Gravitational Energy Conservation
Phil Gibbs
Why does the FAQ say that Energy is not conserved for cosmological redshift?
The FAQ explains that,
> IS ENERGY CONSERVED IN OUR UNIVERSE? NO
>
> Why? Every conserved quantity is the result of some symmetry of
> nature. This is known as Noether's theorem. For example, momentum
> conservation is the result of translation invariance, because position is
> the variable conjugate to momentum. Energy would be conserved due to
> time-translation invariance. However, in an expanding or contracting
> universe, there is no time-translation invariance. Hence energy is not
> conserved. If you want to learn more about this, read Goldstein's
> Classical Mechanics, and look up Noether's theorem.
This doesn't make any sense to me. To say that Noether's theorem does not
apply because there is no time-translation invariance in an expanding universe
is like saying Noether's theorem does not apply to a freely moving particle
because it is moving. The expansion of the universe is part of the solution,
not the equations of motion.
Gravitational equations can be derived from a principle of least action like
any other. The action is not only time translation invariant, it is invariant
under any transformation of the form
t -> f(t) , f(t) positive monotonic C2 function.
Noether's theorem can be applied to derive a tensor for the energy-momentum
of the matter fields and a psuedo-tensor for the energy-momentum of the
gravitational field. From these energy conservation can be derived.
Admittedly the energy is non-local for gravity and coordinate system
dependent, but why should that be a problem?
I can see that in an open universe there might be a problem because the
total energy is going to be infinite unless it is asymptotically flat. But
for a finite closed universe you can integrate the total energy. In a
closed Robertson-Walker-Friedmann cosmology there is an equation,
2
.2 G M 2
M S - --- + Mc = 0
S
where S(t) = scale of universe defined in comoving system.
M = total non-gravitational mass energy
G = gravitational constant
c = speed of light
for a dust radiation mixture M is given by
N
M = M0 + ---
S
where M0 = total rest mass of the dust
N = total radiation particle number
both M0 and N are constants
Why is this not an expression for the conservation of energy including
the effects of cosmological redshift? I know it is a bit strange that
the energy is equated to zero instead of a constant but I thought
that this was a consequence of the more general invariance as I stated
above. I seem to remember some result that in a closed universe with
no closed timelike curves the total energy over a closed spacelike
hypersurface must be zero. Is that right?
I suppose this must of all been discussed to death on sci.physics in
the past otherwise it wouldn't be in the FAQ in which case I apologise
for bringing it up again.
( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (
) ) ) ) ) ) ) ) ) ): ) ) ) ) ) )
( ( :( ( :( ( ( ( ( ( ( ( ( ( ( (
) ) ) ) ) ) ) ) ) ) ) ) ) ): ) )
ph...@strauss.eurocontrol.fr = Phil Gibbs
SCOTT I CHASE
>
>Why does the FAQ say that Energy is not conserved for cosmological redshift?
>The FAQ explains that,
>
>> IS ENERGY CONSERVED IN OUR UNIVERSE? NO
>>
>> Why? Every conserved quantity is the result of some symmetry of
>> nature. This is known as Noether's theorem. For example, momentum
>> conservation is the result of translation invariance, because position is
>> the variable conjugate to momentum. Energy would be conserved due to
>> time-translation invariance. However, in an expanding or contracting
>> universe, there is no time-translation invariance. Hence energy is not
>> conserved. If you want to learn more about this, read Goldstein's
>> Classical Mechanics, and look up Noether's theorem.
>
>This doesn't make any sense to me. To say that Noether's theorem does not
>apply because there is no time-translation invariance in an expanding universe
>is like saying Noether's theorem does not apply to a freely moving particle
>because it is moving. The expansion of the universe is part of the solution,
>not the equations of motion.
The FAQ is somewhat deficient in this regard, and I should update it.
What it means to say is that the Einstein equations have no static
(time-independent) solutions, and that Noether's theorem does not apply
because the equations themselves have no time-translation invariance,
which is reflected in the absence of any static solutions.
Your complaint is a fair one. One of these days (soon!) I'll change it.
-Scott
-------------------- i hate you, you hate me
Scott I. Chase let's all go and kill barney
SIC...@CSA2.LBL.GOV and a shot rang out and barney hit the floor,
no more purple dinosaur.
john baez
>This doesn't make any sense to me. To say that Noether's theorem does not
>apply because there is no time-translation invariance in an expanding universe
>is like saying Noether's theorem does not apply to a freely moving particle
>because it is moving. The expansion of the universe is part of the solution,
>not the equations of motion.
Everything you say is true. The point is that there are a couple of
different things one can mean by the "total energy of the universe."
Let's just talk about a closed universe. One approach is to
work out the integral of T_{00} over one of the standard spacelike
hypersurfaces (at "constant cosmic time"). One gets something that
depends on time. The reason is that there is no time translation
symmetry of this particular spacetime metric -- no "timelike Killing
vector field."
As you note, any reasonable attempt to define the energy of a closed
universe from Noether's theorem (applying the theorem to the
gravity/matter system, NOT to the matter alone treating the metric as
fixed) gives zero. Indeed, it must be zero by general principles,
because it is basically just the integral of G_{00} - 8piT_{00} over one
of the standard spacelike hypersurfaces, which vanishes by Einstein's
equation. But as you note, that's not so bad, it's actually
interesting.
>I suppose this must of all been discussed to death on sci.physics in
>the past otherwise it wouldn't be in the FAQ in which case I apologise
>for bringing it up again.
Yes indeed, all this has been argued about ad infinitum. But perhaps
the FAQ isn't quite clear enough about the subtleties of defining the
"energy of the universe" in the context of GR. And it's always good to
remind people about this interesting issue.
Here's what I said about it in "week11":
2) An algebraic approach to the quantization of constrained systems:
finite dimensional examples, by Ranjeet S. Tate, Syracuse University
physics department PhD dissertation, August 1992, SU-GP-92/8-1. (Tate
is now at rst...@cosmic.physics.ucsb.edu, but please don't ask him for
copies unless you're pretty serious, because it's big.)
Both the technical problems of "canonical" quantum gravity and one of the
main conceptual problems - the problem of time - stem from the fact that
general relativity is a system in which the initial data have
constraints. So improving our understanding of quantizing constrained
classical systems is important in understanding quantum gravity.
Let me say a few words about these constraints and what I mean by
"canonical" quantum gravity.
First consider the wave equation in 2 dimensions. This is
an equation for a function from R^2 to R, say phi(t,x), where t is a
timelike and x is a spacelike coordinate. The equation is simply
d^2 phi/dt^2 - d^2phi/dx^2 = 0.
Now this equation can be rewritten as an evolutionary equation for
initial data as follows. We consider pairs of functions (Q,P) on R -
which we think of phi and d phi/dt on "space", that is, on a surface t =
constant. And we rewrite the second-order equation above as a
first-order equation:
d/dt (Q,P) = (P, d^2Q/dx^2). 1)
This is a standard trick. We call the space of pairs (Q,P) the "phase
space" of the theory. In canonical quantization, we treat this a lot
like the space R^2 of pairs (q,p) describing the initial position and
momentum of a particle. Note that for a harmonic oscillator we have an
equation a whole lot like 1):
d/dt (q,p) = (p, -q).
This is why when we quantize the wave equation it's a whole lot like the
harmonic oscillator.
Now in general relativity things are similar but more complicated.
