# 物理代写|广义相对论代写General relativity代考|PHYS3100

## 物理代写|广义相对论代写General relativity代考|Action Integral for the Gravitational Field

Most of the fundamental physical equations of the classical field can be derived from a variational principle. Therefore, it is expected that Einstein’s field equation can also be derived from a variational principle. To derive it, we follow the same notion of the variational principle in classical mechanics and electromagnetism.

Now, we seek a Lagrangian $L_G$ for the gravitational field, which has a geometrical origin that contains derivatives of fundamental tensor $g_{\mu v}$. Since $\sqrt{-g} d^4 x$ is an invariant, the action integral for the gravitational field will take the following form
$$\int \sqrt{-g} L_G d^4 x$$
(Here integration is done over all spacetime coordinates between two given points.)

There are several scalars apart from constant, however, the simplest scalar as Lagrangian, which we can consider is the scalar curvature $L_G=R$. Here $R$ contains metric tensor and Christoffel symbols $\Gamma_{\mu v}^\alpha$ (i.e., derivatives of $g_{\mu v}$ ). One can also think other scalars like $R^2, f(R), R_{i k} R^{i k}, R_{i j k l} R^{i j k l}$, etc., as Lagrangian, which can be treated as modified gravity. Later, we will discuss these modified gravity theories.
So the action integral for the gravitational field will be taken as $\int \sqrt{-g} R d^4 x$.
This is known as the Einstein-Hilbert action. In addition to the gravitational field, one should add to this integral another one which takes care of different fields presented in our physical system.

## 物理代写|广义相对论代写General relativity代考|Einstein’s Equation from Variational Principle

The Einstein’s equation can be derived from the principle of least action $\delta I=0$ where
$$I=\int \sqrt{-g}\left[L_G+2 k L_F\right] d^4 x .$$
Here, we choose $L_G=R$ as the Lagrangian for the gravitational field with $R=g_{\mu v} R^{\mu v}$. $L_F$ is the Lagrangian for all the other fields. $k=$ Einstein’s gravitational constant $=\frac{8 \pi G}{c^4}$. The integral has been taken over the whole spacetime. Here, the action is usually assumed to be a functional of the metric and matter fields. This integral contains the metric tensor, which determines the curvature of spacetime in the presence of the gravitational field either in the absence or the presence of matter. We are varying the first part of the integral (3.1), which yields
\begin{aligned} \delta \int \sqrt{-g} R d^4 x &=\delta \int \sqrt{-g} g^{\mu v} R_{\mu v} d^4 x, \ &=\int \sqrt{-g} g^{\mu v} \delta R_{\mu v} d^4 x+\int R_{\mu v} \delta\left(\sqrt{-g} g^{\mu v}\right) d^4 x . \end{aligned}
Since
$$\delta R_{\mu v}=\left[\left(\delta \Gamma_{\mu v}^\rho\right){; \rho}-\left(\delta \Gamma{\mu \rho}^\rho\right){; v}\right],$$ we have \begin{aligned} \sqrt{-g} g^{\mu v} \delta R{\mu v} &=\sqrt{-g}\left[\left(g^{\mu v} \delta \Gamma_{\mu v}^\rho\right){; \rho}-\left(g^{\mu v} \delta \Gamma{\mu \rho}^\rho\right){; v},\right] \ &=\sqrt{-g}\left[\left(g^{\mu v} \delta \Gamma{\mu v}^\alpha\right){; \alpha}-\left(g^{\mu \alpha} \delta \Gamma{\mu \rho}^\rho\right){; \alpha},\right] \ &=\sqrt{-g}\left(V^\alpha\right){; \alpha}=\frac{\partial}{\partial x^\alpha}\left(\sqrt{-g} V^\alpha\right), \end{aligned}
(by result (8) in Section (3.4))
where
$$V^\alpha=g^{\mu v} \delta \Gamma_{\mu v}^\alpha-g^{\mu \alpha} \delta \Gamma_{\mu \rho}^\rho,$$
is contravariant vector.

## 物理代写|广义相对论代写General relativity代考|Action Integral for the Gravitational Field

$$\int \sqrt{-g} L_G d^4 x$$
(这里的积分是在两个给定点之间的所有时空坐标上完成的。)

$R^2, f(R), R_{i k} R^{i k}, R_{i j k l} R^{i j k l}$ 等，作为拉格朗日，可以被视为修正重力。稍后，我们将讨论这些修改后 的引力理论。

## 物理代写|广义相对论代写General relativity代考|Einstein’s Equation from Variational Principle

$$I=\int \sqrt{-g}\left[L_G+2 k L_F\right] d^4 x .$$

$$\delta \int \sqrt{-g} R d^4 x=\delta \int \sqrt{-g} g^{\mu v} R_{\mu v} d^4 x, \quad=\int \sqrt{-g} g^{\mu v} \delta R_{\mu v} d^4 x+\int R_{\mu v} \delta\left(\sqrt{-g} g^{\mu v}\right) d^4 x .$$

$$\delta R_{\mu v}=\left[\left(\delta \Gamma_{\mu v}^\rho\right) ; \rho-\left(\delta \Gamma \mu \rho^\rho\right) ; v\right],$$

$$\sqrt{-g} g^{\mu v} \delta R \mu v=\sqrt{-g}\left[\left(g^{\mu v} \delta \Gamma_{\mu v}^\rho\right) ; \rho-\left(g^{\mu v} \delta \Gamma \mu \rho^\rho\right) ; v,\right] \quad=\sqrt{-g}\left[\left(g^{\mu v} \delta \Gamma \mu v^\alpha\right) ; \alpha-\left(g^{\mu \alpha} \delta \Gamma \mu \rho^\rho\right) ; \alpha,\right]$$
（由第 (3.4) 节中的结果 (8) ) 其中
$$V^\alpha=g^{\mu v} \delta \Gamma_{\mu v}^\alpha-g^{\mu \alpha} \delta \Gamma_{\mu \rho}^\rho,$$

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