Physical meaning of Lagrangian?

[Erland]

Does the Lagrangian L in classical mechanics have any physical meaning?

In classical mechanics, the Lagrangian is defined as L=T-V, the difference between the kinetic and potential energy of the system. Does this quantity have any meaning apart from that it can be plugged into Euler-Lagrange's equations to find a stationary path?

Can we say anything about the difference in behavior for systems with high and low values of L, respectively?

 

[dextercioby]

To help you reach conclusions on your own, I may ask you: Is the Lagrangian for a system (let us say free Galilean particle) unique?

 

[Erland]

To help you reach conclusions on your own, I may ask you: Is the Lagrangian for a system (let us say free Galilean particle) unique?

No, not unique. We can, in this case, add an arbitrary function f(t), depending only upon time, explicitly, but not upon position and velocity.

 

[samalkhaiat]

We can, in this case, add an arbitrary function f(t), depending only upon time, explicitly, but not upon position and velocity.

No, you can not do that. Two Lagrangians $L_{1}(q, \dot{q})$ and $L_{2}(q, \dot{q})$ are equivalent to each other, if and only if they differ by total time derivative of some arbitrary function of the coordinates only $\frac{d}{dt}F(q)$. This, actually, is a provable theorem.

 

[Geofleur]

Well, it can be the total time derivative of a function of coordinates and time, so $\frac{d}{dt}F(q,t)$. For example, if we add such a term to $\mathcal{L}$, the action becomes

$S = \int_{t_i}^{t_f} \left[ \mathcal{L} + \frac{d}{dt}F(q,t) \right] dt = \int \mathcal{L}dt + F(q_f,t_f)-F(q_i,t_i)$.

The last two terms are constants and thus vanish upon variation of $S$.

 

[samalkhaiat]

Well, it can be the total time derivative of a function of coordinates and time, so $\frac{d}{dt}F(q,t)$.

Yes, you can stick $t$ in $L_{1}$, $L_{2}$ and in $F$ and the theorem still hold. A function of $q_{a}(t)$, $\dot{q}_{a}(t)$, and $t$ satisfies the Euler-Lagrange’s equations identically (i.e., independent of the $q_{a}(t)$’s) if, and only if, it is the total time derivative $dF/dt$ of some function $F(q(t) , t)$:
$$\Big\{ G(q , \dot{q} , t) = \frac{d}{dt}F(q , t) \Big\} \ \Leftrightarrow \ \Big\{ \frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) - \frac{\partial G}{\partial q_{a}} \equiv 0 \Big\}$$
Okay, I leave you to prove it.

 

[Erland]

Not hard to prove, samalkhaiat. So, the conclusion is that since the Lagrangian is not unique, that it can be altered to such a high degree, there is little hope to find any meaningful physical interpretation of it. It's only useful in calculations.

 

[samalkhaiat]

Not hard to prove, samalkhaiat.

It will be nice if you can share your “not hard” proof with us. I’m interested to see your proof. Just show us how to prove the “only if” part. That is,
$$\Big\{ \frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) \equiv \frac{\partial G}{\partial q_{a}} \Big\} \ \Rightarrow \ \Big\{ G(q,\dot{q},t) = \frac{d}{dt}F(q,t) \Big\} .$$

So, the conclusion is that since the Lagrangian is not unique, that it can be altered to such a high degree, there is little hope to find any meaningful physical interpretation of it. It's only useful in calculations.

If you understand the above little theorem, you will have no problem understanding the fundamental importance (or the “meaning”) of the Lagrangian in physics.

 

[Erland]

Ok, not so easy. It is the other way that is quite easy. But please, don't force me to write down the proof. It would take hours getting the TeX-notation right, with all partial derivatves, indices, etc. Just to indicate some important points: using that LHS holds for all possible paths we can use that L doesn't depend upon the accelerations of the coordinates to infer that L is a linear function of the velocities, with coefficient functions depending only on positions and time. Plugging this into the LHS, we obtain in similar way that $\partial A_i/\partial q_j=\partial A_j/\partial q_i$ ($A_i$ are the coefficient functions), from which we can find a function $F(q_1,\dots,q_n,t)$ whose partial derivatives are the $A_i$ and its total time derivative is $L$.

I suspect that one can also prove this by variational principles.

Ok, I know it is considered to be of fundamental importance, but this nonuniqueness seems to indicate that its only use is to be plugged into equations, that it is not meaningful to think of it as a measurable quantity...

 

[samalkhaiat]

Ok, not so easy. It is the other way that is quite easy. But please, don't force me to write down the proof. It would take hours getting the TeX-notation right, with all partial derivatves, indices, etc. Just to indicate some important points: using that LHS holds for all possible paths we can use that L doesn't depend upon the accelerations of the coordinates to infer that L is a linear function of the velocities, with coefficient functions depending only on positions and time. Plugging this into the LHS, we obtain in similar way that $\partial A_i/\partial q_j=\partial A_j/\partial q_i$ ($A_i$ are the coefficient functions), from which we can find a function $F(q_1,\dots,q_n,t)$ whose partial derivatives are the $A_i$ and its total time derivative is $L$.

What do you mean by “$L$”? The theorem in #6 does not mention Lagrangian or any other dynamical concepts. It is just a mathematical theorem.
Let us start again. I will restate the theorem and use it to demonstrate the importance of the Lagrangian. And please pay attention to the under-lined words.
“A function $G(q,\dot{q},t)$ satisfies Euler-Lagrange equation identically, if and only if, it can be written as a total time derivative $dF/dt$ of the arbitrary function $F(q,t)$.”
The crucial point in here is the statement that the Euler-Lagrange “equation” of $G$ is an identity, i.e., it does not lead to differential equations. This implies that
$$\frac{\partial^{2}G}{\partial \dot{q}_{a}\partial \dot{q}_{b}} = 0 . \ \ \ \ \ \ \ \ \ (1)$$ This in turn means that $G$ is at most linear in the $\dot{q}_{a}$’s
$$G(q,\dot{q},t) = \sum_{a} f_{a}(q,t) \dot{q}_{a} + g(q,t) . \ \ \ \ \ \ (2)$$
Let’s compare the above with the definition of Lagrangian: “The function $L(q,\dot{q},t)$ is a genuine Lagrangian, if and only if the Euler-Lagrange equation of $L$ leads to differential equations of order no higher than the second in time derivatives.”
This means that, almost always
$$\frac{\partial^{2}L}{\partial \dot{q}_{a}\partial \dot{q}_{b}} \neq 0 . \ \ \ \ \ \ \ \ \ (3)$$
This implies that $L$ is, almost always, a quadratic function of the velocities $\dot{q}_{a}$.
$$L(q,\dot{q},t) = \sum_{a,b} g_{ab}(q,t) \dot{q}_{a}\dot{q}_{b} + \sum_{a} f_{a}(q,t)\dot{q}_{a} + g(q,t)$$
Since $G = dF/dt$ leads to no dynamical equation, the theorem implies the followings:
(1) A genuine Lagrangian cannot be a total time derivative of some function of the $q_{a}$’s and time.


(2) Let $A$ and $B$ be two physical systems described by $L_{A}$ and $L_{B}$ respectively. If for any $F(q,t)$
$$L_{A} \neq L_{B} + \frac{d}{dt}F(q,t) ,$$ then $A$ and $B$ are two distinct physical systems.

(3) If there exists some function $\Lambda (q,t)$ such that $$L_{A} = L_{B} + \frac{d}{dt}\Lambda (q,t) ,$$ then $A$ and $B$ are identical physical systems related by a group of symmetry transformations, i.e., $L$ and $L + d\Lambda /dt$ describe one unique physical system whose constants of motion are given by
$$C_{\epsilon} = \frac{\partial L}{\partial \dot{q}_{a}} \delta_{\epsilon}q_{a} + \Lambda_{\epsilon}(q,t) .$$

Ok, I know it is considered to be of fundamental importance, but this nonuniqueness seems to indicate that its only use is to be plugged into equations, that it is not meaningful to think of it as a measurable quantity...

It is, I hope, clear from the above that the correct Lagrangian determines all the measurable dynamical quantities of the physical system. Do physicists wish for a better object than $L$? No, because there is none.
We also cannot measure the electromagnetic potential, the wave function in QM, the (phase-space) density of microstates, the entropy, the fields associated with elementary particles etc. Does this cast doubts on the fundamental importance of these quantities?

 

[Erland]

Ok, so I wrote L instead G, and my proof was along the same lines as yours...

Thanks for your explanations. Yes, the Lagrangian(s) is very useful, it contains all information we can extract. But still, it seems to lack physical intuition, just as the wave function in QM, but at least the electrostatic potential has the interpretation that a positively charged particle tends to move from high to low potential...

So I agree, it is very useful, but there is the pedagogical difficulty that it is hard to grasp...

 

[ShayanJ]

I don't want to hijack this thread, so if Erland is satisfied with the answers, I want to ask this question.

This means that, almost always

$\frac{\partial^2 L}{\partial\dot q_a \partial \dot q_b}\neq0$​

Could you explain about those situations which made you say "almost always"?(You could point to a book or paper in case that's preferable)
Thanks

 

[samalkhaiat]

I don't want to hijack this thread, so if Erland is satisfied with the answers, I want to ask this question.

Could you explain about those situations which made you say "almost always"?(You could point to a book or paper in case that's preferable)
Thanks

Well I don't think you need book or a paper, you should know it. If the coordinates are c-numbers taking values in Grassmann algebra, i.e., anticommuting numbers, the Lagrangian (which is bosonic) must be linear in the "velocities" $\dot{q}_{a}$, just like the classical Dirac Lagrangian.

