Re: Hobba's misconceptions


"Bilge" <dubious@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> schrieb
> Ilja Schmelzer:
> >"Bilge" <dubious@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> schrieb
> >> Ilja Schmelzer:
> >> >"Bilge" <dubious@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> schrieb
> >> >> Ilja Schmelzer:
> >> >> >"shuba" <tim.shuba@xxxxxxxxxxxxx> schrieb
> >> >> >Its defined by the same procedure known as "Einstein
> >> >> >synchronization"
> >> >> >which is used in SR. And was done already by Lorentz and
Poincare.
> >> >>
> >> >> So why all the difficulty and obfuscation in trying to define it
> >> >> mathematically?
> >> >
> >> >I don't understand why you have any difficulty understanding it.
> >>
> >> Because you've defined LET as a trimetric theory with two different
> >> definitions of simultaneity, but have not bothered to say what that
> >> means.
> >
> >Events which have zero distance in the metric dt^2 are absolutely
> >simultanous.
>
> In that case, (t,x,y,z) define a galilean universe,

Again, there is no symmetry of galilean boosts in LET, therefore
it is false to name the LET universe a galilean universe.

> and once you get
> through trying to obfuscate the relationship of your ``metrics,'' it
> ought to be easy to see the difference between relativity and LET.

Of course, it is easy to see. The description of reality is different.
And it is as well easy to see that for the observable effect there is
no difference.

> >Events which have zero distance in the metric dx^2
> >define the same absolute position. Events which have zero distance
> >in the metric dtau^2 are light-like as in SR.
> >
> >The behaviour of clocks is defined by the metric dtau^2.
>
> So, does that mean clocks move at the speed of light or what?

No, it means, as in SR, that clocks which move on a time-like
trajectory x^a(s) (means, with velocity below speed of light)
show a clock time defined by the formula

tau = int_s0^s1 sqrt(dtau^2)
= int_s0^s1 sqrt(dt^2-dx^2-dy^2-dz^2)
= int_s0^s1 sqrt((dt/ds)^2-(dx/ds)^2-(dy/ds)^2-(dz/ds)^2) ds

which for s=t, c=1 reduces to

tau = int_t0^t1 sqrt(1-v^2) dt

> Being deliberaltely vague only results in me assuming whatever
> I want to assume fits the definition. I promise that I will
> assume something as ridiculous as your delibrate vagueness permits,
> since my point is that your absolute frame amounts to wishful
> thinking supported by bull***.

And I promise you to specify whatever you consider to be vague,
(probably because you simply don't have sufficient education about
the meaning of certain well-defined things) as long as the number
of words like the last remains sufficiently small.

> >Using light rays and clocks, above defined by the metric dtau^2,
> >you can define oberver-dependend inertial frames using Einstein's
> >synchronization procedure.

> Draw a diagram. I'm tired of seeing a lot of bull*** about your
> interpretation of ``Einstein's dynchronization procedure.'' If you
> can't figure out how to draw it, I'll obtain the definition from
> einstein's paper and draw it for you.

Too lazy to draw ASCII diagrams. Draw it yourself from
Einstein's paper.

> >> So, the ether is something which observes itself?
> >
> >In some sense, yes.
> >Not much different from field theory - the fields also observe themself.
> >Except you want to postulate that human beings are not described, in
QFT,
> >as certain states of quantum fields.
>
> That is not what I postulate. I postulate that field theory is nothing
> like your ether, which makes your comparison fallacious.

I don't care what you postulate. I have made a comparison with
standard mainstream understanding of field theory, which is that
we, human beings, consist of usual matter, which may be
described by states of quantum fields. Thus, the property
of my ether that observers may be described by waves
of the ether is comparable with a standard property of
standard science. (Thus, if you don't like this property
of my ether theory, I don't care about this.)

> >> So, if you rely on symmetry, you get the results of using symmetry
> >> and call it the ether.
> >
> >I call it ether because a lot (but not all) of its properties remember
the
> >classical ether.
>
> So far, I haven't seen a property that resembles a classical anything.

Means, you have not looked into my paper.

Else, you would have seen equations you know from condensed
matter theory, namely continuity and Euler equations.

(Or, the other solution is that you have never had a good course
in condensed matter theory.)

> >I rely on translational symmetry in space and time, because these
> >are also symmetry properties of the classical ether.

> Sorry, but the classical translational symmetry combined with the
> lorentz transforms gives a change of scale.

??????????

Translations are part of the standard Poincare group. (There are
two versions of the Poincare group, one is the symmetry group
of the Maxwell equations, which includes scale transformations,
the other is the symmetry group of the Minkowski metric, which
does not include any scale transformations. But it includes
translations in space and time. It seems, you mingle these two
groups. With "Poincare group" I always mean the symmetry
group of the Minkowski metric, which does not include
scale transformations.)

> While that fits the description
> of LET, it doesn't describe special relativity.

The Poincare group is the symmetry group of SR and of
observable effects of LET.