The analog of the pairs (phi, d phi/dt) are pairs (Q,P) where Q is the
metric on spacetime restricted to a spacelike hypersurface - that is,
the "metric on space at a given time" - and P is concocted from the
extrinsic curvature of that hypersurface as it sits in spacetime.
Now the name of the game is to turn Einstein's equation for the metric
into a first-order equation sort of like 1). The problem is, in general
relativity there is no god-given notion of time. So we need to *pick* a
"lapse function" on our hypersurface, and a "shift vector field" on our
hypersurface, which say how we want to push our hypersurface forwards in
time. The lapse function says at each point how much we push it in the
normal direction, while the shift vector field says at each point how
much we push it in some tangential direction. These are utterly
arbitrary and give us complete flexibility in how we want to push the
hypersurface forwards. Even if spacetime was flat, we could push the
hypersurface forwards in a dull way like:
-------------------- new
____________________ old
or in a screwy way like
----
/ \ /\
/ --- \
------- new
____________________ old
Of course, in general relativity spacetime is usually not flat, which
makes it ultimately impossible to decide what counts as a "dull way" and
what counts as a "screwy way," which is why we simply allow all possible
ways.
Anyway, having *chosen* a lapse function and shift vector field, we can
rewrite Einstein's equations as an evolutionary equation. This is a bit
of a mess, and it's called the ADM (Arnowitt-Deser-Misner) formalism.
Schematically, it goes like
d/dt (Q,P) = (stuff, stuff'). 2)
where both "stuff" and "stuff'" depend on both Q and P in a pretty
complex way.
But there is a catch. While the evolutionary equations are equivalent
to 6 of Einstein's equations (Einstein's equation for general relativity
is really 10 scalar equations packed into one tensor equation), there
are 4 more of Einstein's equations which turn into *constraints* on Q
and P. 1 of these constraints is called the Hamiltonian constraint and
is closely related to the lapse function; the other 3 are called the
momentum or diffeomorphism constraints and are closely related to the
shift vector field.
For those of you who know Hamiltonian mechanics, the reason why the
Hamiltonian constraint is called what it is is that we can write it as
H(Q,P) = 0
for some combination of Q and P, and this H(Q,P) acts a lot like a
Hamiltonian for general relativity in that we can rewrite 2) using the
Poisson brackets on the "phase space" of all (Q,P) pairs as
d/dt Q = {P,H(Q,P)}
d/dt P = {Q,H(Q,P)}.
The funny thing is that H is not zero on the space of all (Q,P) pairs,
so the equations above are nontrivial, but it does vanish on the submanifold
of pairs satisfying the constraints, so that, in a sense, "the
Hamiltonian of general relativity is zero". But one must be careful in
saying this because it can be confusing! It has confused lots of people
worrying about the problem of time in quantum gravity, where they
naively think "What - the Hamiltonian is zero? That means there's no
dynamics at all!"
john baez
>The FAQ is somewhat deficient in this regard, and I should update it.
>What it means to say is that the Einstein equations have no static
>(time-independent) solutions, and that Noether's theorem does not apply
>because the equations themselves have no time-translation invariance,
>which is reflected in the absence of any static solutions.
The Einstein equations are invariant under *all* diffeomorphisms (changes
of coordintes). So it's not good to say they lack time-translation
invariance. Also, btw, they DO have static solutions, like the
Schwarzschild metric. However, the solution we live in is not static,
so treating that spacetime metric as a given "background," we can say
that our universe lacks time translation symmetry hence conservation of
energy. It's a matter of how you slice the pie.
john baez
that's very simple. Consider a particle bouncing around inside a
piston. Suppose the piston is gradually being pushed in by a spring. If we
consider only the particle and write down its equations of motion, these
equations are not time-invariant, so its energy will not be conserved.
However, if we include the piston and spring, we see that the energy of
the *whole system* is conserved, and that its laws are time-invariant.
Hannu Poropudas
The mistake must not be made, for example, of supposing that the
energy-momentum principle in its special relativity form would
hold in general relativity over a region of finite size.
For example for perfect fluid in isolated systems (assuption of
positive pressure throughout) the proper energy rho_00 dv_0 of
every element of the fluid is decreasing when the system is
expanding or increasing when the system is contracting.
See: Tolman, R. C., 1962.
Relativity Thermodynamics and Cosmology.
University Press, Oxford, Great Britain.
Pages: 214-235 and 361-444.
Best Regards,
Hannu Poropudas,
hapo...@freenet.hut.fi
"Rule, Force and Honour belongs to God."
--
Hannu Poropudas
I would expect that pressure of all neutrino stars (diamonds)
would compensate all of the loss of energy of cosmic backround
radiadiation's energy loss in red-shift in cosmic background
radiation.
This would mean that we must enlarge our concept of Universe.
Best Regards,
Hannu Poropudas,
hapo...@freenet.hut.fi
"Rule, Force and Honour belongs to God."
"God has only Sons."
"The first creatures of God does not have the doigsts, because
they haved to se farnerd, this is due to that they can sorcelleriete.
They can be invisible or visible."
--
Phil Gibbs
> In article <1994Feb16.1...@fozzie.eurocontrol.fr> ph...@strauss.eurocontrol.fr writes:
>
> As you note, any reasonable attempt to define the energy of a closed
> universe from Noether's theorem (applying the theorem to the
> gravity/matter system, NOT to the matter alone treating the metric as
> fixed) gives zero. Indeed, it must be zero by general principles,
> because it is basically just the integral of G_{00} - 8piT_{00} over one
> of the standard spacelike hypersurfaces, which vanishes by Einstein's
> equation. But as you note, that's not so bad, it's actually
> interesting.
>
It is interesting and also very useful when somebody asks that anoying
question, "Where did all the energy come from if there was nothing
before the big bang?"
Do any of the attempts at unification lead to the implication that the
total electric charge must also be zero?
>
> Here's what I said about it in "week11":
>
> [...]
>
> The funny thing is that H is not zero on the space of all (Q,P) pairs,
> so the equations above are nontrivial, but it does vanish on the submanifold
> of pairs satisfying the constraints, so that, in a sense, "the
> Hamiltonian of general relativity is zero". But one must be careful in
> saying this because it can be confusing! It has confused lots of people
> worrying about the problem of time in quantum gravity, where they
> naively think "What - the Hamiltonian is zero? That means there's no
> dynamics at all!"
>
That thought had ocurred to me but I dismissed it with the reasoning that
the energy is only zero for the classical solution and could fluctuate
about zero in a quantum theory.
to...@cc.usu.edu
By the way, if I may shamelessly plug my own work, it has been shown recently
that the diffeo symmetry (and a trivial scale symmetry) are the *only*
symmetries of the vacuum Einstein equations. This means that there are
essentially no conserved currents which could be used to construct notions of
energy density. ( For isolated systems, one gets a notion of total energy (and
momentum) by virtue of boundary conditions.)
Sean Carroll
>>
>>Why does the FAQ say that Energy is not conserved for cosmological redshift?
>>The FAQ explains that,
>>
>>> IS ENERGY CONSERVED IN OUR UNIVERSE? NO
>>>
>>> Why? Every conserved quantity is the result of some symmetry of
>>> nature. This is known as Noether's theorem.
While the subject is being broached, I think that the FAQ is a little
misleading in these sentences. Noether's theorem tells us that every
symmetry leads to a conserved quantity, not the other way around.
