Sub­sec­tions

### D.3 La­grangian me­chan­ics

This note gives the de­riva­tions for the ad­den­dum on the La­grangian equa­tions of mo­tion.

#### D.3.1 La­grangian equa­tions of mo­tion

To de­rive the non­rel­a­tivis­tic La­grangian, con­sider the sys­tem to be build up from el­e­men­tary par­ti­cles num­bered by an in­dex . You may think of these par­ti­cles as the atoms you would use if you would do a mol­e­c­u­lar dy­nam­ics com­pu­ta­tion of the sys­tem. Be­cause the sys­tem is as­sumed to be fully de­ter­mined by the gen­er­al­ized co­or­di­nates, the po­si­tion of each in­di­vid­ual par­ti­cle is fully fixed by the gen­er­al­ized co­or­di­nates and maybe time. (For ex­am­ple, it is im­plicit in a solid body ap­prox­i­ma­tion that the atoms are held rigidly in their rel­a­tive po­si­tion. Of course, that is ap­prox­i­mate; you pay some price for avoid­ing a full mol­e­c­u­lar dy­nam­ics sim­u­la­tion.)

New­ton’s sec­ond law says that the mo­tion of each in­di­vid­ual par­ti­cle is gov­erned by

where the de­riv­a­tive of the po­ten­tial can be taken to be its gra­di­ent, if you (justly) ob­ject to dif­fer­en­ti­at­ing with re­spect to vec­tors, and in­di­cates any part of the force not de­scribed by the po­ten­tial.

Now con­sider an in­fin­i­tes­i­mal vir­tual dis­place­ment of the sys­tem from its nor­mal evo­lu­tion in time. It pro­duces an in­fin­i­tes­i­mal change in po­si­tion for each par­ti­cle. Af­ter such a dis­place­ment, of course no longer sat­is­fies the cor­rect equa­tions of mo­tion, but the ki­netic and po­ten­tial en­er­gies still ex­ist.

In the equa­tion of mo­tion for the cor­rect po­si­tion above, take the mass times ac­cel­er­a­tion to the other side, mul­ti­ply by the vir­tual dis­place­ment, sum over all par­ti­cles , and in­te­grate over an ar­bi­trary time in­ter­val:

Mul­ti­ply out and in­te­grate the first term by parts:

The vir­tual dis­place­ments of in­ter­est here are only nonzero over a lim­ited range of times, so the in­te­gra­tion by parts did not pro­duce any end point val­ues.

Rec­og­nize the first two terms within the brack­ets as the vir­tual change in the La­grangian due to the vir­tual dis­place­ment at that time. Note that this re­quires that the po­ten­tial en­ergy de­pends only on the po­si­tion co­or­di­nates and time, and not also on the time de­riv­a­tives of the po­si­tion co­or­di­nates. You get

 (D.3)

In case that the ad­di­tional forces are zero, this pro­duces the ac­tion prin­ci­ple: the time in­te­gral of the La­grangian is un­changed un­der in­fin­i­tes­i­mal vir­tual dis­place­ments of the sys­tem, as­sum­ing that they van­ish at the end points of in­te­gra­tion. More gen­er­ally, for the vir­tual work by the ad­di­tional forces to be zero will re­quire that the vir­tual dis­place­ments re­spect the rigid con­straints, if any. The in­fi­nite work done in vi­o­lat­ing a rigid con­straint is not mod­eled by the po­ten­tial in any nor­mal im­ple­men­ta­tion.