 

[EnigmaticField]

The crucial point in here is the statement that the Euler-Lagrange “equation” of $G$ is an identity, i.e., it does not lead to differential equations.

I was puzzled for long before finally understanding what you mean by "not lead to differential equations": $G$ always satisfies the Euler-Lagrange equations, which is not to be solved for $q(t)$ while the Lagrangian $L$ only satisfies the Euler-Lagrange equations when the actual path of motion in the configuration space (q, t) between two fixed time points is chosen, and this actual path $q(t)$ is the solution of the differential equations given by the Euler-Lagrange equations of $L$.

If $\frac{\partial^2 L}{\partial\dot{q}_a\partial\dot{q}_b}=0$, then
$L(q,\dot{q},t) = \sum_{a} f_{a}(q,t)\dot{q}_{a} + g(q,t)$, and then the canonical momentum $p_a=\partial L/\partial\dot{q_a}=f_{a}(q,t)$ isn't a function of the generalized velocity $\dot{q}$. Isn't this weird? (Consider the mechanical momentum is defined to be $p_x=m\dot{x}$, where $x$ is the Cartesian coordinate.)

 

[samalkhaiat]

I was puzzled for long before finally understanding what you mean by "not lead to differential equations": $G$ always satisfies the Euler-Lagrange equations, which is not to be solved for $q(t)$ while the Lagrangian $L$ only satisfies the Euler-Lagrange equations when the actual path of motion in the configuration space (q, t) between two fixed time points is chosen, and this actual path $q(t)$ is the solution of the differential equations given by the Euler-Lagrange equations of $L$.

For $G$, the Euler-Lagrange equation is not a equation. So there is no equation to solve, it simply gives you $X = X$. This is why I used the sign ($\equiv$) instead of the equal sign to write the E-L equation for $G$. On the other hand, for a genuine Lagrangian, the E-L equation produces a set of dynamical differential equations $\Phi_{a}(\ddot{q},\dot{q},q) = 0$ which, in principle, can be solved given an appropriate set of initial conditions $q_{a}(t) = f_{a}(q(0), \dot{q}(0);t)$.

If $\frac{\partial^2 L}{\partial\dot{q}_a\partial\dot{q}_b}=0$, then
$L(q,\dot{q},t) = \sum_{a} f_{a}(q,t)\dot{q}_{a} + g(q,t)$, and then the canonical momentum $p_a=\partial L/\partial\dot{q_a}=f_{a}(q,t)$ isn't a function of the generalized velocity $\dot{q}$. Isn't this weird? (Consider the mechanical momentum is defined to be $p_x=m\dot{x}$, where $x$ is the Cartesian coordinate.)

Nothing is weird about it. The Dirac Lagrangian and all gauge constraint systems have singular Hessian matrix. The treatment of constraint system is a very rich and annoying (i.e., hard) subject.
Let me just few words about this. In detailed form, the E-L equations are
$$\frac{\partial^{2}L}{\partial \dot{q}_{a} \partial \dot{q}_{b}} \ddot{q}_{b} = \frac{\partial L}{\partial q_{a}} - \frac{\partial^{2}L}{\partial \dot{q}_{a} \partial q_{b}} \dot{q}_{b} .$$
From this, we see that the accelerations $\ddot{q}_{b}$ at a given time are uniquely determined by the $\dot{q}_{a}$’s and the $q_{a}$’s at that time if, and only if, the Hessian matrix $$M_{ab} = \frac{\partial^{2}L}{\partial \dot{q}_{a} \partial \dot{q}_{b}} , \ \ a = 1, 2 , …, n \ \ \ (1)$$ can be inverted, i.e., $\det |M| \neq 0$. If, on the other hand, the Hessian is singular, i.e., $\det |M| = 0$, the accelerations will not be uniquely determined and the solution to the equations of motion could contain arbitrary functions of time. For the canonical momenta $$p^{a} = \frac{\partial L}{\partial \dot{q}_{a}} ,$$ singular Hessian is just the condition for non-solvability of the velocities as functions of the coordinates and momenta. This means that the $p^{a}$’s are not all independent, rather, there are some relations
$$\Psi_{i}(q,p) = 0 , \ \ i = 1,2, …, r \ \ \ \ \ \ (2)$$ that follow from the defining relations (1). That is to say that replacing the $p$’s in (2) by their definitions (1) in terms of the $q$’s and $\dot{q}$’s, equation (2) becomes an identity. The relations (2) are called primary constraints to indicate that the equation of motion are not used to obtain these conditions and that they implies no restrictions on the local coordinates $(q_{a},\dot{q}_{b})$ in the configuration space. However, they do define a submanifold smoothly embedded in the $(q_{a},p^{b})$ phase-space. This submanifold is called the primary constraint surface.

 

[Erland]

You don't need to know anything of Grassmann algebras etc. to prove this

$$\Big\{ \frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) \equiv \frac{\partial G}{\partial q_{a}} \Big\} \ \Rightarrow \ \Big\{ G(q,\dot{q},t) = \frac{d}{dt}F(q,t) \Big\} .$$

Ok, so assume that $\frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) \equiv \frac{\partial G}{\partial q_{a}}$.
This can be expanded and rewritten as

$$\sum_b\frac{\partial^2 G}{\partial\dot q_b\partial\dot q_a}\ddot q_b\equiv \frac{\partial G}{\partial q_a}-\sum_b\frac{\partial^2 G}{\partial q_b\partial\dot q_a}\dot q_b-\frac{\partial^2 G}{\partial{t}\partial q_a}.$$
This should hold for all $q_b$:s, $\dot q_b$:s, $\ddot q_b$:s, and $t$. Fix the $q_b$:s, $\dot q_b$:s, and $t$. If we put $\ddot q_b=0$ for all $b$, the LHS vanishes. Thus the RHS vanishes too. But since the RHS does not depend on the $\ddot q_b$:s, this means that the LHS vanishes for all choices of the $\ddot q_b$:s, for example if one of them is $1$ and the all the other $0$, which means that $\frac{\partial^2 G}{\partial\dot q_b\partial\dot q_a}=0$. This holds for all $a$ and $b$. From this one can infer, as indicated in previous posts, that
$$G(q,\dot{q},t) = \frac{d}{dt}F(q,t).$$

 

[Erland]

Ooops, should be
$$\sum_b\frac{\partial^2 G}{\partial\dot {q_b}\partial\dot {q_a} }\ddot {q_b}\equiv \frac{\partial G}{\partial q_a}-\sum_b\frac{\partial^2 G}{\partial q_b\partial\dot {q_a}}\dot {q_b}-\frac{\partial^2 G}{\partial{t}\partial\dot {q_a}}.$$
Continuing, writing
$$G(q_1,\dots,q_n,\dot{q_1},\dots,\dot {q_n},t) = \sum_{b} f_{b}(q_1,\dots,q_n,t) \dot{q_b} + g(q_1,\dots,q_n,t),$$we plug this into the RHS of the former equation, which is identically $0$, and using that the $\dot{q_b}$:s are independent, we obtain $\partial f_b/\partial q_a=\partial f_a/\partial q_b$ and $\partial g/\partial q_a=\partial f_a/\partial t$ for all $a$ and $b$.
This means that there is a function $F(q_1,\dots,q_n,t)$ such that $\partial F/\partial q_a=f_a$ for all $a$, and $\partial F/\partial t=g$.

For this $F$
$$G(q_1,\dots,q_n,\dot{q_1},\dots,\dot {q_n},t) =\frac d{dt}F(q_1,\dots,q_n,t)$$
as desired.

But I have a question. This $F$ might exist only locally, if the domain (phase space) does not have the "right" type of connectedness, right? If e.g. the domain is 2-dimensional, then the domain must be simply connected (without "holes"), otherwise it might not be possible to "glue together" the locally defined $F$:s, right?

 

[samalkhaiat]

You don't need to know anything of Grassmann algebras etc. to prove this

No, you don't, but who said you do need Grassmann numbers to prove the theorem? The question from Shyan to me was about the reason for using the phrase "almost always" in post #10.

 

[EnigmaticField]

Does the Lagrangian L in classical mechanics have any physical meaning?

Though the Lagrangian seems not to correspond to an informative physical observable, the action integrand $Ldt=p_adq^a-Hdt$ represents the energy-momentum 1-form in Minkowski spacetime, where $p_a's$ are the canonical momenta and $H$ is the Hamiltonian, if the energy-momentum 4-vector is given by $P=H\partial/\partial t+p_a\partial/\partial q^a$, which applies to a nonrelativistic system as well as a relativistic system. This may give a little justification for that the equations of motion are given by $\delta\int^{t_2}_{t_1} Ldt=0$.

