Fields That ClickEpisode 3 / Mass is the field's brake ── range is ℏ/mc

In Episode 2 we said "a massless field reaches infinity at c." So then ── what happens when the carrier is heavy?

Mass is the field's brake A field with a heavy carrier runs out of breath as it runs and fades away. Its reach = range is
a single length set by the mass m: \(\lambda=\hbar/mc\). Whether a force gets through or not can be read from the ratio of range to distance alone.

Tools you need: the exponential function e⁻ˣ, plus Episode 2's "add mass → slower and shorter" This episode's ratio: r / λ (distance ÷ range), λ = ℏ/mc

In Episode 2 we said that a field whose carrier has zero mass (the photon) has its speed pinned to the ceiling \(c\) and, moreover, reaches the infinite beyond. That's why we can see the light of galaxies 10 billion light-years away, and why radio waves punch through the cosmos. So conversely, what happens to a field whose carrier is heavy? The answer is beautifully simple ── a heavy field, trying to go far, runs out of strength partway and vanishes. The distance it reaches (the range) is shorter the larger the mass. That range is one length set by the mass \(m\) alone: \(\lambda=\hbar/mc\). This episode reads off "why electromagnetism reaches the edge of the universe while the force that binds the atomic nucleus does not seep outside the nucleus" using just one dimensionless quantity ── the ratio of range to distance \(r/\lambda\).

01A massless field reached infinity ── so what about heavy?

Episode 2's photon (zero mass) has the ratio \(\sqrt{k/m}\) hit its ceiling, runs at \(c\), and reaches anywhere without decaying. This was the truth behind electromagnetism's "to the edge of the universe." But in nature there are also forces whose carrier is heavy. The carriers of the weak force, the W and Z particles, are 80–90 times heavier than the proton. Then the way the force gets through changes dramatically. First, let's see why mass becomes a "brake" through the metaphor of a loan.

02A loan has a time limit ── uncertainty sets the range

Let's borrow ahead from Episode 6's "force is the exchange of carriers." Two particles exert force on each other by playing catch with carriers (field quanta). But if a carrier has mass \(m\), producing it costs energy \(mc^2\). There isn't that much energy on hand ── so it borrows from nature for just an instant. Quantum mechanics' uncertainty principle permits this loan. But there is a limit on how long you can borrow.

Let's try it ── the loan's time × the distance you can cover in it

The time you can borrow energy \(\Delta E\approx mc^2\) (uncertainty \(\Delta E\,\Delta t\approx\hbar\))

$$\Delta t \approx \frac{\hbar}{\Delta E}=\frac{\hbar}{mc^2}$$

The distance the carrier can cover in that time at top speed (light speed c)

$$\lambda \approx c\,\Delta t = c\cdot\frac{\hbar}{mc^2}=\frac{\hbar}{mc}$$

The maximum distance the carrier can deliver within the loan's due date \(\Delta t\) is \(\lambda=\hbar/mc\). This is the force's range. Mass \(m\) is in the denominator ── heavier means a shorter loan term and a shorter range. Conversely, if \(m\to0\) then \(\lambda\to\infty\), the term is infinite, and it reaches anywhere. The photon is exactly that.

This episode's length ── range (the Compton wavelength)
$$\lambda=\frac{\hbar}{mc}$$

The one length set by the mass \(m\). The "braking power" mass gives a field, translated into distance. \(m=0\Rightarrow\lambda=\infty\) (infinite range, light speed). Large \(m\Rightarrow\) small \(\lambda\) (short range).

03The Yukawa potential ── Coulomb with a "vanishing" attached

How does the range \(\lambda\) enter the shape of the force? The massless electromagnetic force (Coulomb) merely thins in inverse proportion to distance ── \(V\propto 1/r\) (the truth of the \(1/r^2\) thinning is Episode 5). In a field with mass, this gets multiplied by an exponential brake \(e^{-r/\lambda}\). This is the form Hideki Yukawa wrote in 1935.

Coulomb (massless) and Yukawa (with mass)
$$V_{\text{Coulomb}}(r)\propto\frac{1}{r}\qquad\Longrightarrow\qquad V_{\text{Yukawa}}(r)\propto\frac{e^{-r/\lambda}}{r}$$

The difference is just the one term \(e^{-r/\lambda}\). When the distance \(r\) exceeds the range \(\lambda\), this factor drops rapidly toward zero ── the force vanishes. What matters is, again, the ratio \(r/\lambda\).

The reading is a single ratio, \(r/\lambda\). If \(r\ll\lambda\) (much closer than the range), then \(e^{-r/\lambda}\approx1\), and it's indistinguishable from Coulomb. If \(r\gg\lambda\) (farther than the range), then \(e^{-r/\lambda}\approx0\), and the force is effectively zero. So a force with mass "works inside the range \(\lambda\), vanishes outside it." The boundary is always \(r/\lambda=1\).

