Symmetries and lagrangian formulation

[vanesch]

I recently noticed there's something that escaped me in Lagrangian mechanics. I recently browsed though the first volume of Landau and L., where it is explained that two systems have identical dynamics if their lagrangians differ only by a total differential to time of a function (because the action will be extremal for the same path in configuration space).

But a bit later, they show that translation symmetry in space gives rise to conservation of linear momentum... by assuming that the Lagrangian is invariant under a space translation. Now, this I knew, of course. However, "translation symmetry" should only imply that under an infinitesimal translation, the lagrangian should at most change by a total time differential of a function (but that function can depend on the infinitesimal translation at hand).

So, instead of dL = \partial L / \partial x_i dx_i = 0

and using the E-L equation: d/dt ( \partial L / \partial x_i' dx_i ) = 0

from which follows : d/dt p_i = 0

and hence momentum conservation

we should write:

dL = d F(x_i, d x_i,t) / dt = d/dt F_i(x_i,t) d x_i

with F an arbitrary function.

Now, this then leads to:

d/dt (p_i - F_i(x_i,t) ) = 0

which means: p_i - F_i(x_i,t) is conserved

but given that F_i is arbitrary, this doesn't mean anything anymore.

Nevertheless, the dynamics of the system is invariant under translation here, in that the solution of the equations of motion will give you the same solution in the slightly translated coordinate system than in the original coordinates system.

So why can we require that the Lagrangian itself should be invariant under translations (for instance) ? Shouldn't it simply be the action which needs to be invariant under the symmetry of the system ?

What am I missing ? I checked Goldstein, but there too is assumed that the Lagrangian itself is invariant under the symmetry at hand.

 

[samalkhaiat]

I recently noticed there's something that escaped me in Lagrangian mechanics.

which means: p_i - F_i(x_i,t) is conserved

but given that F_i is arbitrary, this doesn't mean anything anymore.

F is not arbitrary at all. Its form is determined dy the symmetry transformatiom. It vanishes when we deal with internal (gauge) symmetries. In classical mechanics, your transformation;

$$\bar{x} = e^{a\partial_{x}}x = x + a$$

is a gauge transformation, i.e. it does not change the time coordinate. So, if this is a symmetry transformation then F should be zero.
We will also see that,

$$\frac{\partial F}{\partial p} = 0$$

So why can we require that the Lagrangian itself should be invariant under translations (for instance) ? Shouldn't it simply be the action which needs to be invariant under the symmetry of the system ?

In order to explain the point you raised, I need to consider two different derivations of the Noether number from the variation of the Lagrangian.

The first one consists in comparing the "off-shell" variation with the "on-shell" variation.
Consider an infinitesimal trans

$$\bar{x}(t) = x(t) + \delta x$$

The explicit form for $$\delta x$$ is assumed known (we have in mind a definite, though unspecified transformation). Is this a symmetry transformation? Well, this can be decided by examining what happens to the Lagrangian under our transformation;

$$\delta L = \frac{\partial L}{\partial x}\delta x + \frac{\partial L}{\partial \dot{x}}\frac{d}{dt}\delta x$$

NOW comes the important bit. If without the use of equation of motion (i.e. off-shell) we can write this (change in L) in the form

$$\delta L = \frac{\partial L}{\partial x}\delta x + p\frac{d}{dt}\delta x \equiv\frac{d}{dt}F(x,p,t)$$

then the action is not affected by the transformation (invariant), and the transformation is a symmetry transformation.
On the other hand using the equation of motion (i.e. on-shell) we find,

$$\delta L = \dot{p}\delta{x} + p\frac{d}{dt}\delta x = \frac{d}{dt}(p\delta x)$$

which is always true, regardless whether or not we are dealing with a symmetry transformation.

Thus, for x which satisfy the E-L equation (on-shell) we have

$$\frac{dC}{dt} \equiv \dot{C} = 0$$

where,

$$C = F - p\delta x$$

is the Noether number associated with the symmetry transformation. Such number must generate (through Poisson bracket) the correct infinitesimal transformation on x. But,

$$[C,x]= -\frac{\partial C}{\partial p} = -\frac{\partial F}{\partial p} + \delta x$$

Therefore we must have

$$\frac{\partial F}{\partial p} = 0$$

i.e.

$$F(x,p,t) \equiv F(x,t)$$

No further information can be gained about the structure of F from a model-independent considerations (i.e. without reference to a specific form of L). But we can "cheat". We can derive the Noether number by different, but equivalent, method and compare our results.

In this method, we calculate the change in the action integral caused by the transformation in the structure of the function x(t) as well as variable t;

$$\bar{t} = t + \delta t$$
$$\bar{x}(\bar{t}) = x(t) + \bar{\delta}x$$
$$\bar{x}(t) = x(t) + \delta x$$

From these, we can derive the useful relations

$$\bar{\delta}x = \delta x + \dot{x}\delta t$$
$$\frac{d}{dt}\bar{\delta}x = \bar{\delta}\dot{x} + \dot{x}\frac{d}{dt}(\delta t)$$

So, to the first order in $$\delta(x,t)$$ we can show that the action changes by;

$$\bar{\delta}A = \int dt \frac{\Delta L}{\Delta x}\delta x + \int dt \frac{d}{dt}(L\delta t + p\delta x)$$

where

$$\frac{\Delta L}{\Delta x} \equiv \frac{\partial L}{\partial x} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})$$

is the Euler derivative.

To make this method equivalent to the first one, we demand that the change in the action along the extremals (i.e. on-shell) should vanish;
Thus, putting

$$\bar{\delta}A=0$$

$$\frac{\Delta L}{\Delta x}=0$$

we find

$$\frac{d}{dt}(L\delta t + p\delta x) = 0$$

By comparing this with the previous result, we get

$$C = -L\delta t - p\delta x$$
$$F = -L\delta t$$

or, in terms of the Hamiltonian, the noether number becomes

$$C = H\delta t - p\bar{\delta}x$$

and your F becomes

$$F = (H - p\dot{x})\delta t$$

Note now that F does not depend on $$\delta x$$ and it vanishes for symmetry operations that do not change the time (gauge symmetries).

I hope you find this helpful.

regards

sam

 

[vanesch]

Thanks for your answer. In the mean time, I think I figured out myself, in a slightly different way. For easiness, let us consider a specific symmetry, x -> x + dx, and let us consider that all the other generalized coordinates do not change under this symmetry of the dynamics (it is always possible to choose them that way). The small change in x should result in a change in the langrangian which is a total derivative to time (then the dynamics is identical):

$$dL = \frac{ dF(x,q,t)}{dt} dx$$

EDIT: I write it this way, because dL will be the difference between the old and the new lagrangian for a transformation dx ; so for each different dx, we have a different dL and hence dL = dFF(x,q,t,dx)/dt
However, dx being a small quantity, we can write FF = FF(dx=0) + F.dx + O(dx^2) ;
for dx = 0, dL is clearly 0, so we have that FF = F.dx + O(dx^2), which is the form I used above. END EDIT

We will now see whether it is possible to find an equivalent lagrangian L', such that the lagrangian itself is invariant under the symmetry:
$$L' = L - dG/dt$$ with G(x,q,t), requiring dL' = 0

In this case:
$$dL' = \partial L' / \partial x dx = 0$$

$$= dL - \partial (dG/dt) / \partial x dx = 0$$

so:
$$dF/dt - \partial (dG/dt) / \partial x = 0$$Hence we have to have that:

$$dF/dt = \partial (dG/dt) / \partial x$$

In other words, we need to find a function G, such that, for a given F, the above equation is satisfied.

