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Generating functions and Legendre transforms

  1. Apr 22, 2015 #1
    I am confused about the ''4 basic types'' of generating functions. I have searched for this a bit on google but haven't found anything that truly made the click for me on this concept so I'll try here:

    What I do understand and need no elaboration on:

    1) When considering the Hamiltonian and Kamiltonian for which minimal action still holds, their sum together with the sum of two other terms pq' and PQ', can differ up to a total derivative of a function ##F(Q,P,t)## or ##F(q,p,t)##. So I understand the reasoning behind introducing the concept ''generating function''.

    2) Once you give me one of the four types, I can derive the canonical transformations that follow from this.

    What I don't understand.

    1) Why any generating function different than ##F(q,p,t)## or ##F(Q,P,t)## is introduced because when considering this total derivative as mentioned earlier - we are taking this total derivative inside an action integral that goes over either ##q,p## or ##Q,P## independently. Basically we want that F is the same in both end points of the action path under the integral, but the integral is taken either over variations of ##q,p## or variations of ##Q,P##

    2) When 1 is answeren, and it would becoem clear why it makes sense to introduce different kinds of generating functions I still do not get how exactly they are related by the Legendre transforms.

    I hope someone can enlighten me a bit on this subject.
     
  2. jcsd
  3. Apr 27, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Apr 28, 2015 #3

    vanhees71

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    I don't know what a Kamiltonian is. In the usual Hamiltonian formulation of the Hamilton principle of least (stationary) action, the trajectory of a physical system in phase space is given by the stationary points of the phase-space action functional
    $$S[q,p]=\int_{t_1}^{t_2} \mathrm{d} t [\dot{q} \cdot p-H(q,p)].$$
    The trajectories are unconstraint wrt. the momenta and have fixed boundaries for the position-space variables, ##q(t_1)=q_1=\text{const}## and ##q(t_2)=q_2=\text{const}##.

    Now suppose that ##(Q,P)## are new phase-space variables, i.e., there exists a diffeomorphism ##Q=Q(q,p;t)##, ##P=P(q,p;t)## such that the Hamilton equations of motion are forminvariant. This is the case if the variation of the action functional is invariant, this implies that the variation of
    $$\Delta S=\int_{t_1}^{t_2} \mathrm{d} t \{ [\dot{q} \cdot p-H(q,p)]- [\dot{Q} \cdot P-H'(Q,P)] \}$$
    identically vanishes, which is the case if and only if the integrand is a total differential of a function ##f(q,Q,t)## (because only the boundary conditions for ##q## and ##Q## but not of ##p## and ##P## are fixed!), which implies
    $$\mathrm{d} t (H'-H) + \mathrm{d} q \cdot p - \mathrm{d} Q \cdot P=\mathrm{d} f=\mathrm{d} t \partial_t f+\mathrm{d} q \cdot \partial_q f + \mathrm{d} Q \cdot \partial_Q f.$$
    Comparing the coefficients of the differentials on both sides implies that the canonical transformation is defined by the generating function ##f=f(q,Q,t)## and the transformation is given by the relations
    $$H'=H+\partial_t f, \quad p=\partial_q f, \quad P=\partial_Q f. \qquad (1)$$
    Now you can derive all kinds of other generating functions, each being dependent by one of the old and one of the new canonical variables. This is done by a technique called Legendre transformation. E.g., define
    $$g=f+Q \cdot P.$$
    This gives
    $$\mathrm{d} g = \mathrm{d} t \partial_t f + \mathrm{d} q \cdot \partial_q f + \mathrm{d} Q \cdot \partial_Q f + \mathrm{d} Q \dot P + \mathrm{d} P \dot Q.$$
    Now, because of (1) you have
    $$\mathrm{d} g = \mathrm{d} t \partial_t f + \mathrm{d} q \cdot p + \mathrm{d} P \dot Q,$$
    which implies that ##g=g(t,q,P)## and
    $$\partial_t G=\partial_t f, \quad \partial_q g=p, \quad \partial_P g=Q.$$
    So you get a canonical transformation with a generating function ##g(t,q,P)## with the transformed Hamiltonian
    $$H'=H+\partial_t f=H+\partial_t g.$$
    You can figure out the other two forms of generating functions yourself, just using the other appropriate Legendre transformations.
     