There are conservation laws which do not arise from a symmetry --
by accident, or from topological considerations, and so on.
It would be easy to change this paragraph to say that energy conservation
(specifically) arises from time translation invariance, which does not
apply if you consider the energy of matter on a non-static background.
--Sean
Hannu Poropudas
I would like to correct some language errors of my article, which
is dated 17 Feb 1994 14:50:21 GMT, <2k007d#g...@freenet.hut.fi>,
#=dollar-mark:
I would expect that compression of all neutrino stars (diamonds)
would compensate all of the loss of energy in red-shift of
cosmic background radiation.
This would mean that we should enlarge our view of the Universe
to the negative mass particles. I suppose that negative mass
could have something to do with right-handed neutrinos. This
could be a good start to investigate existence of negative mass,
which I have tried to decsribe in my articles in anonymous
computer FTP.FUNET.FI (pub/doc/misc/HannuPoropudas, articles
are named as README.*).
Best Regards,
Hannu Poropudas,
hapo...@freenet.hut.fi
"Rule, Force and Honour belongs to God."
"God has only Sons."
"First creatures of God, called 'muddleds', does not have fingers,
because they have been checked in growth, this is due that they
can do magic like God does. These creatures are able to be
invisible or visible."
--
James Kibo Parry
Hannu Poropudas <hapo...@freenet.hut.fi> wrote:
>
> "First creatures of God, called 'muddleds', does not have fingers,
> because they have been checked in growth, this is due that they
> can do magic like God does. These creatures are able to be
> invisible or visible."
Hannu, could you please explain the bit about the fingers, especially
the part about "checked in growth"?
Seriously,
-- K.
Hannu Poropudas
Other word to this is 'atrophy' in my poor english vocabulary.
Hannu.
--
john baez
>It is interesting and also very useful when somebody asks that anoying
>question, "Where did all the energy come from if there was nothing
>before the big bang?"
Indeed. As what's-her-name said in King Lear, nothing will come from
nothing.
>Do any of the attempts at unification lead to the implication that the
>total electric charge must also be zero?
No unification needed. Maxwell's equations say that in a closed
universe the total electric charge has to be zero -- it's just Gauss'
law, really -- or if you like, there's nowhere for field lines to go!
I should emphasize that the "no energy" result for general relativity
also only holds for a *closed* universe, and indeed for the same reason.
You do an integration by parts and the boundary terms vanish because
there's no boundary in a closed universe; if the universe were
asymptotically flat one *could* define a nonzero energy.
Emory F. Bunn
>Indeed. As what's-her-name said in King Lear, nothing will come from
>nothing.
Her name was King Lear.
-Ted
Matthew MacIntyre
: >The FAQ is somewhat deficient in this regard, and I should update it.
: >What it means to say is that the Einstein equations have no static
: >(time-independent) solutions, and that Noether's theorem does not apply
: >because the equations themselves have no time-translation invariance,
: >which is reflected in the absence of any static solutions.
: The Einstein equations are invariant under *all* diffeomorphisms (changes
: of coordintes). So it's not good to say they lack time-translation
: invariance.
Au contraire, it is "good" to say that, because it is true. What on earth
could "time translation invariance" mean in this context other than the
existence of a timelike Killing vector? Isn't it strange that after all
these years of telling physicists not to care too much about coordinates,
we now have quantum gravity people undoing all that good work with their
obsession with diffeomorphisms? :)
The essential point is this: Noether tells us that energy conservation
has something to do with the symmetry of spacetime itself. The expansion
of the Universe tells us that spacetime itself is NOT time-symmetric, and
therefore in cosmological contexts we must expect energy conservation to
fail. That's all there is to it.
Also, btw, they DO have static solutions, like the
: Schwarzschild metric. However, the solution we live in is not static,
: so treating that spacetime metric as a given "background," we can say
: that our universe lacks time translation symmetry hence conservation of
: energy. It's a matter of how you slice the pie.
But in another post in this thread, you tell us that G00 should have
something to do with "gravitational energy" and that Einstein's equation
indicates that the "total" energy is zero! Moral of the story: it is not
a good idea to try to explain cosmology using the technicalities of
quantum gravity! You don't really believe that the Hamiltonian=0 business
has anything to do with the non-conservation of energy in cosmology,as
the above passage shows! The people who want to talk about "gravitational
energy" are thinking of gravity propagating on a flat background; in
short, they are your natural enemies, and you should be waging war
against them rather than giving them encouragement....... :)
Matthew MacIntyre
: Sorry to post yet again on this issue but I figured out a good analogy
Oh, I see. You mean that gravity is really just another force that lives
on Minkowski space, and that the apparent failure of energy conservation
in cosmology is due to those stubborn mathematical physicists who insist
otherwise: so that if we compute the gravitational energy and add it into
the other energies, then the total is conserved by applying Noether to
this flat background. It's sad the way mathematicians always obfuscate,
isn't it? :)
john baez
>john baez (ba...@guitar.ucr.edu) wrote:
>: In article <16FEB199...@csa5.lbl.gov> sic...@csa5.lbl.gov (SCOTT I CHASE) writes:
>
>: >The FAQ is somewhat deficient in this regard, and I should update it.
>: >What it means to say is that the Einstein equations have no static
>: >(time-independent) solutions, and that Noether's theorem does not apply
>: >because the equations themselves have no time-translation invariance,
>: >which is reflected in the absence of any static solutions.
>: The Einstein equations are invariant under *all* diffeomorphisms (changes
>: of coordintes). So it's not good to say they lack time-translation
>: invariance.
>Au contraire, it is "good" to say that, because it is true. What on earth
>could "time translation invariance" mean in this context other than the
>existence of a timelike Killing vector?
That is a symmetry of a SOLUTION, not an EQUATION, which is what I was
talking about. You know this all anyway and are just having fun
arguing.
>Isn't it strange that after all
>these years of telling physicists not to care too much about coordinates,
>we now have quantum gravity people undoing all that good work with their
>obsession with diffeomorphisms? :)
Well, *YOU* try quantizing gravity without getting involved in the
diffeomorphism-invariance of the equations!
>The essential point is this: Noether tells us that energy conservation
>has something to do with the symmetry of spacetime itself. The expansion
>of the Universe tells us that spacetime itself is NOT time-symmetric, and
>therefore in cosmological contexts we must expect energy conservation to
>fail. That's all there is to it.
No, that's not all there is to it. It's certainly PART of all there is
to it. But Noether also tells us that the diffeomorphism-invariance of
Einstein's equations gives rise to a Hamiltonian constraint. A balanced
understanding has to keep both of these facts in mind. As I said, it's
a matter of how you cut the pie. Say you are playing with Einstein's
equations coupled to Maxwell's. You can study T_{mu nu}, note that the
integral of T_{00} over a spacelike slice ain't conserved when there's
no Killing vector field, and say energy ain't conserved. Or you can
start with the Lagrangian for Einstein/Maxwell, concoct a Hamiltonian
density by the usual recipe (briefly, H = p qdot - L), and note that
MYSTERIOUSLY (actually because of the general covariance you so dislike)
it equals zero when we integrate it over a compact spacelike slice.
Read Wald's appendix on the Hamiltonian formulation of GR if you think
I'm fibbing.
