Openai/6908ee36-d3b4-8010-a468-28e0753d08ae

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gpt-5-thinking

User: 0.00843043657 is the latency of each state in the Fundamental Dilator coherence.

0.00843043657 is the latency of each state in the Fundamental Dilator coherence. At each state, one expects that a full hydrogen mass (in 4D volume) would appear. This number multiplied by the compton period of a hydrogen atom is the delta t. I would like to know how that compares with the Heisenberg Principle Delta T associated with the appearance of a virtual hydrogen atom.

Assistant: Nice question — here’s a clean apples-to-apples comparison.

Nice question — here’s a clean apples-to-apples comparison.

Compton period (hydrogen atom of mass mHm_HmH​):

TC=hmHc2.T_C=\frac{h}{m_H c^2}.TC​=mH​c2h​. Your HU “state latency” time:

ΔtHU=α TC,α=0.00843043657.\Delta t_{\text{HU}} = \alpha\,T_C,\quad \alpha = 0.00843043657.ΔtHU​=αTC​,α=0.00843043657. Heisenberg time for “borrowing” energy ΔE\Delta EΔE (time–energy uncertainty):

ΔtH≳ℏ2 ΔE.\Delta t_{\text{H}} \gtrsim \frac{\hbar}{2\,\Delta E}.ΔtH​≳2ΔEℏ​. Two reasonable choices for ΔE\Delta EΔE:

1. Full rest energy of a hydrogen atom: ΔE=mHc2\Delta E = m_H c^2ΔE=mH​c2.

⇒ΔtH,rest=ℏ2mHc2=TC4π.\Rightarrow \Delta t_{\text{H,rest}}=\frac{\hbar}{2 m_H c^2}=\frac{T_C}{4\pi}.⇒ΔtH,rest​=2mH​c2ℏ​=4πTC​​.

1. Only the atomic binding energy: ΔE≈13.6 eV\Delta E \approx 13.6\ \text{eV}ΔE≈13.6 eV.

⇒ΔtH,bind=ℏ2 (13.6 eV).\Rightarrow \Delta t_{\text{H,bind}}=\frac{\hbar}{2\,(13.6\ \text{eV})}.⇒ΔtH,bind​=2(13.6 eV)ℏ​.

- TC≈4.4054×10−24 sT_C \approx 4.4054\times 10^{-24}\ \text{s}TC​≈4.4054×10−24 s

• ΔtHU=αTC≈3.714×10−26 s\Delta t_{\text{HU}} = \alpha T_C \approx 3.714\times 10^{-26}\ \text{s}ΔtHU​=αTC​≈3.714×10−26 s
• ΔtH,rest=TC/(4π)≈3.506×10−25 s\Delta t_{\text{H,rest}} = T_C/(4\pi) \approx 3.506\times 10^{-25}\ \text{s}ΔtH,rest​=TC​/(4π)≈3.506×10−25 s
• ΔtH,bind≈2.420×10−17 s\Delta t_{\text{H,bind}} \approx 2.420\times 10^{-17}\ \text{s}ΔtH,bind​≈2.420×10−17 s

- Versus full rest energy: ΔtHUΔtH,rest=4π α≈0.106.\frac{\Delta t_{\text{HU}}}{\Delta t_{\text{H,rest}}} = 4\pi\,\alpha \approx 0.106.ΔtH,rest​ΔtHU​​=4πα≈0.106. Your ΔtHU\Delta t_{\text{HU}}ΔtHU​ is about 10.6%10.6\%10.6% of the Heisenberg time for ΔE=mHc2\Delta E=m_Hc^2ΔE=mH​c2, i.e., ~9.4×9.4\times9.4× shorter.

• Versus 13.6 eV binding energy: ΔtHUΔtH,bind≈1.53×10−9.\frac{\Delta t_{\text{HU}}}{\Delta t_{\text{H,bind}}} \approx 1.53\times 10^{-9}.ΔtH,bind​ΔtHU​​≈1.53×10−9. Your ΔtHU\Delta t_{\text{HU}}ΔtHU​ is ~10910^9109 times shorter than that bound.

- If by “appearance of a virtual hydrogen atom” you mean an off-shell composite with energy uncertainty comparable to its full rest energy, the relevant Heisenberg scale is ΔtH,rest∼TC/(4π)\Delta t_{\text{H,rest}}\sim T_C/(4\pi)ΔtH,rest​∼TC​/(4π). On that basis, your coherence step time is notably smaller than the usual ℏ/(2mHc2)\hbar/(2 m_H c^2)ℏ/(2mH​c2) allowance.

