Light and distances - Revision history

Igor: /* Problem 17: the maximum in Friedman models */

2012-10-07T12:57:49Z

‎Problem 17: the maximum in Friedman models

2012-10-07T12:57:23Z

‎Problem 17: the maximum in Friedman models

2012-10-07T12:54:20Z

‎Problem 17: the maximum in Friedman models

2012-10-07T12:28:08Z

‎Problem 12: angular diameter distance in terms of redshift

2012-10-07T12:25:46Z

‎Problem 6: generalization to arbitrary curvature

2012-10-07T12:21:37Z

‎Problem 6: generalization to arbitrary curvature

2012-10-07T12:06:45Z

‎Problem 6: generalization to arbitrary curvature

2012-10-07T11:59:01Z

‎Problem 5: luminosity distance in a flat Universe

2012-10-07T11:57:14Z

2012-10-07T11:50:02Z

@@ Line 336: / Line 336: @@
   \sim \sin \frac{3\eta-2\eta_o}{2}\sin\frac{\eta}{2}.\]
 So the maximum $d_A$ is realized for $\eta_{e}=\frac{2}{3}\eta_{o}$. The corresponding redshift is found from
-\[1+z_{\max}=\frac{a_o}{a_e}=\frac{1-\cos \eta_o}{1-\cos\eta_e}\frac{1-\cos \eta_o}{1-\cos\frac{2}{3}\eta_o}.\]
+\[1+z_{\max}=\frac{a_o}{a_e}=\frac{1-\cos \eta_o}{1-\cos\eta_e}=\frac{1-\cos \eta_o}{1-\cos\frac{2}{3}\eta_o}.\]
 In the open model, following the same procedure, one obtains the same result with trigonometric functions replaced by hyperbolic ones.</p>

@@ Line 324: / Line 324: @@
    <div style="width:100%;" class="NavContent">
      <p style="text-align: left;">
-Tor the closed model
+For the closed model
-\[\Sigma(\chi)=\sin(\chi)\]
+\[\Sigma(\chi)=\sin\chi\]
 and
 \[a(\eta)=\frac{\alpha}{2}(1-\cos\eta),\]
 where $\alpha$ is some constant. Then
-\[d_{A}(\eta)=a_{e}\Sigma(\chi)\sim (1-\cos\eta_{e})\sin(\eta_{o}-\eta_{e}).\]
+\[d_{A}(\eta)=a_{e}\Sigma(\chi)\sim (1-\cos\eta_{e})\sin(\eta_{o}-\eta_{e}),\]
-and condition
+where subscripts "e" and "o" denote emission and observation respectively, and condition
 \[\frac{dd_{A}}{d\eta_{e}}=0\]
-is transformed to
+is transformed, after renaming $\eta_{e}\equiv \eta$, to
-\[0=\sin\eta \sin(\eta_o -\eta)-(1-\cos\eta)\cos(\eta_o -\eta)\sim \cos(2\eta-\eta_o)-\cos(\eta-\eta_o) \sim \sin \frac{3\eta-2\eta_o}{2}\sin\frac{\eta}{2}.\]
+\[0=\sin\eta\; \sin(\eta_o -\eta)-(1-\cos\eta)\cos(\eta_o -\eta)\sim \cos(2\eta-\eta_o)-\cos(\eta-\eta_o)
 So the maximum $d_A$ is realized for $\eta_{e}=\frac{2}{3}\eta_{o}$. The corresponding redshift is found from
-\[1+z_{\max}=\frac{a_o}{a_e}=\frac{1-\cos \eta_o}{1-\cos\eta_e}.\]
+\[1+z_{\max}=\frac{a_o}{a_e}=\frac{1-\cos \eta_o}{1-\cos\eta_e}\frac{1-\cos \eta_o}{1-\cos\frac{2}{3}\eta_o}.\]
 In the open model, following the same procedure, one obtains the same result with trigonometric functions replaced by hyperbolic ones.</p>
    </div>
 </div></div>
-</div></div>
+</div>

@@ Line 323: / Line 323: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"> solution</p>
+     <p style="text-align: left;">
    </div>
 </div></div>
 </div></div>

@@ Line 240: / Line 240: @@
 \[\delta l_\bot=a(t_e)\Sigma(\chi)\delta \theta\]
 and
-\[d_{A}\equiv\frac{\delta l_{\bot}}{\delta\theta}=\frac{\Sigma(\chi)}{1+z}.\]
+\[d_{A}\equiv\frac{\delta l_{\bot}}{\delta\theta}
 </p>
    </div>
@@ Line 248: / Line 249: @@
 <div id="lad13"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 13: in terms of $q(z)$ ===
 Express the angular diameter distance in terms of $q(z)$

@@ Line 121: / Line 121: @@
 and using that $a=H^{-1}(\Omega-1)^{-1/2}$, we finally obtain
 \begin{align}
-d_{L}(z)&=(1+z)a_{o}\;\chi \;\Sigma(\chi)\\
+d_{L}(z)&=(1+z)a_{o}\;\Sigma(\chi)\\
 &=(1+z)(1-\Omega_0)^{-1/2}\cdot H_{0}^{-1}\Sigma\Big[(1-\Omega_0)^{1/2}H_{0}
 \int\limits_{0}^{z}\frac{dz'}{H(z')}\Big],

@@ Line 117: / Line 117: @@
 \[\Sigma(x)=\left\{\begin{array}{ll}\sin x,&\quad \Omega_{0}>1,\\
 x,&\quad \Omega_{0}=1\\ \sinh x,&\quad \Omega_{0}<1.\end{array}\right.\]
-and $\chi$ is the proper distance to the source. Then
+and $\chi$ is the proper distance to the source. Expressing it through the observed quantities
 \begin{align}
 d_{L}(z)&=(1+z)a_{o}\;\chi \;\Sigma(\chi)\\

@@ Line 113: / Line 113: @@
      <p style="text-align: left;">
 The above derivation holds mostly, with the exception that now the observed luminosity is distributed over the sphere of proper surface
-\[S=4\pi \Sigma(\chi),\]
+\[S=4\pi a_{o}^{2}\Sigma^{2}(\chi),\]
 where
-\[\Sigma(z)=\left\{\begin{array}{ll}\sin x,&\quad \Omega_{0}>1,\\
+\[\Sigma(x)=\left\{\begin{array}{ll}\sin x,&\quad \Omega_{0}>1,\\
 x,&\quad \Omega_{0}=1\\ \sinh x,&\quad \Omega_{0}<1.\end{array}\right.\]
-and $chi$ is the proper distance to the source. Then
+and $\chi$ is the proper distance to the source. Then
-\[d_{L}(z)=(1+z)(1-\Omega_0)^{-1/2}\cdot H_{0}^{-1}\Sigma\Big[(1-\Omega_0)^{1/2}H_{0}
+\begin{align}
-\int\limits_{0}^{z}\frac{dz'}{H(z')}\Big],\]
+d_{L}(z)&=(1+z)a_{o}\;\chi \;\Sigma(\chi)\\
 </p>
    </div>
@@ Line 127: / Line 130: @@
 <div id="lad07"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 7: multi-component flat Universe  ===
 Find the expression for the luminosity distance for the multi-component flat Universe

@@ Line 96: / Line 96: @@
 d_{L}=\frac{a_o}{a_e}R=(1+z)R
 \end{equation}
-In a spatially flat Universe $R=a_{o}\chi$, where $chi$ is the comoving distance to the source, so
+In a spatially flat Universe $R=a_{o}\chi$, where $\chi$ is the comoving distance to the source, so
 \[d_{L}=(1+z)a_{o}\chi.\]
 </p>