>The people who want to talk about "gravitational
>energy" are thinking of gravity propagating on a flat background; in
>short, they are your natural enemies, and you should be waging war
>against them rather than giving them encouragement....... :)
The problem with waging war against ignorance is that the battle is so
lopsided, it gets tiring. Sure, most people wondering about
"gravitational energy" are confused, for them I suppose one should
simply say:
EINSTEIN SAYS G_{MU NU} = T_{MU NU}. THERE IS NO "GRAVITATIONAL TERM"
IN THE RIGHT HAND SIDE.
(should I make that my .sig?) But there are a lot of other interesting
things to say about the concept of "energy" and "Hamiltonian" in general
relativity.
john baez
Yes, that's what I mean, because I am completely confused about general
relativity and constantly need you to remind me how it works.
to...@cc.usu.edu
>> In article <1994Feb16.1...@fozzie.eurocontrol.fr>, ph...@strauss.eurocontrol.fr writes...
>>>
>>>Why does the FAQ say that Energy is not conserved for cosmological redshift?
>>>The FAQ explains that,
>>>
>>>> IS ENERGY CONSERVED IN OUR UNIVERSE? NO
>>>>
>>>> Why? Every conserved quantity is the result of some symmetry of
>>>> nature. This is known as Noether's theorem.
>
> While the subject is being broached, I think that the FAQ is a little
> misleading in these sentences. Noether's theorem tells us that every
> symmetry leads to a conserved quantity, not the other way around.
> There are conservation laws which do not arise from a symmetry --
> by accident, or from topological considerations, and so on.
>
The relation between symmetries and conservation laws is usually rather
tight. Of course, any discussion of this depends on a precise definition of
"symmetry" and "conservation law". With a few minor technical assumptions,
it can be shown that if a system of Euler-Lagrange equations admits a conserved
current, then there must be a corresponding symmetry. See, e.g., "Applications
of Lie Groups to Differerential Equations", by Peter Olver for a careful
treatment of such matters. The upshot is that usually the existence of a sym-
metry is a necessary condition for the existence of a conserved current. So,
often times a system of equations will admit a symmetry, but there will be
no corresponding conservation law!
Phil Gibbs
> In article <2k3l6c$n...@scunix2.harvard.edu>, car...@husc8.harvard.edu (Sean Carroll) writes:
> >> In article <1994Feb16.1...@fozzie.eurocontrol.fr>, ph...@strauss.eurocontrol.fr writes...
> >>>
> >>>Why does the FAQ say that Energy is not conserved for cosmological redshift?
> >>>The FAQ explains that,
> >>>
> >>>> IS ENERGY CONSERVED IN OUR UNIVERSE? NO
> >>>>
> >>>> Why? Every conserved quantity is the result of some symmetry of
> >>>> nature. This is known as Noether's theorem.
> >
> > While the subject is being broached, I think that the FAQ is a little
> > misleading in these sentences. Noether's theorem tells us that every
> > symmetry leads to a conserved quantity, not the other way around.
> > There are conservation laws which do not arise from a symmetry --
> > by accident, or from topological considerations, and so on.
> >
> Perhaps you could give an example of what you mean here.
>
An example of what Sean Carroll means is conservation of Baryon Number
in the Standard Model. There is no symmetry in the Standard Model from
which Baryon Number conservation follows by Noether's Theorem (as far
as I know) and yet it is conserved (in the Standard Model).
Chariya Peterson
How would you interpret this in QFT ?
In the classical theory, can a field line escape through a blackhole ?
chariya
Douglas A. Singleton
>In article <2k3l6c$n...@scunix2.harvard.edu>, car...@husc8.harvard.edu (Sean Carroll) writes:
>>> In article <1994Feb16.1...@fozzie.eurocontrol.fr>, ph...@strauss.eurocontrol.fr writes...
>>>>
>>>>Why does the FAQ say that Energy is not conserved for cosmological redshift?
>>>>The FAQ explains that,
>>>>
>>>>> IS ENERGY CONSERVED IN OUR UNIVERSE? NO
>>>>>
>>>>> Why? Every conserved quantity is the result of some symmetry of
>>>>> nature. This is known as Noether's theorem.
>>
>> While the subject is being broached, I think that the FAQ is a little
>> misleading in these sentences. Noether's theorem tells us that every
>> symmetry leads to a conserved quantity, not the other way around.
>> There are conservation laws which do not arise from a symmetry --
>> by accident, or from topological considerations, and so on.
>>
>Perhaps you could give an example of what you mean here.
An explicit example of a conservation law coming from topological
considerations rather than a symmetry of the Lagrangian can be found
in "Quantum Field Theory" by Lewis Ryder pg. 402-406. The example is
the kink solution to the sine-Gordan equation. The claim is that the
stability of the kink solution and the conservation law are the result
of the boundary conditions of the space.
>The relation between symmetries and conservation laws is usually rather
>tight. Of course, any discussion of this depends on a precise definition of
>"symmetry" and "conservation law". With a few minor technical assumptions,
>it can be shown that if a system of Euler-Lagrange equations admits a conserved
>current, then there must be a corresponding symmetry. See, e.g., "Applications
>of Lie Groups to Differerential Equations", by Peter Olver for a careful
>treatment of such matters. The upshot is that usually the existence of a sym-
>metry is a necessary condition for the existence of a conserved current. So,
>often times a system of equations will admit a symmetry, but there will be
>no corresponding conservation law!
Hope this helps.
Doug
.
john baez
Which part? Certain in the quantized version of Maxwell's theory, the
total charge of a spatially compact universe has to be zero. In the
quantum theory of gravity, which doesn't quite exist yet, one expects
all physical states to be solutions of the Wheeler-DeWitt equation
H psi = 0, i.e., they're all annihilated by the Hamiltonian constraint
(which for pure gravity and no matter is simply another way of stating
the Einstein equation G_{00} = 0). The problem with quantum gravity
for many years was that the perturbative theory a la Feynman diagrams
wasn't renormalizable, and it seemed hopeless to find *exact* --
nonperturbative -- solutions of the Wheeler-DeWitt equation, since it
was so bloody complicated and nonlinear (leading also to nasty
"factor-ordering problems" in trying to make sense of it). The Ashtekar
"new variables" revolution seems to have changed all that.
>In the classical theory, can a field line escape through a blackhole ?
Yes, that's just called a charged black hole.
to...@cc.usu.edu
Hmm. Thanks. I can now wallow in my particle physics ignorance. In what
sense is baryon number conserved (theoretically speaking)? More precisely,
does baryon number have a corresponding conserved current? Noether's theorem
only deals with such conservation laws.
john baez
>Hmm. Thanks. I can now wallow in my particle physics ignorance. In what
>sense is baryon number conserved (theoretically speaking)? More precisely,
>does baryon number have a corresponding conserved current? Noether's theorem
>only deals with such conservation laws.
Let me wallow in my own ignorance here and make a guess. It depends
what sort of theory ones talking about, but if we are talking about the
standard model, it seems that baryon number is just 1/3 the number of
quarks minus 1/3 the number of antiquarks, so it could be expressed as a
linear combination of quark currents. One can then rig up a symmetry
that has this as its conserved current, no? Standard model experts will
note, however, that if one goes beyond perturbative calculations and tries to
include contributions from "sphaeleron" solutions in your path integral
one gets terms that violate baryon number conservation. These haven't
been observed, of course.