• If you only “borrow” the binding energy, the Heisenberg time is vastly larger, and your ΔtHU\Delta t_{\text{HU}}ΔtHU​ is far below that.

If you want, I can also express this entirely in your HU terms: since ΔtH,rest=TC/(4π)\Delta t_{\text{H,rest}} = T_C/(4\pi)ΔtH,rest​=TC​/(4π), your choice α\alphaα effectively sets

ΔtHU=(4πα) ΔtH,rest,\Delta t_{\text{HU}} = (4\pi\alpha)\,\Delta t_{\text{H,rest}},ΔtHU​=(4πα)ΔtH,rest​, so tuning α\alphaα above/below 1/(4π)≈0.07961/(4\pi)\approx 0.07961/(4π)≈0.0796 moves you above/below the rest-energy Heisenberg scale. Your α=0.00843\alpha=0.00843α=0.00843 puts you an order of magnitude below it.

User: The latency of the coherence is not exactly expected to be equal to the creation time of a hydrogen atom.

The latency of the coherence is not exactly expected to be equal to the creation time of a hydrogen atom. The reason is that coherence is a hydrogen atom in 4D all the time. So, it is a tunneling time from one state to the other. I don't know how to estimate that. I guessed it would be of the order of the creation of a virtual particle. and it is.. within a factor of 10.

Would you have better ideas on how to estimate latency of states in a coherence?

Assistant: Love the “tunneling-between-states” framing.

Love the “tunneling-between-states” framing. If the coherence is “a hydrogen in 4D all the time,” the latency you want is a transition (or traversal) time between two internal states of the same composite object, not a creation time. Here are practical ways to estimate it, from most model-agnostic to most HU-specific.

These give lower bounds on how fast any pure state can evolve to an orthogonal (or distinguishable) state, independent of the detailed dynamics.

• Mandelstam–Tamm (MT): τ≥πℏ2 ΔE\tau \ge \frac{\pi\hbar}{2\,\Delta E}τ≥2ΔEπℏ​, where ΔE\Delta EΔE is the energy uncertainty (RMS spread) of your state relative to its Hamiltonian.
• Margolus–Levitin (ML): τ≥πℏ2 ⟨E⟩\tau \ge \frac{\pi\hbar}{2\,\langle E\rangle}τ≥2⟨E⟩πℏ​ for evolution to an orthogonal state with average energy above the ground state ⟨E⟩\langle E\rangle⟨E⟩.

How to use: decide what energy scale actually fluctuates during a state hop.

• If it’s only an internal splitting Δ\DeltaΔ (see §2), set ΔE∼Δ\Delta E \sim \DeltaΔE∼Δ.
• If the hop taps a larger chunk of internal energy (e.g., a fraction of mHc2m_H c^2mH​c2), plug that fraction in.

Your current latency τHU≈3.71×10−26 s\tau_{\text{HU}}\approx 3.71\times 10^{-26}\,\text{s}τHU​≈3.71×10−26s is below the QSL you’d get from using the full rest energy (∼10−25 s\sim 10^{-25}\,\text{s}∼10−25s), but it could be consistent if (i) the hop is not an orthogonalization, or (ii) ΔE\Delta EΔE exceeds the rest-energy bound because the hop is only a small phase advance (see §4). So: QSLs give you a sanity lower bound, not the exact τ\tauτ.

Model two adjacent coherence states ∣L⟩,∣R⟩|L\rangle,|R\rangle∣L⟩,∣R⟩ with

H=Δ2 σx+ε2 σz,(ε≈0 resonant).H=\frac{\Delta}{2}\,\sigma_x+\frac{\varepsilon}{2}\,\sigma_z,\quad (\varepsilon\approx 0\ \text{resonant}).H=2Δ​σx​+2ε​σz​,(ε≈0 resonant). The coherent oscillation (tunnel) frequency is Ω=Δ/ℏ\Omega=\Delta/\hbarΩ=Δ/ℏ and the first-passage/half-Rabi latency is

τhop≃πΩ=πℏΔ.\tau_{\text{hop}}\simeq \frac{\pi}{\Omega}=\frac{\pi\hbar}{\Delta}.τhop​≃Ωπ​=Δπℏ​. So if your measured/assumed latency is τ\tauτ, the implied level splitting is

