Lagrangian mechanics

  • #1
Hey everyone,
So I'm just looking around to get a hold of some lagrangian mechanics for the GRE's coming up. Is the lagrangian always dealing with energy? Basically there was a problem I encountered with trying to find the lagrangian of a rolling ball in some setup, and once I knew that it was dealing with L = T - V I was alright, but I had no idea why to assume we were dealing with energy. Is lagrangian always energy??

Any useful websites that gives a good intro of Lagrangian mechanics? Since it's for the GRE, and I'll be learning it formally next year, I probably don't need too in depth an explanation, just a comprehensive one.

Thanks,
Ari
 

Answers and Replies

  • #2
jambaugh
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The Lagrangian in the standard formulation is the density of action per unit time, hence:
[tex] S[t_1,t_2] = \int_{t_1}^{t_2} L dt[/tex]
Essentially:
[tex] L = \frac{d}{dt}S[t,t_0][/tex]

Since energy is canonically dual to time you get L in energy units.

In principle you could pick a spatial coordinate say z, and use that to parametrize a particle's dynamic path x(z),y(z),t(z). (You'd have to be assured the particle never has zero z velocity so z is a "good parameter"). You'd then work with a z-momentum Lagrangian: L = p_z - A where A is a "potential z-momentum" instead of potential energy.
 
  • #3
fluidistic
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Lagrangian isn't energy.
L = T - V isn't energy. I think T+V would be the energy of some systems, rather than T-V.
 
  • #4
Matterwave
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L=T-V is true in (as far as I know) almost all applications of classical mechanics (I hesitate to say all because I haven't studied all of classical mechanics!). In special relativity; however, L=T-V is no longer true.

In general, L is the Lagrangian which gives you the correct equations of motion. But, the definition like this is not all that helpful practically speaking since we usually use L to GET the equations of motion. Thus, usually, if we want to use L practically speaking, we just remember the correct L for specific cases. For cases like pendulums, balls rolling down inclines (classical mechanics stuff), etc, we just use L=T-V. For special relativity, L=-mc^2/gamma. If we have E&M fields (which must be described relativistically), then we can use L=-mc^2/gamma-q*phi+q*(A,v)/c. Where (A,v) is the dot product of A with the velocity. This should be more than you need for the GRE's. Actually, for the GRE's I think mostly the L=T-V definition is entirely adequate.
 
  • #5
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I think that lagrange is involved in principle of least action, so it is more about action. For example, Newton's Laws, momentums, and even Maxwell's in more advanced ones. I think Hamiltonian is more about energy, as H=2K-L=K+U. Just sharing something that I have learnt.
 
  • #6
vanhees71
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In special relativity the Lagrangian, e.g., for the motion of a point charge in an external em. field is given by

[tex]L=-m c^2 \sqrt{1-(\dot{\vec{x}}/c)^2}-q \Phi(t,\vec{x}) \frac{q}{c} + \frac{q}{c} \dot{\vec{x}}\cdot \vec{A}(t,\vec{x}),[/tex]

where [tex]\Phi[/tex] and [tex]\vec{A}[/tex] are the scalar and vector potential of the electromagnetic field.

PS: Why don't inline formulae work anymore?
 
  • #7
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In special relativity the Lagrangian, e.g., for the motion of a point charge in an external em. field is given by

[tex]L=-m c^2 \sqrt{1-(\dot{\vec{x}}/c)^2}-q \Phi(t,\vec{x}) \frac{q}{c} + \frac{q}{c} \dot{\vec{x}}\cdot \vec{A}(t,\vec{x}),[/tex]

where [tex]\Phi[/tex] and [tex]\vec{A}[/tex] are the scalar and vector potential of the electromagnetic field.

PS: Why don't inline formulae work anymore?

Just asking. Is it followed from gauge invariance? The addition of the vector field? I only learnt a little about that, I want to know more.
 
  • #8
vanhees71
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The form of the interaction of point charges with all kinds of fields (scalars, vectors, tensors) can, to a certain extent, be inferred from Lorentz invariance. The Lagrangian, I wrote down for the em. field, indeed leads to a Lorentz invariant action:

[tex]S=-m c^2 \int \mathrm{d} t \sqrt{1-\vec{v}^2/c^2}-\frac{q}{c} \int \mathrm{d} t \dot{x}^{\mu} A_{\mu}[/tex]

with the four-vector notation

[tex](x^{\mu})=(x^0,\vec{x})=(c t, \vec{x}), \quad (A^{\mu})=(\Phi,\vec{A})[/tex]

for the space-time coordinates and em. potentials.

The kinetic part is the proper time of the particle along its trajectory and thus invariant, and the rest is explicitly invariant since [tex]\mathrm{d} x^{\mu} A_{\mu}[/tex] is invariant.

The equations of motion are also gauge invariant, as you can see when you derive them via the Euler-Lagrange Equations (Hamilton's least-action principle): They contain only the Faraday tensor

[tex]F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu},[/tex]

that are both Lorentz and gauge invariant.
 
  • #9
Matterwave
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In special relativity the Lagrangian, e.g., for the motion of a point charge in an external em. field is given by

[tex]L=-m c^2 \sqrt{1-(\dot{\vec{x}}/c)^2}-q \Phi(t,\vec{x}) \frac{q}{c} + \frac{q}{c} \dot{\vec{x}}\cdot \vec{A}(t,\vec{x}),[/tex]

where [tex]\Phi[/tex] and [tex]\vec{A}[/tex] are the scalar and vector potential of the electromagnetic field.

PS: Why don't inline formulae work anymore?

I think you have an extra q/c term next to your scalar potential.
 
  • #10
vanhees71
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Right it must read

[tex]L=-m c^2 \sqrt{1-(\dot{\vec{x}}/c)^2}-q \Phi(t,\vec{x}) + \frac{q}{c} \dot{\vec{x}}\cdot \vec{A}(t,\vec{x}).[/tex]
 

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