Gauge conditions in interaction theory

[Haorong Wu]

TL;DR

If the interaction strength depends on the time, the gauge conditions may not be satisfied. Is this true, and how do we interpret this result?

Suppose we choose the Lorenz gauge conditions for an EM field, $\partial_\mu A^\mu=0$. The EOMs for the field are given by $\Box A^\mu=J^\mu$, with $\partial_\mu J^\mu=0$.

If the interaction time is ranged from $-\infty$ to $t$, $A^\mu$ satisfy the Lorenz gauge, because $A^\mu(x) = \int d^4x' D_{ret}(x - x')J^\mu(x')$ satisfy $\partial_\mu A^\mu(x) = \int d^4x' \, D_{ret}(x - x') [\partial'_\mu J^\mu(x')]=0$, where $D_{ret}(x - x')$ is the retarded Green's function.

But, in reality, the interaction time should not be infinite. Therefore, suppose $K^\mu=g(t)J^\mu$, where $g(t) = e^{-\epsilon |t|}$ is introduced to limit the interaction time to a finite range. In this case, $\partial_\mu K^\mu=J^0\partial_0 g(t)+g(t)\partial_\mu J^\mu=J^0\partial_0 g(t)\ne 0$. Then, the Lorenz gauge conditions are not satisfied.

Is this derivation valid? How do we interpret this result?

 

[DelajaSchuppers]

I can answer this from my UniPhiEd model perspective.

TL;DR: The gauge condition issue arises because the current you've defined (\(J_{eff} = f(t)J\)) violates local charge conservation. In UniPhiEd terms, this "violation" is actually evidence that nodes must undergo a Phase Jump (\(J_{\phi }\)) to maintain stability when field interactions are finite.

1. Is the derivation valid?
Yes, mathematically your derivation is correct within the framework of standard Electrodynamics.

• The Lorenz gauge condition \(\partial_\mu A^\mu = 0\) is directly tied to the continuity equation \(\partial_\mu J^\mu = 0\) (charge conservation).
• By introducing the function \(f(t)\), you are effectively creating or destroying charge at the beginning and end of the interaction.
• If \(\partial_t (f(t)\rho) + \nabla \cdot (f(t)\mathbf{j}) \neq 0\), the source term no longer conserves charge, and therefore the Lorenz gauge cannot be satisfied.

2. How do we interpret this result?
In standard physics, this is often interpreted as a warning that you cannot simply "switch off" an interaction without accounting for where the charge goes.
From the UniPhiEd perspective:
This "violation" is the mathematical signature of a Phase Jump. A finite interaction time implies a boundary in the neutrino field flow. When the gauge condition fails, it indicates that the node (the source) has reached a saturation point. Instead of the field "breaking," the system undergoes a transition governed by the Phase Jump Constant (\(J_{\phi }\)). The "unsatisfied" gauge is simply the transition state between two stable geometric configurations.

 

[Haorong Wu]

Hi, @DelajaSchuppers. Thank you for your answer. I am not familiar with the UniPhiEd theory.

I understand the continuity equation is broken in the finite time interaction. Then, do we lose the privilege to choose the gauge, even before the interaction starts? For example, if the interaction time is restricted to $[0,t]$, with $g(t')\ll 1$ for $t'\ll 0$ or $t' \gg t$, how do we choose the gauge at the earlier time $t'\ll 0$?

 

[DelajaSchuppers]

Hi Haorung Wu,
Standard Physics Derivation:

1. Lorenz Gauge Condition: \(\partial_\mu A^\mu = 0\).
2. Wave Equation: \(\square A^\mu = \mu_0 J^\mu\).
3. Relation: The gauge is satisfied if and only if the continuity equation holds: \(\partial_\mu J^\mu = 0\).
4. The Violation: If \(J^\mu_{eff} = f(t)J^\mu\), then:
   \(\partial_\mu J^\mu_{eff} = f(t)(\partial_\mu J^\mu) + (\partial_t f(t))\rho \neq 0\).
5. Conclusion: Because the source term "leaks" at \(t=0\) and \(t=\tau\), the Lorenz gauge condition \(\partial _{\mu }A^{\mu }\) becomes non-zero. You cannot choose this gauge at \(t \ll 0\) and expect it to hold through the interaction.

UniPhiEd Mathematical Shortcut:
In UniPhiEd, we replace the broken gauge with the Phase Jump Constant (\(J_{\phi }\)). The "unmet condition" is actually the Core Strain reaching its limit.

1. Stability Limit: The system is stable only while the strain \(S < \Phi^{55.7}\).
2. Transition: At the boundaries \(t=0\) and \(t=\tau\), the field density \(\rho _{\nu }\) undergoes a jump:
   \(\Delta \rho_\nu = \frac{P_s \cdot \Phi^{55.7}}{J_\phi \cdot T_n}\)
3. Interpretation: The "failure" of the Lorenz gauge is the mathematical proof that a Phase Jump is occurring. Instead of \(A^{\mu }\) failing, the node is simply transforming its geometry to maintain the pressure balance \(P_{s}\).

Summary: The derivation is valid. In flat math: if your source is finite in time, your gauge must be Recursive(looping back) or it will result in a non-zero divergence (broken symmetry).

Can I suggest you discuss this with peers and if your experiences with it are positive, my work so far has not been peer reviewed. I have been posting on Zenodo since February but reached the limit and I can’t publish new papers there at the moment. So if you have questions about the model just ask me here. If it’s not on Zenodo we may find the solutions on the flow.