Skip to main content

Can you find the number near 137?

Alpha tells us how strongly light and charged particles interact. One divided by alpha is near 137. Find the trial number where the two calculated sides balance.

Step 1 of 8

Begin with the boundary and carrier.

The boundary supplies a normalization called c₃. The five carrier slots supply charges. Squaring and adding those charges gives 5/6; the 3:2 rank ratio turns it into the response power 5/4.

1. Boundary normalization and the 3 + 2 carrier. Square the five charges; multiply by the rank ratio. c₃ = 1/(8π), response power = 5/4. 2. Three families, one Higgs and the stated field weights. Weight the charge traces and apply the abelian normalization. M = 41 and 3 × 16 = 48. 3. c₃, family norm 3/4 and count 48. Compute the base and the four factor weight. Base = 1/(6π), δ = 48c₃⁴. 4. The base, δ and the carrier power 5/4. Reweight δ, then apply the determinant response. φₛ(α) = base + q(1 − q)⁻⁵/⁴. 5. The full opening φₛ(α), c₃ and M = 41. Assemble the specified electromagnetic closure. F(α) = 0. 6. The fixed function F and its response. Use the analytic sign and convexity argument. Exactly one positive root for the stated equation. 7. The unique positive root of F. Bisect a sign changing bracket, then take the reciprocal. α⁻¹ ≈ 137.0359992168. 8. The calculated inverse alpha and a dated observation. Compare the same low energy quantity on one uncertainty scale. 1.90 quoted experimental uncertainties.
Arrows carry the named inputs into later calculations. Added inputs stay marked. Reduced motion is on. Use Next step or choose a stage below.
Carry forward
Boundary normalization and the 3 + 2 carrier
Operation
Square the five charges; multiply by the rank ratio
Result
c₃ = 1/(8π), response power = 5/4
c3=18π,γ=3(13)2+2(12)2=56,32γ=54\begin{gathered}c_3=\frac1{8\pi},\\[.4em]\gamma=3\left(\frac13\right)^2+2\left(\frac12\right)^2=\frac56,\\[.4em]\frac32\gamma=\frac54\end{gathered}

In words: Square the five charges; multiply by the rank ratio

Why this step works and what it assumes

The normalization belongs to the declared boundary principle. The charge trace and carrier norm identity are exact in the specified five slot dictionary. They do not alone select a physical electromagnetic response.

Read the original calculation
See where this fits in the whole theory
Try the individual controls

Try this. Move the trial number and compare the two sides. Press Find balance to locate where they meet.

Make two sides equal

Move the slider. It changes the trial coupling, so the two sides of the specified equation change. Find balance when both bars have the same length.

First sideSecond sideThe first side is larger.
Both bars use the same fixed scale and the actual equation values.
The first side is larger.Move the trial number, or choose Find balance.
See the equation values

First side 4.47712e-7 · second side 3.81885e-7 · difference 6.583e-8.

This balances the stated mathematical equation. Connecting it to physical electromagnetism is a separate step in the full derivation.

Choose a trial number

Move the slider. The two displayed sides are calculated from the same trial value.

Find the balance

Watch the gap between the sides. Find balance locates the value where the gap closes.

Compare the number

The result near 137 is inverse alpha. The dated measurement and its uncertainty provide the next comparison.

Explore the full interactive calculation

Find where the response balances

The curve is the specified equation F(α). Move the trial coupling to make its residual zero. All equation inputs stay fixed.

Residual of the specified alpha equation versus inverse alpha0130145F(α), scaled to this intervalTrial inverse coupling α⁻¹
The crossing is a root of this function. Its global uniqueness is proved analytically in the original argument.
Residual |F(α)|
< 10⁻¹⁸
Bisection bracket after 0 steps
130.00000000 to 145.00000000

Bisection halves an interval whose endpoints have opposite residual signs. This browser calculation uses ordinary numerical precision. The original analytic proof and certified interval enclosure remain separate evidence.

Follow the complete original derivation and every coefficient

Where does the number near 137 come from?

Alpha measures electromagnetic coupling strength. Follow how TFPT combines its boundary normalization, carrier charges and a specified defect response into one equation. That equation has a proved unique positive solution, with inverse coupling α⁻¹ ≈ 137.035 999 2168… .

Alpha is a dimensionless measure of electromagnetic interaction strength. In familiar SI units it combines the elementary charge, the quantum of action and the speed of light:

α=e24πε0ℏc≈1137\alpha=\frac{e^2}{4\pi\varepsilon_0\hbar c}\approx\frac1{137}

In words: Alpha equals elementary charge squared divided by four pi, vacuum permittivity, reduced Planck constant and the speed of light. It has no units.

