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The Hamiltonian formulation of classical mechanics is such a beautiful way of describing classical motion compared to the Newtonian formulation. See Laundau & L

The Hamiltonian formulation of classical mechanics is such a beautiful way of describing classical motion compared to the Newtonian formulation. See Laundau & Lifschitz book 1! All the Hamilton-Jacobi equations are derived from observing symmetries in space time, even Newton’s 3 principles are derived (F=ma an the rest). All of this has the added benefit of transposing well into quantum mechanics, where forces are anyway replaced with hamiltonians. For fluid mechanics I don’t know if Hamiltonians are the right formulation.

Horffupolde 3y ago
That’s a correct usage of the factorial.
fransje26 3y ago
Although not the same, I remember the first time I encountered the description of dynamic systems using Lagrangian mechanics and it was beautiful. It finally made sense. Instead of doing what seemed like some mumbo-jumbo sarting from a static system, that, eventually and through convoluted ways, lead to the equations we were kind of looking for, here was a clean and logical way to look at what was relevant to the dynamics of the system, with an infaillible, straightforward, mathematical way of getting the equations of motion. I'm not that familiar with Hamiltonian formulations, but its conservation properties could bring some important improvements to the current way the Navier-Stokes equations are treated. Conservation of linear and angular momentum, for a start, could be nice.. Now, let's see if I understand anything from this paper..
- deleted 3y ago
  [deleted]
- om2 3y ago
  I have tried to teach myself both Hamiltonian and Lagrangian classical mechanics and there’s one mental hurdle I have not been able to get over. The problems are generally set up with starting position and momentum known, and ending position and momentum known, and then the math tells you the path taken along the way. But what if the ending position and momentum is unknown? How does one use these formulations of mechanics to predict the future and not just postdict the past? Is this just how beginner problems tend to be set up?
  - joshjob42 3y ago
    Hm, well typically in Lagrangian formulations, you're right that you define a functional called an "action", and this involves some integral from an initial to a final position in configuration space. Then the idea is that the system will evolve in such a way as to minimize this action. A stationary point (like a maximum or minimum) has gradient 0. So a condition for this "principle of least action" to be realized is that the "gradient" (called the variation here) of the action is 0 for the real path. That is, if you take a path the system carves out through the configuration space and perturb that path, and you compute the total effect on the action from that small perturbation, you find it should be 0. You do this by actually taking this derivative and you find that you can guarantee that the differential of the action is 0 if the system takes a path which is the solution to a set of differential equations, and you can generally find the solution to those differential equations only with information about the origin, ie you don't need both the start and end conditions to find a unique solution. So you're right, it's a bit weird conceptually. You sort of start saying "the system obeys a path that minimizes the action between it's initial and final positions" and then find that this produces a set of conditions which form a system of diff eqs that you can find general solutions for and select out a unique solution just with the initial conditions, no need for the final condition.
  - fooker 3y ago
    Good question. The answer is, unfortunately, boring. You solve equations to find the unknowns. If there are too many unknowns there are too many solutions and the whole thing is useless.
bdjsiqoocwk 3y ago
Since you sound like you know what you're talking about, wanna tell us what is the difference between the hamiltonian and lagrangian formulations?
- ajkjk 3y ago
  Not the person you were responding to, but: the Lagrangian formulation describes physics in terms of Least Action (so minimizing (or maximizing) S = ∫ L dt) on a manifold in (x,v) coordinates, and its equations of motion are like L_x - d_t L_v = 0. The second term tends to be second-order in time. The Hamiltonian formulation performs a Legendre transformation[1] on L giving H = v L_v - L, which is essentially a convenient trick: it reparameterizes L in (q,p) coordinates, where p = L_v, and writes S as ∫ (pv - H) dt. This changes the E.o.M. to (qdot, pdot) = (H_p, -H_q), which is (a) first-order in time and therefore easier to deal with and (b) geometrically elegant because it is a rotation in (q,p) space, which is easy to think about. At least those are the reasons everyone gives why it's important. I think the real reason is that QM is formulated in terms of H so you need to know it, and also that this (q,p) thing makes statistical mechanics easier because it has good geometric properties: it amounts to saying that time evolution conserves area in (q,p) space, which means that you can treat the evolution of many-particle systems as being in a whole block of states at once, treated as a geometric object that flows over time. I've never been able to understand if there is something "truly fundamental" about H compared to L, or if H is more of a mathematical convenience for making the equations first-order. [1]: https://blog.jessriedel.com/2017/06/28/legendre-transform/ https://blog.jessriedel.com/2017/06/28/legendre-transform/ is a good exposition, if still pretty tough to understand. Legendre transforms are hard to grok.
  - adrian_b 3y ago
    In classical mechanics the Lagrangian and Hamiltonian formulations are mostly equivalent (though, confusingly the Lagrangian formulation is due to Hamilton, who has shown that the Lagrange equations can be derived from a variational principle, while the complete Hamiltonian formulation is due to some later phycisists; for what has become the Hamiltonian formulation, Hamilton has shown only how to obtain the system of first order equations from the system of second order equations, but that had already been done before by Cauchy in 1831 in a journal that few were reading, so it was ignored). Still, in classical mechanics some consider the Hamiltonian formulation to be more fundamental, because the first order equations can be applicable to some problems that have discontinuities incompatible with second-order equations, though such problems are artificial (real systems are continuous enough, strong discontinuities appear only through approximations). However this changes completely in relativistic mechanics, where the Hamiltonian is not invariant, while the Lagrangian is a relativistic invariant quantity. This makes the Lagrangian formulation a far better choice in relativistic mechanics and it is a strong argument to consider the Lagrangian formulation as the fundamental one and the Hamiltonian formulation as only an approximation that can be used at small velocities or only as a mathematical trick for numeric solutions. When the Lagrangian formulation is used, after a coordinate system is chosen, it is always possible to use the Legendre transformation to obtain a Hamiltonian system of first order equations. However, in the relativistic case the system depends on the coordinate system. Therefore, if the coordinate system is changed, the Hamiltonian equations must be derived again from the invariant Lagrangian formulation. The reason why the Lagrangian is a relativistic invariant is that this scalar value is the projection of the energy-momentum 4-vector on the trajectory curve in space-time. The Hamiltonian is just the temporal component of the 4-vector, which is changed by any coordinate transformation. Therefore L is more fundamental than H, in the same sense that the magnitude of a vector is more fundamental than any of the components that the vector happens to have in some particular coordinate system. The traditional formulation of the quantum mechanics using H is a serious inconvenience for extending it to the relativistic case. Coherent formulations of the relativistic quantum mechanics must also use L instead of H.
ajkjk 3y ago
Is it not the case that F=ma and other Newtonian laws are encoded in the Lagrangian, and therefore not derived from symmetries? After all nature and her symmetries alone do not tell us how her physics works; there has to be some rule for time-evolution as well, and that's what's encoded in L (L's form is essentially a list of pairings of variables and their costs of evolution, which for classical mechanics is L = T - V = ∫ p·dv + ∫ F·dx, which says "the cost of changing v is p and the cost of changing x is F"). (QFT sort-of has an explanation for time evolution in terms of symmetries alone, but it requires a lot more machinery. But afaik classical mechanics does not.)
- om2 3y ago
  > the cost of changing v is p and the cost of changing x is F I’m sure the equation is right and all but this seems sideways in terms of an intuitive explanation - velocity changes position, and force changes momentum. Force doesn’t directly change position (only indirectly via changing momentum) and momentum doesn’t change velocity, having momentum is consistent with a constant velocity. It doesn’t even make much sense to me to think of integrals as being about “costs of changing”, would that not be a derivative?
techas 3y ago
Please read "Story of your life" by Ted Chiang. A beautiful story that involves a discussion of Newtonian and Hamiltonian formulations. One of the best stories I have read.

