© Matt Strassler [September 22, 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
- d2Z/dt2 – c2 d2Z/dx2 = – (2 π νmin)2 (Z-Z0)
where Z(x,t) is the field, Z0 is its equilibrium value, x is space, t is time, d2Z/dt2 represents the change with time of the change with time of Z (and d2Z/dx2 similarly for space), c is the universal speed limit (often called “the speed of light”), and the quantity νmin 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 νmin is zero. The quanta of such fields have mass
- m = h νmin / c2
where h is Planck’s quantum mechanics constant. In other words,
- d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 m2 (Z-Z0)
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 equation, where the equilibrium values S0 and Z0 are zero, but there’s an additional term in the equations involving S(x,t) multiplied by Z(x,t)
- d2S/dt2 – c2 d2S/dx2 = – (2π c2/h)2 (mS2 S + y2 S Z2)
- d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 (mZ2 Z + y2 S2 Z)
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, mS, mZ) is a constant. 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 S0 and Z0 are extremely small. Since we’re currently assuming S0=0 and Z0=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 --- really small, much smaller than either a or b.] Since y isn’t big, the terms y2 S Z2 and y2 S2 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 y2 S Z2 and y2 S2 Z terms, the S and Z fields then both satisfy the simple Class 1 equations of motion we started with, from which we deduce that the S and Z “particles” have mass mS and mZ.
But now imagine a world in which Z0 is zero but S0 is not zero.
- d2S/dt2 – c2 d2S/dx2 = – (2π c2/h)2 (mS2 [S- S0] + y2 S Z2)
- d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 (mZ2 Z + y2 S2 Z)
Again, the S and Z fields are functions of space and time, but everything else, including S0, 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) = S0 + s(x,t)
where s is the variation of S away from its equilibrium value S0. 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 not that S(x,t) is small. Substituting the red equation above into the equations above for S and Z, and remembering that S0 is constant so dS0/dt = 0 and dS0/dx=0, we find the equations become
- d2s/dt2 – c2 d2s/dx2 = – (2π c2/h)2 (mS2 s + y2 [S0+s] Z2)
- d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 (mZ2 Z + y2 [S0+s]2 Z)
As before, if we want to know the mass 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 Z2 or sZ2 or sZ or s2Z. But we cannot ignore the term y2 [S0+s]2 Z in the second equation entirely, because it contains a term of the form y2 S02 Z, and that contains only one field. (Note this is not true of the y2 [S0+s] Z2 term in the first equation, every term of which has two factors of Z.) Consequently, although a quantum of the S field still satisfies the equation
- d2s/dt2 – c2 d2s/dx2 = – (2π c2/h)2 mS2 s
and therefore has mass mS, a quantum of the Z field satisfies a new equation:
- d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 [mZ2 + y2 S02] Z
Consequently the quanta of the Z field now have a new mass! not mZ but
- mZnew = [mZ2 + y2 S02]1/2
(recall that x1/2 means the same as √x.) Because of the simple interaction between the S and Z fields with strength y, when S has a non-zero equilibrium value S0, the mass of the Z quantum shifts by an amount related to y and S0.
And if for some reason the mass mZ of the Z quantum had been zero to start with, then we would say that the S field’s non-zero value has given mass to the Z quantum:
- mZold = 0 → mZnew = y S0
Note this mass is proportional both to the equilibrium value S0 and to the Yukawa parameter y; the larger is y, the larger is the resulting mass for the Z quantum.
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
- yσ2 [H(x,t)]2 Σ(x,t) ,
where yσ 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) = H0, shifts the minimum-frequency of Σ waves, and thus the mass of σ particles, from zero to something non-zero: mσ = yσ H0. 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 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) = H0, 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? 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.