# The Standard Model More Deeply: Masses, Lifetimes and Forces

Today’s post is for readers with a little science/math background:

Last week, I explained, without technicalities, how the various elementary forces of nature can be inferred from the pattern of lifetimes of the known particles.  I did this using an image, repeated below, that organized the particles by their masses and lifetimes.  I’ll add more non-technical posts on the Standard Model in the coming days. But today’s post is a tad more technical, using dimensional analysis (a physicist’s secret weapon) (which I demonstrated here, here and here) to explain key features of the image: the red line, the blue line, and the particles at the upper left, as well as why there is a high-energy and a low-energy version of the weak nuclear force. Figure 1:  An assortment of the known particles particles clustered into classes according to the “force” that causes them to decay. See this recent post for details.

In the figure above, the red dotted line indicates the minimum possible lifetime for a particle of a given mass.  If I told you there might be such a minimum, what would you guess it to be? The answer has to be consistent with dimensional analysis. Without relativity and quantum mechanics, there’s no such formula. But relativity gives me the cosmic speed limit (aka speed of light) c, and quantum physics gives me Planck’s constant ħ.  We saw in this post that ħ has dimensions of mass*length2/time, and so ħ/m has dimensions of length2/time. Dividing by c2 gives us something with dimensions of time. Consequently dimensional analysis suggests that a natural time scale for a particle of mass m is

• T = # ħ/(mc2)

where # is an unknown number that we expect to be not too, too far from 1.

I like to call this timescale a particle’s “heartbeat”; the more mass a particle has, the faster is its pulse (in contrast to mammals!)  A particle that has a lifetime shorter than its heartbeat really doesn’t ever come into existence in the first place.  That’s why, in Figure 1 the red dotted line for the minimum lifetime is set by this formula, with # taken to be 1

However, Figure 1 shows that many, even most, of the known particles, both elementary and composite (i.e. not elementary!), have longer lifetimes than T.  Has dimensional analysis failed us? No. But we do need to improve it, using some physics knowledge and intuition.

### The Strengths of Forces

Each of the elementary forces of nature has a characteristic strength.  Gravity’s is set by Newton’s constant G; the Newtonian gravitational force between two objects, of mass M and m and separated by a distance r, is GMm/r2. If you set G to zero, there’s no force at all. So any formula for a gravitational effect has to have a G in it, to assure the effect would disappear if G were set to zero.

Coulomb’s constant k plays a similar role in electromagnetism.  In the 20th century, particle physicists learned that it’s more useful to move the dimensions around a little, and describe most forces other than gravity by a dimensionless number.  Electromagnetism is better described not by k, which has dimensions, but by a simple number α, which arises from the combination

•  α = k e2 / (ħ c) = 1/137.04… = 0.0073…

This α appears in all electromagnetic processes — emission of light, the pull of a magnet, the decay of positronium — and if α were set to zero, all of these processes would cease.

Similarly, for the weak nuclear force there is a corresponding quantity, often called αw (“w” stands for “weak”) or α2 (where “2” stands for “SU(2)”; if you don’t know that that means, don’t worry, as I’ll explain it in a future post in this series.) The strength of the weak nuclear force is determined by αw . We may also write the strength in terms of a quantity gw which is related to αw by

• αw = gw2/(4π)

and sometimes physicists call gw by the names “g2” or simply “g”. [Professionals use all of these notations somewhat interchangeably; in fact if you’ve been reading my series on the Standard Model you’ve seen this. Sorry to say, there are no completely standard notations in science; one has to be a bit light on one’s feet and pay attention to context.]

It turns out, experimentally, that αw is approximately 0.034, or about 1/30.  Notice: though this is still a small number, it is much larger than α  for electromagnetism! Despite its name, the high-energy weak nuclear force is not intrinsically weak, just as the figure above suggests.  It’s called “weak” because of its low-energy manifestation, which we’ll understand before the end of this post.

### The High-Energy Weak Nuclear Force

Just as all gravitational effects must go away if G goes to zero, and all electromagnetic effects must vanish if α goes to zero, all weak nuclear effects must vanish if αw = 0. Therefore, since W bosons decay through the action of the weak nuclear force, the lifetime TW of the W boson must become infinite if the weak nuclear force is switched off — i.e., if αw goes to zero.  If we simply assumed the W lifetime should be the minimum possible one, so that TW = # ħ/(mWc2), this would not be true! The lifetime would not depend on αw at all!

