Calculate the momentum of a 249 MeV pion

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The discussion revolves around calculating the momentum of a 249 MeV pion, focusing on a problem from "Advanced Physics" regarding the decay of the K+ meson into pions. Participants clarify the quark content of the pions, confirming that the π0 cannot contain strange quarks due to spin considerations, and that its composition is limited to combinations of up and down quarks. The calculations for momentum using the relation E² = m₀²c⁴ + p²c² are debated, with one user initially miscalculating the momentum but later correcting it after realizing a mistake in squaring the mass term. The final consensus is that the momentum calculation is indeed correct, and the numerical coincidence of momentum and energy arises from a misunderstanding of the equations. The conversation emphasizes the importance of careful application of physics principles in particle decay scenarios.
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Homework Statement



The question is from "Advanced Physics" by Adams and Allday. It is from section 9 "The Physics of Particles", "Practice Exam Questions". It is question 10. The level of this study is between school and University.

I am uncertain about the answer to part c, ii (3rd meson) and iv.

The whole part c question is given in case the context is relevant.

c) The K+ is a meson with strangeness +1. One of its common decay modes is K+ → π+ + π0. Pions are not strange particles.

i. Name the type of interaction responsible for the K+ → π+ + π0 decay.

ii. Table 9.4 gives the properties of the relevant quarks. Deduce the possible quark content of the K+, π+ and π0
  • u, charge +2/3, baryon number 1/3, strangeness 0
  • d, charge -1/3, baryon number 1/3, strangeness 0
  • s, charge -1/3, baryon number 1/3, strangeness -1
iii. The rest mass of a proton is 938 MeV/c2. The K+ rest mass = 0.53 proton masses, and the π+ rest mass = π0 rest mass = 0.15 proton masses. Assuming that the K+ is stationary when it decays, show that the total energy of each pion produced in the decay is 249 MeV.

iv.Given that E2 = m02c4 + p2c2, calculate the momentum of each pion. Express your answer in units of MeV/c.

Homework Equations


Given in the question.

The Attempt at a Solution


ii. Only unsure about the possible quark content of the π0. Underlining is used below instead of barover to indicate anti-quark.

The π0 is a meson with charge 0 and strangeness 0.
Being a meson it is composed of a quark and an antiquark.
Strangeness 0 so either no s or ss.
Charge 0 so could be ss, dd. or uu.
Quark content of π0: ss, dd. or uu.

References confirmed dd. or uu but not ss. Is ss wrong and, if so, why?

iv.
E2 = m02c4 + p2c2

Using "natural" units and values given in the question this reduces to

p = sqrt(249^2 - (0.15 * 938))
= 249 MeV/c ct3sf.

Is this correct? I am not familiar with using natural units so have some doubt about whether it's OK just to discard all the "c"s like that. If it is correct, is it just coincidence that the momentum is numerically the same as the energy?

Best

Charles
 
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catkin said:
ii. Only unsure about the possible quark content of the π0. Underlining is used below instead of barover to indicate anti-quark.

(snip)

References confirmed dd. or uu but not ss. Is ss wrong and, if so, why?

It's true that there are no s quarks in the \pi^0 wavefunction. But I suppose that if all you know about the pions is their charge then it's not possible to rule them out. What you really need to know about the pions is that they form an isospin triplet and that the only possible quark content for \pi^+ is u\overline{d}.

iv.
E2 = m02c4 + p2c2

Using "natural" units and values given in the question this reduces to

p = sqrt(249^2 - (0.15 * 938))
= 249 MeV/c ct3sf.

Is this correct?

When I do the above calculation I get p=205 MeV/c.
 
Thanks Tom :smile:

OK with ss not being possible because of spin considerations but can you recheck the calculation? I just pasted sqrt(249^2 - (0.15 * 938)) into gcalctool 5.22.3 and got 248.71730941 again ...

Best

Charles
 
You forgot to square the mass inside the square root sign.
 
Thanks again Tom :smile:

Oops! I tried to work out whether sqrt(E^2 - m0) is E or whether it is just a coincidence that it gives the same number. Presumably coincidence because the erroneous calculation is dimensionally inconsistent.

Best

Charles
 
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