Suppose that $L(q^i, \dot{q}^i)$ is a standard and well behaved lagrangian associated to some Dirichlet boundary conditions : $q^i(t_1) = q_1^i$ and $q^i(t_2) = q_2^i$. Now I have this new lagrangian : \begin{equation} L' = L - \frac{d}{dt}(\lambda \: q^i \, p_i), \end{equation} where $\lambda$ is a positive constant. What are the boundary conditions associated to this ?

The only thing I know is, if $\lambda = 1$, then $L'$ should be associated to fixed canonical momentum at the endpoints : $p_i(t_1) = p_{i1}$ and $p_i(t_2) = p_{i2}$ (the boundary conditions are defined by the canonical momentum).

But then what if $\lambda \ne 1$ ?

  • 1
    $\begingroup$ Guess you actually have $L(t,q^i,\dot{q}^i)$. Does something like the following help: math.odu.edu/~jhh/ch44.PDF ? $\endgroup$ – udrv Nov 18 '15 at 5:20
  • $\begingroup$ Is there a systematic method to define the total derivative to be added to a "Dirichlet" lagrangian, if we know the boundary conditions ? $\endgroup$ – Cham Nov 18 '15 at 13:39
  • $\begingroup$ Try working backwards? If the new Lagrangian is $L' = L - \frac{dF}{dt}$, the corresponding natural boundary conditions would be $\frac{\partial L}{\partial \dot{q}} - \frac{\partial}{\partial \dot{q}}\frac{dF}{dt} = 0$. Separate the likely contribution from $F$ and try to rebuild $F$ around it. For instance for the fixed $p$ condition you mention, it's obvious one term has to be $\sim\dot{q}p$, so the simplest total derivative based on it would be $\sim\frac{d}{dt}(qp)$. $\endgroup$ – udrv Nov 18 '15 at 14:45

I) Hamiltonian interpretation. Given a Hamiltonian $H(z;t)$ with canonical coordinates $$\tag{1} (z^1,\ldots,z^{2n}) ~=~ (q^1, \ldots, q^n;p_1,\ldots, p_n), $$ the Hamiltonian Lagrangian reads $$\tag{2} L_H(z,\dot{z};t) ~=~p_k \dot{q}^k - H(z;t). $$ Then OP's modified Hamiltonian Lagrangian becomes $$\tag{3}\tilde{L}_H(z,\dot{z};t) ~:=~L_H(z,\dot{z};t)- \lambda\frac{d}{dt}(q^k p_k) ~=~(1-\lambda)p_k \dot{q}^k -\lambda q^k\dot{p}_k- H(z;t) .\qquad$$ Note that we can view (3) as a Lagrangian in twice as many variables (1). The corresponding momenta read $$\tag{4} Q_k~:=~ \frac{\partial L_H}{\partial \dot{q}^k}~=~(1-\lambda)p_k, \qquad P^k~:=~ \frac{\partial L_H}{\partial \dot{p}_k}~=~-\lambda q^k. $$ From the general theory (as e.g. reviewed in my Phys.SE answer here), we conclude that when we vary the Hamiltonian action $$\tag{5}\tilde{I}_H[q]~:=~\int_{t_i}^{t_f} \! dt~ \tilde{L}_H $$ the boundary terms are $$\tag{6} \left[ Q_k~\delta q^k + P^k~\delta p_k\right]_{t=t_i}^{t=t_f} ~=~\left[ (1-\lambda)p_k~\delta q^k -\lambda q^k~\delta p_k\right]_{t=t_i}^{t=t_f}. $$ These boundary terms (6) should vanish in order for the functional/variational derivative of the Hamiltonian action (5) to exist.

Now let us discuss possible boundary conditions (BCs). If time is cyclic, so that $t_i$ and $t_f$ represent same instant, we should impose periodic BCs. (We ignore anti-periodic BCs and fermions here.)

Let us for the remainder of this answer assume that time is not cyclic, so that initial BC and final BC are physically independent. Let us just discuss the initial BC, since the final BC is similar. We see that we need to impose at least $n$ initial BCs because there are $2n$ canonical coordinates (1).