Instead, we need an improved dimensional analysis formula for the lifetime, in which αw appears. In fact αw must appear in the denominator, in order that TW → infinity as αw → 0.  The simplest such formula is

• TW = # ħ/(αw mW c2

This is an excellent guess!  The complete answer in the Standard Model agrees with this, with # = 8√2/9, not far from 1, as usual.  [Later we’ll come back and learn something from the “9”.]  Our improved dimensional analysis gives a prediction for the W boson lifetime that agrees with experiment. A similar calculation, prediction and measurement applies for the Z boson, which has a similar lifetime.  So the W and Z boson lie slightly to the right of the red line largely because of the factor of 30 from 1/αw , which makes their lifetimes tens of “heartbeats” long.

This brings us to a subtle point; why, you might ask, did I use 1/αw and not 1/gw in my dimensional estimate? Both are dimensionless and the formula would make dimensional sense with either one.

The answer lies in the details of how quantum physics works.

• A particle’s lifetime is inversely related to its decay rate; the faster its decay rate, the shorter its lifetime.
• A decay rate tells us the probability that a particle will decay during the following second.
• In quantum physics all probabilities involve squaring a “probability amplitude”.

For example, sometimes a W will decay to a charm quark c and a strange anti-quark s. We can depict this diagrammatically as in Figure 2 (left). [This is a first glimpse of something like a Feynman diagram, though I won’t overemphasize that connection, since Feynman diagrams are increasingly obsolete.]  Roughly speaking,

• the diagram in Figure 2 represents the probability amplitude for this W decay, and quantum field theory tells us that this amplitude is proportional to gw;
• then we have to square the amplitude, Figure 2 (right), giving an αw in the W’s decay rate;
• and finally, this gives us the 1/αw in the W’s lifetime that we saw in our estimate for TW. Figure 2: (Left) a diagram describing the probability amplitude associated to the interaction of the W boson, strange quark and charm quark; this amplitude has value #gw. (Right): the decay rate involves the square of the amplitude, which is proportional to αw = gw2/(4π) , giving 1/αw in the W lifetime.

The W boson can decay to other things too, such as a muon and a corresponding anti-neutrino.  This can occur through the diagram in Figure 3.  But a crucial feature of the weak nuclear force, also true of the strong nuclear force and the electromagnetic force, is that it has certain universal properties. This is a point we’ll come back to in future posts, but one consequence is that the strengths of the W-c-s and W-μ-ν interactions are identical!  We get gw for the W → μ+ν probability amplitude, and the same estimate for the decay rate as for W→c+s; only the # may be different. Figure 3: as in Figure 2, for the W, muon and corresponding neutrino.

When the dust settles, the W→c+s decay rate, the W → μ+ν decay rate, and all the W’s other decay rates get similar estimates. Adding them all together to get the total decay rate leaves our original lifetime estimate unchanged; all the different ways that a W can decay are swept into the unknown # that appears in our estimate.

Meanwhile, the top quark (t) decays to a bottom quark (b) and a W boson via the probability amplitude in Figure 4, so a natural guess for the lifetime would again be the analogous formula

Tt = # ħ/(αw mt c2)

It’s a good estimate, but let me just warn you that it’s a bit more subtle than it looks, a point I may come back to in a future post. Figure 4: as in Figure 2, for the W, top and bottom quarks, allow t decay to W+b

### The Higgs Force and Higgs Boson Decay

What about the Higgs boson’s lifetime? Since the Higgs field’s interaction with other known elementary particles gives them their masses, the Higgs boson’s interactions are largest for the known elementary particles with the largest masses.  For this reason, the Higgs boson generally will decay to the heaviest particles available, subject to a principle.  This is a principle that all decays must obey, which I call

• the rule of decreasing rest mass: a decaying particle’s rest mass must exceed the sum of the rest masses of the particles to which it decays

This is explained as rule #2 in this post about rules that govern particle decays.

