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Multichannel particle decay survival probability

by pomaranca
Tags: particle decay, statistics
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pomaranca
#1
Dec12-12, 07:23 AM
P: 16
Particle can decay through many channels with probabilities [itex]p_i[/itex], where in each channel its decay time is different [itex]\tau_i[/itex]. It always decays through one of the channels.

Particle decays according to exponential law where probability to decay in time [itex]t[/itex] is
[tex]
P^{(i)}_d(t)={1\over\gamma\tau_i}\exp\left({-{t\over\gamma\tau_i}}\right)\;.
[/tex]
What is the total probability for a particle to survive a given time [itex]t[/itex] (so it does not decay in any channel)?
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MarekS
#2
Dec12-12, 08:24 AM
P: 34
Why not integrate all the decay probabilities from 0 to t, add them up and then subtract the resulting probability from 1 to find the probability of the particle surviving?
Myphyclassnot
#3
Dec12-12, 09:08 AM
P: 4
Please suggest me a good book of nuclear and particle physics. My syllabus is nuclear reaction, nuclear fission and fusion, elementary particles, particle accelerator and detector, nuclear astrophysics.

mfb
#4
Dec12-12, 09:11 AM
Mentor
P: 11,815
Multichannel particle decay survival probability

Quote Quote by pomaranca View Post
where in each channel its decay time is different [itex]\tau_i[/itex]. It always decays through one of the channels.
Do you mean lifetime here? Decay time is usually the time of a specific particle decay in an event.
How can you have different lifetimes? Mixing of neutral mesons can give this on the level of ~1% in an effective description, but you have to consider both states there.
pomaranca
#5
Dec12-12, 12:26 PM
P: 16
@MarekS
[tex]
P_s(t)=1-\sum_i p_i \int\limits_0^t{1\over\gamma\tau_i}\exp\left(-{t'\over\gamma\tau_i}dt'\right)
[/tex]
Is this what you meant?

@Myphyclassnot: for nuclear and particle physics see e.g. Griffiths, Povh, Perkins, Halzen & Martin, Peskin & Schroeder, Bjorken & Drell ...

@mfb: yes, i meant proper lifetime, e.g. [itex]K^+[/itex] can decay via multiple decay channels, see this for listing of all decay modes.
mfb
#6
Dec12-12, 02:12 PM
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P: 11,815
Quote Quote by pomaranca View Post
@mfb: yes, i meant proper lifetime, e.g. [itex]K^+[/itex] can decay via multiple decay channels, see this for listing of all decay modes.
A particle has a single lifetime. PDG gives decay rates (as inverse lifetimes) for individual decays, you can simply add all of them to get the total inverse lifetime. The probability that a particle lives after time t is always an exponential function, unless you have mixing.
MarekS
#7
Dec13-12, 04:44 AM
P: 34
Quote Quote by pomaranca View Post
@MarekS
[tex]
P_s(t)=1-\sum_i p_i \int\limits_0^t{1\over\gamma\tau_i}\exp\left(-{t'\over\gamma\tau_i}dt'\right)
[/tex]
Is this what you meant?
Jep, that's what I meant.
pomaranca
#8
Dec15-12, 08:53 AM
P: 16
Quote Quote by mfb View Post
A particle has a single lifetime. PDG gives decay rates (as inverse lifetimes) for individual decays, you can simply add all of them to get the total inverse lifetime. The probability that a particle lives after time t is always an exponential function, unless you have mixing.
So the total lifetime of a particle with miltiple decay channels is
[tex]
{1\over\tau_{\rm tot}}=\sum\limits_i{1\over\tau_i}\;.
[/tex]
And the probability for such a particle to live for a time [itex]t[/itex] is
[tex]
P_s(t)=\exp\left(-{t\over\gamma\tau_{\rm tot}}\right)=
\prod_i \exp\left(-{t\over\gamma\tau_i}\right)\;.
[/tex]
This is the probability for a particle to survive all channels at once: the first AND the second AND the third etc. And the probabilities for particular channels are not taken into account.
Is this really the case? The resulting probability is also very low.

Quote Quote by MarekS View Post
Jep, that's what I meant.
This was also my first guess. And the result fits perfectly to my simulation results, but this simulation could be flawed.

This is the way I simulate the process:
  1. Generate a random number [itex]\xi\in[0,1)[/itex] with uniform probability and according to its value choose on of the possible decay channels.
  2. For a chosen channel randomly generate particle's lifetime [itex]t[/itex] from exponential distribution with mean value [itex]\gamma\tau_i[/itex].
  3. Apply kinematics of a decay in a chosen channel.
To achieve an exponential distribution with [itex]\tau_{\rm tot}[/itex] I suppose I should send a particle to decay in the first channel and then if it survives to the second and so on through each channel.
mfb
#9
Dec15-12, 10:19 AM
Mentor
P: 11,815
I do not think ##\tau_i## is used somewhere, as it is misleading - this time has no physical meaning. ##\Gamma_i## is the common way to give decay rates for individual channels, if that is done at all (lifetime of the particle and branching fraction is more common).

Quote Quote by pomaranca View Post
This is the way I simulate the process:
  1. Generate a random number [itex]\xi\in[0,1)[/itex] with uniform probability and according to its value choose on of the possible decay channels.
  2. For a chosen channel randomly generate particle's lifetime [itex]t[/itex] from exponential distribution with mean value [itex]\gamma\tau_i[/itex].
  3. Apply kinematics of a decay in a chosen channel.
Just use the lifetime of the particle (and not some decay-channel related value) in step 2.
pomaranca
#10
Dec19-12, 06:22 PM
P: 16
@mfb: Thank you for your answer, this really seems to be the case. Could you suggest some literature, a standard book or an article on this issue. I just can't seem to find this anywhere. Thanks again!
mfb
#11
Dec20-12, 07:10 AM
Mentor
P: 11,815
I think there should be lecture notes and books for particle physics and monte carlo generators, but I don't know any specific one.


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