Also it seems physically reasonable to assume that the BC treats the position coordinates on same footing. Similarly for the momentum coordinates. In other words, we are going to assume that the theory (including BCs) behaves in a natural way under linear canonical transformation of coordinates $$\tag{7} q^{\prime k}~=~M^k{}_{\ell}~ q^{\ell}, \qquad p_{\ell}~=~p^{\prime}_{k} ~M^k{}_{\ell}, \qquad k,\ell~\in~\{1,\ldots, n\} . $$ (This needs not be the case, but we will assume it.) It follows that the initial BC must fix all the positions or all the momenta.

We conclude that:

  • For any $\lambda$, one can choose initial BC $q(t_i)=0$ or $p(t_i)=0$.

  • For $\lambda=0$, one can also choose initial BC $q(t_i)=q_i$.

  • For $\lambda=1$, one can also choose initial BC $p(t_i)=p_i$.

II) Lagrangian interpretation. Given a first-order Lagrangian $L(q,\dot{q};t)$, the Lagrangian momenta reads $$\tag{8} p_k(q,\dot{q};t)~:=~\frac{\partial L(q,\dot{q};t)}{\partial \dot{q}^k}$$ Then OP's modified Lagrangian $$\tag{9} \tilde{L}(q,\dot{q},\ddot{q};t) ~:=~ L(q,\dot{q};t) - \lambda\frac{d}{dt}\left(q^kp_k(q,\dot{q};t)\right)$$ becomes of second-order for $\lambda\neq 0$. Note that for a higher-order theory, one should in general impose more boundary conditions, cf. e.g. my Phys.SE answer here.

The EL eqs.

$$\tag{10} \frac{\partial \tilde{L}}{\partial q^k} -\frac{d}{dt}\frac{\partial \tilde{L}}{\partial \dot{q}^k}+\left(\frac{d}{dt}\right)^2 \frac{\partial \tilde{L}}{\partial \ddot{q}^k} - \ldots~\approx~0 $$

for the Lagrangian (9) are not changed by a total derivative term, and are still generically of second order.

However, for $\lambda\neq 0$, we have to impose four BCs. Since the EL eqs. are of second order, there will generically not be any stationary solutions. We conclude that OP's variational problem is ill-defined for $\lambda\neq 0$.

  • $\begingroup$ in the Lagrangian formalism we have to specify four BC's to satisfy a second order ODE. If however $\lambda=0$ can we just set the Dirichlet conditions $q(1)=q(2)=0$? Would that lead to stationary solutions? $\endgroup$ – AngusTheMan Nov 19 '15 at 0:17
  • 1
    $\begingroup$ @AngusTheMan: Yes, in the case $\lambda=0$, that's a possible choice. $\endgroup$ – Qmechanic Nov 19 '15 at 0:20
  • $\begingroup$ Qmechanic, I don't understand your conclusion. Under the lagrangian formalism (not hamiltonian), what is the correct lagrangian if the canonical momentum $p_i$ are fixed at both boudaries ? It should be $L'$ with $\lambda = 1$, isn't ? And yet you say that it is an ill-defined variational problem ? $\endgroup$ – Cham Nov 19 '15 at 1:19
  • $\begingroup$ Also, from the boundary terms $\left[ (1-\lambda)p_k~\delta q^k -\lambda q^k~\delta p_k\right]_{t=t_i}^{t=t_f}$, you said that for any $\lambda$, one can choose initial BC $q(t_i)=0$ or $p(t_i)=0$. That is not the only possibility. You could also define a quantity $F^i$ or $G_i$ (as in my answer) as a function of $q^i$ and $p_i$, which could be fixed at the endpoints. While $q^i$ and $p_i$ are not separately fixed, a combination of both could be fixed. $\endgroup$ – Cham Nov 19 '15 at 13:33

The following is rewritten after @Qmechanic's comment. While his observation was correct, I think the main point below holds on its own.

