Download Physics 121 Hour Exam #5 Solution This exam consists of a five

Physics 121 Hour Exam #5
Solution
This exam consists of a five problems on five pages. Point values are given with each
problem. They add up to 99 points; you will get 1 free point to make a total of 100.
In any given problem, points are not necessarily evenly divided between the parts.
Two pages are supplied at the end of the booklet in case you need extra workspace.
1. [12 points] In a supernova explosion, the inner core of a star collapses while
the outer layers of the star are blown away at high speed in all directions, as
suggested by the drawing below. Centuries after such an explosion, the debris
from the outer layer can still be seen to be flying away from the remaining core
of the star.
Before Explosion
After Explosion
Consider (i) the pre-explosion mass of the original star and (ii) the
post-explosion mass consisting of the mass of the core that was left behind
plus the mass of the debris that was blown away after the explosion. Which
of these is greater, pre-explosion mass or post-explosion mass, or are they the
same? Briefly justify your answer.
Energy (the first component of four-momentum) is conserved. Mass is one form
that energy can take.
Before the explosion, all of the energy is in the rest mass of the star.
After the explosion, some of this energy has been transformed into kinetic energy
of the debris. There is less energy available for the mass of the core and the debris.
The post-explosion mass is less.
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2. [12 points] Can a photon (particle of light) decay into a single particle which
has mass (m > 0)? Briefly justify your answer.
No. This would violate the conservation of momentum and energy.
Here’s one way to look at it. Let’s say that the photon is in the Home Frame,
moving to the right, and it decays into a single massive particle. The photon had
momentum, so this particle must have momentum; it is moving to the right at
some speed v. Now consider the rest frame of this particle, which is moving at
speed v relative to the Home Frame. The particle has zero momentum in this
frame. Therefore, by conservation of momentum, the photon must have had zero
momentum before the decay as observed in this frame. But this is impossible:
photons always have momentum. Thus it is impossible for a photon to decay into
a single particle.
Here’s another way to look at it. Let Ei be the energy of the photon. Its
momentum is pi = Ei . Let mf , vf , and γf be the mass, velocity, and Lorentz
factor of the massive particle. Four-momentum conservation gives:
 


γf mf
Ei
 γf mf vf   Ei 
=


  0 

0
0
0
The first component gives Ei = γf mf and the second component gives Ei = γf mf vf .
Comparing these, we find vf = 1. However, as we have discussed in class, vf = 1 is
impossible for a particle with mass. Thus it is impossible for a photon to decay into
a single particle.
Going beyond this problem. A straightforward extension of the above arguments
can be used to show that in addition to it being impossible for a photon to decay into
a single massive particle, it also cannot decay into a group of more than one massive
particles. Conversely, neither a single massive particle nor a group of particles can
transform into a single photon. In any such transformation, there must either be
at least two initial photons, or at least two final photons, or at least one initial and
one final photon.
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3. [25 points] A particle with mass M0 and speed 53 collides with a particle of mass
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M0 at rest. They coalesce to form a single new particle. What is the mass and
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speed of this new particle?
v1i =
3
5
⇒
1
γ1i = q
1−
 
γ1i m1i
γf mf
 γf mf vf   γ1i m1i v1i
=

 

