One exponent decides everything on this page. Differentiate it and you have a microscope; integrate it over a Coulomb barrier and you have a star.

§5.2a ended with a formula and a warning: $T$ depends on the barrier through $e^{-2\beta a}$ , the prefactor cannot matter, and the exponent is the only number worth estimating. That is a statement about arithmetic.

This section is what happens when you put real constants into it. The same formula, evaluated twice — once for an electron crossing a vacuum gap, once for a proton crossing the electrostatic repulsion of another proton — lands on $2\beta a \approx 10$ and on an exponent of about 22. The first number built a microscope that sees individual atoms. The second is why the sun is still burning.

Physics you need first — work functions, kT as an energy, and what two protons cost each other

Three pieces of background the book assumes, none of them quantum:

A work function. Electrons inside a metal are not free; they sit in an attractive well formed by the positive ions, and the well has an edge at the surface. The energy needed to lift the most energetic electron over that edge and out is the work function, a few eV for most metals — the same quantity §1.1 measured with the photoelectric effect. In the language of §5.2a it is exactly $V_B - E$ : the height of the barrier above the energy the particle already has.

Temperature as an energy. A gas at temperature $T$ has particles with typical kinetic energy $\sim k_BT$ , where $k_B = 1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$ . This is a conversion, not a law: $10^7\ \mathrm{K}$ is $0.86\ \mathrm{keV}$ , and the useful landmark is that room temperature is $1/40\ \mathrm{eV}$ . When this page says the solar core sits at $1\ \mathrm{keV}$ , that is a temperature quoted in energy units.

What two protons cost each other. Two charges $+e$ a distance $r$ apart have potential energy $e^2/4\pi\epsilon_0 r$ — positive, so pushing them together costs energy, and it rises without limit as $r \to 0$ . The one number to carry: $e^2/4\pi\epsilon_0 = 1.44\ \mathrm{MeV\,fm}$ , so two protons one femtometre apart cost $1.44\ \mathrm{MeV}$ . Below about $2\ \mathrm{fm}$ the strong nuclear force takes over and the potential plunges; that competition is the whole shape of Fig. 5.5.

One quantum ingredient carries over: §5.2a for tunnelling , the penetration parameter $\beta$ , and Eq. (5.40).

Tunnelling electrons

Put two metal surfaces close together. Each holds electrons in an attractive well; between them is a vacuum gap the electrons have no classical business crossing. That is the barrier of Fig. 5.4 built out of real materials, and Eq. (5.40) applies to it unchanged — except that for a gap of any reasonable width $e^{-2\beta a}$ is very small indeed, so the prefactor may be dropped along with everything else that is not exponential:

T \simeq e^{-2\beta a}\quad\text{with}\quad \beta = \frac{\sqrt{2m_e(V_B - E)}}{\hbar}\tag{5.41}

Differentiating the only thing that matters

Because $a$ sits in an exponent, the fractional change in $T$ for a small change in the gap is a constant times $\Delta a$ :

\frac{\Delta T}{T} = -2\beta\,\Delta a\tag{5.42}

What this really says — take the log first, and the sensitivity is one line

Take the log of Eq. (5.41) before differentiating and the reason is obvious: $\ln T = -2\beta a$ , so $\mathrm d(\ln T) = -2\beta\,\mathrm da$ , and $\mathrm d(\ln T)$ is the fractional change $\Delta T/T$ . An exponential is the one function whose fractional response to an absolute change is the same everywhere — it does not matter whether the gap is $0.4\ \mathrm{nm}$ or $0.9\ \mathrm{nm}$ , closing it by $0.001\ \mathrm{nm}$ changes the current by the same 2%.

That constancy is what makes an instrument possible. A sensor whose sensitivity depended on where it happened to be would need calibrating at every point; this one has a single number, $2\beta$ , and it is set by the work function alone.

