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# Re: [Phys-l] Landau on Lagrangian

Regarding John D's response:

Agreed. Well said. Thanks.

And just to make sure everybody is on the same page:
a) It is one thing to decide in advance we want a
particular equation of motion, and
b) It is quite another thing to decide in advance
on a set of symmetries and to pretend that suffices
for deriving the equation of motion.

Yes. But I don't believe anyone would have made claim b) unless I misremembered L&L's argument. The argument I had in mind imposed *more* than just the symmetries of the inhomogeneous Galilei group. It also explicitly imposed a requirement that the order of the EOM be 2nd order. This last requirement is not one of the set of imposed symmetries. It is an explicit imposition of some formal aspect (namely the order) of the equations one is trying to get the Lagrangian to produce. It actually is inputting some explicit N2 data into the derivation. It is just not inputting the entire form of the EOM. The symmetries imposed (and Hamilton's Principle) essentially do the rest of the winnowing of the resulting EOM.

The derivation argument is *3-fold*, not 1-fold (i.e. not just the symmetries of b) above). The argument also assumes Hamilton's Principle, and it also assumes the EOM be 2nd order. All 3 of these elements result in the desired EOM.

On 01/25/2010 09:09 AM, Hugh Haskell replied:
...
...
How we get there is pretty much irrelevant.

It is irrelevant up to but *not including* the point where we (or
L&L) misrepresent how we got there.

I guess any 'misrepresentation' on L&L's part is probably only in not leveling with the reader and not fully explaining the import of the assumptions they use, and the source from where they got them (said source being the properties of the EOM that they are trying to get the Lagrangian formulation to produce).

Any class has multiple goals. The primary, fundamental, and
overarching goal should be for students to learn to think clearly.
Learning this-or-that detail about the equations of motion is
secondary.

Telling students that symmetry principles suffice to derive the
equations of motion -- when that is not in fact true -- is a huge
step in the wrong direction.

I also think it is a straw man. But maybe one might be mistakenly led to think the L&L argument was just such a claim because of their lack of full transparency in what they are actually doing.

If you want to say "we pulled this Lagrangian out of a hat and we
like it because we know by means of 20/20 hindsight that it will
generate the nonrelativistic equations of motion for a point
particle" ... that's fine. If you want to say that "it has some
symmetries that any physical Lagrangian ought to have" that's OK
too.

I'm glad we all agree on this.

We
can guess,

OK.

or we can impose some abstract symmetries,

Necessary but *not* sufficient.

True. But the sufficiency comes from the symmetries, the use of Hamilton's Principle, *and* the 2nd order requirement.

I haven't read the whole L&L passage, and I'm not particularly
interested in doing so ...

Then we are both at somewhat of a disadvantage regarding the L&L argument. You haven't read it, and although I have read it years ago, I can't locate my copy of their book, and haven't been able to verify my memories of it that are stored in a brain with admittedly less than perfect memory function.

but in any case it is clear that if people interpret the passage
as saying that symmetry principles suffice to derive the
Lagrangian, then either the passage is wrong or the interpretation
is wrong.

Yes. But I doubt that L&L would claim that was the proper interpretation of their argument. But maybe I'm just being nice to them and giving them the benefit of any doubt that may arise about it. It would be nice to find the book and look up what they *do* say about it.

The wrongness should be obvious from the fact that the Galilean
symmetries that we are discussing are a subgroup of the Lorentz
symmetries of special relativity.

Whoa. I don't believe the inhomogeneous Galilei group is any *subgroup* of the Poincare Group. Rather I believe it forms a 10-parameter *tangent space* to the manifold of the 10-parameter Poincare group tangent at zero boost velocity where the manifold is flattened along the boost generator directions in that parameter space, but preserves the structure of the spatial rotations and space-time translations. The Poincare group forms a curved manifold in the parameters denominating the set of inertial reference frames. In particular, it is curved w.r.t. the parameters for the spatial rotations and for the boosts. The inhomogeneous Galilei group is only curved w.r.t. the parameters for spatial rotations but not for the boosts. Neither group is curved w.r.t. the parameters for space-time translations (as long as space-time itself remains flat and we don't start doing GR).

We know that relativistic mechanics is different from
non-relativistic mechanics, so the idea that Galilean symmetries
lead to one and only one equation of motion is dead on arrival.