Δ≃πℏτ.\Delta \simeq \frac{\pi\hbar}{\tau}.Δ≃τπℏ​. With τ=3.71×10−26 s\tau=3.71\times10^{-26}\,\text{s}τ=3.71×10−26s, this back-infers Δ≈5.6×1010 eV\Delta \approx 5.6\times 10^{10}\ \text{eV}Δ≈5.6×1010 eV (∼55 GeV). That’s huge compared with atomic scales, which tells you one of two things must be true:

• The hop is not a full ∣L⟩→∣R⟩|L\rangle\to|R\rangle∣L⟩→∣R⟩ Rabi half-cycle (orthogonalization), or
• The states are separated by an ultra-stiff barrier tied to a much higher internal energy scale than ordinary atomic structure.

This is still useful: once you specify a plausible Δ\DeltaΔ from HU microphysics, the formula gives you τ\tauτ directly.

For tunneling through a barrier of “height” V(x)−EV(x)-EV(x)−E and width LLL,

Δ∼ℏω0 e−S/ℏ,S=∫barrier2meff [V(x)−E] dx,\Delta \sim \hbar\omega_0\,e^{-S/\hbar},\qquad S=\int_{\text{barrier}} \sqrt{2m_\text{eff}\,[V(x)-E]}\,dx,Δ∼ℏω0​e−S/ℏ,S=∫barrier​2meff​[V(x)−E]​dx, so

τ∼πℏΔ∼πω0 eS/ℏ.\tau \sim \frac{\pi\hbar}{\Delta}\sim \frac{\pi}{\omega_0}\,e^{S/\hbar}.τ∼Δπℏ​∼ω0​π​eS/ℏ. How to use: propose a 4D effective coordinate and barrier V(χ)V(\chi)V(χ) that separates adjacent coherence states, pick an effective mass meffm_\text{eff}meff​ for that coordinate, and compute SSS. This immediately yields τ\tauτ. Even order-of-magnitude guesses for LLL, V−EV-EV−E, and ω0\omega_0ω0​ will tell you whether 10−26 s10^{-26}\,\text{s}10−26s is natural or requires extreme parameters.

If coherence states are equally spaced phases on an internal rotation with frequency ωC=mHc2/ℏ\omega_C=m_Hc^2/\hbarωC​=mH​c2/ℏ (Compton clock), then a hop is a phase advance δϕ=2π/N\delta\phi=2\pi/Nδϕ=2π/N:

τ=δϕωC=TCN.\tau=\frac{\delta\phi}{\omega_C}=\frac{T_C}{N}.τ=ωC​δϕ​=NTC​​. Your choice α=0.0084304\alpha=0.0084304α=0.0084304 is exactly this with N≃1/α≈118.6N\simeq 1/\alpha\approx 118.6N≃1/α≈118.6 states per Compton period. This is a clean HU-native parameterization: once you defend NNN by symmetry or topology (e.g., allowed irreps or winding number in your 4D manifold), τ\tauτ follows without invoking huge energy splittings.

Define the coherence transition as a scattering through an internal channel:

• Phase (Wigner) delay: τW=dϕ(E)dE\tau_W = \frac{d\phi(E)}{dE}τW​=dEdϕ(E)​, where ϕ\phiϕ is the energy-dependent scattering phase between states.
• Dwell time: τdwell=∫barrier∣ψ∣2dVJinc\tau_{\text{dwell}}=\frac{\int_{\text{barrier}} |\psi|^2 dV}{J_{\text{inc}}}τdwell​=Jinc​∫barrier​∣ψ∣2dV​.

These can be computed once you postulate an internal S-matrix between states. They’re often within factors of order unity of the Büttiker–Landauer/Larmor times for opaque barriers, and they avoid the “what is orthogonal?” ambiguity.

If the system is swept through an avoided crossing (control parameter λ(t)\lambda(t)λ(t) with rate λ˙\dot{\lambda}λ˙):

Pstay=exp⁡ ⁣(−πΔ22ℏ ∣ε˙∣),ε∝λ,P_{\text{stay}} = \exp\!\Big(-\frac{\pi \Delta^2}{2\hbar\,|\dot{\varepsilon}|}\Big),\quad \varepsilon\propto \lambda,Pstay​=exp(−2ℏ∣ε˙∣πΔ2​),ε∝λ, and the effective transition window (latency) is set by the time interval where ∣ε∣≲Δ|\varepsilon|\lesssim \Delta∣ε∣≲Δ. If your states are driven by a slow HU background “clock,” this can give a natural τ\tauτ once Δ\DeltaΔ and ε˙\dot{\varepsilon}ε˙ are known.