For a metal needing $4\ \mathrm{eV}$ to release an electron — so $V_B$ sits about $4\ \mathrm{eV}$ above $E$ :

\beta = \frac{\sqrt{2m_e(V_B - E)}}{\hbar} \approx 10^{10}\ \mathrm{m^{-1}}\tag{5.43}

Put a number on it — the 2% the book exclaims about

Step 1 — $\beta$ . $V_B - E = 4\ \mathrm{eV} = 6.409\times10^{-19}\ \mathrm{J}$ :

\beta = \frac{\sqrt{2(9.109\times10^{-31})(6.409\times10^{-19})}}{1.055\times10^{-34}} = 1.025\times10^{10}\ \mathrm{m^{-1}}

Step 2 — the sensitivity coefficient. $2\beta = 2.049\times10^{10}\ \mathrm{m^{-1}}$ , or more usefully $20.49$ per nanometre.

Step 3 — feed it a gap change. $\Delta a = 0.001\ \mathrm{nm} = 10^{-12}\ \mathrm{m}$ :

\frac{\Delta T}{T} = -2\beta\,\Delta a = -(2.049\times10^{10})(10^{-12}) = -0.0205

A 2% change in current for a gap change of $0.001\ \mathrm{nm}$ . Hold that number against the length scale it refers to: a carbon atom is about $0.14\ \mathrm{nm}$ across, so $0.001\ \mathrm{nm}$ is roughly one hundredth of an atomic diameter, and 2% is a signal any lock-in amplifier will find in its sleep.

Nothing else in physics converts a picometre into a percent. That single line is the scanning tunnelling microscope .

The instrument

tip at2.00 nm

Constant current: a feedback loop pushes the tip up and down to hold I fixed. Fixing I fixes the gap, and the gap is z − h(x), so the tip height is the surface profile — the plot is now a picture of the atoms rather than a graph of a current. This is what an STM image actually is, and it is why the vacancy at 1.4 nm reads as a hole rather than as a dip in a signal. Note that nothing was measured directly: the height is inferred entirely from Eq. (5.41).

work function (eV)4.0closest gap (nm)0.60

Engineer's bridge — a logarithmic sensor inside a servo loop

The exponential is a log amplifier you did not have to build. The measured quantity is current; the quantity of interest is a distance. Eq. (5.41) says $\ln I$ is linear in $a$ with slope $-2\beta$ , so the instrument’s front end is a perfect logarithmic transducer — and the reason it can resolve a picometre is the same reason a log amp gives you 120 dB of usable range: the fractional resolution is constant. Small-signal analysis, gain about an operating point, $\mathrm d(\ln I)/\mathrm da$ — Eq. (5.42) is a transfer function and nothing else.

Constant-current mode is a feedback loop, and its output is the plant inverse. Set the current setpoint, close the loop on the tip’s $z$ piezo, and the controller drives the error to zero by holding the gap fixed. Since the gap is $z - h(x)$ , the control effort $z(x)$ equals $h(x)$ plus a constant. You are not measuring the surface; you are reading off what the controller had to do, which is the classic trick of getting a measurement out of a servo’s actuator signal rather than its sensor.

The loop gain is enormous — 20 per nanometre inside an exponential — which is exactly why the loop is stable and tight, and also why a loose or slow loop crashes the tip into a step edge.

Where the analogy breaks. A log amp’s noise floor is set by its electronics; this one’s is set by the fact that $I$ is a rate of discrete tunnelling events. The current is shot noise limited from the start, and the “signal” is a Poisson process whose mean is $T$ times an attempt rate. There is no averaging your way past it in zero time — you are counting particles, not measuring a field. And the exponential that gives all that sensitivity in $z$ gives none in $x$ : lateral resolution comes from the geometry of the front atom, not from Eq. (5.41), which is why STM images atoms superbly and measures their diameters badly.

Reproduce it in NumPy

Everything here is Eq. (5.41) evaluated in SI. The last block is the transition to the second half of this page: the same barrier, a particle 1836 times heavier, and an answer that is not small but meaningless.

