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物理代写|ASTR160 general relativity

Posted on 2023年4月28日2023年4月28日 by statistics-lab

Statistics-lab™可以为您提供cornell.edu ASTR160 general relativity广义相对论的代写代考和辅导服务！

ASTR160 general relativity课程简介

One-semester introduction to general relativity that develops the essential structure and phenomenology of the theory without requiring prior exposure to tensor analysis.
General relativity is a fundamental cornerstone of physics that underlies several of the most exciting areas of current research, including relativistic astrophysics, cosmology, and the search for a quantum theory of gravity. The course briefly reviews special relativity, introduces basic aspects of differential geometry, including metrics, geodesics, and the Riemann tensor, describes black hole spacetimes and cosmological solutions, and concludes with the Einstein equation and its linearized gravitational wave solutions. At the level of Gravity: An Introduction to Einstein’s General Relativity by Hartle.

PREREQUISITES

ASTR160 general relativity HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

Show that $J^{+}(p)$ is not an open set.
Hint: Suppose $r$ is a point in $J^{+}(p)-I^{+}(p)$. Let us consider a neighborhood of $r$. Then, some points of this neighborhood lie outside of $J^{+}(p)$ and some points lie within $J^{+}(p)$ (see Fig. 29). Therefore, $r$ is not an interior point. Hence, $J^{+}(p)$ is not an open set.

问题 2.

For all subsets $S \subset M$, show that $J^{+}(p) \subset \overline{I^{+}(p)}$.
Hint: Let us consider a Minkowski space with a point $r$ removed as shown in Fig. 30. The points after the deleted point $r$, i.e., on the dotted line are not in $J^{+}(p)$. However, points on the dotted line are in the closer $\overline{I^{+}(p)}$. Hence, $J^{+}(p) \subset \overline{I^{+}(p)}$.

问题 3.

Let $J^{+}(x)$ be closed in $M$. Show that $M$ is causal if $\dot{I}^{+}(x) \cap I^{-}(x)={x}$ for all $x \in M$.
Hint: Given that $M$ is causal. Now let, if possible, $\exists \mathrm{y}(\neq x)$ such that $y \in I^{+}(x) \cap I^{-}(x)$. We know $\dot{I}^{+}(x)=j^{+}(x)$, therefore, $y \in j^{+}(x)$. Since $J^{+}(x)$ is closed in $M$, therefore, $y \in J^{+}(x)$. This means, $x<y$. Similarly, $y \in J^{-}(x)$ indicates $y<x$. This suggests that there is a closed causal curve through $x$. Definitely, this violates the causality. Hence, $y \neq x$ is not true, i.e., $y=x$.

问题 4.

If $\left{\lambda_n\right}$ be a sequence of future inextendible causal curves with a limit point $p$, then $\exists$ a future inextendible causal curve $\lambda$ passing through $p$, may be a limit curve of the $\left{\lambda_n\right}$. Also if $\left{\lambda_n\right}$ be a sequence of time-like curves, then the limit curve $\lambda$ may be only a causal curve.

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

此图像的alt属性为空；文件名为%E7%B2%89%E7%AC%94%E5%AD%97%E6%B5%B7%E6%8A%A5-1024x575-10.png — 物理代写|ASTR160 general relativity

Statistics-lab™可以为您提供cornell.edu ASTR160 general relativity广义相对论的代写代考和辅导服务！请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。

物理代写|PHYS6553 general relativity

Posted on 2023年4月11日2023年4月11日 by statistics-lab

Statistics-lab™可以为您提供cornell.edu PHYS6553 general relativity广义相对论的代写代考和辅导服务！

PHYS6553 general relativity课程简介

PREREQUISITES

PHYS6553 general relativity HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

A fluorescent tube, stationary in a reference frame $S$, is arranged so as to light up simultaneously (in $S$ ) along its entire length $l_0$ at the time $t$.
(a) By considering as two simultaneous events in $S$ the lighting up of two parts of the tube an infinitesimal distance $\Delta x$ apart, determine the temporal and spatial separation of these two events in another frame of reference $S^{\prime}$ moving with a velocity $v$ parallel to the orientation of the tube. Hence describe what is observed from this other frame of reference.
(b) An onlooker is stationary in $S^{\prime}$, with her eye situated at the origin $x^{\prime}=0$ aligned with and looking towards the tube. Describe what the onlooker sees.

