1
$\begingroup$

Consider a connection on a principal $U(1)$-bundle $A_\mu$ over the flat base manifold $M_4$. The action of the theory is described in terms of the curvatures of such connection coupled to some source $j^\mu$ in the following way \begin{align} S=\int_{M_4}d^4x\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-A_\mu j^\mu\right), \end{align} where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ and with sign convention $(+,-,-,-)$. The associated equations of motion are given by \begin{align} \partial_\mu F^{\mu\nu}=j^\nu. \end{align} Both the action and the equations of motion enjoy an invariance under the gauge transformations \begin{align} A'_\mu(x)=A_\mu(x)+\partial_\mu\Lambda(x). \end{align} We now decide to use this invariance to fix a certain condition on the gauge field. For example, we can choose the gauge fixing condition $$ A'_0=0, $$ which is satisfied by the following Cauchy problem \begin{align}\label{condlambda} \begin{cases} \partial_0\Lambda (x)=- A_0(x),\\ \Lambda(\bar x^0,x^i)=\varphi(x^i). \end{cases} \end{align} This is solved by \begin{align} \Lambda(x)=\varphi(x^i)-\int_{\bar x^0}^{x^0}d\tilde x^0 A_0(\bar x^0,x^i). \end{align} A gauge transformation allowed to fix one component of the gauge field to zero.
A second gauge transformation \begin{align} A_\mu''(x)=A'_\mu(x)+\partial_\mu\alpha(x) \end{align} maintaining the gauge fixing condition $A''_0=0$ is still possible if we require \begin{align} \alpha=\alpha(x^i). \end{align} This residual gauge has no effect in terms of fixing another component of the gauge field.
We then focus on the equations of motion: these can be split into \begin{align} \begin{cases} \partial_j\partial^jA''^0(x)-\partial_j\partial^0A''^j(x)=j^0(x),\\ \partial_0\partial^0A''^i(x)+\partial_j(\partial^jA''^i(x)-\partial^iA''^j(x))=j^i(x). \end{cases} \end{align} The first equation can be further simplified $$\partial_j\partial^0A''^j(x)=-j^0(x)\implies \partial_1\partial^0A''^1(x)=-j^0(x)-\partial_a\partial^0A''^a(x),$$ where $a=2,3$. By integrating two times, we get \begin{align}\nonumber \int_{\bar x^0}^{x^0}d\tilde x^0\int_{\bar x^1}^{x^1}d\tilde x^1&\tilde\partial_1\tilde\partial^0A''^1(\tilde x^0,\tilde x^1,x^a)=\\ \nonumber &=-\int_{\bar x^0}^{x^0}d\tilde x^0\int_{\bar x^1}^{x^1}d\tilde x^1\bigg(j^0(\tilde x^0,\tilde x^1,x^a)+\partial_a\tilde\partial^0A''^a(\tilde x^0,\tilde x^1,x^a)\bigg)\\ &\equiv G(\bar x^0, \bar x^1; x^0,x^1, x^a) \end{align} which implies \begin{align} A''^1(x^0,x^1,x^a)-A''^1(\bar x^0,x^1,x^a)-A''^1(x^0,\bar x^1,x^a)+A''^1(\bar x^0,\bar x^1,x^a)=G(\bar x^0, \bar x^1; x^0,x^1, x^a). \end{align} Since $G(\bar x^0, \bar x^1; x^0,x^1, x^a)$ is a function of the remaining directions of the gauge field and the sources, we seem to have determined $A''^1(x^0,x^1,x^a)$ up to terms evaluated at the boundary. Is it correct?
Are these terms part of the Cauchy problem for our system of partial differential equations $\partial_\mu F^{\mu\nu}=j^\nu$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.