# The tautological 1-form: Lagrange vs. Hamilton formalism

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• PGaccount
In summary, classical mechanics is based on conservation laws that represent the symmetries of spacetime. The lagrangian function L is a function of position and velocity, while the hamiltonian is a function of position and momentum. The velocity and momentum descriptions are related by a legendre transformation, and the lagrangian and hamiltonian descriptions are in the tangent and cotangent bundles, respectively. The symplectic 2-form dθ is dp ∧ dx, and conservation laws are of the form d*j = 0 where j is the conserved current. Momentum is the generator and conserved quantity of spacetime translations, and the forces are generated by potentials. In the hamiltonian description, the equations of motion are found
PGaccount
Classical mechanics is based on conservation laws which represent the symmetries of spacetime. The lagrangian function L is a function of position and velocity while the hamiltonian is a function of position and momentum. The velocity and momentum descriptions are related by a legendre transformation:

H = pv - L

which is the integration by parts formula ∫ p dv = pv - ∫ v dp. The lagrangian description is in the tangent bundle and the hamiltonian description is in the cotangent bundle. The structure of the cotangent bundle is encoded in the tautological one-form θ = p dx. The symplectic 2-form dθ is dp ∧ dx. Conservation laws are of the form d*j = 0 where j is the conserved current. Momentum is the generator and conserved quantity of spacetime translations exp(ix'p)f(x) = f(x + x'). The forces are generated by potentials F = -dU. The lagrangian is the sum of the kinetic energy T and potential energy U:

L = T + U = 1/2 mv2 + U(x)

The hamiltonian is the sum of the kinetic and potential energies, H = T + U. Symmetries of spacetime are represented by killing vector fields. The conserved currents are Tijkj where k are the killing vectors. T generates conformal transformations. The equations of motion are d*T = 0. In the hamiltonain description, the equations of motion are found from the Lie brackets, {H, X} = dX/dt = dH/dp and {H, P} = dP/dt = -dH/dx. The lagrangian equations of motion are found from δS = 0. Because of the difference between time and space, the relationship between energy and momentum is not exactly symmetric. Momentum is the flux of energy. This can be proved from the fact that T is a symmetric tensor. The conservation laws are closely related to the fact dd = 0, which is due to the fact that the boundary of a boundary is zero.

The hamiltonian description is based on the phase space which is the space of (x, p). This phase space is the space of motions or the space of solutions to the equations of motion. This space can be described abstractly by the symplectic 2-form ##d\alpha \wedge d\tilde{\alpha}##. Take a field

##\phi(x) = \sum_k \frac{1}{\sqrt{2E}} \alpha(k) e^{ipx} + \tilde{\alpha}(k) e^{-ipx}##

the phase space description can be made covariant by taking the paths or solutions the be the states, rather than the initial conditions. Take a path ϒ

## \gamma \rightarrow \gamma + \epsilon \frac{\partial \tilde{\alpha}}{\partial \epsilon} \frac{\partial}{\partial \tilde{\alpha}} ##

where ##\alpha = \frac{\delta}{\delta \tilde{\alpha}}##

##\alpha## is a Heisenberg picture operator which evolves as ##\alpha \rightarrow e^{itE} \alpha e^{-itE} ##. The alphas satify the commutation relation [α, α] = 1. ## [\alpha, \tilde{\alpha}] = 1##. A path is a point in phase space. The usual notion of evolution in time f(t) → f(t + x) must be interpreted not as the value of f at a later time t, but a function f translated in time by x. what is the correct way to describe the operator formalism in a covariant way? In my next post I will describe details. The energy momentum T generates conformal transformations. In string theory, the states of particles are always on-shell.

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Delta2 and weirdoguy
PrashantGokaraju said:
Classical mechanics is based on conservation laws which represent the symmetries of spacetime
what "based" does mean?

PrashantGokaraju said:
The lagrangian function L is a function of position and velocity
and the time

Just for information: there are a lot of problems in classical mechanics which have first integrals (conservation laws) and these first integrals do not follow from the Noether theorem.

PrashantGokaraju said:
which is the integration by parts formula ∫ p dv = pv - ∫ v dp.
what does it mean?

PrashantGokaraju said:
In my next post I will describe details.
for what purpose ? As a whole this look like some kind of showoff: a person who does not know these things will understand nothing, a person who has already known these things just sees how inaccurate and inconsistently you narrate

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Delta2 and weirdoguy
This is a contradiction to one of Noether's theorems, which goes in both directions: Any one-parameter Lie symmetry of the variation of the action implies a conserved quantity. If there is a conservation law the corresponding conserved quantity defines a one-parameter Lie symmetry of the variation of the action. In the Hamiltonian formalism this Lie symmetry tranformation is defined by the conserved quantity as the generator of the representation of this symmetry group in terms of a canonical transformation.