The case @Cham considers is that of a Lagrangian $L' = L - \frac{d}{dt}q^ip_i$ modified by a total derivative for the purpose of implementing a change in the boundary conditions. Originally the $p_i$-s are assumed to be the canonical momenta $p_i = \frac{\partial L}{\partial \dot{q}^i}$, which makes the total derivative a function of the $q^i$, $\dot{q}^i$. We can relax the assumption on the particular form of the total derivative and generalize instead to $$ L'(t, q, \dot{q}) = L(t, q, \dot{q}) - \frac{d}{dt}F(t, q, \dot{q}) $$

Here is the problem: The total derivative leaves the EL eqs unchanged, but using natural boundary conditions leads to an implicit condition for $F$: $$ \frac{\partial F}{\partial \dot{q}^i} = 0 $$

The modified Lagrangian must necessarily read $$ L'(t, q, \dot{q}) = L(t, q, \dot{q}) - \frac{d}{dt}F(t, q) $$ In other words, the original choice $F = \lambda q^ip_i(t, q, \dot{q})$ is ill-defined for this purpose: taking the $p_i$ to mean canonical momenta, or even any function such that $\partial p_i / \partial \dot{q}^i \neq 0$, leaves the boundary conditions unresolved. Hence the $p_i$-s must be functions of time and/or the $q^i$-s only, $p_i = p_i(t, q)$.

Why this is so:

The contribution of any $\frac{dF}{dt} = \frac{d}{dt}F(t, q, \dot{q})$ to the variation of the action is confined to boundary terms: $$ \delta I = \int_{t_1}^{t_2}{dt\;\delta\left( L(t, q, \dot{q}) - \frac{dF}{dt}(t, q, \dot{q})\right)} = \int_{t_1}^{t_2}{dt\;\delta L(t, q, \dot{q})} - \delta F(t, q, \dot{q})\Big|_{t_1}^{t_2} =\\ \int_{t_1}^{t_2}{dt\;\left( \frac{\partial L}{\partial q^i} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i}\right)\delta q^i} + \left[\left(\frac{\partial L}{\partial \dot{q}^i} - \frac{\partial F}{\partial q^i} \right)\delta q^i - \frac{\partial F}{\partial \dot{q}^i}\delta \dot{q}^i\right]\Big|_{t_1}^{t_2} $$ We may try to find some $G = G(t, q, \dot{q})$ such that $$ \left(\frac{\partial L}{\partial \dot{q}^i} - \frac{\partial F}{\partial q^i} \right)\delta q^i - \frac{\partial F}{\partial \dot{q}^i}\delta \dot{q}^i = \delta G $$ in which case the boundary conditions would amount to $\delta G\Big|_{t=t_{1,2}} = 0$. But this would require $$ \frac{\partial L}{\partial \dot{q}^i} - \frac{\partial F}{\partial q^i} = \frac{\partial G}{\partial q^i}\\ - \frac{\partial F}{\partial \dot{q}^i} = \frac{\partial G}{\partial \dot{q}^i} $$ which, assuming $\frac{\partial^2 G}{\partial \dot{q}^i \partial q^j} = \frac{\partial^2 G}{\partial q^j \partial \dot{q}^i }$, would lead to $$ \frac{\partial^2 G}{\partial \dot{q}^i \partial q^j} = \frac{\partial^2 L}{\partial \dot{q}^i \partial \dot{q}^j} - \frac{\partial^2 F}{\partial \dot{q}^i \partial q^j} = - \frac{\partial^2 F}{\partial \dot{q}^i \partial q^j} $$ and $$ \frac{\partial^2 L}{\partial \dot{q}^i \partial \dot{q}^j} = 0 $$ This follows even if we try $F = \lambda q^i\frac{\partial L}{\partial \dot{q}^i} = \lambda q^i p_i$. So unless $L$ is linear in the $\dot{q}^i$, the two boundary contributions must vanish simultaneously for any $\delta q$, $\delta \dot{q}$, although only one of them can give boundary conditions for a first order Lagrangian. The other must vanish identically for any $t$, $q$, $\dot{q}$. Requesting this of the first term doesn't work since the 2nd one would impose additional conditions on the Lagrangian. This leaves $$ \frac{\partial F}{\partial \dot{q}^i} = 0 $$ and boundary conditions of the form $$ \left(\frac{\partial L}{\partial \dot{q}^i} - \frac{\partial F}{\partial q^i} \right)\Big|_{t=t_{1,2}} = 0 $$ Incidentally the requirement that $\frac{\partial F}{\partial \dot{q}^i} = 0$ removes the issue that the modified Lagrangian $L' = L - \frac{dF}{dt}$ could become second order.

