Symmetries and lagrangian formulation

1. Nov 23, 2006

vanesch

Staff Emeritus
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.

2. Dec 4, 2006

samalkhaiat

Last edited: Dec 4, 2006
3. Dec 6, 2006

vanesch

Staff Emeritus
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.

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

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

Last edited: Dec 6, 2006
4. Dec 15, 2006

samalkhaiat

Last edited: Dec 15, 2006
5. Dec 15, 2006

vanesch

Staff Emeritus
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...)

6. Dec 16, 2006

vanesch

Staff Emeritus
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 possess 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.

Last edited: Dec 16, 2006
7. Dec 16, 2006

vanesch

Staff Emeritus
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.

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...

Last edited: Dec 16, 2006
8. Jan 6, 2007

samalkhaiat

Last edited: Jan 6, 2007
9. Jan 6, 2007

samalkhaiat

Last edited: Jan 6, 2007
10. Jan 6, 2007

samalkhaiat

Last edited: Jan 7, 2007
11. Jan 6, 2007

samalkhaiat

12. Jan 6, 2007

samalkhaiat

Last edited: Jan 7, 2007
13. Jan 7, 2007

samalkhaiat

Last edited: Jan 7, 2007
14. Jan 7, 2007

samalkhaiat

Last edited: Jan 7, 2007
15. Jan 8, 2007

vanesch

Staff Emeritus
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.

Last edited: Jan 8, 2007
16. Jan 8, 2007

vanesch

Staff Emeritus

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.

17. Jan 8, 2007

vanesch

Staff Emeritus
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).

18. Jan 12, 2007

samalkhaiat

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 learnt from mathematicians) is not a trivial matter. In my case, proving that obvious statement showed that I was using a correct condition for invariance.

:
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

19. Jan 12, 2007

samalkhaiat

Last edited: Jan 12, 2007
20. Jan 14, 2007

vanesch

Staff Emeritus
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.