Re: K.E. and momentum

From: Non Ame (noname_at_nospam.net)
Date: 03/04/05 

Date: Fri, 4 Mar 2005 16:46:35 +0000 (UTC)

"Ken S. Tucker" <dynamics@vianet.on.ca> writes:

>My goodness...

>Non Ame wrote:
>> "Ken S. Tucker" <dynamics@vianet.on.ca> writes:
>>
>> >Non Ame wrote:
>> >> "Ken S. Tucker" <dynamics@vianet.on.ca> writes:
>> >>
>> >> >For the math savy, I hold,
>> >>
>> >> >DU^u=0 always,
>> >>
>> >> >(absolute derivative of 4-velocity).
>> >>
>> >> >It is my view that those who intend to
>> >> >deviate from that equation using,
>> >>
>> >> >DU^u = f^u (f^u==Lorentz force)
>> >>
>> >> >are shekkers, i.e. it should be admitted
>> >> >that the use of f^u is a stop gap.
>> >>
>> >> However, that's just Ken's opinion.
>>
>> >No QT is a prediction of GR, and is
>> >physically verified so far.
>>
>> Most people, when they refer to Quantum Theory, mean Quantum
>Mechanics.

>Well higher level physics distiguishes between
>QT and QM.

But high level physics does not mean by Quantum Theory what you mean by
Quantum Theory.

>> You point out below that you make a distinction between them. So it
>is
>> YOUR responsibility to outline the principles and axioms of the
>theory
>> that you are calling Quantum Theory.

>QT = requires a quantized variation of energy.

This looks like one of the ad hoc rules of the Old Quantum Theory, like
the ad hoc rule in the Old Quantum Theory that the orbital angular
momentum must be an integral multiple of hbar. This rule for the angular
momentum is actually true in Quantum Mechanics, but is derived from the
laws of Quantum Mechanics, rather than being an assumption of the theory.

In Quantum Mechanics, an interaction causes a quantum system to jump in
energy (an instantaneous quantum jump if you are working in an energy
eigenbasis), and a sinusoidal variation, in the interaction part of the
Hamiltonian, of frequency omega, can lead to a change in energy (either an
increase or a decrease) of magnitude approximately equal to hbar omega.
Again, this quantized variation of energy due to the interaction is a
consequence of the laws of Quantum Mechanics, rather than being an
assumption of the theory.

What you have stated above is not an axiomatic foundation for a theory.
For example, what determines these quantum jumps? In quantum mechanics,
certain aspects of these quantum jumps are deterministic, and it is these
deterministic aspects which dictate probabilities upon measurement.

Again, I ask for a proper axiomatic foundation for what you understand by
Quantum Theory.

>Relates to photonics.

>> And if you are going to continue to make the claim that Quantum
>Theory is
>> a prediction of General Relativity, then it is also your
>responsibility to
>> supply a reasonably detailed line of argument which starts off at
>General
>> Relativity, and demonstrates that what you mean by Quantum Theory
>MUST
>> follow from General Relativity. Your assertions here are not
>sufficient
>> for the task. Let's see some proper logical argument.

>f_u U^u = q*F_uv U^v U^u = f = 0.

That is not a detailed line of argument as requested. Try again.

>> >> >I've shown f^u=0 is compatible with QT.
>>
>> >> >((f^0 =/=0 is not QT compatible, as that
>> >> >will lead to spiral trajectories, and
>> >> >continuous radiation, against quanta
>> >> >emission)).
>> >>
>> >> There is no such thing as a trajectory in the sense that you mean
>in
>> >> Quantum Mechanics.
>>
>> >Miss No name
>> >Knowing you're a new poster and a female student,
>> >you should study the difference between QT
>> >and QM, where GR is concerned.
>>
>> That is not specified anywhere but in "The Universe according to Ken
>S.
>> Tucker".
>>
>> I suppose that you might be thinking of the attempts to unify General

>> Relativity and Quantum Theory. If that is the case, these attempts
>do NOT
>> logically follow from General Relativity. This means specifically
>that
>> no unified theory is the prediction of General Relativity.
>>
>> Also, if you are to have any hope of understanding any of the
>attempts to
>> unify General Relativity and Quantum Theory, you have to have a
>thorough
>> grounding in Quantum Mechanics first. As yet, you have not provided
>any
>> evidence that you have any grounding in the theory.

>So you're ok with my take on GR.

I didn't say that. But I am trying to work with a single aspect of your
approach to these matters at a single time.