Sean Carroll
Hmmm. I suppose that, as you say, a lot depends on the precise
definition used. I was thinking mostly of topological conservation
laws which, for example, prevent the decay of a cosmic string or
monopole. These do not correspond to any symmetry of the Lagrangian, or
at least that was my impression. Does this not count as a "conservation
law" under your definitions?
On the other hand, I was also under the impression that Noether's
theorem was just the statement that symmetries imply conservation
laws! Now you are telling me that it's not true. Is it because
by "system of equations" you mean something more general than
a Lagrangian-based dynamical system? Or because there are
technical assumptions behind Noether's theorem that are not
necessarily satisfied?
I'll have to do some digging to find a more precise statement of
the correspondence between symmetries and conservation laws.
I'm certainly interested in knowing the precisely correct formulation.
--Sean
Matthew MacIntyre
: That is a symmetry of a SOLUTION, not an EQUATION, which is what I was
: talking about. You know this all anyway and are just having fun
: arguing.
True. :) But remember that we were talking about possible modifications
of the FAQ. Do you wish to take responsibility for the tsunami of
confusion that will result if you start saying that the Einstein
equations have a time translation symmetry, when all you mean is that
they are tensor equations?
: Well, *YOU* try quantizing gravity without getting involved in the
: diffeomorphism-invariance of the equations!
No thanks.
: >The essential point is this: Noether tells us that energy conservation
: >has something to do with the symmetry of spacetime itself. The expansion
: >of the Universe tells us that spacetime itself is NOT time-symmetric, and
: >therefore in cosmological contexts we must expect energy conservation to
: >fail. That's all there is to it.
: No, that's not all there is to it. It's certainly PART of all there is
: to it. But Noether also tells us that the diffeomorphism-invariance of
: Einstein's equations gives rise to a Hamiltonian constraint. A balanced
: understanding has to keep both of these facts in mind. As I said, it's
: a matter of how you cut the pie. Say you are playing with Einstein's
: equations coupled to Maxwell's. You can study T_{mu nu}, note that the
: integral of T_{00} over a spacelike slice ain't conserved when there's
: no Killing vector field, and say energy ain't conserved. Or you can
: start with the Lagrangian for Einstein/Maxwell, concoct a Hamiltonian
: density by the usual recipe (briefly, H = p qdot - L), and note that
: MYSTERIOUSLY (actually because of the general covariance you so dislike)
: it equals zero when we integrate it over a compact spacelike slice.
I fail to see the relevance of this last fact to the issue at hand, which
is the non-conservation of energy in GR. The diffeomorphism invariance of
the [non-gravitational] action leads to divT=0, and in fact that is the
best way of explaining this last equation [see, I am not that rabidly
anti-diffeomorphism invariance!]. And the physical meaning of divT=0 in
the absence of a timelike Killing vector is: "Energy is not conserved".
Now you say, "concoct" [!] a Hamiltonian density and find that the
integral is zero. I am well aware of that...but what of it? How does this
allow us to
resurrect energy conservation? And who wants to resurrect it anyway?
: Read Wald's appendix on the Hamiltonian formulation of GR if you think
: I'm fibbing.
Actually I read it quite recently, and am in any case familiar with it,
like anyone who reads your posts. :)
: The problem with waging war against ignorance is that the battle is so
: lopsided, it gets tiring. Sure, most people wondering about
: "gravitational energy" are confused, for them I suppose one should
: simply say:
: EINSTEIN SAYS G_{MU NU} = T_{MU NU}. THERE IS NO "GRAVITATIONAL TERM"
: IN THE RIGHT HAND SIDE.
OR ANYWHERE ELSE!
: (should I make that my .sig?) But there are a lot of other interesting
: things to say about the concept of "energy" and "Hamiltonian" in general
: relativity.
There certainly are. Again I urge everyone to read Buchdahl, "Seventeen
simple lectures on GR" for a clear discussion of the non-conservation of
energy in GR.
Matthew MacIntyre
Phil Gibbs
>Hmm. Thanks. I can now wallow in my particle physics ignorance. In what
>sense is baryon number conserved (theoretically speaking)? More precisely,
>does baryon number have a corresponding conserved current? Noether's theorem
>only deals with such conservation laws.
Actually currents only come in when you have a Langrangian density for a
field theory. In a basic mechanical system described by a Langrangian there is
no concept of currents but Noether's theorem still applies.
As for baryon number, Perhaps I should have stuck to purely classical examples.
You could try any integration constant as an example, but perhaps it is
possible to contrive a corresponding symmetry for those.
There are also topologically conserved quantities. For example total
magnetic charge in a classical system with topological magnetic monopoles
is conserved. So is the soliton number in various systems with
solitons. Is there a symmetry corresponding to those? The sine-gordon
equation is a simple example of a system which has topological charges
which are also solitons, but oh no! I vaguely remember that someone
once showed that a quantised sine-gordon equation was dual to a two
dimensional Gross-Neveu model in which the conserved charge could be
attributed to the usual phase change invariance.
Here is a simpler quantity which is conserved for all known systems:
42
Furthermore, there is good reason to believe that it must remain conserved
in all possible physical systems. What symmetry of nature is responsible?
In any case, even if somebody can prove there must be a symmetry behind all
those conserved quantities, that wouldn't be Neother's theorem. It would be
the converse of Neother's theorem, so we could still be nit-picking about
what the FAQ says, which is,
> ... Why? Every conserved quantity is the result of some symmetry of
> nature. This is known as Noether's theorem.
But it wouldn't be right to be nit-picking about the FAQ when you
consider how much hard work people have put into the thankless job of
compiling it and making it 99% (estimation only) correct.
to...@cc.usu.edu
> In article <1994Feb21.0...@cc.usu.edu>, <to...@cc.usu.edu> wrote:
>>In article <2k3l6c$n...@scunix2.harvard.edu>, car...@husc8.harvard.edu (Sean Carroll) writes:
>>>> In article <1994Feb16.1...@fozzie.eurocontrol.fr>, ph...@strauss.eurocontrol.fr writes...
>>>>>
>>>>>Why does the FAQ say that Energy is not conserved for cosmological redshift?
>>>>>The FAQ explains that,
>>>>>
>>>>>> IS ENERGY CONSERVED IN OUR UNIVERSE? NO
>>>>>>
>>>>>> Why? Every conserved quantity is the result of some symmetry of
>>>>>> nature. This is known as Noether's theorem.
>>>
>>> While the subject is being broached, I think that the FAQ is a little
>>> misleading in these sentences. Noether's theorem tells us that every
>>> symmetry leads to a conserved quantity, not the other way around.
>>> There are conservation laws which do not arise from a symmetry --
>>> by accident, or from topological considerations, and so on.
>>>
>>Perhaps you could give an example of what you mean here.
>
> An explicit example of a conservation law coming from topological
> considerations rather than a symmetry of the Lagrangian can be found
> in "Quantum Field Theory" by Lewis Ryder pg. 402-406. The example is
> the kink solution to the sine-Gordan equation. The claim is that the
> stability of the kink solution and the conservation law are the result
> of the boundary conditions of the space.
>
>
that such conservation laws are not associated with a symmetry. I think that
the mathematicians call them "rigid conservation laws", we might call them
"kinematical conservation laws" because they are conserved whether or not *any*
field equations are satisfied. Thus such conservation laws do not refelect the
dynamical content of the theory, but only the types of fields being used,
boundary conditions, etc. Noether's theorem only deals with what we might
call, "dynamical conservation laws", i.e., quantitites that are conserved
*because* the field equations are satisfied. It is for the dynamical
conservation laws that there is a very tight relationship with symmetries.
