Definition of Lagrange function

[kuengb]

I'm learning classical mechanics right now, and I have a question about the "initial definition" of the Lagrange function. In my book, it is only introduced as L=T-V, but in many cases this doesn't help a lot, since it's not obvious what T or V is. For example, how would I come to the Lagrangian of a particle in a el.-magn. field without ever having heard of the vector potential, B=rot(A) ?

This question came to me when I wanted to find the (time-dependent) Lagrangian of a simple one-dim. movement with friction proportional to velocity:
m*d2(x) = f(x) - c*x

thanks for help

Bruno

 

[kuengb]

sorry, the equation should be
m*d2(x) = f(x) - c*d(x)

d( ) and d2( ) are the derivatives, f a force function

 

[Janitor]

"Lagrangian of a simple one-dim. movement with friction proportional..."

Maybe I am talking stupidly (I am going by memory from some reading done about ten years ago), but isn't it the case that extremal methods (such as those involving a Lagrangian) are not useful for solving the dynamics of non-conservative systems (such as those involving friction)?

If I am wrong about this, I am sure somebody here will correct me.

 

[pmb_phy]

Originally posted by kuengb
I'm learning classical mechanics right now, and I have a question about the "initial definition" of the Lagrange function. In my book, it is only introduced as L=T-V, but in many cases this doesn't help a lot, since it's not obvious what T or V is. For example, how would I come to the Lagrangian of a particle in a el.-magn. field without ever having heard of the vector potential, B=rot(A) ?

This question came to me when I wanted to find the (time-dependent) Lagrangian of a simple one-dim. movement with friction proportional to velocity:
m*d2(x) = f(x) - c*x

thanks for help

Bruno

The Lagragian is defined as that which appears under the action integral. It's usually given by L = Kinetic Energy - Generalized Potential. I don't know of a way to derive the Lagrangian for a particle in an EM field - sorry. I only know the Lagrangian itself. For a relativistic particle it's at

http://www.geocities.com/physics_world/em/relativistic_charge.htm

For non-relativistic particle its

L = T - U

where

U = Phi - qv*A

where Phi = Coulomb Potential, q = charge, v = velocity and A = vector potential. Plug this into Lagrange's equations and you get the Lorentz force

 

[jamesrc]

You can tabulate the work done by non-conservative forces in a Lagrangian formulation by including generalized forces.

$$L = T^* - U$$

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = Q_i$$

(That's just the notation I remember: q_i = i^th generalized coordinate, Q_i = i^th generalized force, T^* = kinetic coenergy, U = potential energy, etc.)

Back to the original question, here's a site where the Lagrangian of a particle is found starting with the Lorentz force law. I don't know if you can get around dealing with the magnetic vector potential. Hope that helps.

 

[turin]

Originally posted by kuengb
... how would I come to the Lagrangian of a particle in a el.-magn. field without ever having heard of the vector potential, ...

The electromagnetic potential is necessarily a 4-vector. The only literal answer to your question that I can think of is: "Someone can tell you what the Lagrangian is for a charge in an E&M field."

 

[kuengb]

Originally posted by turin
The electromagnetic potential is necessarily a 4-vector. The only literal answer to your question that I can think of is: "Someone can tell you what the Lagrangian is for a charge in an E&M field."

Well, certainly you end up with the same potential, whatever you do. I meant, how would you come to it by only looking at the equation of movement. In my course, the potential U = Phi - qv*A had simply been "claimed", and we checked whether it is of the correct form for the Lagrange function.

Maybe I am talking stupidly (I am going by memory from some reading done about ten years ago), but isn't it the case that extremal methods (such as those involving a Lagrangian) are not useful for solving the dynamics of non-conservative systems (such as those involving friction)

Well, I'm sure you're right. When attacking that little but frustrating problem I mentioned, I thaught it would be an exception to that rule, but now I think it's one of those excercises where you'd have to show that they cannot be solved .

 

[turin]

Originally posted by kuengb
... how would you come to it by only looking at the equation of movement.