I think that $$G = \int F.dx$$ will do.

So this would mean that, if we have system with "dynamics invariant under space translations", and hence its lagrangian changes only by a total time derivative under a small space translation:

$$dL = dF/dt. dx$$

that we can always write this using another lagrangian L' which generates identical dynamics, by L' = L - dG/dt, and that for this new lagrangian, we have that it is (as is usually required) invariant itself under space translations:

$$dL' = 0$$

from which it follows that the momentum $$p' = dL'/d\dot{x}$$ is conserved.

$$p' = dL'/d\dot{x} = dL/d\dot{x} - \partial (dG/dt)/\partial d\dot{x}$$

$$= p - d G / dx = p - F$$

 

[samalkhaiat]

$$dL = \frac{ dF(x,q,t)}{dt} dx$$

EDIT: I write it this way, because dL will be the difference between the old and the new lagrangian for a transformation dx ; so for each different dx, we have a different dL and hence dL = dFF(x,q,t,dx)/dt
However, dx being a small quantity, we can write FF = FF(dx=0) + F.dx + O(dx^2) ;
for dx = 0, dL is clearly 0, so we have that FF = F.dx + O(dx^2), which is the form I used above. END EDIT

Well, I do not understand your reasoning, but it is clear to me that your final result is wrong.

Let me rewrite "your invariace condition" in the conventional notations (I do not use the differential operator "d" insead of the variation symbol $$\delta$$ , because the two do not, in general, commute);

$$\delta L=\frac{dF(x,t)}{dt}\delta x$$

Before I do any thing with this equation of yours, I must say that I know of NO symmetry transformation that induces such a change in the Lagrangian. i.e. "your F" can not be obtained from any symmetry that I'm aware of.
Now, according to your equation, the transformation:

$$x'-x=\delta x = f(t)$$

with,

$$\frac{dF}{dt}\neq 0$$

can not be a symmetry operation because the integral (i.e.the change in your action),

$$\int \delta L dt = \int \frac{dF}{dt} f(t) dt$$

does not vanish in this case.

However, the transformation:

$$\delta x = g(t)$$

with,

$$\frac{dF}{dt} = 0$$

is a symmetry because it leaves your action invariant:

$$\int \delta L dt = \int \dot{F} g(t) dt \equiv 0$$

But, for such symmetry transformation, "your invariance condition" becomes:

$$\delta L = 0$$

Clearly this can not be the case because the symmetry transformation (we are dealing with) depends on the time. Hence, it does not leave the Lagrangian invariant. For example, the Galilean transformation;

$$\delta x = vt$$

leaves the action invariant:

$$\int \delta L dt = 0$$

but not the Lagrangian:

$$\delta L \neq 0$$

So, I'm afraid, your invariance condition:

$$\delta L = dF/dt \delta x$$

is not correct.

So this would mean that, if we have system with "dynamics invariant under space translations", and hence its lagrangian changes only by a total time derivative under a small space translation:

$$dL = dF/dt. dx$$

No, For translations as well as rotations, the Lagrangian itself is invariant. i.e.,

$$\delta L = L' - L = 0$$

that we can always write this using another lagrangian L' which generates identical dynamics, by L' = L - dG/dt,

I see no reason to have the F as well as the G:
If you are given the equation;

$$L' - L = \dot{G}(x,t)$$

then you can show that the L'-system is equivalent to the L-system.(i.e. L' & L are connected by symmetry transformation),

conversely, if you are given a symmetry transformation;

$$x \rightarrow x' = g(x,t)$$

then you can show that the Lagrangian changes according to:

$$L \rightarrow L' = L + \dot{G}(x,t)$$

So, THE INVARIANCE CONDITION is always given by

$$\delta L = \frac{dG(x,t)}{dt}$$

and NO F is needed.

The great advantage of this equation lies in the fact that it allows us, for a given system, to determined all possible symmetries by solving "simple" partial differential equations.

As a concrete example, let us consider the simplest physical system, that is the motion of a free particle of unit mass in one dimension:

$$L = 1/2 \dot{x}^2$$

and investigate the possible symmetry of this system under a transformation of the form;

$$x' = x + f(t)$$
$$\dot{x'} = \dot{x} + \dot{f}(t)$$

where f is some function of t, to be determined by the invariance condition.
The new Lagrangian is;

$$L'(x',\dot{x'}) \equiv L[x(x'),\dot{x}(x',\dot{x'})] = 1/2 (\dot{x'} - \dot{f})^2$$

Thus,

$$L'(\dot{x'}) = L(\dot{x'}) - \dot{f}\dot{x'} + 1/2 \dot{f}^2$$

This transformation becomes a symmetry transformation provided we can find a function G(x',t), such that

$$\dot{G}(x',t) \equiv \frac{\partial G}{\partial x'} \dot{x'} + \frac{\partial G}{\partial t} = -\dot{f}\dot{x'} + 1/2 \dot{f}^2$$

i.e.,

$$\frac{\partial G}{\partial x'} = -\dot{f}$$

$$\frac{\partial G}{\partial t} = 1/2 \dot{f}^2$$

These imply;

$$\ddot{f} = 0$$

which gives

$$f(t) = x_0 + vt$$

and

$$G(x',t) = -vx' + 1/2 v^2 t$$

where $$x_0 = f(0)$$ and $$v = \dot{f}$$ are constants.

Therefore

$$x' = x + x_0 + vt$$

is the most general symmetry transformation for free particle.
As expected, this symmetry consists of;

1) spatial translation for $$v = \dot{f} = 0$$

2) Galilean transformation for $$v \neq 0$$

Notice that

$$\frac{\partial G}{\partial x'} = -v$$

can be thought of as the momentum of a free particle sitting still in the x-system;
$$\dot{x'} = \dot{x} - \frac{\partial G}{\partial x'}$$


merry X-mass

regards

sam

 

[vanesch]

Before I do any thing with this equation of yours, I must say that I know of NO symmetry transformation that induces such a change in the Lagrangian. i.e. "your F" can not be obtained from any symmetry that I'm aware of.
Now, according to your equation, the transformation:

$$x'-x=\delta x = f(t)$$

with,

$$\frac{dF}{dt}\neq 0$$

can not be a symmetry operation because the integral (i.e.the change in your action),

$$\int \delta L dt = \int \frac{dF}{dt} f(t) dt$$

does not vanish in this case.

