Subsections


1.3 Relativistic Mechanics


1.3.1 Intro to relativistic mechanics

Nonrelativistic mechanics is often based on the use of a potential energy to describe the forces. For example, in a typical molecular dynamics computation, the forces between the molecules are derived from a potential that depends on the differences in position between the atoms. Unfortunately, this sort of description fails badly in the truly relativistic case.

The basic problem is not difficult to understand. If a potential depends only on the spatial configuration of the atoms involved, then the motion of an atom instantaneously affects all the other ones. Relativity simply cannot handle instantaneous effects; they must be limited by the speed of light or major problems appear. It makes relativistic mechanics more difficult.

The simplest way to deal with the problem is to look at collisions between particles. Direct collisions inherently avoid erroneous action at a distance. They allow simple dynamics to be done without the use of a potential between particles that is relativistically suspect.

Figure 1.3: Example elastic collision seen by different observers.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...0,-20){\makebox(0,0)[b]{.}}
\end{picture}
}
\end{picture}
\end{figure}

As a first example, consider two particles of equal mass and opposite speeds that collide as shown in the center of figure 1.3. You might think of the particles as two helium atoms. It will be assumed that while the speed of the atoms may be quite high, the collision is at a shallow enough angle that it does not excite the atoms. In other words, it is assumed that the collision is elastic.

As seen by observer C, the collision is perfectly symmetric. Regardless of the mechanics of the actual collision, observer C sees nothing wrong with it. The energy of the helium atoms is the same after the collision as before. Also, the net linear momentum was zero before the collision and still zero afterwards. And whatever little angular momentum there is, it too is still the same after the collision.

But now consider an observer A that moves horizontally along with the top helium atom. For this observer, the top helium atom comes down vertically and bounces back vertically. Observer B moves along with the bottom helium atom in the horizontal direction and sees that atom moving vertically. Now consider the Lorentz transformation (1.7) of the vertical velocity $v_{y,2}$ of the top atom as seen by observer A into the vertical velocity $v_y$ of that atom as seen by observer B:

\begin{displaymath}
v_y = \sqrt{1 - (v_x/c)^2} v_{y,2}
\end{displaymath}

They are different! In particular, $v_y$ is smaller than $v_{y,2}$. Therefore, if the masses of the helium atoms that the observers perceive would be their rest mass, linear momentum would not be conserved. For example, observer A would perceive a net downwards linear momentum before the collision and a net upwards linear momentum after it.

Clearly, linear momentum conservation is too fundamental a concept to be summarily thrown out. Instead, observer A perceives the mass of the rapidly moving lower atom to be the moving mass $m_v$, which is larger than the rest mass $m$ by the Lorentz factor:

\begin{displaymath}
m_v = \frac{m}{\sqrt{1-(v/c)^2}}
\end{displaymath}

and that exactly compensates for the lower vertical velocity in the expression for the momentum. (Remember that it was assumed that the collision is under a shallow angle, so the vertical velocity components are too small to have an effect on the masses.)

It is not difficult to understand why things are like this. The nonrelativistic definition of momentum allows two plausible generalizations to the relativistic case:

\begin{displaymath}
{\skew0\vec p}= m \frac{{\rm d}{\skew0\vec r}}{{\rm d}t}
...
...rn-1.3pt\strut}{{\rm d}t_0} \mbox{ ?}
\end{array}
\right.
\end{displaymath}

Indeed, nonrelativistically, all observers agree about time intervals. However, relativistically the question arises whether the right time differential in momentum is ${\rm d}{t}$ as perceived by the observer, or the proper time difference ${\rm d}{t}_0$ as perceived by a hypothetical second observer moving along with the particle.

A little thought shows that the right time differential has to be ${\rm d}{t}_0$. For, after collisions the sum of the momenta should be the same as before them. However, the Lorentz velocity transformation (1.7) shows that perceived velocities transform nonlinearly from one observer to the next. For a nonlinear transformation, there is no reason to assume that if the momenta after a collision are the same as before for one observer, they are also so for another observer. On the other hand, since all observers agree about the proper time intervals, momentum based on the proper time interval ${\rm d}{t}_0$ transforms like ${\rm d}\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over r}
\kern-1.3pt$, like position, and that is linear. A linear transformation does assure that if an observer A perceives that the sum of the momenta of a collection of particles $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1,2,...is the same before and after,

\begin{displaymath}
\sum_j \kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hb...
...ghtarrow$\hspace{0pt}}}\over p}
\kern-1.3pt_{jA,{\rm before}}
\end{displaymath}

then so does any other observer B:

\begin{displaymath}
\sum_j \Lambda_{B\leftarrow A}\kern-1pt{\buildrel\raisebox...
...ghtarrow$\hspace{0pt}}}\over p}
\kern-1.3pt_{jB,{\rm before}}
\end{displaymath}

Using the chain rule of differentiation, the components of the momentum four-vector $\kern-1pt{\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{1pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over p}
\kern-1.3pt$ can be written out as

\begin{displaymath}
p_0 = m c \frac{{\rm d}t}{{\rm d}t_0} \quad
p_1 = m \fra...
... m \frac{{\rm d}t}{{\rm d}t_0} \frac{{\rm d}z}{{\rm d}t}\quad
\end{displaymath} (1.14)