import numpy as np

hbar, me, eV, mp = 1.054571e-34, 9.10938e-31, 1.602177e-19, 1.672622e-27

def beta(dV_eV, m=me):
    """Eq. (5.41): the penetration parameter for a barrier dV_eV above E."""
    return np.sqrt(2*m*dV_eV*eV)/hbar

b = beta(4.0)
print("tunnelling electrons: 4 eV to get out of the metal")
print("=" * 64)
print(f"  beta   = {b:.4e} /m    -- Eq. (5.43) says 'about 1e10', and it is")
print(f"  1/beta = {1e9/b:.4f} nm    -- one decay length, about one atom wide")

print("\nEq. (5.42), dT/T = -2 beta da, at the book's da = 0.001 nm")
da = 0.001e-9
lin, exact = -2*b*da, np.exp(-2*b*da) - 1
print(f"  linearized {100*lin:+7.3f} %   <- the book's 'measurable 2 %'")
print(f"  exact      {100*exact:+7.3f} %   (e^(-2 beta da) - 1)")
print(f"  the tangent is good to {abs(lin-exact)/abs(exact)*100:.1f} % of itself here")

print("\nwhat a change in the gap does to the current")
for name, dz in (("the book's 0.001 nm", 0.001e-9),
                 ("a 0.01 nm corrugation", 0.01e-9),
                 ("a monatomic step, 0.235 nm", 0.235e-9)):
    print(f"  {name:<28} -> {100*np.exp(-2*b*dz):8.3f} % of the current")

print("\nwhy the tip does not need to be sharp: the front atom does the work")
for extra in (0.1e-9, 0.2e-9, 0.3e-9):
    print(f"  an atom {1e9*extra:.1f} nm further back carries "
          f"{100*np.exp(-2*b*extra):6.2f} % as much current")

print("\nthe same gap, a heavier particle (4 eV barrier, 0.5 nm gap)")
a = 0.5e-9
for name, m in (("electron", me), ("proton", mp)):
    bm = beta(4.0, m)
    print(f"  {name:<9} beta = {bm:.3e} /m   2*beta*a = {2*bm*a:8.2f}   "
          f"T ~ {np.exp(-2*bm*a):.2e}")

printstunnelling electrons: 4 eV to get out of the metal ================================================================ beta = 1.0246e+10 /m -- Eq. (5.43) says 'about 1e10', and it is 1/beta = 0.0976 nm -- one decay length, about one atom wide Eq. (5.42), dT/T = -2 beta da, at the book's da = 0.001 nm linearized -2.049 % <- the book's 'measurable 2 %' exact -2.028 % (e^(-2 beta da) - 1) the tangent is good to 1.0 % of itself here what a change in the gap does to the current the book's 0.001 nm -> 97.972 % of the current a 0.01 nm corrugation -> 81.471 % of the current a monatomic step, 0.235 nm -> 0.810 % of the current why the tip does not need to be sharp: the front atom does the work an atom 0.1 nm further back carries 12.88 % as much current an atom 0.2 nm further back carries 1.66 % as much current an atom 0.3 nm further back carries 0.21 % as much current the same gap, a heavier particle (4 eV barrier, 0.5 nm gap) electron beta = 1.025e+10 /m 2*beta*a = 10.25 T ~ 3.55e-05 proton beta = 4.391e+11 /m 2*beta*a = 439.06 T ~ 2.08e-191

Tunnelling protons

The centre of the sun is an ionized gas of electrons, protons and light nuclei at about $10^7\ \mathrm{K}$ . Protons collide constantly, and just occasionally two of them fuse and release the energy that eventually leaves the surface as sunlight. The question is how often “occasionally” is — and the classical answer is never.

quantity⇅	value⇅
thermal energy of a proton at the solar centreⓘ	$E \approx k_{B} T \approx 1 keV$ ⓘ
height of the Coulomb barrier they must crossⓘ	$\approx 1 MeV$ ⓘ
classical distance of closest approachⓘ	$r_{C} = 1440 fm$ ⓘ
temperature the sun would need, classicallyⓘ	$8 \times 1 0^{9} K$ ⓘ

Figure 5.5, and the shape of the problem

E = \frac{e^2}{4\pi\epsilon_0 r_C}\tag{5.44}

V(r) = e²/4πε₀r, and the nuclear well below r_N
energy of approach E

energy E (keV)1.00

r_C, Eq. (5.44) 1440 fm

r_C / r_N 720

√(E_G/E) 22.21

T, Eq. (5.50) 2.27e-10

Both axes are logarithmic, which is the only way both landmarks fit on one picture: at 1.00 keV the turning point r_C sits 720× further out than the nuclear radius r_N. The book’s Fig. 5.5 draws them a factor of about four apart, because on a linear axis it has no choice. On these axes the Coulomb potential is a straight line of slope −1 — that is what 1/r is — and Eq. (5.44) is no longer a formula to trust but simply the point where that line crosses your energy. Everything hatched between the two is territory a classical proton can never enter, and the protons must cross 1438 fm of it to reach the nuclear well on the left, where fusion happens.