Note however that assuming $F=F(t, q, \dot{q})$ does not imply by itself that the EL equations for $L'$ are necessarily higher order. This is because although $\frac{dF}{dt}$ is linear in the $\ddot{q}^i$, when taking the variation under the action integral the terms in $\delta \ddot{q}^i$ get integrated by parts twice and cancel the remaining contributions from $F$, leaving only the EL terms in $L$ and the corresponding boundary terms.

  • $\begingroup$ There's a mistake after the line "We may try to find some $G = G(t, q, \dot{q})$ such that". The requirement that the left expression is equal to $\delta G$ is much too strong. You need to introduce two functions of $q$ and $\dot{q}$, such that the right part reads $F \; \delta G$ instead. It is more general, since the left part isn't necessarily an exact differential. $F(t, q, \dot{q})$ would be an integrating factor. And then don't forget to add the indices : $F^i \; \delta G_i$ or $P_i \; \delta Q^i$. Then the calculations after that are completely changed. $\endgroup$ – Cham Nov 22 '15 at 14:23
  • $\begingroup$ By the way, using equation numbers with the \tag{} command would be usefull. $\endgroup$ – Cham Nov 22 '15 at 14:26
  • $\begingroup$ @Cham I see what you mean. But say I do replace $\delta G \rightarrow D^i\delta G_i$. Then from $\frac{\partial L}{\partial \dot{q}^k} - \frac{\partial F}{\partial q^k} = D^i\frac{\partial G_i}{\partial q^k}$ and $-\frac{\partial F}{\partial \dot{q}^k} = D^i\frac{\partial G_i}{\partial \dot{q}^k}$ it follows that $\delta F = \left( \frac{\partial L}{\partial \dot{q}^k} - D^i\frac{\partial G_i}{\partial q^k}\right)\delta q^k - D^i\frac{\partial G_i}{\partial \dot{q}^k}\delta \dot{q}^k$ and we get differentiability conditions. $\endgroup$ – udrv Nov 23 '15 at 0:44
  • $\begingroup$ @Cham These turn out to involve only (sums of products of) 1st derivatives of the $D^i$ and $G_i$, depend on $L$ alone, and can be looked at as a system for the $\frac{\partial D^i}{\partial q^k}$, $\frac{\partial G_i}{\partial q^k}$, $\frac{\partial D^i}{\partial \dot{q}^k}$, $\frac{\partial G_i}{\partial \dot{q}^k}$. The system is underdetermined, so if there are solutions there is likely a continuum of them. The problem I see now is: there should be a 1-to-1 correspondence between $F$ (up to an additive constant) and the boundary conditions it induces. $\endgroup$ – udrv Nov 23 '15 at 0:44
  • $\begingroup$ @Cham So how to prove that the solution for $D^i\frac{\partial G_i}{\partial q^k}$ and $D^i\frac{\partial G_i}{\partial \dot{q}^k}$ is unique? $\endgroup$ – udrv Nov 23 '15 at 0:44

I'm posting here an extraction from a paper written by T. Padmanabhan (http://arxiv.org/abs/hep-th/0608120) :

Consider a dynamical variable $q(t)$ in point mechanics described by a lagrangian $L_q(q,\dot q)$. Varying the action obtained from integrating this Lagrangian in the interval $(t_1,t_2)$ and keeping $q$ fixed at the endpoints, gives the Euler Lagrange equations for the system $(\partial L_q/\partial q)=dp/dt$, where we have defined a function $p(q,\dot q)\equiv (\partial L_q/\partial \dot q)$. (The subscript $q$ on $L_q$ is an indicator of the variable that is kept fixed at the end points.) The lagrangian contains only up to first derivatives of the dynamical variable and the equations of motion are - in general - second degree in the time derivative.

When the lagrangian $L_q$ depends on $\ddot q$ as well, the theoretical formulation becomes more complicated. For example, if the equations of motion become higher order, then more initial conditions are required to pose a well-defined initial value problem and the corresponding definition of path integral in quantum theory, using the lagrangian, is nontrivial. Interestingly enough, there exists a wide class of lagrangians $L(\ddot q,\dot q,q)$ which depend on $\ddot q$ but still lead to equations of motion which are only second order in time.