Concerning the fact that strings are always on-shell, The diffeomorphism invariant action for a particle is ##\int e-1(\dot{x}^2 + em^2) dτ## . The equation of motion for e is ##\dot{x}^2 + e^2 m^2 = 0## This is the mass-shell condition in curved space. The conjugate momentum with the gauge choice e = 1/m is pμ = m gμν x'2. This action has τ reparametrization invariance under the transformations δx = ξx' and δe = d/dτ (ξe). The constraint ## \dot{x}^2 + e^2 m^2 = 0 ## is the analogue of the mass-shell condition in curved space. The equation p2 = m2 becomes pipj = m2 gijgik vj vk where ##v = \dot{x}## . In electrodynamics we have the equation p2 gij - pi pj = 0.

The conservation laws in general relativity are expressed in the form d*T = 0, where *T = eμ Tμν d3ν. The Einstein equation is G = 8πT, where *G = eμ Gμν d3ν

weirdoguy
vanhees71 said:
This is a contradiction to one of Noether's theorems, which goes
Are you speaking about my post?

if so: the Noether theorem provides first integrals which are the polynomials of first or second degree in generalized velocities. The Kovalevskaya top has the first integral of 4th degree

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Then the Lagrangian must be of higher order than the first derivatives. Of course then you have to adapt the standard methods to such Lagrangians. I'm pretty sure that then you can also derive Nother's theorems for such higher-order Lagrangians. For standard Hamiltonian systems there's a one-to-one correspondence between the existence of conserved quantities and one-parameter Lie symmetry groups.

BTW: What's a Kovalevskaya top?

vanhees71 said:
Then the Lagrangian must be of higher order than the first derivatives.
classical mechanics as I said
vanhees71 said:
BTW: What's a Kovalevskaya top?
Kovalevskaya, Sofia (1889), "Sur le problème de la rotation d'un corps solide autour d'un point fixe", Acta Mathematica (in French), 12: 177–232

and also

But then Kovalevskaya treats the standard rigid body dynamics, which is a usual Hamiltonian system, and of course for this Noether's theorems are valid in their standard form. So again: If there is a conserved quantity it generates a corresponding symmetry transformation in terms of a canonical transformation in phase space. Where can I find, what you are referring to (preferrably in English or German; my French is close to not existent unfortunately).

vanhees71 said:
If there is a conserved quantity it generates a corresponding symmetry transformation in terms of a canonical transformation in phase space.
Sure, but not any such a symmetry can be obtained by the Noether theorem.
ok let's clarify the terms. What do you call "Noether theorem"?
I call the Noether theorem the following assertion. Assume we have a Lagrangian system ##L=L(x,\dot x)## here ##x## is a point on the configuration manifold ##M## and correspondingly ##(x,\dot x)\in TM##. Assume also we have one parametric group ##g^s:M\to M## which is generated by a vector field ##v(x)\in T_xM##.
Then if
$$L\Big(g^s(x),\frac{\partial g^s}{\partial x}(x)\dot x\Big)=L(x,\dot x)\quad \forall s$$
then there is a first integral $$F(x,\dot x)=\frac{\partial L}{\partial{\dot x}}v$$

vanhees71 and weirdoguy
This is only a very special case.

The most beautiful formulation is in the Hamilton formalism on extended phase space. If you have a Hamiltonian ##H(x,p,t)## and a conserved quantity ##g## then ##g## generates a one-parameter Lie symmetry of the system, i.e., the variation of the action is invariant. The other way is also true, i.e., if there's a one-parameter Lie symmetry group represented by the corresponding canonical transformation then its generator is conserved.

Your case is the special case of this: (a) it assumes not only that the variation of the action is invariant but even the Lagrangian itself is invariant under the symmetry transformation and (b) that the transformation is expressible as a transformation in configuration space (which is a subset of the more general canonical transformations).