0
0
0
0

3 2
5
=
5
4
  5
γ2i m2i
4 M0
  γ2i m2i v2i   5 M0 3
5
= 4
+
  0
 
0
0
0

⇒
v2i = 0



 
+
 
7
4 M0
0
0
0
γ2i = 1

3M0
  3 M0 
= 4

  0 
0


The first component gives
γf mf = 3M0 .
The second component gives
3
M0
4
3
3
3
M0
M0
vf = 4
vf = 4
= 4
γf mf
3M0
3
γf mf vf =
vf =
1
= 0.25
4
This gives Lorentz factor
1
4
= q = √ = 1.033
2
15
15
1 − 14
16
γf = q
1
So the mass is
γf mf = 3M0
⇒
mf =
3M0
3M0
=
γf
1.033
mf = 2.90 M0
Alternative method. After setting up the momentum four-vectors as above, use
m2 = E 2 − p2 to calculate
s
2
q
3
2
2
mf = Ef − pf = (3M0 )2 −
M0
= 2.90 M0 .
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Then solve γf mf = 3M0 to find γf = 1.033, from which you can calculate vf = 0.25.
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4. [25 points] A particle of initial mass mi moves in the +x direction at speed
vi , corresponding to Lorenz factor γi . A photon of energy E moves in the +y
direction. The photon collides with the particle and is absorbed by it. (This
is like two things hitting and sticking together, except one of those things is a
photon of light). After absorbing the photon, the particle moves off at some
angle θ, as shown.
What is θ?
(Your answer should be a formula which depends on E, mi , vi , and/or γi ,)
Before Absorption
After Absorption
vf
vi
θ
photon
Write out conservation of four-momentum. As will become evident when we
do the trigonometry below, the calculation is easiest if we write the x- and
y-components of momentum in terms of vf x and vf y rather than substituting
vf x = vf cos θ and vf y = vf sin θ.

 
Ef
γi mi
 pfx   γi mi vi
 

 pfy  = 
0
0
0



E
  0 
+

  E 
0
The second and third components give
pfx = γi mi vi
pfy = E.
Trigonometry gives
tan θ =
pfy
E
=
pfx
γi mi vi
⇒
θ = tan−1
E
γi mi vi
Note: this same reasoning works if you write the second and third components of
the final four-momentum as γf mf vfx and γf mf vfy and then use tan θ = vfy /vfx . The
factors of γf mf cancel out and the math is similar to that done above.
θ
pxf
4
pyf
5. [25 points] A spaceship is moving at speed v = 0.866, which corresponds to
γ = 2, toward the Earth. A laser beam is emitted by the spaceship in the
direction of Earth. The beam consists of photons each of which has energy E 0
as measured in the spaceship reference frame.
Earth
v=0.866
(a) What is the energy of one of these photons in the Earth reference frame?
(Your answer should be a formula which depends on E 0 .)
(b) In quantum mechanics, it is claimed that the frequency of a photon is
proportional to its energy. This can be written E = hf , where E is the
photon energy, f is its frequency, and h is a physical constant which is the
same number for all photons. Consider the Doppler shift of the frequency
of the a photon emitted in the spaceship reference frame and received in the
Earth reference frame. Is your answer to (a) consistent with this Doppler
shift? Justify your answer using formulas or a calculation. (Don’t just
answer “yes” or “no”).
(a) For a photon, momentum equals energy, so the four-vector of this photon in the rocket ship is
p0 = [E 0 , E 0 , 0, 0].
˜
The key point is that the x-component of momentum equals the energy, p0x = E 0 . Use the
four-momentum transform equations to transform these to the earth frame:
`
´
E = γ E 0 + βp0x = 2(E 0 + 0.866E 0 ) = 2(1 + 0.866)E 0 = 3.732E 0
Incidentally, since E = px for a rightward-moving photon in the Home Frame, you could also
use the transformation equation for px to get the same result:
`
´
px = γ βE 0 + p0x = 2(0.866E 0 + E 0 ) = 2(0.866 + 1)E 0 = 3.732E 0
E = px = 3.732E 0
(b) For a signal emitted at speed v = β (spaceship) moving toward the receiver (Earth), the
frequencies are related by
r
1−v
fR =
fE
1+v
If E = hf in general, then in this problem we have E = hfR (received signal, Earth frame) and
E 0 = hfE (emitted signal, Spaceship frame). Re-writing these as fR = E/h and fE = E 0 /h
and plugging them into the Doppler formula gives
E
=
h
r
1 − v E0
1 + v h0
r
⇒
E=
1−v 0
E.
1+v
For emitter moving toward receiver, v is negative, so v = −0.866:
s
E=
1 − (−0.866) 0
E = 3.732E 0
1 + (−0.866)
The answer to (a) is consistent with the Doppler shift.
They both give the same relation
between E and E 0 . It is stunning that two very different approaches to this problem, one
in which the light is treated as a single particle carrying a certain amount of energy, and the
other in which the light is treated as a wave of a certain frequency, give exactly the same result.
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