The W boson’s mass is 80.4 GeV/c2. If the Higgs boson’s mass had been just over twice that, then most Higgs bosons would have decayed to two W bosons, through the weak nuclear force, and then our estimate for its lifetime would have been similar to those for the W, Z and top. [Again, as for the top quark, this statement is a bit naive; I’ll try to come back to it.] But with its mass only 125 GeV/c2, it can’t decay to two W’s, because of the rule of decreasing rest mass. Instead, it has to decay mostly to the next-lightest particle — to bottom quark/anti-quark pairs. Figure 5: as in Figure 2, for the Higgs and bottom quarks.

The amplitude for a Higgs boson to decay to a bottom quark/anti-quark pair is shown in Figure 4, and it is proportional something we will call yb , the “Yukawa coupling” of the bottom quark (named after the great physicist Hideki Yukawa). Because the Higgs particle is a ripple in the Higgs field, the bottom quark’s interaction with the Higgs boson (a.k.a. Higgs particle) has to be of the same strength as its interaction with the Higgs field. The latter determines the the bottom quark’s mass, and so this Yukawa coupling is related to the bottom quark’s mass, as follows:

• yb = (mb/v) (c2/√2)

where

• yb is the bottom quark Yukawa coupling
• mb is the bottom quark’s rest mass, approximately 5 GeV/c2
• v is the “value” (or “vacuum expectation value”) of the Higgs field throughout the universe, approximately 246 GeV.

So yb is approximately 0.02, quite small.

The quantity yb is analogous to gw – it appears in a probability amplitude – so the quantity which appears in a decay rate or lifetime is

• αb = yb2/(4π) = 0.00002

That is really small!  And so the Higgs lifetime

• TH = # ħ/(αb mH c2)

is tens of thousands of times longer than its minimum possible lifetime. [In fact this prediction was just recently verified, approximately, for the first time, using a clever indirect method invented in 2013 by these people.]

Thus, despite having a similar mass to the W and Z, the Higgs boson’s lifetime is notably longer — tens of thousands of heartbeats — because its decay is controlled by its interaction with the bottom quark, which we know is small, since the bottom quark has a much lower mass compared to the top quark, W or Z.

### The Low-Energy Weak Interaction

Finally, let’s explain the blue dashed line. What happens when low-mass particles try to decay via the weak nuclear force? How does the “Low-Energy Weak Nuclear Force” in Figure 1 differ from the High-Energy version we’ve already discussed?

Why, for instance, doesn’t the charm quark decay to a strange quark and a W boson via Figure 2, just as the W boson decays to a charm quark and strange anti-quark? It cannot do so, because such a decay would violate the rule of decreasing rest mass. The charm quark’s mass is only about 1.5 GeV/c2, less than 2% of a W’s mass. And so the high-energy weak nuclear force process that allows the W boson and top quark to decay to two particles of lower mass is not available to a charm quark.

Instead what it does is decay to three particles, such as a strange quark, a muon and an anti-neutrino.  How does this process occur?

Through the same diagram of Figure 2, a charm quark can decay to a strange quark and a low-energy disturbance in the W field, often called a “virtual W particle” —  but it is not a particle.  (For instance, unlike a true particle, a virtual particle can have any mass, even an imaginary mass, and it can have zero energy.)  This part of the probability amplitude for a charm quark’s decay comes with a factor of gw, as before.

The disturbance in the W field can only exist for a brief instant; it is immediately transformed into something else, such as a muon and an anti-neutrino (of muon type).  This process, too, requires the weak nuclear force, via Figure 3.  This will give the probability amplitude a second factor of gw.

