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When we enter into the scope of Analytical mechanics we usually start with these two primary notions: Lagrangian function & Hamiltonian function

And usually textbooks define Lagrangian as $L=T-V$ and Hamiltonian as $H=T+V$ where $T$ is Kinetic energy and $V$ is Potential energy. But as we proceed it turns out Lagrangian and Hamiltonian may not always have these values and this occurs in just some special cases and there is a more general definition.

My question is:

  1. What is the generalized definitions of these two functions?

  2. Is there a general defining equation where the mentioned values could be extracted as a special case?

EDIT: A third question

Do we know about the notions of "Kinetic energy" and "Potential energy" beforehand or are they defined after the Lagrangian gets its famous form $L=T-V$ and then we assign $T$ and $V$ their respective definitions?

EDIT 2: A fourth question

What if there were non-conservative fields? Then obviously $V$ term couldn't show their effects(since Potentials are always associated with conservative fields). Then how could we bring the effects of non-conservative fields into play?

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  1. This is a huge topic. How general do you want to be? We will only be schematic here. Perhaps the most general setting is a physical system with a set of equations of motion. (The word equations of motion refer to a coupled set of differential equations that governs time evolution. E.g. in Newtonian mechanics, it's Newton's 2nd law.)

  2. There are no general algorithms to determine whether a variational formulation exists or not, cf. e.g. this Phys.SE post. In other words, finding a variational formulation is in general an art.

  3. Generically, there does not exist a variational formulation for dissipative and non-conservative systems. However, see e.g. this and this Phys.SE posts. For a discussion of the non-conservative forces, see also this Phys.SE post.

  4. The Lagrangian $L$ need not be of the form $T-U$, cf. e.g. this Phys.SE post. Of course, classes of physical systems do obey $L=T-U$, cf. e.g. this Phys.SE post.

  5. Next assume that a Lagrangian $L$ is given. To find the Hamiltonian $H$, perform a (possibly singular) Legendre transformation between the generalized velocity and momentum variables. (If the Lagrangian depends on higher time derivatives, cf. e.g. this Phys.SE post, see the Ostrogradsky formalism.)

  6. For the condition when the Hamiltonian $H$ & the total energy $E$ agree, see e.g. this & this Phys.SE posts & links therein.

  7. For books on Lagrangian & Hamiltonian formulations, see e.g. this & this resource recommendation lists & links therein.

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  • $\begingroup$ OK if you don't mind I wanna ask some more basic questions: 1. What is the exact definition of an equation of motion? An equation giving the path the particle/system takes in the actual physical space we observe or the path it takes in the generalized coordinates? 2. Unfortunately my knowledge about "Calculus of variations" and the "Theory of differential forms" is not much so I didn't understand what you said in the mentioned PSE post. Can you explain it more simply? 3. How could it be possible for Lagrangian to depend on higher time derivatives? Don't we always say $L=L(q,\dot{q},t)$? $\endgroup$ – Hamed Begloo Nov 25 '16 at 9:42
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    $\begingroup$ 1. I updated the answer. 2 & 3. Those questions belong more to the linked posts. $\endgroup$ – Qmechanic Nov 25 '16 at 12:12
  • $\begingroup$ Thank you. Just two other things: 1. Can you give me the title of some source textbooks which study Lagrangian and Hamiltonian formalisms in its most general forms? (My whole knowledge of analytical mechanics is based on chapter 10th of "Fowles-Cassiday"!) 2. And also can you give me the title of some source textbooks which talks about the concept of "Time evolution" in classical mechanics(from beginner to advanced)? $\endgroup$ – Hamed Begloo Nov 25 '16 at 17:45
  • $\begingroup$ Another question just came into my mind: Are these formalisms axiomatized correctly? I mean what are the axioms of Analytical mechanics? If it depends on the theory being formulated by the formalism, for example what are the axioms in classical mechanics? $\endgroup$ – Hamed Begloo Nov 25 '16 at 17:46
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    $\begingroup$ That sentence discusses existence of an action; Whether it has a simple form or not (say, in terms of standard functions) is another issue. $\endgroup$ – Qmechanic Nov 26 '16 at 16:21
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In general - No there is not. In the corresponding formalism, the Lagrange or the Hamilton-function will be the starting point to describe the system, and that means that you have to shape the Lagrange function in a suitable way to describe the system. There are no other properties of the system at that point, that have allready been translated into a mathematical object (Like the kinetic Energy, or the Potential energy), you start with the Lagrange function, and in a suitable system you can afterwards identify terms within the Lagrange-function with $V$ or with $T$.

