Lagrangian with generalized positions

  • #1
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Summary:
dependence on generalized velocities
Hi Pfs
When instead of the variables x,x',t the lagrangiean depends on the trandformed variables q,q',t , time may be explicit in this lagrangian and q' (the velocity of q) may appear outside. I am looking for a toy model in which tine is not explicit in L but where the velocities appear somhere else than in the kinetic term.
i would then calculate the conjugated momentum and the hamilttonian (is it consrrained?)
thanks
 

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  • #2
vanhees71
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In most textbook applications time is not explicit in ##L##. The most simple example is a particle in some external potential,
$$L=T-V=\frac{m}{2} \dot{\vec{x}}^2 -V(\vec{x}),$$
where ##\vec{x}## can be expressed in arbitrary generalized coordinates ##q=(q^i)##, ##\vec{x}=\vec{x}(q)##.
 
  • #3
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yes but here there is no velocity dependence in V.
 
  • #4
Dale
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Summary:: dependence on generalized velocities

where the velocities appear somhere else than in the kinetic term
I am not sure exactly what you are looking for here. For example, in rotating coordinates the Lagrangian would be $$L=\frac{1}{2}m \dot r^2 + \frac{1}{2}m r^2 \dot \theta^2 +\frac{1}{2}m r^2 \omega^2 +m r^2 \dot \theta \omega - V$$ Would you consider any of those to be velocities somewhere else than in the kinetic term, and if not then why not?
 
  • #5
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my quesstion comes from this sentence that i read:
Note that Jacobi’s generalized energy and the Hamiltonian do not equal the total energy E. However, in the special case where the transformation is scleronomic, then T1 = T0 = 0 and if the potential energy U does not depend explicitly of the velocities of the q variables then the generalized energy (Hamiltonian) equals the total energy, that is, H = E Recognition of the relation between the Hamiltonian and the total energy facilitates determining the equations of motion.

So i am looking to a counter exemple where the velocity appear in the potential energy U.
 
  • #6
Dale
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So why don’t you just write ##L=T(q,\dot q)-U(q,\dot q)##
 
  • #7
vanhees71
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yes but here there is no velocity dependence in V.
An example for a Lagrangian, where in the interaction term is a velocity is that for the motion in an electromagnetic field with a potential and vector potential ##\Phi## and ##\vec{A}##,
$$L=\frac{m}{2} \dot{\vec{x}}^2 -q \Phi + \frac{q}{c} \vec{v} \cdot \vec{A}.$$
If the potentials are also explicitly time-dependent that's an example for an explicitly time-dependent Lagrangian.
 
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  • #8
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can we say then that the corresponding hamilton H does not equal the total energy of the system (time does not appear explicily in the hamiltonian but v appear out side the kiniétic term)?
 
  • #9
vanhees71
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In this case you get the total energy for the case of static fields, i.e., for ##\vec{A}=\vec{A}(\vec{x})## and ##\Phi=\Phi(\vec{x})##, which is
$$H=\frac{m}{2} \dot{\vec{x}}^2 + q \Phi(\vec{x}).$$
Note that the magnetic part of the Lorentz force is ##\vec{F}_B=q/c \vec{v} \times \vec{B}##, i.e., it doesn't do any work on the charged particle, and then you also have a conserved energy, and it's the conserved quantity due to time-translation invariance and thus called "energy" by definition (Noether's theorem).

For the general case of time dependent fields the Hamiltonian has no direct physical interpretation since it is gauge dependent and also isn't conserved.
 
  • #10
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So here we have a static field with a velocity term outside the kinetic term but where H = E. so H = E do not need no velocity in the potential term. is it correct?
What i would like to get is a model where there time is not explicit and velocities in the potential term but where H does not equal the total energy.
 
  • #11
Dale
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I think that for the Hamiltonian to not equal the energy requires explicit time, no?
 
  • #12
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Here i call total energy the sum of kinetic and potential energy of the particle(s) = T + U
take the simple Lagrancien L = m/2 v^2 - U(x)
we first calculate the momentum = dL/dv = mv and get the Hamiltonian pv - L = T + U. no problem.
let us take another Lagrangian where time is not explicit:
L = m/2 v^2 - U(x,v)
the momentum dL/dv = mv - dU/dv
and the Hamiltonian is T + U - v dU/dv
(the derivatives are partial)
so H does not equal the total energy in this case.
is there a physical system with such a Lagrangian?
 
  • #13
vanhees71
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But for a particle in a static electromagnetic field ##H## is the total energy. Of course, it's not ##T+U##, because the magnetic forces don't "do work on the particle".
 
  • #14
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i found the sentence in this book
did i misanderstand what Douglas Clide wrote in section 7?
 
  • #15
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What i call here total energy is the energy of the particles in an external field, not the energy of particles plus fields. and it is something concrete . that i can store in batteries to use later not a mathematical object. if i get H = T + U + another thing, cans i use that 3rd thing in real life?
 
