time of flight between two scintillators - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-18T22:19:23Z https://physics.stackexchange.com/feeds/question/241362 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/241362 1 time of flight between two scintillators TheStrangeQuark https://physics.stackexchange.com/users/93711 2016-03-04T03:04:21Z 2016-03-04T10:00:34Z <p>I found this page on Wikipedia about finding distance between time of flight of two particles passing past two scintillators, <a href="https://en.wikipedia.org/wiki/Time_of_flight_detector" rel="nofollow">https://en.wikipedia.org/wiki/Time_of_flight_detector</a>, but I can't find any other proofs or anything other than simply showing this equation. Does anyone know where this is derived? Or possibly shown for relativistic speeds.</p> https://physics.stackexchange.com/questions/241362/-/241369#241369 1 Answer by Floris for time of flight between two scintillators Floris https://physics.stackexchange.com/users/26969 2016-03-04T04:24:57Z 2016-03-04T04:24:57Z <p>The equation is</p> <p>$$\delta t = L\left(\frac{1}{v_1}-\frac{1}{v_2}\right)$$</p> <p>This is just saying that the difference in time recorded between the "start" and the "stop" depends on the velocity of the particle: the time it takes a particle with $v_1$ to travel a distance $L$ is $\frac{L}{v_1}$, obviously.</p> <p>The next bit is trickier. We are supposed to prove that this reduces to</p> <p>$$\delta t \approx \frac{Lc}{2p^2}\left(m_1^2 - m_2^2\right)$$</p> <p>when the particles have the same momentum $p$, and we are told in the article that this is a relativistic approximation. </p> <p>Looking at the units, we have $L L T^{-1}M^{-2}L^{-2}T^2M^2 = T$ - which is encouraging.</p> <p>Relativistically,</p> <p>$$E^2 = m^2c^4 = p^2 c^2 + m_0^2 c^4$$</p> <p>Assuming that the mass in the equation is the relativistic mass, and that this is much larger than the rest mass, we find $$p^2 = m^2 c^2 - m_0^2 c^2$$</p> <p>Now the relationship between the mass $m$ and the rest mass $m_0$ is of course given by</p> <p>$$m= \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$</p> <p>Or</p> <p>$$m^2~\left(1-\frac{v^2}{c^2}\right)=m_0^2$$</p> <p>and the momentum reduces to</p> <p>$$p^2 = m^2 v^2$$</p> <p>which is the same as the classical equation.</p> <p>Right now I'm blocking on how to take the last step here... I am marking this "community wiki" and leaving this for someone else to complete. Maybe I will be able to do it myself with fresh eyes in the morning.</p> https://physics.stackexchange.com/questions/241362/-/241406#241406 1 Answer by Farcher for time of flight between two scintillators Farcher https://physics.stackexchange.com/users/104696 2016-03-04T08:33:34Z 2016-03-04T10:00:34Z <p>What you need is $v$ in terms of $m$., $p$ and $c$.</p> <p>Starting from $E^2 = p^2c^2+m^2c^4 = \gamma^2 m^2c^4$ or $p=\gamma mv$ where $\gamma^2 = \dfrac{1}{\left ( 1-\frac{v^2}{c^2}\right)}$ </p> <p>some algebra gives $\dfrac 1 v = \dfrac 1 c \cdot \sqrt{\left (1+\dfrac{m^2c^2}{p^2}\right) }$</p> <p>Because $p\gg mc$ the square root can be expanded to the first term </p> <p>$\sqrt{\left (1+\dfrac{m^2c^2}{p^2}\right) } \approx 1+ \dfrac{m^2c^2}{2p^2} + . . .$ </p> <p>and this leads to the required relationship because the momentum $p$ is the same for both particles.</p> <p>PS<br> I am afraid I do not understand what @Floris meant by "I am marking this "community wiki" and leaving this for someone else to complete." An explanation would be appreciated.</p>