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Ref. 1 mentons that you can achieve the momentum space Lagrangian by doing a so called double Legendre transform. It goes on to write:$$ K(p,\dot{p},t) ~=~ L(q,\dot{q},t) - p\dot{q} - q\dot{p},\tag{5.63}$$ where $K(p,\dot{p},t)$ is the momentum space Lagrangian. I don't see how this is a Legendre transform as there seems to be a sign error in the way the Legendre transform has been done. In general when we say a Legendre transform we mean $H(p,q,t) = p\dot{q} - L(q,\dot{q},t)$, so according to that for our momentum space Lagrangian it should be $K(p,\dot{p},t) = p\dot{q}+q\dot{p} - L(q,\dot{q},t)$ but that doesn't seem to be the case. Can someone point me in the right direction because the sign matters when we try to get different forms of the generating functions.

References:

  1. Hand & Finch, Analytical mechanics, Ch 5, pg 190, eq. (5.63).
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    $\begingroup$ The overall sign really doesn't matter. You can multiply a Lagrangian by any nonzero number and it won't change the results. $\endgroup$ – knzhou Jul 9 '18 at 9:37
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This means to apply two consecutive Legendre transforms : $$K(p, \dot p, t) = J(q, p, t) - q \dot p = L(q, \dot q, t) - q \dot p - p \dot q$$

Now the Hamiltonian is defined as $H = p \dot q - L = - J$. In fact, this doesn't matter, because multiplying the Hamiltonian / Lagrangian by a non-zero constant gives the same dynamics equations.

I think that the reason one poses $K(p, \dot p, t) = L(q, \dot q, t) - q \dot p - p \dot q$ and not $K(p, \dot p, t) = q \dot p + p \dot q - L(q, \dot q, t)$ is that one wants the momentum space Lagrangian to resemble the original Lagrangian. The reason why one has $H = p \dot q - L$ must be that we want it to match with energy when it is possible.

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I) In this answer, we would like to incorporate the double Legendre transformation (5.63) via extended action principles in the spirit of my Phys.SE answer here. Compare the two extended actions $$\begin{align} L_E(q,\dot{q},v,p,t)~:=~& p_i(\dot{q}^i-v^i)+L(q,v,t) \tag{LE}\cr \stackrel{\text{int. out } v^i}{\longrightarrow}&\quad L_H(q,\dot{q},p,t)~=~p_i\dot{q}^i-H(q,p,t), \tag{LH}\cr &\qquad\qquad\qquad H(q,p,t)~:=~\sup_v \left(p_iv^i-L(q,v,t)\right),\cr \stackrel{\text{int. out } p_i}{\longrightarrow}&\quad L(q,\dot{q},t)\tag{L}\end{align}$$ and $$\begin{align} K_E(q,\dot{q},p,f,t)~:=~& -q^i(\dot{p}_i-f^i) +K(p,f,t) \tag{KE}\cr \stackrel{\text{int. out } f_i}{\longrightarrow}&\quad K_H(q,\dot{q},p,t)~=~-q^i\dot{p}_i-H(q,p,t), \tag{KH}\cr &\qquad\qquad\qquad H(q,p,t)~:=~\sup_f \left(-q_if^i-K(p,f,t)\right), \cr \stackrel{\text{int. out } q^i}{\longrightarrow}&\quad K(p,\dot{p},t) \tag{K}\end{align}$$

II) Notice that the difference $$L_H-K_H~=~\frac{d(p_iq^i)}{dt} $$ is a total time derivative. This shows that $L_H$ and $K_H$ have the same Euler-Lagrange (EL) equations (namely Hamilton's equations). When we integrate out $p_i$ xor $q^i$ we arrive at $L$ and $K$, respectively, thereby implementing the double Legendre transformation (5.63).

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Well, one may argue that eq. (5.63) is in fact precisely two consecutive Legendre transformations of the standard form$^1$

$$\text{ function } + \text{ Legendre transformed function } = \text{ product of Legendre variables }.\tag{0}$$

In detail, we have

$$\begin{align} H(-q,p,t) ~+~L(-q,v,t) &=~ p_iv^i \quad\quad\longleftarrow\text{ (Legendre transf. in }p_i\leftrightarrow v^i) \tag{1}\cr\cr H(-q,p,t) ~+~K(p,f,t) &=~ -\!q^if_i \quad\longleftarrow\text{ (Legendre transf. in } -\! q^i\leftrightarrow f_i)\tag{2}\cr\cr & \Downarrow\cr\cr L(-q,v,t)~-~K(p,f,t) &~=~ p_iv^i + q^if_i \qquad\longleftarrow\text{ (Double Legendre transf.)} \tag{3}\end{align}$$ Here we have identified $$v^i~=~\dot{q}^i \quad\text{and}\quad f_i~=~\dot{p}_i.\tag{4}$$ The minus sign on the rhs. of eq. (2) is introduced in order to ensure the standard sign convention $$ \dot{p}_i~=~f_i~=~\frac{\partial H}{\partial(-q^i)} ~\equiv~ -\frac{\partial H}{\partial q^i} \tag{5}$$ for the Hamilton's equations. In order for eq. (2) to be of the form (0), we therefore have introduced the argument $-q^i$ rather than $+q^i$. Of course, every function of $+q^i$ can be re-expressed as a function of $-q^i$, and vice-versa, so this is not a limitation. [Alternatively, one may work with the argument $+q^i$, and accept a sign difference between eqs. (0) & (2).]

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$^1$ One can of course introduce various signs in the definition (0), but one should keep in mind that it will have consequences for the intertwining relations.

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