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--- formulas:gauss_law [2018/03/30 15:41]
jakobadmin [Concrete]
+++ formulas:gauss_law [2018/05/13 09:19] (current)
jakobadmin ↷ Page moved from formulas:yang_mills_equations:gauss_law to formulas:gauss_law
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 ====== Gauss Law ======
+//also called "Gauss constraint"//
@@ Line 6: / Line 6: @@
   * https://www.physicsforums.com/insights/partial-derivation-gausss-law/
 <tabbox Concrete>
 **Gauss' Law for Yang-Mills Theories**
-Gauss' law for Yang-Mills theories reads
-$$  I(x) \equiv D_i G^{0i} =\partial_i E^i + [A_i,E^i]=0 $$
+Gauss law is the $\nu=0$ component of the [[equations:yang_mills_equations|Yang-Mills equation]]
-and is one of the field equations that are derivable from the [[equations:yang_mills_equations|Yang-Mills Lagrangian]].
+$$ (\partial_\mu F_{\mu \nu})^a = g j_\nu^a  $$
+$$ \rightarrow (\partial_i F_{i 0})^a = g j_0^a  $$
+which is exactly analogous to the inhomogeneous Maxwell equation in the presence of matter fields.
+It does not contain second order time derivatives is therefore not an equation that governs the time development (=an equation of motion), but rather a constraint on the initial conditions.
+//(Source: "Classical Theory of Gauge Fields" by Rubakov. See also the discussion there for more details.)
+//
+We can rewrite it in terms of the gauge potentials as
+$$  I(x) \equiv D_i G^{0i} =\partial_i E^i + [A_i,E^i]=0 .$$
-It does not contain second order time derivatatives is therefore not an equation that govers the time development (=an equation of motion), but rather a constraint on the initial conditions.
 This equation contradicts the commutator relations and thus is not an operator equation. Instead, we use it as a condition for physical states, which have to satisfy:
@@ Line 50: / Line 61: @@
 <tabbox Abstract>
+**Gauss law generates gauge transformations**
+Those gauge transformations that approach identity at infinity and are contractible to the identity are generated by Gauss' law $((\vec D\cdot \vec E) -\rho) |\text{phys}\rangle =0$, where $\rho$ describes a possible contribution from matter.
+[[https://books.google.de/books?id=0oNj1AdbFq8C&lpg=PA143&ots=Ubbuq2J-jQ&dq=gauss%20law%20generator%20of%20noether%20symmetry&hl=de&pg=PA143#v=onepage&q&f=false|Source]]
+<blockquote>
+The choice $A_0= 0$ does not exhaust all gauge freedom; time-independent gauge transformations are still permitted. We must examine the precise content auf gauge invariance under such transformations, in the quantum field theoretical context. In the canonical Hamiltonian procedure in the presence of constraints, this freedom due to time-independent gauge transformations can be traced to Gauss's law:
+\begin{align}
+I(x) &= Div E(x) -j_0(x) \notag \\
+&= dE_x/dx-\frac{1}{2}ie(\phi^\star D_0\Phi-\Phi(D_0\Phi)^\star)=0.
+\end{align}
+Recall that this equation is one of the Euler-Lagrange equations arising from the Lagrangian (10.31). But it involves no time derivatives of the electromagnetic field and is a constraint equation rather than a genuine equation of motion. In the quantized theory, it cannot be considered an operator equation since it conflicts with canonical commutation rules. This difficulty is well known (see for example Bjorken and Drell 1965). Equal time commutation rules give
+\begin{align}
+&[E_x(x_1),A_x(x_2)]_{t=0}=i \delta(x_1-x_2); \notag \\
+&[\phi,A_\mu]_{t=0}=[D_0\Phi,A_\mu]_{t=0}=0.
+\end{align}
+Hence
+$$[I(x_1),A_x(x_2)]_{t=0}=i \partial/\partial_{x_1} [ \delta(x_1-x_2)]\neq 0 $$
+and $I(x_1)$, as an operator equation, cannot vanish. On therefore imposes Gauss's law as a constraint on 'physical' states. These are defined as that subset of states which obey
+$$ I(x) |\Psi\rangle _{phys}=0. $$
+<cite>page 321 in Solitons by Rajaraman</cite>
+</blockquote>
+<blockquote>
+Next, consider the operator
+$$ U_\Lambda = exp\left( \frac{i}{e} \int_{-\infty}^\infty \Lambda(x) I(x)dx \right), $$
+where \Lambda(x) is any c-number function. Then
+$$ U_\Lambda |\Psi\rangle _{phys} = |\Psi\rangle _{phys} \tag{10.43} $$
+Now, suppose we restrict ourselves to that subset of functions, labelled $\tilde \Lambda(x)$, which satisfies $\tilde \Lambda (x) \to 0$ as $x\to \infty$.
+[...]
+We see that $U_{\tilde \Lambda}$ is just the operator which executes time-independent gauge transformations, in the quantum context [...]
+Equation (10.43) thus tells us that physical states must be invariant under such transformations. This is hardly an unfamiliar result, but in deriving it this way we notice the important restriction that $\tilde \Lambda( \pm \infty) =0$. Gauge transformations certainly exist for a general $\Lambda (x)$ not satisfying $\Lambda (\pm \infty)=0$ and they do leave the Lagrangian and the Hamiltonian invariant. But Gauss's law does not necessarily force all wavefunctionals to be invariant under them. Gauge transformations by the restricted subset of functions $\tilde \Lambda (x)$ may be called, for a want of a better name, "small gauge transformation". The name implies not that $\tilde \Lambda (x)$ is small everywhere, but that $\Lambda (\pm \infty)=0$. Gauge transformations that are not "small", will be called large.
+<cite>page 323 in Solitons by Rajaraman</cite>
+</blockquote>
+----
 <blockquote>Additional complications arise when defining quantised gauge field theories. This is because the restriction of a theory to be invariant under a gauge group symmetry $\mathcal{G}$, which corresponds to local invariance under some global symmetry group $G$, leads to a strengthened form of the Noether current conservation condition called the //local Gauss law// [11]:
@@ Line 66: / Line 132: @@
 <cite>[[https://arxiv.org/abs/1408.3233|Boundary terms in quantum field theory and the spin structure of QCD]] by Peter Lowdon</cite></blockquote>
 <tabbox Why is it interesting?>
+Gauss law in an important constraint on the initial data in gauge theories. It is particularly important if we want to understand how we can [[advanced_tools:gauge_symmetry:gauge_fixing|fix the gauge]] consistently.
 <blockquote>

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