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Why Legendre transform?

  1. May 12, 2009 #1
    The Legendre transformation creates a new function which contains the same information as the old, but is of a different variable. This is used to obtain the Hamiltonian from the Lagrangian. My question is, why is there more advantageous than simply rearranging the q's for p's and plugging them into the Lagrangian? What was the motivation for using a Legendre transform rather than a simple change of variable?
  2. jcsd
  3. May 13, 2009 #2
    No one seems to be answering. Does the question not make sense?
  4. May 13, 2009 #3


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    Maybe you only see the workings of a Legendre transformation for a very specific case. Have you seen the thermodynamics application of the Legendre transformation in going from one set of thermodynamics state equation to another? I'm not sure if you can do that simply by "rearranging".

    See this: http://www.ruf.rice.edu/~ctlee/chapter4.pdf

  5. Dec 15, 2009 #4
    This is a valid question which so far has not received a valid answer.
    The application of the Legendre Transform (LT) in thermodynamics is best considered as ScienceFiction: the oldest suggestion that the functions U,H,F,G are transforms of each other goes back to 1950: but the functions themselves are defined 1850-1882. The LT assumes that (dU/dV)=p for ALLl physical systems while it is well known that for ideal gases U is not a function of V, and for solids the only p around is the external pressure, which is independent from U altogether.
    The idea that the Hamiltonian is the LTof the Lagrangean could not traced back by me to before 1950, but it seems certain the it was not Hamilton who conceived it. The interesting question is rather how Lagrange arrived at the minus sign in his equation: upon transform this sign changes, which is more relevant than the substitution of velocity by momentum.
    The question remains open as far ia I am concerned.
  6. Dec 17, 2009 #5
  7. Dec 18, 2009 #6
    It empowers you to work with intensive variable rather than extensive variable
    http://www.actionurl.com/b1oc6 [Broken]
    Last edited by a moderator: May 4, 2017
  8. Dec 21, 2009 #7
    Why the Legendre transform (LT) in thermodynamics ??.

    Callen, one of the first to convert to the LT-approach of thermodynamics, writes in 1987 (Thermodynamics etc, p138): "the introduction of the transformed representations is purely a matter of convenience". Zia etal (in the above link provided), state more or less the same (p10): Physically, we have motivated the LT and as facilitating a way of thinking about a complex problem in terms of a more naturally controllable independent variable.
    Zia etal add a condition to the existence /usefulness of the LT a function F(x) with respect to x: F(x) must be strictly convex in x. Strictly convex means F"(x)>0. To test for convexity we need the specification of F(x), eg F(x) = ½mx2. If we do not know this specification we cannot differentiate and and cannot test for strict convexity.

    If we limit the discussion at this point to thermodynamics (because the devastation there is more profound then in mechanics), we have now at least two criteria to test the LT-hypothesis: Convenience and convexity.
    1. First of all: there is nothing convenient about the Legendre transform. It is more simple to define H as H=U+pV as the sum of two forms of energy (where H and U can be expressed in variables of choice), then to define H(p,S) = (dU/dV)SV -U(V,S), ie as an LT of U(S,V), where one has to prove that (dU/dV)S = -/+p for all physical systems, and needs U written as a function of S and V. Zia etal do not report that the LT of U(x,y) does not yield H,F or G, unless U(x,y)=U(S,V), so we need the function U(S,V), which is very inconvenient. Note that there is also a sign-problem to make H a true LT and make the 2 approaches match.
    2. In general (for an arbitrary, but well defined system), both U(V,S) and H(p,S) are unknown (unspecified) functions: This means that it is not known whether these functions are strictly convex, and also that the actual LT cannot be performed.
    3. If we want to know the enthalpy H of 100 g water (example of a well defined system), we look it up in a table (where it arrived through calorimetric measurements, not through a LT) and find it to be -15.9 kJ, at T=298K and p=1 atm. We can calculate H as function of T by using the heat-capacity at constant pressure. Hence H(100g water, 1 atm) = -15.9 + 0.418(T-298). This is a convenient function, already expressed in an intensive variable. Applying a LT on this function will get you absolutely nowhere.
    4. Zia writes: For historic reasons, LT variables (in thermodynamics) are not always chosen as conjugate pairs. This defeats the whole idea of the LT. In addition one has to fiddle with signs, which destroys the property of duality, inherent in a true LT.

    It can be concluded that there is no reason to assume that H is a LT of U, and that it surely would be very inconvenient if it would (Zia etal note problems in teaching). For F and G (as LT) we apply one of the principles of Roman Law: Quod gratis assertibus gratis negatur. =What is presented without proof can be denied without proof. Note that U,H, F and G have physical meaning: they are not mathematical auxiliary functions, as suggested by LT.