By the way, this discussion started (I think) with considerations of
conservation laws (specifically energy) in general relativity. There is a
kinematical conservation law, analogous to that found for the kink, in
relativity. It arises because there is a non-trivial third cohomology class in
the bundle of *Lorentzian* metrics over spacetime. Finkelstein and Misner came
across this conservation law in 1959 and called it "metricity". Like the
topological conservation law of the sine-gordon equation, the metricity is
conserved whether or not *any* field equations hold for the spacetime metric,
so it really carries no dynamical content.
Hannu Poropudas
Hi,
That Crackpot fouder guy Baez don't understand General Relativity
as I thought. His writings should be checked at least from the year
1992, when he started as far as I know in sci.physics news group.
John Baez says in his article "Re: Gravitational Energy Conservation",
which was dated Thu Feb 17 01:45:50 1994. :
>The Einstein equations are invariant under *all* diffeomorphisms (changes
>of coordinates). So it's not good to say they lack time-translation
>invariance. ...
From the year 1992 he has this same error in his thoughts, which he
seem not to understand properly as General Relativity has thought us.
I still think that he has tried all the time confused us with his
mathematical strange words, which has little to do with reality in
physics. We must more carefull in future in listening these strange
guys of mathemathics, when they tries to explain us what is reality
and what is not in physics. I suppose that we have better to believe
what we see, and what we listen ourselves than that what others tries
to force us to believe.
Best Regards,
Hannu Poropudas,
hapo...@freenet.hut.fi
"All people who don't love the truth more than anything else in their
life' are evil and they are servants of the monster space."
--
to...@cc.usu.edu
> In article <1994Feb22.1...@cc.usu.edu> to...@cc.usu.edu writes:
>
>>Hmm. Thanks. I can now wallow in my particle physics ignorance. In what
>>sense is baryon number conserved (theoretically speaking)? More precisely,
>>does baryon number have a corresponding conserved current? Noether's theorem
>>only deals with such conservation laws.
>
> Actually currents only come in when you have a Langrangian density for a
> field theory. In a basic mechanical system described by a Langrangian there is
> no concept of currents but Noether's theorem still applies.
of current specializes as well...one simple drops the spatial components of the
vector, i.e., mechanics is treated as field theory with zero spatial
dimensions.
> As for baryon number, Perhaps I should have stuck to purely classical examples.
>
> You could try any integration constant as an example, but perhaps it is
> possible to contrive a corresponding symmetry for those.
>
> There are also topologically conserved quantities. For example total
> magnetic charge in a classical system with topological magnetic monopoles
> is conserved. So is the soliton number in various systems with
> solitons. Is there a symmetry corresponding to those? The sine-gordon
> equation is a simple example of a system which has topological charges
> which are also solitons, but oh no! I vaguely remember that someone
> once showed that a quantised sine-gordon equation was dual to a two
> dimensional Gross-Neveu model in which the conserved charge could be
> attributed to the usual phase change invariance.
>
> Here is a simpler quantity which is conserved for all known systems:
>
> 42
>
> Furthermore, there is good reason to believe that it must remain conserved
> in all possible physical systems. What symmetry of nature is responsible?
Noether's theorem only deals with quantities that are conserved because certain
equations of motion are satisfied.
SCOTT I CHASE
>
>Hmm. Thanks. I can now wallow in my particle physics ignorance. In what
>sense is baryon number conserved (theoretically speaking)?
It is better to not talk theoretically at all. :-) But if you must, then
do so only after thinking about the facts:
Baryon number is conserved *empirically*. That's just a fact of life.
We have never seen a reaction, or inferred the existence of a reaction,
in which baryon number is not conserved.
Now the question is "Can we develop some theoretical insight as to
why this is the case?"
>More precisely,
>does baryon number have a corresponding conserved current? Noether's theorem
>only deals with such conservation laws.
Baryon number does not *seem* to be associated with a conserved current.
As others have discussed, there are other reasons for conservation laws.
But we just don't know the original of baryon conservation. In the
absence of a conserved current, people suspect that the mechanism for
baryon number conservation is probably not exact - that under suitable
conditions, there will, in fact, be *some* violation. But the conditions
under which this might happen are well out of experimental range. There
was some speculation that the SSC *might* detect such a violation, because
in the one popular model it turns out that the suppression of such a
violation is ameliorated (baryon number violation is desuppressed) at
high energy density. This is just one more example of physics we will
not be able to study without the SSC.
-Scott
-------------------- i hate you, you hate me
Scott I. Chase let's all go and kill barney
SIC...@CSA2.LBL.GOV and a shot rang out and barney hit the floor,
no more purple dinosaur.
to...@cc.usu.edu
>> On the other hand, I was also under the impression that Noether's
> theorem was just the statement that symmetries imply conservation
> laws! Now you are telling me that it's not true. Is it because
> by "system of equations" you mean something more general than
> a Lagrangian-based dynamical system? Or because there are
> technical assumptions behind Noether's theorem that are not
> necessarily satisfied?
>
> I'll have to do some digging to find a more precise statement of
> the correspondence between symmetries and conservation laws.
> I'm certainly interested in knowing the precisely correct formulation.
>
> --Sean
Sorry if I caused confusion. Symmetries of Euler-Lagrange equations, i.e.,
transformations mapping solutions to solutions lead to conservation laws, if
and only if the transformation preserves the Lagrangian up to a total
derivative. If there is a current (or function on phase space in
mechanics) that is conserved by virtue of the E-L equations, then there will be
a corresponding symmetry of the Lagrangian. The topological-type conservation
laws are conserved whether or not any E-L equations are satisfied, and so do
not fit in with Noether's theorem. But they are, as you point out,
conservation laws.
In this forum it does not seem a good idea to get too rigorous, but there are
of course technical assumptions lurking behind the above statements. Again, I
would suggest the standard text: Applications of Lie Groups to Differential
Equatios, by P. Olver.
john baez
>But remember that we were talking about possible modifications
>of the FAQ. Do you wish to take responsibility for the tsunami of
>confusion that will result if you start saying that the Einstein
>equations have a time translation symmetry, when all you mean is that
>they are tensor equations?
Oh, I wasn't suggesting that my off-the-cuff remarks be enshrined in the
FAQ to mislead novice physicists for generations to come. Even apart
from the fact that I *never* take responsibility for tsunamis of
confusion, I agree that a careful FAQ shouldn't blithely say "Einstein's
equations have time translation invariance." I don't like your guess
about "all I meant," but never mind that.
>: No, that's not all there is to it. It's certainly PART of all there is
>: to it. But Noether also tells us that the diffeomorphism-invariance of
>: Einstein's equations gives rise to a Hamiltonian constraint. A balanced
>: understanding has to keep both of these facts in mind. As I said, it's
>: a matter of how you cut the pie. Say you are playing with Einstein's
>: equations coupled to Maxwell's. You can study T_{mu nu}, note that the
>: integral of T_{00} over a spacelike slice ain't conserved when there's
>: no Killing vector field, and say energy ain't conserved. Or you can
>: start with the Lagrangian for Einstein/Maxwell, concoct a Hamiltonian
>: density by the usual recipe (briefly, H = p qdot - L), and note that
>: MYSTERIOUSLY (actually because of the general covariance you so dislike)
>: it equals zero when we integrate it over a compact spacelike slice.