  • #16
vanhees71
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That's what I'm also talking about. The total energy of a particle in a static electromagnetic field is
$$E=\frac{m}{2} \dot{\vec{x}}^2+q \Phi(\vec{x}).$$
The important point of course is that the Hamiltonian is this expression expressed in terms of canonical momenta,
$$\vec{p}_{\text{can}}=\frac{\partial L}{\partial \dot{\vec{x}}}=m \dot{\vec{x}} + \frac{q}{c} \vec{A},$$
and thus
$$\hat{H}=\frac{1}{2m} \left (\vec{p}_{\text{can}}-\frac{q}{c} \vec{A} \right)^2 +q \Phi(\vec{x}).$$
The canonical momentum in this case is a gauge dependent quantity and thus not the physical momentum of the particle!
 
  • #17
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thank you,
so the we have an example of Hamiltonian where time is not explicit and wherer H does not give immediately the physical energy of the particle but it contains information about the motion of the charged particle in the field.
 
  • #18
vanhees71
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No ##H## does give the total energy of the particle in this case. At least if you define energy as the conservation law resulting via Noether's theorem from time-translation symmetry, and this is a symmetry if the Lagrangian is not explicitly time-dependent.

For time-dependent electromagnetic fields, I don't think that ##H## has a physical meaning at all. If you take into account the closed system of all charges and the em. field, of course you can define an energy via Noether's theorem, and this energy (mechanical + field energy) will be conserved.
 
  • #19
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there is a notation problem
E is the total energy ot the particle and we have H = E + 2 termq conrainig A
what do you mean when you say that H gives the total energy of the particle? with or without these two terms?
 
  • #20
vanhees71
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No we have
$$L=\frac{m}{2} \dot{\vec{x}}^2 -q \Phi + \frac{q}{c} \dot{\vec{x}} \cdot \vec{A},$$
from which the canonical momentum is
$$\vec{p}_{\text{can}} = \frac{\partial L}{\partial \dot{\vec{x}}} =m \dot{\vec{x}}+ \frac{q}{c} \vec{A}.$$
Now
$$H=\vec{p}_{\text{can}} \cdot \dot{\vec{x}} - L = \frac{m}{2} \dot{\vec{x}}^2 + q \Phi=E.$$
To derive the correct equations of motion from the canonical equations with the Hamiltonian, you have to express the Hamiltonian as a function of ##\vec{x}## and ##\vec{p}_{\text{can}}## to get
$$H=\frac{1}{2m} \left (\dot{\vec{p}}_{\text{can}}^2-\frac{q}{c} \vec{A} \right)^2+ q \Phi.$$
 
  • #21
Dale
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H does give the total energy of the particle in this case. At least if you define energy as the conservation law resulting via Noether's theorem from time-translation symmetry, and this is a symmetry if the Lagrangian is not explicitly time-dependent.
Yes, that is my understanding too. I would always consider the conserved quantity to be the energy, even if it is not T+V
 
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  • #22
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I agree with you, Vanhees71.
All that is fine but may be too fine. i explain:
the Hamiltonian pv - L is perfect with the quadratic term in velocity.
having the kinetic term , pv gives it twice and you subtract if . you get it unchanged from the lagrangian to the hamiltonian.
take the magnetic term vA which do not increase the energy of the particle. pv let in unchanged vA -> A -> vA
and in pv - L it cancels...
in my first post i was looking for cases in which things do not work so fine (if they exist) may be with velocities appearing in powers greater than 2 or in other functions or v.
my initial question arised when i read a sentence written by Feynman in his thesis. it was about cases where the hamiltonian formalism is problematic. i'll try to find what he wrote exactly.

Edit: I found what it was about. In the Wheeler Feynman absorber theory there was an action but no hamiltonian describing the physical process.
 
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  • #23
vanhees71
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I'm still not sure, what you are after. An example, where the Lagrangian has not much in common with the kinetic and "potential" energy is the motion of a relativistic particle in an electromagnetic field. Here the Lagrangian (in the (1+3)-formalism with coordinate time of a Lorentzian reference frame as independent paraemter) reads
$$L=-m c^2 \sqrt{1-\vec{\dot{x}}^2/c^2}-q \Phi + \dot{\vec{x}} \cdot \vec{A}/c.$$
Then you have
$$\vec{p} = m \gamma \dot{\vec{x}}+\frac{q}{c} \dot{\vec{x}} \quad \text{with} \quad \gamma=\frac{1}{\sqrt{1-\dot{\vec{x}}^2/c^2}}.$$
Then
$$H=\vec{p}_{\text{can}} \cdot \dot{\vec{x}}-L = m c^2 \gamma + q \Phi.$$
In the case of static fields, where ##\Phi=\Phi(\vec{x})## and ##\vec{A}=\vec{A}(\vec{x})##, that's the total energy of the particle. Of course, again you have to express ##H## in terms of the canonical momenta and ##\vec{x}##:
$$H=c \sqrt{m^2 c^2 + (\vec{p}-q \vec{A}/c)^2}+q \Phi.$$
Of course to derive the equations of motion both the Lagrangian and the Hamiltonian formulation lead to the correct equations also in the case of time-dependent fields, but in this case of time-dependent fields, I'd not interpret ##H## as the "total energy". [Italics edited in view of #24]
 
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  • #24
Dale
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I'd not interpret H as the "total energy".
Not even in the case of static fields where H is the Noether conserved quantity from time symmetry?
 
  • #25
vanhees71
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No, then it's fine. I referred to my final remark on the general case of time-dependent fields only. I'll edit this sentence to make that clear.
 

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