Setting up the quantum problem

Two protons with relative energy $E$ are described by an eigenfunction $\psi(\mathbf r)$ obeying

\left[-\frac{\hbar^2}{2\mu}\nabla^2 + V(r)\right]\psi(\mathbf r) = E\psi(\mathbf r)\tag{5.45}

where $\mu = m_p/2$ is the reduced mass of the pair. For low-energy protons the relevant eigenfunction has no angular dependence and can be written

\psi(\mathbf r) = \frac{u(r)}{r}\tag{5.46}

and then $u(r)$ obeys a one-dimensional equation:

-\frac{\hbar^2}{2\mu}\frac{\mathrm d^2u}{\mathrm dr^2} + V(r)u = Eu\tag{5.47}

The book prints Eq. (5.47) with its right-hand side broken

Page 102 has $-\dfrac{\hbar^2}{2\mu}\dfrac{\mathrm d^2u}{\mathrm dr^2} + V(r)u = E$ — with a bare $E$ where $Eu$ belongs. It is a dimensional impossibility: the left-hand side is an energy multiplied by $u$ , the right-hand side is an energy, so the equation cannot balance unless $u \equiv 1$ . Verified from the page image, so it is the book’s slip and not an artefact of the text extraction.

This is the third of five errors found in chapter 5’s printed text on this site. It follows the “Section 4.7” cross-reference that §5.1b and §5.2a both flag, and the energy misprinted in problem 3; two more turn up in Problems 5 — problem 6’s transmission formula and problem 1(c)‘s bound-state count.

The correction is not cosmetic in the way a wrong cross-reference is. With $Eu$ restored, Eq. (5.47) is exactly the one-dimensional energy eigenvalue equation of §4.3 with $x \to r$ and $m \to \mu$ — which is the entire justification for reusing §5.2a’s one-dimensional result on a three-dimensional problem. Printed as it stands, the equation says nothing at all.

Why the substitution ψ = u(r)/r is the right move

A three-dimensional radial problem looks harder than a one-dimensional one, and it is — but only because of the $1/r^2\,\mathrm d/\mathrm dr(r^2\,\mathrm d/\mathrm dr)$ in the Laplacian, which is not a plain second derivative. Writing $\psi = u/r$ is the change of variable that cancels exactly that clutter: it converts the radial Laplacian into $\dfrac{1}{r}\dfrac{\mathrm d^2u}{\mathrm dr^2}$ , and the leftover $1/r$ divides out of every term.

What is left, Eq. (5.47), is literally the equation of §5.2a with $x$ renamed $r$ . That is the licence for everything that follows: the protons’ problem is three-dimensional, but $u(r)$ tunnels through a barrier in one dimension exactly as before. The one price is that $u(0) = 0$ is now required — otherwise $\psi = u/r$ blows up at the origin — and that is the boundary condition worked out in problem 2.

From a square barrier to a Coulomb one

The book does not solve Eq. (5.47). It reasons from §5.2a instead, in two steps.

Step one: pretend the barrier is square. A barrier of constant height $V_B$ and width $r_C - r_N$ gives $u \propto e^{\beta r}$ in the forbidden region — the wave decays as $r$ decreases, going inward — so the probability of getting from $r_C$ in to $r_N$ is the ratio $|u(r_N)|^2/|u(r_C)|^2$ :

T \simeq \left|\exp[-\beta(r_C - r_N)]\right|^2\tag{5.48}

Step two: admit that it is not square. The Coulomb barrier’s height falls with $r$ , so $\beta$ is a function of position, set at each radius by how far $V(r)$ exceeds $E$ there. A varying decay rate accumulates as an integral rather than a product, so the exponent becomes a sum over the barrier:

T \simeq \left|\exp\left[-\int_{r_N}^{r_C}\beta\,\mathrm dr\right]\right|^2\tag{5.49}

This step is a plausible generalization, not a derivation

The book is explicit that it is writing down “a plausible form” rather than solving anything, and it is worth being clear about what has been assumed: that $\beta(r)$ may be treated as locally constant, so the wave function decays at whatever rate the local barrier height dictates.

