An oscillation changes the translation response of the conditional surface-bending model. It can give a positive response for one mode, even though the unexcited tension background has a negative longitudinal coefficient. With the assigned electron and muon modes at one common radius and canonical wave action, ordinary tension cannot give both positive inertia and the required frequency ratio within the leading wave calculation.

This conclusion concerns the stated geometric surface action. It does not establish a failure of the proposed inertial origin of mass or of multidimensional particles. The full supporting field may carry additional energy, stress and momentum. Those contributions have not been derived from this surface model. Mass remains the selected particle frequency in equivalent units.

Derive the complete graph momentum first

Use the codimension-two graph X(theta)=(r(theta)theta,z(theta)), where theta belongs to the unit m-sphere and d=m+2. The geometric symplectic form is the one already derived in The simplex spectrum from surface bending and Nonlinear response of the bending modes. Its translation momentum is the oriented vector volume of the cone over the surface, multiplied by Gamma. A direct exterior product gives its density, before any spatial restriction:

(m+1) dP⊥/Γ=[−zrmθ+zrm−1∇r−rm∇z]dΩ,(m+1) dPz/Γ=rm+1dΩ.(1) \begin{aligned} (m+1)\,dP_\perp/\Gamma &=\left[-zr^m\theta+zr^{m-1}\nabla r-r^m\nabla z\right]d\Omega,\\ (m+1)\,dP_z/\Gamma&=r^{m+1}d\Omega. \end{aligned} \tag{1}

The gradient is the full unit-sphere tangent gradient. Integration by parts in the first line gives the useful equivalent expressions

P⊥=Γ∫zrm−1(∇r−rθ)dΩ,Pz=Γm+1∫rm+1dΩ.(2) \boxed{ P_\perp=\Gamma\int z r^{m-1}(\nabla r-r\theta)d\Omega, \qquad P_z=\frac\Gamma{m+1}\int r^{m+1}d\Omega.} \tag{2}

Both radial and axial graph variations have entered this result. It includes all ordinary translation directions, rather than only the chosen surface axis.

Use the exact canonical radial coordinate

rm+1=Rm+1+(m+1)Rmq.(3) r^{m+1}=R^{m+1}+(m+1)R^m q. \tag{3}

If q has zero mean, the axial momentum is exactly P=Gamma A_m R^(m+1)/(m+1). The assigned degrees 14, 36 and 24 are even. For their full real or complex harmonic patterns, r(-theta)=r(theta) and z(-theta)=z(theta). The transverse integrand in (2) is odd, so its complete integral vanishes. This remains true for any even graph deformation with zero mean q, while the graph is regular. It is not an assumption that the oscillation carries no momentum.

Average the kinetic term without fitting a magnitude to mass

For fixed R the normal kinetic form is Gamma R^m integral dq wedge dv. A single normalized harmonic has canonical action I and rotating coordinates

q=2IΓRmYLcos⁡ϕ,v=2IΓRmYLsin⁡ϕ.(4) q=\sqrt{\frac{2I}{\Gamma R^m}}Y_L\cos\phi, \qquad v=\sqrt{\frac{2I}{\Gamma R^m}}Y_L\sin\phi. \tag{4}

The action I is the canonical normalization of the oscillation. Its frequency is not inferred from its magnitude. Both the normalization and the radius must be common when comparing the assigned modes.

Start from the full one-form Gamma integral r^(m+1) dz/(m+1). Its pullback contains P dZ and the oscillator one-form. Radius variations generate terms proportional to sin(phi)cos(phi) or cos(2phi). Their averages over a complete oscillation vanish. The averaged translational and internal terms are therefore

Θav=P dZ+I dϕ,(5) \Theta_{\rm av}=P\,dZ+I\,d\phi, \tag{5}

with the appropriate orientation variables when the whole surface is rotated. They may carry their own angular Berry terms. Equation (5) is the leading averaged radial translation/action reduction, not a full particle boost or an assertion that every geometric collective coordinate is independent.

For small wave action, the averaged energy is

H(P,I)=E0+T(R)+IωL(R)+O(I2),T(R)=τAmRm.(6) H(P,I)=E_0+T(R)+I\omega_L(R)+O(I^2),\qquad T(R)=\tau A_mR^m. \tag{6}

The round conformal bending energy E_0 is constant in R. It remains part of the energy although its radial derivative vanishes. At leading weak tension the frequency is proportional to R^(-m-2). For a finite tension the complete linear normal-pair frequency must be used, as below.

