Set
$$
a=\frac{v}{2r},
\qquad b=\frac1r,
\qquad c=\frac vr,
\qquad d=\frac{\cot\theta}{r}.
$$
Direct exterior differentiation gives
$$
de^0=0,
\qquad de^1=a,e^0\wedge e^1,
$$
$$
de^2=b,e^1\wedge e^2-c,e^0\wedge e^2,
$$
$$
de^3=b,e^1\wedge e^3-c,e^0\wedge e^3
+d,e^2\wedge e^3.
$$
Solving <Cartan's first structure equation> and imposing $\omega_{\mu\nu}=-\omega_{\nu\mu}$ gives the independent mixed-index <connection 1-forms>
$$
\boxed{\omega^0{}_1=a e^1,
\quad\omega^0{}_2=-c e^2,
\quad\omega^0{}_3=-c e^3},
$$
$$
\boxed{\omega^1{}_2=-b e^2,
\quad\omega^1{}_3=-b e^3,
\quad\omega^2{}_3=-d e^3}.
$$
The remaining forms follow from Lorentz-signature antisymmetry: $\omega^1{}_0=\omega^0{}_1$, $\omega^2{}_0=\omega^0{}_2$, $\omega^3{}_0=\omega^0{}_3$, and $\omega^j{}_i=-\omega^i{}_j$ for spatial indices.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-309/4/b/solution