Transcript Module 11 - FacStaff Home Page for CBU

Module 11 Relativistic Particles
Energy Analysis
We now have developed the necessary relationships to model the behavior of matter particles moving at a relativistic speed
of u with respect to an observer. These include the expressions for mass and momentum developed in Module 9 and the
expressions for rest energy, kinetic energy, and total energy developed in Module 10. Let’s list these equations again for
convenience.
Relativistic Mass m  mo
(11.1)
Momentum



p  mu  mo u
(11.2)
Rest Energy
mo c 2
(11.3)
Kinetic Energy
K  (   1)mo c 2
(11.4)
Total Energy
E  mc 2  mo c 2
(11.5)
The following equation relating energy to momentum, while not fundamental, is sometimes useful. You derived this
equation in one the Module 10 homework problems.
E 2  ( pc) 2  (mo c 2 ) 2
Module 11
Relativistic Particles
(11.6)
1
What are some of the things that we should notice about these expressions for these physical quantities? First of all,
remember that all of them revert to the classical, Newtonian expressions if u << c. We can get by just fine with saying that
mass doesn’t change with speed and that kinetic energy is (1/2)mu2 if the speed is much less than the speed of light.
When can’t we get away this? The easiest way to answer that question is by looking at the expression for relativistic mass.
It differs from the rest mass by the factor of , just like how a nonproper time interval differs from a proper time interval.
Recall in the homework of Module 4 that you showed that there is a 0.1% difference between nonproper and proper time
intervals ( = 1.001) when u = 0.0445c which is about 1.3x107 m/s. Thus, only particles traveling greater than about 107 m/s
have to be treated relativistically.
Also note that as the speed of a particle approaches the speed of light, its mass, momentum, and energy approach infinity
(as uc, ). This suggests that a particle made out of matter cannot reach a speed of c.* Indeed, this is one of the
more interesting implications of the theory. It would take an infinite amount of work (and hence energy) to accelerate a
particle to c. The amount of work required to accelerate a particle from an initial speed of ui to a final speed of uf is easily
found. Since the work done on the particle is the change in its kinetic energy, we have from Eq. (11.4)
Work  K  (  f   i )mo c 2
(11.7)
In fact, since the gamma factor increases superlinearly for high speeds, it takes an ever increasing amount of work to
increase the speed by the same increment. For example, to accelerate an electron from 0.90c to 0.92c takes 0.1315 MeV of
energy. To accelerate it to 0.94c requires not another 0.1315 MeV, but rather 0.1939 MeV more energy. (You should
confirm these calculations for yourself!)
The reason for this limitation of never reaching the speed of light can be explained physically by noting the increase in
mass as the particle accelerates. An ever increasing amount of the energy put in to speed up the particle is converted to
mass, resulting in the particle “bulking up” at the expense of experiencing an ever decreasing rate of acceleration.
* The theory of special relativity implies that a particle made out of mass cannot reach the speed of light if it begins its
existence at a speed that is less than c. However, it doesn’t rule out the possibility that there may be particles in the
universe that were created traveling greater than c. If they do exist, these particles, termed tachyons, have some strange
properties which we’ll briefly investigate in the module on time travel.
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The prediction of the conversion of energy to matter and vice versa is arguably the most important part of relativity theory.
We see examples of this conversion in many physical interactions. In nuclear physics, this conversion explains why a
bound nucleus is less massive than the individual nucleons that comprise it. Some of the nucleons’ mass is converted into
the binding energy that holds the nucleus together. (The same thing happens with a nucleus and electrons when they
combine to form an atom. Some of their mass is converted to the binding energy of the atom.) We can also understand
nuclear decay since we now realize that some of the mass of the nucleus is converted into kinetic energy of the escaping
alpha particle or beta particle. And, of course, we can model the fission and fusion of nuclei in nuclear reactions.
One of the more interesting examples of mass-to-energy conversion is pair
annihilation, where a particle and its antimatter counterpart destroy one another
and produce pure energy in the form of gamma radiation. Recall that the
antimatter particle has the same mass as its matter counterpart but the opposite
charge. Such an encounter is depicted in Figure 11-1 where a proton and
antiproton meet in the “before” picture, interact, and two gamma rays are
produced in the “after” picture. (You will examine this problem in more detail
in one of the homework problems for this module.)
+
before
after
Fig. 11-1
The flip side of pair annihilation is pair production where a gamma ray, when interacting with a matter particle, is
converted into a matter-antimatter pair. This is predicted by theory and observed in nature. An extreme case of pair
production is the postulated “vacuum fluctuation” that is common in cosmological models. This concept combines
relativity and quantum mechanics and says that a particle and antiparticle can pop out of vacuum provided that they
recombine and pop out of existence in an amount of time that is shorter than that allowed by the Heisenberg Uncertainty
Principle. (In other words, they can exist for a maximum time of t as long as the energy E that they momentarily bring
into our universe obeys E ·t = ħ.) These vacuum fluctuations could be going on all of the time. Some suggest that it
was a vacuum fluctuation in another universe that begat our universe. The energy that was brought out of that universe
when the particle pair recombined began our universe. By this line of reasoning, you could argue that there are universes
being produced all of the time!
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Acceleration Analysis
Analyzing the momentum and energy of high speed particles is sufficient for modeling their behavior. It isn’t necessary to
examine their position, velocity, and acceleration as functions of time. But we would miss some interesting results if we
didn’t take just a little peek.
Let’s examine the acceleration. Suppose a particle is subjected to a net force of size F. By Newton’s Second Law, this force
is the time rate of change in the particle’s momentum:
 dp d (mu )
F

