Fall 2012 - Revision history

Tom Neiser: /* Question 12 */

2014-05-13T01:14:56Z

‎Question 12

Tom Neiser at 17:40, 21 April 2014

2014-04-21T17:40:32Z

Tom Neiser: /* Question 12 */

2014-04-17T23:00:14Z

‎Question 12

Tom Neiser at 22:57, 17 April 2014

2014-04-17T22:57:16Z

Tom Neiser: /* Question 12 */

2014-04-17T22:53:36Z

‎Question 12

Tom Neiser: /* Question 12 */

2014-04-17T22:52:30Z

‎Question 12

Tom Neiser: /* Question 12 */

2014-04-17T22:40:40Z

‎Question 12

Tom Neiser: /* Question 12 */

2014-04-17T22:39:47Z

‎Question 12

Tom Neiser: /* Question 12 */

2014-04-17T22:39:02Z

‎Question 12

Tom Neiser at 22:36, 17 April 2014

2014-04-17T22:36:24Z

@@ Line 28: / Line 28: @@
 \[\frac{1}{v_g} = \frac{\partial k}{\partial \omega} = \frac{\omega}{c \sqrt{\omega^2- \omega_p^2}} =\frac{1}{c \sqrt{1- \frac{\omega_p^2}{\omega^2}}} \approx \frac{1}{c} \left[1+\frac{\omega_p^2}{2\omega^2} \right]\]
 \[\tau= s \left[\frac{1}{v_{g,2}} - \frac{1}{v_{g,1}}\right] \approx \frac{s \omega_p^2}{2c} \left[\frac{1}{\omega_2^2} - \frac{1}{\omega_1^2}\right]\]
-\[s = \frac{2c\tau}{\omega_p^2}\left[\frac{\omega_1^2\omega_2^2}{\omega_1^2-\omega_2^2}\right]\]
+\[s \approx \frac{2c\tau}{\omega_p^2}\left[\frac{\omega_1^2\omega_2^2}{\omega_1^2-\omega_2^2}\right]\]
 </div>
 <br/>

← Older revision		Revision as of 17:40, 21 April 2014
Line 1:		Line 1:
−	Official Questions and Solutions for [~~Fall 2012~~.pdf].	+	Official Questions and Solutions for [http://www.physwiki.com/w/images/9/90/F%2712.pdf Fall 2012].
	<div class="fb-share-button" data-href="http://developers.facebook.com/docs/plugins/" data-type="button_count"></div>		<div class="fb-share-button" data-href="http://developers.facebook.com/docs/plugins/" data-type="button_count"></div>

@@ Line 14: / Line 14: @@
 \[\epsilon_r =1+\chi_e = 1 + \frac{P}{\epsilon_0 E}=  1 + \frac{n_e p_e}{\epsilon_0 E} =1+\frac{n_e ex}{\epsilon_0 E}\]
-, where $E= E_0 e^{-i\omega t}$ and $n_e =$ electron number density.
+, where $E= E_0 e^{-i\omega t}$, $P= n_e p_e =$ Polarization vector, $n_e =$ electron number density and $p_e =$ electron dipole moment.
 $\rightarrow$ Need $x$.

← Older revision		Revision as of 22:57, 17 April 2014
Line 1:		Line 1:
−	Official Questions and Solutions for [Fall 2012].	+	Official Questions and Solutions for [Fall 2012.pdf].
	<div class="fb-share-button" data-href="http://developers.facebook.com/docs/plugins/" data-type="button_count"></div>		<div class="fb-share-button" data-href="http://developers.facebook.com/docs/plugins/" data-type="button_count"></div>

@@ Line 20: / Line 20: @@
 Have $F= eE= m_e \ddot{x}$, so
 \[\ddot{x} = \frac{e}{m_e} E_0 e^{-i\omega t}\]
-Looking at the above equation, we must have $\ddot{x} = -\omega^2 x$, so
+Therefore we must have $\ddot{x} = -\omega^2 x$, so
 \[x = -\frac{\ddot{x}}{\omega^2}= -\frac{e}{m_e \omega^2} E_0e^{-i\omega t} = -\frac{e E}{m_e \omega^2}\]
 \[\epsilon_r = 1 - \frac{n_e e^2 E}{\epsilon_0 m_e \omega^2 E}= 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2} =1 - \frac{\omega_p^2}{\omega^2}\], where $\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$.

