are often not super accurate. So, in an effort to be more accurate I'm gonna try to walk you through some of the actual math. Don't worry if you don't know what individual terms mean-- I'm gonna give you a broad idea of what the equations are saying, and any detail I leave out can be safely ignored for this level of an explanation. Disclamer: I'm not a particle physicist, so take this with a grain of salt. I have taken a class in this stuff before but I would definitely defer to someone with a particle physics background if they disagree with something I've said. Alright, let's get started shall we?

TL;DR-- In the area where you used this weapon you might still have bound matter that is simlar to atoms but it would almost assuredly radically alter chemistry so as to kill everything living and completely change how matter behaves.

Terminology and Fundamentals

First off, let's start with some of the terminology used in particle physics. There's a lot, and I don't wan't to say stuff that I'm not explaining so strap in-- this is gonna be a long one. The most important term is the lagrangian of a system, denoted by $\mathcal{L}$. Basically, this is an expression that encodes all the dynamics of the universe. In case you're curious, it does so via the Feynman path integral, but the specifics of this aren't really necessary-- you just need to know that $\mathcal{L}$ contains all the information about how the universe works.

Now, the total lagrangian of the universe can be split into the sum of smaller lagrangians, each one of which is governing the dynamic of some field. I realize the end of that sentence might be mumbo jumbo, but fret not! Fields are basically the "stuff" that fills the universe-- every particle that we see is an oscillation in some field. From a quantum field theory perspective, when we see two electrons, what we're really seeing is two oscillations in the electron field.

Now, fermions are the types of particles that we most often think of when we think of matter-- stuff like electrons and the quarks that make up protons/neutrons (to be precise, it's the half integer spin particles, but that's not important right now). Bosons are the other broad class of particles and are a bit more removed from every day experience with the exception of the photon. Broadly speaking, they tend to control the interactions between fermions, with note notable exception (you guessed it, the higgs). The three interactions that let fermions interact with each other via bosons are:

• electromagnetism, whose theory is called QED and whose mediating boson is the photon
• the weak interaction, whose theory is usually combined with electromagnetism for reasons I'll get into. The combined theory is called EWT, and the weak interaction has the $W^+$, $W^-$, and $Z$ bosons as mediators.
• the strong interaction, whose theory is called QCD and whose mediating bosons are the 8 gluons.

(Gravity is also considered a fundamental interaction, but is left off due to complications when you try to combine it with quantum mechanics).

With all that out of the way, let's start looking at some actual lagrangians! We'll start with the QED Lagrangian because it's the simplest, and because the Lagrangians for other interactions have very similar forms. I'm gonna write what the QED lagrangian looked like before Higg's theory, and then I'll revisit it to make a slight change after the next section. Wihout any further ado, the QED lagrangian is given by:

$$\mathcal{L}_{QED} = \overline{\psi}_f i \gamma^{\mu}D_{\mu}\psi_f- \overline{\psi}_f m_f\psi_f-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}$$

Don't worry, I'll explain everything! The $\overline{\psi}_f$ and $\psi_f$ are terms that basically just describe the state of some electrically charged, spin 1/2 fermion field $f$. The $\gamma^{\mu}$ are a set of matrices that more or less encode the wonky structure of space time in special relativity, while the $D_{\mu}$ is an operator called the gauge co-variant derivative that acts on $\psi_f$ and basically encodes how the fermion field changes, both thanks to the electromagnetic interaction and the momentum of particles in the field. Meanwhile, the $m_f$ is the mass of the particles of the fermion field $f$. So taken together, the first two terms basically represent how particles of field $f$ move around and respond to the electromagnetic field. In reality, you have a copy of this pair of terms for every electrically charged, spin 1/2 particle field.

Finally, the $F^{\mu \nu}$ term is the electromagnetic field strength tensor which fancy words aside is simply information about the photon field (which recall, is the field that mediated the electromagnetic interaction). When taken together with the first two terms (or more accurately the sum of all such terms for all charged particle types as mentioned above), it describes how the photon field is affected by the various fermion fields. Woo, that about does it! Now let's move onto where the higgs boson comes in.

Why the Higgs boson?

One important constant of nature is that physicists are lazy. So, given how successful the QED lagrangian was at predicting nature, they figured why not keep basically the same form for the QCD and Weak Lagrangians with some slight modifications due to handedness of the weak force and the fact that there are multiple mediating bosons? And such an approach did actually work pretty well for QCD. But with the weak force, they ran into the problem that if you structure the lagrangian similar to the QED lagrangian, it becomes a mathematical necessity for the mediating gauge bosons to be massless. But it was already well established experimentally that the $W$ bosons had mass, so what gives?

