*© Matt Strassler [September 22, 2012; revised October 10, 2012]*

*This is article 1 in the sequence entitled How the Higgs Field Works: with Math.*

If you have read my series of articles on Particles and Fields (with math), you know that all the elementary “particles” are really quanta *(i.e. waves whose amplitude and energy are the minimum allowed by quantum mechanics)* in relativistic quantum fields. Such fields typically satisfy Class 1 equations of motion (or generalizations thereof, as we’ll see) of the form

- d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= – (2 π ν_{min})^{2}(Z-Z_{0})

where Z(x,t) is the field, Z_{0} is its equilibrium value, x is space, t is time, d^{2}Z/dt^{2} represents the change with time of the change with time of Z (and d^{2}Z/dx^{2} similarly for space), c is the universal speed limit (often called “the speed of light”), and the quantity ν_{min} is the minimum frequency allowed for waves in this field. *(For a review of these equations, see this article.)* A few fields satisfy Class 0 equations, which are just Class 1 equations where the quantity ν_{min} is zero. The quanta of such fields have mass

- m = h ν
_{min }/ c^{2}

where h is Planck’s quantum mechanics constant. In other words,

- d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= – (2π c^{2}/h)^{2}m^{2 }(Z-Z_{0})

Now all of this is only true to a point. Really, if all that fields did was satisfy Class 0 and Class 1 equations, nothing would ever happen in the universe. Their quanta would just pass by each other and they wouldn’t do anything… no scattering, no smashing, no formation of interesting things like protons or atoms. So let’s put in a modification that is common, interesting, and required by what we know about nature from experiment. We’ll explore it more later.

Let’s think for a moment about two fields, S(x,t) and Z(x,t). Imagine that the equations of motion for S(x,t) and Z(x,t) are modified versions of the Class 1 and Class 0 equation, which means S particles are massive but Z particles are massless. We’ll assume (for now) that the equilibrium values S_{0} and Z_{0} are zero.

- d
^{2}S/dt^{2}– c^{2}d^{2}S/dx^{2}= – (2π c^{2}/h)^{2}m_{S}^{2}S - d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= 0

But now let’s make the equations more complicated, in a way that is ubiquitous in for the fields of nature. Specifically, there are additional terms in the equations involving S(x,t) multiplied by Z(x,t)* [and yes, the two terms are supposed to be different -- it's not a typo, and the difference will be important]*

- d
^{2}S/dt^{2}– c^{2}d^{2}S/dx^{2}= – (2π c^{2}/h)^{2}(m_{S}^{2}S + y^{2}S Z^{2}) - d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= – (2π c^{2}/h)^{2}y^{2}S^{2}Z

Remember there’s a little shorthand here, so read carefully. S and Z are shorthand for the fields S(x,t) and Z(x,t), which can vary over space and time. Everything else (c, h, y, m_{S}) is a constant that does not depend on space or time. The parameter y is a number, typically between 0 and 1, and called a “Yukawa parameter” or “Yukawa coupling”, for historical reasons. We’ll see how it comes into the story in a few moments.

* In almost all circumstances in particle physics, the deviations of fields like S(x,t) and Z(x,t) from their equilibrium values S_{0} and Z_{0} are extremely small.* Since we’re currently assuming S

_{0}=0 and Z

_{0}=0, this means S and Z themselves are extremely small: any waves in S and Z typically have small amplitude

*(they are typically made from a single quantum)*and although there are always spontaneous quantum disturbances going on

*(often referred to as virtual particles, and discussed in the Particles and Fields articles as a sort of quantum jitter)*these are also rather small in amplitude (though sometimes big in importance.) And if S is small and Z is small, then S times Z is

**really**small.

*[Consider two numbers: if a = 0.01, and b = 0.03, then a times b is 0.0003 ---*Since y isn’t big, the terms y

**really**small, much smaller than either a or b.]^{2}S Z

^{2}and y

^{2}S

^{2}Z are small enough to ignore under many circumstances.

Specifically, we can ignore them in figuring out the mass of the S and Z “particles” (i.e., quanta). To figure out what an S particle is like, we need to consider a wave in S(x,t), with Z(x,t) assumed to be very small. To figure out what an Z particle is like, we need to consider a wave in Z(x,t), with S(x,t) assumed very small. Once we ignore the extra y^{2} S Z^{2} and y^{2} S^{2} Z terms, the S and Z fields then both satisfy the simple Class 0 or 1 equations of motion we started with, from which we deduce that the S particles have mass m_{S} and the Z particles have zero mass.