  5. Apr 28, 2015 #4
    The Kamiltonian is what you use as ##H'##, I've read some authors use it like that.

    Thanks for the explanation, it is now very clear why it is exactly the ##f(q,Q,t)## that comes out of the integral in the beginning. I guess that my problem is understanding what the Legendre transformations really do on an intuitive level, so I will read up on that first and then come back to the second part of your explanation.
     
  6. May 31, 2015 #5
    Hello this is an old threat but I have some additional questions in regards to this topic if you don't mind:

    1) Consider the second integral you wrote down. You didn't do it explicitly but the reasoning to go from that to the conclusion about being a total differential happens when taking the variation of that expression (second integral). We ask that the variation of that expression is zero. However, this variation, does it allow for independent variations in all 4 coordinates. That is, do we have to say ##\delta q (t)## implies ##\delta Q(t)##? Or we really can pick ##\delta q(t)## and ##\delta Q(t)## freely?

    2) If so, what's up with treating q,p,Q,P as independent until you arrive at the final partial derivatives relations of the generating function? Can you elaborate on why that's okay on an intuitive level.
     
  7. May 31, 2015 #6

    vanhees71

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    In the Hamiltonian formalism the phase-space variables ##(q,p) \in \mathbb{R}^{6n}## (##n##=number of particles) are independent variables. The variation in the Hamilton principle of the action,
    $$A[q,p]=\int_{t_1}^{t_2} \mathrm{d} t [\dot{q}^{\mu} p_{\mu} - H(t;q,p)]$$
    is understood as being unrestricted concerning the variation of the canonical momenta and for fixed boundary conditions for the configuration variables, i.e., ##\delta q(t_1)=\delta q(t_2)=0##.
    A canonical transformation is a (locally) invertible sufficiently smooth mapping
    $$(Q,P)=[Q(t;q,p),P(t,q,p)]$$
    that keeps the canonical Poisson-bracket relations
    $$\{q^{j},q^{k} \}=\{p_j,p_k \}=0, \quad \{q^j,p_k \}=\delta^{j}_{k}$$
    unchanged. In short: A canonical transformation is a symplectomorphism.

    Using the generating functions as described above, which are always a function of one set of new and one set of old phase-space coordinates, guarantees the invariance of the canonical Poisson-bracket relations or, equivalently, the invariance of the variation of the action and thus the forminvariance of the Hamilton canonical equations of motion of the particles in phase space.
     
  8. May 31, 2015 #7
    We only proved that those Poisson-bracket relations remain undependent on the coordinate system if the canonical transformations are time independent, ie. when ##H(q,p,t)=H'(Q,P,t)##. The way you write it seems that it holds even for time dependent canonical transformations given by a ##F(q,Q,t)## explicit time dependent generating function?
     
  9. May 31, 2015 #8

    vanhees71

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    No, the only difference is that then you have an additional term in the Hamiltonian
    $$H'(Q,P,t)=H(q,p,t)+\partial_t g.$$
    See #3.
     
  10. May 31, 2015 #9
    So just to confirm, because it's important to get it right. The following statements are all equivalent?

    1) The transformations ##Q(q,p,t)## and ##P(q,p,t)## are called canonical if, for any old Hamiltonian ##H(q,p,t)## where the old coordinates satisfy the equations, there exists a new hamiltonian ##H'(Q,P,t)## for which the new coordinates satisfy the equations and give the same results. The old and new hamiltonians are not necessarily the same, and the precise form can be analyzed using generating functions.

    2) The transformations ##Q(q,p,t)## and ##P(q,p,t)## are such that for any functions ##f(q,p,t)## and ##g(q,p,t)## the following holds : ##\{ f,g \} _{q,p}=\{ f,g\} _{Q,P}##

    3) The transormations ##Q(q,p,t)## and ##P(q,p,t)## obey the following brackets: ##\{ Q_i,Q_j\} _{q,p}=0## , ##\{ P_i,P_j\} _{q,p}=0##, ##\{Q_i,P_j\} _{q,p}=\delta_{ij}## and vice versa the same.

    I'm pretty sure we've proven the equivalence of these statements for transformations ##Q(q,p)## and ##P(q,p)## where there is no time dependence. So just confirming that the case is the same for time dependent c.transformations.
     
