Definition of Lagrange function

In summary: NM - m) } where NM is the Newtonian potential (again, a 4-vector). From here, you can use the Euler-Lagrange equation to write the Lagrangian as:L = (En+...(NM-m))T where T is the kinetic energy.
  • #1
kuengb
106
0
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-dependant) Lagrangian of a simple one-dim. movement with friction proportional to velocity:
m*d2(x) = f(x) - c*x

thanks for help

Bruno
 
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  • #2
sorry, the equation should be
m*d2(x) = f(x) - c*d(x)

d( ) and d2( ) are the derivatives, f a force function
 
  • #3
"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.
 
  • #4
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-dependant) 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
 
  • #5
You can tabulate the work done by non-conservative forces in a Lagrangian formulation by including generalized forces.

[tex] L = T^* - U [/tex]

[tex] \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = Q_i [/tex]

(That's just the notation I remember: qi = ith generalized coordinate, Qi = ith 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.
 
  • #6
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."
 
  • #7


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:smile: .
 
  • #8


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 + εnml (dxm/dt) Bl }

and that, in general:

En = -∂nφ - ∂tAn

and

Bl = εlkj∂kAj

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

Putting the potential expressions into the equation of motion gives:

d2xn/dt2 = (e/m) { ( -∂nφ - ∂tAn ) + εnml (dxm/dt) ( εlkj∂kAj ) }
= (e/m) { -∂nφ - ∂tAn + εnmlεlkj (dxm/dt) ∂kAj }
= (e/m) { -∂nφ - ∂tAn + ( δnkδmj - δnjδmk ) (dxm/dt) ∂kAj }
= (e/m) { -∂nφ - ∂tAn + ( (dxm/dt) ∂nAm - (dxm/dt) ∂mAn ) }
= (e/m) { -∂nφ + (dxm/dt) ∂nAm - ( ∂tAn + (dxm/dt) ∂mAn ) }
= (e/m) { -∂nφ + (dxm/dt) ∂nAm - (dAn/dt) }

=>

d2xn/dt2 + (e/m) (dAn/dt) = (e/m) { -∂nφ + (dxm/dt) ∂nAm }

=>

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

=>

(d/dt){ m dxn/dt + e An } + ∂n{ e φ - e (dxm/dt) Am } = - e ∂n[ (dxm/dt) ] Am

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 (dxm/dt) Am } = 0

which can be identified with the Euler-Lagrange equation:

(d/dt){ ∂L/∂vn } - ∂n{ L } = 0

to give

∂L/∂vn = m dxn/dt + e An

and

L = -e φ + e (dxm/dt) Am + f(vi,t)

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

L = f(vi,t) - e ( φ - (dxm/dt) Am )
= (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.
 
Last edited:
  • #9
Originally posted by jamesrc
You can tabulate the work done by non-conservative forces in a Lagrangian formulation by including generalized forces.

[tex] L = T^* - U [/tex]

[tex] \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = Q_i [/tex]

(That's just the notation I remember: qi = ith generalized coordinate, Qi = ith 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 .
 
  • #10
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.
 
  • #11
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.
 

1. What is the Lagrange function and what does it represent?

The Lagrange function, also known as the Lagrangian, is a mathematical function used in the field of optimization to find the minimum or maximum value of a system. It represents the difference between the system's kinetic and potential energy at any given point.

2. How is the Lagrange function used in physics?

In physics, the Lagrange function is used in the Lagrangian mechanics approach to describe the dynamics of a system. It helps to determine the equations of motion for a system by considering the system's potential and kinetic energy.

3. What is the relationship between the Lagrange function and the Euler-Lagrange equations?

The Euler-Lagrange equations are a set of differential equations that are derived from the Lagrange function. They are used to find the critical points of the Lagrange function and are essential in solving optimization problems.

4. Can the Lagrange function be used to solve constrained optimization problems?

Yes, the Lagrange function can be used to solve constrained optimization problems by adding constraints to the function. These constraints ensure that the solution to the problem satisfies specific conditions, making it a powerful tool in optimization.

5. Are there any real-world applications of the Lagrange function?

Yes, the Lagrange function has many real-world applications, including in physics, economics, and engineering. It is used to solve optimization problems in various fields, such as finding the shortest path between two points or maximizing profit in a business.

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