> >> You rely on asserting the equivalence.
> >
> >I have proven it.
>
> No, you haven't and I have yet to find _any_ article which claims
> to prove any such thing with respect to identical particles.

Seems, you have switched again to BM? In case of BM, its not
me who has proven the equivalence.

BTW, there is a trick to describe fermions as particles using
a non-maximal set of commuting operators to define the configuration space.
For such non-maximal sets, the guiding equation

d_t Q = <Psi | J | Psi> / <Psi | Psi>

also allows to prove the equilibrium hypothesis. But I don't
like this trick and prefer Bohmian field theory, as I have already
explained. The clear lecture of QFT (at least on curved background)
is that the field picture is more fundamental.

> All you've
> proven (or that the quantum equilibrium hypothesis proves), is that
> you can pretend there is no indeterminacy in quantum mechanics by shifting
> the indeterminacy to a postulate of indeterminacy for determinate
> measurements.

In the trivial variant, to an assumption about an equilibrium state as the
initial state. In the more sophisticated variant, with decoherence, I need
a subsystem of the universe which has had sufficient interaction with
its environment in the past.

> It says nothing about why identical particles should be
> treated differently than non-identical particles.

Non-identical particles are a quantum effect of fields. I see absolutely
no reason to consider such particles as something fundamental and
to base a Bohmian theory on such particles. Argue about this question
with somebody who believes particles are more fundamental than fields.

> >The things I have proven I can, of course, use as I like, rely on them.
> >That's the purpose of proving theorems -
> >it allows us to rely on them later.
> >
> >It seems, you have a similar problem here and in BM.
>
> The problem appears to be you and bohmian mechanics. Bohmian
> mechanics treats two particles as independent particles on different
> trajectories and nowhere is there a proof for why it should it
> should matter whether the particles are identical or not identical.

As I have explained you, I don't care about Bohmian theories
which handle identical particles.

> >In BM, we derive (once) the general rules of QM. Then we can
> >later (always) rely on them.

> No, you derive once a small subset of the rules and then declare
> total equivalence. Then, you simply use the non-bohmian version of
> quantum mechanics beause you can't actually use bohmian mechanics
> and chant the bohmian mantra. I don't suppose you ever wondered why
> bohm and hiley were never supporters of bohmian mechanics.

Should I? There are good reasons to consider Bohmian mechanics
only as an intermediate step, not as a final theory. BTW, I also prefer
Nelsonian stochastics. Which is IMHO, nonetheless, also only an
intermediate step, not the final theory.

> Even the
> name ``bohmian'' is an attempt to equivocate two inequivalent theories,
> since it certainly is not bohm's theory nor one bohm ever advocated
> (at least that is what basil hiley has said about he, bohm and their
> relatiion to bohmian mechanics). Hiley has a much more sane view
> of bohm's theory.

All I'm doing is to use BM as an existence proof for a certain class of
theories (namely realistic theories) and to defend it against cranks who
claim nonsense like that "BM predictions differ from QM predictions"
which disagrees with well-known theorems.

> >> You haven't proven any such thing, since the observables are frame
> >> dependent. Regardless of what you call your ether frame, there are
> >> observables in different frames which are not observables in your
> >> preferred frame and vice-versa.
> >
> >This shows a gross misunderstanding of relativity.
>
> Oh, really? So, let's see... Observer A measures the momentum for
> some experiment. Observer B measures a position for the same experiment.
> If both observers are in frames which differ from the ether frame by a
> lorentz transform, then the lorentz transforms are good observables

Lorentz transforms as quantum observables? That's nonsense.

Seems, you don't even know that there is no observable for time.
Already in non-relativistic quantum theory, and, of course, such an
observable does not appear in relativistic quantum theory too.

> in the ether frame and therefore the lorentz transforms commute with
> measurements of both p and x. If p and x both commute with the lorentz
> transforms, then p and x commute.

Why this? If two operators commute with a third one, they have
to commute with each other? LOL. (Just for fun: Every operator
commutes with the operator 1. Thus, all operators commute.)

> I claim that p and x don't commute
> and that there is no transformation that can make them commute. You
> claim I don't understand relativity. Care to prove that?

I have, as you see.

> >Everything what
> >happens in reality may be described in a single inertial frame (in GR
> >in a single system of coordinates). Whatever we define in other
> >inertial frames (other coordinates) we can translate into our
> >inertial frame (our coordinates) using well-defined transformation
> >rules. These transformation rules transform observables into
> >observables.
>
> Thank you for disproving the results of the epr experiment using nothing
> but logic.

First, this was a consideration about classical theory, not related with
EPR. As you would know if you have had a good course in QT,
there is no self-adjoint operator for time. Therefore the quantum situation
with Lorentz transforms is much less understood. It is not even clear how
to define the wave function in a Lorentz-invariant way. (A weak formulation.
I could as well say impossible.)

Second, the EPR experiment may be described in BM in a single frame
(the preferred frame) and predicts, as required, the violation of Bell's
inequality. Just for your information.

Ilja


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