The beauty of the Hamiltonian formulation is that you have a Lie algebra defined by the Poisson bracket and thus a Lie derivative defined by the Poisson bracket. Then you get a formalism very similar to the realization of the Lie algebras of the symmetries in quantum theory in terms of commutators of self-adjoint operators.

fresh_42
##t## in the Hamiltonian is not essential. Non autonomous case is easily reduced to the autonomous one.

vanhees71 said:
If you have a Hamiltonian and a conserved quantity then generates a one-parameter Lie symmetry of the system,
I understand this part
vanhees71 said:
, the variation of the action is invariant.
and I do not understand this part. Please be more detailed. What does it mean in the Hamiltonian statement "the variationof the action is invariant" ?

vanhees71 said:
The most beautiful formulation is in the Hamilton formalism on extended phase space. If you have a Hamiltonian and a conserved quantity then generates a one-parameter Lie symmetry of the system, i.e., the variation of the action is invariant. The other way is also true, i.e., if there's a one-parameter Lie symmetry group represented by the corresponding canonical transformation then its generator is conserved.

Your case is the special case of this: (a) it assumes not only that the variation of the action is invariant but even the Lagrangian itself is invariant under the symmetry transformation and (b) that the transformation is expressible as a transformation in configuration space (which is a subset of the more general canonical transformations).
More formally: Assume we have a Hamiltonian system ##H=H(z)## here ##z## belongs to the phase space ##N##. Assume also that we have a one-parametric group of symplectic diffeomorphisms ##g_F^s:N\to N##. This group is generated by a Hamiltonian vector field with Hamiltonian ##F##. Then $$\{F,H\}=0\Longleftrightarrow g_F^sg^t_H=g^t_Hg^s_F,\qquad(*)$$ where ##g_H^t## is the phase flow of the system with Hamiltonian ##H##
I never knew that somebody calls this "Noether theorem" but if you mean that ,then ok

UPD:
Formula (*) is a direct consequence of the following general fact from ODE
Let we have two vector fields ##v,u## with phase flows ##g_v^t,g_u^s## then
$$[v,u]=0\Longleftrightarrow g^t_vg^s_u=g^s_ug^t_v$$
Is this Nother theorem too?

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wrobel said:
##t## in the Hamiltonian is not essential. Non autonomous case is easily reduced to the autonomous one. I understand this part

and I do not understand this part. Please be more detailed. What does it mean in the Hamiltonian statement "the variationof the action is invariant" ?
It's the same as in the Lagrangian statement. You have already a symmetry if the variation of the action is invariant, because then the Euler-Lagrange equations and thus the equations of motion are invariant. This enables more symmetries than the more restrictive demand that the Lagrangian itself is invariant under the symmetry transformation. It boils finally down to the fact that equivalent Lagrangians are only defined up to a total time derivative of a function ##\Omega=\Omega(q,t)##.

This allows, e.g., a "Noetherian treatment" of Galilei-boost invariance.

wrobel said:
More formally: Assume we have a Hamiltonian system ##H=H(z)## here ##z## belongs to the phase space ##N##. Assume also that we have a one-parametric group of symplectic diffeomorphisms ##g_F^s:N\to N##. This group is generated by a Hamiltonian vector field with Hamiltonian ##F##. Then $$\{F,H\}=0\Longleftrightarrow g_F^sg^t_H=g^t_Hg^s_F,\qquad(*)$$ where ##g_H^t## is the phase flow of the system with Hamiltonian ##H##
I never knew that somebody calls this "Noether theorem" but if you mean that ,then ok

UPD:
Formula (*) is a direct consequence of the following general fact from ODE
Let we have two vector fields ##v,u## with phase flows ##g_v^t,g_u^s## then
$$[v,u]=0\Longleftrightarrow g^t_vg^s_u=g^s_ug^t_v$$
Is this Nother theorem too?
It's ##\{F,H \}+\partial_t F=0##. As I said you need this more general case to also include Galilei (or also Lorentz) boosts as Noether symmetries.

vanhees71 said:
It's the same as in the Lagrangian statement. You have already a symmetry if the variation of the action is invariant,
I hoped to see the formulas and precise assertions

vanhees71
Well, that's a bit lengthy for a forum posting. A very good book which contains the general form of Noether's theorem is

F. Scheck, Mechanics, Springer (2010)

fresh_42 said:
Here is how P.Olver (Applications of Lie Groups to Differential Equations) phrased them:
this book does not contain Noether theorem in context of least action principle in the Hamiltonian formalism which is vanhees71 claimed .

vanhees71 said:
Well, that's a bit lengthy for a forum posting
sure that is!

wrobel said:
this book does not contain Noether theorem in context of least action principle in the Hamiltonian formalism which is vanhees71 claimed .
No, it is Euler-Lagrange, but aren't they equivalent?

fresh_42 said:
No, it is Euler-Lagrange, but aren't they equivalent?
Actually not, but it is another song ( is such an expression in English? :) )
Let ##M## be a configuration space then the tangent bundle ##Q=TM## is a phase space of a Lagrangian system ##L=L(x,\dot x)##.
if you consider Euler-Lagrange you can only use diffeomorphisms of##Q## such that they take fiber to fiber in the tangent bundle (else you destroy the Euler-Lagrange shape of equations) . Particularly such diffeomorphisms take a polynomial ##\sum_k a_k(x)\dot x^k## to the polynomial of the same degree. That is what I begun with.

