Jump to content
Main menu
Main menu
move to sidebar
hide
Navigation
Main page
Recent changes
Random page
freem
Search
Search
Appearance
Create account
Log in
Personal tools
Create account
Log in
Pages for logged out editors
learn more
Contributions
Talk
Editing
Openai/6908ee36-d3b4-8010-a468-28e0753d08ae
(section)
Add languages
Page
Discussion
English
Read
Edit
Edit source
View history
Tools
Tools
move to sidebar
hide
Actions
Read
Edit
Edit source
View history
General
What links here
Related changes
Special pages
Page information
Appearance
move to sidebar
hide
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
=== 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∼ω0πeS/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(πsω0TC) 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=mHc2/ℏ and ω0TC=h/ℏ=2π\omega_0T_C = h/\hbar=2\piω0TC=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+εεmHc2. * 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)mHc2 to first order, while the exact expression above handles the finite-ε\varepsilonε case.
Summary:
Please note that all contributions to freem are considered to be released under the Creative Commons Attribution-ShareAlike 4.0 (see
Freem:Copyrights
for details). If you do not want your writing to be edited mercilessly and redistributed at will, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource.
Do not submit copyrighted work without permission!
Cancel
Editing help
(opens in new window)