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\centerline{\bf Chapter 2: Relativistic Mechanics} 

\centerline{\bf Introduction}

We know that Newton's laws of motion hold and are invariant in inertial
frames of reference that are related by the Galilean transformation.
But it is the Lorentz transformation rather than the Galilean transformation
that is correct. That means that the laws of classical mechanics
cannot be correct, and we must find new laws of relativistic mechanics.

In doing so, we will be guided by 3 principles:
\begin{enumerate}
\item A correct relativistic law must hold in all inertial frames, i.e., it
must be invariant under the Lorentz transformation.
\item Relativistic definitions and laws must reduce to their 
nonrelativistic counterparts when applied to systems moving much 
slower than the speed of light.
\item Our relativistic laws must agree with experiment.
\end{enumerate}

\centerline{\bf Mass}
How do we define mass $m$? We know that at slow speeds, mass can be defined
classically by $m=F/a$. In relativity we will measure the mass $m$ of an 
object in its rest frame. This is called the {\bf rest mass} or the 
{\bf proper mass}. Observers in different rest frames could measure the mass
of an object by bringing the object to rest in their frame of reference,
and then measuring its mass. The mass they would measure would be 
the same in all inertial reference frames.

\centerline{\bf Relativisitic Momentum} 

Classically, momentum is defined by
\be
{\vec p}=m{\vec u}
\label{eq:pmv}
\ee
where $m$ and ${\vec u}$ the mass and velocity of an object. What is
the correct relativistic definition of momentum? We have the freedom to 
define momentum any way we like, but to be useful, we would like to
define it such that total momentum is conserved in all inertial frames.

Recall the law of momentum conservation. If there are $n$ bodies with
momenta ${\vec p}_1$, ..., ${\vec p}_n$, then, in the absence of
external forces, the total momentum is given by
\be
\sum{\vec p}={\vec p}_1+...+{\vec p}_n
\ee
cannot change. We would like to define momentum relativistically to
preserve conservation of momentum. 

As your book describes in section 2.3, we can't use the classical definition 
Eq. (\ref{eq:pmv}) because of the way velocity transforms between 
reference frames. For example, the way $u_y$ transforms in going from
$S$ to $S^{\prime}$ depends on $u_x$. The classical definition of momentum
is
\be
{\vec p}=m{\vec u}=m\frac{d{\vec r}}{dt}
\ee
The problem is that both ${\vec r}$ and $t$ are subject to the Lorentz
transformation and that makes things messy. 

The relativistic way to define momentum is
\be
{\vec p}=m\frac{d{\vec r}}{dt_0}
\ee
where $t_0$ is the proper time, i.e., the time measured in the rest
frame of the particle or object. Notice that with this definition,
\be
p_y=m\frac{dy}{dt_0}
\ee
which is invariant when one goes between reference frames that are
moving relative to one another along the $x-$axis. That is because
$t_0$ is invariant and $y=y^{\prime}$.
Similarly for $p_z$.

Since $dt=\gamma dt_0$ where $\gamma=1/\sqrt{1-u^2/c^2}$, we can write
\be
{\vec p}=\frac{m{\vec u}}{\sqrt{1-u^2/c^2}}=\gamma m{\vec u}
\ee
 
A constant force changes the momentum and hence the velocity.
Notice that as $u\rightarrow c$, $\gamma$ and hence $p$
increase without limit, but the speed $u$ never reaches the speed
of light (see plot of $u$ versus $p$ in Fig 2.2 of your book). 
Thus no object can go faster than the speed of light. 
\begin{figure}
\begin{center}
\leavevmode
\epsfxsize=3.0 true in
\epsfbox{u_vs_p.eps}
\end{center}
\caption{Velocity versus momentum.}
\end{figure}

We can define a relativistic mass $M(u)$ by
\be
M(u)=\frac{M}{\sqrt{1-u^2/c^2}}=M\gamma
\label{eq:relMass}
\ee
Your book calls this a variable mass $m_{var}$, but it doesn't want to use
this. However, it does make it easy to write the momentum in the
traditional form
\be
{\vec p}=M(u){\vec u}
\ee
Notice from Eq. (\ref{eq:relMass}) that if the speed $u$ of an object with
nonzero mass is equal to the speed of light, the relativistic mass is 
infinity. If the object's speed exceeds the speed of light, then the 
relativistic mass $M(u)$ is imaginary. Clearly, this is absurd and so, 
once again, the velocity of an object cannot exceed the speed of light. 
If the object has mass, it cannot travel at the speed of light; its speed
must be less than the speed of light.