But that's the whole point, it doesn't have to vanish, in order for there to be a symmetry of the dynamics (that means, that the set of dynamical solutions q(t) is invariant under the symmetry). The action itself doesn't need to be invariant, only the extremals of the action.
Exactly the same reasoning is used to show that if L is a Lagrangrian, then L+dF/dt is an equivalent lagrangian, because generating the same set of dynamical solutions q(t). This is what I wanted to point out: the symmetry doesn't need to be a symmetry of the lagrangian (or the action), it only needs to be a symmetry of the set of solutions. As such, applying an infinitesimal symmetry element doesn't need to imply that the $$\delta L$$ is zero ; it only needs to imply that the overall set of solutions {q(t)} remains the same.

(sorry, have to go, will continue this later...)

 

[vanesch]

As an example, let us consider the lagrangian:
$$L = \frac{\dot{x}^2}{2} + \frac{\dot{y}^2}{2} + 2 y x^3 t \dot{y} + 3 y^2 x^2 t \dot{x} + y^2 x^3$$

and let us consider the (time-independent) symmetry operation:
$$x \rightarrow x + \delta x$$

If we work out the variation of the Lagrangian under this operation, we find:
$$\delta L = 2 y t \dot{y} 3 x^2 \delta x + 6 x \delta x y^2 t \dot{x} + y^2 3 x^2 \delta x = (6 y t \dot{y} x^2 + 6 x y^2 t \dot{x} + 3 x^2 y^2) \delta x$$

which clearly doesn't vanish, and
EDIT:
which takes on the form $$dF/dt \delta x$$ with $$F(x,y,t) = 3 x^2 y^2 t$$
end edit.

Nevertheless, as you can see, the lagrangian is equivalent to the following lagrangian:

$$L' = \frac{\dot{x}^2}{2} + \frac{\dot{y}^2}{2}$$

by the simple transformation $$L' = L - \frac{dG}{dt}$$

with $$G = y^2 x^3 t$$

from which we deduce that the set of solutions generated by L is the same set of solutions as the one generated by L', and from that last one, we know that it is invariant under the proposed symmetry operation.

EDIT:
Note that, as I said, if we integrate the $$F(x,y,t) = 3 x^2 y^2 t$$ wrt x, then we find:
$$\int dx F(x,y,t) = x^3 y^2 t$$, which is exactly the function G found above.
end edit.So from this (artificial, admitted) example, we see that we have a lagrangian which is itself not invariant under a symmetry operation, but which generates nevertheless a family of dynamical solutions (in this case {x(t),y(t)}) which DOES possesses said symmetry. The symmetry which has physical meaning is the one of the set of dynamical solutions, and not of the Lagrangian per se (of course, a symmetry of the lagrangian itself will of course induce a symmetry of the family, but not, a priori,vice versa).

As such, when postulating a *physical* symmetry, one shouldn't, a priori, assume that the Lagrangian itself has that symmetry. Now, as I think I showed, in the end it doesn't matter, because there seems to always exist an equivalent lagrangian which IS invariant under any symmetry that the solution space possesses. But this is a priori not evident.

 

[vanesch]

Let me rewrite "your invariace condition" in the conventional notations (I do not use the differential operator "d" insead of the variation symbol $$\delta$$ , because the two do not, in general, commute);

$$\delta L=\frac{dF(x,t)}{dt}\delta x$$

Before I do any thing with this equation of yours, I must say that I know of NO symmetry transformation that induces such a change in the Lagrangian.

Well, I just gave you an (artificial) example in the other post. But let us get something straight: the $$\delta x$$ is not supposed to be dependent on time: it is the infinitesimal generator of the symmetry. I was considering - as a special case - the translations in space, and $$\delta x$$ was an infinitesimal generator of these translations.

i.e. "your F" can not be obtained from any symmetry that I'm aware of.
Now, according to your equation, the transformation:

$$x'-x=\delta x = f(t)$$

So, no, this was not supposed to be dependent on time. We should eventually consider symmetries of the form $$x'-x = f(t) \delta x$$ (which I didn't do), for a given f(t), as a specific generator of a symmetry.

A Galilean boost would then give rise to the following symmetry operation:
$$x \rightarrow x + t \delta v$$

If we start with the lagrangian $$L = \frac{\dot{x}^2}{2}$$
then this infinitesimal transformation gives rise to:
$$\delta L = \dot{x} \delta v$$
so my associated F(x,t) is simply x, because $$\frac{dF}{dt} = \dot{x}$$ in that case, and we find that $$\delta L = \frac{dF}{dt} \delta v$$

In this case, however, the trick doesn't work with subtracting $$G = \int F dx$$ to make the Lagrangian invariant under the proposed symmetry...

 

[samalkhaiat]

But let us get something straight: the $$\delta x$$ is not supposed to be dependent on time: it is the infinitesimal generator of the symmetry. I was considering - as a special case - the translations in space, and $$\delta x$$ was an infinitesimal generator of these translations.

Why generator? Continuous transformations contain adjustable PARAMETERS.
Physicists write them as

$$x'=T(p) \cdot x$$

Mathematicians prefer:

$$x'=f(x;p)$$

For particular values $$p^{a}_{0}$$ of the parameters the transformations reduce to the identity:

$$x'=T(p_{0}) \cdot x = e \cdot x = x$$

$$x' = f(x;p_{0}) = x$$

transformations for which the parameters are infinitesimally close to $$p_0$$ :

$$p^{a} = p^{a}_{0} + \delta p^{a}$$

and for which the coordinates x' differ infinitesimally from x, are called infinitesimal transformations.
By (Taylor) expanding T or f to the 1st order around the identity, we get:

$$x' = e \cdot x + \frac{\partial T}{\partial p}|_{0} \cdot \delta p \cdot x$$

or

$$x' = f(x;p_{0}) + \frac{\partial f}{\partial p}|_{0} \cdot \delta p$$

Thus;

$$\delta x^{i} = J^{ij}_{a} x^{j} \delta p^{a}$$

where the differential operators (vector fields) on $$R^3$$

$$J^{ij}_{a} = \frac{\partial T^{ij}}{\partial p^{a}}|_{0}= \frac{\partial f^j}{\partial p^a}|_{0}\frac{\partial}{\partial x_i}$$

are the generators of the infinitesimal transformations associated with a given parameters $$\delta p^a$$ .

In order to get rid of the parameters from their equations, some peopel write

$$\delta = \delta_{a} \delta p^{a}$$

so that

$$\delta_{a} x^{i} \equiv J^{ij}_{a} x^j$$

So, except for global translations:

$$\delta x^{i} = \delta p^{i}$$

$$J^{ij}_{a} = \delta^{j}_{a} \frac{\partial}{\partial x_i}$$

we always have

$$\frac{d}{dt} \delta x^{i} = \delta \dot{x}^{i} = J^{ij}_{a} \dot{x}^{j}\delta p^{a}\neq 0$$

So clearly, $$\delta x$$ depends on t at least through x.