However, choosing a Lagrange-function is not entirely abitrary: Your Lagrange function $L(\vec{x}, \dot{\vec{x}})$ is supposed to describe your system. That means for example, that $L$ has the same symmetries as you observe at your system.

For example, consider the free space: Every region is the same to a flying particle, the laws of physics are the same everywhere. That means your have translational invariance, and hence $L(\vec{x} + \vec{p}, \dot{\vec{x}})=L(\vec{x}, \dot{\vec{x}}) $, for every $\vec{p}$ you can think of. That means $L$ doesn't depend on x. Do the same with Rotational invariance, then you know L can just depend on $|\dot{\vec{x}}|$. By now L has the form $$ L= c f(|\dot{\vec{x}}|) $$ If you demand that every inertial frame of reference will yield the same equations of motion (galileian invariance), then you demand that $$L(\dot{\vec{x}}) = L(\dot{\vec{x}}+\vec{v}) + D(\dot{\vec{x}}) \\ D(\vec{x},\dot{\vec{x}}) = \nabla g(x)\dot{\vec{x}}$$ This is a less rigid condition, the Term $D$ is a gauge-term and doesn't change the equations of motion. One way to satisfy this condition is to choose: $$ L= c \dot{\vec{x}^2} $$ Then $$ L(\dot{\vec{x}}+\vec{v}) = c \dot{\vec{x}^2} + c \vec{v}^2 + 2 c \vec{v} \dot{\vec{x}} = \tilde{L}(\dot{\vec{x}}) $$ $L$ and $\tilde{L}$ will lead to the same equations of motion in two different frames of reference (as you demanded it). L is now shaped in a way that regards homogenity of time, of space, isotropy of space, and free choice of your frame of reference to be symmetries of the system. Now look at the outcome:$$ L= c \dot{\vec{x}^2} $$ Up to constants (which will come in to play when you define units), you have the kinetic Energy, and you know this function to correctly describe the motion of a free particle. But you did't know in the first place, that there is something like newtons axioms, or something like energy. Everything you knew where the symmetries, and using them, you arrived at this point. In the same way you can demand gauge invariance, then you will arrive at the electromagnetic coupling terms in the Lagrange-function, and so on.

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  • $\begingroup$ Thank you. What I get from your answer is our axioms and the supplementary conditions given by the problem shapes the Lagrangian and Hamiltonian functions. Now the question is what are the axioms? Are they homogeneity and isotropy of space plus homogeneity of time and galilean invariance? And the other question is what are the additional conditions needed to form Lagrangian and Hamiltonian or in which way these conditions show themselves to us? $\endgroup$ – Hamed Begloo Nov 25 '16 at 9:38
  • $\begingroup$ Besides one thing that bothers me is the $V$ term would not necessarily exist since Potentials are always associated with conservative fields. What if there existed a non-conservative field? Then how the field shows its effects on the Lagrangian function? $\endgroup$ – Hamed Begloo Nov 25 '16 at 9:39
  • $\begingroup$ In this case the axioms are: Principle of Action, homogenity of time and space, isotropy and galileian invariance. For your field question: In case your field can be described by a lagrangian too, then you have one big lagrangian for field and particles. It goes straight forward, you plug in everything you know (or demand) and the outcome is the relativistic lorenz-force + maxwell equations. $\endgroup$ – Quantumwhisp Nov 25 '16 at 10:15
  • $\begingroup$ "In case your field can be described by a lagrangian too..." Wait a minute. Do you mean there are fields that couldn't be described at all(of course at theoretical grounds)? $\endgroup$ – Hamed Begloo Nov 25 '16 at 10:30

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