@@ Line 13: / Line 13: @@
 $\rightarrow$ Need $\epsilon_r$.
-Have $\epsilon_r =1+\chi_e = 1 + \frac{P}{\epsilon_0 E}=  1 + \frac{n_e p_e}{\epsilon_0 E} =1+\frac{n_e ex}{\epsilon_0 E}$, where $E= E_0 e^{-i\omega t}$ and $n_e =$ electron number density.
+\[\epsilon_r =1+\chi_e = 1 + \frac{P}{\epsilon_0 E}=  1 + \frac{n_e p_e}{\epsilon_0 E} =1+\frac{n_e ex}{\epsilon_0 E}\]
 $\rightarrow$ Need $x$.
@@ Line 19: / Line 20: @@
 Have $F= eE= m_e \ddot{x}$, so
 \[\ddot{x} = \frac{e}{m_e} E_0 e^{-i\omega t}\]
-Have $\ddot{x} = -\omega x$, so
+Looking at the above equation, we must have $\ddot{x} = -\omega^2 x$, so
-\[x = x_0 e^{-i\omega t}= -\frac{e}{m_e \omega^2} E_0e^{-i\omega t}\]
+\[x = -\frac{\ddot{x}}{\omega^2}= -\frac{e}{m_e \omega^2} E_0e^{-i\omega t} = -\frac{e E}{m_e \omega^2}\]
-\[\epsilon_r = 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2} =1 - \frac{\omega_p^2}{\omega^2}\], where $\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$.
+\[\epsilon_r = 1 - \frac{n_e e^2 E}{\epsilon_0 m_e \omega^2 E}= 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2} =1 - \frac{\omega_p^2}{\omega^2}\], where $\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$.
-From $v_p$ above we get:
+From $v_p=\frac{\omega}{k}\approx \frac{c}{\sqrt{\epsilon_r}}$ we get:
 \[k = \frac{\omega \sqrt{\epsilon_r}}{c} = \frac{1}{c} \sqrt{ \omega^2- \omega_p^2}\]
 \[\frac{1}{v_g} = \frac{\partial k}{\partial \omega} = \frac{\omega}{c \sqrt{\omega^2- \omega_p^2}} =\frac{1}{c \sqrt{1- \frac{\omega_p^2}{\omega^2}}} \approx \frac{1}{c} \left[1+\frac{\omega_p^2}{2\omega^2} \right]\]

@@ Line 4: / Line 4: @@
 == Crowdsourced Answers ==
 === Question 12 ===
-A pulsar emits bursts of radio waves at two frequencies $\omega_1$, $\omega_2$. The pulses arrive at different times $t_1$, $t_2$ due to interaction with the interstellar medium, which is a dilute Hydrogen plasma. Find the distance $s$ from pulsar to earth, given $\tau= t_2 - t_2$.<div class="mw-collapsible mw-collapsed" style="width:750px">
+A pulsar emits bursts of radio waves at two frequencies $\omega_1$, $\omega_2$. The pulses arrive at different times $t_1$, $t_2$ due to interaction with the interstellar medium- a dilute Hydrogen plasma. Find the distance $s$ from pulsar to earth, given $\tau= t_2 - t_2$.<div class="mw-collapsible mw-collapsed" style="width:750px">
 ''Method'':

← Older revision		Revision as of 22:36, 17 April 2014
Line 1:		Line 1:
−	Official Questions and Solutions for [~~http://physwiki.com/w/images/9/90/F%2712.pdf~~ Fall 2012].	+	Official Questions and Solutions for [Fall 2012].
	<div class="fb-share-button" data-href="http://developers.facebook.com/docs/plugins/" data-type="button_count"></div>		<div class="fb-share-button" data-href="http://developers.facebook.com/docs/plugins/" data-type="button_count"></div>