  11. Jun 2, 2015 #10

    vanhees71

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    Yes, a possible time-dependence of the generating functional doesn't make much trouble. It only occurs in the formula for the new Hamiltonian
    $$H'(Q,P,t)=H[q(Q,P),p(Q,P),t]+\partial_t g.$$
     
  12. Jun 2, 2015 #11
    Yeah but the Hamiltonians being different changes everything about the proof of the three equivalencies above. At least the way we did it in class. We were looking for transformations that have for any hamiltonian ##\{ Q,H\} _{q,p}=\{ Q,H\} _{Q,P}## which is the same as asking that both obey the equations of motion for their respective coordaintes. This results in asking for transformations that keep the poisson bracket of two functions invariant which in turn results in the latter equivalence. If the Hamiltonians are not the same, the first assumption is not correct and that method of proof fails.
     
  13. Jun 2, 2015 #12

    vanhees71

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    Where is your proof from?

    The mathematics underlying the canonical transformations is that they are symplectomorphisms. That means you can define them as those leaving the fundamental Poisson brackets invariant. E.g., if you take the "original" choice of coordinates for the generating function
    $$g=g(q,Q,t)$$
    you have
    $$p_k=\frac{\partial g}{\partial q^k}, \quad P_k=-\frac{\partial g}{\partial Q^k}.$$
    This implies
    $$\frac{\partial p_k}{\partial Q^j}=\frac{\partial^2 g}{\partial q^k \partial Q^j}=-\frac{\partial P_j}{\partial q^k}.$$
    With the other three choices of variables for the generating function, which are obtained by Legendre transformations from the original one, you get more such constraints on the transformations, and those allow you to prove that for any phase-space distribution functions ##A## and ##B## the Poisson brackets ##\{A,B \}## are the same, no matter whether you use the old or the new coordinates to evaluate them.

    Now, for the equations of motion, it becomes of course important whether the transformations are explicitly time dependent since time derivatives are involved, and the time-derivatives in the Hamilton principle are to be read as total time derivatives. So you have
    $$\frac{\mathrm{d} Q^k}{\mathrm{d} t}:=\dot{Q}^k=\dot{q}^j \frac{\partial Q^k}{\partial q^j} + \dot{p}_j \frac{\partial Q^k}{\partial p_j} + \frac{\partial Q^k}{\partial t}.$$
    To get the same equations of motion you exactly need the addition time derivative of ##g## for the new Hamiltonian. It's of course much more elegant to use Hamilton's principle of least action to derive this than to do all the partial derivatives, but you can do it, of course also in this way. You always need the constraints of the canonical transformation being a symplectomorphism.
     
  14. Jun 2, 2015 #13
    Oh right, I think your post gave me an idea for how to tweek our proof to include time transformations. The proof I'm talking about isn't really from a book or anything, the instructor just did it on the blacbkoard. It wasn't all that formal since this is an introductory undergrad class, but the core of the reasoning of the proof we used was contained in my last post. (1) Noticing that asking for the EOM to hold for both hamiltionians is similar to asking for {f,g} being invariant for any functions. (2) Then looking at when {f,g} is invariant and finding that this is only the case if relations like {Q_i,Q_j}= 0 are fufilled. It doesn't matter if the transformations are time dependent or not for (2), but for (1) it does. However I think seeing the final expression in your post made me see that (1) still holds for time dependence.
     
  15. Jun 2, 2015 #14

    vanhees71

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    That's true, but if the transformations are explicitly time-dependent you get the additional time derivative from the explicit time dependence (which I denoted by a partial time derivative), and this you don't get with the Poisson brackets. The dot over a phase-space function in the equations of motion is the total time derivative along the trajectory in phase space. So you get for any such function ##A(q,p,t)##
    $$\dot{A}=\{A,H \}+\partial_t A.$$
    So the explicit time dependence has to be taken care of seprately and is not taken care of by the Poisson bracket. So you have
    $$\dot{Q}^k=\{Q^k,H \} + \partial_t Q^k=\frac{\partial H}{\partial P_k}+\partial_t Q^k.$$
    Now using the generating functional in the form ##g=g(q,P,t)## you get
    $$Q^k=\frac{\partial g}{\partial P_k} \; \Rightarrow \; \frac{\partial Q^k}{\partial t}=\frac{\partial^2 g}{\partial t\partial P_k}.$$
    Thus to make
    $$\dot{Q}^k=\{Q^k,H' \}=\frac{\partial H'}{\partial P_k},$$
    you need to set
    $$H'=H+\partial_t G.$$
    It's easy to show that this also fits the equation of motion for ##P_k##.
     