\centerline{\bf Relativistic Energy}
 
Now we need to define a relativistic energy. We will use 2 criteria:
\begin{enumerate}
\item When applied to slowly moving bodies, the new definition reduces to
the classical definition.
\item The total energy $\sum E$ of an isolated system should be conserved
in all reference frames.
\end{enumerate}

The relativistic energy that satisfies these requirements turns out to be
\be 
E=\frac{mc^2}{\sqrt{1-u^2/c^2}}=\gamma mc^2
\label{eq:Rel_E}
\ee
This applies to any single body, no matter how big or how small.
Notice that the units are right. (It's always important to check units.)
$\gamma$ is dimensionless, and $mc^2$ has units of energy. (Recall that
kinetic energy is $(1/2)mv^2$.)

Now let's take the nonrelativistic (slow) limit $u\ll c$. Then 
\be
\gamma=\left(1-\frac{u^2}{c^2}\right)^{-1/2} \approx
1 + \frac{1}{2}\frac{u^2}{c^2}
\ee
Therefore, when $u\ll c$,
\be
E=\gamma mc^2\approx\left(1 + \frac{1}{2}\frac{u^2}{c^2}\right)mc^2
=mc^2+\frac{1}{2}mu^2
\ee
The first term is a constant independent of the speed $u$, 
and we can always add a constant to the energy since we can set the zero
of the energy ($E=0$) anywhere. The second term is just the usual
classical expression for the kinetic energy. So our definition of the
relativistic energy satisfies the criterion of reducing to the classical
kinetic energy in the nonrelativistic (slow) limit.

Now let's look at that constant $mc^2$.
If the object is at rest, then $u=0$, $\gamma=1$, and Eq. 
(\ref{eq:Rel_E}) reduces to the famous equation
\be
E=mc^2
\label{eq:E=mc2}
\ee
This is called the rest energy of the mass $m$. If we could convert a mass
$m$ completely into energy, then this is the amount of energy we would get.
It's a lot of energy because the speed of light $c$ is so large, and
$c^2$ is even larger. 
For example, the rest energy of a 1 kg lump of metal is
\be
E=mc^2=(1\;{\rm kg})\times(3\times 10^8\;{\rm m/s})^2=9\times 10^{16}
\;{\rm joules}
\ee
which is about the energy generated by a large power plant in one year.
$E=mc^2$ is the basic equation behind nuclear power, and atomic bombs.
These are powered by nuclear fission in which, typically, uranium 235 
nuclei are split apart, converting about 1/1000 of the rest energy 
into heat. Thus 1 kg of $^{235}$U can yield a fantastic amount of heat
($9\times 10^{13}$ J).

When an object is not at rest, its energy $E=\gamma mc^2$ is 
the sum of its rest energy $mc^2$ plus its kinetic energy
$K=(E-mc^2)$:
\be
E=mc^2+K
\ee
where
\be
K=E-mc^2=(\gamma-1)mc^2
\label{eq:KE}
\ee
In the nonrelativistic limit $K\approx\frac{1}{2}mu^2$, the classical kinetic
energy. However, in the relativistic limit of $u\rightarrow c$, we get 
something quite different. Because $\gamma$ can approach infinity, $K$
can approach infinity even though the speed $u$ can never reach the speed
of light $c$. Notice that $K$, like its classical counterpart, is always
positive ($K\geq 0$).

\centerline{Two Useful Relations}
Let's derive 2 useful relations starting from
\be
{\vec p}=\gamma m{\vec u}
\label{eq:p}
\ee
and
\be
E=\gamma mc^2
\label{eq:E}
\ee
If we divide Eq. (\ref{eq:p}) by Eq. (\ref{eq:E}), we obtain
\be
\frac{{\vec p}}{E}=\frac{{\vec u}}{c^2}
\label{eq:pE}
\ee
or
\be
{\vec\beta}=\frac{{\vec u}}{c}=\frac{{\vec p}c}{E}
\label{eq:betaPE}
\ee
which give the dimensionless velocity ${\vec\beta}={\vec u}/c$ in terms of
${\vec p}$ and $E$. Notice that ${\vec p}c$ has dimensions of energy, so the
ratio ${\vec p}c/E$ is dimensionless, just like ${\vec\beta}$. 
Eq. (\ref{eq:betaPE}) is one useful relation. 