If there exists some function $\Lambda (q,t)$ such that $$L_{A} = L_{B} + \frac{d}{dt}\Lambda (q,t) ,$$ then $A$ and $B$ are identical physical systems related by a group of symmetry transformations, i.e., $L$ and $L + d\Lambda /dt$ describe one unique physical system whose constants of motion are given by
$$C_{\epsilon} = \frac{\partial L}{\partial \dot{q}_{a}} \delta_{\epsilon}q_{a} + \Lambda_{\epsilon}(q,t) .$$

Why is this called a group of symmetry transformations? This is just a shift of the boundary term of the action integral. In gravitational field theories I learned one can utilize the adjustment of the boundary term of the action integral or the Hamiltonian integral to change the boundary condition, such as from the Dirichlet to Neumann boundary conditions, because different boundary terms correspond to different boundary conditions, which is in some sense like the Legendre transformation. But $\Lambda(q, t)$ here seems not to have this function because $\Lambda$ is not a function of $\dot{q}$. In gravitational field theories the added boundary term can be a function of $\partial_\mu{Q}$ as well as ${Q}$. Is it possible to extend $\Lambda(q, t)$ to $\Lambda(q, \dot{q}, t)$ here? Since Hamilton's principle has restricted the generalized coordinates at the time endpoints $t_1$ and $t_2$ not to be varied, the variation of the added boundary term should vanish even $\Lambda$ is a function of $\dot{q}$ as well: $\delta\big(\Lambda(q(t_2), \dot{q}(t_2) ,t_2)-\Lambda(q(t_1), \dot{q}(t_1), t_1)\big)=0$.

A function of $q_{a}(t)$, $\dot{q}_{a}(t)$, and $t$ satisfies the Euler-Lagrange’s equations identically (i.e., independent of the $q_{a}(t)$’s) if, and only if, it is the total time derivative $dF/dt$ of some function $F(q(t) , t)$:
$$\Big\{ G(q , \dot{q} , t) = \frac{d}{dt}F(q , t) \Big\} \ \Leftrightarrow \ \Big\{ \frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) - \frac{\partial G}{\partial q_{a}} \equiv 0 \Big\}$$

Then in field mechanics, do we have
$$\Big\{ G(Q , \partial_\mu{Q} , x^\mu) = \frac{d}{dt}F(Q , x^\mu) \Big\} \ \Leftrightarrow \ \Big\{ \frac{d}{dx^\mu} \left( \frac{\partial G}{\partial(\partial_\mu{Q}_a)} \right) - \frac{\partial G}{\partial Q_{a}} \equiv 0 \Big\}$$?

 

[samalkhaiat]

Why is this called a group of symmetry transformations? This is just a shift of the boundary term of the action integral.

The reason is in the word "some".
Which means that not all $F(q,t)$, but rather some function $\Lambda(q.t)$, i.e., you obtain the Lagrangian $L_{B}$ plus $d\Lambda_{\epsilon}/dt$ when you transform $L_{A}$ under some transformation $T(\epsilon)$ of the coordinates and time.

In gravitational field theories

That is completely different story, diffeomorphism is a "gauge" group which is the symmetry group in the second Noether theorem.

Is it possible to extend $\Lambda(q, t)$ to $\Lambda(q, \dot{q}, t)$ here?

If by "here" you mean general relativity and diffeomorphism invariance, then my answer is Yes. You can split the H-E action of GR this way.

Then in field mechanics, do we have
$$\Big\{ G(Q , \partial_\mu{Q} , x^\mu) = \frac{d}{dt}F(Q , x^\mu) \Big\} \ \Leftrightarrow \ \Big\{ \frac{d}{dx^\mu} \left( \frac{\partial G}{\partial(\partial_\mu{Q}_a)} \right) - \frac{\partial G}{\partial Q_{a}} \equiv 0 \Big\}$$?

Yes.

 

[lightarrow]

Does the Lagrangian L in classical mechanics have any physical meaning?
In classical mechanics, the Lagrangian is defined as L=T-V, the difference between the kinetic and potential energy of the system. Does this quantity have any meaning apart from that it can be plugged into Euler-Lagrange's equations to find a stationary path?
Can we say anything about the difference in behavior for systems with high and low values of L, respectively?

I've seen a lot of learned answers, but mine will be very poor and can only have just an "heuristic" value (if it have at all).
$L = T - V = 2T - (T + V) = 2T - E = p v - \hbar \omega = \hbar k v - \hbar \omega = \hbar (k v - \omega) = \hbar d(k r - \omega t)/dt = \hbar d \phi /dt$ where p is momentum, v is velocity, $\phi$ is the classical phase of a wavenction in QM; so L would represent the time derivative of a phase, multiplied by $\hbar$. Indeed, integrating L in dt we get the action S, which, as known, represents the phase I've written up.

--
lightarrow

 

[EnigmaticField]

The reason is in the word "some".
Which means that not all $F(q,t)$, but rather some function $\Lambda(q.t)$, i.e., you obtain the Lagrangian $L_{B}$ plus $d\Lambda_{\epsilon}/dt$ when you transform $L_{A}$ under some transformation $T(\epsilon)$ of the coordinates and time.

Do you mean if two Lagrangians $L_{A}$ and $L_{B}$ happen to differ by a total time derivative of some function $\Lambda(q, t)$, that is, $L_{A}-L_{B}=\frac{d\Lambda(q, t)}{dt}$, (and thus then $L_{A}$ and $L_{B}$ describe the same physical system,) because $\Lambda(q, t)$ is not unique, for some $\Lambda(q, t)$, say, $\Lambda_1(q, t)$, $\frac{\partial L}{\partial \dot{q}^a}\delta q^a$ and $\frac{\partial L}{\partial \dot{q}^a}\delta q^a+\delta\Lambda_1(q, t)$ are the same conservative quantity expressed in two different sets of coordinates, and for some $\Lambda(q, t)$, say, $\Lambda_2(q, t)$, $\frac{\partial L}{\partial \dot{q}^a}\delta q^a$ and $\frac{\partial L}{\partial \dot{q}^a}\delta q^a+\delta\Lambda_2(q, t)$ aren't related by a coordinate transformation? If that's the case, what's the relation between the two conservative quantities in the latter case?

If by "here" you mean general relativity and diffeomorphism invariance, then my answer is Yes. You can split the H-E action of GR this way.

Excuse me, by "here" I mean the current case, the Lagrangian formalism for the particle mechanics. I just checked and found $\frac{d\Lambda(\dot{q}, q, t)}{dt}$, with $\Lambda(\dot{q}, q, t)$ being an arbitrary function of $\dot{q}, q, t$, doesn't satisfy the Euler-Lagrange equations. But if we vary $\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt$, then interchange $\delta$ with $\frac{d}{dt}$, we get $$\delta\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt=\int^{t_2}_{t_1}\frac{d(\delta\Lambda(\dot{q}, q, t))}{dt}dt=\int^{t_2}_{t_1}d(\delta\Lambda(\dot{q}(t), q(t), t))=\delta\Lambda(\dot{q}(t_2), q(t_2), t_2)-\delta\Lambda(\dot{q}(t_1), q(t_1), t_1)=0.$$ However, the Euler-Lagrange equations of $\frac{d\Lambda(\dot{q}, q, t)}{dt}$ result from varying $\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt$ with the assumption that the variation of $q$ at the endpoints vanishes, but $\delta\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt=\int^{t_2}_{t_1}d\large(\delta\Lambda(\dot{q}(t), q(t), t)\large)$ is a pure boundary integral so vanishes by assumption. Therefore, $\delta\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt=0$ is reached by the assumption that variation on the boundary vanishes, not by the Euler-Lagrange equations, so why can't we just say $L(\dot{q}, q, t)$ and $L(\dot{q}, q, t)+\frac{d\Lambda(\dot{q}, q, t)}{dt}$ (which is more general than $L(\dot{q}, q, t)+\frac{d\Lambda(q, t)}{dt}$) describe the same physical system?

I've seen a lot of learned answers, but mine will be very poor and can only have just an "heuristic" value (if it have at all).
$L = T - V = 2T - (T + V) = 2T - E = p v - \hbar \omega = \hbar k v - \hbar \omega = \hbar (k v - \omega) = \hbar d(k r - \omega t)/dt = \hbar d \phi /dt$ where p is momentum, v is velocity, $\phi$ is the classical phase of a wavenction in QM; so L would represent the time derivative of a phase, multiplied by $\hbar$. Indeed, integrating L in dt we get the action S, which, as known, represents the phase I've written up.

--
lightarrow

I think your answer is a good interpretation, but a good interpretation of the action integral, not the Lagrangian, in quantum mechanics. Indeed, in quantum mechanics, the probability amplitude for two spacetime points, $(x_1, t_1)$ and $(x_2,t_2)$, is proportional to the exponential of the classical action for the path between the two spactime points, that is, $$<x_2,t_2\mid x_1,t_1> \varpropto\textrm{exp}\frac{\textrm{i}}{\hbar}\int^{t_2}_{t_1}Ldt.$$ The phase, as expressed by the action integral divided by $\hbar$ here, can indeed be observed, e.g. through the interference pattern in the Aharonov-Bohm experiment.

So it's like though the Lagrangian has no informative physical interpretation, the action integrand can be interpreted as the energy-momentum 1-form, as I stated in the post # 19, and the action integral divided by $\hbar$ can be interpreted as the phase of the probability amplitude for two spacetime points in quantum mechanics. However, when one composes a Lagrangian for a physical to study its mechanics, one criterion being referred is that the action integrand needs to be parameterization invariant. So I think it's more justified to ask the physical interpretation of the action integrand or the action integral than that of the Lagrangian, which is parameterization variant.