The components $p_1,p_2,p_3$ can be written in the same form as in the nonrelativistic case by defining a moving mass
\begin{displaymath}
m_v = m \frac{{\rm d}t}{{\rm d}t_0} = \frac{m}{\sqrt{1-(v/c)^2}}
\end{displaymath} (1.15)

How about the zeroth component? Since it too is part of the conservation law, reasonably speaking it can only be the relativistic equivalent of the nonrelativistic kinetic energy. Indeed, it equals $m_vc^2$ except for a trivial scaling factor 1$\raisebox{.5pt}{$/$}$$c$ to give it units of momentum.

Figure 1.4: A completely inelastic collision.
\begin{figure}
\centering
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\begin{picture}(...
...0)[r]{$M$}}
\put(330,10){\vector(0,-1){12}}
\end{picture}\par
\end{figure}

Note that so far, this only indicates that the difference between $m_vc^2$ and $mc^2$ gives the kinetic energy. It does not imply that $mc^2$ by itself also corresponds to a meaningful energy. However, there is a beautifully simple argument to show that indeed kinetic energy can be converted into rest mass, [21]. Consider two identical rest masses $m$ that are accelerated to high speed and then made to crash into each other head-on, as in the left part of figure 1.4. In this case, think of the masses as macroscopic objects, so that thermal energy is a meaningful concept for them. Assume that the collision has so much energy that the masses melt and merge without any rebound. By symmetry, the combined mass $M$ has zero velocity. Momentum is conserved: the net momentum was zero before the collision because the masses had opposite velocity, and it is still zero after the collision. All very straightforward.

But now consider the same collision from the point of view of a second observer who is moving upwards slowly compared to the first observer with a small speed $v_{\rm {B}}$. No relativity involved here at all; going up so slowly, the second observer sees almost the same thing as the first one, with one difference. According to the second observer, the entire collision process seems to have a small downward velocity $v_{\rm {B}}$. The two masses have a slight downward velocity $v_{\rm {B}}$ before the collision and so has the mass $M$ after the collision. But then vertical momentum conservation inevitably implies

\begin{displaymath}
2 m_v v_{\rm {B}} = M v_{\rm {B}}
\end{displaymath}

So $M$ must be twice the moving mass $m_v$. The combined rest mass $M$ is not the sum of the rest masses $m$, but of the moving masses $m_v$. All the kinetic energy given to the two masses has ended up as additional rest mass in $M$.


1.3.2 Lagrangian mechanics

Lagrangian mechanics can simplify many complicated dynamics problems. As an example, in this section it is used to derive the relativistic motion of a particle in an electromagnetic field.

Consider first the nonrelativistic motion of a particle in an electrostatic field. That is an important case for this book, because it is a good approximation for the electron in the hydrogen atom. To describe such purely nonrelativistic motion, physicists like to define a Lagrangian as

\begin{displaymath}
{\cal L}= {\textstyle\frac{1}{2}} m \vert\vec v\vert^2 - q \varphi
\end{displaymath} (1.16)

where $m$ is the mass of the particle, $\vec{v}$ its velocity, and $q$ its charge, while $q\varphi$ is the potential energy due to the electrostatic field, which depends on the position of the particle. (It is important to remember that the Lagrangian should mathematically be treated as a function of velocity and position of the particle. While for a given motion, the position and velocity are in turn functions of time, time derivatives must be implemented through the chain rule, i.e. by means of total derivatives of the Lagrangian.)

Physicists next define canonical, or generalized, momentum as the partial derivative of the Lagrangian with respect to velocity. An arbitrary component $p^{\rm {c}}_i$ of the canonical momentum is found as

\begin{displaymath}
p^{\rm {c}}_i = \frac{\partial{\cal L}}{\partial v_i}
\end{displaymath} (1.17)

This works out to be simply component $p_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mv_i$ of the normal momentum. The equations of motion are taken to be
\begin{displaymath}
\frac{{\rm d}p^{\rm {c}}_i}{{\rm d}t} = \frac{\partial{\cal L}}{\partial r_i}
\end{displaymath} (1.18)

which is found to be

\begin{displaymath}
\frac{{\rm d}p_i}{{\rm d}t} = - q \frac{\partial\varphi}{\partial r_i}
\end{displaymath}

That is simply Newton’s second law; the left hand side is just mass times acceleration while in the right hand side minus the spatial derivative of the potential energy gives the force. It can also be seen that the sum of kinetic and potential energy of the particle remains constant, by multiplying Newton’s equation by $v_i$ and summing over $i$.

Since the Lagrangian is a just a scalar, it is relatively simple to guess its form in the relativistic case. To get the momentum right, simply replace the kinetic energy by an reciprocal Lorentz factor,

\begin{displaymath}
- m c^2 \sqrt{1-(\vert\vec v\vert/c)^2}
\end{displaymath}

For velocities small compared to the speed of light, a two term Taylor series shows this is equivalent to $mc^2$ plus the kinetic energy. The constant $mc^2$ is of no importance since only derivatives of the Lagrangian are used. For any velocity, big or small, the canonical momentum as defined above produces the relativistic momentum based on the moving mass as it should.