But now imagine a world in which Z_{0} is zero * but S_{0} is not zero*. We change the equations just slightly:

- d
^{2}S/dt^{2}– c^{2}d^{2}S/dx^{2}= – (2π c^{2}/h)^{2}(m_{S}^{2}[S- S_{0}] + y^{2}S Z^{2}) - d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= – (2π c^{2}/h)^{2}y^{2}S^{2}Z

Again, the S and Z fields are functions of space and time, but everything else, including S_{0}, is a constant. In this case Z(x,t) itself very small, but S(x,t) is not! Instead it is useful to write

- S(x,t) = S
_{0}+ s(x,t)

where **s is the variation of S away from its equilibrium value S _{0}**. We can say that s(x,t) is a shifted version of the S(x,t) field. The statement that fields in particle physics stay very near their equilibrium values most of the time is the statement that

**s(x,t) is very small**, and

*that S(x,t) is small. Substituting the red equation above into the equations above for S and Z, and remembering that S*

**not**_{0}is constant so dS

_{0}/dt = 0 and dS

_{0}/dx=0, we find the equations become

- d
^{2}s/dt^{2}– c^{2}d^{2}s/dx^{2}= – (2π c^{2}/h)^{2}(m_{S}^{2}s + y^{2}[S_{0}+s] Z^{2})

- d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= – (2π c^{2}/h)^{2}y^{2}[S_{0}+s]^{2}Z

= – (2π c^{2}/h)^{2}y^{2}(S_{0}^{2 }+ 2 sS_{0}+s^{2}) Z

As before, if we want to know the masses of the quanta of the S and Z fields, we can drop any term in these equations that involves a product of two or more small fields — terms like Z^{2} or sZ^{2} or sZ or s^{2}Z. I’ve marked all the fields in purple so you can count them easily. So let’s look at what remains if we only keep the terms involving one field:

- d
^{2}s/dt^{2}– c^{2}d^{2}s/dx^{2}= – (2π c^{2}/h)^{2}m_{S}^{2}s + …

- d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= – (2π c^{2}/h)^{2}y^{2}S_{0}^{2}Z + …

*(The “+ …” is there to remind us that we dropped some terms.)* Note that there is no significant change in the s field’s equation, because the new terms, y^{2} [S_{0}+s] Z^{2} all contain at least two powers of Z. But for the Z field’s equation, we could **not** ignore entirely the term y^{2} [S_{0}+s]^{2} Z entirely, because it contained a term of the form y^{2} S_{0}^{2} Z, which contains * only one field. *Consequently, although a quantum of the S field still satisfies a Class 1 equation and has mass m

_{S}, a quantum of the Z field no longer satisfies a Class 0 equation!

**It now satisfies a Class 1 equation:**

- d
^{2}Z/dt^{2}– c^{2}d^{2}Z/dx^{2}= – (2π c^{2}/h)^{2}y^{2}S_{0}^{2}Z

Consequently the quanta of the Z field now have a *mass*!

- m
_{Z}= y S_{0}

**Because of the simple interaction between the S and Z fields with strength y, a non-zero equilibrium value S _{0} for the S field gives the Z quantum a mass proportional both to y and to S_{0}.**

**The S field’s non-zero value has given mass to the particle of the Z field!**

*Fine point: Even if for some reason the mass m _{Z} of the Z particle had been non-zero to start with, then the mass of the Z particle would still be shifted.*

*m*_{Z}^{new}= [m_{Z}^{2}+ y^{2}S_{0}^{2}]^{1/2 }

*(recall that x ^{1/2} means the same as √x.)*

Well, this is basically how the Higgs field H(x,t) gives mass to particles. It turns out that for each known particle σ *[except the Higgs itself]*, the equation of motion for its corresponding field Σ(x,t) is a Class 0 equation, which naively would imply the σ particle is massless. But for many of these fields there are extra terms in the equation of motion, including a term of the form

- y
_{σ}^{2}[H(x,t)]^{2}Σ(x,t) ,

where y_{σ} is a Yukawa parameter, *different for each field*, that represents the strength of the interaction between the H field and the Σ field. In such a circumstance, a non-zero average value for the Higgs field, H(x,t) = H_{0}, shifts the minimum-frequency of Σ waves, and thus the mass of σ particles, from zero to something non-zero: m_{σ} = y_{σ} H_{0}. Diversity among the Yukawa parameters for the various fields of nature leads to the diversity of masses among the “particles” (more precisely, the “quanta”) of nature.