Suppose the tube is lying along the $x$ axis, at rest in the reference frame $S$, with one end at $x=0$, and the other at $x=l_0$. Further suppose the origin of the second reference frame, moving with a speed $v$ with respect to the first, coincides with the origin of the first at a time $t=t^{\prime}=0$.
(a) We can identify two events: the lighting up of the tube at $x$ at time $t$ and at $x+\Delta x$ also at time $t$ in $S$. The coordinates of these two events in $S^{\prime}$ will then be $x^{\prime}$ at time $t^{\prime}$ and $x^{\prime}+\Delta x^{\prime}$ at time $t^{\prime}+\Delta t^{\prime}$.
These two sets of coordinates can be related by the Lorentz transformation:
$$
\begin{array}{ll}
x^{\prime}=\gamma(x-v t) & t^{\prime}=\gamma\left(t-v x / c^2\right) \
x^{\prime}+\Delta x^{\prime}=\gamma(x+\Delta x-v t) & \left.t^{\prime}+\Delta t^{\prime}=\gamma\left(t+v(x+\Delta x) / c^2\right)\right)
\end{array}
$$
Subtracting the two sets of equations then gives
$$
\Delta x^{\prime}=\gamma(\Delta x-v \Delta t) \quad \Delta t^{\prime}=-\gamma v \Delta x / c^2
$$
Although some conclusions can be drawn from each of these equations, it is best to combine them to find the speed with which the lighted up region spreads along the tube, at least as far as it is measured in $S^{\prime}$. This speed is given by
$$
u_{\text {measured }}=\frac{\Delta x^{\prime}}{\Delta t^{\prime}}=\frac{\Delta x}{-v \Delta x / c^2}=-\frac{c^2}{v}
$$
which is a velocity that will always be greater than the speed of light!
(b) The light from the first event at $\left(x^{\prime}, t^{\prime}\right)$ in $S^{\prime}$ will reach the eye of the onlooker (at the origin of the coordinates of $S^{\prime}$ ) at a time
$$
t_r^{\prime}=t^{\prime}+x^{\prime} / c
$$
while the light from the event $\left(x^{\prime}+\Delta x^{\prime}, t^{\prime}+\Delta t^{\prime}\right)$ will reach the onlooker’s eye at a time
$$
t_r^{\prime}+\Delta t_r^{\prime}=t^{\prime}+\Delta t^{\prime}+\left(x^{\prime}+\Delta x^{\prime}\right) / c
$$
This light will arrive from a section of the tube of length $\Delta x^{\prime}$, over a time interval $\Delta t_r^{\prime}$ given by
$$
\Delta t_r^{\prime}=\Delta t^{\prime}+\Delta x^{\prime} / c
$$
so that the apparent speed with which the tube lights up will be
$$
\frac{\Delta x^{\prime}}{\Delta t_r^{\prime}}=\frac{\Delta x^{\prime}}{\Delta t^{\prime}+\Delta x^{\prime} / c}=\frac{\Delta x^{\prime} / \Delta t^{\prime}}{1+\left(\Delta x^{\prime} / \Delta t^{\prime}\right) / c}
$$

where appearing here is just the speed $\Delta x^{\prime} / \Delta t^{\prime}$ with which the tube lights up as measured in $S^{\prime}$. Substituting for this then leads to the final result
$$
u_{\text {apparent }}=\frac{c^2}{v-c}
$$
which is also faster than the speed of light – an example of a phenomenon that apears to occur at a superluminal speed. There is no conflict with relativity as the events that give rise to this apparent superluminal velocity were prearranged to occur simultaneously in $S$, i.e. the simultaneous lighting up of the tube was not due to the passage of some kind of physical signal along the tube.

问题 2.

Two Star Wars killer satellites are stationary in the $S$ frame at points on the $X$-axis separated by a distance $d$. They fire laser pulses at one another simultaneously. From the point of view of the frame of reference of an observer space shuttle moving with a velocity $u$ relative to $S$, show that one satellite fires a time $\gamma u d / c^2$ before the other.

Suppose the Star Wars satellites are positioned at $x_1$ and $x_2$ as measured in $S$, and that they both fire their lasers at the instant $t$. These two events will then have the spacetime coordinates $\left(x_1^{\prime}, t_1^{\prime}\right)$ and $\left(x_2^{\prime}, t_2^{\prime}\right)$ as measured in $S^{\prime}$, the frame of reference of the space shuttle.

The times at which these events occur according to $S$ are then given in terms of their coordinates in $S^{\prime}$ are then
$$
t_n^{\prime}=\gamma\left(t-u x_n / c^2\right) ; \quad n=1,2
$$
The time difference between the events as measured in $S$ is then
$$
t_2-t_1=-\gamma u\left(x_2-x_1\right) / c^2=-\gamma u d / c^2
$$
In other words, one satellite is observed to fire a time $\gamma u d / c^2$ before the other. The order depends on the sign of $u$, and on the sign of the difference $x_2-x_1$.

Textbooks

Statistics-lab™可以为您提供cornell.edu PHYS6553 general relativity广义相对论的代写代考和辅导服务！请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。