The potential energy part of the Lagrangian is a bit trickier. The previous section showed that momentum is a four-vector including energy. Therefore, going from one observer to another mixes up energy and momentum nontrivially, just like it mixes up space and time. That has consequences for energy conservation. In the classical solution, kinetic energy of the particle can temporarily be stored away as electrostatic potential energy and recovered later intact. But relativistically, the kinetic energy seen by one observer becomes momentum seen by another one. If that momentum is to be recovered intact later, there should be something like potential momentum. Since momentum is a vector, obviously so should potential momentum be: there must be something like a vector potential $\skew3\vec A$.

Based on those arguments, you might guess that the Lagrangian should be something like

\begin{displaymath}
\fbox{$\displaystyle
{\cal L}= - m c^2 \sqrt{1-(\vert\ve...
...}\over A}
= \Big(\frac{1}{c}\varphi,A_x,A_y,A_z\Big)
$} %
\end{displaymath} (1.19)

And that is in fact right. Component zero of the potential four-vector is the classical electrostatic potential. The spatial vector $\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(A_x,A_y,A_z)$ is called the “magnetic vector potential.”

The canonical momentum is now

\begin{displaymath}
p^{\rm {c}}_i = \frac{\partial{\cal L}}{\partial v_i} = m_v v_i + q A_i
\end{displaymath} (1.20)

and that is no longer just the normal momentum, $p_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_vv_i$, but includes the magnetic vector potential.

The Lagrangian equations of motion become, the same way as before, but after clean up and in vector notation, {D.6}:

\begin{displaymath}
\frac{{\rm d}{\skew0\vec p}}{{\rm d}t} = q \skew3\vec{\cal E}+ q \vec v \times \skew2\vec{\cal B}
\end{displaymath} (1.21)

The right-hand side in this equation of motion is called the Lorentz force. In it, $\skew3\vec{\cal E}$ is called the electric field and $\skew2\vec{\cal B}$ the magnetic field. These fields are related to the four-vector potential as

\begin{displaymath}
\skew3\vec{\cal E}= - \nabla \varphi - \frac{\partial \ske...
... t}
\qquad
\skew2\vec{\cal B}= \nabla \times \skew3\vec A
\end{displaymath}

where by definition

\begin{displaymath}
\nabla = {\hat\imath}\frac{\partial}{\partial x}
+ {\hat...
...c{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}
\end{displaymath}

is the vector operator called nabla or del.

Of course, if the Lagrangian above is right, it should apply to all observers, regardless of their relative motion. In particular, all observers should agree that the so-called action integral $\int{\cal L}{\rm d}{t}$ is stationary for the way that the particle moves, {A.1.3}, {D.3.1} That requires that ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}$ transforms according to the Lorentz transformation.

(To see why, recall that dot products are the same for all observers, and that the square root in the Lagrangian (1.19) equals ${\rm d}{t}_0$$\raisebox{.5pt}{$/$}$${\rm d}{t}$ where the proper time interval ${\rm d}{t}_0$ is the same for all observers. So the action is the same for all observers.)

From the Lorentz transformation of ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}$, that of the electric and magnetic fields may be found; that is not a Lorentz transformation. Note that this suggests that ${\buildrel\raisebox{-1.5pt}[0pt][0pt]
{\hbox{\hspace{2.5pt}$\scriptscriptstyle\hookrightarrow$\hspace{0pt}}}\over A}$ might be more fundamental physically than the more intuitive electric and magnetic fields. And that is in fact exactly what more advanced quantum mechanics shows, chapter 13.1.

It may be noted that the field strengths are unchanged in a “gauge transformation” that modifies $\varphi$ and $\skew3\vec A$ into

\begin{displaymath}
\varphi' = \varphi - \frac{\partial\chi}{\partial t}
\qquad
\skew3\vec A' = \skew3\vec A+ \nabla \chi
\end{displaymath} (1.22)

where $\chi$ is any arbitrary function of position and time. This might at first seem no more than a neat mathematical trick. But actually, in advanced quantum mechanics it is of decisive importance, chapter 7.3, {A.19.5}.

The energy can be found following addendum {A.1} as

\begin{displaymath}
E = \vec v \cdot {\skew0\vec p}^{\,\rm c} - {\cal L}= m_v c^2 + q\varphi
\end{displaymath}

The Hamiltonian is the energy expressed in terms of the canonical momentum ${\skew0\vec p}^{\,\rm {c}}$ instead of $\vec{v}$; that works out to

\begin{displaymath}
H = \sqrt{(mc^2)^2+\big({\skew0\vec p}^{\,\rm c}-q\skew3\vec A\big)^2c^2} + q\varphi
\end{displaymath}

using the formula given in the overview subsection. The Hamiltonian is of great importance in quantum mechanics.