Difference between revisions of "Fall 2012"
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| − | Official Questions and Solutions for [Fall 2012]. | + | Official Questions and Solutions for [http://www.physwiki.com/w/images/9/90/F%2712.pdf Fall 2012]. |
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== Crowdsourced Answers == | == Crowdsourced Answers == | ||
=== Question 12 === | === 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 | + | 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'': | ''Method'': | ||
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$\rightarrow$ Need $\epsilon_r$. | $\rightarrow$ Need $\epsilon_r$. | ||
| − | + | \[\epsilon_r =1+\chi_e = 1 + \frac{P}{\epsilon_0 E}= 1 + \frac{n_e p_e}{\epsilon_0 E} =1+\frac{n_e ex}{\epsilon_0 E}\] | |
| + | , where $E= E_0 e^{-i\omega t}$, $P= n_e p_e =$ Polarization vector, $n_e =$ electron number density and $p_e =$ electron dipole moment. | ||
$\rightarrow$ Need $x$. | $\rightarrow$ Need $x$. | ||
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Have $F= eE= m_e \ddot{x}$, so | Have $F= eE= m_e \ddot{x}$, so | ||
\[\ddot{x} = \frac{e}{m_e} E_0 e^{-i\omega t}\] | \[\ddot{x} = \frac{e}{m_e} E_0 e^{-i\omega t}\] | ||
| − | + | Therefore we must have $\ddot{x} = -\omega^2 x$, so | |
| − | \[x = | + | \[x = -\frac{\ddot{x}}{\omega^2}= -\frac{e}{m_e \omega^2} E_0e^{-i\omega t} = -\frac{e E}{m_e \omega^2}\] |
| − | \[\epsilon_r = 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2} =1 - \frac{\omega_p^2}{\omega^2}\], 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 $v_p$ | + | From $v_p=\frac{\omega}{k}\approx \frac{c}{\sqrt{\epsilon_r}}$ we get: |
\[k = \frac{\omega \sqrt{\epsilon_r}}{c} = \frac{1}{c} \sqrt{ \omega^2- \omega_p^2}\] | \[k = \frac{\omega \sqrt{\epsilon_r}}{c} = \frac{1}{c} \sqrt{ \omega^2- \omega_p^2}\] | ||
\[\frac{1}{v_g} = \frac{\partial k}{\partial \omega} = \frac{\omega}{c \sqrt{\omega^2- \omega_p^2}} =\frac{1}{c \sqrt{1- \frac{\omega_p^2}{\omega^2}}} \approx \frac{1}{c} \left[1+\frac{\omega_p^2}{2\omega^2} \right]\] | \[\frac{1}{v_g} = \frac{\partial k}{\partial \omega} = \frac{\omega}{c \sqrt{\omega^2- \omega_p^2}} =\frac{1}{c \sqrt{1- \frac{\omega_p^2}{\omega^2}}} \approx \frac{1}{c} \left[1+\frac{\omega_p^2}{2\omega^2} \right]\] | ||
\[\tau= s \left[\frac{1}{v_{g,2}} - \frac{1}{v_{g,1}}\right] \approx \frac{s \omega_p^2}{2c} \left[\frac{1}{\omega_2^2} - \frac{1}{\omega_1^2}\right]\] | \[\tau= s \left[\frac{1}{v_{g,2}} - \frac{1}{v_{g,1}}\right] \approx \frac{s \omega_p^2}{2c} \left[\frac{1}{\omega_2^2} - \frac{1}{\omega_1^2}\right]\] | ||
| − | \[s | + | \[s \approx \frac{2c\tau}{\omega_p^2}\left[\frac{\omega_1^2\omega_2^2}{\omega_1^2-\omega_2^2}\right]\] |
</div> | </div> | ||
<br/> | <br/> | ||
Latest revision as of 01:14, 13 May 2014
Official Questions and Solutions for Fall 2012.
[edit] Crowdsourced Answers
[edit] 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- a dilute Hydrogen plasma. Find the distance $s$ from 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 $\epsilon_r$.
\[\epsilon_r =1+\chi_e = 1 + \frac{P}{\epsilon_0 E}= 1 + \frac{n_e p_e}{\epsilon_0 E} =1+\frac{n_e ex}{\epsilon_0 E}\] , where $E= E_0 e^{-i\omega t}$, $P= n_e p_e =$ Polarization vector, $n_e =$ electron number density and $p_e =$ electron dipole moment.
$\rightarrow$ Need $x$.
Have $F= eE= m_e \ddot{x}$, so \[\ddot{x} = \frac{e}{m_e} E_0 e^{-i\omega t}\] 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 $\omega_p^2 = \frac{n_e e^2}{\epsilon_0 m_e}$.
From $v_p=\frac{\omega}{k}\approx \frac{c}{\sqrt{\epsilon_r}}$ we get: \[k = \frac{\omega \sqrt{\epsilon_r}}{c} = \frac{1}{c} \sqrt{ \omega^2- \omega_p^2}\] \[\frac{1}{v_g} = \frac{\partial k}{\partial \omega} = \frac{\omega}{c \sqrt{\omega^2- \omega_p^2}} =\frac{1}{c \sqrt{1- \frac{\omega_p^2}{\omega^2}}} \approx \frac{1}{c} \left[1+\frac{\omega_p^2}{2\omega^2} \right]\] \[\tau= s \left[\frac{1}{v_{g,2}} - \frac{1}{v_{g,1}}\right] \approx \frac{s \omega_p^2}{2c} \left[\frac{1}{\omega_2^2} - \frac{1}{\omega_1^2}\right]\] \[s \approx \frac{2c\tau}{\omega_p^2}\left[\frac{\omega_1^2\omega_2^2}{\omega_1^2-\omega_2^2}\right]\]
