forked from ernestyalumni/Gravite
-
Notifications
You must be signed in to change notification settings - Fork 29
/
lecture13.tex
234 lines (189 loc) · 8.24 KB
/
lecture13.tex
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
\section{Lecture 13: Relativistic spacetime}
Recall, from Lecture 9, the definition of Newtonian spacetime
\[
(M, \mathcal{O}, \mathcal{A}, \nabla, t) \quad \quad \quad \, \begin{aligned}
& \nabla \text{ torsion free } \\
& t \text{(the so-called absolute time) } \in C^{\infty}(M) \\
& dt \neq 0 \\
& \nabla dt = 0 \quad \, \text{ (uniform time) }
\end{aligned}
\]
and the definition of relativistic spacetime (before Lecture 1)
\[
(M, \mathcal{O}, \mathcal{A}^{\uparrow}, \nabla, g, T ) \quad \quad \quad \, \begin{aligned}
& \nabla \text{ torsion-free } \\
& g \text{ Lorentzian metric} (+---) \\
& T \text{ time-orientation }
\end{aligned}
\]
The role played by $t$ in Newtonian spacetime is played by the interplay of two new structures $g$ and $T$ in relativistic spacetime.
\subsection{Time orientation}
\begin{definition}
$(M,\mathcal{O},\mathcal{A}^{\uparrow},g)$ a Lorentzian manifold. Then a time-orientation is given by a vector field $T$ that
\begin{enumerate}
\item[(i)] does \textbf{not} vanish anywhere
\item[(ii)] $g(T,T)>0$
\end{enumerate}
\end{definition}
Newtonian vs. relativistic \\
Recall that a vector $X$ in Newtonian spacetime was called future-directed if $dt(X) > 0$.
$\forall \, p \in M$, take half plane, half space of $T_pM$ \\
also stratified atlas so make planes of constant $t$ straight \\
relativistic \\
half cone $\forall \, p, q \in M$, half-cone $\subseteq T_pM$ \\
This definition of \underline{spacetime}
Question \\
I see how the cone structure arises from the new metric. I don't understand however, how the $T$, the time orientation, comes in \\
Answer \\
$(M,\mathcal{O}, \mathcal{A},g)$ $g \xleftarrow (+---)$
requiring $g(X,X)>0$, select cones \\
$T$ chooses which cone \\
This definition of \underline{spacetime} has been made to enable the following physical postulates:
\begin{enumerate}
\item[(P1)] The worldline $\gamma$ of a \underline{massive} particle satisfies
\begin{enumerate}
\item[(i)] $g_{\gamma(\lambda)}(v_{\gamma, \gamma(lambda)} , v_{\gamma,\gamma(\lambda)} ) >0$
\item[(ii)] $g_{\gamma(\lambda)}(T, v_{\gamma,\gamma(\lambda)}) >0$
\end{enumerate}
\item[(P2)] Worldlines of \underline{massless} particles satisfy
\begin{enumerate}
\item[(i)] $g_{\gamma(\lambda)}(v_{\gamma,\gamma(\lambda)}, v_{\gamma,\gamma(\lambda)}) = 0$
\item[(ii)] $g_{\gamma(\lambda)}(T,v_{\gamma,\gamma(\lambda)}) >0$
\end{enumerate}
\underline{picture}: spacetime:
\end{enumerate}
Answer (to a question) $T$ is a smooth vector field, $T$ determines future vs. past, ``general relativity: we have such a time orientation; smoothness makes it less arbitrary than it seems'' -FSchuller,
\underline{Claim}: $9/10$ of a metric are determined by the cone
spacetime determined by distribution, only one-tenth error
\subsection{Observers} $(M,\mathcal{O}, \mathcal{A}^{\uparrow},\nabla ,g, T)$
\begin{definition}
An \underline{observer} is a worldline $\gamma$ with
\[
\begin{aligned}
& g(v_{\gamma}, v_{\gamma}) > 0 \\
& g(T,v_{\gamma}) > 0
\end{aligned}
\]
together with a choice of basis
\[
v_{\gamma,\gamma(\lambda)} \equiv e_0(\lambda) , e_1(\lambda), e_2(\lambda), e_3(\lambda)
\]
of each $T_{\gamma(\lambda)}M$ where the observer worldline passes, if $g(e_a(\lambda), e_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab}$
\underline{precise}: observer $=$ \underline{smooth} curve in the frame bundle $LM$ over $M$
\end{definition}
\subsubsection{Two physical postulates}
\begin{enumerate}
\item[(P3)] A \textbf{clock} carried by a specific observer $(\gamma, e)$ will measure a \textbf{time}
\[
\tau := \int_{\lambda_0}^{\lambda_1} d\lambda \sqrt{ g_{\gamma(\lambda)}(v_{\gamma,\gamma(\lambda)}, v_{\gamma,\gamma(\lambda)}) }
\]
between the two ``\underline{events}''
\[
\gamma(\lambda_0) \quad \quad \quad \, \text{ ``start the clock'' }
\]
and
\[
\gamma(\lambda_1) \quad \quad \quad \, \text{ ``stop the clock'' }
\]
\underline{Compare} with Newtonian spacetime:
\[
t(p)=7
\]
Thought bubble: \underline{proper time/eigentime} $\tau$
\underline{Application/Example.}
$\begin{aligned}
& M = \mathbb{R}^4 \\
& \mathcal{O} = \mathcal{O}_{\text{st}} \\
& \mathcal{A} \ni (\mathbb{R}^4, \text{id}_{\mathbb{R}^4} ) \\
& g : g_{(x)ij} = \eta_{ij} \quad \, ; \quad \quad \, T_{(x)}^i =(1,0,0,0)^i
\end{aligned}
$
\[
\Longrightarrow \Gamma_{(x) \, \, jk }^i = 0 \text{ everywhere }
\]
$\Longrightarrow (M,\mathcal{O}, \mathcal{A}^{\uparrow},g,T,\nabla)$ \quad \, $\text{Riemm}=0$ \\
$\Longrightarrow $ spacetime is flat
This situation is called special relativity.
