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Re: [Phys-l] Landau on Lagrangian

Regarding Bob S's response:

I wrote: . . .
But even If I grant you these requirements and accept that they
imply that the mechanical L = L(v^2), you still have to show that
this implies: KE(translation) is proportional to v^2.

To which David Bowman responded:
That was already done (more than once). That comes from the
invariance of the EOM under Galilean boosts once it is granted
that the resulting equations be no higher than 2nd order and
the symmetries of space & time homogeneity and spatial isotropy
are imposed. Please reread the part of the argument about how
only the expression v^2 obeys v^2 = v'^2 + dF/dt when
v = v' + v_0 and F = F(r',v',t). Other nonlinear functions of
v^2 do not have this property.

Those arguments only attempt to show that L = L(v^2).

No, they do not. Please pay attention.

The homogeneity in space and time constrain L to not depend explicitly on t or r. The 2nd order of the EOM requirement constrains L to now depend only on v and not on any higher derivatives of r WRT t. The isotropy of space then forces L to depend only on the square magnitude v^2. Now get this: *The invariance under Galilean boosts constrains this L(v^2) to now be PROPORTIONAL TO THE FIRST POWER of v^2* (up to an irrelevant additive constant). This is because any other function of v^2 that is nonlinear (i.e. not proportional to it) will not have the needed property that L(v^2) = L(v'^2) + dF/dt where F is some function F(r',v',t) when v = v' + v_0. This property is needed to make the EOM invariant under Galilean boosts. We then define the mass as twice this proportionality constant. If Hamilton's principle is taken to be a *minimum* principle rather than merely a principle of stationarity then the constant mass must be positive as well.

They do not address the proposition: KE(translation) is
proportional to v^2.

That is *precisely* what the point of the invariance-under-boosts part of the argument does address.

How does the concept of KE even enter the development?

It is the resulting Lagrangian of a free particle after the assumption of 2nd order EOM and all the symmetries are imposed (after subtracting off any possible arbitrary irrelevant additive constant that one may have included, i.e. the 'rest Lagrangian').

Bob Sciamanda

David Bowman