Shunyaya Symbolic Mathematical Chemistry – Sign Lemma & Monotonicity (3.3, 3.4)

What this page shows (plain language):

  • Direction without tables: the sign of RSI follows the sign of the energetic contrast e. If formation is favorable (e > 0), RSI > 0; if not, RSI < 0.
  • Monotone response: with fixed stoichiometry and a positive global slope c, RSI strictly increases as e increases.
  • Scale-safe: positive rescalings of units, slope, or weights do not change the sign; duplicating a balanced reaction (scaling all coefficients) does not change RSI.
  • Edge cases and guards: e = 0 ⇒ RSI = 0. A valid reaction must have positive total reactant weight; clamp alignments before any atanh to keep numerics safe.

Plain ASCII formulas (copy-ready)

Claim (Sign Lemma)

Given:
  w = abs(m)^gamma  with gamma >= 0
  W_r = sum_over_reactants(w_r) > 0
  W_p = sum_over_products(w_p)
  e = (E_formed - E_broken) / E_unit   with E_unit > 0
  a_p = tanh(+c * e), a_r = tanh(-c * e)  with c > 0

Then:
  sign(RSI) = sign(e)

Proof (ASCII)

u_p = atanh(a_p) = +c * e
u_r = atanh(a_r) = -c * e

U_r = sum_over_reactants( w_r * u_r )
V_p = sum_over_products(  w_p * u_p )
W_r = sum_over_reactants( w_r )

V_p - U_r = c * e * ( W_p + W_r )
s = ( V_p - U_r ) / W_r = c * e * ( W_p + W_r ) / W_r
RSI = tanh(s)

Since W_r > 0 and c > 0, (W_p + W_r)/W_r > 0.
tanh is odd and strictly increasing ⇒ sign(RSI) = sign(e).  QED

Consequences

E_formed > E_broken (e > 0)  =>  RSI > 0
E_formed < E_broken (e < 0)  =>  RSI < 0
Sign is unchanged by positive rescalings of E_unit, c, or gamma
Stoichiometric scaling by any k > 0 leaves RSI unchanged
Edge: e = 0 => RSI = 0 ; if W_r = 0, reaction statement invalid (RSI undefined)

Notes (guards)

Clamp alignments so |a| <= 1 - eps_a before atanh(a)
Use a single lens assignment; for multi-step paths, accumulate contrast first, then assign once


Claim (Monotonicity)

For fixed stoichiometry and c > 0:  d(RSI)/de > 0

Proof (ASCII)

Let RSI = tanh(s) with s = ( V_p - U_r ) / W_r.
From the symmetric assignment:
  atanh(a_p) = +c * e  =>  d/de[atanh(a_p)] =  c
  atanh(a_r) = -c * e  =>  d/de[atanh(a_r)] = -c

Thus:
  ds/de = (1/W_r) * ( sum_p w_p * c - sum_r w_r * (-c) )
        = c * ( W_p + W_r ) / W_r  > 0  (since c>0, W_r>0)

And:
  d(RSI)/de = ( 1 - tanh(s)^2 ) * ds/de  > 0.   QED

Notes (guards)

Requires a single lens and fixed stoichiometry with w = abs(m)^gamma, gamma >= 0 => W_r > 0
Clamp |a| <= 1 - eps_a to keep atanh(a) finite
Invariant to positive rescalings of E_unit and c, and to uniform scaling of all coefficients (k > 0)
W_r_safe with eps_w is a numerical guard; proof uses W_r (>0)


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