Let's back up a minute.

I think we can justify, using a L&L-esque-type of argument for a free particle, that requiring: 1) Hamilton's Principle, 2) the EOM respect the symmetries of the inhomogeneous Galilei group, and 3) the resulting EOM be 2nd order will indeed give us N1 & N2. But if we replace 2) here with: 2') the EOM respect the symmetries of the inhomogeneous Poincare group then we will get a Lagrangian of the form L = -m*c^2*sqrt(1 - v^2/c^2). And this latter Lagrangian gives us the analogous SR equations and expressions. But the justification needs to be understood and explained to students, as Stefan J said, as only a *plausibility argument* and not some kind of fully mathematically rigorous derivation.

The way the modified argument would go for the 2') version is to realize that substituting a Lorentz-boosted velocity (relating v to v' & v_0 by the SR/Lorentz velocity addition formula) into no simple function of v^2 (including a linear one like with the Galilean/Newtonian version) will result in the same function of v'^2 plus a total time derivative dF/dt for F = F(r',v',t) *unless* the function F is precisely the zero function, and the action is actually a scalar under Lorentz/Poincare transformations, and that the action is just proportional to the proper time of the path. Imposing this scalar requirement on the action means that the Lagrangian needs to be proportional to d[tau]/dt (i.e. 1/[gamma]).

The value of the proportionality constant is found by taking the Newtonian limit and matching things with the Newtonian expressions at speeds tiny compared to c. What this argument also discovers along the way is that the idea that the action for the physical path being minimal is equivalent to the proper time for the physical (time-like) path being maximal. The negative proportionality constant -m*c^2 between them enforces this. Also expanding the Lagrangian in inverse powers of c^2 in the Newtonian/Galilean limit nicely makes the rest energy offset m*c^2 automatically pop out of the derivation as a non-Newtonian zero level for the energy scale.

Of course this procedure needs to be invoked only *after* the 2nd order EOM requirement is imposed on the situation so that our action and Lagrangian will not have to contend with expressions involving derivatives of higher order than the velocity.

The idea that the elapsed proper time is maximal for the physical time-like path of a free particle is very useful in doing particle mechanics in general relativity because in a curved space-time a maximally elapsed (i.e. longest) proper time over a time-like path defines the notion of geodesic straightness for time-like paths the same way that minimal (i.e. shortest) proper distance defines the notion of geodesic straightness for space-like paths. In GR the path a massive free particle (i.e. a particle subject to no non-gravitational influences) follows a time-like geodesic in space-time. And this means that the Lagrangian for such a particle in GR is just like what it is in SR, namely, L = -m*c^2*d[tau]/dt where t is the coordinate time used in labeling space-time events, and [tau] is the increment of proper time for the particle (d[tau]/dt = 1/[gamma]). But d[tau] = sqrt(d[tau]^2) encodes in it the metric properties of the curved space-time manifold and the idiosyncrasies of the coordinate system used to describe it.

.... Neither Lagrange nor Newton are fundamental to nature.

Agreed!

What is fundamental is the symmetries. If we go one way around
the circle from the symmetries, we get Lagrange, and that leads
to Newton.

No, it does not. Symmetries are not sufficient to deduce the
equations of motion.

Well, the symmetries, Hamilton's Principle, and the 2nd order EOM assumption do determine it. But that is a pretty restrictive set of requirements that may be considered as taken equivalent to assuming the form of the EOM.

What that tells me is that the Lagrangian method is good if we
are clever enough to guess the Lagrangian (i.e., the one that
leads to equations that predict what is actually observed). It
seems rather apparent to me that when we find the correct
Lagrangian, we get Einstein, or Heisenberg/Schroedinger, or
Dirac/Feynman, ....

Agreed. When expressed as a guessing game, validated by 20/20
hindsight, that's entirely correct. That is how it is often
presented.

I'm OK with it, too.

=================================

Symmetries *are* related to the equations of motion.

Once you know the Lagrangian, then Noether's theorem says that for
every continuous symmetry of the action, there is a corresponding
conserved quantity. But once again, you have to know the
Lagrangian first.

This is what we should be discussing! Noether's theorem is real
physics, beautiful physics.

OK, what about Noether's theorem do you want to discuss?

Exegesis of some Landau/Lifshitz snafu, not so much.

Fine.

David Bowman