    The why-question for application of the LT in mechanics (Lagrangean<=>Hamiltonian) remains unresolved here, but must be addressed as well.
  9. Dec 21, 2009 #8
  10. Jan 3, 2011 #9
    The assumption is that (dU/dV)=p at constant entropy. At worst, you can't say that U is not a function of V, just that it is a constant function of V. But that's U(T,V). If you look at U(S,V) it has an explicit volume dependence.
  11. Jan 4, 2011 #10
    mathematically 'constant entropy' is easy (dS=0): physically it is next to impossible.

    Its interesting that (for an ideal gas) the property U is independent from V as shown experimentally by Guy-Lussac and Joule: now by changing the variables this property becomes a function of volume ??? Can you do this trick with any quantity ?
    Hence you can make U a function of the height of the Eifel tower? (assumed constant).
    Note that the Legendre transform of U(T,V) is not a known thermodynamic quantity.

    I would like to see U(S,V) and H(S,p) for a glass of water. If there is no such function, it will not be of much practical value.
  12. Jan 4, 2011 #11


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    U most certainly is a function of V when the entropy is held fixed; your partial derivative for a closed system should be written

    [tex]\left(\frac{\partial U}{\partial V}\right)_S=-P[/tex].

    It doesn't bode well for your argument that all the thermo texts are wrong when you have an error in your first quantitative statement.
  13. Jan 5, 2011 #12
    Thanks for pointing out this error in my first post in this thread (dec15 09).
    However this error does not affect my arguments against the Legendre transform in my second post (dec21 09).
  14. Jan 10, 2011 #13
    Since we're talking about the mathematics, the truth value of the second part of your statement is irrelevant.

    Yes, because we all know the height of the Eiffel tower is at least as relevant to thermodynamics of a gas as the volume that contains that gas.... Seriously though, if you had put even a little thought into what it would take to replace T in U(T,V) to get a function U(S,V) you would have seen the answer.

    Btw, see the Wikipedia page on the ideal gas: http://en.wikipedia.org/wiki/Ideal_gas#Thermodynamic_potentials
    They write down explicit forms for U(S,V) and H(S,P).

    I think we both know you won't find explicit functions derived from first principles for a glass of water for U and H of any variables. You'll find tables which can be used to manually integrate to find changes in U, H, or S, or perhaps model functions.

    And yes, functions of entropy don't have a whole lot of practical value, because it's easier to measure temperature and pressure. Which is the point of doing the Legendre transformations, btw, since we can change from using functions of S, E, V which are difficult to measure to functions of N, T, P which are easy to measure. But you can go to the tables and do the integrations yourself and change variables to plot H(S,P) at a specific pressure if you really wanted to.
  15. Jan 12, 2011 #14
    Zemansky-Dittman(Seventh Edition) traces this use of formalism back to Williard Gibbs(1870s)
  16. May 10, 2013 #15
    If the LT is the only reversible procedure to change the coordinates of a function, why don't we see it being applied more often? Does the physicist not need this nice behaviour of differentials under every change of variable? Without giving it a lot of thinking, the LT should play a role in fields like differential geometry where changing the coordinates/variables of a function defined on a manifold is a routine.
  17. May 11, 2013 #16
    My understanding is if we simply rearrange the generalised velocity in the Lagrangian, then generalised momentum is not an independent variable which it becomes under a LT to Hamilton. And then it has a number of advantages for example it is not restricted to different representations of momentum in different coordinate systems(as in Lagrangian), but abstract quantities under suitable generating functions.
  18. May 11, 2013 #17
    Let us consider a trivial example. The functions [itex]f_1(x)=ax^2[/itex] and [itex]f_2(x)=a(x+c)^2[/itex] have the same transformation [itex]\tilde{f}(u)=\frac{u^2}{4a}[/itex] under a variable change: [itex]x→u=\frac{df_i}{dx}[/itex] , [itex]i=1,2[/itex]. We see that the transformation is not reversible.
    As a matter of fact, the LT yields a proper result, by definition, to those transformations that aim to replace a variable with the derivative of the function with respect to this variable. So on this account, I could partially undestand why we don't encounter this special transformation in other fields of physics like GR.
  19. Nov 26, 2016 #18
    No, the LT of [itex]f_2[/itex] is [itex]\tilde{f}(u)=\frac{u^2}{4a}-cu[/itex]. The LT is invertible, and is its own inverse.
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