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Distinct particles detection

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose teo particles with masses m1 and m2 enter a detector both with momentum p. Calculate the difference Δt of the times they need to cross a distance L.

    Suppose Δt can be measured with a 300 ps resolution. How long must L be if we want to distinguish particles with m1 from particles with m2 with two standard deviations if their momentums are p?

    2. Relevant equations


    3. The attempt at a solution

    The first part is easy. I used [itex] \beta _1 = \frac{p}{E_1} = \frac{p}{\sqrt{p^2+m_1^2}}[/itex] and [itex] t_1 = \frac{L}{\beta_1}[/itex], and similar for the second particle. I'm not sure if that formula for the time is entirely correct, but I get to the same result if I integrate the velocity.

    But I am lost in the second part. What exactly does it mean for the detector to have a 300 ps resolution? And how does it relate to the standard deviations?

    If there was no mention of the s.d. I would suppose that for the detector to distinguish between two particles they have to enter it with at least a 300 ps difference. But what about the s.d.?
     
    Last edited: Sep 27, 2015
  2. jcsd
  3. Sep 27, 2015 #2

    mfb

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    The measured time will deviate from the actual arrival time in a random way for each measurement, and the standard deviation is 300 ps. As an example, the measurement will often be wrong by 100 or 200 ps, but rarely by 1000 ps because this would be more than 3 standard deviations away from the mean.
     
  4. Sep 27, 2015 #3
    Oh, so the time resolution just means the standard deviation of the Gaussian? I haven't been given a definition either in the lectures or in the class notes, and I don't find any (useful) definition on any book or in the internet.

    Thanks for your time.
     
  5. Sep 27, 2015 #4

    mfb

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    Well... probably. Sometimes other definitions are used for "resolution", like FWHM, but without more context I think it is save to assume it is the standard deviation of a Gaussian.
     
  6. Sep 27, 2015 #5
    Well, I'll have to make do with that one and see how it goes. Many thanks!
     
  7. Sep 27, 2015 #6
    One more thing, just to get it right: my approach now is that the probability for a particle that enters at time t1 to be detected at time t is
    [itex]p_1(x)=\frac{1}{\sigma\sqrt{2\pi}}exp\left[-\frac{1}{2}\left(\frac{t-t_1}{\sigma}\right)\right][/itex],
    and likewise for the second particle that enters at time [itex]t_1 + \Delta t[/itex]. Now the question is: how much needs [itex] \Delta t [/itex] to be so that the two curves can be resolved within two standard deviations? That is, they must intersect at a distance [itex]2\sigma[/itex] from each peak.

    So I set [itex] p_1(2\sigma)=p_2(-2\sigma)[/itex] and I get [itex]\Delta t = 4 \sigma [/itex], so there should be a delay of 1200 ps between their arrivals. Right?
     
  8. Sep 27, 2015 #7

    mfb

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    The two standard deviations are the full separation already, so they should be 600 ps apart I think.
     
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