Writing its inverse gives the number near 137. The comparison here concerns the low energy Thomson limit. Matching a calculated boundary response to that physical observable is a specific part of the task.

There are three kinds of step below: exact counts and identities, a response calculated under a named contract, and the remaining identification with physical electromagnetism. Keeping those steps visible explains both the number and what its calculation establishes.

  1. Begin with the boundary and carrier

    P1 fixes the seam normalization. The marked five slot carrier supplies the 3+2 charge split; its even representation has 16 components. Three family directions then give 48 admissible components. These are internal representation counts.

    c3=18π,φbase=16π,δtop=3⋅16 c34.\begin{aligned}c_3&=\frac1{8\pi},\\\varphi_{\rm base}&=\frac1{6\pi},\\\delta_{\rm top}&=3\cdot16\,c_3^4.\end{aligned}

    In words: The boundary normalization is one over eight pi. The base opening is one over six pi. Three families of sixteen give the topological weight forty eight times the fourth power of the seam constant.

    Original architecture: normalization, carrier count and contact grammar

  2. Let the defect respond to alpha

    In the original defect sector, an additive action shift αJ changes the relative weight by exp(−αJ). On the stated double cover, primitive spectral flow one per sheet gives J = 2. This is the source of the exponential, including its factor two.

    q(α)=δtop e−2αq(\alpha)=\delta_{\rm top}\,e^{-2\alpha}

    In words: The alpha dependent defect weight is the topological weight multiplied by the exponential of minus two alpha.

    The original circle also determines the single defect line: its winding changes the polarization by a rank one projector Π. Consequently its ordered relative Fredholm determinant is exactly 1−q. The later marked D₅ grade calculation realizes the fractional response (1−uq)−5/4 in the existing source. The common time diagonal has exponent 2, so the marked measurement matters.

    φs(α)=16π+q(α) [1−q(α)]−5/4\varphi_s(\alpha)=\frac1{6\pi}+q(\alpha)\,[1-q(\alpha)]^{-5/4}

    In words: The seam opening equals its base plus the defect weight times the inverse determinant response with exponent five quarters.

    The exponent has an exact charge trace origin: Bγ = (3/2)(5/6) = 5/4. Choosing the EM coupling q(α), the outer factor q and this additive baseline in that same physical channel remains part of the response contract.

    Original exponential reweightingReconstructed circle defectMarked D₅ response proof

  3. Read the shared charge and flavour budget

    The compiler gives a total flavour transport length of 40. One Higgs contribution makes 41. The same three family charge content gives the normalized U(1) index b₁ = 41/10. This is why the electromagnetic equation reads a budget already present in the matter construction.

    M=∑f,jLf,j+NΦ=40+1=41,8b1=8⋅4110=45⋅41.\begin{aligned}M&=\sum_{f,j}L_{f,j}+N_\Phi\\&=40+1=41,\\8b_1&=8\cdot\frac{41}{10}=\frac45\cdot41.\end{aligned}

    In words: Forty transport units plus one Higgs give forty one. Eight times b one equals four fifths times forty one.

    The value 40 is connected to the charge pattern through the stated compiler identity below. Here X = 6Y is integer hypercharge and the trace sums over the sixteen components of one family. The division by three is this compiler relation, rather than an average over three families; the flavour operators and their transport remain its prerequisites.

    Tr⁡X2=120,∑f,jLf,j=Tr⁡X23=40.\begin{aligned}\operatorname{Tr}X^2&=120,\\\sum_{f,j}L_{f,j}&=\frac{\operatorname{Tr}X^2}{3}=40.\end{aligned}

    In words: The squared integer hypercharges sum to one hundred twenty. The stated compiler relation divides that trace by three to give the flavour transport budget forty.

    Inspect the charge trace and budget identityRead its registered statement

    Original budget and hypercharge indexFollow the charge construction

  4. Assemble the stated electromagnetic closure

    The closure balances a local cubic term, a quadratic sheet term and a logarithmic transport response. The paper names these the Maxwell moment, Calderón sheet contribution and C₆ transport determinant.

    F(α)=α3−2c33α2−45 Mc36log⁡1φs(α),M=41,F(α)=0.\begin{aligned}F(\alpha)&=\alpha^3-2c_3^3\alpha^2\\&\quad-\frac45\,M c_3^6\log\frac1{\varphi_s(\alpha)},\\M&=41,\qquad F(\alpha)=0.\end{aligned}

    In words: F of alpha equals alpha cubed minus twice c three cubed times alpha squared minus four fifths times forty one times c three to the sixth times the logarithm of one over the seam opening. The closure sets F to zero.