Horffupolde 3y ago

That’s a correct usage of the factorial.

fransje26 3y ago

Although not the same, I remember the first time I encountered the description of dynamic systems using Lagrangian mechanics and it was beautiful. It finally made sense. Instead of doing what seemed like some mumbo-jumbo sarting from a static system, that, eventually and through convoluted ways, lead to the equations we were kind of looking for, here was a clean and logical way to look at what was relevant to the dynamics of the system, with an infaillible, straightforward, mathematical way of getting the equations of motion. I'm not that familiar with Hamiltonian formulations, but its conservation properties could bring some important improvements to the current way the Navier-Stokes equations are treated. Conservation of linear and angular momentum, for a start, could be nice.. Now, let's see if I understand anything from this paper..

deleted 3y ago
[deleted]
om2 3y ago
I have tried to teach myself both Hamiltonian and Lagrangian classical mechanics and there’s one mental hurdle I have not been able to get over. The problems are generally set up with starting position and momentum known, and ending position and momentum known, and then the math tells you the path taken along the way. But what if the ending position and momentum is unknown? How does one use these formulations of mechanics to predict the future and not just postdict the past? Is this just how beginner problems tend to be set up?
- joshjob42 3y ago
  Hm, well typically in Lagrangian formulations, you're right that you define a functional called an "action", and this involves some integral from an initial to a final position in configuration space. Then the idea is that the system will evolve in such a way as to minimize this action. A stationary point (like a maximum or minimum) has gradient 0. So a condition for this "principle of least action" to be realized is that the "gradient" (called the variation here) of the action is 0 for the real path. That is, if you take a path the system carves out through the configuration space and perturb that path, and you compute the total effect on the action from that small perturbation, you find it should be 0. You do this by actually taking this derivative and you find that you can guarantee that the differential of the action is 0 if the system takes a path which is the solution to a set of differential equations, and you can generally find the solution to those differential equations only with information about the origin, ie you don't need both the start and end conditions to find a unique solution. So you're right, it's a bit weird conceptually. You sort of start saying "the system obeys a path that minimizes the action between it's initial and final positions" and then find that this produces a set of conditions which form a system of diff eqs that you can find general solutions for and select out a unique solution just with the initial conditions, no need for the final condition.
- fooker 3y ago
  Good question. The answer is, unfortunately, boring. You solve equations to find the unknowns. If there are too many unknowns there are too many solutions and the whole thing is useless.

