Download Phys 801 - Homework 1 Special and General relativity, Relativistic

Phys 801 - Homework 1
Special and General relativity, Relativistic kinematics
1. Show that Newtonʼs laws of motion are not invariant under a transformation to
a frame that is uniformly accelerated with respect to an inertial frame of Newton
mechanics. What is the equation of motion in the accelerated frame?
[Hint: consider transformations between 2 frames with aligned x-axes and
acceleration along x].
2. A resonance of significance for experiments on ultra-high energy neutrinos in
astrophysics is the Glashow resonance: ν e + e − →W − .
Assuming that the target electrons in the cosmos are at rest, show that this
resonance would be excited for antineutrino energies of about 6,400 TeV. Some
expressions for the cross sections are given below. Chose at least one of them
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and plot it (you should get something similar to the figure below). Then derive the
value of the cross section around the resonance in an interval equal to the boson
W width. Use any software tool you like for numerical integrations and plots
and the values of the constants in the table.
3. Calculate the minimum photon energy in the rest frame of the proton for direct
+
pion production in the reaction p-gamma→π + X, given that the mass of the
charged pion is 139.57 MeV. Infer what can be X from the fact that baryon
number and charge must be conserved in the reaction.
4. Demonstrate that the neutrino takes ~1/4 of the pion energy (rest mass) in the
+
+
charged pion decay: π → µ + ν µ , working in the rest frame of the pion. The
muon mass is 105.66 MeV.
5. Demonstrate that an isolated electron cannot absorb a photon.
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6. [Optional for undergraduate students] Find the total time to the Big
crunch in the case of a closed universe with a curvature term with k =
+1 and mass 1023 MSun, where MSun = 2 x 1030 kg.
Hints: we found in class the value of the maximum radius
. You need to calculate the time to reach this radius using
the change of variable:
and then calculate the time to the Big Crunch where the universe collapses to
Rr = 0.
7. [Optional] Calculate the minimum projectile energy in the rest frame of the
target proton for a pp reaction that produces a single π0 (remember Baryon
conservation!). The rest mass of the π0 is 134.97 MeV, and the mass of the
proton is mp = 938.27 MeV.
Suggested solutions
1.
We consider the uniform acceleration along the x axis so that
transformations are:
⎧
1 2
⎪ x' = x − 2 at
⎪
y' = y
⎨
⎪
z' = z
⎪
t' = t
⎩
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Calculating the second derivatives in time we do not get zero from the first one so
(xʼyʼzʼ) frame is not inertial.
d 2 x' d 2 x
We get
= 2 − at ⇒ F' = ma − mat
dt 2
dt
€ 2. Glashow.C script available on request. I get 2.2e-31 cm^2 GeV for formula 6
+
3. p+gamma→π + X
X can be a neutron for baryon number and charge conservation and
because it is lighter than other baryons. In the rest frame of the proton (the
target) we calculate the minimum photon energy of the photon projectile:
(mn + mπ ) 2 − m 2p (939.57 + 139.57) 2 − 938.27 2
Eγ =
=
= 151.45MeV
2m p
2 × 938.27
→ µ + + ν µ In the rest frame of the decaying pion: Energy conservation (and considering the neutrino as a massless particle so that E =| p |: 4. π
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+
ν
ν
2
p µ + mµ2 + Eν
mπ =
and for momentum conservation:
p = -­‐p
ν
µ
2
(mπ − Eν ) 2 = p µ + mµ2 = Eν2 + mµ2 ⇒
mπ2 + Eν2 − 2 Eν mπ = Eν2 + mµ2 ⇒ Eν =
mπ2 − mµ2
2mπ
= 29.8MeV
So the fraction of pion energy is
2
2
Eν mπ − mµ
=
= 0.21
mπ
2mπ2
This is approximately 1/4 of pion energy.
5. 4-momentum conservation:
( pγ )µ + ( pe )µ = ( p'e )µ
⇒ ( pγ )µ ( pe )µ = 0
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E r
in the frame of the electron initially at rest:
( pe ) µ = (me c,0), ( pγ ) µ = ( , pγ ) c
r
me E − 0⋅ pγ = 0
so energy-momentum conservation requires a photon with 0 energy, hence the
process is impossible.
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6.
From the Friedmann equation:
For k = 1
� �2
Ṙ
R
Substituting:
Differentiating:
And the extreme of integration are: for
=
8πGρ
3
−
c2
r 2 R2
7.
The reaction is (conservation of baryon number and charge):
pp→π0pp
In the rest frame of the proton:
(2m p + mπ ) 2 − (2m p ) 2 (2 × 938.27 + 134.97) 2 − 4 × 938.27 2
Ep =
=
= 279.65MeV
2m p
2 × 938.27