Difference between revisions of "Fall 2012"
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=== Question 12 === | === Question 12 === | ||
| + | A pulsar emits bursts of radio waves at two frequencies $\omega_1$ and $\omega_2$. 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 the pulsar to earth, given $\tau= t_2 - t_2$.<div class="mw-collapsible mw-collapsed" style="width:750px"> | ||
| + | ''Method'': | ||
| + | |||
| + | Need $v_g = \frac{\partial \omega}{\partial k}$, so we have to find $\omega(k)$. | ||
| + | |||
| + | Have $v_p = \frac{\omega}{k} = \frac{c}{n(\omega)} \approx \frac{c}{\sqrt{\epsilon_r}}$ | ||
| + | |||
| + | $\rightarrow$ Need $\epsilon_r$. | ||
| + | |||
| + | Have $\epsilon_r =1+\chi_e = 1+\frac{n_e ex}{\epsilon_0 E}$, where $E= E_0 e^{-i\omega t}$ and $n_e =$ electron number density. | ||
| + | |||
| + | $\rightarrow$ Need $x$. | ||
| + | |||
| + | 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 | ||
| + | \[x = x_0 e^{-i\omega t}= -\frac{e}{m_e \omega^2} E_0e^{-i\omega t}\] | ||
| + | \[\epsilon_r = 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2} =1 - \frac{\omega_p^2}{\omega^2}\], where $\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$. From $v_p$ above we get: | ||
| + | \[k = \frac{\omega \sqrt{\epsilon_r}}{c} = \frac{1}{c} \sqrt{ \omega^2- \omega_p^2}\] | ||
| + | \[\frac{1}{v_g} = \frac{\partial k}{\partial \omega} = \frac{\omega}{c \sqrt{\omega^2- \omega_p^2}} =\frac{1}{c \sqrt{1- \frac{\omega_p^2}{\omega^2}}} \approx \frac{1}{c} \left[1+\frac{\omega_p^2}{2\omega^2} \right]\] | ||
| + | \[\tau= s \left[\frac{1}{v_{g,2}} - \frac{1}{v_{g,1}}\right] \approx \frac{s \omega_p^2}{2c} \left[\frac{1}{\omega_2^2} - \frac{1}{\omega_1^2}\right]\] | ||
| + | \[s = \frac{2c\tau}{\omega_p^2}\left[\frac{\omega_1^2\omega_2^2}{\omega_1^2-\omega_2^2}\right]\] | ||
| + | |||
| + | See [[Fall_2012#Question_12| F'12Q12]]. | ||
| + | </div> | ||
| + | <br/> | ||
=== Question 13 === | === Question 13 === | ||
=== Question 14 === | === Question 14 === | ||
Revision as of 04:39, 17 April 2014
Contents |
Existing Questions and Answers
Questions and Solutions for F'12: File:F'12.pdf.
Crowdsourced Answers
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
Question 11
Question 12
A pulsar emits bursts of radio waves at two frequencies $\omega_1$ and $\omega_2$. 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 the pulsar to earth, given $\tau= t_2 - t_2$.Method:
Need $v_g = \frac{\partial \omega}{\partial k}$, so we have to find $\omega(k)$.
Have $v_p = \frac{\omega}{k} = \frac{c}{n(\omega)} \approx \frac{c}{\sqrt{\epsilon_r}}$
$\rightarrow$ Need $\epsilon_r$.
Have $\epsilon_r =1+\chi_e = 1+\frac{n_e ex}{\epsilon_0 E}$, where $E= E_0 e^{-i\omega t}$ and $n_e =$ electron number density.
$\rightarrow$ Need $x$.
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 \[x = x_0 e^{-i\omega t}= -\frac{e}{m_e \omega^2} E_0e^{-i\omega t}\] \[\epsilon_r = 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2} =1 - \frac{\omega_p^2}{\omega^2}\], where $\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$. From $v_p$ above we get: \[k = \frac{\omega \sqrt{\epsilon_r}}{c} = \frac{1}{c} \sqrt{ \omega^2- \omega_p^2}\] \[\frac{1}{v_g} = \frac{\partial k}{\partial \omega} = \frac{\omega}{c \sqrt{\omega^2- \omega_p^2}} =\frac{1}{c \sqrt{1- \frac{\omega_p^2}{\omega^2}}} \approx \frac{1}{c} \left[1+\frac{\omega_p^2}{2\omega^2} \right]\] \[\tau= s \left[\frac{1}{v_{g,2}} - \frac{1}{v_{g,1}}\right] \approx \frac{s \omega_p^2}{2c} \left[\frac{1}{\omega_2^2} - \frac{1}{\omega_1^2}\right]\] \[s = \frac{2c\tau}{\omega_p^2}\left[\frac{\omega_1^2\omega_2^2}{\omega_1^2-\omega_2^2}\right]\]
See F'12Q12.
