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Given the following Lagrangian,

$L=\frac{1}{2} g_{i j} \dot{q}^{i} \dot{q}^{j}+b_{i j} \dot{q}^{i} q^{j}-U(q)$

I am told that,

The eom depend only on the anti-symmetric part of $b_{i j} .$ The reason is that the symmetric part of $b_{i j}$ contributes a total time derivative to $L .$ Indeed, \begin{aligned} b_{i j} \dot{q}^{i} q^{j} &=\frac{1}{2}\left(b_{i j}-b_{j i}\right) \dot{q}^{i} q^{j}+\frac{1}{2}\left(b_{i j}+b_{j i}\right) \dot{q}^{i} q^{j} \\ &=\frac{1}{2}\left(b_{i j}-b_{j i}\right) \dot{q}^{i} q^{j}+\frac{1}{4} \frac{d}{d t}\left(\left(b_{i j}+b_{j i}\right) q^{i} q^{j}\right) \end{aligned}

I know how to obtain all of the above, but I don't understand why contributing a total time derivative would mean that it does not appear in the eom.

Frobenius
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David
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2 Answers2

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The definition of the action is: $S = \int dt L$ if one considers the modified action with $L\rightarrow L + \frac{df}{dt}$ then the action changes to: $S\rightarrow S + K$ for some cosntant $K$, since the shift in the ation is a constant one the minimization conditions remain unchanged.

Diógenes Figueroa
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  • Is there perhaps a simpler and more direct approach? – David Jul 04 '20 at 14:14
  • All I'm saying is that in general shifting a Lagrangian by a total time derivative does not change the equations of motion, since the change in the action was to shift it by a constant and the equations of motion are minimization conditions on the action. So in your case both Lagrangians would yield the same equations of motion. – Diógenes Figueroa Jul 04 '20 at 14:24
  • @David : This is the simpler and more direct approach. – Frobenius Jul 04 '20 at 19:17
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    @Diógenes Figueroa : it's very important to note that "if $,f(q^j,t),$ is a function of the generalized coordinates $,q^j,$ and time only'' then a total time derivative of ... – Frobenius Jul 04 '20 at 19:30
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From $^{\prime\prime}$Classical Mechanics$^{\prime\prime}$ by H.Goldstein-C.Poole-J.Safko, Third Edition 2000 :

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

...the Eqs. (1.53) become \begin{equation} \dfrac{d\hphantom{t}}{dt}\left(\dfrac{\partial L}{\partial \dot{q}_j}\right)\boldsymbol{-}\dfrac{\partial L}{\partial q_j}\boldsymbol{=}0 \tag{1.57}\label{1.57} \end{equation} expressions referred to as $^{\prime\prime}$Lagrange's equations$^{\prime\prime}$.

Note that for a particular set of equations of motion there is no unique choice of Lagrangian such that Eqs.\eqref{1.57} lead to the equations of motion in the given generalized coordinates. Thus, in Derivations 8 and 10 it is shown that if $\,L(q, \dot{q}, t)\,$ is an approximate Lagrangian and $\,F(q,t)\,$ is any differentiable function of the generalized coordinates and time, then \begin{equation} L^{\prime}(q, \dot{q}, t)\boldsymbol{=}L(q, \dot{q}, t)\boldsymbol{+}\dfrac{dF}{dt} \tag{1.57'}\label{1.57'} \end{equation} is a Lagrangian also resulting in the same equations of motion.

$\vdots$

$\vdots$

DERIVATIONS

1.$\cdots$

$\vdots$

8. If $\,L\,$ is a Lagrangian for a system of $\,n\,$ degrees of freedom satisfying Lagrange's equations, show by direct substitution that \begin{equation} L^{\prime}\boldsymbol{=}L\boldsymbol{+}\dfrac{d F\left(q_1,\cdots,q_n,t\right)}{dt} \tag{[1.57']}\label{[1.57']} \end{equation} also satisfies Lagrange's equations where $\,F\,$ is any arbitrary, but differentiable, function of its arguments.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

Derivation 8

$\boldsymbol{\S} \bf A.$ By direct substitution

Consider the function \begin{equation} \mathcal M\boldsymbol{\equiv}\dfrac{\mathrm d F\left(q_1,\cdots,q_n,t\right)}{\mathrm dt} \tag{A-01}\label{A-01} \end{equation} We have \begin{equation} \mathrm d F\boldsymbol{=}\dfrac{\partial F}{\partial t}\mathrm dt\boldsymbol{+}\sum\limits_{k\boldsymbol{=}1}^{k\boldsymbol{=}n}\dfrac{\partial F}{\partial q_k}\mathrm dq_k\quad \boldsymbol{\Longrightarrow} \quad \dfrac{\mathrm d F}{\mathrm dt}\boldsymbol{=}\dfrac{\partial F}{\partial t}\boldsymbol{+}\sum\limits_{k\boldsymbol{=}1}^{k\boldsymbol{=}n}\dfrac{\partial F}{\partial q_k}\dot{q}_k \tag{A-02}\label{A-02} \end{equation} so \begin{equation} \!\!\!\!\!\!\boxed{\:\:\mathcal M\left(q_1,\dot{q}_1,\cdots,q_n,\dot{q}_n,t\right)\boldsymbol{=}\dfrac{\partial F\left(q_1,\cdots,q_n,t\right)}{\partial t}\boldsymbol{+}\sum\limits_{k\boldsymbol{=}1}^{k\boldsymbol{=}n}\dfrac{\partial F\left(q_1,\cdots,q_n,t\right)}{\partial q_k}\dot{q}_k\:\:} \tag{A-03}\label{A-03} \end{equation} Note that while the function $\,F\,$ is a function of the generalized coordinates $\,q_j\,$ and time $\,t$, the fun- ction $\,\mathcal M\,$ is a function of the generalized coordinates $\,q_j$, the generalized momenta $\,\dot{q}_j\,$ and time $\,t$.