For each infinitesimal parameter of a symmetry transformation a number C can be constructed (Noether theorem) such that

$$\dot{C}\equiv\dot{C}_{a}\delta p^{a} = 0$$

$$\delta_{a} x^{i}= \{C_{a} , x^{i} \}$$

and

$$\{C_{a} , C_{b} \} = f_{abc}C_{c}$$

I.e., the time independent Noether numbers (constants of motiom) do generate the correct symmetry transformations on the dynamical variables and satisfy the Lie algebra of the symmetry group.
We call these conserved numbers the generators of the symmetry transformations.

I hope this help you see the differences between $$\delta x$$ ,the generators and the parameters.

will continue later

happy new year

sam

 

[samalkhaiat]

But that's the whole point, it doesn't have to vanish, in order for there to be a symmetry of the dynamics (that means, that the set of dynamical solutions q(t) is invariant under the symmetry). The action itself doesn't need to be invariant, only the extremals of the action.

In the passive view of the symmetry transformation (i.e.,when we have 2 equivalent observers looking at the same curve) the action is always invariant;

$$\delta A= A'(c) - A(c) = 0$$

but, in the active view (i.e.,when we have one observer looking at 2 neighbouring curves) the action changes by at most a term which gives zero under all fixed end variations.i.e.,

$$\delta A= A(c') - A(c)= [F(x,t) \delta x]^{2}_{1}$$

for some F(x,t)(in this way you get your "invariant dynamics").
So, my objection is still valid even in the active view of symmetry because(according to you);

$$\delta A =\delta \int^{2}_{1} L dt = \int^{2}_{1} \dot{F} \delta x dt$$

Obviously, this does not vanish when $$\delta x$$ is a fixed end variation;

$$\delta x(t_{1}) = \delta x(t_{2}) = 0$$

i.e., you will not get "your invariant dynamics" from

$$\delta L = \dot{F}(x,t) \delta x$$

If you insist on having this F, then I would suggest to take your invariance condition in the form

$$\delta L = \frac{d}{dt}\{F(x,t) \delta x \}$$

which is the invariance condition that I keep taking about.


Using mathematics, and mathematics alone, I have shown before that under any continuous transformation, the action changes according to

$$\delta A = \int \{ \dot{C} + \frac{\Delta L}{\Delta q} \delta q \} dt$$

where
$$C=C(q,t;\delta q)$$
and
$$\frac{\Delta L}{\Delta q} = \frac{\partial L}{\partial q} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})$$

is Euler derivative.

Thus $$\delta A = 0$$ if, and only if,

$$\dot{C} + \frac{\Delta L}{\Delta q} \delta q = 0$$

identically. This identity holds true for all symmetry transformations. It is ,indeed, the statement of the Noether theorem: dC/dt vanishes along a trajectory that obeys the E-L equations.

So, The invariance of dynamics (whatever that means) does not drop down from the sky. The fact that extremal curves corresponding to the transformed action is identical with those belonging to the orignal action, is just a colourful way of saying that the quantity of action is the same for all EQUIVALENT observers.

will continue later

sam

 

[samalkhaiat]

Exactly the same reasoning is used to show that if L is a Lagrangrian, then L+dF/dt is an equivalent lagrangian, because generating the same set of dynamical solutions q(t). This is what I wanted to point out: the symmetry doesn't need to be a symmetry of the lagrangian (or the action), it only needs to be a symmetry of the set of solutions. As such, applying an infinitesimal symmetry element doesn't need to imply that the $$\delta L$$ is zero ; it only needs to imply that the overall set of solutions {q(t)} remains the same.

I think one needs to say few words about symmetry transformations. The subject though needs a separate thread, but unfortunatly I do not have the time...

1) passive/active symmetry transformations define equivalent observers/systems.i.e, all frames, which can be reached by such transformations, are equivalent for the description of a given system.

2)Equivalent observers should be able to communicate,i.e., there should be a definite rules to transform the dynamical variables from one coordinate system to any equivalent coordinate system. This is equivalent to the mathematical problem of finding all the representations of the group under which the observers are equivalent.

3)Transforming a physically possible satuation should also be physically possible.i.e., a possible motion in one frame should appear possible in any equivalent frame.

4)The above condition (3) should be identical for all equivalent observers, i.e., the equations of motion should have the same form in all equivalent frames.

5) 3 and 4 hold true when the action has the same value for all equivalent observers and is stationary.

more to come later...

sam

 

[samalkhaiat]

As an example, let us consider the lagrangian:
$$L = \frac{\dot{x}^2}{2} + \frac{\dot{y}^2}{2} + 2 y x^3 t \dot{y} + 3 y^2 x^2 t \dot{x} + y^2 x^3$$

Your "Lagrangian" is

$$L = (... ) + \frac{d}{dt}(y^2 x^3 t)$$

so you do not need to introduce F inorder to get G. And, for that, you do not need any transformation (let alone symmetry one). this G is completely arbitrary. Nothing would change if you take

$$G = 10^x^{3} e^y^{66} t^{108}$$

this is because of the theorem:

A function $$g(x,\dot{x},t)$$ satisfies the E-L equation identically (i.e., independent of x) if, and only if,

$$g(x,\dot{x},t) = \frac{d}{dt}G(x,t)$$

The equation

$$L_2 = L_1 + \dot{G}(x,t)$$

expresses the fact that, from the point of view of the action principle, Lagrangians are not uniquely defined, i.e., we have at our disposal an infinite number of (mathematically) different Lagrangians to play with. Such a freedom though does not spoil or change the the symmetry of the system (I will prove this statement when I formulate (again) the invariace condition).

more to come soon

sam

 

[samalkhaiat]

So from this (artificial, admitted) example, we see that we have a lagrangian which is itself not invariant under a symmetry operation, but which generates nevertheless a family of dynamical solutions (in this case {x(t),y(t)}) which DOES possesses said symmetry. The symmetry which has physical meaning is the one of the set of dynamical solutions, and not of the Lagrangian per se (of course, a symmetry of the lagrangian itself will of course induce a symmetry of the family, but not, a priori,vice versa).

Symmetry is a statement about equivalent observers. So let us deal with 2 observers connected by some continuous and invertable transformation

$$x'=f(x;p)$$
$$x= g(x;p^{-1})$$

where p is the parameter of the transformation. Looking at the same system (L), the prime observer uses the Lagrangian:

$$L'(x',\dot{x}') = L[g(p^{-1};x'), \dot{g}(p^{-1};x';\dot{x}')]$$

This equation says: solve the transformation equations to express the unprime variables in $$L(x,\dot{x})$$ in terms of the prime variables; this gives the Lagrangian L' as used by the prime observer.
If, by arranging the terms on the RHS, we find

$$L'(x',\dot{x}') = L(x',\dot{x}') + \dot{G}(x',t;p)$$

then the 2 observers are equivalent to each other and the transformation is a symmetry operation. this means; the equation of motion has the same form for both observers. For example, if the L-observer sees

$$\ddot{x} + x =0$$

,the L'-observer will see;

$$\ddot{x}' + x' =0$$

with solutions connected by the symmetry transformations.