That assumption is exact only if $V(r)$ changes slowly over one decay length $1/\beta$ — and it has a name outside this book, the WKB approximation. It is not a small point: it is the reason a formula derived for a square barrier can be applied to a smooth one at all. Where it fails is near $r = r_C$ , where $\beta \to 0$ and the “local decay length” becomes infinite. Fortunately, that is also where the integrand vanishes, so the damage is limited — which the widget above quantifies.

Doing the integral

step 1 of 8

The book compresses this into 'by substituting r = r_C cos²θ, we find that'. It is six lines, and the only approximation is in step 5.

1Start from Eq. (5.49). The modulus-squared just doubles the exponent, exactly as it did for e^(−2βa) in §5.2a — the probability is the amplitude squared.
$T ≃ exp [- 2 \int_{r_{N}}^{r_{C}} β (r) d r]$
So everything from here is about one integral.
2Write β from its definition. Inside the barrier V(r) > E, so the bracket is positive and β is real — the same condition that made the wave decay rather than oscillate throughout §5.2.
$β (r) = \frac{2 μ ( \frac{e ^{2}}{4 π ϵ _{0} r} - E )}{ℏ}$
3Factor E out of the bracket. This is the move that makes the problem tractable: r_C was DEFINED by e²/4πε₀r_C = E, so the bracket collapses into the pure ratio r_C/r and every constant leaves the integral.
$β (r) = k \frac{r _{C}}{r} - 1, k \equiv \frac{2 μ E}{ℏ}$
k is the wave number the protons have when they are far apart and free.
4Substitute r = r_C cos²θ. The point of this particular substitution — and the reason the book names it — is that it turns the awkward square root into a tangent exactly, with no leftover.
$\frac{r _{C}}{r} - 1 = sec^{2} θ - 1 = tan θ, d r = - 2 r_{C} cos θ sin θ d θ$
5THE ONE APPROXIMATION. Assume r_C ≫ r_N and push the lower limit down to r = 0, i.e. θ = π/2. Everything before this is exact; everything after inherits it. At solar energies r_C/r_N = 720, which is what makes it defensible.
$2 k \int_{0}^{r_{C}} \frac{r _{C}}{r} - 1 d r = 4 k r_{C} \int_{0}^{π /2} sin^{2} θ d θ$
tan θ · cos θ sin θ = sin²θ, and the two minus signs (from dr and from swapping the limits) cancel.
6Do the elementary integral: the mean value of sin²θ is ½, over an interval of length π/2.
$4 k r_{C} \cdot \frac{π}{4} = π k r_{C}$
7Put k and r_C back in terms of E. Both depend on E — k as √E, r_C as 1/E — so the product carries E^(−1/2), and multiplying and dividing by c produces the fine-structure constant.
$π k r_{C} = π \frac{2 μ E}{ℏ} \cdot \frac{e ^{2}}{4 π ϵ _{0} E} = π (\frac{e ^{2}}{4 π ϵ _{0} ℏ c}) \frac{2 μ c ^{2}}{E}$
8Finally, pull the π inside the square root, since π√X = √(π²X). Everything that is not E gathers into a single constant with the dimensions of an energy — and that constant is worth naming.
$= \frac{( \frac{e ^{2}}{4 π ϵ _{0} ℏ c} ) ^{2} 2 π ^{2} μ c ^{2}}{E} = \frac{E _{G}}{E}$
Which is Eqs. (5.50) and (5.51) together.

