---
title: "Building an electric-flux chain with Gauss’s law"
description: "Calculate electric flux site by site, distinguish staggered charge from occupation, and test why total charge alone misses local violations."
canonical_url: "https://sandboxphysics.com/stories/lattice-gauge/local-gauss/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/lattice-gauge/local-gauss/"
last_modified: "2026-10-27"
---

[Lattice gauge fields and string breaking](https://sandboxphysics.com/stories/lattice-gauge/)

L1 · Chapter 1

# Building an electric-flux chain with Gauss’s law

Sandbox Physics · 2026-10-27

Electric flux increases when you cross a positive charge and decreases when you cross a negative charge. A valid chain must satisfy this rule at every site. Checking only its total charge can miss a pair of local mistakes.

## Before you begin

You need addition with signed numbers. Treat a site as a position and a link as the segment between neighboring positions. Quantum states and the Hamiltonian enter in the next chapter.

- Infer a local charge from a difference in flux.
- Reconstruct every link from the left boundary.
- Explain how a neutral chain can still violate a local constraint.

## Walk along the chain

A chain has no sideways direction for the field to spread into. Choose rightward flux as positive. Start with zero flux at the left boundary. Passing a positive charge adds one unit. Neutral sites leave it unchanged. An equal negative charge at the far end brings the flux back to zero.

Label the sites $n=0,1,\ldots,N-1$. Write $E_n$ for the flux on the link from site $n$ to its right neighbor. At each site, outgoing rightward flux minus incoming rightward flux equals the total charge there.

$$
E_n-E_{n-1}=Q_n+q_n^{\mathrm{ext}}
$$

$Q_n$ is the mobile matter charge and $q_n^{\mathrm{ext}}$ is a fixed external charge. The fixed left and right boundary fluxes enter the first and last site equations.

Fixed sources at the endpoints · zero boundary flux

$q_0^{\mathrm{ext}}=+1$ $q_3^{\mathrm{ext}}=-1$

Intact string

$0$

$1$

$1$

$1$

$0$

$1$

$1$

One link screened

$0$

$1$

$0$

$0$

$1$

$1$

$1$

Original configuration diagram. Circles show matter occupation; links show rightward electric flux. Removing the particle from odd site one and placing it on even site two changes the middle flux from one to zero.

Reading: David Tong · Sandbox Physics FR04 finite Hamiltonian

## An occupied site need not carry positive charge

Staggered fermions use alternating reference occupations: empty on even sites, occupied on odd sites. Charge is the actual occupation minus this reference. The alternating product configuration is therefore neutral. It is a convenient counting reference; finding the interacting ground state is a separate calculation.

Each actual occupation $f_n$ is zero or one. Relative to the background $b_n$, filling an even site produces positive charge. Emptying an odd site produces negative charge. The missing particle is called a hole. Moving one particle between those sites can create a positive-negative pair without changing total particle number.

$$
b_n=\frac{1-(-1)^n}{2},\qquad Q_n=f_n-b_n
$$

The four-site reference reads empty, occupied, empty, occupied. Charge counts departures from that pattern.

A neutral reference and one neighboring hop, listed from left to right.

| Configuration | Occupations | Matter charges |
| --- | --- | --- |
| Neutral reference | $(0,1,0,1)$ | $(0,0,0,0)$ |
| Middle particle hops right | $(0,0,1,1)$ | $(0,-1,+1,0)$ |

## Work through all four sites

Set both boundary fluxes to zero. Put a fixed positive source on site zero and a fixed negative source on site three. Start the matter in its neutral reference configuration. The first source raises the flux to one; the two middle sites leave it unchanged; the final source lowers it to zero. The internal links carry $(1,1,1)$.

Let the middle particle hop right, giving occupations $(0,0,1,1)$. A negative hole appears on site one and a positive particle on site two. Adding the charges again gives $(1,0,1)$. The new matter charges remove the field from the middle link. This is local screening.

If you changed the occupations but left all three fluxes at one, total charge would still vanish. Two sites would nevertheless fail Gauss’s law. A global test cannot detect where the error occurred.

$$
G_n=E_n-E_{n-1}-Q_n-q_n^{\mathrm{ext}}
$$

The residual $G_n$ measures the mismatch at one site. Every residual must vanish in an allowed configuration.

## The boundary restricts the allowed configurations

Sum Gauss’s law over the chain. Every internal flux appears once with each sign and cancels. Right-boundary flux minus left-boundary flux equals total charge. With both boundaries fixed, adding one unbalanced charge must fail the last boundary test.

The same observation gives a useful construction. Start at the left boundary and add each site’s matter charge and external source. This determines every internal flux. Check the right boundary at the end, then check whether each flux lies inside the retained electric range. The next chapter uses this construction to build a smaller physical state space.

$$
E_n=E_{-1}+\sum_{j=0}^{n}(Q_j+q_j^{\mathrm{ext}})
$$

On an open one-dimensional chain, accumulated charge fixes the links. A periodic chain also needs a winding-flux choice, so its construction is different.

## Try the prediction

### [Make a local violation and repair it](https://sandboxphysics.com/labs/gauss-law-builder/)

1. Choose “An allowed string” and predict each link flux between the two external sources.
2. Change one occupation or one link and identify which sites acquire a residual.
3. Rebuild the electric field from left to right, then check whether the fixed right boundary permits your configuration.

## Exercises and explanations

1. Why does increasing one internal link flux by one change the Gauss residual at two sites?

   Read the explanation

   That link is the right-hand flux of one site and the left-hand flux of its neighbor. Its contributions have opposite signs. One residual increases by one and the other decreases by one, so their sum can remain unchanged.

2. In the four-site example, change the occupations to $(1,0,0,1)$. Which fluxes follow? Can an electric cutoff of one retain them?

   Read the explanation

   The matter charges become $(+1,-1,0,0)$. Including the fixed endpoint sources gives internal fluxes $(2,1,1)$ and returns the right boundary to zero. The configuration satisfies Gauss’s law but exceeds the cutoff on the first link.

[Next: Turning allowed configurations into a quantum model](https://sandboxphysics.com/stories/lattice-gauge/hamiltonian-basis/)

## Sources and calculation scope

1. [Sandbox Physics FR04 finite Hamiltonian, conventions and calculation scope](https://github.com/yangzichao/sandboxphysics/blob/main/docs/frontier/fr04/model-specification.md)

   Defines the course’s zero-based staggered charge, open boundaries, integer-flux hard cutoff, source geometry and deliberate gauge-breaking error. The accompanying paper-reproduction and numerical-validation notes specify which finite-model calculations were checked independently. This is our model definition and calculation record.

2. [David Tong, Gauge Theory (2018 lecture notes)](https://www.damtp.cam.ac.uk/user/tong/gaugetheory/gt.pdf)

   Sections 2.5.1 and 2.7 discuss static probes and dynamical matter; sections 4.2.1 and 4.3.1 introduce lattice links and two-dimensional staggered fermions; section 7.1 gives the electrostatic energy in one spatial dimension. Our finite Hamiltonian, indexing and hard cutoff are defined explicitly in this course.