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 $j$. 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 $j$ is gov­erned by

\begin{displaymath}
m_j \frac{{\rm d}^2 {\skew0\vec r}_j}{{\rm d}t^2} =
- \frac{\partial V}{\partial {\skew0\vec r}_j} + \vec F'_j
\end{displaymath}

where the de­riv­a­tive of the po­ten­tial $V$ can be taken to be its gra­di­ent, if you (justly) ob­ject to dif­fer­en­ti­at­ing with re­spect to vec­tors, and $\vec{F}'_j$ 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 $\delta{\skew0\vec r}_j(t)$ for each par­ti­cle. Af­ter such a dis­place­ment, ${\skew0\vec r}_j+\delta{\skew0\vec r}_j$ 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 ${\skew0\vec r}_j$ above, take the mass times ac­cel­er­a­tion to the other side, mul­ti­ply by the vir­tual dis­place­ment, sum over all par­ti­cles $j$, and in­te­grate over an ar­bi­trary time in­ter­val:

\begin{displaymath}
0 = \int_{t_1}^{t_2} \sum_j \left[- m_j \frac{{\rm d}^2 {\s...
...r}_j}
+ \vec F'_j\right]\cdot\delta{\skew0\vec r}_j{ \rm d}t
\end{displaymath}

Mul­ti­ply out and in­te­grate the first term by parts:

\begin{displaymath}
0 = \int_{t_1}^{t_2} \sum_j
\left[
m_j \frac{{\rm d}{\ske...
... r}_j
+ \vec F'_j \delta {\skew0\vec r}_j
\right] { \rm d}t
\end{displaymath}

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

\begin{displaymath}
0 = \delta \int_{t_1}^{t_2} {\cal L}{ \rm d}t
+ \int_{t_1...
...j \left[\vec F'_j\cdot\delta{\skew0\vec r}_j\right] { \rm d}t
\end{displaymath} (D.3)

In case that the ad­di­tional forces $\vec{F}'_j$ 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 $V$ 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 $\delta{q}_k(t)$ in a sin­gle generic gen­er­al­ized co­or­di­nate. Rep­re­sent the change in the La­grangian in the ex­pres­sion above by its par­tial de­riv­a­tives, and the same for $\delta{\skew0\vec r}_j$:

\begin{displaymath}
0 =
\int_{t_1}^{t_2}
\left[
\frac{\partial{\cal L}}{\par...
...ial{\skew0\vec r}_j}{\partial q_k}\delta q_k\right] { \rm d}t
\end{displaymath}

The in­te­grand in the fi­nal term is by de­f­i­n­i­tion the gen­er­al­ized force $Q_k$ mul­ti­plied by $\delta{q}_k$. In the first in­te­gral, the sec­ond term can be in­te­grated by parts, and then the in­te­grals can be com­bined to give

\begin{displaymath}
0 =
\int_{t_1}^{t_2}
\left[
\frac{\partial{\cal L}}{\par...
...{\partial\dot q_k}\right)
+ Q_k
\right]\delta q_k { \rm d}t
\end{displaymath}

Now sup­pose that there is any time at which the ex­pres­sion within the square brack­ets is nonzero. Then a vir­tual change $\delta{q}_k$ 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 ${\rm d}{\cal L}$,

\begin{displaymath}
{\rm d}H =
\left(\sum_k p^{\rm {c}}_k{\rm d}\dot q_k\right...
...ot q_k\right)
- \frac{\partial {\cal L}}{\partial t} {\rm d}t
\end{displaymath}

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,

\begin{displaymath}
\frac{\partial H}{\partial q_k} =
- \frac{\partial {\cal L...
...rtial H}{\partial t} =
- \frac{\partial {\cal L}}{\partial t}
\end{displaymath}

Now con­sider an ac­tual mo­tion. For an ac­tual mo­tion, $\dot{q}_k$ is the time de­riv­a­tive of $q_k$, 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

\begin{displaymath}
\sum_{j} \sum_{{\underline k}=1}^K\sum_{\underline{{\underl...
...}\dot q_{\underline{{\underline k}}}
-V(q_1,q_2,\ldots,q_K,t)
\end{displaymath}

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

\begin{displaymath}
\sum_k \dot q_k \frac{\partial{\cal L}}{\partial\dot q_k} - {\cal L}
\end{displaymath}

