In more advanced areas of physics, you really start seeing the power of these formulations and their differences quite well.eval(ez_write_tag([[728,90],'profoundphysics_com-large-mobile-banner-2','ezslot_8',115,'0','0'])); For example, the Hamiltonian is a fundamental part of quantum mechanics as it can be used to calculate the total energy. [I find that, in physics, I often have to remind myself of what the problem actually is. Earlier though, we calculated the Hamiltonian with velocities. (Simple Proof). And because energy is conserved, potential energy (V) and kinetic energy (T) should add up to some constant. ( Log Out /  We have kinetic versus potential energy. And then, of course, you’ll usually see a second subscript i = 1, 2,… m to keep track of every qj for each and every particle in the system, so we’ll have n×m qij‘s in our model and so, yes, good to stick to Lagrange in that case. In fact, that’s what distinguishes this ‘Newtonian’ approach from the Lagrangian and Hamiltonian approach. So you’ll wonder what’s the difference then between Newtonian and Lagrangian mechanics? Does Flour Dissolve In Water (And Why)? So… Yes. Also in quantum mechanics, velocity usually does not have quite as clear of a meaning, while momentum is much easier to calculate and describe. Hence, we – social scientists – would surely expect them to develop approaches that are much more intricate and abstract than the ones that are being used by us, wouldn’t we? The goal of Profound Physics is to create a helpful and comprehensive internet resource aimed particularly for anyone trying to self-learn the essential concepts of physics (as well as some other science topics), with all of the fundamental mathematical concepts explained as intuitively as possible through lots of concrete examples and applications. This is how our configuration space looks like in this example: Now, imagine you start moving the particles along the line of length L. As they move from one of the boundaries to the other (from one side to the other), we can track both of the particles’ position on the configuration space and then get a representation for all of the possible configurations this system can have: This yellow square in the video then represents all of the possible configurations, meaning all of the possible ways you can have the two particles positioned on this line L. While this concept of configuration space and tracking the positions of things is useful in many cases, it really lacks to tell you anything about the motion of these particles. OP said simple.Come on. […] Well… When checking it out, I found that the answer is: yes, and no. Anyway, sometimes working with simple first order derivatives might be easier even if there are two separate equations. Huh? :-)]. If we’d want to do stuff in the classical world only, the answer seems to be: yes! Functions like yours are often referred to as "Lagrangians" in economic textbooks and such, but in the context of physics a Lagrangian is a functional, not just a function, and implies the concept of action, which in turn implies a dynamic situation. More accurately speaking, they are the generalized position and velocity coordinates, but this isn’t especially important for our purposes here. So… Let’s go… . In this notation, j = 1, 2,… n, and n is the number of degrees of freedom, i.e. In that case, the Lagrangian approach will do and may actually seem much easier, because we don’t have a set of equations to solve. The Lagrangian is in the form of kinetic minus potential, T – V, while the Hamiltonian is T + V. While this difference may seem quite insignificant, it actually has some pretty important consequences in terms of conservation laws, which are discussed more later in this article. In addition, the horizontal line where the particles can move has two walls on each side, so that the particles can only move within a length L between the walls. We know that’s equal to V = kx2/2. Just click the link. Not at all, really. […] Well… At least that’s what I know about it from googling stuff here and there. In any case, I’ll let you judge for yourself. Change ), Newtonian, Lagrangian and Hamiltonian mechanics, II. Hence, I will not try to explain why this approach gives the correct answer (i.e. The red curve represents kinetic energy (T) as a function of the displacement x: T is zero at the turning points, and reaches a maximum at the x = 0 point. What about potential energy (V)? The Lagrangian is simply a tool to describe motion (a very useful tool in all areas of physics for that matter), but it doesn’t represent any particular physical phenomena like the Hamiltonian does. The Lagrangian is in the form of kinetic minus potential, T – V, while the Hamiltonian is T + V. While this difference may seem quite insignificant, it actually has some pretty important consequences in terms of conservation laws, which are discussed more later in this article. Click on the button below. Just check it: it’s an alternative formula for T really. A detailed derivation and explanation of the Euler-Lagrange equation can be found in one of my articles here. Let me explain this through a simple example. So, instead of a point representing only the position, in phase space a point represents the position as well as momentum. I’ve added it, a year later or so, because, before you continue to read, you should note I am not going to explain the Hamiltonian matrix here, as it’s used in quantum physics. Yes, you’re right: we’re indeed solving the same second-order differential equation here. the so-called Lagrangian and Hamiltonian functions) look very similar: we write the Lagrangian as the difference between the kinetic and potential energy of a system (L = T – V), while the Hamiltonian is the sum of both (H = T + V). Of course, another reason for physicists to prefer the Hamiltonian approach may well that they think social science (like economics) isn’t real science. In case you wonder what that is: it’s the modeling approach that we’ve been using all along. Hence, we don’t really need the Lagrangian or Hamiltonian. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Change ), You are commenting using your Facebook account. Of course, we already know all of the relevant equations for this system just from applying Newton’s laws (so that’s Newtonian mechanics). We went over that when discussing kinetic gas theory, so I won’t say more about that here. eval(ez_write_tag([[250,250],'profoundphysics_com-large-leaderboard-2','ezslot_0',111,'0','0']));eval(ez_write_tag([[250,250],'profoundphysics_com-large-leaderboard-2','ezslot_1',111,'0','1'])); Okay, you might ask what the point of that even is. the displacement), and the direction of the force is towards the zero point–not away from it, and so we have a minus sign. The steps that are involved in the Lagrangian approach are the following: So, yes, we’re solving the same differential equation here. The difference between these two is that configuration space is the representation of all of the possible spacial positions of a system, while phase space is more like a representation of all the possible motion states of a system as phase space includes both momentum and position. In fact, you’ll usually see the spatial variables noted as qj. link to Why Do Photons Have No Mass? We also already know from previous posts (or, more likely, because you already know a lot about physics) that A is related to the energy of the system. Generally, the kinetic energy term in the Hamiltonian is replaced by something that rather includes momentum. One of the key differences that becomes explicitly important in for example quantum mechanics, is the fact that Hamiltonian mechanics uses position and momentum to describe motion and Lagrangian mechanics mainly deals with position and velocity. That’s the topic of another post, which involves far more advanced mathematical concepts. (Simple Proof), link to Does Flour Dissolve In Water (And Why)? The correspondence between Lagrangian and Hamiltonian mechanics is achieved with the tautological one-form. Some general tools for singular Legendre transformations are available, such as Dirac-Bergmann analysis, Faddeev-Jackiw method, etc. The second main difference is how you can transform problems. Change ), You are commenting using your Twitter account. = T = mv2/2 = m(dx/dt)2/2 = p2/2m. […] OK… Sorry, but I must move on. However, in this case, the why is really rather complicated.