The equation

$$L\rightarrow L' = L + \dot{G}(x,t;p)$$

(the transformation law of L) can be used to check whether or not a given transformation

$$x\rightarrow g(p) x$$

is a symmetry of our theory.
The most important fact about this G(p) is that there is no way of removing it for all the transformations $$g_{i}$$ of the symmetry group by adding a total derivative to the Lagrangian(I will prove this later)
Note that this G is not arbitrary. It is determined completely by the symmetry transformation,i.e., it is a function of the parameters of the transformation. This is why the Noether numbers (constants of motion) include G in their definition:

$$C= \frac{\partial L}{\partial \dot{x}} \delta x + G(x,t;p)$$

It is easy to prove the theorem: $$\dot{c}=0$$ on critical trajectories if, and only if,
$$\delta L = \dot{G}(x,t;p)$$

i.e., if and only if the invariance(symmetry) condition is satisfied.

As such, when postulating a *physical* symmetry, one shouldn't, a priori, assume that the Lagrangian itself has that symmetry. Now, as I think I showed, in the end it doesn't matter, because there seems to always exist an equivalent lagrangian which IS invariant under any symmetry that the solution space possesses. But this is a priori not evident

In real life we either
1) have a theory (L) and some transformation. So in order to say that our theory is invariant under the transformation at hand, we need to prove
$$L\rightarrow L + \dot{G}(x,t;p)$$
(as I explained above), or
2)do not have a theory but (we) believe that nature is invariant under certain group of transformation. In this case, our task is to construct a (symmetry)-respecting L, i.e., we find L that transforms like

$$L \rightarrow L + \dot{G}(x;p)$$

under our beloved symmetry.


will continue soon...


regards

sam

 

[samalkhaiat]

Such a freedom though does not spoil or change the the symmetry of the system (I will prove this statement when I formulate (again) the invariace condition).

Now we are in a position to prove that adding a total derivative (of some arbitrary function) to the Lagrangian does not change the symmetry of the system.In other words; the invariance(symmetry) condition is invariant under a redefinition of the Lagrangian.
Let $$\mathcal{G}(g_{1},g_{2},...)$$ be a symmetry group of our theory $$L_{1}$$ , and assume that under:

$$x \rightarrow g_{1} x = x + \delta x$$

our Lagrangian transforms according to

$$L_{1} \rightarrow L_{1}' = L_{1} + \dot{G}_{1}(x;g_{1})$$

Now we define a new Lagrangian $$L_{2}$$ by

$$L_{2} = L_{1} + \dot{G}(x)$$

where G(x) is some arbitrary function of x. Under the transformation
$$g_{1} \in \mathcal{G}$$
we will have

$$\delta L_{2} = \delta L_{1} + \delta \dot{G}(x)$$

Since the transformation commute with the time derivative, the change in $$L_{2}$$ , $$L_{2}' - L_{2}$$ , is given by

$$\delta L_{2} = \frac{d}{dt} [G_{1}(x;g_{1}) + \{G(g_{1}x) - G(x) \}] \equiv \frac{d}{dt} G_{2}(x;g_{1})$$

which is (again) an invariance condition. end of proof.

Of cource the redefinition of the Lagrangian will modify the Noether number by$$\delta G(x) = G(g_{1}x) - G(x) \approx \frac{\partial G}{\partial x} \delta x$$

Now a question arises as to whether we can make $$\delta L_{2}$$ to vanish for all $$g \in \mathcal{G}$$ . The answer is NO. We can not remove the function $$G_{2}(x;g)$$ for all the elements of the symmetry group.

However, there exists a $$g_{1} \in \mathcal{G}$$ , for which we can find G(x) such that ;

$$\delta_{g_{1}} L_{2} = 0$$

To see what I means by this, let us go back to the free particle case. Before this, I need to say few words about the Galilei group:

The elements of the ten-parameter Galilie group are parametrized by

$$g = g(T, \alpha^{i} ,v^{i},R(\theta))$$
with i=1,2,3.

i.e., ageneral Galilean transformatio consists of time translation, 3space translations, 3boosts and 3 rotations:

$$g: x \rightarrow gx = Rx + \alpha + vt$$
$$g: t \rightarrow gt = T + t$$

Ignoring time translation and rotations (i.e., putting T= 0, R(0)= I), we find

$$g(0, \alpha ,0,1): x \rightarrow g(\alpha )x = x + \alpha$$
for translation,

$$g(0,0,v,1): x \rightarrow g(v)x = x + vt$$
for boosts, and

$$g(0, \alpha ,v,1) = g(\alpha) \cdot g(v) : x \rightarrow g(\alpha ,v) x = x + \alpha + vt$$
for translation and boost.

Now for the free particle:

$$L_{1} = 1/2 \dot{x}^{2}$$

we have

$$\delta_{g(\alpha)} L_{1} = 0$$

but for boosts, we find (see post #4)

$$\delta_{g(v)} L_{1} = \delta_{g} L_{1} = \frac{d}{dt} (vx + \frac{1}{2} tv^{2}) \equiv \dot{G_{1}}(x,t;v)$$

so for a general transformation g, $$G_{1}(x;g)$$ can not be removed.
Now define a new Lagrangian:

$$L_{2} = L_{1} + \dot{G}(x) = L_{1} + \frac{d}{dt}(-x^{2}/2t) = 1/2 ( \dot{x} - x/t)^{2}$$

(I will later show you how to calculate this G(x) for boosts. and I will do that using the "correct version" of your method )

Now, in contrast with $$L_{1}$$ , $$L_{2}$$ is invariant under the Galilean boosts:

$$\delta_{g(v)} L_{2} = 0$$

But, under the transformation $$g(0, \alpha ,v,1)$$ , we find:

$$\delta_{g} L_{2} = \delta_{g(\alpha )} L_{2} = \frac{d}{dt}(-x \alpha /t) \equiv \frac{d}{dt}G_{2}(x,t; \alpha )$$

Thus, although $$L_{2}$$ is invariant under boosts $$g(v)$$ ,

$$\delta_{g} L_{2} \neq 0$$
for a general transformation g.