We can draw a whole diagram for this, joining Figure 2 and 3 by a line, which represents the W disturbance.  But there is a price for this disturbance, whose energy is somewhere around the charm quark’s mass-energy mcc2, far below the mass of the W field’s particle, mW. It turns out the price to be paid in the probability amplitude is roughly mc2/mW2 .  Why? Figure 6: Sewing together Figures 2 and 3 by a “virtual W” (a non-particle-like disturbance in the W field) allows the charm quark to decay; this is the low-energy manifestation of the weak nuclear force. The probability amplitude behaves as gw2(mc2/mW2).
• If the W mass were infinite, then it would be as though there were no W at all, and this process could not occur. Therefore the probability amplitude must go to zero as mW becomes infinite, which requires we put a power of 1/mW in the probability amplitude.
• But since 1/mW has dimensions of 1/mass, if we include it in our formulas then there must be something with dimensions of mass that accompanies it; otherwise our formulas won’t be dimensionally consistent. Although any of the masses of the particles in the decay could appear, the largest is mc, so the most optimistic estimate – giving the fastest possible decay and shortest lifetime – is mc/mW.
• Finally, some inside information: in a quantum field theory, one never sees a boson’s mass, only its mass-squared.  (Einstein says: E2=m2c4 for bosons.) Therefore mc/mW cannot appear in a probability amplitude; only mc2/mW2  will do.

Our estimate for the probability amplitude is then gw2(mc2/mW2). To get a probability we must square this, giving αw2(mc2/mW2)2, to which the decay rate is proportional. Then to get the charm quark lifetime, we invert it, giving us the estimate

• Tc = # (mW2/mc2)2 ħ/(αW2 mc c2) = # ħ mW4 /(αw2 mc5 c2)

which, at about a trillionth of a second, is roughly 10 billion times longer than its minimum possible lifetime!

Notice the factor of mW4 /(αw2mc5) that appears here, to be compared with the W boson’s lifetime where what appears is 1/(αwmW). This turns out to be the key distinction between the High-Energy and Low-Energy versions of the weak nuclear force in particle decays:

• First, the weak nuclear force has to act only once in the high-energy case, whereas here it has to act twice; hence the 1/(αw2).
• Second, the need for a virtual W that carries such low energy comes with a huge price, lengthening the lifetime by mW4 /mc4. Figure 7: as in Figure 6, but for muon decay. The two neutrinos are of different types, a detail that plays no role in this post.

We can apply the same type of argument to other particles. For instance, a muon μ decays to a muon neutrino and a disturbance in the W field, which in turn is converted to an electron and an electron anti-neutrino, diagrammed in Figure 7. The argument and the resulting formula looks the same

• Tμ = # ħ mW4 /(αW2 mμ5 c2)

though the # will be different.  Since the muon has a mass about 15 times smaller than the charm quark, its lifetime is longer by roughly 155 , nearly a million times. Our estimate agrees with experiment, which tells us the muon’s lifetime is about 2 millionths of a second.

More generally, for any particle x with mass well below the W’s mass, we expect

• Tx = # ħ mW4 /(αW2 mx5 c2)

This formula (for a fixed choice of #) gives the dashed blue line in Figure 1. Since the # in front varies from one particle to the next, we expect that particles decaying via the low-energy weak interaction will lie close to, but not necessarily on, this line.

[Finally, for the neutron n decaying to a proton p, an electron e, and an anti-neutrino, if we replace mn with mn-mp-me (the anti-neutrino’s mass is tiny and can be ignored), we get an answer which is in agreement with data.  But to show this is the correct thing to do requires more effort than I will make here.]

As you can see, the further you go, the more subtle an improved dimensional analysis can become — but by the time you’re done reasoning carefully, it almost always works!  Very rarely is the result of a calculation truly mysterious. This is even clearer in some final remarks below; but if you’re tired, you can safely skip these. They take us a bit deeper into the weeds, and I’ve already explained the most important aspects of Figure 1.

### Final Brief Comments on Exceptions and Electromagnetic Forces

If you’re still reading, let me point out a few issues that I’ve swept under the rug til now.

First, we saw that the neutron’s dot ended up an outlier because there of the tiny mass difference between the initial neutron and the final proton, electron and neutrino. The effect was to greatly lengthen the neutron’s lifetime, and thus shift the neutron’s dot far to the right of the dashed blue line. A few other particles’ dots suffer the same fate: the D* in the purple oval, the KL0 in the blue oval, and (less obviously from the way I’ve drawn things, but see below) the Σ0 (“Sigma-zero”) in the orange oval are all shifted rightward by having small mass differences in their decays. When you account for these mass differences, then dimensional analysis works again: just as the neutron ends up close to the blue line, so does the KL0 , while the D* ends up close to the red line.