And I really do not understand what is the Noether theorem in Hamiltonian statement in context of least action principle

fresh_42
vanhees71 said:
As you quoted the original paper by Noether, maybe you read German? Then you find it in my manuscripts on mechanics (both in the Lagrangian and the Hamiltonian version)

https://itp.uni-frankfurt.de/~hees/publ/matherg2.pdf
I do not read German but tell me the pages in your book I will try to look through the formulas at least

The Hamiltonian version is on p. 42 onward (it's far from being a book; it's just a manuscript/lecture notes).

vanhees71 said:
he Hamiltonian version is on p. 42 onward
now I feel I am starting to understand what you mean. By the way why did you choose such a language? I mean why do not you use the invariant language of differential geometry? Functions like ##G(q,P,t)## are terribly non invariant and locally defined.

The energy of a particle is

E = p2/2m + U

which means that

dE = p/m dp + dU

The action for a relativistic particle is -m ∫ds, where ds2 = gij dxi dxj.

## S = -m \int ds = -m \int \frac{ds}{d\lambda} ##

The momentum is related to the action by

## \frac{\partial S}{\partial x} = p = -m \int \frac{\partial}{\partial x} \frac{ds}{d\lambda} d\lambda ##

The conservation of momentum is a consequence of the bianchi identity dF = 0. The equation of motion p2 - m2 = 0 just projects out the unphysical degrees of freedom. Not all ways to translate a state in the configuration space are physical degrees of freedom. In a covariant picture, the equation of motion is equivalent to a specification of the physical degrees of freedom.

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weirdoguy
wrobel said:
now I feel I am starting to understand what you mean. By the way why did you choose such a language? I mean why do not you use the invariant language of differential geometry? Functions like ##G(q,P,t)## are terribly non invariant and locally defined.
It's a lecture for physicists, not mathematicians ;-).

This thread has come to an end. The discussion about Hamiltonian versus Lagrangian - although interesting - uses too many non English references, and the OP does not respond to any posts addressed at him.

The thread will therefore be closed.

If you want to continue the discussion about Noether's theorems, please open a new thread. I think this would be an insightful debate.

Here's a hint for any readers who want to read the links of this thread but can't because it's not English. If the links aren't to a photography as the original papers of Noether had to be, then Chrome via "translate this page" often produces a reasonable version in English. It's not perfect, but in most cases with a technical language and many formulas, it will at least provide the context.

berkeman and vanhees71

## 1. What is the tautological 1-form in physics?

The tautological 1-form is a mathematical concept used in the Lagrange and Hamilton formalisms of classical mechanics. It is a differential form that represents the total energy of a system, and it is defined as the sum of the kinetic and potential energies.

## 2. How is the tautological 1-form used in the Lagrange formalism?

In the Lagrange formalism, the tautological 1-form is used to formulate the equations of motion for a system. It is multiplied by the infinitesimal change in the system's coordinates, and then integrated over time to find the path of the system.

## 3. How is the tautological 1-form used in the Hamilton formalism?

In the Hamilton formalism, the tautological 1-form is used to define the Hamiltonian, which is a function that represents the total energy of a system. It is also used to derive the Hamilton's equations of motion, which describe the evolution of a system over time.

## 4. What is the difference between the Lagrange and Hamilton formalisms in terms of the tautological 1-form?

The main difference between the two formalisms is the approach to finding the equations of motion. In the Lagrange formalism, the tautological 1-form is integrated over time to find the path of the system, while in the Hamilton formalism, it is used to define the Hamiltonian and derive the equations of motion.

## 5. What are the advantages of using the tautological 1-form in classical mechanics?

The tautological 1-form provides a concise and elegant way to express the total energy of a system in both the Lagrange and Hamilton formalisms. It also allows for a systematic approach to finding the equations of motion and can be easily extended to more complex systems. Additionally, it is a fundamental concept in symplectic geometry, which has many applications in physics and mathematics.

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