Notice that for both Lagrangians, we still have an invariance (symmetry) condition given by

$$\delta_{(g=g(\alpha) \cdot g(v))} \{ L_{1},L_{2} \} = \frac{d}{dt} \{ G_{1}(x,t;v) , G_{2}(x,t; \alpha ) \}$$

Thus, as I mentioned before, there is no way of removing the functions $$G_{i}(x,t;g)$$ for all the transformations g of the Galilie group by adding a total derivative. The reason for this can be shown to be the non-trivial cohomology of the Galilie group. This is why the projective representation of the Galilie group in QM cannot be transformed into an ordinary one. Remember that the invariance of the Schrodinger equation:

$$i\frac{\partial \Psi}{\partial t} = -1/2 \nabla^{2} \Psi$$

under the Galilie group can be achieved if the wavefunction transforms in the projective representation:

$$\Psi' (x',t') = e^{iG_{1}(x,t;g)} \Psi (x,t)$$
with
$$G_{1}(x,t;v) = vx + \frac{1}{2} v^{2}t$$

as in the case of the free (classical) particle, and this G cannot vanish for all the group elements. This is why the old Lande' paradox does not arise in Galilean QM.

will continue again

sam

 

[samalkhaiat]

If we start with the lagrangian $$L = \frac{\dot{x}^2}{2}$$
then this infinitesimal transformation gives rise to:
$$\delta L = \dot{x} \delta v$$
so my associated F(x,t) is simply x, because $$\frac{dF}{dt} = \dot{x}$$ in that case, and we find that $$\delta L = \frac{dF}{dt} \delta v$$

But this is garbage. Sir you are using a wrong equation.

In this case, however, the trick doesn't work with subtracting $$G = \int F dx$$ to make the Lagrangian invariant under the proposed symmetry..

Put your F in the garbage can and follow the following "correct" reasoning which, I think, it speaks your Language.
In your "method", you have one observer playing with 2 Lagrangians:

$$L_{2}(x, \dot{x}) = L_{1}(x, \dot{x}) + \frac{d}{dt}G(x,t)$$

then you consider an infinitesimal transformation

$$x \rightarrow x + \delta x$$

(here, as it should be, I take $$\delta x$$ to be a function of x, t and the infinitesimal parameter of the transformation.In this way "your" method will not be restricted to the time-independent translation)

Such transformation will induce the changes:

$$\delta L_{2} = \delta L_{1} + \frac{d}{dt} \delta G(x)$$

since G does not depend on $$\dot{x}$$ , we can write

$$\delta L_{2} = \delta L_{1} + \frac{d}{dt}(\frac{\partial G}{\partial x} \delta x)$$

Now "you" say: if, for this $$\delta x$$ , we can find G(x) such that

$$\delta L_{2} = 0$$

for this transformation, i.e., for a given $$\delta x$$ you want to find G(x) from the equation

$$\delta L_{1} = -\frac{d}{dt}(\frac{\partial G}{\partial x} \delta x)$$

Well, this is nothing but my good old symmetry condition:

$$\delta L_{1} = \frac{d}{dt}G_{1}(x,t;\delta g)$$

with

$$G_{1}(x,t;\delta g) = -\frac{\partial G}{\partial x} \delta x$$

Indeed, you can integrate this to find G(x):
For

$$L_{1} = 1/2 \dot{x}^{2}$$
and
$$\delta x = \delta v t$$

we saw (post#4)

$$G_{1}(x,t;\delta v) = x \delta v$$

thus
$$\frac{\partial G}{\partial x}t = -x$$

therefore
$$G(x) = -\frac{x^2}{2t}$$
This is the G that your (incorrect) method could not determine.

Also, "you" can find G from

$$\delta L_{1} = - \frac{d}{dt}(\frac{\partial G}{\partial x}\delta x)$$

For free particle:

$$\dot{x}\delta \dot{x} = -\frac{d}{dt}(\frac{\partial G}{\partial x}\delta v t)$$

Thus
$$\dot{x} + \frac{d}{dt}(t\frac{\partial G}{\partial x})= 0$$

and $$G(x) = -\frac{x^2}{2t}$$

*****************************
you could put F instead of

$$-\frac{\partial G}{\partial x}$$

and get "your" (now correct) invariance condition:

$$\delta L_{1} = \frac{d}{dt}(F\delta x)$$
but I see no reason for this F, so I will leave it for you
*********************************

regards


sam

 

[vanesch]

Hi Sam,

thanks for all those posts, I will have to go through them 1 by 1. However, I think we are talking next to each other. We define "symmetry" differently and from there of course we obtain different results. Let us recall that I'm working entirely within classical mechanics. The "reality" of a physical situation there, is given by the flows over configuration space, that is, the set of all solutions to the equations of motion. What I call a symmetry, is a symmetry over this set of solutions, that is, a group of operations - which can be parametrized - which transforms this set of solutions into itself.

Now the way I see it, "action" and "Lagrangian" are a trick to find this set of solutions to a given problem, but have no "reality" to themselves in this view. As such, a symmetry of the "reality of the physical situation" as I defined above has a priori not much to do with any invariance of a Lagrangian or the like. But of course there must be some relationship, given that, from a Lagrangian descriptions, we can reconstruct the set of solutions, a symmetry of this set of solutions must imply somehow some property of the Lagrangian, or the action.

There's one thing I take as established, and that is: if there is a set S of all solutions (that is q(t)) to a given problem, and if there is L(q,q-dot,t), a Lagrangian generating this set of solutions S, then the set of solutions S is also generated by any other lagrangian L'(q,q-dot,t), such that there exists a function F(q,t), with L' = L + dF/dt.

So thinking of functions F is a cheap way to find different lagrangians (in casu L and L') that generate the same set of solutions S (and that's what I did in my example).

Now, I also take it that IF there are two lagrangians L and L' that generate the same set S, then they must necessarily be linked by such an equation, in other words, if L and L' are two lagrangians generating the set of solutions S, then there must exist an F(q,t), such that L' = L + dF/dt (if we fix somehow a scale for L, because there might also be a trivial multiplicative constant).

My initial point was: consider a symmetry operation over S. That means, there is a T, which takes a solution q(t) from S, and transforms this into another function T(q)(t) = q'(t). T is a symmetry operation iff q'(t) is also an element of S. Consider now that T is a parametrised group of symmetry operations over S, which we parametrise in our example with $$\delta x$$.

So $$T_{\delta x}$$ forms a Lie group of symmetry operations over S. We have of course a precise prescription of what exactly $$T_{\delta x}$$ does to a solution (or even a non-solution) $$q(t)$$.

The question is: what can we say about any lagrangian L, for which this group $$T_{\delta x}$$ is a symmetry group over the set of solutions S generated by the Lagrangian. Now, the point I was making, was:

Pick a given parameter $$\delta x$$, and with it, a transformation T. Apply it to a solution $$q(t)$$ to become $$q'(t)$$. We can consider now a lagrangian expression in the new variables q', on the condition that the transformation T is a point transformation (such as a global translation). This excludes of course quite some potential symmetries, but is already sufficient to make my point. Properties of more general symmetry classes must of course also be satisfied by more restricted observations of symmetry. So we only consider symmetries such that $$q' = f(q,\delta x)$$

What we know of the dynamics expressed in the new variables q' and q'-dot, is that it must be describable by a lagrangian L'(q',q'-dot,t). We can apply the point transformation to L' to go back to the variables q, and hence we will find a lagrangian L"(q,q-dot,t) which has to describe exactly the same dynamics as the original lagrangian L(q,q-dot,t), in other words, the set of solutions S to L" must be identical to the set of solutions L. We know that, in that case, L and L" can differ by at most a function dF/dt, with F(q,t).