(11.8)
dt
dt
Since both relativistic mass and velocity change with time,


du  dm
F m
u
dt
dt
(11.9)
The kinetic energy is the total energy minus the rest energy or
K  mc 2  mo c 2
(11.10)
The rest energy does not change with time so
dK dm 2

c
dt
dt
(11.11)
Substituting the mass derivative from Eq. (11.11) into Eq. (11.9) gives



du u dK
F m

dt c 2 dt
Module 11
Relativistic Particles
(11.12)
4
Once again we invoke the work-energy theorem, this time to replace the small change in kinetic energy, dK, with the
amount of work done by the force over the small displacement of size ds,
 
dK  F  ds
(11.13)
Differentiating both sides of this equation with respect to time gives
 

dK d ( F  ds )  ds  

F
 F u
dt
dt
dt
(11.14)
Notice that we have assumed that the force is constant and that we identified the displacement derivative as the velocity of
the particle. Substituting this expression for dK/dt into Eq. (11.12) gives



du u  
F m

(F  u )
dt c 2
(11.15)
We now identify that the time derivative of the velocity is the acceleration and rearrange this equation to find

 
u
 F

a 
(F  u )
m mc 2
(11.16)
 
This relativistic expression for acceleration should be compared to the low-speed, Newtonian result of a  F / mo .
In Newtonian mechanics, the acceleration is always parallel to the direction of the net force. Not
necessarily so in relativistic mechanics! Notice that, in general, the acceleration is not in the same
F
a
direction as the force as depicted in Figure 11-2.
u
Fig. 11-2
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There are two exceptions to the acceleration direction being different that the force direction. The first occurs if the force is
perpendicular to the velocity. In this case, the dot product of the force and the velocity is zero so that
a
F
F

m mo
(11.17)
A charged particle moving in a magnetic field would experience this acceleration. (Recall that
the magnetic force is always perpendicular to the velocity.) If the field is perpendicular to the
velocity, then the particle will traverse a circular arc as shown in Figure 11-3. The particle will
respond to the magnetic force as if it has a mass of mo. This mass is sometimes called the
transverse mass of a particle. Note that it is the same as the relativistic mass of the particle.
F
u
Fig. 11-3
The other exception is where the force is parallel to the velocity. Then the acceleration will be parallel to this same
direction. Consider the case where the force is in the same direction as the velocity as depicted in Figure 11-4. Here,
 
F  u  Fu
u
F
 
 F uFu
a 
m mc 2
so that
Fig. 11-4
Since all of these vectors are parallel, we can write that the equation in component form or
a
F Fu 2

m mc 2
(11.18)
Replacing m with mo and remembering that  = (1 – u2/c2)-1/2 gives
a
Module 11
Relativistic Particles
F
F
(1  u 2 / c 2 ) 
m o
 3m o
(11.19)
6
In this case, the particle responds to the force as if it has a mass of 3mo. This mass is referred to as the longitudinal mass of
the particle. Notice that it is greater that the relativistic mass and, therefore the transverse mass, by a factor of 2. This is
very strange indeed! A particle traveling along at some speed acts as if its mass is mo in a collision (where momentum is
transferred) or if you bend it in a circle with a magnetic field. But try to speed it up with a collinear force and suddenly it
acts as if its mass is 3mo. This increase can be quite dramatic at high speeds where the gamma factor is significantly larger
than one.
This case of speeding up a particle with a collinear force is so interesting that we’ll take a closer look at it. Let’s assume
that the particle starts from rest. Since a = du/dt, we can write Eq. (11.19) as
du
F
F


(1  u 2 / c 2 ) 3 / 2
3
dt  m
mo
o
(11.20)
We can solve this equation for the velocity as a function of time by cross multiplying and integrating both sides:
u
t F

dt


2 2 3/ 2
0 (1  u / c )
0 mo
du
(11.21)
Notice that we integrate from t=0 where u=0 (starts from rest) to time t where the velocity is u. The integral on the right
side is easy since the force and rest mass are constants. The integral on the left side isn’t too bad (look it up if you’re lazy)
and our equation becomes
u
(1  u 2 / c 2 )1 / 2