Seen in the field equation ── mass is a "brake term" A massless field obeys \(\nabla^2\phi=0\), whose solution is \(1/r\). A field with mass obeys \(\big(\nabla^2-\tfrac{1}{\lambda^2}\big)\phi=0\), whose solution is \(e^{-r/\lambda}/r\). The added \(-\,1/\lambda^2\) term is exactly the brake mass adds to the field. Since \(1/\lambda^2=(mc/\hbar)^2\), the larger the mass, the stronger the brake, and the faster the field decays. This is the truth of the "shorter" that Episode 2 foreshadowed when it said "a field given mass becomes slower and shorter."

04Let's play with it ── make the carrier heavy and the force stops reaching

The figure below plots the force strength \(V(r)\) against distance \(r\). Blue is the massless Coulomb (\(1/r\), reference), plum is the massive Yukawa (\(e^{-r/\lambda}/r\)). Try raising the carrier's mass \(mc^2\) (in MeV) with the slider.

At zero mass (far left), the two curves overlap exactly and the force reaches the right edge = far away. The more you raise the mass, the more the range \(\lambda=\hbar/mc\) shrinks, and the plum curve drops to nothing at short distance. The vertical dotted line marks the range \(\lambda\) ── you can see the force wither once you cross it. Also confirm that at short distance it still overlaps with Coulomb (for \(r\ll\lambda\) they're indistinguishable).

Figure: force strength V(r). Blue = Coulomb (1/r, massless), plum = Yukawa (e^(−r/λ)/r, with mass). Raise the carrier's mass mc²[MeV] and the range λ=ℏ/mc shrinks, the force vanishing at short distance. The vertical dotted line is the range λ
Coulomb 1/r (massless, infinite range) Yukawa e^(−r/λ)/r (with mass)

05Hitting the numbers ── from the range, read the carrier's mass

This \(\lambda=\hbar/mc\) is not just a pretty formula. From the actual range of a force, you can predict the mass of a still-unseen carrier. That is exactly what Yukawa did. The range of the nuclear force (which binds protons and neutrons) is, from experiment, about \(1.4\) fm (femtometers, \(10^{-15}\) m). Using the handy conversion \(\hbar c\approx197\) MeV·fm ──

Let's try it ── the nuclear force's range → the carrier's mass $$mc^2=\frac{\hbar c}{\lambda}\approx\frac{197\ \text{MeV·fm}}{1.4\ \text{fm}}\approx 140\ \text{MeV}$$

"There should be an unknown particle of mass about \(140\) MeV carrying the nuclear force" ── Yukawa predicted this in 1935 from the range alone. Twelve years later, a pion of mass \(140\) MeV was really found in cosmic rays. Working backward from the ratio called range dug up a new particle.

The same formula also reads off the "weakness" of the weak force. Its carrier, the W particle, has \(mc^2\approx80{,}400\) MeV. The range is \(\lambda=197/80400\approx0.0025\) fm ── less than 1/300 of the proton (about 0.8 fm). The weak force is "weak" not because the force's essence is weak, but because the carrier is too heavy and the range is nearly zero. Seen at everyday scales \(r\), \(r/\lambda\) is astronomically large, and \(e^{-r/\lambda}\) is utterly dead. The reason we hardly feel the weak force is this ratio.

The bridge onward ── where does mass itself come from? Up to here we used the mass \(m\) as a "given number." But where does \(m\) itself come from? The masses of the carriers (W, Z), electrons, and quarks are set by the strength of their coupling to the Higgs field that fills the vacuum (Episode 6). And that "strength of coupling" too is a relative of the unitless coupling constant handled next episode (Episode 4). The \(m\) that sets the range, the \(\alpha\) that sets the force's strength ── the numbers characterizing a field all end up, in the end, at dimensionless ratios.
◇ ◇ ◇
The honest line ── how true are "the loan" and "the range"?

That the range \(\lambda=\hbar/mc\) (the Compton wavelength) and the Yukawa potential \(V\propto e^{-r/\lambda}/r\) are the static solution of a field with mass (the Klein–Gordon field) is established physics. That the pion's mass was predicted from the nuclear force's range and hit the mark, and that the weak force is short-range because the W is heavy, are also standard physics.

However, "playing catch with virtual particles by borrowing energy through uncertainty" is a powerful mnemonic for estimating the range in your head, not a rigorous picture. Virtual particles are not real little spheres but a computational device (the propagator), and \(\Delta E\,\Delta t\approx\hbar\) is not itself the rigorous time–energy inequality. Precisely: "the solution of a field equation with a mass term \(1/\lambda^2\) is \(e^{-r/\lambda}/r\)" ── that is the body of the brake. One more point: short range does not mean the force is slow. Range is about how far it reaches; the ceiling on the speed at which change (news) travels is \(c\), as in Episode 1. Don't confuse "short range" with "low speed."