Consider two observers:
\[
\begin{aligned} &
\begin{aligned}
& \gamma : (0,1) \to M \\
& \gamma_{(x)}^i = (\lambda , 0 ,0 ,0 )^i \end{aligned} \\
&
\begin{aligned}
& \delta :(0,1) \to M \\
\alpha \in (0,1) : & \delta_{(x)}^i = \begin{cases} ( \lambda , \alpha \lambda , 0 , 0)^i & \lambda \leq \frac{1}{2} \\
(\lambda, (1-\lambda)\alpha, 0,0)^i & \lambda > \frac{1}{2} \end{cases}
\end{aligned}
\end{aligned}
\]
let's calculate:
\[
\begin{aligned}
& \tau_{\gamma}:= \int_0^1 \sqrt{ g_{(x)ij} \dot{\gamma}^i_{(x)} \dot{\gamma}^j_{(x)} } = \int_0^1 d\lambda 1 = 1 \\
& \tau_{\delta} := \int_0^{1/2} d\lambda \sqrt{ 1- \alpha^2} + \int_{1/2}^1 \sqrt{ 1^2 - (-\alpha)^2 } = \int_0^1 \sqrt{ 1 - \alpha^2 } = \sqrt{ 1 - \alpha^2}
\end{aligned}
\]
Note: piecewise integration
Taking the clock postulate (P3) seriously, one better come up with a realistic clock design that supports the postulate.
\underline{idea}.
2 little mirrors
\item[(P4)] \underline{Postulate}
Let $(\gamma, e)$ be an observer, and \\
$\delta$ be a \emph{massive} particle worldline that is parametrized s.t. $g(v_{\gamma}, v_{\gamma})=1$ (for parametrization/normalization convenience)
Suppose the observer and the particle \emph{meet} somewhere (in spacetime)
\[
\delta(\tau_2) = p = \gamma(\tau_1)
\]
\emph{This} observer measures the 3-velocity (spatial velocity) of this particle as
\begin{equation}\label{Eq:spatialv}
v_{\delta}: \epsilon^{\alpha}( v_{\delta, \delta(\tau_2)} ) e_{\alpha} \quad \quad \quad \, \alpha =1,2,3
\end{equation}
where $\epsilon^0, \boxed{ \epsilon^1,\epsilon^2,\epsilon^3}$ is the unique dual basis of $e_0,\boxed{ e_1,e_2,e_3}$
\end{enumerate}
EY:20150407
There might be a major correction to Eq. (\ref{Eq:spatialv}) from the Tutorial 14 : Relativistic spacetime, matter, and Gravitation, see the second exercise, Exercise 2, third question:
\begin{equation}
v := \frac{ \epsilon^{\alpha}({v}_{\delta} ) }{ \epsilon^0({v}_{\delta}) } e_{\alpha}
\end{equation}
\underline{Consequence}:
An observer $(\gamma, e)$ will extract quantities measurable in his laboratory from objective spacetime quantities always like that.
\underline{Ex}: $F$ Faraday $(0,2)$-tensor of electromagnetism:
\[
F(e_a,e_b) = F_{ab} = \left[ \begin{matrix} 0 & E_1 & E_2 & E_3 \\
-E_1 & 0 & B_3 & -B_2 \\
-E_2 & -B_3 & 0 & B_1 \\
-E_3 & B_2 & -B_1 & 0 \end{matrix} \right]
\]
observer frame $e_a,e_b$
$E_{\alpha} := F(e_0,e_{\alpha})$ \\
$B^{\gamma}:= F(e_{\alpha},e_{\rho})\epsilon^{\alpha \beta \gamma}$
where
$\epsilon^{123} = +1$ totally antisymmetric
\subsection{Role of the Lorentz transformations}
Lorentz transformations emerge as follows: \\
Let $(\gamma,e)$ and $(\widetilde{\gamma},\widetilde{e})$ be observers with $\gamma(\tau_1) = \widetilde{\gamma}(\tau_2)$
(for simplicity $\gamma(0) = \widetilde{\gamma}(0)$
Now
\[
\begin{gathered}
e_0 , \dots , e_1 \quad \quad \quad \, \text{ at } \tau = 0 \\
\text{ and }
\widetilde{e}_0 , \dots , \widetilde{e}_1 \quad \quad \quad \, \text{ at } \tau = 0 \\
\end{gathered}
\]
both bases for the same $T_{\gamma(0)}M$
\underline{Thus}: $\widetilde{e}_a = \Lambda^b_{ \, \, a} e_b $ \quad \quad \, $\Lambda \in GL(4)$
Now:
\[
\begin{aligned}
\eta_{ab} = g(\widetilde{e}_a, \widetilde{e}_b) & = g(\Lambda^m_{ \, \, a}e_m, \Lambda^n_{ \, \, b} e_n ) = \\
& = \Lambda^m_{ \, \, a} \Lambda^n_{ \, \, b} \underbrace{g(e_m,e_n)}_{ \eta_{mn}}
\end{aligned}
\]
i.e. $\Lambda \in O(1,3)$
\underline{Result}: Lorentz transformations relate the \emph{frames} of \emph{any two observers} at the same point.
``$\widetilde{x}^{\mu} - \Lambda^{\mu}_{ \, \, \nu} x^{\nu}$'' is utter nonsense
\subsection*{Tutorial}
I didn't see a tutorial video for this lecture, but I saw that the Tutorial sheet number 14 had the relevant topics. Go there.