To understand the Σ0 we need to say a bit more about the particles in Figure 1 that decay electromagnetically; each one is an exception of sorts, and requires its own unique improvement to the simplest dimensional analysis. Recall the electromagnetic force strength is given by α; let’s call it αem to avoid any confusion. The naive guess would be that all lifetimes of particles that decay electromagnetically should be roughly 1/αem ~ 137 times to the right of the red line. But in fact the lifetimes are longer.  To show this, in Figure 8 I’ve redrawn Figure 1, adding dotted orange lines where particles would be expected to sit if their lifetimes are extended by 1/αem, 1/αem2, 1/αem3, 1/αem4 and 1/αem5. Nothing decaying in this way actually sits on the 1/αem line. That’s in contrast to the weak nuclear force; the t, W and Z do sit about 1/αW to the right of the red line. What happened? In each case there’s a reason.

• The Σ0 decays to a single photon plus a particle called a Lambda (Λ), so its lifetime is enhanced by 1/αem, but as mentioned above its lifetime is further enhanced by the small mass difference between Σ0 and Λ;
• The eta η and pion π each decay to two photons, which means the electromagnetic force has to act twice; we get a 1/αem for each photon, and so their lifetimes are enhanced by 1/αem2, roughly 20000. But that’s still not quite enough; these decays only can occur through a quantum effect, which further enhances the lifetime by about 100. [A similar quantum effect reduces the probability that Higgs bosons decay to two photons; more on that in a later post.]
• The positronium atom (ee), made from an electron and a positron, also decays to two photons. Decay occurs when the electron and positron meet and annihilate. Just as for the η and π, a decay to two photons enhances the lifetime by 1/αem2. But in contrast to η and π, whose size is set by the strong nuclear force, the radius of positronium, like that of hydrogen, is proportional to 1/αem, and so the volume of positronium is proportional to 1/αem3. The bigger the positronium atom, the less likely the electron and positron are to run into each other, meaning they are less likely to annihilate, and so the lifetime is enhanced by a factor of the volume! Combining all the factors gives us 1/αem5; notice the positronium dot does lie near that line.

Finally, a last detail: the states ηc and J/ψ are, like positronium, little atoms of a charm quark and a charm anti-quark, while ηb and Υ are similarly made from a bottom quark/anti-quark pair. Their lifetimes are enhanced by a combination of effects, which we can discuss in the comments if someone wants to ask. Figure 8: As in Figure 1, with some details added: green arrows indicate some particles whose lifetimes are notably lengthened by a small mass difference between the decaying particle and their decay products; orange dashed lines show expectations for lifetimes enhanced by up to five powers of 1/αem.

### 23 responses to “The Standard Model More Deeply: Masses, Lifetimes and Forces”

1. hansvanleunen

A huge difference exists between the Standard Model of experimental particle physicists and the theoretical Standard Model that the promoters of QFT, QED, and QCD put forward. These theories lack a proper foundation and therefore mislead the crowd. The theories do not explain why the shortlist of electric charges exists. They do not explain the origin of gravity. and they do not explain the existence of the dynamic universe field and its relation to the electrical fields. Please take a look at [Editors’ note: link removed, I do not allow commenters to advertise their personal theories of the universe on this site; that’s what academic journals are for. Repeat offenders will be blocked.]

2. Angelo Dimitris

Great article, once again, but this one in particular because you gave us a sense of how a physicist’s “intuition” is applied with the “knowledge” to create a theory that is consistent with experiments.

“Decay”. Very difficult for me, anyway, to visualize what is going on when a particle decays. I once saw an interview with Leonard Susskind and he used his thumb and middle finger to repeatedly touch and open, and he said it could come down to a “string” that opens and closes. I guess he was referring to the most fundamental particle and hence the fundamental force that all other forces are derived from, the unified theory.

My question is, in this opening and closing of strings, (most likely clouds, like the electron), could the “minimum” angular momentum be the threshold that creates and decays (“evaporites”) particles in and out of existence?

Is Susskind talking about an aether? I could visualize a vortex that gets trapped, i.e. its angular momentum cannot reduce below the threshold that will decay it, and vaporizes it.

BTW, I can’t wait to see what James Webb Space Telescope has in store for us. today at 5 pm we’ll get a teaser by President Biden with the first photo!