Practice problems (solvable with only this episode's formulas. ℏc ≈ 197 MeV·fm)
  1. If the carrier's mass is doubled, by what factor does the range \(\lambda=\hbar/mc\) change?
    See the answer
    Since \(\lambda\propto 1/m\), it is \(1/2\). Double the mass, halve the range. Mass is the range's "reciprocal brake."
  2. A certain force had a range \(\lambda=2.0\) fm. How many MeV is the carrier's mass \(mc^2\)?
    See the answer
    \(mc^2=\hbar c/\lambda=197/2.0\approx99\) MeV. Just by measuring the range, you know the carrier's mass (Yukawa's back-calculation).
  3. At distance \(r=3\lambda\) (three times the range), roughly what is the Yukawa factor \(e^{-r/\lambda}\)? What has happened to the force?
    See the answer
    \(e^{-3}\approx0.05\). At three times the range it's already only about 5%. Once \(r/\lambda\) exceeds 1, the force withers rapidly. At a few times the range, it's effectively zero.
  4. The photon has zero mass. What happens to the range \(\lambda=\hbar/mc\) and the Yukawa factor \(e^{-r/\lambda}\), and what does the force's shape become?
    See the answer
    At \(m\to0\), \(\lambda\to\infty\), \(e^{-r/\lambda}\to e^{0}=1\). The brake vanishes, and the force becomes \(V\propto1/r\) (Coulomb) = infinite range. That's why electromagnetism reaches the edge of the universe.

Episode 3 summaryMass is a brake, range is ℏ/mc ── what matters is the ratio r/λ

When a carrier has mass \(m\), the field runs out of breath far away and vanishes. The distance it reaches = range is a length set by mass alone, \(\lambda=\hbar/mc\) (the Compton wavelength). By uncertainty, the time you can borrow \(mc^2\) is \(\Delta t\approx\hbar/mc^2\), and the distance covered in that time at light speed is \(\lambda\). Because mass is in the denominator, heavier means shorter range, and zero mass means infinite range (= photon, electromagnetism). The force's shape is Yukawa \(e^{-r/\lambda}/r\), Coulomb \(1/r\) with an exponential brake attached, and in the field equation the brake term \(1/\lambda^2=(mc/\hbar)^2\) is its truth.

What decides the effect is always the ratio \(r/\lambda\). For \(r\ll\lambda\) it's indistinguishable from Coulomb; for \(r\gg\lambda\) the force withers. Back-calculation from the range, \(mc^2=\hbar c/\lambda\), predicted the pion (140 MeV) from the nuclear force's \(1.4\) fm and hit the mark. Because the W is heavy (80 GeV), the weak force is ultra-short-range ── the truth of "weak" was the hugeness of the ratio \(r/\lambda\). The unit-bearing \(m,\,r,\,\lambda\) are stage sets; what decides how a force gets through is the dimensionless ratio \(r/\lambda\). The third step in reading a field through ratios.

This document is Episode 3 of the "Fields That Click" series, a reading piece for high-schoolers and undergrads who love physics. That the static solution of a field with mass (the Klein–Gordon field) is the Yukawa potential \(V(r)\propto e^{-r/\lambda}/r\), with the range given by the Compton wavelength \(\lambda=\hbar/mc\); that the field equation takes the form \((\nabla^2-\lambda^{-2})\phi=0\) with the mass term \(\lambda^{-2}=(mc/\hbar)^2\) producing the decay; that Yukawa predicted the pion's mass (about 140 MeV) from the nuclear force's range (about 1.4 fm) and it was confirmed by experiment; and that the large mass of the W boson (about 80.4 GeV) yields the short range of the weak interaction, are all established standard physics (\(\hbar c\approx197.3\) MeV·fm). "Borrowing a virtual particle via uncertainty (\(\Delta E\,\Delta t\approx\hbar\))" is a heuristic explanation for estimating the range; virtual particles are internal lines (propagators) in perturbative calculation and differ from real particles. Range is the reach of a force, not a signal propagation speed; the ceiling on the speed of propagating change is \(c\) (Episode 1). The figure is a schematic comparison of the static Yukawa/Coulomb potential shapes, with coefficients normalized for display convenience. ── To print, use your browser's "Print" → "Save as PDF" (in the print version the sliders and answers are frozen or hidden). Adjacent episodes: Episode 2 Why c / Contents.

Print / make PDF: ⌘+P (Ctrl+P on Windows). On screen, the carrier's mass slider changes the range λ=ℏ/mc and how the force gets through. "See the answer" opens the solutions.