1. Pick your ontology for the hop: - Phase step on an internal rotation? → Use §4. Your current α\alphaα corresponds to N≈119N\approx 119N≈119. That makes the “factor of 10” vs. the rest-energy Heisenberg time unsurprising and physically harmless (it’s not an orthogonalization). - Literal tunneling between quasi-degenerate wells? → Use §2 + §3. Write a minimal 1D (or radial-4D) barrier, compute SSS, get Δ\DeltaΔ, then τ\tauτ.

1. Cross-check with QSLs (§1): ensure your τ\tauτ respects MT/ML given the actual ΔE\Delta EΔE (level spread) of your coherence manifold, not mHc2m_Hc^2mH​c2 unless the hop truly accesses that scale.
2. If dissipation exists: include a dephasing time T2T_2T2​ and relaxation T1T_1T1​. The observable hop time is then τobs∼min⁡{πℏ/Δ, T2}\tau_{\text{obs}}\sim \min\{\pi\hbar/\Delta,\ T_2\}τobs​∼min{πℏ/Δ, T2​}.

- Compton period TC≈4.41×10−24 sT_C\approx 4.41\times 10^{-24}\,\text{s}TC​≈4.41×10−24s.

• Your latency τ=αTC≈3.71×10−26 s\tau=\alpha T_C\approx 3.71\times10^{-26}\,\text{s}τ=αTC​≈3.71×10−26s.
• If (and only if) we forced a full Rabi hop interpretation, it implies Δ∼πℏ/τ≈5.6×1010 eV\Delta\sim \pi\hbar/\tau\approx 5.6\times 10^{10}\,\text{eV}Δ∼πℏ/τ≈5.6×1010eV (tens of GeV) — clearly signaling that the Rabi-half-cycle picture is not the right one for your number.
• Interpreted as a phase step on an internal clock, it’s perfectly natural: N≈119N\approx 119N≈119 equal steps per TCT_CTC​.

- Your “within a factor of 10” intuition against a naive Heisenberg rest-energy time is fine if the hop is a small phase advance on an internal rotation, not an orthogonalization.

• The best path to a principled estimate is to (a) decide whether the hop is a phase step or a true tunnel between wells, then (b) use either phase-step kinematics (§4) or WKB → splitting → Rabi (§§3–2), and finally (c) validate with QSL bounds (§1).

If you want, outline your preferred internal potential (even schematically: barrier height V0V_0V0​, width LLL, effective mass meffm_\text{eff}meff​), and I’ll compute τ\tauτ both via WKB (instanton) and via Wigner–Smith/dwell-time to see which regime lands closest to your α=0.00843\alpha=0.00843α=0.00843.

User: The way I see it, it that the Tunneling period is equal to the compton wavelenght of a hydrogen atom.

The way I see it, it that the Tunneling period is equal to the compton wavelenght of a hydrogen atom. maybe the extra energy is the spinning energy (which is smaller than the binding energy. This latency time or delay is being added to the compton wavelength. If there were no latency, the "distance between steps in the universe expansion would be exactly the compton wavelength of the hydrogen atom. My impression MB andML use measures that are mapped to the states themselves and might be too big or too low. That I am measuring is related to spin I suspect.

Just for fun, recalculate the hop using the Compton Period of a Hydrogen atom. Will we get the mass of a hydrogen atom?

Assistant: Fun check—and it neatly splits into two interpretations.

Fun check—and it neatly splits into two interpretations.

Compton period of H:

TC=hmHc2≈4.405×10−24 s.T_C=\frac{h}{m_H c^2}\approx 4.405\times10^{-24}\ \text{s}.TC​=mH​c2h​≈4.405×10−24 s. If you define the hop time τhop=TC\tau_{\text{hop}}=T_Cτhop​=TC​, then inverting gives

m=hτhop c2=hTCc2=mH.m=\frac{h}{\tau_{\text{hop}}\,c^2}=\frac{h}{T_C c^2}=m_H.m=τhop​c2h​=TC​c2h​=mH​. So yes—you recover the hydrogen mass exactly (it’s true by construction).