>
>I fail to see the relevance of this last fact to the issue at hand, which
>is the non-conservation of energy in GR. The diffeomorphism invariance of
>the [non-gravitational] action leads to divT=0, and in fact that is the
>best way of explaining this last equation [see, I am not that rabidly
>anti-diffeomorphism invariance!]. And the physical meaning of divT=0 in
>the absence of a timelike Killing vector is: "Energy is not conserved".
I was not trying to stick to the "issue at hand," a hopeless dream in any
event on sci.physics; I was simply rambling on about various notions of
energy and Hamiltonian in general relativity, and noting that if one
uses the standard Lagrange/Hamilton recipe to figure out the energy of a closed
universe, one gets something that is indeed conserved, namely,
0
which is a satisfying result, but not one of course that says too much
about what the background radiation will do.
>Now you say, "concoct" [!] a Hamiltonian density and find that the
>integral is zero. I am well aware of that...but what of it? How does this
>allow us to
>resurrect energy conservation? And who wants to resurrect it anyway?
See above. Note that the "concoction" of the Hamiltonian density is not
an arbitrary, ad hoc one, but simply a matter of following the standard
cookbook recipe, appropriately taking into account that there is no preferred
time parameter here, so that one needs to specify a lapse and shift.
>: Read Wald's appendix on the Hamiltonian formulation of GR if you think
>: I'm fibbing.
>Actually I read it quite recently, and am in any case familiar with it,
>like anyone who reads your posts. :)
I suppose you are alluding to the fact that practically nobody really
reads them? That's not very nice. :-)
>There certainly are. Again I urge everyone to read Buchdahl, "Seventeen
>simple lectures on GR" for a clear discussion of the non-conservation of
>energy in GR.
Hmm, I haven't read these, but will.
john baez
>Here is a simpler quantity which is conserved for all known systems:
>
> 42
>
>Furthermore, there is good reason to believe that it must remain conserved
>in all possible physical systems. What symmetry of nature is responsible?
Symmetry under the group generated by the self-adjoint operator
42
that is, phase symmetry. Work it out.
>But it wouldn't be right to be nit-picking about the FAQ when you
>consider how much hard work people have put into the thankless job of
>compiling it and making it 99% (estimation only) correct.
No, it's good to nitpick about the FAQ, at least within reason, so that
it keeps improving.
john baez
Poropudas) writes, of yours truly (JB):
>I still think that he has tried all the time confused us with his
>mathematical strange words, which has little to do with reality in
>physics. We must more carefull in future in listening these strange
>guys of mathemathics, when they tries to explain us what is reality
>and what is not in physics. I suppose that we have better to believe
>what we see, and what we listen ourselves than that what others tries
>to force us to believe.
I'm afraid what you wrote is false, because diffeomorphisms of a
spin manifold do *not* lift to the spin bundle in a canonical
way. The point is, these spin structures are in one-to-one
correspondence with elements of H^2(M,Z/2), but *not* canonically, so
one can't just use the natural action of Diff(M) on the cohomology
groups to get a Diff(M) action on the set of spin structures.
john baez
>....spin structures are in one-to-one
>correspondence with elements of H^2(M,Z/2)......
whoops, I meant H^1, but luckily this doesn't affect the conclusion.
Sean Carroll
Baryon number conservation can be thought of as arising from a
symmetry of the Standard Model Lagrangian. The symmetry is basically
\psi_j -> exp(i\theta)\psi_j (for all j), where the \psi_j are the
quark fields. The important points are that the parameter \theta is
global (independent of spacetime), and that you must rotate all of the
quarks simultaneously. (It's always bothered me that Noether's theorem
gives you the same number of conservation laws for a global symmetry
as for a gauge symmetry, which is much bigger. Is there a deep reason
for this?) Same holds for lepton number.
Of course, in the path integral the Lagrangian is not quite the
whole story, and 't Hooft showed that non-perturbative processes
violate baryon number. On the other hand, the difference of baryon
number and lepton number is conserved.
I have tried and failed to come up with an example of a conservation
law which results neither from a symmetry nor from topology. What
about the energy-momentum tensor in GR? In flat space it's
divergenceless because of Poincare invariance, but in curved space
it's because of the Bianchi identity. Can that be thought of as
arising somehow from Noether's theorem?
--Sean
Hannu Poropudas
And that KILL-file guy SCOTT I CHASE seem not to understand
General Relativity either. In his article 'Re: Gravitational
Energy Conservation', which was dated: Wed Feb 16 22:25:00 1994
he writes:
>The FAQ is somewhat deficient in this regard, and I should update it.
>What it means to say is that the Einstein equations have no static
>(time-independent) solutions, and that Noether's theorem does not apply
>because the equations themselves have no time-translation invariance,
>which is reflected in the absence of any static solutions.
>Your complaint is a fair one. One of these days (soon!) I'll change it.
As far as Einstein equations are concerned I know at least three
static solutions of them, namely one which is decribed by the Einstein
line element, one which is described by the de Sitter line element and
one which is described by the special relativity line element.
Best Regards,
Hannu Poropudas,
hapo...@freenet.hut.fi
--
Hannu Poropudas
Again you mr. Baez tried to confuse us:
The question was not about diffeomorphisms. The error, which I
and Matthew MacIntyre noted was about time-translation invariance.
You wrote:
>of coordinates). So it's not good to say they lack time-translation
>invariance. ...
I say also it is 'good' to say that, because it is true in case
of Einstein's equations in General Relativity.
Mr S.C. Woon
Lie groups.
So in addition to topological invariants,
all conservation laws of quantities with respect to discrete transformations
(like parity in EM and symmetries of other finite groups)
also do not arise from Noether's theorem.
Right?
--
see-chin woon
s.w...@ic.ac.uk or wo...@vxcern.cern.ch
(UK) (Switzerland)
SCOTT I CHASE
>
>And that KILL-file guy SCOTT I CHASE seem not to understand
>General Relativity either. In his article 'Re: Gravitational
>Energy Conservation', which was dated: Wed Feb 16 22:25:00 1994
For the unititiated, I will explain that Hannu, that lovable wood-duck,
seems to think that I have absolute power on the Internet, that I am
capable of establishing global kill files that permit individuals to
see only those messages which I endorse(*), and that, of course, I uniformly
suppress his messages. This explains, in Hannu's world-view, why
so few people respond to his messages.
(*) The world would obviously be a much better place if this were so.
Everyone knows that my judgement is far superior to all others. The
very fact that the Net is so chaotic is a clear demonstration that, alas,
I have no such power. Perhaps Hannu can intercede with God to arrange
this for me. I would be eternally grateful. If, however, God does
not exist, then I will be disappointed, but only for a few more decades.
:-)
-Scott
-------------------- Physics is not a religion. If
Scott I. Chase it were, we'd have a much easier
SIC...@CSA2.LBL.GOV time raising money. -Leon Lederman
john baez
>
>Again you mr. Baez tried to confuse us:
>The question was not about diffeomorphisms. The error, which I
>and Matthew MacIntyre noted was about time-translation invariance.
