Re: About Lagrange in GR II

Ken S. Tucker wrote:
My elementary understanding of the Lagrange "L"
first in a Newtonian context is
L = T-V == kinetic energy - potential energy.
So I throw a ball up and the "L" remains constant,
(neglecting air force), is that right?

No. When you throw a ball up, its _energy_ remains constant. For that case (neglecting air resistance), E=T+V, which is quite different from the Lagrangian. Remember that the action is the integral of L over the path of the ball. The motion of the ball is such that the action is _extremized_; that is, the action is considered to be a functional of the ball's path and the path actually taken is the one for which the action is either a minimum or a maximum (or a saddle). In classical mechanics it is invariably a minimum, hence the "principle of least action". But in GR the other possibilities arise.


The "L" is in units of energy right?

Yes, but remember that's the classical Lagrangian, not the Lagrangian density of GR. In classical mechanics it is integrated over time to yield the action, which has units energy*time. In GR the Lagrangian is really a scalar density with units energy/length^3, which when integrated over all spacetime gives units energy*time.


In a Newtonian context a unit of energy, like an erg,
is absolute, however relativity introduces the concept
of the relativity of mass/energy, and this appears as
a tensor density in GR Lagrange's, and it's reasonable
that the "L" being in units of energy will become a
local and coordinate dependant unit, is that right?

The action itself is a scalar. The Lagrangian density of GR is a scalar density which when integrated over all spacetime yields a scalar.

Units are a whole different thing than coordinates, and you seem still confused about the distinction: coordinates are _arbitrary_ assignments of labels to points in the manifold (the labels happen to be N-tuples of real numbers); units are standards of measure -- these are QUITE different. It does not make sense to discuss a "coordinate dependent unit", because such a thing would not be a "unit" at all; a standard of measure that is not standard would be useless.

Think of it this way: a second is so many cycles of a Cesium atom, and a meter is so many atoms of Platinum lined up in a row -- those are definitive measures that are completely independent of any coordinates, because all one needs to do is count. These units can thus be used anywhere and anywhen in the universe.

The current ISO definition of the meter is different.
The principle I stated above stands, however.


I should ask if the "L" becomes a "running constant"
in GR because it's actually a "scalar density"?
Here I'm wondering about the terminology,
"running constant".

I suspect you are confusing the context of GR with the context of Quantum Field Theory in which there are often "running coupling constants".

The term "running coupling constant" is an historical
artifact. What was originally a constant in the "bare"
theory was discovered to "run" (i.e. be a function of
energy) in the "dressed" theory. I'm glossing over
a lot here....

Those two contexts are incommensurate.


OTOH, units of *action* such as Planck's constant,
are global invariants, such as h == ergs*seconds.
(I think that's agreeable).

As I said, action is a scalar, and the Lagrangian density is a scalar density. Neither is independent of one's choice of units, and one is free to select any convenient set of units, including one in which hbar=1 (i.e. hbar itself is a pure number with no associated units; e.g. energy measured in cm^-1 and time in cm).


IMO the density that occurs in the Lagrange can be
alleviated by multiplying by "seconds" to convert
the units of the Lagrange (ergs) to a global invariant,
expressed in units of action.

I think you are confused.


Tom Roberts
.

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