(Spatially, the same logic holds with the Compton wavelength λC=h/(mc)\lambda_C=h/(m c)λC​=h/(mc): if the “step length” equals λC\lambda_CλC​, then m=h/(λCc)m=h/(\lambda_C c)m=h/(λC​c).)

Using τhop=πℏ/Δ\tau_{\text{hop}}=\pi\hbar/\Deltaτhop​=πℏ/Δ. Plugging τhop=TC\tau_{\text{hop}}=T_Cτhop​=TC​ gives

Δ=πℏTC=12 mHc2≈4.69×108 eV,\Delta=\frac{\pi\hbar}{T_C}=\frac{1}{2}\,m_H c^2\approx 4.69\times10^8\ \text{eV},Δ=TC​πℏ​=21​mH​c2≈4.69×108 eV, i.e., half the rest energy. That’s not an atomic/binding/spin scale—it just tells you the 2-level/Rabi picture isn’t the right lens for a Compton-paced step.

Say your measured step-time is T⋆=TC+ε=(1+α)TCT^\star = T_C + \varepsilon = (1+\alpha)T_CT⋆=TC​+ε=(1+α)TC​ with α=0.0084304\alpha=0.0084304α=0.0084304.

Then the mass you’d infer from step-time would be

m⋆=hT⋆c2=mH1+α≈0.99164 mH,m^\star=\frac{h}{T^\star c^2}=\frac{m_H}{1+\alpha}\approx 0.99164\,m_H,m⋆=T⋆c2h​=1+αmH​​≈0.99164mH​, i.e., lower by ≈0.836%\approx 0.836\%≈0.836%. Numerically,

• mH≈1.6735×10−27 kgm_H \approx 1.6735\times10^{-27}\ \text{kg}mH​≈1.6735×10−27 kg
• m⋆≈1.6595×10−27 kgm^\star \approx 1.6595\times10^{-27}\ \text{kg}m⋆≈1.6595×10−27 kg
• difference Δm≈1.40×10−29 kg\Delta m \approx 1.40\times10^{-29}\ \text{kg}Δm≈1.40×10−29 kg ⇒\Rightarrow⇒ ΔE≈7.85 MeV\Delta E\approx 7.85\ \text{MeV}ΔE≈7.85 MeV.

So: adding a small positive latency to TCT_CTC​ makes the “mass-from-period” slightly smaller (fractional shift −α-\alpha−α to first order).

If the latency were truly set by a spin-scale energy, that energy is tiny (e.g., the 21 cm hyperfine splitting is ∼5.9 μeV\sim 5.9\,\mu\text{eV}∼5.9μeV), which corresponds to a much longer characteristic time (∼0.7 ns\sim 0.7\ \text{ns}∼0.7 ns), not the ∼10−24\sim 10^{-24}∼10−24–10−26 s10^{-26}\ \text{s}10−26 s regime. So spin-scale energies lengthen times, they don’t shorten them into the Compton regime.

Bottom line:

• Setting the hop equal to the Compton period by definition returns the hydrogen mass.
• Your observed “Compton + small latency” picture maps to a slightly reduced effective mass m⋆≈mH/(1+α)m^\star \approx m_H/(1+\alpha)m⋆≈mH​/(1+α).
• If you want a dynamics-based estimate rather than a definitional one, avoid the 2-level/Rabi mapping at Compton times; treat the hop as an internal phase step on the Compton clock (clean and HU-consistent), or build a WKB/instanton barrier and compute the latency from the action.

User: The way to think is that the tunneling process spinning + tunneling subtracts a little of the mass and extends the per…

The way to think is that the tunneling process (spinning + tunneling) subtracts a little of the mass and extends the period by the total amount 4* the 0.008.. That might be the right way of thinking about this.

Assistant: I like that framing.

I like that framing. If “spinning + tunneling” adds a small delay to each step, you can model it as a multiplicative stretch of the Compton period—and that maps one-to-one to an effective mass defect for the coherence mode.

Let α=0.00843043657\alpha=0.00843043657α=0.00843043657. You’re saying the total added latency per full step is 4α4\alpha4α of the Compton period, so

T⋆=(1+4α) TC,TC=hmHc2.T_\star=(1+4\alpha)\,T_C,\qquad T_C=\frac{h}{m_Hc^2}.T⋆​=(1+4α)TC​,TC​=mH​c2h​. Inferring mass from the period (m=h/(Tc2)m=h/(Tc^2)m=h/(Tc2)) gives the coherence “effective” mass

meff=hT⋆c2=mH1+4α.m_{\rm eff}=\frac{h}{T_\star c^2}=\frac{m_H}{1+4\alpha}.meff​=T⋆​c2h​=1+4αmH​​.

