# 物理代写|广义相对论代写General relativity代考|Perfect Cosmological Principle

## 物理代写|广义相对论代写General relativity代考|Perfect Cosmological Principle

The perfect cosmological principle states that the universe appears to be the same not only at all points and in all directions but in all epochs. This hypothesis leads to a steady state model of the universe. According to the perfect cosmological principle, the observable parameter, Hubble constant, $\frac{a}{a}$ must be independent of the present time $t_0$, i.e., its value remains constant throughout the evolution of the universe. Let us represent the fixed value, $H$ of the Hubble constant. Then we have,
$$\frac{\dot{a}}{a}=H, \forall t$$
This implies
$$a(t)=a\left(t_0\right) \exp [H t]$$
In this model the deceleration parameter assumes the fixed value as
$$q=-\frac{a \ddot{a}}{\dot{a}^2}=-1$$
Thus, the perfect cosmological principle provides the model of the universe with the following line element
$$d s^2=d t^2-a^2\left(t_0\right) e^{2 H t}\left[d r^2+r^2\left(d \theta^2+\sin ^2 \theta d \phi^2\right)\right]$$

## 物理代写|广义相对论代写General relativity代考|Particle and Event Horizon

Consider a galaxy $G_1$ at $\left(r_1, \theta_1, \phi_1\right)$ emitting light waves towards us at time $t_1$. Let the light wave arrive at $r=0$ at time $t=t_0$. Then null geodesic equation $(d s=0)$ for $\mathrm{R}-\mathrm{W}$ metric
$$\int_{t_1}^{t_0} \frac{d t}{a(t)}=\int_0^{r_1} \frac{d r}{\left(1-k r^2\right)^{1 / 2}}$$
Then redshift as well as luminosity distance are given by
$$z=\frac{a\left(t_0\right)}{a\left(t_1\right)}-1, \quad D_1=r_1 a\left(t_0\right)(1+z)$$
Also one can write the proper distance from $r=0$ to $r=r_1$ as
$$d_p(t)=\int_0^{r_1} \sqrt{g_{r r}} d r=a(t) \int_0^{r_1} \frac{d r}{\sqrt{1-k r^2}}=a(t) \int_{t_1}^{t_0} \frac{d t}{a(t)} .$$
The limit on the proper distance up to which we can observe is called the particle horizon. Note that if the $t$-integral converges, then our vision is restricted by the particle horizon. Let the limiting value of $r_1$ as $z \rightarrow \infty$, be $r_l$. Hence the limiting proper distance is
$$R_l=a(t) \int_0^{r_l} \frac{d r}{\left(1-k r^2\right)^{1 / 2}}$$

If one gets a finite value of $R_l$, then we say that the universe has a particle horizon. It is not possible to see the particles at present with $r_1>r_l$. Note that the particle horizon creates a barricade to communication from the past.

As above let a galaxy at $r=r_1, t=t_0$ send light signal to an observer at $r=0$. Suppose $t_1$ is the time of arrival. Then, we have
$$\int_{t_0}^{t_1} \frac{d t}{a(t)}=\int_0^{r_1} \frac{d r}{\left(1-k r^2\right)^{1 / 2}} .$$
Suppose the left-hand side of the above integral converges to a finite value as $t_1 \rightarrow \infty$. This finite value is achieved by the right-hand side integral for $r_1=r_H$ (say). Hence it is obvious that for $r_1>r_H$ the above relation does not hold good. As a result, no signal from $r_1>r_H$ will come to the observer at $r_0$. Hence no light from a distant galaxy beyond a proper distance
$$R_H=a(t) \int_{t_0}^{\infty} \frac{d t}{a(t)}$$
will reach the observer at $r_0$. This limit is known as the event horizon. Friedmann models do not possess an event horizon. However, the de Sitter model contains an event horizon at $\frac{1}{H_0^2}$. Note that the event horizon creates a barricade to communication from the future.

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Perfect Cosmological Principle

$$\frac{\dot{a}}{a}=H, \forall t$$

$$a(t)=a\left(t_0\right) \exp [H t]$$

$$q=-\frac{a \ddot{a}}{\dot{a}^2}=-1$$

$$d s^2=d t^2-a^2\left(t_0\right) e^{2 H t}\left[d r^2+r^2\left(d \theta^2+\sin ^2 \theta d \phi^2\right)\right]$$

## 物理代写|广义相对论代写General relativity代考|Particle and Event Horizon

$$\int_{t_1}^{t_0} \frac{d t}{a(t)}=\int_0^{r_1} \frac{d r}{\left(1-k r^2\right)^{1 / 2}}$$

$$z=\frac{a\left(t_0\right)}{a\left(t_1\right)}-1, \quad D_1=r_1 a\left(t_0\right)(1+z)$$

$$d_p(t)=\int_0^{r_1} \sqrt{g_{r r}} d r=a(t) \int_0^{r_1} \frac{d r}{\sqrt{1-k r^2}}=a(t) \int_{t_1}^{t_0} \frac{d t}{a(t)} .$$

$$R_l=a(t) \int_0^{r_l} \frac{d r}{\left(1-k r^2\right)^{1 / 2}}$$

$$\int_{t_0}^{t_1} \frac{d t}{a(t)}=\int_0^{r_1} \frac{d r}{\left(1-k r^2\right)^{1 / 2}} .$$

$$R_H=a(t) \int_{t_0}^{\infty} \frac{d t}{a(t)}$$

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