Ft
mo
(11.22)
We can now solve for u in terms of t. It’s not pretty so we’ll skip the algebra here. Do it at home to confirm that you get
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u (t ) 
Fct
mo2 c 2  F 2 t 2
(11.23)
Now that we have the velocity function, we can differentiate to get the acceleration as a function of time and integrate it to
get the position as a function of time. Again, we’ll skip the math details (as usual, try it at home!) and write the results.
a(t ) 
x(t ) 
Fmo2 c 3
(mo2 c 2  F 2 t 2 ) 3 / 2
c 2 2
mo c  F 2t 2  mo c


F
(11.24)
(11.25)
Okay, we just have to plot these three quantities versus time. In fact, to make things really interesting, let’s plot them along
with what classical, Newtonian theory would give where the acceleration would be constant. Newton’s theory would give
a (t )  F / mo
u (t ) 
F
t
mo
x(t ) 
F 2
t
2 mo
These are just the equations of motions for constant acceleration where the object starts at rest at x=0.
In order to crunch some numbers, let’s take the specific case of an electron that is accelerated through a 750 kiloVolt difference
over a distance of 10 meters. (Maybe we have built a van de Graff accelerator with these specifications in our basement.)
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1.5
2
m/s )
Acceleration
As expected, we see the acceleration decreasing in size as time
goes on since the longitudinal mass of the particle is increasing.
In fact, the acceleration asymptotically approaches zero.
1
16
a (x10
Let’s start with the acceleration plot shown in Figure 11-5. The
constant acceleration of Newton’s theory is shown in green; the
relativistic result is shown in red.
What speed is being asymptotically approached as the
acceleration tends towards zero? Yes, you guessed it, the speed
of light. This is confirmed if we look at the velocity motion
plot in Figure 11-6. The speed continues to increase but at a
slower and slower rate. Once again we see that a particle that
starts out below the speed of light cannot reach c. (Of course,
Newtonian theory doesn’t have a problem with the speed limit
of c. It predicts the speed to increase linearly with the electron
reaching c at around 23 nanoseconds after take-off.)
0.5
0
0 10 20 30 40 50 60
t (ns)
Fig. 11-5
u/c
Velocity
1.2
1
0.8
0.6
0.4
0.2
0
0 10 20 30 40 50 60
t (ns)
Fig. 11-6
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Position
Notice that very early on in the motion, both theories give
similar results. However, as the electron reaches relativistic
speeds, the divergence grows.
x (m)
Finally, we examine the position plot. We see the familiar
parabola from Newton and a curve that doesn’t rise as steeply
from relativity theory.
It’s fun to compare the final results of the two theories at the
finish line of the accelerator. According to Newton, it takes the
electron 39 nanoseconds to travel the 10 meters and its speed is
1.71c. According to Einstein, the electron takes about 51 ns to
travel the 10 m and its speed is about 0.92c. While Newton was
a great physicist, we have to go with Einstein’s results here.
12
10
8
6
4
2
0
0 10 20 30 40 50 60
t (ns)
Fig. 11-7
There is one last observation that we’ll make before we leave and it involves the speed of the electron at which the
change in its acceleration is greatest. Look again at the acceleration motion plot in Figure 11-5. Notice that there is a
critical point where the change in the acceleration is greatest. It’s marked with a gold dot in the figure. (For this
particular example, it occurs around 11 ns.)
To examine this point in more detail, let’s differentiate the acceleration function of Eq. (11-24) with respect to time to
find the jerk (the rate of change of the acceleration):
 3F 3 mo2 c 3t
da
j (t ) 

dt (mo2 c 2  F 2 t 2 ) 5 / 2
Module 11
Relativistic Particles
(11.26)
10
Jerk
0
3
j (x10 m/s )
0
10 20 30 40 50 60
-2
23
The plot of the jerk versus time is shown in Figure 11-8. (The
jerk is negative in this example since the acceleration is
decreasing.) The peak in the size of the jerk corresponds to the
critical point of the acceleration plot. You will show in a
homework problem that this peak occurs at a time given by
t max  mo c / 2F
and that the acceleration, velocity, and position of the electron
at this time are given by
8 F
a (t max ) 
5 5 mo
u (t max ) 
-4
-6
t (ns)
Fig. 11-8
c
5
 5  mo c 2
x(t max )  
 1
2

 F
Note that the time of maximum jerk, the acceleration at this time, and the position at this time all depend on the mass of the
electron and the force it experiences. The force depends on the electron’s charge and the size of electric field that does the
accelerating. What is interesting is that the velocity at this time of maximum jerk is independent of all of these parameters!
If we change the electric field or if we even change the type of particle that we accelerate, the speed at which the
acceleration changes the most is always the same c/5. No matter the particle type, it is at this speed where the particle
“bulks up” the quickest and its acceleration decreases the quickest.
Module 11
Relativistic Particles
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