Numbers

• 1+4α=1.033721746281+4\alpha=1.033721746281+4α=1.03372174628 → period is +3.372% longer.
• meff/mH=11+4α=0.9673783139m_{\rm eff}/m_H=\frac{1}{1+4\alpha}=0.9673783139meff​/mH​=1+4α1​=0.9673783139 → mass is −3.262% lower.
• Using mH≈1.6735×10−27 kgm_H\approx1.6735\times10^{-27}\,\text{kg}mH​≈1.6735×10−27kg: Δm=mH−meff≈5.46×10−29 kg,\Delta m = m_H-m_{\rm eff}\approx5.46\times10^{-29}\ \text{kg},Δm=mH​−meff​≈5.46×10−29 kg, ΔE=Δm c2≈30.6 MeV.\Delta E=\Delta m\,c^2\approx 30.6\ \text{MeV}.ΔE=Δmc2≈30.6 MeV.

Interpretation

• Mathematically, your “4α4\alpha4α” latency subtracts an energy ΔE  =  mHc2 ⁣(1−11+4α)  =  mHc24α1+4α,\Delta E \;=\; m_Hc^2\!\left(1-\frac{1}{1+4\alpha}\right) \;=\; m_Hc^2\frac{4\alpha}{1+4\alpha},ΔE=mH​c2(1−1+4α1​)=mH​c21+4α4α​, which for your α\alphaα is ∼\sim∼30 MeV—about a few percent of mHc2m_Hc^2mH​c2.
• Conceptually, that ΔE\Delta EΔE is the budget tied up in the internal spin+tunneling “machinery,” so the coherence’s clock runs slower by 1+4α1+4\alpha1+4α and the mode appears with an effective mass reduced by the same factor’s inverse.

If you decide the delay is some other multiple k αk\,\alphakα (instead of 4), just replace 4α4\alpha4α by kαk\alphakα: meff=mH/(1+kα)m_{\rm eff}=m_H/(1+k\alpha)meff​=mH​/(1+kα), ΔE=mHc2 (kα)/(1+kα)\Delta E=m_Hc^2\,(k\alpha)/(1+k\alpha)ΔE=mH​c2(kα)/(1+kα).

Want me to package these into a tiny HU “state-latency ↔ mass-defect” lemma you can drop into the paper, with the general formulas and the α=0.00843043657\alpha=0.00843043657α=0.00843043657 worked example?

User: I think we can call that a binding energy or even a barrier.

I think we can call that a binding energy or even a barrier. Between the two states, there's a barrier that takes extra time to overcome.

Assistant: That works nicely as a clean “barrier/binding” parametrization.

That works nicely as a clean “barrier/binding” parametrization. You can wrap the whole idea in two equivalent, portable formulas.

A. Period-stretch ↔ mass-defect (binding energy)

Let the added delay per step be a fraction ε\varepsilonε of the Compton period TC=h/(mHc2)T_C=h/(m_Hc^2)TC​=h/(mH​c2). Your “4×0.008430436574\times 0.008430436574×0.00843043657” case means ε=4α≈0.033721746\varepsilon=4\alpha\approx 0.033721746ε=4α≈0.033721746.

• Period: T⋆=(1+ε) TCT_\star=(1+\varepsilon)\,T_CT⋆​=(1+ε)TC​.
• Effective mass: meff=hT⋆c2=mH1+εm_{\rm eff}=\dfrac{h}{T_\star c^2}=\dfrac{m_H}{1+\varepsilon}meff​=T⋆​c2h​=1+εmH​​.
• Binding/barrier energy (the “subtracted mass”):