Notice, by the way, that the Higgs **particle** has nothing to do with this. A Higgs particle is a quantum of the Higgs field — a ripple of minimum energy in H(x,t), a little wave that * depends on space and time*. What gives mass to the other known particles of nature is the non-zero equilibrium

**constant**value for the Higgs field, H(x,t) = H

_{0}, all across the universe; this timeless and universally present constant is very different from Higgs particles, which are ripples that vary over space and time, and are both localized and ephemeral.

That’s the basic idea. I’ve left lots of obvious questions unanswered here: why should we expect there to be terms in the equations that involve the product of two or more fields? *[Read about the profound importance of such terms here.] *Why would the known particles be massless if there were no Higgs field? Why is the Higgs field’s equilibrium value non-zero while this is not true of most other fields? How does the Higgs particle enter the story? The ensuing articles will try to address these and other issues.

Matt, a question:

In the article “What If the Higgs Field Were Zero” you show that the EM and Weak force would no longer exist, instead there would be Hypercharge and Isospin force. My question is: How does the Higgs Field interfere in the electromagnetic interaction if the photon is massless??? That sounds paradoxal to me.

Matt,

It seems that the Higgs theory replaces the arbitrary particle masses with an equal number of arbitrary Yukawa parameters, as well as an arbitrary non-zero equilibrium value for the Higgs field itself.

How does the Higgs theory simply things?

The answer is on slide 8 here

http://www2.ph.ed.ac.uk/sussp61/lectures/01_Sachradjda_StandardModel/standrews2.pdf

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Thanks so much for taking the (considerable) trouble to explain this basic physics to the less-competent, like me. As an ex-physicist who has spent much of his life working with microwaves, the idea of a lower cut-off frequency is a familiar one. Does this mean that the energy of a standing-wave field in a waveguide at cut-off gives an effective mass to aphoton inside the waveguide?

Hi Robert,

The existence of a lower cut-off frequency in a cavity with microwaves is due to the geometry of the cavity — the length of the cavity sets the longest allowed wavelength for a standing wave that the cavity can support. The standing waves inside the cavity are still standard electromagnetic waves, which are of type “Class 0″ as defined by Prof. Stassler. For “Class 1″ waves, on the other hand, the cutoff is, in some sense, intrinsic (please compare the equations of motion of the two classes of waves written down by Prof. Strassler). For example, by making your cavity longer, you can decrease the cut-off. Had it been a “Class 1″ wave inside your cavity, this would not have been possible.

Hope this helps.

JB

Hi Matt,

Have you considered using Mathjax for the math formulas on your pages? It’s a simple WordPress plugin that lets you use inline Latex syntax.

I also vote for Mathjax or something similar

Matt,

Some clarification would be greatly appreciated here (!):

You show that a particle of the Z field gains mass due to S having a non-zero equilibrium value. To begin, however, we assume that quanta of the S field `already’ have non-zero mass (before any interaction whatsoever?). How do we explain this phenomenon? (Maybe I missed something absolutely crucial there… ) I imagine that quanta of the S field indeed would NOT have mass unless there was, to begin, some interaction between Z and S. Then, interaction of Z and S is indeed what gives quanta of BOTH fields their mass? This seams reasonable experimentally even… since if there was ONLY one field in existence, I couldn’t imagine a way to go about measuring the mass of the field quanta. So, who is to say what the mass is in this case, and what difference would it make if there were no interactions anyway? (Unless, the field could interact with itself? i.e. self-coupling?)

Furthermore, when we add a term like y^2 S Z^2 to the right side of the S field equation, this seems to me to effectively immediately alter the mass of the S field quanta. ie. L ~ (m_S)^2 S + y^2 S Z^2 = [(m_S)^2 + y^2 Z^2]S …

(where `L’ is short for the `left-hand side of the equation’)

so now the S field quanta has effective mass [(m_S)^2 + y^2 Z^2] ?

* mass-SQUARED, sorry.

… quanta *have*… (sorry again.)

Is seems like good, but it’s based on a very useful eq., hv (nu) = mc2, what can do mass for everything, he-he, because photon is massles, and the commonly used E (or H in some cases) = mc2 (what means the T+V). Funny. But this term anybody can use, maybe before the Higgs-term born, so I think it’s wouldn’t enough to understand “how the Higgs-field gives mass for particles” term.