3. Joseph

the rule of decreasing rest mass: a decaying particle’s rest mass must exceed the sum of the rest masses of the particles to which it decays

Is this because (maybe I missed it in the article) mass is not really conserved, but energy is. So, the decaying particle must account for the energy of the rest mass of the new particles….plus any kinetic energy given to the particles ?

• Matt Strassler

Correct: or more precisely, **rest mass** is not conserved; it always decreases in a decay. Only energy, momentum, angular momentum, electric charge, and a couple of other things are conserved; rest mass is not.

4. Doug McDonald

I like how you show how you can get rough estimates of these numbers from dimensional analysis. I looked at the and agree that the calculations are correct and do in fact pick out the correct reasons that determine the experimental facts. BUT … at this point I will compare you to Sheldon Glashow. My comparison method makes you his equal … which is neither overall compliment or complaint. You and he are very very good at this sort of stuff. By stuff I mean cherry-picking.

In grad school (1966-1971) at Harvard (I was a chemist … my office was under the entrance foyer of Mallinckrodt) I sat in on Glashow’s first year grad quantum mechanics (not field theory) class. He absolutely loved to do exactly what you just did to field theory, to plain quantum mechanics. Things like “explaining” why in the Schrodinger equation the space derivative is second but time is first, or why we need a Lamb Shift.. Most of these explanations
were clearly motivated by what he knew about relativistic field theory (which, of course, was rudimentary because he was nowhere near finished inventing the Standard Model just quite yet.) Yet when we, his listeners, got home and thought … his explanations simply collapsed. There simply HAD to be some more pieces to the puzzle, like explaining why cherry-pick that particular hand-wavy argument (which of course CAN be done with what we know today).

I think this is, or is close to, the standard method of such, called “lies to children”. I think people should admit that. I certainly did when I used such methods in my QM classes … at least when I realized that’s what I was doing! (My students groaned when I said that … they just wanted algorithms to by rote solve each and every test question.)

• Matt Strassler

I understand your complaint, but it’s neither entirely fair nor entirely wrong. I don’t have a choice but to cherry pick to a degree; otherwise I just have to give up on explaining things. BUT! The methods I am using here are exactly the same methods I use when teaching graduate students: I teach an entire course on particle physics using almost nothing but dimensional analysis, which is appropriate when students haven’t learned quantum field theory yet. Moreover, these are exactly the methods that *I use* (and other professionals use) when trying to understand whether a new idea has any chance of explaining data, or when trying to decide whether a hard calculation is worth doing. Yes, it is true that, if you do not have professional experience, you cannot invent these methods, or apply them on your own to particle physics. You do need experience to know which things matter and which ones don’t. But you’ll see, again and again, that this type of reasoning works… and **just knowing this fact** is an important part of understanding both how physics is done and why it is so reliable.

• Matt Strassler

p.s. I was trained really well in these methods by Jeff Kuhn in sophomore year of college. After I did my first spontaneous back-of-the-envelope calculation, performed on my own in junior year, I felt I’d finally learned how to do physics. I still have that page of notes, framed.

5. Doug McDonald

You replied to me with “Moreover, these are exactly the methods that *I use* (and other professionals use) when trying to understand whether a new idea has any chance of explaining data, or when trying to decide whether a hard calculation is worth doing. ”

Exactly! and that’s exactly what readers of your pieces must realize! The key words are “any chance”! Then you start going through all the alternatives.
Its how science works! (No apology for four exclamation marks.) Its probably the absolute most important thing any reader of your blog needs to know.

6. Kudzu

I, for one, am interested in the effects extending the lifetimes of the ηc and J/ψ particles. I do not know if I will be able to comprehend the explanations, but the articles in this series have proven to be remarkably enlightening in regards to things I know but have not understood until now.

• Matt Strassler

Ok; in brief. A key point is that the strong interaction is not so incredibly strong for energies above about 1 GeV, and that’s important for charm and bottom quarks, which have mass above 1 GeV/c^2.

Now why is that relevant? Recall positronium. Its lifetime is long because electromagnetism is a weak force. First, the volume of positronium is large; then there is parapositronium (spin 0) which decays to two photons with a moderately long lifetime, and orthopositronium (spin 1) which decays to three photons with a somewhat longer lifetime.