After a few years of collectively banging their heads into walls, physicists realized that if you introduce a spin 0 boson field $\phi$ (the famous higgs field) with certain special properties, then after all of the dust settles you're left with the actual physical mediators of the weak force being massive bosons! Yay! The actual terms in the electroweak lagrangian that enable such a thing are:

$$\mathcal{L}_h = (D^{\mu}\phi)(D_{\mu}\phi)^* - \lambda(\phi\phi^* - \frac{v^2}{2})^2$$

The first term again through the $D_{\mu}$ operator describes how the higgs field responds and interacts with the mediator bosons. The second term describes the mass of the higgs and self interaction terms, but more importantly, it has the important property that if you plot it as a function of $\phi \phi^*$, it has infinitely many minima. This is important because the way we work with quantum field theory is by looking at the vacuum state where no particles are present and then adding small changes to that. But since this term has infinitely many minima, the system has to make an arbitrary choice as to which vacuum state it must settle in, which is known as spontaneous symmetry breaking. It turns out that this choice affects the mediator bosons of the weak force through the $(D^{\mu}\phi)(D_{\mu}\phi)^*$ term and leads to the desired result of a massless mediator for the electromagnetic force and massive ones for the weak force.

We're still left with one small problem-- the weird handedness properties of the weak force screws up our ability to use the simple $\overline{\psi}_f m_f\psi_f$ term for lepton masses that we were using before (specifically, it wreck gauge covariance when using terms like this). But luckily, the higgs boson provides us an out-- instead we can write the mass terms as something roughly like

$$g\overline{\psi}_f \phi \psi_f$$

where $g$ is a constant describing the coupling strength of the particle to the higgs field. It can then be shown that when spontaneous symmetry breaking occurs, this term results in a mass for the fermion field that doesn't cause any mathematical issues. And bam, that's the standard model in a nutshell! I salute you for reading this far.

Answering your question

Now, to answer your question, we have to be more specific about what you mean when you say that you "switch off" the higgs field, cause there's a couple interpretations. One would be that the higgs field is deleted from the Lagrangian entirely in a certain region of space. Another could be that $v$ becomes zero, so there's only one vacuum state and spontaneous symmetry breaking doesn't occur. Finally, you could mean that $g$ goes to zero for all the higgs coupling constants mentioned before. These all have different implications and conceptual difficulties (especially the first), but they should all achieve the same basic effect that fermion masses go to zero, so that's what I'll try to explore.

Now, one thing you hopefully took away from the previous discussion is that the higgs boson is responsible for the mass of all fundamental fermions (except maybe neutrinos). However, not all the mass we usually interact with is the mass of these fermions. In fact, the majority isn't-- If you look up the mass of the up and down quarks, you'll see that sum of the masses of the consituent quarks of a proton make up only about 1% of the proton mass! So where's all this extra mass coming from? The answer is that most of the mass comes from the energy caused by the strong interaction that binds the nucleus together. Now QCD is a super difficult mess so I'm gonna be honest and say that I'm not entirely sure what would happen if quarks in a nucleon became massless, but my guess is that if anything they would become more strongly bound, so I think you would still have a massive nucleus of some sort.

As for electrons, If you look at the $m_e \rightarrow 0$ limit of the Schrodinger equation solution to the hydrogen atom you see that it doesn't change very much. This isn't entirely valid because the non-relativistic Schrodinger equation wont be accurate for massless electrons, but since I don't feel like delving into the dirac equation I'm going to say there's a not terrible chance you'll still have matter resembling atoms with a strongly bound nucleus of quarks and a shell of electromagnetically bound electrons in the region that you use this weapon. However, life is very fragile so it's almost certain that even if you still have "atom-like" matter, the structure and chemistry will be so foreign to our universe's that it will instantly kill any living creature in an area affected by this weapon.

As a final remark, I've seen that several comments and answers have suggested that using this weapon would result in a massive release in energy due to disappearing rest mass. However, this isn't necessarily true-- any of the definitions you use for what "turning off the higgs field" means requires that you alter the Lagrangian of the universe in space and time. This means that we no longer have an action invariant in time and we can't use Noether's theorem to reason that energy is conserved.

posted over 4 years ago

el duderino‭

31 reputation 0 3 3 0