How do we find L" and L' ? Well, we know that L, expressed in coordinates q, generates the set S. We also know that after transformation T, S remains itself. So, S, expressed in the variables q', is exactly the same set, as S, expressed in the variables q. So it is not difficult to find a lagrangian, expressed in q', which generates the set S: take L'(q',q'-dot,t) to be exactly the same formal prescription, but using the variables q', as L was, but using the variables q. In other words, keep the formal expression of L(q,q-dot,t), but simply replace everywhere q by q' and q-dot by q'-dot.
Now, how do we find L" ? That's easy. We now express q' as a function of q, using the point transformation, in the expression of L'.

The whole point I was trying to make, since the beginning of this thread, was:
there is no reason why, using this procedure, the functional expression L"(q,q-dot,t) should be equal to the functional expression L(q,q-dot,t).

In order to illustrate my point, I restricted myself to a specific class of symmetry operations, namely those that correspond to point transformations. I know that one can do more general things. But even with this restricted class of symmetry transformations, we see that a symmetry of S does not need to imply automatically a symmetry of the lagrangian functional expression that generates the set S.

Now, in the case of a point transformation, we now have two different functional expressions L(q,q-dot,t) and L"(q,q-dot,t) which generate the same set S, so they must be linked by a dF/dt. However, we fixed a specific parameter of the transformation, $$\delta x$$. For another value of that parameter, we will find another dF/dt of course. But if we work with infinitesimal parameters, we can hope to write:
$$F(q,t,\delta x)$$, and expanding: $$dF/dt = \frac{d}{dt}\left(F_1(q,t) \delta x + O(\delta x^2) \right)$$
Given that $$\delta x$$ is a parameter, and hence doesn't depend on time, we can rewrite this as:
$$\frac{dF_1(q,t)}{dt}\delta x$$ and we drop the 1 subscript, leading to:

$$L''(q,\dot{q},t) = L(q,\dot{q},t) + \delta x \frac{dF(q,t)}{dt}$$

Now, it is traditional to write $$\delta L = L''(q,\dot{q},t) - L(q,\dot{q},t)$$ hence:

$$\delta L = \delta x \frac{dF(q,t)}{dt}$$

Now, as I said, this only works when we have a symmetry of S which corresponds to a point transformation, but that's good enough. I only wanted to discuss this point, because it is already sufficient to show that not just any old lagrangian L, which generates an S which possesses a certain symmetry, must be invariant under that symmetry ; however (in the case of point transformations for the symmetry of S), the infinitesimal change in L must take on the form given above.

When the symmetry is not a point transformation, as you pointed out, we cannot do this and things are a bit more involved. But already in the case of point transformations, we see that $$\delta L$$ is not zero under the infinitesimal transformation.

Nevertheless, and that was the beginning of this thread, this is the assumption that is made by Landau and Lif****z, when they try to derive the conservation of momentum by assuming translational invariance: they assume that the Lagrangian itself is invariant, and hence assume delta L to be zero.

Now as an illustration, I just picked a lagrangian of which I knew that the solutions had a certain symmetry, and added an arbitrary dF/dt. This gave me some new lagrangian, with the same set of solutions (and hence with the same symmetry). When applying naively the substitution q -> q + \delta q in this new lagrangian, I found what I wanted to illustrate (of course): that delta L was not zero, but rather of the form $$\delta L = \delta x \frac{dF}{dt}$$

Of course this was a cheap trick, but it illustrated that a symmetry of the solution set S must not necessarily be a symmetry of the Lagrangian, which seemed to be the point of departure of the reasoning in L&L.BTW, I think, but I'm not clear on this, that it is always possible to reduce a symmetry of S to a point transformation, however, that is, after picking the right description which might imply a genuine canonical transformation (in the Hamiltonian sense). This is what I vaguely had in mind in the beginning of this thread: we can (probably - I don't know for sure) always choose a configuration space and coordinates in such a way that every symmetry operation corresponds to just a global translation of one of the coordinates. But I agree that this is beyond what I was discussing here and maybe erroneous.

 

[vanesch]

In real life we either
1) have a theory (L) and some transformation. So in order to say that our theory is invariant under the transformation at hand, we need to prove
$$L\rightarrow L + \dot{G}(x,t;p)$$
(as I explained above), or
2)do not have a theory but (we) believe that nature is invariant under certain group of transformation. In this case, our task is to construct a (symmetry)-respecting L, i.e., we find L that transforms like

$$L \rightarrow L + \dot{G}(x;p)$$

under our beloved symmetry.

But that was exactly the point I was also making. So we are saying about the same things here. The point made by L&L is what was disturbing me: they required "symmetry by space translation" and derived momentum conservation assuming that the lagrangian L had to be invariant under space translations. As you point out, the only requirement we can have, is that after symmetry transformation applied to the lagrangian, we obtain an equivalent lagrangian (your case (2)), which, in the case of space translations, or more general, point transformations, reduces to the form I gave for the dF/dt (but which is not entirely general, as yours is, because my expression is essentially restricted to point transformations).
However we both seem to agree upon the fact that there is no requirement for the Lagrangian expression itself to be invariant under the symmetry, which is what L&L (and also Goldstein and others) take for granted.

 

[vanesch]

The equation

$$L_2 = L_1 + \dot{G}(x,t)$$

expresses the fact that, from the point of view of the action principle, Lagrangians are not uniquely defined, i.e., we have at our disposal an infinite number of (mathematically) different Lagrangians to play with. Such a freedom though does not spoil or change the the symmetry of the system (I will prove this statement when I formulate (again) the invariace condition).

This is almost trivial: given that two lagrangians linked by the above equation (which we can call, equivalent lagrangians) generate exactly the same set of solutions S (the set of all q(t) satisfying the E-L equations), this set S will have the same symmetries in both cases (given that it is the same set).

 

[samalkhaiat]

]This is almost trivial

Hi,

My statement can be described as "obvious", but not trivial. Well this is exactly what I meant to say: It is "obvious" that redefining the Lagrangian by adding a total derivative to it does not spoil the symmetry.
However, Proving-the-obvious (a bad habit I have learned from mathematicians) is not a trivial matter. In my case, proving that obvious statement showed that I was using a correct condition for invariance.

:

given that two lagrangians linked by the above equation (which we can call, equivalent lagrangians)

This is an obvious statement too, but to show the eqivalence, you need to prove the theorem that I have stated in post#11. The proof though (as you might know) is neither trivial nor easy.

regards

sam

 

[samalkhaiat]

Hi Sam,

This is what I vaguely had in mind in the beginning of this thread: we can (probably - I don't know for sure) always choose a configuration space and coordinates in such a way that every symmetry operation corresponds to just a global translation of one of the coordinates. But I agree that this is beyond what I was discussing here and maybe erroneous.

