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PHYSICS262课程简介
General relativity is a theory of gravitation developed by Albert Einstein between 1907 and 1915. The theory of general relativity says that the observed gravitational effect between masses results from their warping of spacetime.
By the beginning of the 20th century, Newton’s law of universal gravitation had been accepted for more than two hundred years as a valid description of the gravitational force between masses. In Newton’s model, gravity is the result of an attractive force between massive objects. Although even Newton was troubled by the unknown nature of that force, the basic framework was extremely successful at describing motion.
Prerequisites
Einstein’s General Theory of Relativity is a basis for modern ideas of fundamental physics, including string theory, as well as for studies of cosmology and astrophysics. The course begins with an overview of special relativity, and the description of gravity as arising from curved space. From Riemannian geometry and the geodesic equations, to curvature, the energy-momentum tensor, and the Einstein field equations. Applications of General Relativity: topics may include experimental tests of General Relativity and the weak-field limit, black holes (Schwarzschild, charged Reissner-Nordstrom, and rotating Kerr black holes), gravitational waves (including detection methods), and an introduction to cosmology (including cosmic microwave background radiation, dark energy, and experimental probes). Prerequisite: PHYSICS 121 or equivalent including special relativity.
Terms: Aut | Units: 3
Instructors: Graham, P. (PI) ; Abdel-Aziz, A. (TA)
Schedule for PHYSICS 262
PHYSICS262 General Relativity HELP（EXAM HELP， ONLINE TUTOR）
Give the components of $\left(\begin{array}{l}2 \ 0\end{array}\right)$ tensor $M^{\alpha \beta}$ as the matrix
$$
\left[\begin{array}{cccc}
0 & 1 & 0 & 0 \
1 & -1 & 0 & 2 \
2 & 0 & 0 & 1 \
1 & 0 & -2 & 0
\end{array}\right]
$$
find:
the components of $M^\alpha \beta$
the components of $M_{\alpha \beta}$
Solution:
This can be written with the metric tensor as
$$
M_\beta^\alpha=g_{\sigma \beta} M^{\alpha \sigma}=\left[\begin{array}{cccc}
0 & 1 & 0 & 0 \
-1 & -1 & 0 & 2 \
-2 & 0 & 0 & 1 \
-1 & 0 & -2 & 0
\end{array}\right]
$$
Solution:
The previous tensor can be used to calculate this
$$
M_{\alpha \beta}=g_{\sigma \beta} M_\alpha{ }^\sigma=\left[\begin{array}{cccc}
0 & -1 & & 0 \
-1 & -1 & 0 & 2 \
-2 & 0 & 0 & 1 \
-1 & 0 & -2 & 0
\end{array}\right]
$$
(Schutz 3.30) In some $\mathcal{O}$, the vector $U$ and $D$ have the components
$$
\begin{array}{r}
U \rightarrow\left(1+t^2, t^2, \sqrt{2} t, 0\right) \
D \rightarrow(x, 5 t x, \sqrt{2} t, 0)
\end{array}
$$
and the scalar $\rho$ has the value
$$
\rho=x^2+t^2-y^2
$$
(a) Find $U \cdot U, U \cdot D, D \cdot D$. Is $U$ suitable as four-velocity field? Is D?
Solution:
The components of $U_\mu$ are $U_\mu=\left(-\left(1+t^2\right), t^2, \sqrt{2} t, 0\right)$ and the components of $D_\mu$ are $D_m u=$ $(-x, 5 t x, \sqrt{2} t, 0)$ so the dot products are
$$
\begin{aligned}
U \cdot U=U^\mu U_m u & =\left(-\left(1+t^2\right)^2+t^4+2 t^2+0\right)=-1-2 t^2-t^4+t^4+2 t^2=-1 \
D \cdot D=D^\mu D_\mu & =\left(-x^2+25 t^2 x^2+2 t^2+0\right)=x^2\left(25 t^2-1\right)+2 t^2 \
U \cdot D=U^\mu D_\mu & =-x\left(1+t^2\right)+5 t^3 x+2 t^2=x\left(5 t^3-t^2-1\right)+2 t^2
\end{aligned}
$$
Since the inner product of $U$ with itself is -1 its is suitable for a four velocity while $D$ is not (except possibly for fixed values of $x$ and $t$ ).
Find the spatial velocity $v$ of a particle whose four-velocity is $U$, for arbitrary $t$. What happens to it in the limits $t \rightarrow 0$ and $t \rightarrow \infty$ ?
Solution:
$$
v^i=\frac{U^i}{U^0}=\left(\frac{t^2}{1+t^2}, \frac{\sqrt{2} t}{1+t^2}, 0\right)
$$
In the limit $t \rightarrow \infty$ we get $\boldsymbol{v}=(1,0,0)$ and in the limit $t \rightarrow 0$ we get $v=(0,0,0)$
Find $U_\alpha$ for all $\alpha$
Solution:
With the Minkowski metric the values of $U_\alpha$ is $U_\alpha=\left(-(1+t)^2, t^2, \sqrt{2} t, 0\right)$
Show that $U_{\alpha, \beta}^\alpha=0$ for all $\beta$. Show that $U^\alpha U_{\alpha, \beta}=0$ for all $\beta$.
Solution:
For various values of $\beta U_\alpha U_{, \beta}^\alpha$ is
$$
\begin{aligned}
& \beta=0:: U_\alpha U_{, 0}^\alpha=\frac{\partial}{\partial t}\left(-\left(1+t^2\right)^2+t^4+2 t\right)=-2\left(1+t^2\right) \cdot 2 t+4 t^3+4 t=0 \
& \beta=1:: U_\alpha U_{, 1}^\alpha=\frac{\partial}{\partial x}\left(-\left(1+t^2\right)^2+t^4+2 t\right)=0 \
& \beta=2:: U_\alpha U_{, 2}^\alpha=\frac{\partial}{\partial y}\left(-\left(1+t^2\right)^2+t^4+2 t\right)=0 \
& \beta=3:: U_\alpha U_{, 3}^\alpha=\frac{\partial}{\partial z}\left(-\left(1+t^2\right)^2+t^4+2 t\right)=0
\end{aligned}
$$
We have $U^\alpha U_\alpha$ is the inner product of $U \cdot U$ and so $U \cdot U=U^\alpha U_\alpha=U_\alpha U^\alpha$ so the expression
$$
U^\alpha U_{\alpha, \beta}=\left(U_\alpha U^\alpha\right)_{, \beta}=0, \forall \beta
$$
MY-ASSIGNMENTEXPERT™可以为您提供STANFORD.EDU PHYSICS262 GENERAL RELATIVITY广义相对论课程的代写代考和辅导服务！