Now, to prove that the addition of the function $\,\mathcal M\,$ to a Lagrangian $\,L\,$ doesn't change the Lagrange equations \eqref{1.57} we proceed as follows : Considering that only $\,\mathcal M\,$ is a Lagrangian of the system we'll prove that the resulting Lagrange equations \eqref{1.57} \begin{equation} \dfrac{\mathrm d\hphantom{t}}{\mathrm d t}\left(\dfrac{\partial \mathcal M}{\partial \dot{q}_j}\right)\boldsymbol{-}\dfrac{\partial\mathcal M}{\partial q_j}\boldsymbol{=}0 \tag{A-04}\label{A-04} \end{equation} are identically zero, that is the left hand side is zero for any degree of freedom $\,j$.

Indeed, from \eqref{A-03} \begin{equation} \dfrac{\partial \mathcal M}{\partial \dot{q}_j}\boldsymbol{=}\dfrac{\partial F}{\partial q_j} \tag{A-05}\label{A-05} \end{equation} so \begin{equation} \dfrac{\mathrm d\hphantom{t}}{\mathrm d t}\left(\dfrac{\partial \mathcal M}{\partial \dot{q}_j}\right)\boldsymbol{=}\dfrac{\mathrm d\hphantom{t}}{\mathrm d t}\left(\dfrac{\partial F}{\partial q_j}\right)\boldsymbol{=}\dfrac{\partial\hphantom{q_j} }{\partial q_j}\left(\dfrac{\mathrm d F}{\mathrm dt}\right) \boldsymbol{=}\dfrac{\partial\mathcal M}{\partial q_j} \tag{A-06}\label{A-06} \end{equation} quod erat demonstrandum ($\dot{o}\pi\epsilon\rho\:\:\epsilon\delta\epsilon\dot{\iota}\chi\theta\eta$).

$\boldsymbol{\S} \bf B.$ By calculus of variations

Lagrange equations \eqref{1.57} follow from Hamilton's principle stated as

The motion of the system from time $\,t_1\,$ to time $\,t_2\,$ is such that the line integral (called the action or the action integral) \begin{equation} S\boldsymbol{=}\int\limits_{t_1}^{t_2}L\mathrm d t \tag{B-01}\label{B-01} \end{equation} has a stationary value for the actual path of the Motion.

Hamilton's principle is summarized by saying that the motion is such that the variation of the line integral $\,S\,$ for fixed $\,t_1\,$ and $\,t_2\,$ is zero: \begin{equation} \delta S\boldsymbol{=}\delta\int\limits_{t_1}^{t_2}L\left(q_1,\cdots,q_n,\dot{q}_1,\cdots,\dot{q}_n,t\right)\mathrm d t\boldsymbol{=}0 \tag{B-02}\label{B-02} \end{equation} For a new Lagrangian $\,L^{\prime}\,$ according to equation \eqref{[1.57']}
\begin{equation} L^{\prime}\boldsymbol{=}L\boldsymbol{+}\dfrac{\mathrm d F\left(q_1,\cdots,q_n,t\right)}{\mathrm d t} \tag{B-03}\label{B-03} \end{equation} we have a new action $\,S^{\prime}\,$ \begin{equation} S^{\prime}\boldsymbol{=}\int\limits_{t_1}^{t_2}\left(L\boldsymbol{+}\dfrac{\mathrm d F}{\mathrm d t}\right)\mathrm d t\boldsymbol{=}\int\limits_{t_1}^{t_2}L\mathrm d t\boldsymbol{+}\int\limits_{t_1}^{t_2}\mathrm d F\boldsymbol{=}S\boldsymbol{+}\underbrace{F\left(t_2\right)\boldsymbol{-}F\left(t_1\right)}_{\boldsymbol{=}C\boldsymbol{=}\text{constant}} \tag{B-04}\label{B-04} \end{equation} so \begin{equation} \delta S^{\prime}\boldsymbol{=}\delta S\boldsymbol{+}\underbrace{\delta\left[F\left(t_2\right)\boldsymbol{-}F\left(t_1\right)\right]}_{\boldsymbol{=}\delta C\boldsymbol{=}0} \tag{B-05}\label{B-05} \end{equation} that is \begin{equation} \delta S^{\prime}\boldsymbol{=}\delta S \tag{B-06}\label{B-06} \end{equation} From the equality of variations of the actions equation \eqref{B-06} we have the same Lagrange equations.

Frobenius
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