Eb≡mHc2−meffc2=mHc2 ⁣(1−11+ε)=mHc2 ε1+ε≈ε mHc2(ε≪1).E_b \equiv m_Hc^2-m_{\rm eff}c^2 = m_Hc^2\!\left(1-\frac{1}{1+\varepsilon}\right) = m_Hc^2\,\frac{\varepsilon}{1+\varepsilon} \approx \varepsilon\,m_Hc^2\quad(\varepsilon\ll1).Eb​≡mH​c2−meff​c2=mH​c2(1−1+ε1​)=mH​c21+εε​≈εmH​c2(ε≪1). Numbers (Hydrogen): with ε=0.033721746\varepsilon=0.033721746ε=0.033721746,

meffmH=11+ε≈0.967378⇒Eb≈ε1+ε mHc2≈30–31 MeV.\frac{m_{\rm eff}}{m_H}=\frac{1}{1+\varepsilon}\approx 0.967378\quad\Rightarrow\quad E_b \approx \frac{\varepsilon}{1+\varepsilon}\,m_Hc^2 \approx 30\text{–}31\ \text{MeV}.mH​meff​​=1+ε1​≈0.967378⇒Eb​≈1+εε​mH​c2≈30–31 MeV. (Using mHc2≈938.8 MeVm_Hc^2\approx 938.8\ \text{MeV}mH​c2≈938.8 MeV for the rest-energy scale.)

This captures exactly your statement: “a barrier subtracts a little mass and stretches the clock.”

B. Map the stretch to a tunneling action (WKB/instanton)

If you picture a true tunnel between adjacent coherence states with an internal attempt frequency ω0\omega_0ω0​, then a standard estimate is

τ∼πω0 eS/ℏ,\tau \sim \frac{\pi}{\omega_0}\,e^{S/\hbar},τ∼ω0​π​eS/ℏ, so the stretch factor s≡T⋆/TC=1+εs\equiv T_\star/T_C = 1+\varepsilons≡T⋆​/TC​=1+ε gives

  Sℏ  ≈  ln⁡ ⁣(s ω0 TCπ)  \boxed{\;\frac{S}{\hbar}\;\approx\;\ln\!\Big(\frac{s\,\omega_0\,T_C}{\pi}\Big)\;}ℏS​≈ln(πsω0​TC​​)​ Choose ω0\omega_0ω0​ according to your ontology:

• If the internal clock is the Compton clock, ω0=ωC=mHc2/ℏ\omega_0=\omega_C=m_Hc^2/\hbarω0​=ωC​=mH​c2/ℏ and ω0TC=h/ℏ=2π\omega_0T_C = h/\hbar=2\piω0​TC​=h/ℏ=2π. Then

  Sℏ≈ln⁡ ⁣(2s)=ln⁡ ⁣(2(1+ε))  \boxed{\;\frac{S}{\hbar}\approx \ln\!\big(2s\big)=\ln\!\big(2(1+\varepsilon)\big)\;}ℏS​≈ln(2s)=ln(2(1+ε))​ For your ε=0.033721746\varepsilon=0.033721746ε=0.033721746: S/ℏ≈ln⁡(2.0674)≈0.726S/\hbar\approx \ln(2.0674)\approx 0.726S/ℏ≈ln(2.0674)≈0.726. That’s a pleasantly modest barrier (action of order ℏ\hbarℏ)—intuitively, just enough to add a few-percent dwell time each step.

Practical use in your paper (drop-in)

• Definition: Each HU step is delayed by a fractional latency ε\varepsilonε due to an internal spin+tunnel barrier.
• Consequences: T⋆=(1+ε)TC,meff=mH1+ε,Eb=ε1+ε mHc2.T_\star=(1+\varepsilon)T_C,\qquad m_{\rm eff}=\frac{m_H}{1+\varepsilon},\qquad E_b=\frac{\varepsilon}{1+\varepsilon}\,m_Hc^2.T⋆​=(1+ε)TC​,meff​=1+εmH​​,Eb​=1+εε​mH​c2.
• If ω0=ωC\omega_0=\omega_Cω0​=ωC​: the barrier action is S/ℏ=ln⁡ ⁣(2(1+ε))S/\hbar=\ln\!\big(2(1+\varepsilon)\big)S/ℏ=ln(2(1+ε)).

If you later want to separate the “spin” part from the “barrier” part, you can write ε=εspin+εbar\varepsilon=\varepsilon_{\rm spin}+\varepsilon_{\rm bar}ε=εspin​+εbar​ (for small pieces adding linearly) so that

Eb≃(εspin+εbar) mHc2E_b \simeq (\varepsilon_{\rm spin}+\varepsilon_{\rm bar})\,m_Hc^2Eb​≃(εspin​+εbar​)mH​c2 to first order, while the exact expression above handles the finite-ε\varepsilonε case.