Let consider the following question: We want to modify the lagrangian $L_q$ such that the same equations of motion are obtained when - instead of fixing $q$ at the end points - we fix some other (given) function $C(q,\dot q)$ at the end points. This is easily achieved by modifying the lagrangian by adding a term $-df(q,\dot q)/dt$ which depends on $\dot q$ as well. (The minus sign is just for future convenience.) The new lagrangian is: \begin{equation}\tag{1} L_C(q,\dot q,\ddot q)=L_q(q,\dot q)-\frac{df(q,\dot q)}{dt} \end{equation} We want this lagrangian $L_C$ to lead to the same equations of motion as $L_q$, when some given function $C(q,\dot q)$ is held fixed at the end points. We assume $L_q$ and $C$ are given and we need to find $f$. The standard variation gives \begin{equation}\tag{2} \delta A_C =\int_{t_1}^{t_2} dt\left[\left(\frac{\partial L}{\partial q}\right)-\frac{dp}{dt}\right]\delta q -\int_{t_1}^{t_2} dt \frac{d}{dt}\left[\delta f -p \; \delta q \right] \label{elf} \end{equation} We will now invert the relation $C=C(q,\dot q)$ to determine $\dot q=\dot q(q,C)$ and express $p(q,\dot q)$ in terms of $(q,C)$ obtaining the function $p=p(q,C)$. In the boundary term in (2) we treat $f$ as a function of $q$ and $C$, so that the variation of the action can be expressed as: \begin{align} \delta A_C=&\int_{t_1}^{t_2} dt\left[\left(\frac{\partial L}{\partial q}\right)-\frac{dp}{dt}\right]\delta q+ \left[p(q,C)-\left(\frac{\partial f}{\partial q}\right)_C\right]\delta q\Big|_{t_1}^{t_2} -\left(\frac{\partial f}{\partial C}\right)_q\delta C\Big|_{t_1}^{t_2} \\ =&\int_{t_1}^{t_2} dt\left[\left(\frac{\partial L}{\partial q}\right)-\frac{dp}{dt}\right]\delta q+ \left[p(q,C)-\left(\frac{\partial f}{\partial q}\right)_C\right]\delta q\Big|_{t_1}^{t_2} \tag{3} \end{align} since $\delta C=0$ at the end points by assumption. To obtain the same Euler-Lagrange equations, the second term should vanish for any $\delta q$. This fixes the form of $f$ to be: \begin{equation}\tag{4} f(q,C)=\int p(q,C)dq +F(C) \label{f} \end{equation} where the integration is with constant $C$ and $F$ is an arbitrary function.

Thus, given a lagrangian $L_q(q,\dot q)$ which leads to certain equations of motion when $q$ is held fixed, one can construct a family of lagrangians $L_C(q,\dot q,\ddot q)$ which will lead to the {\it same} equations of motion when an arbitrary function $C(q,\dot q)$ is held fixed at the end points. This family is remarkable in the sense that $L_C$ will be a function of not only $q,\dot q$, but will also involve $\ddot q$. In spite of the existence of the $\ddot q$ in the lagrangian, the equations of motion are still of second order in $q$ because of the special structure of the lagrangian. (The results obtained above have an interpretation in terms of canonical transformations etc. which we purposely avoid since we want to stay within the lagrangian framework). So, even though a general lagrangian which depends on $\ddot q$ will lead to equations of higher order, there is a host of lagrangians with a special structure which will not.

  • $\begingroup$ At least I got the 2nd derivative idea and kinda danced around the b.c. one, lmho. Nice. I held off on setting $F=q \partial L / \partial \dot{q}$ because I couldn't get the transition $\dot{q} \rightarrow p$ right. So for now either $\partial F / \partial \dot{q}^i = 0$ or $C_i$ are known and $\partial F / \partial \dot{C}_i = 0$ (or the covariant version). $\endgroup$ – udrv Nov 23 '15 at 4:05

