Lagrangian V Hamiltonian

[ydoaPs]

I've not really studied much about this concept, but Lagrangian Mechanics seems pretty cool. In fact, the most I've really dealt with it is in Klaynos's blog. However, I don't quite understand why the Lagrangian is used. What is the physical significance? Why T-V? If I were to go about making equations of motion, I think I'd use the Hamiltonian (H=T+V). It just seems like the sum of the potential and kinetic energies would be more useful than the difference.

[Tsadi]

I need to go out for half an hour, or i would talk about it more... but do you mean

 

H = T + U

 

T is a function only of velocity and from the conditions before U is a function only of position.

[swansont]

I need to go out for half an hour, or i would talk about it more... but do you mean

 

H = T + U

 

T is a function only of velocity and from the conditions before U is a function only of position.

 

 

U and V are both used, depending on who's writing down the equations.

[Tsadi]

Very well. I had to check up on that just now.

 

The quantity in question, the (T-V) part, is the Langrangian itself. It can also be expressed as;

 

[math]\delta \int Ldt=0[/math]

 

So, T and V are position, velocity and time, and may not necesserily be expressed by any standard lengths, but can be generalized.

 

To answer your question, but not by much.

[ydoaPs]

Swansont, care to give my question a go?

 

The quantity in question, the (T-V) part, is the Langrangian itself. It can also be expressed as;

 

[math]\delta \int Ldt=0[/math]

 

I was under the impression that that integral was the Action integral.

[swansont]

I haven't used Lagrangians in ages. Maybe someone a little closer to having taken the class.

[Klaynos]

The Lagrangian itself has no physical significance (or so we were told and I've never happened across one) we use it because it is useful for solving equations of motion with simpler differential equations than newtonian mechanics... It does drop out of a derivation though, it's not just thought up... but even so it's unusual not to have some kind of physical meaning...

[Tsadi]

Swansont, care to give my question a go?

 

 

 

I was under the impression that that integral was the Action integral.

 

Why yes, very important too.

 

Reason why, is because Hamiltons Principle is also known as the principle of least action, which basically means under an integral of action, everything is minimized to such an extend to use the least Kinetic Energy minus the potential energy so;

 

[math]\delta \int T-U dt=0[/math]

 

Remember, T-U part here has been substituted for L, which was used previously, since L=T-U.

Edited October 5, 2008 by Tsadi
multiple post merged

[Severian]

In my view, the Lagrangian is more fundamental than the Hamiltonian. The Hamiltonian is basically just the energy operator, while it is the Lagrangian which tells you the physical behaviour of the system.

 

Remember too that the identification of H = T + U and L = T - U is a special case and is not necessarily true. The fact that people think T+U is more fundamental is because of the way we teach physics - we look at systems where this is a constant, which gives the Hamiltonian a special significance.

 

I would also disagree with Klaynos that the Lagrangian is not physical. While in some sense, only observables are physical, since they are the only things we can measure (and thus the Hamiltonian corresponds to a physical observable, the energy), when building theories of fundamental physics, we think in terms of Lagrangians, not Hamiltonians.

 

The usual way people construct theories nowadays is to think up when particle content they want or need, decide on the symmetries they want to impose, and write down every term that they can think of using the particles which obeys the symmetries. The resulting Lagrangian is then the most general that it can be, but still provides physical laws which respect the symmetries. For example, making the Lagrangian invariant under time translations makes the Hamiltonian a conserved quantity (energy conservation).

 

As for Hamiltonian mechanics more generally, one normally thinks of it as a consequence of Lagrangian mechanics and defines the Hamiltonian as the Legendre transform of the Lagrangian:

 

[math]{\cal H} = p \dot x - {\cal L}[/math]

 

While this isn't necessary (strictly speaking you can define Hamiltonian mechanics independently) it is how most people think of it. Indeed, this is where the U+V thing comes from: e.g. if p = mv, then [math]p \dot x = mv^2 = 2T[/math] so [math]H = 2T - T + U = T+U[/math].

[Tsadi]

Wonderfully put.