The other is 
\be 
E^2=\left(pc\right)^2+\left(mc^2\right)^2
\ee
This has the same form as the Pythagorean relation for a right triangle
with sides $pc$ and $mc^2$ and hypotenuse $E$, though there isn't any deep 
meaning to this. To show that this is correct, we can plug in 
$E=\gamma mc^2$ and $p=\gamma mv$, solve for $\gamma^2$ and show that 
this reduces to the definition $\gamma^2=(1-v^2/c^2)^{-1}$.

\centerline{\it Units}
Since we are dealing with small things, we should use appropriate units.
For example, if I ask how tall you are, you would tell me so many feet and
so many inches. You could give me your height in miles or light years, but
those are not appropriate units. Similarly, joules is not an appropriate 
unit for atomic and subatomic particles like electrons. The commonly
used unit is the {\bf electron volt} or eV. This is defined as the amount 
of work needed to move an electron ($q=-e=-1.60\times 10^{-19}$ Coulomb)
through a voltage drop of 1 volt ($\Delta V=-1$ volt); thus
\be
1\; {\rm eV}=q\Delta V=\left(-1.60\times 10^{-19}\;{\rm C}\right)\times
(-1\;{\rm Volt})=1.60\times 10^{-19}\;{\rm J}
\ee
Typical atomic energies are on the order of 1 eV or so, while those in
nuclear physics are on the order of 10$^{6}$ eV = 1 MeV.

Masses of particles are often given in eV with the understanding that
they are referring to the rest energy $mc^2$. Techically, the mass is
properly in units of eV/$c^2$. So, for example, the ``mass'' of the electron
is given as 0.511 MeV, meaning $mc^2$ = 0.511 MeV. Similarly, when
momentum $p$ is given in units of MeV, they really mean the quantity
$pc$, or the units of $[p]$ is eV/$c$.

If you look up a table for the mass of atoms, the mass is often given
in atomic mass units (denoted u). The conversion is
\be
1\;{\rm u}=1.66\times 10^{-27}\;{\rm kg}
\ee

\centerline{\bf Conversion of Mass to Energy}

The famous equation $E=mc^2$ implies that if you can convert mass into 
energy, then you would get a lot of energy because the speed of light
squared is so big. So does matter get converted to energy? The answer
is yes. An example where this happens for atomic nuclei is when 
atomic nuclei break apart, either because they are unstable 
(radioactive) or because they are forced apart by an incoming
particle (e.g., a neutron) in a process called fission. This is the
principle behind atomic power plants like San Onofre. 

Let's suppose that a nucleus at rest has a mass $M$, so that the initial
energy is $Mc^2$. Then suppose the nucleus breaks apart due to radioactive
decay in which pieces of the nucleus go flying apart. The energy then
is a combination of the kinetic energy $K$ of the pieces and their rest energy. 
Suppose that the nucleus breaks into 2 pieces.
Since total energy (including the rest mass energy) is conserved, we have
\begin{eqnarray}
Mc^2 &=& E_1+E_2\nonumber\\
&=& (K_1+m_1c^2)+(K_2+m_2c^2)\nonumber\\
&=& (K_1+K_2)+(m_1c^2 + m_2c^2) 
\end{eqnarray}
The sum of the masses of the pieces ($m_1+m_2$) is less than the initial
mass $M$, so there is a difference between the initial mass and the total
final mass:
\be
\Delta Mc^2=Mc^2-(m_1+m_2)c^2
\ee
The missing mass is converted into the kinetic energy of the fragments flying
apart:
\be
\Delta Mc^2=K_1+K_2
\ee

