Point gravitational lenses - Revision history

Cosmo All: /* Problem 9: source of finite size */

2014-04-23T00:36:09Z

‎Problem 9: source of finite size

Cosmo All: /* Problem 9: source of finite size */

2014-04-23T00:35:43Z

‎Problem 9: source of finite size

Cosmo All: /* Problem 9: source of finite size */

2014-04-23T00:34:26Z

‎Problem 9: source of finite size

Cosmo All: /* Problem 9: source of finite size */

2014-04-23T00:33:34Z

‎Problem 9: source of finite size

Cosmo All: /* Problem 9: source of finite size */

2014-04-23T00:32:47Z

‎Problem 9: source of finite size

Cosmo All: /* Problem 9: source of finite size */

2014-04-23T00:31:59Z

‎Problem 9: source of finite size

Cosmo All: /* Problem 9: source of finite size */

2014-04-23T00:31:32Z

‎Problem 9: source of finite size

Cosmo All: /* Problem 9: source of finite size */

2013-01-15T13:10:50Z

‎Problem 9: source of finite size

Cosmo All at 13:10, 15 January 2013

2013-01-15T13:10:04Z

Cosmo All: /* Problem 9: source of finite size */

2013-01-15T11:27:49Z

‎Problem 9: source of finite size

@@ Line 268: / Line 268: @@
 <br/>
 		In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain
-$$\vec p_{1,2} \simeq \vec \tilde{\rho_S} \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde{\rho_S}}}\right\}.$$
+$\vec p_{1,2} \simeq \vec \tilde{\rho_S} \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde{\rho_S}}}\right\}.$
 		For disk of the projected radius $\tilde R_S$ Einstein ring has finite width of

@@ Line 268: / Line 268: @@
 <br/>
 		In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain
-		$$
+$$\vec p_{1,2} \simeq \vec \tilde{\rho_S} \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde{\rho_S}}}\right\}.$$
-		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde\rho_S}}\right\}.
-		$$
 		For disk of the projected radius $\tilde R_S$ Einstein ring has finite width of

@@ Line 269: / Line 269: @@
 		In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain
 		$$
-		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde{\rho_S}}}\right\}.
+		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde\rho_S}}\right\}.
 		$$

@@ Line 269: / Line 269: @@
 		In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain
 		$$
-		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2  {\tilde{l} \over \tilde{\rho_S}}\right\}.
+		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde{\rho_S}}}\right\}.
 		$$

@@ Line 269: / Line 269: @@
 		In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain
 		$$
-		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2 \pm {\tilde{l} \over \tilde{\rho_S}}\right\}.
+		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2  {\tilde{l} \over \tilde{\rho_S}}\right\}.
 		$$

@@ Line 269: / Line 269: @@
 		In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain
 		$$
-		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2 \pm {\tilde{l} \over \tilde\rho_S}\right\}.
+		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{\frac 1 2 \pm {\tilde{l} \over \tilde {\rho_S}}\right\}.
 		$$
@@ Line 284: / Line 284: @@
 <div id="ocgl10"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 10: source of finite size  ===
 Qualitatively consider the general situation, when a source of finite size, lens and observer are not on one line. Estimate the angular sizes of the observed images.

@@ Line 11: / Line 11: @@
      <p style="text-align: left;">
 	When light is deflected in optical lens, its deflection angle is proportional to impact parameter. In contrast, gravitational field of gravitational lens decreases with distance, and so does deflection angle.
+<br/>
 	Considering opaque lenses and small angles, the light ray deflection angle is inversely proportional to impact parameter.
+<br/>
 	Two main conclucions can be drawn from these differences in deflection angle dependencies for optical and gravitational lenses. First, unlike optical lens, gravitational lens has no focal point. Indeed, rays, which passed through the gravitational lens, can not intersect in one point, since rays with larger impact parameters would cross the axis of the system at a greater distance from lens, than rays with smaller impact parameters $p$.
+<br/>
 	Instead, gravitational lens has focal semiaxis. Moreover, opacity of the lens leads to the existence of shadow area behind the lens, which location is determined by the rays, passed close to the lenses edge.
+<br/>
 	Second, in contrast to optical lens, the intensity of registered radiation after passing the gravitational lens is infinite not in focal point, but on entire focal semiaxis.  However, this singularity exists only in small angles approximation and dissapears if we account for finite size of the source.
+<br/>
 	It's important to understand the notion of ''small angles'' in the context of gravitational lensing. Rays near the axis of optical lens have small deflection angles, so that their impact parameter is small. Small deflection angle of light ray in gravitational lens means that impact parameter is large (much larger, than gravitational radius of the lens, to be more precise).</p>
    </div>
@@ Line 219: / Line 219: @@
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 7: multiple images  ===
-Show that when the gravitational lens is placed between source and observer in the general case the two images of the source would be observed. How are the images placed relative to the lens and observer?<div class="NavFrame collapsed">
+Show that when the gravitational lens is placed between source and observer in the general case the two images of the source would be observed. How are the images placed relative to the lens and observer?
 <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
      <p style="text-align: left;"></p>
    </div>
-</div></div>
+</div>--></div>
@@ Line 232: / Line 233: @@
 === Problem 8: Einstein ring  ===
-How should source, gravitational lens and observer be placed relative to each other in order to observe the Einstein ring? Calculate the radius of the ring.<div class="NavFrame collapsed">
+How should source, gravitational lens and observer be placed relative to each other in order to observe the Einstein ring? Calculate the radius of the ring.
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
@@ Line 336: / Line 338: @@
 === Problem 12: double Einstein ring  ===
 Recently the exceptional phenomenon was observed using the Hubble space telescope: the double Einstein ring, formed by the influence of the gravitational field of the galaxy on the light from two other more distant galaxies. What conditions are necessary for the observation of this phenomenon?
-<div class="NavFrame collapsed">
+<!--<div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
      <p style="text-align: left;"></p>
    </div>
-</div></div>
+</div>--></div>
@@ Line 371: / Line 373: @@
 		These expression are valid when $x>x_{\min}$, while $\mu = 0$ for $x<x_{\min}$.
+<br/>
 		When observer is located on the source-lens axis, the amplification tends to infinity $\mu \to\infty$. Thus, unlike optical lens, gravitational lens produces image on any distance from it. When moving away from the lens the intenity increases as $\sqrt {D_d}$.</p>
    </div>
 </div></div>

@@ Line 267: / Line 267: @@
 		In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain
 		$$
-		\vec p_{1,2} \simeq \vec \tilde \rho_S \left\{{1\over 2} \pm {\tilde l \over \tilde \rho_S}\right\}.
+		\vec p_{1,2} \simeq \vec \tilde\rho_S \left\{{1\over 2} \pm {\tilde{l} \over \tilde\rho_S}\right\}.
 		$$