Physics:Melvin metric

Short description: Exact solution for the Einstein field equations

The Melvin metric describes the geometry of empty space having cylindrical symmetry containing a magnetic field pointing in the $z$-direction. The Melvin metric is given by

$d{s}^2=f^2(-d{t}^2+d{z}^2+d{r}^2)+\frac{r^2}{f^2}d{\varphi }^2,f=1+\frac{B^2{r}^2}{4},$

with $t,z\in (-\mathrm{\infty },+\mathrm{\infty }),r\in [0,\mathrm{\infty }),\varphi \in [0,2\pi )$. The Melvin magnetic solution is used in astrophysical models as a background for more complicated spacetimes.

Einstein tensor

The nonzero components of the Einstein tensor are

$G_{tt}=G_{rr}=\frac{B^2}{f^2},G_{zz}=-\frac{B^2}{f^2},G_{\varphi \varphi }=\frac{B^2{r}^2}{f^6}$

Stress-energy tensor

The field strength tensor is

$F=e^{i\psi }B(dz\wedge dt)$

$\psi$ is the duality rotation parameter. For $\psi =\pi /2$, we have $F=B(1+B^2{r}^2/4)rdr\wedge d\varphi$, a magnetic field oriented in the $z$-direction. The nonzero components of field strength tensor are

$F_{r\varphi }=-F_{\varphi r}=\frac{Br}{f^2}$

Recall the form of the stress-energy tensor for a source-free electromagnetic field

$T_{\mu \nu }=\frac{1}{4\pi }(F_{\mu \rho }{F}_{\nu }^{\rho }-\frac{1}{4}{g}_{\mu \nu }{F}_{\rho \sigma }{F}^{\rho \sigma })$

Knowing the form of the Einstein tensor, we know that $T_{\mu \nu }$ is nonzero for

$T_{tt}=T_{rr}=-T_{zz}=\frac{B^2}{8\pi f^2},T_{\varphi \varphi }=\frac{B^2{r}^2}{8\pi f^6}$