 

[samalkhaiat]

Do you mean if two Lagrangians $L_{A}$ and $L_{B}$ happen to differ by a total time derivative of some function $\Lambda(q, t)$, that is, $L_{A}-L_{B}=\frac{d\Lambda(q, t)}{dt}$, (and thus then $L_{A}$ and $L_{B}$ describe the same physical system,) because $\Lambda(q, t)$ is not unique, for some $\Lambda(q, t)$, say, $\Lambda_1(q, t)$, $\frac{\partial L}{\partial \dot{q}^a}\delta q^a$ and $\frac{\partial L}{\partial \dot{q}^a}\delta q^a+\delta\Lambda_1(q, t)$ are the same conservative quantity expressed in two different sets of coordinates, and for some $\Lambda(q, t)$, say, $\Lambda_2(q, t)$, $\frac{\partial L}{\partial \dot{q}^a}\delta q^a$ and $\frac{\partial L}{\partial \dot{q}^a}\delta q^a+\delta\Lambda_2(q, t)$ aren't related by a coordinate transformation? If that's the case, what's the relation between the two conservative quantities in the latter case?

No, I meant when you apply the transformation $T(\epsilon)$ to $L_{A}$ you obtain $L_{B} + d\Lambda / dt$. In this case, $\Lambda$ IS unique and totally determined by the type of the transformation $T$. For example, If the transformation is an infinitesimal time translation $\delta t$, then $\Lambda = L \delta t$.

Excuse me, by "here" I mean the current case, the Lagrangian formalism for the particle mechanics. I just checked and found $\frac{d\Lambda(\dot{q}, q, t)}{dt}$, with $\Lambda(\dot{q}, q, t)$ being an arbitrary function of $\dot{q}, q, t$, doesn't satisfy the Euler-Lagrange equations. But if we vary $\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt$, then interchange $\delta$ with $\frac{d}{dt}$, we get $$\delta\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt=\int^{t_2}_{t_1}\frac{d(\delta\Lambda(\dot{q}, q, t))}{dt}dt=\int^{t_2}_{t_1}d(\delta\Lambda(\dot{q}(t), q(t), t))=\delta\Lambda(\dot{q}(t_2), q(t_2), t_2)-\delta\Lambda(\dot{q}(t_1), q(t_1), t_1)=0.$$ However, the Euler-Lagrange equations of $\frac{d\Lambda(\dot{q}, q, t)}{dt}$ result from varying $\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt$ with the assumption that the variation of $q$ at the endpoints vanishes, but $\delta\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt=\int^{t_2}_{t_1}d\large(\delta\Lambda(\dot{q}(t), q(t), t)\large)$ is a pure boundary integral so vanishes by assumption. Therefore, $\delta\int^{t_2}_{t_1}\frac{d\Lambda(\dot{q}, q, t)}{dt}dt=0$ is reached by the assumption that variation on the boundary vanishes, not by the Euler-Lagrange equations, so why can't we just say $L(\dot{q}, q, t)$ and $L(\dot{q}, q, t)+\frac{d\Lambda(\dot{q}, q, t)}{dt}$ (which is more general than $L(\dot{q}, q, t)+\frac{d\Lambda(q, t)}{dt}$) describe the same physical system?

Since, I did not understand any of this, can you give me an example? Let say I give you the simplest Lagrangian $L = \frac{1}{2} (\frac{dx}{dt})^{2}$, and you give me that function you are talking about.

 

[EnigmaticField]

No, I meant when you apply the transformation $T(\epsilon)$ to $L_{A}$ you obtain $L_{B} + d\Lambda / dt$. In this case, $\Lambda$ IS unique and totally determined by the type of the transformation $T$.

Then I misunderstood what you meant by $``$ The reason is in the word "some".
Which means that not all F(q,t), but rather some function Λ(q.t) $"$.
I thought your "some" means "a few". Your "some" should mean "a certain".

If I understand your ideas correctly this time, you mean whenever a given Lagrangian $L$ is subject to a symmetry transformation $T(\varepsilon)$, there is a uniquely corresponding $\Lambda(q, t)$ so that the transformed Lagrangian is $L+\frac{d\Lambda(q, t)}{dt}$ and the transformed conservative quantity is $\frac{\partial L}{\partial \dot{q}^a}\delta q^a+\delta\Lambda(q, t)$. Then if conversely, two Lagrangians happen to differ by a total time derivative of a function $\Lambda(q, t)$, that is, $L_A-L_B=\frac{d\Lambda(q, t)}{dt}$, does there exist a unique transformation $T(\varepsilon)$ to account for it?
In the former case $T(\varepsilon)\Rightarrow\Lambda(q,t)$, must $T(\varepsilon)$ be restricted to a transformation which doesn't change the Euler-Lagrangian equations? And in the latter case $\Lambda(q,t)\Rightarrow T(\varepsilon)$, $T(\varepsilon)$should automatically keep the Euler-Lagrangian equations invariant?

But isn't it possible to choose a variation to be a symmetry transformation (fall in a 1-parameter subgroup, say, $S(\lambda)$)? So in this case the conservative quantity $\frac{\partial L}{\partial \dot{q}^a}\delta q^a+\delta\Lambda$ is the result of two symmetry transformations $T(\varepsilon)$ and $S(\lambda)$? (I guess your $\Lambda_\varepsilon$ is just $\delta\Lambda$.)

Your associating the change of the Lagrangian with a symmetry transformation recollects me the gauge field theory, where the Lagrangian of a matter field is variant under the transformation of a local symmetry group $G(\alpha)$, so a $G$ gauge field which transforms in a certain way is introduced into the Lagrangian to make the resultant Lagrangian invariant under the local $G$ symmetry transformation. But here introducing a time-dependent function to cancel the change of the Lagrangian $\frac{d\Lambda}{dt}$ resulting from the transformation $T(\varepsilon)$ seems to have no good interpretation.

For example, If the transformation is an infinitesimal time translation $\delta t$, then $\Lambda = L \delta t$.

What does $``L"$ represent?

Since, I did not understand any of this, can you give me an example? Let say I give you the simplest Lagrangian $L = \frac{1}{2} (\frac{dx}{dt})^{2}$, and you give me that function you are talking about.

I was just stating a well-known notion in the Lagrangian formalism. I'm sorry if I didn't make my explication perspicuous to you. Since you couldn't understand what I said, I try to elaborate more as follows. Hope you can understand it now.

What I want to say is the Euler-Lagrange equations are derived by varying the action $\int Ldt$ with the assumption that the variation $\delta q$ on the boundary (in the current 1-dimensional case, the boundary is just the two end points) vanishes, so the addition of a term which is a total time derivative of any function to the Lagrangian would lead to the same Euler-Lagrange equations as those derived from the original Lagrangian and therefore the new Lagrangian and original Lagrangian should describe the same physical system. The assumption that the variation $\delta q$ on the boundary vanishes is in order to assure that the derived Euler-Lagrange equations are well-defined. (Actually such an assumption is over-restrictive: requiring the boundary integral of the action variation to vanish is sufficient.) Now what you want to add to the Lagrangian $L$, $\frac{d\Lambda}{dt}$, is exactly a total time derivative term, so no matter what function $\Lambda$ you choose, the new Lagrangion $L+\frac{d\Lambda}{dt}$ and the original Lagrangian $L$ should describe the same physical system. Indeed, such an addition would only contribute to the boundary integral of the action variation (not the bulk integral of the action variation, which determines the Euler-Lagrange equations), which would, when we withdraw the assumption that variation on the boundary vanishes and assume the Euler-Lagrange equations are already satisfied, lead to the change of the conservative quantity.

Since $\Lambda$ can be chosen arbitrarily, why must you restrict $\Lambda$ to be a function of only $t, q$? I think it can even be a function of $t, q, \dot{q}, \frac{d^2q}{dt^2}...\frac{d^nq}{dt^n}$, $n$ being any positive integer. But since here we are talking about the Lagrangian which is a function of only $t, q, \dot{q}$, I ask you why we can't choose $\Lambda$ to be a function of $t, q, \dot{q}$ instead of a function of just $t, q$, that is, why we can't just say $L(\dot{q}, q, t)$ and $L(\dot{q}, q, t)+\frac{d\Lambda(\dot{q}, q, t)}{dt}$ describe the same physical system.

Based on the example you give, $L=\frac{1}{2} (\frac{dx}{dt})^{2}$, the simplest representive example which differs by $\frac{d\Lambda(dx/dt, x, t)}{dt}$ from $L$ is $L'=\frac{1}{2}(\frac{dx}{dt})^{2}+\frac{d}{dt}(\frac{dx}{dt}+x+t)=\frac{1}{2}(\frac{dx}{dt})^{2}+\frac{d^2x}{dt^2}+\frac{dx}{dt}+1$, where $\Lambda(dx/dt, x, t)=\frac{dx}{dt}+x+t$. It's easy to check that $L$ and $L'$ give the same equation of motion.In addition, I still have some doubt regarding your following statement.

(2) Let $A$ and $B$ be two physical systems described by $L_{A}$ and $L_{B}$ respectively. If for any $F(q,t)$

$$L_{A} \neq L_{B} + \frac{d}{dt}F(q,t) ,$$ then $A$ and $B$ are two distinct physical systems.(3) If there exists some function $\Lambda (q,t)$ such that $$L_{A} = L_{B} + \frac{d}{dt}\Lambda (q,t) ,$$ then $A$ and $B$ are identical physical systems related by a group of symmetry transformations, i.e., $L$ and $L + d\Lambda /dt$ describe one unique physical system whose constants of motion are given by

$$C_{\epsilon} = \frac{\partial L}{\partial \dot{q}_{a}} \delta_{\epsilon}q_{a} + \Lambda_{\epsilon}(q,t) .$$

According to your statement, the Lagrangians describing the same physical system can only differ by a total time derivative term. But on page 21 of the book Classical Mechanics, Third Edition, Goldstein, Poole & Safko, it says $``...$ if $L(q, \dot{q}, t)$ is an approximate Lagrangian and $F(q,t)$ is any differentiable function of the generalized coordinates and time, then $L'(q, \dot{q}, t)= L(q, \dot{q}, t)+\frac{dF}{dt}$ is a Lagrangian also resulting in the same equations of emotion. It is also often possible to find alternative Lagrangians beside those constructed by this prescription.$..."$ What's your comment about the underlined statement?