For example, your book considers the radioactive
decay of $^{232}Th$ via the reaction
\be
^{232}Th\rightarrow ^{228}Ra+^{4}He
\ee
The sum of the masses of $^{228}Ra$ and $^{4}He$ nuclei is 0.004 u
smaller than the original $^{232}Th$ nucleus. Since
1 u = $1.66\times 10^{-27}$ kg, this corresponds to 
$7\times 10^{-30}$ kg. This missing mass is converted into 
kinetic energy because the products ($^{228}Ra$ and $^{4}He$) 
fly apart. The amount of kinetic energy is
\begin{eqnarray}
\Delta Mc^2 &=&\left(7\times 10^{-30}\;{\rm kg}\right)
\left(3\times 10^8\;{\rm m/s}\right)^{2}\nonumber\\
&=& 4\;{\rm MeV}
\end{eqnarray}
If we reversed the process and forced the 2 pieces together to form
the original nucleus, it would take work. The amount of energy
we would have to expend would be $\Delta Mc^2$. In other words,
energy we put in would be converted into the additional mass so that
the rest mass $M$ of the final nucleus would be greater than the sum
$m_1+m_2$.

So sometimes we need to add energy to merge or fuse 2 pieces together. However,
sometimes, the opposite can happen. In particular,
energy can be released when 2 things are brought together
to form a more stable (lower energy) thing. For example, if we bring
together an electron and a proton to form a hydrogen atom, then 13.6 eV
is released. This is called the {\bf binding energy} since is the
amount of energy that would be required to pull apart the proton and 
electron in a hydrogen atom. If we add the mass $m_e$ of a free electron
(free means it is floating around, and is not bound inside an atom), and
the mass $m_p$ a free proton, the sum is 13.6 eV/$c^2$ greater than the 
mass $M_H$ of a 
hydrogen atom. This is another example of mass being converted into energy.
In symbols:
\be
m_ec^2+m_pc^2=M_Hc^2+B
\ee
where $B$ is the binding energy. So the binding energy corresponds to the
rest energy of the missing mass:
\be
B=\left(m_ec^2+m_pc^2\right)-Mc^2
\ee
Similarly, when atoms are combined to
form stable molecules, binding energy is released. 
Energy is also released when 
nuclei are forced together to form bigger stable nuclei in a process called
fusion. For example, 4 hydrogen atoms can be fused to make 1 helium
atom. This releases binding energy. This is what powers the sun and 
hydrogen bombs. 

\centerline{\bf Example: $\Lambda$ Decay}

The $\Lambda$ particle is a subatomic particle that can spontaneously
decay into a proton and a negatively charged pion.
\be
\Lambda\rightarrow p+\pi^{-}
\ee
The sum of the mass of the proton and the pion will be less than that of the
$\Lambda$. In a certain experiment, the outgoing particles were traveling in
the $+x$ direction with momenta
$p_p=$ 581 MeV/$c$ and $p_{\pi}$=256 MeV/$c$. The rest masses are
$m_p$ = 938 Mev/$c^2$ and $m_{\pi}=140$ MeV/$c^2$. Find the rest mass
$m_{\Lambda}$ of the $\Lambda$.

To solve this problem, let us start with 
\be
E^2=(pc)^2+(mc^2)^2
\label{eq:E2}
\ee
Solving for the mass of the $\Lambda$ gives
\be
m_{\Lambda}=\frac{\sqrt{E_{\Lambda}^2-\left(p_{\Lambda}c\right)^2}}{c^2}
\label{eq:massLambda}
\ee
So we need the energy and the momentum of the $\Lambda$ to find the mass.
First we find the energy using Eq. (\ref{eq:E2})
\be
E^2=(pc)^2+(mc^2)^2
\ee
This gives $E_p$ = 1103 MeV and $E_{\pi}$ = 292 MeV. So
\be
E_{\Lambda}=E_p+E_{\pi}=1395\;{\rm MeV}
\ee
Conservation of momentum gives
\be
{\bf p}_{\Lambda}={\bf p}_{p}+{\bf p}_{\pi}=837\;{\rm MeV}/c
\ee
where the momenta are all pointing along the positive $x$ axis. Now that
we know $E_{\Lambda}$ and $p_{\Lambda}$, we can Eq.~(\ref{eq:massLambda})
to find the mass:
\be
m_{\Lambda}=\frac{\sqrt{E_{\Lambda}^2-\left(p_{\Lambda}c\right)^2}}{c^2}
=1116\;{\rm MeV}/c^2
\ee
This is how many masses of unstable subatomic particles are measured.