Yes, but not a global translation and not the whole symmetry group. You see, Configuration space does carry an irreducible representation of the symmetry group. Therefore, it must preserve the group structure (multiplication law).
In general, symmetry groups can be non-abelian, like the rotation group O(3) and the Lorentz group SO(1,3). These groups can not be put in 1-to-1 correspondence with any global abelian group like translation.
However, an interesting (though not surprising) property of all symmetry transformations of the "coordinates" is that they can be written formally as local translations. Actually, this is a property of the translation generators (momentum operators), whose action on the coordinates is given by the Kronecker delta:

$$i P_{b}x^{a} = \partial_{b} x^{a} = \delta^{a}_{b}$$

Now, in terms of its generators, a global infinitesimal translation can be written in the form:

$$\delta_{(GT)}x^{a} = i \alpha^{b} P_{b}x^{a}$$

We say GLOBAL when the infinitesimal parameters $$\alpha^{a}$$ are constants (dont depend on x,t or both).
Now consider (for example) the infinitesimal Lorentz transformation:

$$\delta_{(L)}x^{a} = \omega^{a}{}_{b}x^{b}$$

where

$$\omega^{ab}=-\omega^{ba}$$

are the six infinitesimal Lorentz parameters.

Using the Kronecker-delta trick, this becomes:

$$\delta_{(L)}x^{a} = i \alpha^{c}(x) P_{c} x^{a}$$

which is a local translation with

$$\alpha^{a}(x) = \omega^{a}{}_{b}x^{b}$$

as the translation parameters. In other words, an infinitesimal Lorentz transformation of the coordinates is equivalent to translation with

$$\alpha^{a}(x) = \delta_{(L)}x^{a}$$

as the parameters.

The reason for this eqivalence is that, because the Minkowski spacetime is transitive under translation, any two points connected by a Lorentz transformation can also be connected by translation.
Notice that:
1) the reverse is not trure.
2) it is not true for the whole Lorentz group.
3) the infinitesimal parameters of the translation depend linearly on the spacetime coordinates.
4) similar tricks can also be applied to other symmetry transformations like the SO(3) rotations:
$$\delta_{(R)}x^{i} = \theta^{ij} x^{j} = i \alpha^{j}(x) P_{j}x^{i}=\delta_{(LT)}x^{i}$$

and Galilean transformation

$$\delta_{(G)}x^{i} = v^{i}t = i \alpha^{j}(t) P_{j}x^{i}$$


regards

sam

 

[vanesch]

Yes, but not a global translation and not the whole symmetry group. You see, Configuration space does carry an irreducible representation of the symmetry group. Therefore, it must preserve the group structure (multiplication law).
In general, symmetry groups can be non-abelian, like the rotation group O(3) and the Lorentz group SO(1,3). These groups can not be put in 1-to-1 correspondence with any global abelian group like translation.

Well, I was actually considering only a single symmetry parameter: you can do them 1 by 1 and obtain the conservation law of the quantity that goes with each of them. A 1-parameter group is Abelian, no ?

BTW, conserning the "trivial" remark: I was already accepting that L + dF/dt = L' gave a lagrangrian with an identical set of solutions, but that needs to be proved of course. However, once this is established, I would consider it trivial that the set of solutions to L has the same symmetry group as the set of solutions to L', given that we know that it is exactly the same set.

 

[samalkhaiat]

Well, I was actually considering only a single symmetry parameter: you can do them 1 by 1 and obtain the conservation law of the quantity that goes with each of them. A 1-parameter group is Abelian, no ?

vanesch,this is how we do things anyway! I thought, I said that there are as many Noether numbers as prameters in a symmetry group (= the dimension of the group).
So yes, each Noether number generates a 1-parameter group of symmetry transformations. But, what is it that makes you think that each one of these groups "corresponds" to a global translation of some coordinate? I can understand that your "theorem" is correct when the symmetry group is the rotation group SO(3), because, then you can take one of its three 1-parameter (abelian ) subgroups SO(2), and because of the isomorphism

$$SO(2) \approx U(1)$$

you can actually show that any SO(2) transformation is equivalent to a global "translation" of some polar angle. But can we do the same thing with the symmetry transformation x' = x + vt or with t' = t + T ?

regards

sam

 

[Hurkyl]

I recently noticed there's something that escaped me in Lagrangian mechanics. I recently browsed though the first volume of Landau and L., where it is explained that two systems have identical dynamics if their lagrangians differ only by a total differential to time of a function (because the action will be extremal for the same path in configuration space).

But a bit later, they show that translation symmetry in space gives rise to conservation of linear momentum... by assuming that the Lagrangian is invariant under a space translation. Now, this I knew, of course. However, "translation symmetry" should only imply that under an infinitesimal translation, the lagrangian should at most change by a total time differential of a function (but that function can depend on the infinitesimal translation at hand).
...
What am I missing ? I checked Goldstein, but there too is assumed that the Lagrangian itself is invariant under the symmetry at hand.

Say that two Lagrangians are equivalent if they yield the same dynamics.

It sounds like there's probably a theorem to the effect of: if the dynamics of a Lagrangian L are symmetric under the action of a group G, then there exists a Lagrangian L' equivalent to L, such that L' is invariant under G.

In fact, I have a decent guess as to how to construct L': take the average value of s(L) over all s in G.

 

[vanesch]

Say that two Lagrangians are equivalent if they yield the same dynamics.

It sounds like there's probably a theorem to the effect of: if the dynamics of a Lagrangian L are symmetric under the action of a group G, then there exists a Lagrangian L' equivalent to L, such that L' is invariant under G.

That was exactly what I was asking for, and what I finally demonstrated myself in my post number 3. However, as samalkhaiat points out, this is not true for the case of symmetries which involve time. As, however, my initial problem was about rotational and translational symmetry of space (as used in L&L) then this does indeed work out. But not for more general symmetries.

 

[vanesch]

But can we do the same thing with the symmetry transformation x' = x + vt or with t' = t + T ?

Probably not. This was beyond my initial problem with L&L, in which they simply ASSUME, without any extra statements, that from the translational symmetry of space, must follow that L is INVARIANT under space translations - while my idea was that it is only the set of solutions, and not L itself, which needs to be invariant. In the case that we can transform said symmetry to just a global translation of a single coordinate (which is the case for the problem at hand) I think what I said was more or less correct, while you considered a more general treatment when the symmetries also involve time.

 

[samalkhaiat]

if the dynamics of a Lagrangian L are symmetric under the action of a group G, then there exists a Lagrangian L' equivalent to L, such that L' is invariant under G

This is wrong. If G has non-trivial cohomology, then the best thing we can do about the construction of L' is to make it invariant under a subgroup of G(see post#13).

In fact, I have a decent guess as to how to construct L': take the average value of s(L) over all s in G

I am not a good mathematician, so I would like you to apply your construction method on the following simple problem:
The dynamics of the free particle(in one dimension)

$$L=1/2 (\frac{dx}{dt})^2$$

is (as we know) symmetric under the action of the Galilie group (translations+boosts).
How can you construct an equivalent, Galilean-invariant Lagrangian L'?

regards

sam

 

[Hurkyl]

Ah, I see the particular construction I suggested requires the symmetry group to be a compact Lie group.