 

[samalkhaiat]

Then if conversely, two Lagrangians happen to differ by a total time derivative of a function $\Lambda(q, t)$, that is, $L_A-L_B=\frac{d\Lambda(q, t)}{dt}$, does there exist a unique transformation $T(\varepsilon)$ to account for it?

That depends on the resulting $\Lambda$. Each symmetry transformation $T(\epsilon)$ leads to a unique $\Lambda_{\epsilon}(q,t)$ and, therefore unique constant of motion. Of course, the same Lagrangian can have different symmetries and therefore different $\Lambda$. To avoid ambiguity caused by ordinary language, I will explain this in few examples:
1) Particle (of unit mass) moving in the uniform gravitational field $g$ can be described by an action principle with anyone of the following Lagrangians
$$L_{1} = \frac{1}{2} (\frac{dz}{dt})^{2} - gz$$
$$L_{2} = \frac{1}{2} (\frac{dz}{dt})^{2} - gz + g \alpha ,$$
$$L_{3} = \frac{1}{2} (\frac{dz}{dt})^{2} - gz -\beta \frac{dz}{dt} + \frac{1}{2}\beta^{2} + \beta gt ,$$
where $\alpha$ and $\beta$ are constants.
Now,
$$L_{2} - L_{1} = g\alpha = \frac{d}{dt}(gt \alpha ) .$$
Thus $\Lambda_{\alpha} = gt \alpha$. This implies that $z \to z + \alpha$ is a symmetry and the corresponding conserved quantity in this case is simply the initial velocity $\dot{z}_{0} = \dot{z} + gt$:
$$Q = \frac{\partial L_{1}}{\partial \dot{z}} \delta z + \Lambda_{\alpha} = \alpha (\dot{z} + gt) .$$
Now consider the difference
$$L_{3} - L_{1} = \frac{d}{dt}\left( - (z - \frac{1}{2} gt^{2}) \beta + \frac{1}{2} \beta^{2} t \right) .$$
Thus, for infinitesimal $\beta$, we have
$$\delta \Lambda_{\beta} = (- z + \frac{1}{2}gt^{2}) \beta ,$$
and the infinitesimal symmetry transformations are
$$z \to z + \beta t , \ \ \dot{z} \to \dot{z} + \beta .$$
In this case, the constant of motion is just the initial position
$$z_{0} \beta = \frac{\partial L_{1}}{\partial \dot{z}} \delta z + \delta \Lambda_{\beta} = (\dot{z}t - z + \frac{1}{2}gt^{2}) \beta .$$
2) Free particle (of unit mass) can be described by an action integral with any of the following Lagrangians
$$L_{1} = \frac{1}{2} (\dot{x})^{2} ,$$
$$L_{2} = \frac{1}{2} (\dot{x})^{2} - \dot{x} \dot{a} + \frac{1}{2} (\dot{a})^{2} ,$$
where $a = a(t)$ is a time dependent parameter (not a dynamical variable). Now
$$L_{2} - L_{1} = - \dot{x} \dot{a} + \frac{1}{2} (\dot{a})^{2} = \frac{d \Lambda}{dt} .$$
Therefore
$$\frac{\partial \Lambda}{\partial t} = \frac{1}{2} (\dot{a})^{2} , \ \ \ \frac{\partial \Lambda}{\partial x} = - \dot{a} .$$
Together these imply that $\ddot{a} = 0$ which in turn leads to
$$a = \alpha + \beta t , \ \ \Lambda_{\beta} = - \beta x + \frac{1}{2} \beta^{2}t .$$
Thus, for infinitesimals $\alpha$ and $\beta$, i.e., $\delta \Lambda_{\beta} = - x \beta$, the free particle action is invariant under the infinitesimal spatial displacement by $\alpha$ and Galilean boost by $\beta$:
$$x \to x + \alpha + \beta t$$
Again, associated with this infinitesimal symmetry transformation is the following constant of motion
$$\dot{x} (\alpha + \beta t) - x \beta = \dot{x} \alpha + (\dot{x} t - x) \beta .$$
Here we have two constants of motion: the linear momentum $\dot{x}$ associated with spatial translation by the parameter $\alpha$, and the initial position $-x_{0} = \dot{x}t - x$ associated with the Galilean boost parameter $\beta$.

But isn't it possible to choose a variation to be a symmetry transformation (fall in a 1-parameter subgroup, say, $S(\lambda)$)? So in this case the conservative quantity $\frac{\partial L}{\partial \dot{q}^a}\delta q^a+\delta\Lambda$ is the result of two symmetry transformations $T(\varepsilon)$ and $S(\lambda)$? (I guess your $\Lambda_\varepsilon$ is just $\delta\Lambda$.)

$S(\lambda)$ has its own $\Lambda_{\lambda}$ and correspondingly its own constant of motion $Q_{\lambda}$. And similarly, $T(\epsilon)$ has its own $\Lambda_{\epsilon}$ and $Q_{\epsilon}$. This should have become clear from the above examples. If $T$ and $S$ are different symmetries, why should they lead to the same conserved quantity?

Your associating the change of the Lagrangian with a symmetry transformation recollects me the gauge field theory, where the Lagrangian of a matter field is variant under the transformation of a local symmetry group $G(\alpha)$, so a $G$ gauge field which transforms in a certain way is introduced into the Lagrangian to make the resultant Lagrangian invariant under the local $G$ symmetry transformation. But here introducing a time-dependent function to cancel the change of the Lagrangian $\frac{d\Lambda}{dt}$ resulting from the transformation $T(\varepsilon)$ seems to have no good interpretation.

I have been working on gauge field theories for long long time, yet I did not understand a word from this. Gauge transformations are internal symmetries. Internal symmetries are defined by $\Lambda = 0$ or the possibility to remove it by an appropriately chosen total divergence.

What does $``L"$ represent?

What do you think it is? Time translation $t \to t + \delta t$ induces two types of variation on the coordinates
$$\bar{\delta} q = \bar{q}(t + \delta t) - q(t) = \delta q + \dot{q} \delta t$$
where, $\delta q = \bar{q}(t) - q(t)$. So
$$Q = \frac{\partial L}{\partial \dot{q}} (\bar{\delta}q - \dot{q}\delta t ) + L \delta t$$
This can be rewritten as
$$Q = \frac{\partial L}{\partial \dot{q}} \bar{\delta}q - \left(\frac{\partial L}{\partial \dot{q}} \dot{q} - L \right) \delta t .$$
The expression in the bracket is a Legendre transform. What does Legendre transformation do to the $\dot{q}$-dependence? Is that what you were intended to say?

I was just stating a well-known notion in the Lagrangian formalism... I ask you why we can't choose $\Lambda$ to be a function of $t, q, \dot{q}$ instead of a function of just $t, q$, that is, why we can't just say $L(\dot{q}, q, t)$ and $L(\dot{q}, q, t)+\frac{d\Lambda(\dot{q}, q, t)}{dt}$ describe the same physical system.

I think you should study the calculus of variation. Suppose that $L$ is a legitimate Lagrangian, i.e., for fixed ends variation $\delta q(t_{1}) = \delta q(t_{2}) = 0$, we have
$$\int_{t_{1}}^{t_{2}} dt \ \delta L(\dot{q} , q) = 0 . \ \ \ \ \ \ (1)$$
Question: Can
$$L_{1} = L + \frac{d}{dt} \Lambda( q , \dot{q}) ,$$
be another legitimate Lagrangian?
Answer: Big No.
Proof:
$$\delta L_{1} = \delta L + \frac{d}{dt} \left( \frac{\partial \Lambda}{\partial q} \delta q + \frac{\partial \Lambda}{\partial \dot{q}} \delta \dot{q} \right)$$
Now, integrating this and using (1) and $\delta q(t) |_{t_{1}}^{t_{2}} = 0$, we find
$$\int_{t_{1}}^{t_{2}} dt \ \delta L_{1} = \Big \{ \frac{\partial \Lambda}{\partial \dot{q}} \delta \dot{q} \Big \}_{t_{1}}^{t_{2}}$$
Thus, $L_{1}$ is an other legitimate Lagrangian if and only if
$$\frac{\partial \Lambda}{\partial \dot{q}} = 0,$$
which means that $\Lambda = \Lambda (q , t)$. You can not put $\delta \dot{q} = 0$ at the end points. If you do, the whole of the variation calculus become garbage.

Based on the example you give, $L=\frac{1}{2} (\frac{dx}{dt})^{2}$, the simplest representive example which differs by $\frac{d\Lambda(dx/dt, x, t)}{dt}$ from $L$ is $L'=\frac{1}{2}(\frac{dx}{dt})^{2}+\frac{d}{dt}(\frac{dx}{dt}+x+t)=\frac{1}{2}(\frac{dx}{dt})^{2}+\frac{d^2x}{dt^2}+\frac{dx}{dt}+1$, where $\Lambda(dx/dt, x, t)=\frac{dx}{dt}+x+t$. It's easy to check that $L$ and $L'$ give the same equation of motion.