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

\begin{displaymath}
{\cal L}= \Lag_0 + \int \pounds { \rm d}^3{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
\Lag_0 = \Lag_0(\ldots; q_k,\dot{q}_k;\ldots)
\end{displaymath}

where $q_k$ $\vphantom0\raisebox{1.5pt}{$=$}$ $q_k(t)$ de­notes dis­crete vari­able num­ber $k$. The dot in­di­cates the time de­riv­a­tive of that vari­able. The La­grangian den­sity also de­pends on the fields

\begin{displaymath}
\pounds = \pounds (\ldots;\varphi_\alpha,\varphi_\alpha\str...
...\strut_2,\varphi_\alpha\strut_3;
\ldots;q_k;\dot{q}_k;\ldots)
\end{displaymath}

where $\varphi_\alpha$ is field num­ber $\alpha$. A sub­script $t$ in­di­cates the par­tial time de­riv­a­tive, and 1, 2, or 3 the par­tial $x$, $y$ or $z$ de­riv­a­tive.

The ac­tion is

\begin{displaymath}
{\cal S}= \int_{t_1}^{t_2} \left( \Lag_0 + \int \pounds {\,\rm d}^3{\skew0\vec r}\right) {\,\rm d}t
\end{displaymath}

where the time range from $t_1$ to $t_2$ 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 $\delta{q}_k$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta{q}_k(t)$ in a dis­crete vari­able $q_k$. The change in ac­tion that must be zero is then

\begin{displaymath}
0 = \delta{\cal S}=
\int_{t_1}^{t_2} \left(
\frac{\partia...
...,\rm d}^3{\skew0\vec r}\; \delta \dot{q}_k
\right) {\,\rm d}t
\end{displaymath}

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,

\begin{displaymath}
0 = \delta{\cal S}=
\int_{t_1}^{t_2} \left[
\frac{\partia...
...q}_k} {\,\rm d}^3{\skew0\vec r}
\right] \delta q_k {\,\rm d}t
\end{displaymath}

This can only be zero for what­ever you take $\delta{q}_k$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta{q}_k(t)$ if the ex­pres­sion within square brack­ets is zero. That gives the fi­nal La­grangian equa­tion for the dis­crete vari­able $q_k$ as
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\frac{{\rm d}}{{\rm d}t} \lef...
... \frac{\partial\pounds }{\partial q_k} {\,\rm d}^3{\skew0\vec r}
$\ \hfill(1)}$

Next con­sider an in­fin­i­tes­i­mal de­vi­a­tion $\delta\varphi_\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta\varphi_\alpha({\skew0\vec r};t)$ in field $\varphi_\alpha$. The change in ac­tion that must be zero is then

\begin{displaymath}
0 =\delta S = \int_{t_1}^{t_2} \int \left(
\frac{\partial\...
...hi_\alpha\strut_i
\right) { \rm d}^3{\skew0\vec r}{ \rm d}t
\end{displaymath}

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

\begin{displaymath}
0 = \delta S = \int_{t_1}^{t_2} \int \left[
\frac{\partial...
...ight] \delta\varphi_\alpha { \rm d}^3{\skew0\vec r}{ \rm d}t
\end{displaymath}

Here $r_i$ for $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, or 3 stands for $x$, $y$, or $z$. If the above ex­pres­sion is to be zero for what­ever you take the small change $\delta\varphi_\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta\varphi_\alpha({\skew0\vec r};t)$ 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 $\varphi_\alpha$:
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\frac{\partial}{\partial t}
...
...rut_i}\right)
= \frac{\partial\pounds }{\partial\varphi_\alpha}
$ \hfill(2)}$

The canon­i­cal mo­menta are de­fined as

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
p^{\rm{c}}_k \equiv \frac{\pa...
...a \equiv \frac{\partial\pounds }{\partial\varphi_\alpha\strut_t}
$\ \hfill(3)}$
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 $h$:

\begin{displaymath}
H = \int h { \rm d}^3{\skew0\vec r}\qquad
h = \sum_\alpha \pi^{\rm {c}}_\alpha \varphi_\alpha\strut_t - \pounds
\end{displaymath}

Take a dif­fer­en­tial of the Hamil­ton­ian den­sity

\begin{displaymath}
{\rm d}h = \sum_\alpha \left[
\pi^{\rm {c}}_\alpha {\rm d}...
...ounds }{\partial\varphi_\alpha} {\rm d}\varphi_\alpha
\right]
\end{displaymath}

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

\begin{displaymath}
\frac{\partial h}{\partial\pi^{\rm {c}}_\alpha} = \varphi_\...
...i_\alpha} =
- \frac{\partial\pounds }{\partial\varphi_\alpha}
\end{displaymath}

The first of these ex­pres­sions gives the time de­riv­a­tive of $\varphi_\alpha$. 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
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\frac{\partial\varphi_\alpha}...
...
\left(\frac{\partial h}{\partial\varphi_\alpha\strut_i}\right)
$ \hfill(4)}$