T \simeq \exp\left[-\left(\frac{E_G}{E}\right)^{1/2}\right]\tag{5.50}

E_G = \left(\frac{e^2}{4\pi\epsilon_0\hbar c}\right)^2 2\pi^2\mu c^2\tag{5.51}

The constant $E_G$ is the Gamow energy , and for two protons it comes to $493\ \mathrm{keV}$ .

symbol: $\frac{e ^{2}}{4 π ϵ _{0} ℏ c}$
is: the fine-structure constant α = 1/137.036 — the same dimensionless number ch01 met in the Bohr atom. Its appearance here is the tell that this is an electromagnetic problem with a quantum length scale in it.
units: dimensionless
type: dimensionless, ≈ 0.0073

Put a number on it — how often two protons in the sun actually fuse

Step 1 — the Gamow energy. With $\alpha = 1/137.036$ and $\mu c^2 = 469.1\ \mathrm{MeV}$ :

E_G = \alpha^2\,2\pi^2\mu c^2 = (5.325\times10^{-5})(19.74)(469.1\ \mathrm{MeV}) = 493.1\ \mathrm{keV}

matching the book’s $493\ \mathrm{keV}$ .

Step 2 — the exponent at solar energies. With $E = 1\ \mathrm{keV}$ :

\sqrt{\frac{E_G}{E}} = \sqrt{493.1} = 22.21

Step 3 — the probability.

T \simeq e^{-22.21} = 2.3\times10^{-10}

About one collision in four billion. The book quotes $e^{-22} \approx 3\times10^{-10}$ , rounding the exponent to a whole number — which is exactly the right precision here, because $r_N$ itself is known no better than that.

Now put that against the classical answer, which was zero, and against the temperature the sun would need to do it classically, which was $8\times10^9\ \mathrm{K}$ instead of $1.5\times10^7$ . One in four billion is tiny; it is also infinitely more than nothing, and the sun has an enormous number of protons and a great deal of time.

Reproduce it in NumPy

The integral of Eq. (5.49) is done two independent ways — numerically, and by the closed form that keeps the r_N cutoff — so the widget above is checked against something other than itself. E_G is computed from Eq. (5.51), never typed in.

import numpy as np

hbar, c, eV = 1.054571e-34, 2.997925e8, 1.602177e-19
mp, eps0, kB = 1.672622e-27, 8.854188e-12, 1.380649e-23
keV, MeV = 1e3*eV, 1e6*eV
ke = eV*eV/(4*np.pi*eps0)      # e^2/4 pi eps0
mu = mp/2                       # reduced mass of two protons, footnote 4
rN = 2e-15                      # range of the nuclear force

alpha = ke/(hbar*c)
EG = alpha**2 * 2*np.pi**2 * mu*c**2                       # Eq. (5.51)

print("the barrier two protons face at the centre of the sun")
print("=" * 68)
print(f"  e^2/4 pi eps0 hbar c = {alpha:.6f} = 1/{1/alpha:.3f}  (fine structure)")
print(f"  Coulomb barrier at rN = 2 fm : {ke/rN/MeV:.3f} MeV   book: 'about 1 MeV'")
print(f"  thermal energy there         : E = kT = 1 keV")
rC = ke/(1*keV)
print(f"  closest approach rC          : {rC/1e-15:.0f} fm      Eq. (5.44)")
print(f"  rC / rN                      : {rC/rN:.0f}       'three orders of magnitude'")
print(f"  to clear it classically the sun needs {ke/rN/kB:.2e} K, not 1.5e7 K")
print(f"  -- short by a factor of {ke/rN/kB/1.5e7:.0f}.  Classically, the sun cannot shine.")

print(f"\nEq. (5.51):  EG = alpha^2 * 2 pi^2 * mu c^2 = {EG/keV:.1f} keV   book: 493 keV")

def exponent_exact(E):
    """2 int_rN^rC beta dr, Eq. (5.49), keeping the cutoff. Substituting
    r = rC(1 - s^2) turns it into 4 k rC int_0^S s^2/sqrt(1-s^2) ds."""
    rC = ke/E
    k = np.sqrt(2*mu*E)/hbar
    S = np.sqrt(1 - rN/rC)
    s = np.linspace(0, S, 200_001)
    return 4*k*rC*np.trapezoid(s**2/np.sqrt(1 - s**2), s)

def exponent_closed(E):
    """The same integral in closed form: 2 k rC [arcsin S - S sqrt(1-S^2)]."""
    rC = ke/E
    k = np.sqrt(2*mu*E)/hbar
    S = np.sqrt(1 - rN/rC)
    return 2*k*rC*(np.arcsin(S) - S*np.sqrt(1 - S*S))