What makes you think that you can treat $\dot{x}$, $\ddot{x}$ as well as $x$ as independent variables? $\ddot{x}=0$ is fixed by your original Lagrangian. So, your $\Lambda$ is simply equal to $x + t$

According to your statement, the Lagrangians describing the same physical system can only differ by a total time derivative term. But on page 21 of the book Classical Mechanics, Third Edition, Goldstein, Poole & Safko, it says $``...$ if $L(q, \dot{q}, t)$ is an approximate Lagrangian and $F(q,t)$ is any differentiable function of the generalized coordinates and time, then $L'(q, \dot{q}, t)= L(q, \dot{q}, t)+\frac{dF}{dt}$ is a Lagrangian also resulting in the same equations of emotion.

Why there is a “But”? I don’t see any contradiction between my statement and theirs, except for their use of the words “approximate Lagrangian” which I do not know what it means and in what context they used it.

It is also often possible to find alternative Lagrangians beside those constructed by this prescription.$..."$ What's your comment about the underlined statement?

Again, it is not clear to me what they mean by that statement. I can only assume that they have the method of Lagrange multipliers in mind.

 

[EnigmaticField]

Thank you very much for giving the three examples to show me the connection between symmetry transformations and conservative quantities in discrete-system mechanics so that I can get a clear idea of what you were talking about previously in the posts of this thread. The related books I have read never talk about the Noether theorem in discrete-system mechanics, where they just assume the variation at the boundary vanishes so that the boundary integral of the action integral vanishes; therefore I was never aware that the action boundary integral still carries treasurable information of a discrete physical system until you talk about it. But in discrete-system mechanics, the dynamical variables are just the coordinates, so the conservative quantities can only include energy-momentum, angular momentum, center-of-mass moment and in addition, maybe the supertranslation charges of the discrete physical system, right?

$S(\lambda)$ has its own $\Lambda_{\lambda}$ and correspondingly its own constant of motion $Q_{\lambda}$. And similarly, $T(\epsilon)$ has its own $\Lambda_{\epsilon}$ and $Q_{\epsilon}$. This should have become clear from the above examples. If $T$ and $S$ are different symmetries, why should they lead to the same conserved quantity?

I did think $S(\lambda)$ and $T(\varepsilon)$ can be different symmetry transformations. But in all your three examples, they seem to be the same symmetry transformations$-$your variations are just your symmetry transformations, being, respectively, in the first example the spatial translation transformation parameterized by $\alpha$, in the second example the Galilean boost transformation parameterized by $\beta$, and in the third example both the spatial translation transformation parameterized by $\alpha$ and the Galilean boost transformation parameterized by $\beta$, aren't they?

I have been working on gauge field theories for long long time, yet I did not understand a word from this. Gauge transformations are internal symmetries. Internal symmetries are defined by $\Lambda = 0$ or the possibility to remove it by an appropriately chosen total divergence.

OK, then what is the gauge transformation for a gravitational system in the theory of general relativity? The variation of the metric at a fixed spacetime point? Is this also counted as internal symmetries?
Also, I see in your note Noether Theorem, Noether Charge and All That, you mention $``$the orbital part of the (symmetry) transformation$"$ $\delta^*\varphi_r(x)-\delta\varphi_r(x)=\delta x^\mu\partial_\mu \varphi_r(x)$. Is $\delta x^\mu\partial_\mu \varphi_r(x)$ exactly all possible external (in contrast to internal) symmetry transformations? And the conservative currents corresponding to them are just the energy-momentum, angular momentum, center-of-mass moment currents of matter, that is, the field source (in contrast to the field $\varphi_r(x)$)? Further, is $\delta\varphi_r(x)$ exactly all possible internal symmetry transformations (gauge transformations)? Can a transformation only be either internal or external?

What do you think it is? Time translation $t \to t + \delta t$ induces two types of variation on the coordinates
$$\bar{\delta} q = \bar{q}(t + \delta t) - q(t) = \delta q + \dot{q} \delta t$$
where, $\delta q = \bar{q}(t) - q(t)$. So
$$Q = \frac{\partial L}{\partial \dot{q}} (\bar{\delta}q - \dot{q}\delta t ) + L \delta t$$
This can be rewritten as
$$Q = \frac{\partial L}{\partial \dot{q}} \bar{\delta}q - \left(\frac{\partial L}{\partial \dot{q}} \dot{q} - L \right) \delta t .$$
The expression in the bracket is a Legendre transform.

Excuse me, I don't quite understand the above. Does $\delta q = \bar{q}(t) - q(t)$ represent the change in the functional form of $q(t)$ at a fixed time point (that is, without considering the time translation $\delta t$)? Then what do you say $``$time translation induces two types of variation on the coordinates$"$ since $\delta q$ is irrelevant to the $\delta t$?
And how to get $Q = \frac{\partial L}{\partial \dot{q}} (\bar{\delta}q - \dot{q}\delta t ) + L \delta t$? Is $Q$ defined as $Q := \frac{\partial L}{\partial \dot{q}} \delta q +\delta \Lambda_t$? What isn't $Q$ defined as $Q := \frac{\partial L}{\partial \dot{q}} \bar{\delta} q +\delta\Lambda_t$?
So $L$ is some function of $q, \dot{q}, t$ such that $L \delta t=\delta\Lambda_t$, the difference of the conservative quantity resulting from the time transformation? According to your context, it's like $L$ is just the untransformed Lagrangian. Why is so? I can't see how to infer from $L \delta t=\delta\Lambda_t$ to get that $L$ is the untransformed Lagrangian.

What does Legendre transformation do to the $\dot{q}$-dependence? Is that what you were intended to say?

Excuse me, I don't know what you mean. I know the Legendre transformation is a variable transformation which will transform the $\dot{q}$-dependence of $L$ into the $p$-dependence of $\frac{\partial L}{\partial \dot{q}} \dot{q} - L$.

You can not put $\delta \dot{q} = 0$ at the end points. If you do, the whole of the variation calculus become garbage.

Previously when I read this I didn't and couldn't figure out the reason but just kept it in suspense in my mind. Recently when I read a paper about gravitational theories, I see it says fixing the differential of variables at the boundary is forbidden by the symplectic structure. Then I started to realize this is a big issue. Previously I usually considered fixing the conjugate momentum $P$ to be fixing $\dot Q$ (the capitalized $P, Q$ represent the variables in field mechanics in contrast to the lower-cased counterpart in discrete-system mechanics), but now I become aware that this is not always correct. (But what made me think so is that the papers in gravitational theories usually refer to fixing the $P$ at the boundary as the Neumann condition. I wonder why they use that terminology.)
So why can't one fix the $\dot Q$ or $\dot q$ at the boundary? Because we have fixed the $Q$ at the boundary for ensuring that the field equations are well defined, then if we further fix the $\dot Q$ at the boundary, the variation at the boundary will be fixed completely and thus there is no variation at the boundary at all. Recalling we have two kinds of boundary conditions, Dirichlet and Neumann conditions, we can only choose either Dirichlet condition or Neumann condition, but can't choose the both at the same time, which would be over restrictive. Am I correct?

What makes you think that you can treat $\dot{x}$, $\ddot{x}$ as well as $x$ as independent variables? $\ddot{x}=0$ is fixed by your original Lagrangian. So, your $\Lambda$ is simply equal to $x + t$

You are correct. Sorry I didn't pay attention to that.
So in the transformation $L=\frac{1}{2}(\frac{dx}{dt})^2 \rightarrow L'=\frac{1}{2}(\frac{dx}{dt})^2+\frac{dx}{dt}+1$ with $\Lambda=x + t$ is reached by the transformation $\frac{dx}{dt} \rightarrow \frac{dx}{dt}+1$, which gives rise to $x \rightarrow x+t$, followed by the shift of $L$ by $\frac{1}{2}$. What is the transformation $T(\varepsilon)$ invoked to account for the shift of $\frac{1}{2}$? (Recall I previously asked you whether every total-derivative shift of the Lagrangian has a transformation to account for it.) A scale transformation? If we drop the $\frac{1}{2}$ and instead consider $L'=\frac{1}{2}(\frac{dx}{dt})^2+\frac{dx}{dt}+\frac{1}{2}$ with $\Lambda=x +\frac{1}{2}t$, we then have the conservative quantity $Q=-\frac{dx}{dt}t+x+\frac{1}{2}t$, what is the interpretation of $Q$ here? A Galilean boost should correspond to a conservative quantity about position, but what position does $Q=-(\frac{dx}{dt}-\frac{1}{2})t+x$ represent?

 

[EnigmaticField]

Let me correct some mistake I made and update my information according to my latest understanding.

Also, I see in your note Noether Theorem, Noether Charge and All That, you mention $``$the orbital part of the (symmetry) transformation$"$ $\delta^*\varphi_r(x)-\delta\varphi_r(x)=\delta x^\mu\partial_\mu \varphi_r(x)$. Is $\delta x^\mu\partial_\mu \varphi_r(x)$ exactly all possible external (in contrast to internal) symmetry transformations? And the conservative currents corresponding to them are just the energy-momentum, angular momentum, center-of-mass moment currents of matter, that is, the field source (in contrast to the field $\varphi_r(x)$)?

I think I was wrong in the question statement $``$And the conservative currents corresponding to them are just the energy-momentum, angular momentum, center-of-mass moment currents of matter, that is, the field source (in contrast to the field $\varphi_r(x)$)?$"$. The correct one should be $``$And the conservative currents corresponding to them are just the energy-momentum, angular momentum, center-of-mass moment currents of the field $\varphi_r(x)$?$"$, because when the Lagrangian density for the field is invariant under an external symmetry transformation, that indicates the absence of the field source.

