# Definition of Lagrange function

1. Mar 9, 2004

### 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

2. Mar 9, 2004

### kuengb

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

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

3. Mar 9, 2004

### 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)?

4. Mar 9, 2004

### pmb_phy

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. Mar 9, 2004

### 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: 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. Mar 10, 2004

### 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."

7. Mar 11, 2004

### kuengb

Re: Re: Definition of Lagrange function

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.

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 .

8. Mar 11, 2004

### turin

Re: Re: Re: Definition of Lagrange function

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 + &epsilon;nml (dxm/dt) Bl }

and that, in general:

En = -&part;n&phi; - &part;tAn

and

Bl = &epsilon;lkj&part;kAj

where the Latin indices represent Cartesian coordinate indices, &part;n is the partial derivative with respect to the coordinate with the given index, and &epsilon;nml/&epsilon;lkj is the Levi-Civita symbol.

Putting the potential expressions into the equation of motion gives:

d2xn/dt2 = (e/m) { ( -&part;n&phi; - &part;tAn ) + &epsilon;nml (dxm/dt) ( &epsilon;lkj&part;kAj ) }
= (e/m) { -&part;n&phi; - &part;tAn + &epsilon;nml&epsilon;lkj (dxm/dt) &part;kAj }
= (e/m) { -&part;n&phi; - &part;tAn + ( &delta;nk&delta;mj - &delta;nj&delta;mk ) (dxm/dt) &part;kAj }
= (e/m) { -&part;n&phi; - &part;tAn + ( (dxm/dt) &part;nAm - (dxm/dt) &part;mAn ) }
= (e/m) { -&part;n&phi; + (dxm/dt) &part;nAm - ( &part;tAn + (dxm/dt) &part;mAn ) }
= (e/m) { -&part;n&phi; + (dxm/dt) &part;nAm - (dAn/dt) }

=>

d2xn/dt2 + (e/m) (dAn/dt) = (e/m) { -&part;n&phi; + (dxm/dt) &part;nAm }

=>

(d/dt){ dxn/dt + (e/m) An } + (e/m) { &part;n&phi; - &part;n[ (dxm/dt) Am ] + &part;n[ (dxm/dt) ] Am } = 0

=>

(d/dt){ m dxn/dt + e An } + &part;n{ e &phi; - e (dxm/dt) Am } = - e &part;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 } - &part;n{ -e &phi; + e (dxm/dt) Am } = 0

which can be identified with the Euler-Lagrange equation:

(d/dt){ &part;L/&part;vn } - &part;n{ L } = 0

to give

&part;L/&part;vn = m dxn/dt + e An

and

L = -e &phi; + 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 ( &phi; - (dxm/dt) Am )
= (1/2)mv2 - e ( &phi; - v.A )

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

Last edited: Mar 11, 2004
9. Mar 12, 2004

### pmb_phy

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

turin wrote
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. Mar 12, 2004

### 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.

11. Mar 12, 2004

### turin

Yes, that's exactly what I meant.