print("\nEq. (5.50) against the integral it approximates, Eq. (5.49)")
print(f"  {'E/keV':>6} {'exact':>9} {'closed form':>12} {'sqrt(EG/E)':>11} "
      f"{'T exact':>10} {'T Eq.(5.50)':>12}")
for E_keV in (0.5, 1.0, 2.0, 5.0, 10.0):
    E = E_keV*keV
    Gn, Gc, Gl = exponent_exact(E), exponent_closed(E), np.sqrt(EG/E)
    print(f"  {E_keV:6.1f} {Gn:9.4f} {Gc:12.4f} {Gl:11.4f} "
          f"{np.exp(-Gn):10.2e} {np.exp(-Gl):12.2e}")
print("  columns 2 and 3 agree, so the closed form is right; column 4 is what")
print("  dropping the rN cutoff costs.")
d = np.sqrt(EG/keV) - exponent_closed(keV)
print(f"  at 1 keV that is {d:.2f} in the exponent = a factor of {np.exp(d):.1f} in T,")
print(f"  against a T that swings by 10^11 across this table.  Noise.")

print("\nthe number the sun runs on")
G = np.sqrt(EG/(1*keV))
print(f"  T = exp[-sqrt(EG/E)] = exp[-{G:.2f}] = {np.exp(-G):.2e}")
print(f"  = one collision in {1/np.exp(-G):.1e}.  Book: 'exp[-22] = 3e-10'.")

print("\nwhy a star is slow, and why it is stable")
for T_K in (1.0e7, 1.5e7, 2.0e7):
    E = kB*T_K
    print(f"  {T_K:.1e} K -> E = {E/keV:5.2f} keV -> T = {np.exp(-np.sqrt(EG/E)):.2e}")
print("  doubling the temperature buys three orders of magnitude of fusion.")

printsthe barrier two protons face at the centre of the sun ==================================================================== e^2/4 pi eps0 hbar c = 0.007297 = 1/137.036 (fine structure) Coulomb barrier at rN = 2 fm : 0.720 MeV book: 'about 1 MeV' thermal energy there : E = kT = 1 keV closest approach rC : 1440 fm Eq. (5.44) rC / rN : 720 'three orders of magnitude' to clear it classically the sun needs 8.36e+09 K, not 1.5e7 K -- short by a factor of 557. Classically, the sun cannot shine. Eq. (5.51): EG = alpha^2 * 2 pi^2 * mu c^2 = 493.1 keV book: 493 keV Eq. (5.50) against the integral it approximates, Eq. (5.49) E/keV exact closed form sqrt(EG/E) T exact T Eq.(5.50) 0.5 30.3511 30.3511 31.4047 6.59e-14 2.30e-14 1.0 21.1530 21.1530 22.2065 6.51e-10 2.27e-10 2.0 14.6491 14.6491 15.7024 4.34e-07 1.52e-07 5.0 8.8785 8.8785 9.9310 1.39e-04 4.86e-05 10.0 5.9710 5.9710 7.0223 2.55e-03 8.92e-04 columns 2 and 3 agree, so the closed form is right; column 4 is what dropping the rN cutoff costs. at 1 keV that is 1.05 in the exponent = a factor of 2.9 in T, against a T that swings by 10^11 across this table. Noise. the number the sun runs on T = exp[-sqrt(EG/E)] = exp[-22.21] = 2.27e-10 = one collision in 4.4e+09. Book: 'exp[-22] = 3e-10'. why a star is slow, and why it is stable 1.0e+07 K -> E = 0.86 keV -> T = 4.08e-11 1.5e+07 K -> E = 1.29 keV -> T = 3.29e-09 2.0e+07 K -> E = 1.72 keV -> T = 4.51e-08 doubling the temperature buys three orders of magnitude of fusion.

Why the smallness of T is the point, not a problem

It is tempting to read $2\times10^{-10}$ as a disappointing number — tunnelling nearly fails. Read it the other way round.