What do you think it is? Time translation $t \to t + \delta t$ induces two types of variation on the coordinates
$$\bar{\delta} q = \bar{q}(t + \delta t) - q(t) = \delta q + \dot{q} \delta t$$
where, $\delta q = \bar{q}(t) - q(t)$. So
$$Q = \frac{\partial L}{\partial \dot{q}} (\bar{\delta}q - \dot{q}\delta t ) + L \delta t$$
This can be rewritten as
$$Q = \frac{\partial L}{\partial \dot{q}} \bar{\delta}q - \left(\frac{\partial L}{\partial \dot{q}} \dot{q} - L \right) \delta t .$$
The expression in the bracket is a Legendre transform. What does Legendre transformation do to the $\dot{q}$-dependence?

I just read the final part of the book the book Classical Mechanics, Third Edition, Goldstein, Poole & Safko $-$13.7 Noether's theorem, where there is a brief statement of Noether's theorem for discrete systems. I did read this part long time ago but I had almost put it into oblivion because I have never seen any book or paper giving a concrete example of it; it's like most literature about Noether's theorem concentrates on its application to field mechanics, at least this is the case with all those I have read.
After reading it, I think I know how your $\delta\Lambda_t =L \delta t$ comes from: if the Lagrangian $L$ is invariant under a time translation $\delta t$, then when we vary the action, we get
$$\delta\int^{t_2}_{t_1} Ldt=\int^{t_2}_{t_1}\delta Ldt+\int^{t_2}_{t_1}L\delta dt=0 \\ \Rightarrow \int^{t_2}_{t_1}\{[\frac{\partial L}{\partial q}-\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})]\delta qdt+\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\delta q)dt+d(L\delta t)-dL\delta t\} \\ \doteq\int^{t_2}_{t_1}\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\delta q+L\delta t) dt=0.$$
Therefore the conservative quantity is $Q=\frac{\partial L}{\partial \dot{q}}\delta q+L\delta t$, which is to be compared with $\frac{\partial L}{\partial \dot{q}}\delta q+\delta \Lambda_t$, so $\delta \Lambda_t=L\delta t$.
Then writing $\delta q$, which is the functional form change, in terms of the total variation $\bar{\delta} q$ through $\bar{\delta} q=\delta q+\dot{q}\delta t$, your result $Q = \frac{\partial L}{\partial \dot{q}} \bar{\delta}q - \left(\frac{\partial L}{\partial \dot{q}} \dot{q} - L \right) \delta t$ is obtained. But I still think $\delta q = \bar{q}(t) - q(t)$ is not induced by the time translation $\delta t$.
Further, I think there is no difference between a variation of time and a symmetry translation of time $-$ they can always be unified together for time has only one dimension so that all the symmetry transformation we can do to it is translation. Am I right?

That depends on the resulting $\Lambda$. Each symmetry transformation $T(\epsilon)$ leads to a unique $\Lambda_{\epsilon}(q,t)$ and, therefore unique constant of motion. Of course, the same Lagrangian can have different symmetries and therefore different $\Lambda$.

Do you imply that the converse, that is, the existence of a certain $\Lambda_{\epsilon}(q,t)$ and a constant of motion $\Rightarrow$ the existence of a certain symmetry, is not always true?
I saw in Noether's own statement of her two theorems says the converse, the existence of conservative currents $\Rightarrow$ the existence of symmetries of the action, is true, as quoted as follows:

I. If the integral I is invariant with respect to a $G_\rho$ then $\rho$ linearly independent combinations of the Lagrange expressions become divergences $-$ and from this, conversely, invariance of I with respect to a $G$ will follow. The theorem holds good even in the limiting case of infinitely many parameters.
II. If the integral I is invariant with respect to a $G_{\infty\rho}$ in which the arbitrary functions occur up to the $\sigma$-th derivative, then there subsist $\rho$ identity relationships between the Lagrangian expressions and their derivatives up to the $\sigma$-th order. In this case also, the converse holds.

But I saw in Classical Mechanics, Third Edition, Goldstein, Poole & Safko, on page 596 it says

Note that while Noether's theorem proves that a continuous symmetry property of the Lagrangian density leads to a conservation condition, the converse is not true. There appear to be conservation conditions that cannot correspond to any symmetry property. The most prominent examples at the moment are the fields that have soliton solutions, for example, are described by the sine-Gordon equation or the Korteweg-de Vries equation.

Why are the two statements not in harmony?
So is your implication in accord with what Classical Mechanics, Third Edition, Goldstein, Poole & Safko says but not what Noether says?

 

[EnigmaticField]

I get one question again.

You can not put $\delta \dot{q} = 0$ at the end points. If you do, the whole of the variation calculus become garbage.

Then in the Hamiltonian formalism, can we put either $\delta{q}^i=0$ or $\delta p_i=0$ at the endpoints?
If we vary the Hamiltonian $H=p_i\dot{q}^i-L$, we get $$\delta H=\delta p_i\dot{q}^i+p_i\delta\dot{q}^i+(\frac{\partial p_i}{\partial t}\dot{q}^i+p_i\frac{\partial \dot{q}^i}{\partial t})\delta t-[\frac{\partial L}{\partial q^i}-\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}^i})]\delta q^i-\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}^i}\delta q^i)-\frac{\partial L}{\partial t}\delta t \\ =\delta p_i\dot{q}^i+\frac{d}{dt}(p_i\delta q^i)-dp_i\delta q^i+(\frac{\partial p_i}{\partial t}\dot{q}^i+p_i\frac{\partial \dot{q}^i}{\partial t})\delta t-\frac{d}{dt}(p_i\delta q^i)-\frac{\partial L}{\partial t}\delta t \\ =\delta p_i\dot{q}^i-dp_i\delta q^i+(\frac{\partial p_i}{\partial t}\dot{q}^i+p_i\frac{\partial \dot{q}^i}{\partial t}-\frac{\partial L}{\partial t})\delta t.$$
There is no boundary integral in the above, so putting either $\delta{q}^i=0$ or $\delta p_i=0$ at the endpoints seems to make no difference. But in field mechanics the Hamiltonian is defined as $\int dV\mathcal{H}:=\int dV(P_i\dot{Q}^i-\mathcal{L})$, in which $P_i:=\frac{\partial L}{\partial\dot{Q}^i}$, where putting $\delta{Q}^i=0$ or $\delta P_i=0$ at the boundary does make a difference because even if the Hamiltonian inherited from the Lagrangian doesn't include a boundary integral, one can add one boundary integral whose integrand is $\mathcal{F}(Q^i,P_i,x^j)$, any function of $Q^i,P_i, x^j$ with $j=1, 2, 3$ denoting the spatial coordinates, to it like in the Lagrangian formalism since the boundary integral doesn't affect the field equations. So here can we put either $\delta{Q}^i=0$ or $\delta P_i=0$ at the boundary or $\delta{Q}^i=0$ and $\delta P_i=0$ at the boundary (unlike in the Lagrangian formalism, where we can only put $\delta{Q}^i=0$ at the boundary)?

 

[samalkhaiat]

You asked me to reply to your posts, but when I looked at them I found many unconventional terms, questions irrelevant to this thread and other incorrect understanding of physics.
1) What is the meaning of “external symmetry transformation”? I have never heard of such transformations. We have space-time transformation and internal transformation. Space-time transformations are those for which $\delta x^{\mu} \neq 0$. Internal transformations are those for which $\delta x^{\mu} = 0$. These internal transformation act in the internal “charge” space of the fields, i.e., they act on the field indices that are not space-time indices. The transformation of a compact gauge group is, in this sense, internal transformation.
2) General relativity is “similar” to the non-abelian gauge theories because its symmetry group (the group of general coordinate transformations) is non-abelian. However, unlike ordinary non-abelian gauge theories, the symmetry group of general relativity is not compact.
3) $\delta q = \bar{q}(t) - q(t)$ does represent the functional change of $q(t)$, but it is NOT at fixed point. Explicitly, it means
$$\delta q(t) = \bar{q}(\bar{t})|_{\bar{t} = t} - q(t) .$$ Have you heard of the Lie derivative? $\mathcal{L}_{\delta t} q(t) \equiv \delta q(t)$.
4) In the Lagrangian formalism, the trajectory is given by the curve $q(t)$. While in the Hamiltonian formalism, the trajectory is given by a curve $X(t) = (q,p)(t)$ in phase space. When you apply the variation principle, in any formalism, you impose a fixed-end variation of the trajectories. So, in the Lagrangian formalism, this means setting $\delta q(t) = 0$ at the end points. And, in the Hamiltonian formalism, you set $\delta X (t) = 0$ at the end points, which is equivalent to setting to zero both $\delta q(t)$ and $\delta p(t)$.
5) Where did I say that the converse of Noether theorem is not true? You need to understand the correct statement of Noether theorem first. The theorem says: Invariance of the action under continuous transformation leads to a current that is conserved on actual trajectories, i.e., on the solutions of the equations of motion. Conversely, a current which is conserved only on actual trajectories is a symmetry current. This does not mean that every conserved current is a symmetry current. Global topological considerations may allow you to construct conserved currents. For example, in 2D space-time and for any scalar field $\varphi (x)$, the following current is conserved identically $$J^{\mu} = \epsilon^{\mu\nu}\partial_{\nu}\varphi .$$ This is not symmetry current because the conservation statement $\partial_{\mu}J^{\mu}= 0$ is true even when $\varphi(x)$ is not a solution of the field equation.