Lets define the following action : \begin{equation}\tag{1} S' = \int_{t_1}^{t_2} L' \, dt, \end{equation} where \begin{equation}\tag{2} L' = L - \lambda \, \frac{d}{dt} (q^k \, p_k). \end{equation} An arbitrary variation $\delta q^k$ gives the following : \begin{align} \delta S' &= \int_{t_1}^{t_2} \Big( \frac{\partial L}{\partial q^i} \; \delta q^i + \frac{\partial L}{\partial \dot{q}^i} \; \delta \dot{q}^i \Big) \, dt - \lambda \, \delta (q^k \, p_k) \Big|_{t_1}^{t_2} \\[12pt] &= \int_{t_1}^{t_2} \Big[ \frac{\partial L}{\partial q^i} - \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q}^i} \Big) \Big] \, \delta q^i \; dt + \int_{t_1}^{t_2} \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q}^i} \; \delta q^i \Big) \, dt - \lambda \, \delta (q^k \, p_k) \Big|_{t_1}^{t_2} \\[12pt] &= (\text{usual Euler-Lagrange variation of } S) + \big(p_i \; \delta q^i - \lambda \, \delta (q^k \; p_k)\big)\Big|_{t_1}^{t_2} \\[12pt] &= (\ldots) + \big( (1 - \lambda) \, p_k \; \delta q^k - \lambda \, q^k \, \delta p_k \big) \Big|_{t_1}^{t_2}. \tag{3} \end{align} The first term cancels from the Euler-Lagrange equations. To find a proper quantity that is fixed at the endpoints $t_1$ and $t_2$, the last term need to be written like this (look at the position of the indices. The bar is here on purpose, since there are two solutions) : \begin{equation}\tag{4} (1 - \lambda) \, p_k \; \delta q^k - \lambda \, q^k \, \delta p_k = -\; Q^k \; \delta P_k \quad \text{or} \quad \bar{P}_k \; \delta \bar{Q}^k. \end{equation} We could then fix $P_k$ or $\bar{Q}^k$ at $t_1$ and $t_2$ : $\delta P_k(t_1) = 0$ and so on. The two solutions are apparently a consequence of the symmetry between the variables $q^k$ and $p_k$ in the product $q^k \, p_k$. From the indices position, we can guess the following : \begin{align} Q^k &= \mathcal{Q}(q, p) \, q^k, & P_k &= \mathcal{P}(q, p) \, p_k, \tag{5} \\[18pt] \bar{Q}^k &= \bar{\mathcal{Q}}(q, p) \, q^k, & \bar{P}_k &= \bar{\mathcal{P}}(q, p) \, p_k. \tag{6} \end{align} From dimensional ground and indices position, we could expect that $\mathcal{P}(q, p) \propto (q^i \, p_i)^a$ and $\bar{\mathcal{Q}}(q, p) \propto (q^i \, p_i)^b$, where $a$ and $b$ are unknown exponents to be found as functions of $\lambda$. The variation of $P_k$ and $\bar{Q}^k$, substituted into the equation above, give us some simple algebraic equations that are easy to solve : \begin{equation} -\; a \, \mathcal{Q} \, (q^i \, p_i)^a \, p_k \; \delta q^k - (1 + a) \, \mathcal{Q} \, (q^i \, p_i)^a \, q^k \, \delta p_k = (1 - \lambda) \, p_k \; \delta q^k - \lambda \, q^k \, \delta p_k, \end{equation} or : \begin{equation} (1 + b) \, \bar{\mathcal{P}} \, (q^i \, p_i)^a \, p_k \; \delta q^k + b \, \bar{\mathcal{P}} \, (q^i \, p_i)^a \, q^k \, \delta p_k = (1 - \lambda) \, p_k \; \delta q^k - \lambda \, q^k \, \delta p_k. \end{equation} The complete algebraic solutions are then $a = \lambda - 1$, $b = - \lambda$ and \begin{align} \mathcal{Q}(q, p) &= (q^i \, p_i)^{1 - \lambda}, & \mathcal{P}(q, p) &= (q^i \, p_i)^{\lambda - 1}, \\[18pt] \bar{\mathcal{Q}}(q, p) &= (q^i \, p_i)^{- \lambda}, & \bar{\mathcal{P}}(q, p) &= (q^i \, p_i)^{\lambda}. \end{align} Finally, the quantities that are fixed at the endpoints are these : \begin{equation}\tag{7} P_i = (q^k \, p_k)^{\lambda - 1} \, p_i, \end{equation} or : \begin{equation}\tag{8} \bar{Q}^i = (q^k \, p_k)^{- \lambda} \, q^i. \end{equation} Now I'm still puzzled as why there are two solutions. Also, since $q^i \, p_i$ may turn out negative, these quantities may be ill-defined for some values of $\lambda$ ($\lambda = \tfrac{1}{2}$, for example).