Stars adjust. A star contracts under gravity until it is hot enough to release energy fast enough to hold itself up, and then it stops. The equilibrium temperature is therefore set by whatever value of $T$ makes the fusion rate match the leakage of light from the surface — and because $T$ depends on temperature so ferociously, that balance is struck at an energy far below the barrier, where tunnelling is rare.

So the rarity is not an obstacle the sun overcomes; it is the property that sets the sun’s temperature in the first place, and it is why the fuel lasts for billions of years rather than going up at once. A universe with a smaller $E_G$ would have stars that burned out before anything interesting happened on the planets. The sun shines slowly for the same reason it shines at all.

Engineer's bridge — an activation law with the wrong exponent, and a thermostat with absurd gain

You already know this functional form, and the difference matters. Reliability engineering runs on the Arrhenius law: failure rate $\propto e^{-E_a/k_BT}$ , the basis of every temperature-acceleration factor in a qualification plan. Compare Eq. (5.50), written against temperature: rate $\propto e^{-\sqrt{E_G/k_BT}}$ .

Same shape, different exponent, and the difference is exactly quantum mechanical. Arrhenius counts particles that have classically got over a barrier of height $E_a$ , so the exponent is linear in $1/T$ . The Gamow factor counts particles that went through, and because a Coulomb barrier gets both lower and narrower as the energy rises, the exponent goes as $1/\sqrt T$ — a much weaker dependence. That weakness is the sun’s licence to operate five hundred times colder than the classical requirement.

And it is a thermostat. The loop is: gravity compresses → temperature rises → fusion rate rises very steeply → pressure rises → expansion → temperature falls. A feedback loop whose forward gain runs to ~10³ per doubling of the input settles into a very tight regulation band, which is why the sun’s core temperature is stable to a few percent over billions of years without anything regulating it. High loop gain buys tight regulation; that is the same trade as in any op-amp.

Where it breaks. The analogy is a rate law, and rate laws hide their statistics. In an op-amp the gain is a property of the circuit; here the “gain” is the tail of a distribution, and the protons that actually fuse are not typical ones — they are the rare fast ones where the falling Maxwell–Boltzmann tail meets the rising Gamow factor. The product of those two exponentials peaks at an energy well above $k_BT$ (it has a name, the Gamow peak), so “the fusion rate at 1 keV” is a useful estimate and not a real mechanism. This page, like the book, computes the estimate.

Where this goes next

Two numbers, from one formula. $2\beta a \approx 10$ makes a microscope that maps single atoms; $\sqrt{E_G/E} \approx 22$ makes a star that burns for ten billion years. Nothing separates them except what went into $\beta$ .

Problems 5 pushes the second case twice more: problem 7 works the proton–proton estimate through in detail, and problem 8 applies the same Coulomb-barrier integral in reverse — to alpha particles escaping a nucleus — where the identical exponent explains why radioactive half-lives span twenty orders of magnitude.

Chapter 6 leaves square potentials behind for a smooth one, the harmonic oscillator; the exponential tails into the classically forbidden region that §5.1a found on a bound state reappear there unchanged, and so does the technique of reading a wave function’s shape off the sign of $E - V(x)$ .

Check yourself

0 / 6 answered

Switch the microscope widget to constant-current mode and scan across the surface.
1.The lower trace stops looking like a current and starts looking like a row of atoms. What is actually being plotted, and why does it reproduce the surface?
2.A monatomic step on the surface is $0.235 nm$ high. With $β = 1.025 \times 1 0^{10} m^{- 1}$ , by what factor does the tunnelling current change as a constant-height tip passes over it?
3.The Gamow exponent is $E_{G} / E$ . Why a square root, rather than the $E_{G} / E$ an activation law would give?
4.Two protons meet with relative energy $E = 1 keV$ , and $E_{G} = 493 keV$ . What does Eq. (5.50) give?
5.Page 102 prints Eq. (5.47) as $- \frac{ℏ ^{2}}{2 μ} \frac{d ^{2} u}{d r ^{2}} + V (r) u = E$ . Without consulting any other source, how can you tell it is a misprint?
Step 5 of the derivation replaces the lower limit $r_{N}$ with 0, on the grounds that $r_{C} ≫ r_{N}$ .
6.What does that assumption cost at solar energies, and how would you find out?