The oscillation can repair one branch's response

Write W=I omega_L. In the leading bending limit, as functions of P proportional to R^(m+1), T has exponent p=m/(m+1) and W has exponent q=-(m+2)/(m+1). Differentiating the complete (6) gives the longitudinal and transverse inverse response coefficients

I∥−1=H′′(P)=p(p−1)T+q(q−1)WP2,I⊥−1=H′(P)P=pT+qWP2.(7) \begin{aligned} \mathcal I_\parallel^{-1} &=H''(P)=\frac{p(p-1)T+q(q-1)W}{P^2},\\ \mathcal I_\perp^{-1} &=\frac{H'(P)}P=\frac{pT+qW}{P^2}. \end{aligned} \tag{7}

There is one longitudinal coefficient and d-1 transverse ones. These are derivatives of the steady translation momentum-velocity relation within this averaged family. They do not yet include the supporting field or a complete frequency-dependent shape response.

Both coefficients in (7) are positive precisely when

m(m+2)(2m+3)<WT<mm+2.(8) \boxed{\frac{m}{(m+2)(2m+3)} <\frac WT<\frac m{m+2}.} \tag{8}

For m=4 this interval is 2/33<W/T<2/3. The background alone has the negative longitudinal coefficient found earlier; the wave can overcome it. Too much wave energy instead makes the transverse coefficient negative. At W/T=m/(3m+4), the two coefficients happen to agree. That algebraic equality at a translating point is not by itself a relativistic rest-mass test; a moving relativistic object need not have equal longitudinal and transverse responses.

The positive interval spans a factor 2m+3. At a common radius, tension and I, the assigned wave energies have ratio W_mu/W_e=3838380/18564=206.7647059. For m=4 this exceeds the factor 11 available in (8). Therefore no common wave action or tension in this leading model makes both inertia tensors positive. This comparison does not retune a separate magnitude for each mass.

The weak-tension approximation is controlled when wave action and tension are both scaled small. Their energies in (6) are then first order, while tension's correction to the wave energy and nonlinear wave terms enter at second order. It must not be extrapolated to a large tension without the following calculation.

Retain the exact linear tension correction

Define x=tau R^m/kappa, lambda=L(L+m-1) and p_L=(L-1)L...(L+m). The exact linear two-normal frequency from the same surface action is

ωL=κmΓRm+2[pL+mx(λ−m)][pL+mxλ].(9) \omega_L=\frac{\kappa}{m\Gamma R^{m+2}} \sqrt{[p_L+mx(\lambda-m)][p_L+mx\lambda]}. \tag{9}

At fixed tau, kappa and Gamma, write a_L=d log(omega_L)/d log R and b_L=d^2 log(omega_L)/d(log R)^2. Put u=p_L+mx(lambda-m) and v=p_L+mx lambda. Direct differentiation gives

aL=−(m+2)+m2(u−pLu+v−pLv),bL=m22((u−pL)pLu2+(v−pL)pLv2).(10) \begin{aligned} a_L&=-(m+2)+\frac m2\left(\frac{u-p_L}{u}+\frac{v-p_L}{v}\right),\\ b_L&=\frac{m^2}{2}\left(\frac{(u-p_L)p_L}{u^2} +\frac{(v-p_L)p_L}{v^2}\right). \end{aligned} \tag{10}

The complete leading-in-action response is now positive when

maL2−(m+1)aL+bL<IωLT<m−aL.(11) \frac{m}{a_L^2-(m+1)a_L+b_L} <\frac{I\omega_L}{T}<\frac m{-a_L}. \tag{11}

Intersecting the two intervals in I/T tests every common normalization at that x. Near the simplex spectrum the intersection is empty. Using the earlier illustrative fractional ratio budget of 2e-5, x is at most about 0.29156255, and the upper common bound is only about 0.0532 times the required lower bound.

The first point where the two bounds meet is approximately x=7.85163034e6. At that boundary one response coefficient vanishes; strict positivity begins beyond it. The corresponding frequency ratio is only 6.3023954. Thus increasing ordinary tension enough to permit joint positivity changes the ratio by about 97 percent. It does not preserve the mass relationship.

Verification and physical limit

oscillating-surface-inertia-20260914T025641_942833Z.json passes 120 checks. Complete exterior products independently verify every translation component in (1). Actual graph quadrature at the assigned even degrees verifies (2)-(3), and the full pulled-back symplectic form verifies (5). Independent energy derivatives check (7)-(8). The exact tension frequency and interval intersection give the threshold in (11).

A separate part of the report evaluates the finite first-order spinor parent's radius derivatives. Perron derivatives of its exact scaled inverse agree with independent 90-digit singular-value differences. These are canonical radius-response coefficients of that parent. Inserting them into the surface momentum law would require an additional derivation connecting the two kinetic actions; the report does not assert such a connection or call it a complete particle inertia.

The result strengthens the earlier surface test by retaining the oscillation's momentum and energy. A different supporting same-field stress, normal profile or kinetic reduction remains possible. The actual master-field momentum and boost response must be derived before identifying this geometric feature with a physical particle.