Now, it’s not an accident that ηc and J/ψ are referred to as “charmonium” states; they are different states of a charm-anticharm atom. Because the strong interaction isn’t so strong, ηc is like parapositronium and J/ψ is like orthopositronium. Both have volume somewhat larger than, say, the proton or rho meson. Also, ηc decays to two gluons while J/ψ decays to three; not that one observes these gluons, but one can infer that’s fundamentally what’s happening. That’s why both objects have longer lifetimes than most hadrons, and why the J/ψ has the longer of the two — so long that, in fact, decay via a virtual photon to electron-positron competes with its three-gluon decay.

The same logic applies to the two bottomonium states I listed, and even more so, since the strong nuclear force is even a little weaker at that mass scale.

7. Mousa Jafar

Informative presentation that gives good grasping of the Standard Model and the features besides interrelationships between mass and lifetime of particles decay.

8. Jake

If you were to plot all of the sufficiently known non-elementary particles, would the two weak nuclear force clusters start to grow into each other? Or is that gap really there?

• Matt Strassler

The gap is really there; there are no known particles between about 11 GeV/c^2 and 80.4 GeV/c^2. (And no elementary ones between 5 and 80.4). In principle there could have been others in between, and then I could have drawn a curve connecting the two regimes, but for whatever reason, nature made a gap. No idea why it is there.

9. 4gravitons

As someone in the “attempting to make Feynman diagrams obsolete” subfield, I have to say that saying

“[This is a first glimpse of something like a Feynman diagram, though I won’t overemphasize that connection, since Feynman diagrams are increasingly obsolete.]”

in this context feels more than a little premature/overoptimistic. There are a lot of things one shouldn’t use Feynman diagrams for at this point, but they’re not bad for something like this.

• Matt Strassler

There’s far too much emphasis on Feynman diagrams in books that explain particle physics to non-experts, as though the diagrams actually describe the physics. Moreover, the diagrams in this post resemble but are not Feynman diagrams (since I’m leaving out a lot, such as arrows, vertex functions, correct propagators, etc.) So I’m quite clear about my purposes here.

10. John Duffield

Hi Matt. This is an interesting and thoughtful post involving considerable effort. However I’ve now reached the conclusion that the Standard Model is something of a “patchwork quilt”, with significant foundational issues. So much so, that it’s actually done harm to fundamental physics, which is rather moribund these days. I’d love to see people like yourself investigating these issues and trying to put things right.

11. zeynel

«If you set G to zero, there’s no force at all.»

Setting G to zero does not mean “there is no force” because G=(4pi^2)(R^3/T^2). Setting G to zero means setting R=0 which means there is no orbit. How can we claim there is no force when we set the radius of the orbit to zero?

• Matt Strassler

No, that’s not the way it works. The formula you used, G=(4pi^2)(R^3/T^2), assumed there is a circular orbit. If you *assume* a circular orbit of radius R, then as G becomes very small,the object travels more and more slowly and takes longer and longer to make the orbit, so T becomes larger and larger. Once G = 0, the object stops moving, and T is infinity.

However, **much more generally than any circular orbit**, if G is zero then all gravitational effects are zero too, no matter what the nature of the objects or how they are moving. There are no black holes, no solar systems, no planets, no gravitational waves, and no orbits of any kind.

12. Graham Dungworth

Re- the lack of particles between 11GeV/c^2 and 80.4 GeV/c^2
A chemist might ponder; noticing the value of vev,

Is the Higgs particle, 2H, a summation of (W+ W- and Z) ?
or a composite boson of 2 chiralities?

• Graham Dungworth

Yes

• Matt Strassler

If the Higgs boson were a composite object made of known particles, we would certainly have seen signs of this in data. It is possible that it is composite and made of new types of particles, but it must be quite small compared to its mass (unlike an atom or even a proton, which are large compared to their masses — the comparison is made using the uncertainty principle and dimensional analysis, as in https://profmattstrassler.com/2022/06/30/the-size-of-an-atom-how-scientists-first-guessed-its-about-quantum-physics/ .)