    The powers of c₃ follow the stated seam insertion grammar: three in the quadratic coefficient, six in the transport coefficient. The logarithm belongs to the determinant description. The carrier identities fix the displayed coefficient values within this closure; they do not alone derive its local functional form.

    The older scalar potential makes the construction explicit: it supplies the local derivative α³−2c₃³α² and integrates the desired response to obtain a stationary action. The present physical question is whether the original local EM operator and its renormalized measure produce this same variation.

    Original definition of the closure potentialLocal measure and Quillen scope

  5. Prove that the equation selects one positive root

    This is a theorem for the specified function, whose seam opening depends on α. For 0 < α ≤ 2c₃³ the function is negative. Log convexity of the seam response gives strict convexity of F beyond that point. The cubic eventually grows without bound, so F crosses zero exactly once, with positive derivative.

    Why do those facts rule out a second root?

    Set t = 2c₃³ and C = (4/5)c₃⁶M. The response satisfies 0 < φs < 1, so its logarithm is negative. Before t, both contributions to F are negative. Beyond t, exact log convexity gives a strictly increasing derivative:

    F(x)=x2(x−t)+Clog⁡φs(x),F′′(x)=6x−4c33+C(log⁡φs)′′>0(x≥t).\begin{aligned}F(x)&=x^2(x-t)+C\log\varphi_s(x),\\F''(x)&=6x-4c_3^3\\&\quad+C(\log\varphi_s)''\\&>0\quad(x\ge t).\end{aligned}

    F(t) is negative and the cubic eventually tends to infinity. At its first zero r, convexity makes the derivative greater than the positive secant slope from (t,F(t)) to (r,0). The derivative then keeps increasing, so the function cannot turn back through zero. The source proves log convexity by an exact differentiated identity, without a small term approximation or a sampled sign scan.

    The argument holds for every positive budget M. It proves uniqueness after the inputs and response are fixed; it does not select 41 in place of another positive budget. An interval enclosure then locates the root.

    The function is fixed. The slider changes the trial value, not the equation.

    Residual of the specified electromagnetic equationThe horizontal axis is inverse alpha. The curve crosses zero near 137.0359992168. This plot illustrates a root and does not prove global uniqueness or physical identification.130145Inverse alpha0F(α), scaled

    Residual |F(α)| < 10⁻¹⁸. This display uses ordinary browser arithmetic; the source proof and interval enclosure are separate evidence.

    Read the analytic proof and interval enclosure

  6. Compare the result with measurement

    How close is the calculated value to the measurement?

    Calculated closure root137.035 999 216 8…
    CODATA 2022 recommended value137.035 999 177 (21)
    -3-2-10+1+2+3Difference in experimental standard uncertainties (σ)
    CODATA ±1σ Calculated root1.90σ above the CODATA recommended value. This dated comparison uses the experimental uncertainty. The measured value was known before the formula freeze.
    NIST: CODATA 2022 recommended values, published 2025

    This dated comparison uses the CODATA 2022 recommended inverse coupling, 137.035 999 177 (21). The closure differs by about 1.9 experimental standard deviations. Alpha had already been measured when the formula was constructed; this comparison is a calculated relation to an existing measurement.

    The mathematical root, the reconstructed circle line, the marked D₅ response and the finite determinant line and KMS results remain achieved. The remaining connection is specific: the original local EM source and operator map must yield the relative real renormalized determinant norm, its coupling and variation, then match the physical Thomson coupling.