bdjsiqoocwk 3y ago

Since you sound like you know what you're talking about, wanna tell us what is the difference between the hamiltonian and lagrangian formulations?

ajkjk 3y ago
Not the person you were responding to, but: the Lagrangian formulation describes physics in terms of Least Action (so minimizing (or maximizing) S = ∫ L dt) on a manifold in (x,v) coordinates, and its equations of motion are like L_x - d_t L_v = 0. The second term tends to be second-order in time. The Hamiltonian formulation performs a Legendre transformation[1] on L giving H = v L_v - L, which is essentially a convenient trick: it reparameterizes L in (q,p) coordinates, where p = L_v, and writes S as ∫ (pv - H) dt. This changes the E.o.M. to (qdot, pdot) = (H_p, -H_q), which is (a) first-order in time and therefore easier to deal with and (b) geometrically elegant because it is a rotation in (q,p) space, which is easy to think about. At least those are the reasons everyone gives why it's important. I think the real reason is that QM is formulated in terms of H so you need to know it, and also that this (q,p) thing makes statistical mechanics easier because it has good geometric properties: it amounts to saying that time evolution conserves area in (q,p) space, which means that you can treat the evolution of many-particle systems as being in a whole block of states at once, treated as a geometric object that flows over time. I've never been able to understand if there is something "truly fundamental" about H compared to L, or if H is more of a mathematical convenience for making the equations first-order. [1]: https://blog.jessriedel.com/2017/06/28/legendre-transform/ https://blog.jessriedel.com/2017/06/28/legendre-transform/ is a good exposition, if still pretty tough to understand. Legendre transforms are hard to grok.
- adrian_b 3y ago
  In classical mechanics the Lagrangian and Hamiltonian formulations are mostly equivalent (though, confusingly the Lagrangian formulation is due to Hamilton, who has shown that the Lagrange equations can be derived from a variational principle, while the complete Hamiltonian formulation is due to some later phycisists; for what has become the Hamiltonian formulation, Hamilton has shown only how to obtain the system of first order equations from the system of second order equations, but that had already been done before by Cauchy in 1831 in a journal that few were reading, so it was ignored). Still, in classical mechanics some consider the Hamiltonian formulation to be more fundamental, because the first order equations can be applicable to some problems that have discontinuities incompatible with second-order equations, though such problems are artificial (real systems are continuous enough, strong discontinuities appear only through approximations). However this changes completely in relativistic mechanics, where the Hamiltonian is not invariant, while the Lagrangian is a relativistic invariant quantity. This makes the Lagrangian formulation a far better choice in relativistic mechanics and it is a strong argument to consider the Lagrangian formulation as the fundamental one and the Hamiltonian formulation as only an approximation that can be used at small velocities or only as a mathematical trick for numeric solutions. When the Lagrangian formulation is used, after a coordinate system is chosen, it is always possible to use the Legendre transformation to obtain a Hamiltonian system of first order equations. However, in the relativistic case the system depends on the coordinate system. Therefore, if the coordinate system is changed, the Hamiltonian equations must be derived again from the invariant Lagrangian formulation. The reason why the Lagrangian is a relativistic invariant is that this scalar value is the projection of the energy-momentum 4-vector on the trajectory curve in space-time. The Hamiltonian is just the temporal component of the 4-vector, which is changed by any coordinate transformation. Therefore L is more fundamental than H, in the same sense that the magnitude of a vector is more fundamental than any of the components that the vector happens to have in some particular coordinate system. The traditional formulation of the quantum mechanics using H is a serious inconvenience for extending it to the relativistic case. Coherent formulations of the relativistic quantum mechanics must also use L instead of H.

ajkjk 3y ago

Is it not the case that F=ma and other Newtonian laws are encoded in the Lagrangian, and therefore not derived from symmetries? After all nature and her symmetries alone do not tell us how her physics works; there has to be some rule for time-evolution as well, and that's what's encoded in L (L's form is essentially a list of pairings of variables and their costs of evolution, which for classical mechanics is L = T - V = ∫ p·dv + ∫ F·dx, which says "the cost of changing v is p and the cost of changing x is F"). (QFT sort-of has an explanation for time evolution in terms of symmetries alone, but it requires a lot more machinery. But afaik classical mechanics does not.)

om2 3y ago
> the cost of changing v is p and the cost of changing x is F I’m sure the equation is right and all but this seems sideways in terms of an intuitive explanation - velocity changes position, and force changes momentum. Force doesn’t directly change position (only indirectly via changing momentum) and momentum doesn’t change velocity, having momentum is consistent with a constant velocity. It doesn’t even make much sense to me to think of integrals as being about “costs of changing”, would that not be a derivative?

techas 3y ago

Please read "Story of your life" by Ted Chiang. A beautiful story that involves a discussion of Newtonian and Hamiltonian formulations. One of the best stories I have read.