Some special cases :

If $\lambda = 0$, the natural solution is given by the second one : $\bar{Q}^i = q^i$ (Dirichlet conditions). Why is there another solution (first one above) : \begin{equation}\tag{9} P_i = \frac{p_i}{q^k \, p_k} \; ? \end{equation}

If $\lambda = 1$, then the natural solution is given by the first one : $P_i = p_i$. Why is there another solution (second one above) : \begin{equation}\tag{10} \bar{Q}^i = \frac{q^i}{q^k \, p_k} \; ? \end{equation}

Someone confirms the calculations above, and have an interpretation of the dual solutions ? And what about the sign problem, for some specific values of $\lambda$ ?

EDIT 1 : It is funny to notice that the $Q^i$ and $P_i$ (or $\bar{Q}^i$ and $\bar{P}_i$) defined above (expressions (5) or (6)) are canonical transformations of the $q^i$ and $p_i$ : \begin{align}\tag{11} \{ Q^i, \, Q^j \} &= 0, &\{ P_i, \, P_j \} &= 0, &\{ Q^i, \, P_j \} &= \delta^i_j. \end{align}

EDIT 2 : The expression $q^k \, p_k$ is invariant under the canonical transformation \begin{align}\tag{12} Q^k &= (q^i p_i)^{-1} \, q^k, & P_k &= (q^i p_i) \, p_k, \\[12pt] q^k &= (Q^i P_i) \, Q^k, & p_k &= (Q^i P_i)^{-1} \, P_k, \tag{13} \end{align} so $q^k \, p_k \equiv Q^k \, P_k$, and it's easy to verify the identity \begin{equation}\tag{14} - \; q^k \; \delta p_k \equiv P_k \; \delta Q^k. \end{equation} So it is not surprising after all that there's a second solution to the problem. This $Q^k$ may be interpreted as another quantity that is fixed at the boundaries, in the case of $\lambda = 1$ ($p_k$ and $Q^k$ are both "natural" quantities that are fixed in this case).

  • $\begingroup$ When you define the total derivative in the Lagrangian, are the $p_k$-s explicitly canonical momenta or just functions? $\endgroup$ – udrv Nov 19 '15 at 14:40
  • $\begingroup$ They are functions of $q^i$ and $\dot{q}^i$, since by definition $p_i = \partial L \,/\, \partial \dot{q}^i$. $\endgroup$ – Cham Nov 19 '15 at 14:44
  • $\begingroup$ Note that when we write $\delta p_i$ in equations (3) and (4) of my answer, you could also express that as a linear combination of $\delta q^i$ and $\delta \dot{q}^i$, but then it would give the same results at the end. $\endgroup$ – Cham Nov 19 '15 at 14:46
  • $\begingroup$ This may be a stupid question, but is it necessary to define the $p_i$-s like this? Asking because if you want to retrieve boundary conditions for the canonical momenta $\partial L / \partial \dot{q}^i = p_i$ for $\lambda = 1$, I think it suffices to take $p_i = p_i(t)$ as functions that impose the boundary values. $\endgroup$ – udrv Nov 19 '15 at 15:21
  • 1
    $\begingroup$ No, no, I didn't mean as in defining $\partial L / \partial \dot{q}^i$ as a function of time, that defeats the purpose. Maybe a change in notation to drive the point: Instead of $L-\frac{d}{dt}\lambda q^i p_i$ take $L - \frac{d}{dt} \lambda q^i \xi_i$, where $\xi_i = \xi_i(t)$. The EL equations remain the same (no $\xi$-s), while the boundary condition becomes $((\partial L /\partial \dot{q}^i)\delta q^i - \delta (\lambda q^i \xi_i)) \Big|_{t=t_{1,2}} = (\partial L /\partial \dot{q}^i - \lambda \xi_i)\delta q \Big|_{t=t_{1,2}} = 0 $. The $\xi$-s would define boundary values only, not the EL. $\endgroup$ – udrv Nov 19 '15 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.