\centerline{\bf Force in Relativity}
There are 2 definitions of force in classical mechanics:
\be
{\bf F}=m{\bf a}
\ee
and
\be
{\bf F}=\frac{d{\bf p}}{dt}
\ee
These are equivalent because ${\bf p}=m{\bf u}$, so $d{\bf p}/dt=m{\bf a}$.
But in relativity, ${\bf p}=\gamma m{\bf u}$, so $d{\bf p}/dt\neq m{\bf a}$.
In relativity, the most convenient definition of force turns out to be
\be
{\bf F}=\frac{d{\bf p}}{dt}
\ee
Example 2.9 shows that this definition of force allows the work-energy
theorem to remain valid. Namely,
\be
{\bf F}\cdot d{\bf r}=dE
\ee
This says that if a mass $m$, acted on by a force ${\bf F}$, moves a small
distance $d{\bf r}$, the change in its energy, $dE$, equals the work
done by ${\bf F}$.

This definition of force allows us to keep the Lorentz force law intact:
\be
{\bf F}=q({\bf E}+{\bf u}\times {\bf B})
\label{eq:Lorentz}
\ee
One easy application of this equation is when a particle with charge
$q$ moves with velocity ${\bf u}$ perpendicular to a uniform field
{\bf B}. The particle's energy is constant because no work is done on it.
To see this, look at the work-energy theorem. It turns out that the
force is perpendicular to the direction of motion of the particle because
the particle moves in a circle and the force points radially in towards
the center of the circle. Therefore, since the energy 
($E=\gamma mc^2$) is constant, the 
velocity is constant in magnitude and only changes direction.
To see that it moves in a circle, start with
\be
\frac{d{\bf p}}{dt}=q{\bf u}\times {\bf B}
\ee
which we can write as
\be
\gamma m\frac{d{\bf u}}{dt}=q{\bf u}\times{\bf B}. 
\ee
Use acceleration ${\bf a}=d{\bf u}/dt$
and the fact that ${\bf u}$ is perpendicular to ${\bf B}$ to write
\be 
\gamma ma=quB
\ee
For motion in a circle of radius $R$, the centripetal acceleration $a=u^2/R$.
The reason is the same in relativity as in classical mechanics.
Using $\gamma mu=p$, we can solve for $R$ and write
\be
R=\frac{p}{qB}
\ee
or
\be
p=qBR
\label{eq:circle}
\ee
This provides a convenient way to measure the momentum of a particle of
known charge $q$. We know the applied field $B$ and we can measure
the radius $R$ of the circle.

For example, suppose we want to find the momentum of a proton moving 
perpendicular to a uniform magnetic field $B=1.0$ tesla (T) in a circle 
of radius 1.4 m. We plug this into Eq. (\ref{eq:circle}) using
$q=e=1.6\times 10^{-19}$ C. This gives
\begin{eqnarray}
p&=&qBR=(1.6\times 10^{-19}\;{\rm C})\times (1.0\;{\rm T})\times(1.4\;{\rm m})
\nonumber\\
&=& 2.24\times 10^{-19}\;{\rm kg-m/s}
\end{eqnarray}
Using the conversion 1 MeV/$c$=$5.34\times10^{-22}$ kg-m/s, we can change
units:
\be
p=(2.24\times 10^{-19}\;{\rm kg-m/s})\times\frac{1\;{\rm MeV}/c}
{5.34\times 10^{-22}\;{\rm kg-m/s}}=420\;{MeV}/c.
\ee

Since the proton's rest mass is known to be $m=938$ MeV/$c^2$, we can 
find its energy from the ``Pythagorean relation'' to be
\be
E=\sqrt{(pc)^2+(mc^2)^2}=1030\;{\rm MeV}
\ee

\centerline{\bf Massless Particles}
Is it possible to have massless particles? Classically, the answer is ``no.''
Momentum (${\bf p}=m{\bf u}$) and kinetic energy ($KE=mu^2/2$) are 
proportional to the mass $m$. If $m=0$, then $p=E=0$. One might think
the same would be true relativistically since
${\bf p}=\gamma m{\bf u}$ and $E=\gamma mc^2$. Again, $p$ and $E$
are proportional to $m$ and $m=0$. But what if the massless particle
travels at the speed of light? Then 
$\gamma=\infty$, and $\gamma m=0\cdot\infty$ which is ill-defined.

