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# 物理代写|广义相对论代写General Relativity代考|Computing curvature and geodesics with forms

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## 物理代写|广义相对论代写General Relativity代考|Computing curvature and geodesics with forms

As a simple example of computing with forms, let us compute the Ricci tensor of the unit sphere. The diad of a sphere can be taken to be
$$e^1=d \theta, \quad e^2=\sin \theta d \phi$$
which yields (3.34). Writing equation (3.87) explicitly gives
\begin{aligned} & d e^1+\omega^{12} \wedge e^2=0 \ & d e^2+\omega^{21} \wedge e^1=0 \end{aligned}
that is, using the value of the diad explicitly,
\begin{aligned} \omega^{12} \wedge \sin \theta d \phi & =0, \ \cos \theta d \theta \wedge d \phi+\omega^{21} \wedge d \theta & =0 . \end{aligned}
The first equation gives $\omega_\theta^{12}=0$, the second, $\omega_\phi^{12}=-\cos \theta$. That is,
$$\omega^{12}=-\cos \theta d \phi$$
From the definition of the curvature, we have
$$R^{12}=d \omega^{12}+\omega^{1 k} \wedge \omega^{k 2}=\sin \theta d \theta \wedge \mathrm{d} \phi$$
while the second term vanishes. We easily see that the diagonal components of $R^{i j}$ vanish as well. Finally, the Ricci scalar is
$$R=e_i^a e_j^b R_{a b}^{i j}=e_1^1 e_2^2 R_{12}^{12}+e_2^2 e_1^1 R_{21}^{21}=2 \frac{\sin \theta}{\sin \theta}=2$$

## 物理代写|广义相对论代写General Relativity代考|GEOMETRY

Consider a 3d space with triad field $e_a^i(\mathrm{x})$ and metric $g_{a b}(\mathrm{x})$. Geometrical quantities can be expressed as follows.

As we have already seen, the length of a line $\gamma$ defined by $x^a(\tau)$ is
$$L[\gamma]=\int_\gamma \sqrt{g_{a b} \frac{d x^a}{d \tau} \frac{d x^b}{d \tau}} d \tau=\int_\gamma \sqrt{e^i(\dot{x}) e^j(\dot{x}) \delta_{i j}} d \tau .$$
The volume of a finite region $R$ is given by
$$V[R]=\int_R \sqrt{\operatorname{det}[g]} d^3 x=\int_R \operatorname{det}[e] d^3 x=\frac{1}{3 !} \int \epsilon_{i j k} e^i \wedge e^j \wedge e^k$$
Consider next a 2 d surface $S$ immersed in the space coordinated by the coordinates $\sigma, \tau$ and defined by the functions $x^a(\sigma, \tau)$. The tangents to the surface are

$$\dot{x}^a=\frac{\partial X^a}{\partial \sigma}, \quad \hat{X}^a=\frac{\partial x^a}{\partial \tau} .$$
The area of the surface is given by the integral of the square root of the determinant $g^{(2)}$ of the metric induced on the surface
\begin{aligned} A[S] & =\int_S \sqrt{\operatorname{det} g^{(2)}} d \sigma d \tau \ & =\int_S \sqrt{\left(g_{a c} g_{b d}-g_{a b} g_{c d} \mid \dot{\mathbf{X}}^a \hat{\mathbf{X}}^b \dot{\mathrm{X}}^c \hat{\mathbf{X}}^d\right.} d \sigma d \tau \ & =\int_S \sqrt{E_{a b}^i E_{c d}^i \dot{\mathbf{X}}^a \hat{\mathrm{X}}^b \dot{\mathrm{X}}^c \hat{\mathrm{X}}^d} d \sigma d \tau \ & \equiv \int_S|E|, \end{aligned}
where the two-form
$$E^i=E_{a b}^i d x^a \wedge d x^b=\frac{1}{2} \epsilon_{i j k} e^i \wedge e^k$$
is called the ‘Ashtekar’s electric field’, or the ‘gravitational electric field’. Hence we can say that ‘the area is the norm of the gravitational electric field’.

## 物理代写|广义相对论代写General Relativity代考|Computing curvature and geodesics with forms

$$e^1=d \theta, \quad e^2=\sin \theta d \phi$$

$$d e^1+\omega^{12} \wedge e^2=0 \quad d e^2+\omega^{21} \wedge e^1=0$$

$$\omega^{12} \wedge \sin \theta d \phi=0, \cos \theta d \theta \wedge d \phi+\omega^{21} \wedge d \theta \quad=0$$

$$\omega^{12}=-\cos \theta d \phi$$

$$R^{12}=d \omega^{12}+\omega^{1 k} \wedge \omega^{k 2}=\sin \theta d \theta \wedge \mathrm{d} \phi$$

$$R=e_i^a e_j^b R_{a b}^{i j}=e_1^1 e_2^2 R_{12}^{12}+e_2^2 e_1^1 R_{21}^{21}=2 \frac{\sin \theta}{\sin \theta}=2$$

## 物理代写|广义相对论代写General Relativity代考|GEOMETRY

$$L[\gamma]=\int_\gamma \sqrt{g_{a b} \frac{d x^a}{d \tau} \frac{d x^b}{d \tau}} d \tau=\int_\gamma \sqrt{e^i(\dot{x}) e^j(\dot{x}) \delta_{i j}} d \tau$$

$$V[R]=\int_R \sqrt{\operatorname{det}[g]} d^3 x=\int_R \operatorname{det}[e] d^3 x=\frac{1}{3 !} \int \epsilon_{i j k} e^i \wedge e^j \wedge e^k$$

$$\dot{x}^a=\frac{\partial X^a}{\partial \sigma}, \quad \hat{X}^a=\frac{\partial x^a}{\partial \tau}$$

$$A[S]=\int_S \sqrt{\operatorname{det} g^{(2)}} d \sigma d \tau \quad=\int_S \sqrt{\left(g_{a c} g_{b d}-g_{a b} g_{c d} \mid \dot{\mathbf{X}}^a \hat{\mathbf{X}}^b \dot{\mathrm{X}}^c \hat{\mathbf{X}}^d\right.} d \sigma d \tau=\int_S \sqrt{E_{a b}^i E_{c d}^i \dot{\mathbf{X}}^a \hat{\mathbf{X}}^b \dot{\mathrm{X}}^c \hat{\mathbf{X}}^d d \sigma d \tau} \quad \equiv \int_S|E|$$

$$E^i=E_{a b}^i d x^a \wedge d x^b=\frac{1}{2} \epsilon_{i j k} e^i \wedge e^k$$

## MATLAB代写

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