 

[EnigmaticField]

1) What is the meaning of “external symmetry transformation”? I have never heard of such transformations. We have space-time transformation and internal transformation. Space-time transformations are those for which $\delta x^{\mu} \neq 0$. Internal transformations are those for which $\delta x^{\mu} = 0$. These internal transformation act in the internal “charge” space of the fields, i.e., they act on the field indices that are not space-time indices. The transformation of a compact gauge group is, in this sense, internal transformation.

I consider whenever a special term is coined for the purpose of referring to a class of things differing from their original counterparts in a certain way, the original counterparts are automatically assigned some contrast term when confusion is likely to occur or emphasis is intended. For example, when ''pseudovector" is coined, "true vector" is automatically used (I read a post several days ago here in which somebody said (s)he has never heard of the term "true vector", but in my impression I have heard of it in some texts when "pseudovector" is also mentioned.), and when "virtual displacement" is coined, "real displacement" is automatically used. So I consider if there is "internal transformation" coined, "external transformation" can be automatically used to refer to "spacetime transformation" though I'm not sure if there are people using it. I do have heard of "internal transformation' in some texts$-$which seem to always use "internal transformation" to refer to the group transformation associated to spinor. Are there other internal transformations besides the spinor group transformation?

2) General relativity is “similar” to the non-abelian gauge theories because its symmetry group (the group of general coordinate transformations) is non-abelian. However, unlike ordinary non-abelian gauge theories, the symmetry group of general relativity is not compact.

Isn't SI(2, C) also non-compact? Is there a gauge theory whose gauge group is Sl(2, C)?

3) $\delta q = \bar{q}(t) - q(t)$ does represent the functional change of $q(t)$, but it is NOT at fixed point. Explicitly, it means
$$\delta q(t) = \bar{q}(\bar{t})|_{\bar{t} = t} - q(t) .$$ Have you heard of the Lie derivative? $\mathcal{L}_{\delta t} q(t) \equiv \delta q(t)$.

I'm very familiar with the Lie derivative, which is the dfference between the derivative of the variation derived from the fundamental variable variation and the derivative of the Lie transfer, namely $(\mathcal{L}_\xi\phi)^r=\rm{lim}_{\epsilon\rightarrow 0}\frac{d\phi^r-\Delta\phi^r}{\epsilon}$, where $\xi$ is the vector resulting from the flow parameterized by $\epsilon$ and $\Delta\phi^r$ is the Lie transfer of $\phi^r$ along the $\epsilon$-flow. Thus I think the Lie derivative should have included the variation derived from the fundamental variable variation ( here it's the time variation) so I wonder why $\bar{\delta}q=\delta q+\dot{q}\delta t$ since $\delta q$ represents the Lie derivative (which should have included $\dot{q}\delta t$). Actually I was puzzled very much for the same reason when I saw in your note, Noether Theorem, Noether Charge and All That, you said $\delta\varphi_r(x)$ in $\delta^∗\varphi_r(x)=\delta\varphi_r(x)+\delta x^{\mu}\partial_{\mu}\varphi_r(x)$ is nothing but the Lie derivative of $\varphi_r(x)$. Now you mention it again, so it puzzles me again. This time after thinking for a long time, I think of a possibility$-$you are talking about a nonautonomous flow, a parameter-dependent flow (I recall for this case the Lie derivative is applied on a differential form $\alpha$ in the way $\mathcal{L}_{\bf{v}}\alpha=i_{\bf{v}}\bf{d}\alpha+\bf{d}i_{\bf{v}}\alpha+\frac{\partial\alpha}{\partial t}$ for the flow parameterized by $t$ corresponding to the vector $\bf{v}(\bf{x}, t)=\frac{d\bf{x}}{dt}$, where $\bf{d}$ is the derivative with respect to the spatial coordinates and $\bf{x}$ is the spatial coordinates.). But I think a parameter-dependent flow gives rise to a local symmetry transformation. So are you talking about a local symmetry transformation, aren't you?

5) Where did I say that the converse of Noether theorem is not true? You need to understand the correct statement of Noether theorem first. The theorem says: Invariance of the action under continuous transformation leads to a current that is conserved on actual trajectories, i.e., on the solutions of the equations of motion. Conversely, a current which is conserved only on actual trajectories is a symmetry current. This does not mean that every conserved current is a symmetry current. Global topological considerations may allow you to construct conserved currents. For example, in 2D space-time and for any scalar field $\varphi (x)$, the following current is conserved identically $$J^{\mu} = \epsilon^{\mu\nu}\partial_{\nu}\varphi .$$ This is not symmetry current because the conservation statement $\partial_{\mu}J^{\mu}= 0$ is true even when $\varphi(x)$ is not a solution of the field equation.

You didn't say the converse of Noether theorem is not true, but you implied (implicitly expressed) the conservative current won't necessarily lead to the symmetry, based on firstly, when I asked you $``$Then if conversely, two Lagrangians happen to differ by a total time derivative of a function $\Lambda(q, t)$, that is, $L_A-L_B=\frac{d\Lambda(q, t)}{dt}$, does there exist a unique transformation $T(\varepsilon)$ to account for it?$"$, you replied $``$That depends on the resulting $\Lambda$. Each symmetry transformation $T(\epsilon)$ leads to a unique $\Lambda_{\epsilon}(q,t)$ and, therefore unique constant of motion. Of course, the same Lagrangian can have different symmetries and therefore different $\Lambda."$, and secondly, the three examples you gave seem to be ad hoc$-$ in each example you deliberately manipulated a $\Lambda"$ for which a symmetry transformation can be invoked to account and for which the corresponding constant of motion has a good interpretation. If you make an arbitrary $\Lambda"$, it's very possible that it doesn't correspond to an attributable symmetry transformation (that is, attributable to the symmetry transformations applied on $q, \dot{q}, t$) with a corresponding constant of motion which has a good interpretation. For instance, in my aforementioned (post #26) example $L=\frac{1}{2}(\frac{dx}{dt})^2 \rightarrow L'=\frac{1}{2}(\frac{dx}{dt})^2+\frac{dx}{dt}+\frac{1}{2}$ with $\Lambda=x +\frac{1}{2}t$ reached by the symmetry transformation $\frac{dx}{dt} \rightarrow \frac{dx}{dt}+1$, the resultant constant of motion $Q=-(\frac{dx}{dt}-\frac{1}{2})t+x$ seems to have no good interpretation.

Thank you very much for your reply, but you didn't reply the part I desire you to reply the most. You said

You can not put $\delta\dot{q}=0$, at the end points. If you do, the whole of the variation calculus become garbage.

then my response to this is

Previously when I read this I didn't and couldn't figure out the reason but just kept it in suspense in my mind. Recently when I read a paper about gravitational theories, I see it says fixing the differential of variables at the boundary is forbidden by the symplectic structure. Then I started to realize this is a big issue. Previously I usually considered fixing the conjugate momentum $P$ to be fixing $\dot Q$ (the capitalized $P, Q$ represent the variables in field mechanics in contrast to the lower-cased counterpart in discrete-system mechanics), but now I become aware that this is not always correct. (But what made me think so is that the papers in gravitational theories usually refer to fixing the $P$ at the boundary as the Neumann condition. I wonder why they use that terminology.)
So why can't one fix the $\dot Q$ or $\dot q$ at the boundary? Because we have fixed the $Q$ at the boundary for ensuring that the field equations are well defined, then if we further fix the $\dot Q$ at the boundary, the variation at the boundary will be fixed completely and thus there is no variation at the boundary at all. Recalling we have two kinds of boundary conditions, Dirichlet and Neumann conditions, we can only choose either Dirichlet condition or Neumann condition, but can't choose the both at the same time, which would be over restrictive. Am I correct?

Can you reply the above, especially the underlined texts?

In addition, I think I may be wrong somewhere in the following$-$time should also be varied when we want to derive the conservative quantity. And afterward I read a book in which the multipilcation of $d[...]$ and $\delta[...]$ is not neglected, so maybe here $dL\delta t$ can't be neglected, since $\delta dt$ is not neglected, either. Also, since the variation of time and the symmetry transformation of time can always be unified, I can just write both time variation and time transformation altogether as $\delta t$.

if the Lagrangian $L$ is invariant under a time translation $\delta t$, then when we vary the action, we get
$$\delta\int^{t_2}_{t_1} Ldt=\int^{t_2}_{t_1}\delta Ldt+\int^{t_2}_{t_1}L\delta dt=0 \\ \Rightarrow \int^{t_2}_{t_1}\{[\frac{\partial L}{\partial q}-\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})]\delta qdt+\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\delta q)dt+d(L\delta t)-dL\delta t\} \\ \doteq\int^{t_2}_{t_1}\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\delta q+L\delta t) dt=0.$$

The correct calculation should be
$$\delta\int^{t_2}_{t_1} Ldt=\int^{t_2}_{t_1}\delta Ldt+\int^{t_2}_{t_1}L\delta dt=0 \\ \Rightarrow \int^{t_2}_{t_1}\{[\frac{\partial L}{\partial q}-\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})]\delta qdt+\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\delta q)dt+[\frac{dL}{dt}\delta t+ L\frac{d\delta}{dt}]dt\} \\ =\int^{t_2}_{t_1}\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\delta q+L\delta t) dt=0.$$

 

[S Morita]

If anyone is still seeing this, take a look at:
"Intuitive Concept or Physical Meaning of Lagrangian"
http://file.scirp.org/Html/3-4900404_65112.htm