>> >> >AE suggested that f_u=0 in his GR1916 paper
>> >> >just past Eq. 65a, and I agree.
>> >>
>> >> You have ignored the possibility that something may have been
>missed
>> >in
>> >> the translation from the German language to the English language.
>>
>> >We went threw that before, I've paid translation
>> >experts to translate from German to English and
>> >you'll need to find fault with them, not me.
>>
>> Have you considered the possibility that Einstein was discussing the
>> contribution of the electromagnetic field to kappa_sigma in Equation
>> (65a), and then his later comment was regarding the TOTAL kappa_sigma

>> from both matter and electromagnetic field? It could be that
>Einstein
>> meant that, but that somebody, possibly even Einstein himself, may
>have
>> been a little sloppy in the exposition.

>Look, if you want to read between the lines
>see just after GR1916 Eq(65a)
>"If the electric masses are free...kappa_sigma will vanish"

>Do you want to discuss the "if" ?

You didn't answer my suggestion. What is to stop Einstein from meaning
the nett kappa_sigma when he made that statement? As I previously
remarked, somebody may have just been a bit sloppy in the exposition.

>> Incidentally, I notice that you are still persisting in spelling
>'through'
>> as 'threw'. Does this mean that you want to throw something. You
>seem to
>> be obsessed with this misspelling. Obviously a proper vocabulary is
>not
>> your strong point. Maybe you should invest in a dictionary to help
>with
>> this problem. While the words 'through' and 'threw' are pronounced
>> *exactly* the same, but the two words mean totally different things.

>ok

>> > Do your own work, get a grant.
>>
>> Do you know how hard it is to get a grant? Money for grants is not
>in
>> plentiful supply.

>> >Miss NoName, if you're really ambitious, compute
>> >the norm of Lorentz's
>>
>> >f_u = qF_uv U^v
>>
>> >i.e. f (or f^2). I won't give hint's since you
>> >might use that as a basis for your Ph.d thesis.
>>
>> Why should I want to do that? I already have a Ph.D. And don't be
>so
>> bloody condescending. Do *you* have a Ph.D. degree?

>I don't know or care, I was expelled from gov to
>the commercial sector in Gr.6. (age 10).

How come you don't know whether, or not, you have a Ph.D degree? I can
only assume that you don't have a Ph.D because you would definitely know
if you DID have such a degree. Nor would you be so off-hand about the
degree, if you had actually done the work involved in receiving one. It
takes years of hard work and dedication to get a Ph.D. Have you even ever
attended a University (or College) and received a degree? If so, which
academic degrees do you hold? Or do you just read books about Physics and
base your 'knowledge' on what you read in them. If this is so, it is not
the way to learn the subject on Grade 6 knowledge only. Nor is it a way
to acquire a proper or complete understanding of it on that level of
knowledge, You need more.

Why were you expelled?

>>Do you have any
>> idea of just how much work is required to be able to get a Ph.D? I
>bet
>> you don't. I also object to the slur that I would ever want to steal
>> anybody's work. Does your insulting inference mean that you would
>> consider doing such a base thing in order help you with a thesis -
>after
>> all, it was your comment, not mine, which means that it is something
>in
>> your mind. Do not attribute actions or ideas to other people what
>you,
>> yourself, think.
>>
>> Also, I should think that a doctorate from a university that would be

>> willing to employ you as a physics lecturer would be worthless.

>Not really, I've lectured lots of times,

One would hope that it wasn't in physics at a proper university.

>the thing I enjoy the most is being embarassed
>by questions that define my ignorance...I blush,
>and everybody makes fun of me. I really enjoy it
>when the students find out I don't the answer,
>and then spit-ball me.

So why don't you admit the mistakes that you make here? People would
respect you better if you were honest about your difficulties with the
subject. Your grandiose stance does not engender respect for you, nor
does it make people accept your word. You expect other people to treat
you with deference and as a 'wise and sage mentor', but life is not like
that.

>> >Ps: Show your work.
>>
>> I will take the signature to be (+---), as opposed to the only other
>> viable alternative (-+++).

>Your presumption of that signature produces
>artifacts. You'll need to go to the basis
>vector foundation of your metric. Recall
>how tensors relate to CS's, then issue
>your Kronecker's.

There is a theorem in real linear algebra which states that a vector space
with a nondegenerate bilinear form has a unique signature, which cannot be
altered by an change of basis. In the case of Minkowski space, this
signature is either 3 + and 1 -, or 1 + and 3 -, depending on the
convention. This means that it is mathematically impossible for Minkowski
space to have signature (++++). Light follows null paths in Minkowski
space. There are no null paths in any space with signature (++++).

>> I have set epsilon_0 = 1 and mu_0 = 1 (as opposed to epsilon_0 = 1/(4
>pi)
>> and mu_0 = 4 pi, which also has its merits).
>>
>> If rho^2 > J^2 (here, J denote the three-current density), then
>transform
>> to a frame in which J' = 0 and rho' = rho_0, where rho_0^2 = rho^2 -
>J^2.
>> In that frame, f^2 = - rho_0^2 (E')^2, so that
>>
>> f^2 = - rho_0^2 ((E')^2/2 + (B')^2/2) - rho_0^2 ((E')^2 -
>(B')^2)/2.
>>
>> Here, f is the density of the three-force (spatial component of f)
>and the
>> density of the power (temporal component of f), so that f is a
>> four-vector.

