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.
\end{picture} }

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:

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

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:

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

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:

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

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, the same be­fore and af­ter,

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

then so does any other ob­server B:

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

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

p_0 = m c \frac{{\rm d}t}{{\rm d}t_0} \quad
p_1 = m \frac{...
...= 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
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.

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

2 m_v v_{\rm {B}} = M v_{\rm {B}}

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

{\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

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
\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

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

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,

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

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

{\cal L}= - m c^2 \sqrt{1-(\vert\vec ...
...t}}}\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

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}:

\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

\skew3\vec{\cal E}= - \nabla \varphi - \frac{\partial \skew...
...ial t}
\skew2\vec{\cal B}= \nabla \times \skew3\vec A

where by de­f­i­n­i­tion

\nabla = {\hat\imath}\frac{\partial}{\partial x}
+ {\hat\j...{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}

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

\varphi' = \varphi - \frac{\partial\chi}{\partial t}
\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

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

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

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

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