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.

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 of the top atom as seen by ob­server A into the ver­ti­cal ve­loc­ity of that atom as seen by ob­server B:

They are dif­fer­ent! In par­tic­u­lar, is smaller than . 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 , which is larger than the rest mass by the Lorentz fac­tor:

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:

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

A lit­tle thought shows that the right time dif­fer­en­tial has to be . 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 trans­forms like , 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 1,2,...is the same be­fore and af­ter,

then so does any other ob­server B:

Us­ing the chain rule of dif­fer­en­ti­a­tion, the com­po­nents of the mo­men­tum four-vec­tor can be writ­ten out as

 (1.14)

The com­po­nents can be writ­ten in the same form as in the non­rel­a­tivis­tic case by defin­ing a mov­ing mass
 (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 ex­cept for a triv­ial scal­ing fac­tor 1 to give it units of mo­men­tum.

Note that so far, this only in­di­cates that the dif­fer­ence be­tween and gives the ki­netic en­ergy. It does not im­ply that 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 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 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 . 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 . The two masses have a slight down­ward ve­loc­ity be­fore the col­li­sion and so has the mass af­ter the col­li­sion. But then ver­ti­cal mo­men­tum con­ser­va­tion in­evitably im­plies

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

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

 (1.16)

where is the mass of the par­ti­cle, its ve­loc­ity, and its charge, while 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 of the canon­i­cal mo­men­tum is found as

 (1.17)

This works out to be sim­ply com­po­nent of the nor­mal mo­men­tum. The equa­tions of mo­tion are taken to be
 (1.18)

which is found to be

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 and sum­ming over .

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,

For ve­loc­i­ties small com­pared to the speed of light, a two term Tay­lor se­ries shows this is equiv­a­lent to plus the ki­netic en­ergy. The con­stant 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 .

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

 (1.19)

And that is in fact right. Com­po­nent zero of the po­ten­tial four-vec­tor is the clas­si­cal elec­tro­sta­tic po­ten­tial. The spa­tial vec­tor is called the “mag­netic vec­tor po­ten­tial.”

The canon­i­cal mo­men­tum is now

 (1.20)

and that is no longer just the nor­mal mo­men­tum, , 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}:

 (1.21)

The right-hand side in this equa­tion of mo­tion is called the Lorentz force. In it, is called the elec­tric field and the mag­netic field. These fields are re­lated to the four-vec­tor po­ten­tial as

where by de­f­i­n­i­tion

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 is sta­tion­ary for the way that the par­ti­cle moves, {A.1.3}, {D.3.1} That re­quires that 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 where the proper time in­ter­val is the same for all ob­servers. So the ac­tion is the same for all ob­servers.)

From the Lorentz trans­for­ma­tion of , that of the elec­tric and mag­netic fields may be found; that is not a Lorentz trans­for­ma­tion. Note that this sug­gests that 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 and into

 (1.22)

where 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

The Hamil­ton­ian is the en­ergy ex­pressed in terms of the canon­i­cal mo­men­tum in­stead of ; that works out to

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