  16. Jun 7, 2015 #15
    Hello, don't feel obligated to answer because I feel like you have already given tons of help. However there remains a bit of confusion so it never hurts to ask.

    It seems to me (correct me if I'm wrong) that your two posts above just prove the generating function method in a different way than the variational theorem. However what I'm looking for is slightly different, namely the following:

    Assume that we know all the theory there is that is given by the variational method. So we know the 4 types of generating functions, we know how these functions give us relations between the coordiantes, we know that for these relations the transformations are canonical. What I want to prove now is:

    1) Pick a generating function. (So now we know that the transformations are canonical and so on)

    2) Show that this implies that for any functions ##f## and ##g## the Poisson bracket ##\{ f , g \}## remains invariant relative to both coordinate systems.

    I have tried this for a bit and I'm using the earlier said method to first show that for ##H'## must hold that ##\{ Q , H' \}## is invariant relative to both coordinate systems. Because since Q and H' are arbirtary functions of (q,p,t) it also should hold for any two functions ##f## and ##g##.

    At this point all I know is that the transformations are canonical and that the equations of motion hold for both hamiltonians for their respective coordinates. I have tried to start from this and show that ##\{ Q , H' \}## is invariant but this always results to showing that ##\frac{\partial Q}{\partial t}= \{ Q, \frac{\partial F}{\partial t} \} _{q,p}## where ##F## is the generating function. After writing this out it seems to me that it is not always an equality. Where am I wrong?

    EDIT: So if you see post #9 :

    I have no problems proving (2) <=> (3)

    What this post is about is proving (1) => (2)
     
  17. Jun 7, 2015 #16

    vanhees71

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    For this way to prove it you need the relations between old and new canonical variables, I started to derive in #12. It's just a bit work to write them down, using all possible combinations of old and new coordinates for the generating functions. If you have time-dependent generating functions, also write down the relation for partial time derivatives and derivatives wrt. the phase-space coordinates. This is then a complete set of constraints for the transformation to be canonical. My logic of proof would be

    Hamilton principle (in Hamilton formulation) -> definition of canonical transformations as those which leave the variation of the action invariant -> variational calculus then leads to the existence of the generating function in terms of ##g(q,Q,t)## -> Legendre transformation lead to all other kinds of writing the generating functions like ##g(p,Q,t)##, etc. -> relations between partial derivatives of the old and new variables -> covariance of Poisson brackets

    Now from the covariance of Poisson brackets you should be able to derive everything also going backwards, and then you have for sure the equivalence between all these steps.
     
  18. Jun 7, 2015 #17
    So I'm trying to prove a subcase of the theorem. Trying to prove that ##\{ Q, H \}## is invariant.This is as far as I get, once I start plugging in one of the 4 functions it doesn't really progress much.:

    http://imgur.com/0rdHKmY

    So basically if the final line you see is equal to ##\frac{\partial Q}{\partial t}## then I've proven what I wanted.


    Note: I'm even doing one degree of freedom to make it better looking.
     
  19. Jun 8, 2015 #18

    vanhees71

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    It's not clear to me which generating functional you use. It should work with all 4 forms. Let's do it with the most used one ##F(q,P,t)##. Then you have
    $$p=\partial_q F, \quad Q=\partial_P F, \quad H'(Q,P,t)=H(q,p,t)+\partial_t F.$$
    Now you get
    $$\{Q,H'\}=\frac{\partial H}{\partial P}+\frac{\partial^2 F}{\partial t \partial P}=\frac{\partial H}{\partial q} \frac{\partial q}{\partial P} + \frac{\partial H}{\partial p} \frac{\partial p}{\partial P}+\frac{\partial Q}{\partial t}.$$
    Now from the conditions for the transformation to be canonical you take
    $$\frac{\partial q}{\partial P}=-\frac{\partial Q}{\partial p}, \quad \frac{\partial p}{\partial P}=\frac{\partial Q}{\partial q}$$
    to find
    $$\{Q,H' \}=[Q,H]+\frac{\partial Q}{\partial t},$$
    which is precisely what one wants to show. The time derivative of ##Q## is given by the Poisson bracket with the original ##H## plus the partial derivative from the explicit time dependence.
     
  20. Jun 8, 2015 #19
    I see now. I'd like to thank you for all the help. In an ideal world I should have struggled on to find it myself, however I still have to learn these things for classes and sometimes don't have the time to ponder on a problem for a week. Really great to find some tips and guidance then.
     
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