Single displacement reaction

28 year-old Painting Investments Worker Truman from Regina, usually spends time with pastimes for instance interior design, property developers in new launch ec Singapore and writing. Last month just traveled to City of the Renaissance. In Riemannian geometry, we can introduce a coordinate system over the Riemannian manifold (at least, over a chart), giving n coordinates

$x_{i}\;{\text{,}}\qquad i=1,\dots ,n$

for an n-dimensional manifold. Locally, at least, this gives a basis for the 1-forms, dxⁱ where d is the exterior derivative. The dual basis for the tangent space T is e_i.

Now, let's choose an orthonormal basis for the fibers of T. The rest is index manipulation.

Example

Take a 3-sphere with the radius R and give it polar coordinates α, θ, φ.

e(e_α)/R,

e(e_θ)/R sin(α) and

e(e_φ)/R sin(α) sin(θ)

form an orthonormal basis of T.

Call these e₁, e₂ and e₃. Given the metric η, we can ignore the covariant and contravariant distinction for T.

Then, the dreibein (triad),

$e_{1}=R\,d\alpha$

$e_{2}=R\,\sin {(\alpha )}d\theta$

$e_{3}=R\,\sin {(\alpha )}\sin {(\theta )}d\phi$.

So,

$de_{1}=0$

$de_{2}=R\cos {(\alpha )}d\alpha \wedge d\theta$

$de_{3}=R(\cos {(\alpha )}\sin {(\theta )}d\alpha \wedge d\phi +\sin {(\alpha )}\cos {(\theta )}d\theta \wedge d\phi )$.

from the relation

$d_{\mathbf {A} }e=de+A\wedge e=0$,

we get

$A_{12}=-\cos {(\alpha )}\,d\theta$

$A_{13}=-\cos {(\alpha )}\,\sin {(\theta )}d\phi$

$A_{23}=-\cos {(\theta )}\,d\phi$.

(d_Aη=0 tells us A is antisymmetric)

So, $\mathbf {F} =d\mathbf {A} +\mathbf {A} \wedge \mathbf {A}$,

$F_{12}=\sin {(\alpha )}d\alpha \wedge d\theta$

$F_{13}=\sin {(\alpha )}\sin {(\theta )}d\alpha \wedge d\phi$

$F_{23}=\sin ^{2}{(\alpha )}\sin {(\theta )}d\theta \wedge d\phi$