Un­chang­ing ac­tion is an in­te­gral equa­tion in­volv­ing the La­grangian. To get or­di­nary dif­fer­en­tial equa­tions, take the vir­tual change in po­si­tion to be that due to an in­fin­i­tes­i­mal change in a sin­gle generic gen­er­al­ized co­or­di­nate. Rep­re­sent the change in the La­grangian in the ex­pres­sion above by its par­tial de­riv­a­tives, and the same for :

The in­te­grand in the fi­nal term is by de­f­i­n­i­tion the gen­er­al­ized force mul­ti­plied by . In the first in­te­gral, the sec­ond term can be in­te­grated by parts, and then the in­te­grals can be com­bined to give

Now sup­pose that there is any time at which the ex­pres­sion within the square brack­ets is nonzero. Then a vir­tual change that is only nonzero in a very small time in­ter­val around that time, and every­where pos­i­tive in that small in­ter­val, would pro­duce a nonzero right hand side in the above equa­tion, but it must be zero. There­fore, the ex­pres­sion within brack­ets must be zero at all times. That gives the La­grangian equa­tions of mo­tion, be­cause the ex­pres­sion be­tween paren­the­ses is de­fined as the canon­i­cal mo­men­tum.

#### D.3.2 Hamil­ton­ian dy­nam­ics

To de­rive the Hamil­ton­ian equa­tions, con­sider the gen­eral dif­fer­en­tial of the Hamil­ton­ian func­tion (re­gard­less of any mo­tion that may go on). Ac­cord­ing to the given de­f­i­n­i­tion of the Hamil­ton­ian func­tion, and us­ing a to­tal dif­fer­en­tial for ,

The sums within paren­the­ses can­cel each other be­cause of the de­f­i­n­i­tion of the canon­i­cal mo­men­tum. The re­main­ing dif­fer­ences are of the ar­gu­ments of the Hamil­ton­ian func­tion, and so by the very de­f­i­n­i­tion of par­tial de­riv­a­tives,

Now con­sider an ac­tual mo­tion. For an ac­tual mo­tion, is the time de­riv­a­tive of , so the sec­ond par­tial de­riv­a­tive gives the first Hamil­ton­ian equa­tion of mo­tion. The first par­tial de­riv­a­tive gives the sec­ond equa­tion when com­bined with the La­grangian equa­tion of mo­tion (A.2).

It is still to be shown that the Hamil­ton­ian of a clas­si­cal sys­tem is the sum of ki­netic and po­ten­tial en­ergy if the po­si­tion of the sys­tem does not de­pend ex­plic­itly on time. The La­grangian can be writ­ten out in terms of the sys­tem par­ti­cles as

where the sum rep­re­sents the ki­netic en­ergy. The Hamil­ton­ian is de­fined as

and straight sub­sti­tu­tion shows the first term to be twice the ki­netic en­ergy.

#### D.3.3 Fields

As dis­cussed in {A.1.5}, the La­grangian for fields takes the form

Here the spa­tial in­te­gra­tion is over all space. The first term de­pends only on the dis­crete vari­ables

where de­notes dis­crete vari­able num­ber . The dot in­di­cates the time de­riv­a­tive of that vari­able. The La­grangian den­sity also de­pends on the fields

where is field num­ber . A sub­script in­di­cates the par­tial time de­riv­a­tive, and 1, 2, or 3 the par­tial , or de­riv­a­tive.

The ac­tion is

where the time range from to must in­clude the times of in­ter­est. The ac­tion must be un­changed un­der small de­vi­a­tions from the cor­rect evo­lu­tion, as long as these de­vi­a­tions van­ish at the lim­its of in­te­gra­tion. That re­quire­ment de­fines the La­grangian. (For sim­ple sys­tems the La­grangian then turns out to be the dif­fer­ence be­tween ki­netic and po­ten­tial en­er­gies. But it is not ob­vi­ous what to make of that if there are fields.)

Con­sider now first an in­fin­i­tes­i­mal de­vi­a­tion in a dis­crete vari­able . The change in ac­tion that must be zero is then

Af­ter an in­te­gra­tion by parts of the sec­ond and fourth terms that be­comes, not­ing that the de­vi­a­tion must van­ish at the ini­tial and fi­nal times,

This can only be zero for what­ever you take if the ex­pres­sion within square brack­ets is zero. That gives the fi­nal La­grangian equa­tion for the dis­crete vari­able as

Next con­sider an in­fin­i­tes­i­mal de­vi­a­tion in field . The change in ac­tion that must be zero is then

Now in­te­grate the de­riv­a­tive terms by parts in the ap­pro­pri­ate di­rec­tion to get, not­ing that the de­vi­a­tion must van­ish at the lim­its of in­te­gra­tion,

Here for 1, 2, or 3 stands for , , or . If the above ex­pres­sion is to be zero for what­ever you take the small change to be, then the ex­pres­sion within square brack­ets will have to be zero at every po­si­tion and time. That gives the equa­tion for the field :

The canon­i­cal mo­menta are de­fined as

These are the quan­ti­ties in­side the time de­riv­a­tives of the La­grangian equa­tions.

