hwsol2012_11

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Physics 249 Homework 11
Due Dec 14th
1) Two charged pions at rest annihilate according to the interaction πœ‹ + + πœ‹ − → 𝛾 + 𝛾.
a) Why must the energies of the two gamma rays be equal?
b) Find the energy of each gamma ray.
c) Can two pions annihilate to one gamma ray? Explain your answer.
a,b)
In rest frame the four vectors are:
for the initial state
π‘πœ‹− = (πΈπœ‹ , π‘βƒ‘πœ‹ )
π‘πœ‹+ = (πΈπœ‹ , −π‘βƒ‘πœ‹ )
where πΈπœ‹ = (|π‘βƒ‘πœ‹ |2 + π‘šπœ‹2 )1/2
for the final state:
𝑝𝛾1 = (𝐸𝛾1 , 𝑝⃑𝛾1 )
𝑝𝛾2 = (𝐸𝛾2 , −𝑝⃑𝛾1 )
2
1/2
where 𝐸𝛾1 = (|𝑝⃑𝛾1 | + π‘šπ›Ύ2 )
2
1/2
= |𝑝⃑𝛾1 | = 𝐸𝛾2 = (|−𝑝⃑𝛾1 | + π‘šπ›Ύ2 )
= |𝑝⃑𝛾1 |
c)
For the final state in this problem
𝑝𝛾 = (2πΈπœ‹ , π‘βƒ‘πœ‹ + (−π‘βƒ‘πœ‹ )) = (2πΈπœ‹ , 0)
then
𝑣𝛾 = 0
and
π‘šπ›Ύ = ((2πΈπœ‹ )2 + 0)1/2 = 2πΈπœ‹ ≠ 0
both of which are unphysical properties for a real photon. Note they are fine for a virtual
photon.
2) Antiproton can be made via the interaction 𝑝 + 𝑝 → 𝑝 + 𝑝 + 𝑝 + 𝑝̅ .
a) Derive the minimum kinetic energy for this interaction of one proton is at rest in this
interaction.
b) Why can’t we produce antiprotons via 𝑝 + 𝑝 → 𝑝 + 𝑝̅? Give one reason.
a)
In the lab frame of the proton at rest.
For the initial state
𝑝𝑖1 = (π‘šπ‘ , 0)
𝑝𝑖2 = (𝐸𝑖2 , 𝑝⃑𝑖2 )
1/2
𝐸𝑖2 = (|𝑝⃑𝑖2 |2 + π‘šπ‘2 )
For the final state in the lab frame of the proton at rest. The minimum energy occurs
when there is not extra momentum. i.e. the four particles are going in the direction of the
original proton with no negative contributions. In fact should have the same momentum
or there would be extra contributions to the energy in their rest own frame.
1/2
𝑝𝑇 = (𝐸𝑖2 + π‘šπ‘ , 𝑝⃑𝑖2 ) = ((|𝑝⃑𝑖2 |2 + π‘šπ‘2 )
where
1/2
𝐸𝑖2 + π‘šπ‘ = (|𝑝⃑𝑖2 |2 + 16π‘šπ‘2 )
1/2
(|𝑝⃑𝑖2 |2 + π‘šπ‘2 )
+ π‘šπ‘ , 𝑝⃑𝑖2 )
1/2
+ π‘šπ‘ = (|𝑝⃑𝑖2 |2 + 16π‘šπ‘2 )
1/2
|𝑝⃑𝑖2 |2 + π‘šπ‘2 + 2π‘šπ‘ (|𝑝⃑𝑖2|2 + π‘šπ‘2 )
1/2
2π‘šπ‘ (|𝑝⃑𝑖2 |2 + π‘šπ‘2 )
+ π‘šπ‘2 = |𝑝⃑𝑖2 |2 + 16π‘šπ‘2
= 14π‘šπ‘2
1/2
(|𝑝⃑𝑖2 |2 + π‘šπ‘2 ) = 7π‘šπ‘
|𝑝⃑𝑖2 |2 + π‘šπ‘2 = 49π‘šπ‘2
|𝑝⃑𝑖2 | = (48)1/2 π‘šπ‘
then
1/2
1/2
𝐸𝑖2 = (|𝑝⃑𝑖2 |2 + π‘šπ‘2 ) = (48π‘šπ‘2 + π‘šπ‘2 )
and for kinetic (non mass energy)
= 7π‘šπ‘
𝐾𝐸𝑖2 = 7π‘šπ‘ − π‘šπ‘ = 6π‘šπ‘ = 5.6𝐺𝑒𝑉
b)
Conservation or charge or conservation of baryon number
3) Text book 12-12
a) EM
b) Weak
c) EM
d) Weak
e) Strong
f) Weak
4) Text book 12-17
a) Conservation of energy, though this can be allowed in a atom where the energy can
come from the bound state energy of the nucleus.
b) Conservation of energy. The pion is too massive.
c) Conservation of momentum would be violated to give the photon non zero momentum
and thus velocity in the center of mass frame.
d) This is fine
e) Conservation of leptons number
f) Conservation of baryon number
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