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

\begin{displaymath}
H = \sum_k p^{\rm {c}}_k \dot{q}_k
+ \int \sum_\alpha \pi^...
...kew0\vec r}
- \Lag_0 - \int \pounds {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

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 ${\rm d}{\cal V}$. Then by ap­prox­i­ma­tion

\begin{displaymath}
\int \pounds { \rm d}^3{\skew0\vec r}\approx \sum_n \pounds _n {\rm d}{\cal V}
\end{displaymath}

Here $n$ num­bers the small pieces and $\pounds _n$ stands for the value of $\pounds $ at the cen­ter point of piece $n$. 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

\begin{displaymath}
H_{\rm app} = \sum_k p^{\rm {c}}_k \dot{q}_k
+ \sum_{\alph...
... {\rm d}{\cal V}
- \Lag_0 - \sum_n \pounds _n {\rm d}{\cal V}
\end{displaymath}

The dif­fer­en­tial of this ap­prox­i­mate Hamil­ton­ian is

\begin{eqnarray*}
{\rm d}H_{\rm app} & = & \sum_k \dot{q}_k {\rm d}p^{\rm {c}}_...
...ha}_i\strut_n}
{\rm d}{\cal V}{\rm d}{\varphi_\alpha}_i\strut_n
\end{eqnarray*}

The ${\rm d}\dot{q}_k$ and ${\rm d}{\varphi_\alpha}_t\strut_n$ 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 ${\rm d}{p}^{\rm {c}}_k$ term al­lows the time de­riv­a­tive of $q_k$ to be iden­ti­fied with the par­tial de­riv­a­tive of $H_{\rm {app}}$ with re­spect to $p^{\rm {c}}_k$. And the La­grangian ex­pres­sion for the time de­riv­a­tive of $p^{\rm {c}}_k$, as given in (1), may be rewrit­ten in terms of cor­re­spond­ing de­riv­a­tives of the ap­prox­i­mate Hamil­ton­ian. To­gether that gives, in the limit ${\rm d}{\cal V}\to0$,

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\frac{{\rm d}q_k}{{\rm d}t} =...
...rm d}p^{\rm{c}}_k}{{\rm d}t} = - \frac{\partial H}{\partial q_k}
$ \hfill(5)}$

For the field, con­sider an po­si­tion ${\skew0\vec r}$ cor­re­spond­ing to the cen­ter of an ar­bi­trary lit­tle vol­ume $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline n}$. Then the ${\rm d}\pi^{\rm {c}}_\alpha\strut_{\underline n}$ term al­lows the time de­riv­a­tive of $\varphi_\alpha$ at this ar­bi­trary po­si­tion to be iden­ti­fied in terms of the par­tial de­riv­a­tive of the ap­prox­i­mate Hamil­ton­ian with re­spect to $\pi^{\rm {c}}_\alpha$ at the same lo­ca­tion. And the La­grangian ex­pres­sion for the time de­riv­a­tive of $\pi^{\rm {c}}_\alpha$, as given by (2), may be rewrit­ten in terms of cor­re­spond­ing de­riv­a­tives of the ap­prox­i­mate Hamil­ton­ian. To­gether that gives, in the limit ${\rm d}{\cal V}\to0$, and leav­ing ${\underline n}$ away since it can be any po­si­tion,

$\parbox{400pt}{\hfill$\displaystyle
\frac{\partial\varphi_\alpha}{\partial t}
...
...V}}
\frac{\partial H_{\rm app}}{\partial\varphi_\alpha\strut_i}
$ \hfill(6)}$

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

\begin{displaymath}
\frac{\partial}{\partial\varphi_\alpha}\int{\pounds }{ \rm d}^3{\skew0\vec r}
\end{displaymath}

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

\begin{displaymath}
\frac{\partial}{\partial\varphi_\alpha\strut_{\underline n}...
...}{\partial\varphi_\alpha\strut_{\underline n}} {\rm d}{\cal V}
\end{displaymath}

where ${\underline n}$ in­di­cates the po­si­tion that you are con­sid­er­ing the field at. And you are go­ing to di­vide out the vol­ume ${\rm d}{\cal V}$. That then boils down to

\begin{displaymath}
\frac{\partial}{\partial\varphi_\alpha}\int{\pounds }{ \rm...
...ghtarrow\quad
\frac{\partial\pounds }{\partial\varphi_\alpha}
\end{displaymath}

even though the left hand side would math­e­mat­i­cally be non­sense with­out dis­cretiza­tion and di­vi­sion by ${\rm d}{\cal V}$.