It's better to go back to 
\be
E^2=(pc)^2+(mc^2)^2
\ee
and 
\be 
\beta=\frac{u}{c}=\frac{pc}{E}
\ee
If $m=0$, then these equations reduce to
\be
E=pc
\ee
and
\be
\beta=\frac{pc}{E}=1=\frac{u}{c}\quad\quad {\rm or}\quad\quad u=c
\ee
So the massless particle travels at the speed of light. Conversely, if
we discovered a particle that traveled at speed $c$, then
$\beta=u/c$ would equal 1=$pc/E$. That would mean $E=pc$ and hence,
that $m=0$, i.e., it would be a massless particle. So massless particles
travel at the speed of light, and particles that travel at $c$ must be
massless. Particles with nonzero mass must travel at less than the speed
of light.

Do massless particles exist? Yes. The best known massless particle is the
{\bf photon} which is a particle of light. As we will discuss more later,
light can be described in 2 ways: (1) as an electromagnetic wave, and (2)
as particles or packets of energy called photons. 

To find the energy and momentum of a massless particle, we look at a reaction
that it is involved in, e.g., photoionization of hydrogen in which a photon
hits the electron in a hydrogen atom, causing the electron to be ejected:
\be
\gamma + H\rightarrow e+p
\ee
where $\gamma$ is the symbol for a photon.
We measure the momentum and energy of the particles with mass, then use
conservation of energy to deduce the energy and momentum of the massless
particle.

Another example of a massless particle is the graviton but there is no
experimental evidence for the graviton. It used be accepted that the
neutrino had no mass, but now there is experimental evidence that it
does have a little bit of mass. 

\centerline{Example: Electron-Positron Annihilation}

You've probably heard that when matter and antimatter collide, they annihilate
and release a lot of energy. The {\bf positron} is an anti-electron, i.e., it's
the antimatter version of an electron. It is exactly like an electron 
except that it has a positive charge ($+e$). The positron has the same
mass as an electron. Suppose a positron and an electron are both at rest 
with no momentum. So they only have their rest mass energies:
$m_ec^2$ and $m_pc^2=m_ec^2$. Then when they annihilate, 2 photons are produced.
Conservation of momentum requires that their total momentum is zero, and
the sum of their total energy is $2m_ec^2$:
\be
{\bf p}_1+{\bf p}_2=0
\ee
and
\be
E_1+E_2=2m_ec^2
\ee
So the photons that are created must have equal and opposite momenta:
\be
{\bf p}_1=-{\bf p}_2
\ee
Since $E=pc$ for photons, the photons must have equal energies: $E_1=E_2$,
and
\be
E_1=E_2=mc^2=0.511\;{\rm MeV}
\ee
So all the rest mass energy is converted into electromagnetic energy.

Positrons are used in PET scans. PET stands for Positron Emission Tomography.
This is a medical diagnostic probe in which a patient is injected with a
radioactive positron emitter (like carbon 11) in a suitably chosen
solution like glucose (sugar) that will tend to collect in the brain, for
example. Once in the brain, the emitted positrons annihilate with 
nearby electrons, and photons are emitted that are detected by a ring of
detectors. In this way, it is possible to create an accurate map 
of the areas of interest.

\centerline{\bf When is nonrelativistic mechanics good enough?}

The relativistic equations all reduce to their classical counterparts
in the limit of low speeds ($v\ll c$). When should one use the 
relativistic formulae and when is it good enough to use the classical
equations? There is no clear-cut answer. It depends on what you need it
for. For example, GPS (Global Positioning System) needs an accuracy of 1 
ns, so relativistic effects must be taken into account. Table 2.4
of your book compares the values of the kinetic energy
calculated with relativistic and classical formulas for various values
of the velocity. As a general rule of thumb, you can ignore
relativisitic effect if the speeds are 0.1$c$ or less, or if a
particle's kinetic energy $K$ is much less than 1\% of its rest energy. 
(Recall that $K$ goes as $u^2$, so this is why a speed of
0.1$c$ corresponds to 0.01\% of the energy.) For speeds greater than
0.1$c$ or kinetic energies greater than 0.01$mc^2$, relativistic
effects probably need to be taken into account, and relativistic
equations should probably be used. 

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