For Hamil­ton’s equa­tions, as­sume at first that there are no dis­crete vari­ables. In that case, the Hamil­ton­ian can be writ­ten in terms of a Hamil­ton­ian den­sity :

Take a dif­fer­en­tial of the Hamil­ton­ian den­sity

The first and third terms in the square brack­ets can­cel be­cause of the de­f­i­n­i­tion of the canon­i­cal mo­men­tum. Then ac­cord­ing to cal­cu­lus

The first of these ex­pres­sions gives the time de­riv­a­tive of . The other ex­pres­sions may be used to re­place the de­riv­a­tives of the La­grangian den­sity in the La­grangian equa­tions of mo­tion (2). That gives Hamil­ton’s equa­tions as

If there are dis­crete vari­ables, this no longer works. The full Hamil­ton­ian is then

To find Hamil­ton’s equa­tions, the in­te­grals in this Hamil­ton­ian must be ap­prox­i­mated. The re­gion of in­te­gra­tion is men­tally chopped into lit­tle pieces of the same vol­ume . Then by ap­prox­i­ma­tion

Here num­bers the small pieces and stands for the value of at the cen­ter point of piece . Note that this is es­sen­tially the Rie­mann sum of cal­cu­lus. A sim­i­lar ap­prox­i­ma­tion is made for the other in­te­gral in the Hamil­ton­ian, and the one in the canon­i­cal mo­menta (3). Then the ap­prox­i­mate Hamil­ton­ian be­comes

The dif­fer­en­tial of this ap­prox­i­mate Hamil­ton­ian is

The and terms drop out be­cause of the de­f­i­n­i­tions of the canon­i­cal mo­menta. The re­main­der al­lows ex­pres­sions for the par­tial de­riv­a­tives of the ap­prox­i­mate Hamil­ton­ian to be iden­ti­fied.

The term al­lows the time de­riv­a­tive of to be iden­ti­fied with the par­tial de­riv­a­tive of with re­spect to . And the La­grangian ex­pres­sion for the time de­riv­a­tive of , as given in (1), may be rewrit­ten in terms of cor­re­spond­ing de­riv­a­tives of the ap­prox­i­mate Hamil­ton­ian. To­gether that gives, in the limit ,

For the field, con­sider an po­si­tion cor­re­spond­ing to the cen­ter of an ar­bi­trary lit­tle vol­ume . Then the term al­lows the time de­riv­a­tive of at this ar­bi­trary po­si­tion to be iden­ti­fied in terms of the par­tial de­riv­a­tive of the ap­prox­i­mate Hamil­ton­ian with re­spect to at the same lo­ca­tion. And the La­grangian ex­pres­sion for the time de­riv­a­tive of , as given by (2), may be rewrit­ten in terms of cor­re­spond­ing de­riv­a­tives of the ap­prox­i­mate Hamil­ton­ian. To­gether that gives, in the limit , and leav­ing away since it can be any po­si­tion,

Of course, in real life you would not ac­tu­ally write out these lim­its. In­stead you sim­ply dif­fer­en­ti­ate the nor­mal Hamil­ton­ian un­til you have to start dif­fer­en­ti­at­ing in­side an in­te­gral, like maybe,

Then you think to your­self that you are not re­ally eval­u­at­ing this, but ac­tu­ally

where in­di­cates the po­si­tion that you are con­sid­er­ing the field at. And you are go­ing to di­vide out the vol­ume . That then boils down to

even though the left hand side would math­e­mat­i­cally be non­sense with­out dis­cretiza­tion and di­vi­sion by .