Again you mr. Poropudas are confused. Think of it this way: if R is a
left noetherian ring and M is a left R-module with a finite free
resolution of length less than or equal to n, then every free resolution
of M each of whose terms is finitely generated has a stably free nth
syzygy. To see this you just use the generalized Schanuel lemma. I
hope this clears things up. If not, you obviously need to brush up on
your homological algebra.
Chris Metzler
|> In article <2ki1q3$l...@freenet.hut.fi>, hapo...@freenet.hut.fi (Hannu Poropudas) writes...
|> >
|> >And that KILL-file guy SCOTT I CHASE seem not to understand
|> >General Relativity either. In his article 'Re: Gravitational
|> >Energy Conservation', which was dated: Wed Feb 16 22:25:00 1994
|> >he writes...
|>
|> For the unititiated, I will explain that Hannu, that lovable wood-duck,
|> seems to think that I have absolute power on the Internet, that I am
|> capable of establishing global kill files that permit individuals to
|> see only those messages which I endorse(*), and that, of course, I uniformly
|> suppress his messages. This explains, in Hannu's world-view, why
|> so few people respond to his messages.
|>
Hannu should feel honored. Only a few people may bother to respond to
his messages, but Kibo is one of them!
--
Chris Metzler
Department of Physics, University of Michigan 313-764-4607 (office)
Randall Lab, 500 E. University 313-996-9249 (home)
Ann Arbor, MI 48109-1120 USA E-MAIL: met...@pablo.physics.lsa.umich.edu
"Well, I had a dream, and in it I went to a little town, and all the girls
in town were named . . .Betty." -- Laurie Anderson
Matt McIrvin
>An example of what Sean Carroll means is conservation of Baryon Number
>in the Standard Model. There is no symmetry in the Standard Model from
>which Baryon Number conservation follows by Noether's Theorem (as far
>as I know) and yet it is conserved (in the Standard Model).
I think there actually is a U(1) symmetry to the Standard Model
Lagrangian (more precisely, to the terms which are *not* total
divergences) which insures perturbative baryon number conservation.
Just rotate the phases of all fields by an angle proportional to the
baryon number, and terms in a B-conserving Lagrangian ought not to
change.
What may cause confusion on this point is that the symmetry is not
a *gauge* symmetry; there is no gauge symmetry insuring baryon number
conservation in the Standard Model, the way there is for electric
charge conservation or color conservation. Also, the symmetry is
broken by an anomalous term that is a total divergence, so that it
cannot have any effects in perturbation theory, but at high
temperatures it may induce baryon number violation via the field
configuration called the "sphaleron." So I'd say that to the extent
baryon number *is* conserved in the Standard Model, it follows from a
symmetry.
--
Matt 01234567 <-- The original Indent-o-Meter
McIrvin ^ Someday, tab damage will light our homes!
Matt McIrvin
>Baryon number is conserved *empirically*. That's just a fact of life.
>We have never seen a reaction, or inferred the existence of a reaction,
>in which baryon number is not conserved.
In a way, we have inferred the existence of such a reaction, but only
via a chain of theoretical arguments so long you can hardly call it
inferring from empirical evidence! It involves assuming the Standard
Model, renormalizing it with triangle diagrams, then taking the resulting
anomaly and seeing what it does in topologically nontrivial situations.
>>More precisely,
>>does baryon number have a corresponding conserved current? Noether's theorem
>>only deals with such conservation laws.
>Baryon number does not *seem* to be associated with a conserved current.
>As others have discussed, there are other reasons for conservation laws.
But (modulo the anomalies described above) it *is* associated with a
conserved current. Anything that is conserved in a local manner (and
in a Lorentz-invariant universe it can hardly be otherwise, provided
that you can define a local density of the stuff) has such a
current. It just isn't a current that happens to be coupled
to any gauge field, and anomalies violate its conservation as well in
theory, so we have ample reason to find its conservation suspect.
Everything you said about this is correct, provided that you replace
"currents" with "gauge field source currents."
James Kibo Parry
Chris Metzler <met...@pablo.physics.lsa.umich.edu> wrote:
>
> Hannu should feel honored. Only a few people may bother to respond to
> his messages, but Kibo is one of them!
Of course. Once upon a time Hannu introduced me to the Space Potato and
John_-_Winston sent me a gallon of Space Syrup, and I put two and two
together and came up with a lovely German potato dumpling.
Also, I now have a Mr. Space Potato Head who looks like Hannu, only with
his nose where his ear should be.
-- K.
Matthew MacIntyre
: For the unititiated,
OK, I am male.
I will explain that Hannu, that lovable wood-duck,
: seems to think that I have absolute power on the Internet, that I am
: capable of establishing global kill files that permit individuals to
: see only those messages which I endorse(*), and that, of course, I uniformly
: suppress his messages. This explains, in Hannu's world-view, why
: so few people respond to his messages.
: (*) The world would obviously be a much better place if this were so.
: Everyone knows that my judgement is far superior to all others. The
: very fact that the Net is so chaotic is a clear demonstration that, alas,
: I have no such power.
You are obviously ignorant of theology.
Perhaps Hannu can intercede with God to arrange
: this for me.
Unfortunately, Hannu is a "muddled". Ever since "god's sack of space
potatoes" fell on his head.
Matthew MacIntyre
: I still think that he has tried all the time confused us with his
: mathematical strange words, which has little to do with reality in
: physics. We must more carefull in future in listening these strange
: guys of mathemathics, when they tries to explain us what is reality
: and what is not in physics. I suppose that we have better to believe
: what we see, and what we listen ourselves than that what others tries
: to force us to believe.
You have been talking to Howard Georgi again, haven't you?
Hannu Poropudas
Mr. SCOTT I CHASE,
You seem to be real 'despot' of the NET. You sent me E-Mail, where
you said:
>I now extend my warning to any messages posted by you to the Net, and
>directed specially to me. If you contact me by *any* means, including
>but not limited to posting directed messages to my via the Net, I will
>take this as an indication that you are continuing to harass and threaten
>me, and I will send the complete text of our correspondence, beginning
>with your death threat, to all of the above-named organizations as well
>as the administrator of your current USENET site. This is your only
>warning. The next time you send any message to USENET addressed to me,
>I will turn over all of your writings to the authorities.
What can a little man else do in this case than to pray God's Curse
to this kind of 'despot' of the NET, if he does not regret what he
have said and what he have done to me.
Hannu Poropudas,
hapo...@freenet.hut.fi
--
Gregory Stewart-Nicholls
sic...@csa5.lbl.gov "SCOTT I CHASE" writes:
> In article <2ki1q3$l...@freenet.hut.fi>, hapo...@freenet.hut.fi (Hannu
> Poropudas) writes...
> >
> >And that KILL-file guy SCOTT I CHASE seem not to understand
> >General Relativity either. In his article 'Re: Gravitational
> >Energy Conservation', which was dated: Wed Feb 16 22:25:00 1994
> >he writes...
>
> For the unititiated, I will explain that Hannu, that lovable wood-duck,
> seems to think that I have absolute power on the Internet, that I am
> capable of establishing global kill files that permit individuals to
> see only those messages which I endorse(*), and that, of course, I uniformly
> suppress his messages. This explains, in Hannu's world-view, why
> so few people respond to his messages.
Hannu ???? Who is Hannu ??? is he this guy who was in the kill file that
came with my snews distribution .... the one you sent me ....
--
Vidi | Gregory Stewart-Nicholls
Vici | ni...@olympus.demon.co.uk
Veni | TeknoLogika ltd