>Well I see you're from Australia, so happens
>one of my favorite authors is Derek Lawden,
>(a New Zealander) especially his "Tensor
>Calculus...Relativity).

>(1) see his (27.14) note "is only a tensor...
>rectangular".

I would have to check the library to find this book.

>(2) Lorentz force is NOT a density.

pho E + [J,B] is not a tensor density (in the sense of Weinberg's
"Gravitation and Cosmology", for example). But rho E + [J,B] is a force
per unit volume, so it is the volume density of the force. Similarly, J.E
is not a scalar density, but it is the volume density of the power.
(J.E, rho E + [J,B]) is a four-vector, so it is not a vector density (or,
alternatively, (J.E, rho E + [J,B]) is a vector density of weight zero).

>(That's why we stopped calling relative
>tensors density's it's a ***-up word.
>Let's stay with relative tensors ok?)

>I've read the balance of your work
>several times, did you want the to
>now the answer.

Since you insist that we accept that f = 0, then you presumably demand
that f^2 = 0, otherwise you would be inconsistent. You are the only
person who is convinced that f = 0, so none of the rest of us require that
f^2 = 0.

Because you evidently object to my being anonymous, I decided to mention
that my initials are H.P. (this is the only information I intend to give).

>> first term in the above expression is
>>
>> - rho_0^2 ((E')^2/2 + (B')^2/2) = - J_u J_v T^{uv},
>>
>> in this frame, where T^{uv} are the components of the stress-energy
>> density tensor for the electromagnetic field. The second term,
>>
>> - rho_0^2 ((E')^2 - (B')^2)/2 = J_u J^u F_{vw} F^{vw}/4,
>>
>> in this frame. It follows that
>>
>> f^2 = - J_u J_v T^{uv} + J_u J^u F_{vw} F^{vw}/4,
>>
>> in this frame, and so, since this is a tensor equation, it is true in

>> every frame, i.e.
>>
>> f^2 = - J_u J_v T^{uv} - (rho^2 - J^2) (E^2 - B^2)/2
>>
>> in every frame. Since both sides are polynomial in the components
>J^u and
>> F_{uv}, and since the equality holds if J^u J_u > 0, then the
>equality holds
>> in all cases. Proof is left as an exercise for the reader.
>>
>> Therefore, in ALL cases,
>>
>> f^2 = - J_u J_v T^{uv} - (rho^2 - J^2) (E^2 - B^2)/2.
>>
>> See if you can work out how to modify this equality if we take
>>
>> epsilon_0 = 1/(4 pi), mu_0 = 4 pi.
>>
>> Alternatively, if you want to deal with the case of a point charge
>(so
>> that f is the four-force, i.e. the derivative of the four-momentum
>with
>> respect to the proper time), then in the frame in which the charge is

>> stationary,
>>
>> f^2 = - q^2 E^2 = - q^2 (E^2/2 + B^2/2) - q^2 (E^2 - B^2)/2
>>
>> = - q^2 U_u U_v T^{uv} - q^2 (E^2 - B^2)/2,
>>
>> where T^{uv} are the components of the stress-energy density tensor
>for
>> the electromagnetic field, and so, since this equation is true in one

>> frame, and both sides are invariant, it follows that
>>
>> f^2 = - q^2 U_u U_v T^{uv} - q^2 (E^2 - B^2)/2
>>
>> in all frames.
>>
>> See if you can work out how to modify this equality if we take
>>
>> epsilon_0 = 1/(4 pi), mu_0 = 4 pi.
>>
>> The components of the stress-energy density tensor for the
>electromagnetic
>> field are:
>>
>> T^{00} = E^2/2 + B^2/2,
>>
>> T^{01} = T^{10} = E_y B_z - E_z B_y,
>>
>> T^{02} = T^{20} = E_z B_x - E_x B_z,
>>
>> T^{03} = T^{30} = E_x B_y - E_y B_x,
>>
>> T^{11} = (E^2 + B^2)/2 - E_x^2 - B_x^2,
>>
>> T^{12} = T^{21} = - E_x E_y - B_x B_y,
>>
>> T^{13} = T^{31} = - E_x E_z - B_x B_z,
>>
>> T^{22} = (E^2 + B^2)/2 - E_y^2 - B_y^2,
>>
>> T^{23} = T^{32} = - E_y E_z - B_y B_z,
>>
>> T^{33} = (E^2 + B^2)/2 - E_z^2 - B_z^2.
>>
>> Note that T^{01} = T^{10}, T^{02} = T^{20} and T^{03} = T^{30} are
>the
>> x, y and z components, respectively, of the Poynting vector.

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