    Read the current reviewed physical scopeOpen the dated empirical comparison

Trace every ingredient: value, meaning, original source and scope
Provenance of the specified alpha closure
ValueWhat it meansOriginal sourceWhat the step establishes
c3=18πc_3=\frac1{8\pi}The P1 boundary normalization. The paper writes 8π as two sheets times the Gauss–Bonnet integral 4π.Architecture, EM closureAccepted oriented boundary principle and normalization. Its physical realization remains a separate question.
φbase=16π=c33/4\varphi_{\rm base}=\frac1{6\pi}=\frac{c_3}{3/4}The unperturbed seam opening. The denominator 3/4 is the A₃ discriminant norm, paired with the D₅ norm 5/4.Architecture, determinant lineExact relation in the named normalization. Why the physical EM response has this additive baseline is part of the source map.
48=3×16,δtop=48c34=3256π448=3\times16,\quad\delta_{\rm top}=48c_3^4=\frac3{256\pi^4}Three families of the 16 dimensional even carrier give the admissible count. The four seam factors give the specified topological weight.Architecture, contact unit and responseThe count and algebra are exact in this carrier branch. Using this weight in the EM response belongs to its stated coupling contract.
e−2α,2=d SF=2×1e^{-2\alpha},\quad 2=d\,\mathrm{SF}=2\times1Adding αJ to a Euclidean defect action multiplies its relative weight by exp(−αJ). The primitive spectral flow is one on each of two identical sheets.v2.7, defect sector reweightingExact reweighting under the stated additive action and index normalization. It specifies a topological coupling, not automatically an energy or heat parameter.
54=Bγ=32 ⁣(39+24)=q(D5)\frac54=B\gamma=\frac32\!\left(\frac39+\frac24\right)=q(D_5)B is the 3:2 carrier rank ratio; γ is the squared hypercharge trace on the five slots. The same value is the D₅ discriminant norm.v4.2, charge trace identitiesExact carrier identity. The marked D₅ grade calculation additionally realizes this power as an actual response in the specified E₈ source.
q(1−q)−5/4q(1-q)^{-5/4}The rank one relative determinant is 1−q. Its chosen inverse susceptibility has power 5/4; the seam opening adds the resulting response with one outer q.v4.2, minimal seam transferThe determinant identity is proved. The inverse response, the outer q and the physical parameter map must be realized in the same EM channel.
M=40+1=41,b1=4110M=40+1=41,\quad b_1=\frac{41}{10}The specified flavour transport lengths sum to 40; one Higgs contribution gives 41. The hypercharge content independently expresses this as the normalized abelian index 10b₁.Architecture, flavour budget and U(1) indexExact budget and charge trace in the selected three family, one Higgs branch. Root uniqueness itself does not select this budget.
8b1=45 418b_1=\frac45\,41Eight times the abelian index equals (4/5) times the budget. The 4/5 is this exact normalization identity; it is not another fitted parameter.Architecture, Ward readingCoefficient identity. The use of rank(E₈) × b₁ in the local EM functional still has its stated physical interpretation.
α3−2c33α2\alpha^3-2c_3^3\alpha^2The local part of the defined closure. Two names the sheet factor; c₃³ means three seam insertions. Its normalized primitive is α⁴/4 − (2/3)c₃³α³.v4.2, normalized closure potentialIntegrating the supplied derivative is exact. That operation does not derive the derivative from an original coupled EM operator. The cubic moment and exact local normalization need that identification.
c36log⁡ ⁣1φsc_3^6\log\!\frac1{\varphi_s}The transport term uses six seam insertions, associated in the compiler with the C₆ hexagon and six positive A₃ roots. Logarithms enter the determinant and effective action description.Architecture, Ward and contact grammarSpecified transport and determinant grammar. A real renormalized local determinant must produce this full logarithmic response, including its measure and variation.
Inspect the numerical enclosure, full precision and response series
0.007 297 352 562 2097<α⋆,α⋆<0.007 297 352 562 2100.\begin{gathered}0.007\,297\,352\,562\,2097<\alpha_\star,\\\alpha_\star<0.007\,297\,352\,562\,2100.\end{gathered}

In words: The proved root lies strictly between 0.0072973525622097 and 0.0072973525622100.

The original scalar closure gives the following higher precision approximation. These extra digits describe the calculated equation; the displayed interval above states the cited certified enclosure.

α⋆−1≈137.035 999 216 840+7.1250353786030380388037⋅10−13\begin{gathered}\alpha_\star^{-1}\approx137.035\,999\,216\,840\\{}+7.1250353786030380388037\cdot10^{-13}\end{gathered}

In words: Calculated inverse alpha is approximately 137.03599921684071250353786030380388037.

q(1−q)−5/4=q+54q2+4532q3+⋯q(1-q)^{-5/4}=q+\frac54q^2+\frac{45}{32}q^3+\cdots

In words: The response series starts with q, plus five quarters q squared, plus forty five thirty seconds q cubed.

The series fixes higher response coefficients once this transfer is specified. The earlier linear, exponential, two defect and resummed calculations retain their own contracts; their similar factors do not identify all their physical actions.

Read the reviewed reconstruction of the original alpha factors

How this fits into the whole theory

The two sides reuse ingredients from the matter and E₈ constructions. Finding their balance calculates the number for those supplied ingredients.

What this experiment establishes

Uniqueness is a theorem for the fixed response and every positive budget. Selecting the physical budget and identifying the full local electromagnetic variation with the renormalized Thomson coupling require the original source and observable map.