I've never been completely satisfied with the Lagrangian formulation as a fundamental formulation of mechanics. I think that you have to already have a good understanding of E&M in Newtonian mechanics. You would then know of course that the equation of motion can be written in component form as:

d²xⁿ/dt² = (e/m) { Eⁿ + ε^nml (dx_m/dt) B_l }

and that, in general:

Eⁿ = -∂ⁿφ - ∂_tAⁿ

and

B_l = ε_lkj∂^kA^j

where the Latin indices represent Cartesian coordinate indices, ∂ⁿ is the partial derivative with respect to the coordinate with the given index, and ε^nml/ε_lkj is the Levi-Civita symbol.

Putting the potential expressions into the equation of motion gives:

d²xⁿ/dt² = (e/m) { ( -∂ⁿφ - ∂_tAⁿ ) + ε^nml (dx_m/dt) ( ε_lkj∂^kA^j ) }
= (e/m) { -∂ⁿφ - ∂_tAⁿ + ε^nmlε_lkj (dx_m/dt) ∂^kA^j }
= (e/m) { -∂ⁿφ - ∂_tAⁿ + ( δⁿ_kδ^m_j - δⁿ_jδ^m_k ) (dx_m/dt) ∂^kA^j }
= (e/m) { -∂ⁿφ - ∂_tAⁿ + ( (dx_m/dt) ∂ⁿA^m - (dx_m/dt) ∂^mAⁿ ) }
= (e/m) { -∂ⁿφ + (dx_m/dt) ∂ⁿA^m - ( ∂_tAⁿ + (dx_m/dt) ∂^mAⁿ ) }
= (e/m) { -∂ⁿφ + (dx_m/dt) ∂ⁿA^m - (dAⁿ/dt) }

=>

d²xⁿ/dt² + (e/m) (dAⁿ/dt) = (e/m) { -∂ⁿφ + (dx_m/dt) ∂ⁿA^m }

=>

(d/dt){ dxⁿ/dt + (e/m) Aⁿ } + (e/m) { ∂ⁿφ - ∂ⁿ[ (dx_m/dt) A^m ] + ∂ⁿ[ (dx_m/dt) ] A^m } = 0

=>

(d/dt){ m dxⁿ/dt + e Aⁿ } + ∂ⁿ{ e φ - e (dx_m/dt) A^m } = - e ∂ⁿ[ (dx_m/dt) ] A^m

I can't remember off the top of my head how to argue that the R.H.S. vanishes, so this is the one flaw that I leave to you. Anyway, assuming that this has been argued, we have:

(d/dt){ m dxⁿ/dt + e Aⁿ } - ∂ⁿ{ -e φ + e (dx_m/dt) A^m } = 0

which can be identified with the Euler-Lagrange equation:

(d/dt){ ∂L/∂v_n } - ∂ⁿ{ L } = 0

to give

∂L/∂v_n = m dxⁿ/dt + e Aⁿ

and

L = -e φ + e (dx_m/dt) A^m + f(vⁱ,t)

where f(vⁱ,t) is an arbitrary function of velocity and time. Therefore:

L = f(vⁱ,t) - e ( φ - (dx_m/dt) A^m )
= (1/2)mv² - e ( φ - v.A )

Clearly then, e ( φ - v.A ) is the potential energy, and φ - v.A is the electromagnetic potential.

 

[pmb_phy]

Originally posted by jamesrc
You can tabulate the work done by non-conservative forces in a Lagrangian formulation by including generalized forces.

$$L = T^* - U$$

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = Q_i$$

(That's just the notation I remember: q_i = i^th generalized coordinate, Q_i = i^th generalized force, T^* = kinetic coenergy, U = potential energy, etc.)

It should be noted that the L in that equation may not obey Hamilton's principle.


turin wrote

The electromagnetic potential is necessarily a 4-vector.

What is is that you're calling "electromagnetic potential"?

The 4-potential is defined as a 4-vector whose time component is proportional to the Coulomb potential and whose spatial portion is the vector potential .

 

[outandbeyond2004]

The Lag. is not always derived. It's like finding a closed formula for the integral of a new function. You guess at a definition and try it to see if it works. Maybe in some cases, if you are unlucky, you do too much trial-and-error work.

 

[turin]

Originally posted by pmb_phy
What is is that you're calling "electromagnetic potential"?

The 4-potential is defined as a 4-vector whose time component is proportional to the Coulomb potential and whose spatial portion is the vector potential .

Yes, that's exactly what I meant.