Undergrad Generalized coordinates basic question

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The discussion clarifies the distinction between the Hamiltonian and Lagrangian formulations in classical mechanics, emphasizing that the Hamiltonian must be expressed in terms of generalized coordinates and their corresponding momenta, not velocities. The confusion arises from the equation H(qi, pi, t) = ∑i=1npiq⋅i - L(qi, q⋅i,t), which appears to include velocities, but is correct as it relates to the Hamiltonian's definition. The process of deriving the Hamiltonian involves multiple steps, including computing momentum from the Lagrangian and expressing it solely in terms of position and momentum. The conversation also touches on the historical context of mixed formulations, noting their rarity in modern texts. Ultimately, the Hamiltonian plays a crucial role in determining the equations of motion, reinforcing the importance of its proper formulation.
nomadreid
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From "A Student's Guide to Langrangins and Hamiltonians", Patrick Hamill, Cambridge, 2017 edition.
Apologies: since I do not know how to put dots above a variable in this box, I will put the dots as superscripts. Similarly for the limits in a sum.
On page 6,
"we denote the coordinates by qi and the corresponding velocities by qi."
Further he terms pi the generalized momenta.
Then, on page 97, using L as the Lagrangian,
" H(qi, pi, t) = ∑i=1npiqi - L(qi, qi,t)" [equation (4.8)]
The function H is called the Hamiltonian...Keep in mind that the Hamiltonian must be expressed in terms of the generalized momentum. An expression for H involving velocities is wrong." (Italics in the original.)

But it is not clear to me why equation 4.8 is not also in terms of velocities, i.e., qi

(Perhaps this should be a mathematics threads rather than a QM thread; I place it here because of the context.)
 
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I don't get that sentence either, since there is a 1 to 1 correspondence between momenta and velocities.
 
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By the way, a dot is obtained by the code ##\dot{q}_i##, "\dot{q}_i".
 
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But the book is in fact right. For the Hamiltonian formulation of Hamilton's principle the Hamiltonian must be expressed in terms of generalized coordinates and their canonical momenta, not generalized velocities. The Lagrangian formulation deals with the generalized coordinates and their assiciated generalized velocities. Of course you can use any form of "mixture", which is due to Routh. It's rarely found in modern textbooks, and I've never found any application which is easier solved using this most general description than with Lagrange or Hamilton. If you are interested in it, you find it in Sommerfeld's Lectures on Theoretical Physics, vol. 1 (which book series I cannot recommend enough; for me it's the master piece in theoretical-physics textbook writing which sets the very high standard of any text-book writing in this field).
 
Obviously, you can convert back and forth between momenta and velocities. But in Hamiltonian dynamics, the Hamiltonian serves a role beyond its numerical value (which is of course, not changed by substituting velocities for momenta or vice-versa). The Hamiltonian equations of motion are (I'm just going to use a single variable, x and the corresponding momentum, p. The generalization to multiple generalized coordinates and momenta is straight-forward.)
  1. \frac{dx}{dt} = \frac{\partial H}{\partial p}
  2. \frac{dp}{dt} = - \frac{\partial H}{\partial x}
In order to compute the partial derivatives on the right side of the equalities, you have to write H in terms of x and p

So there's really four steps in obtaining the hamiltonian:
  1. First, compute the momentum as a function of position and velocity, using the Lagrangian: p = \frac{\partial L}{\partial \dot{x}}
  2. Use that to compute the velocity as a function of position and momentum: \dot{x} = F(x,p)
  3. Then, compute the "mixed" expression: H(x,p,\dot{x}) = p \dot{x} - L
  4. Finally, use the result of 2 to rewrite all occurrences of \dot{x} in terms of p and x, to get an expression: H(p,x) that only involves p and x
 
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Thanks, stevendaryl, for the very helpful and complete answer.
Thanks also vanhees71; I will keep the recommended book in mind if I find a way to freely access it.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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