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Certainly the "usual" system would not behave in this manner, being
in a state of mechanical but not thermodynamic equilibrium. The
specific system I had in mind is the usual system with a gimmick.
Place a second heat-incapacious thermally-insulating piston in the
middle, and put different gases, say one monatomic and the other
diatomic, on the two sides of that piston. One can now carry out
reversible processes on this system that leave the two sides at
different temperatures. Consider an adiabatic compression of the
system from an initial state of thermodynamic equilibrium, for
example. Which gas gets warmer?
Leigh
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).
They do not address the proposition: KE(translation) is
proportional to v^2.
How does the concept of KE even enter the development?
Bob Sciamanda
On 01/25/2010 08:39 PM, curtis osterhoudt wrote:
http://mitworld.mit.edu/video/540/
Horrifying. Macabre.
Some professors weigh in on teaching the 2nd law of Thermo.
Apparently no physics professors. Also nobody from
the cryptography, communications, machine learning,
or statistics communities.
Interesting takes on where students get hung up.
Interesting? I found it infuriating. My reaction
was, "Duh, if you run through a cactus plantation
at night, you're going to get hung up." That is
to say, the nine panelists described at least eight
different ways of making the subject seem infinitely
more complicated and less useful than it really is.
No wonder the students get hung up.
I had no idea the situation was so bad. Maybe I'm
naive, but I would have preferred to stay naive.
Now I need to go watch The Exorcist or The Shining,
something to get the image of that MIT video out of
my head. Otherwise I'm gonna have nightmares.
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
>Those arguments only attempt to show that L = L(v^2).
No, they do not. Please pay attention.
Some professors weigh in on teaching the 2nd law of Thermo. Interestingtakes on where students get hung up.
Before I fell asleep at the wheel, I heard one speaker offer that
http://mitworld.mit.edu/video/540/
/************************************
Down with categorical imperative!
flutzpah@yahoo.com
************************************/
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
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
... why not define the unit 'cycle' = 2pi radians, and then make
1 Hz = 1 cycle/s = 2pi radians/s = 2pi/s ?
I've always been mildly disturbed by the definition of the unit hertz as
equivalent to 1/second. (Such as can be found, for example, at
<http://physics.nist.gov/cuu/Units/units.html> and
<http://physics.nist.gov/cuu/Units/SIdiagram.html>.
Can anyone see any problems with this [radian] option, other than the near
impossibility of changing tradition?