Re: The need of geodesics in physics

Igor <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
>On 5 Nov 2003 09:18:59 -0800, [EMAIL PROTECTED] (Ken S. Tucker)
>wrote:
>>Igor <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
>>>>Igor <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
>>>>>On 28 Oct 2003 10:28:28 -0800, [EMAIL PROTECTED] (Ken S. Tucker)
>>>>>wrote:
>>>>>>Igor <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
>>>>>>>On 27 Oct 2003 08:54:55 -0800, [EMAIL PROTECTED] (Ken S. Tucker)
>>>>>>>wrote:
>>>>>>>>That's a bit confusing, one can set the Lorentz 
>>>>>>>>Force to generally vanish ie. 
>>>>>>>>f_u = q*F_uv *U^v =0
>>>>>>>>and in effect decribe EM geodesics, then put
>>>>>>>>q*F_uv in the metric forming a non-symmetrical 
>>>>>>>>metric.
>>>>>>>The problem with that is that both sides of that equation are tensors.
>>>>>>>If they vanish in one frame, then they must vanish in all frames.
>>>>>>>Geodesics are based on affine connections, which are not tensors, so
>>>>>>>they have non-homogeneous transformation properties.  
>>>>>>Ah, but the geodesic equation is the absolute derivative
>>>>>>of the 4 velocity U^u = dx^u/ds, ie
>>>>>>DU^u/ds = U^u;v * U^v =0.
>>>>> True, but how does that relate to the LF?
>>>>Well multiply by mass (scalar invariant) then, 
>>>>m*DU^u/ds =0 
>>>>but QT requires m to change discretely, and 
>>>>since dm >0 requires a continuous change in m, 
>>>>dm=0, therefore set P^u = m*U^u (the 4-
>>>>momentum) and find 
>>>>DP^u/ds =0    (P^u = m*U^u, dm=0).
>>>I don't know where you got that notion from.  QT says nothing about m
>>>changing discretely.  In fact, the rules of QT maintain continuous
>>>change of mechanical variables (of course under the restrictions of
>>>the operator algebra).  What is discrete in QT are the quantized
>>>values of the so-called stationary states, but these tend to be the
>>>exception rather than the rule.  Continuous fluctuation tends to occur
>>>everywhere in the quantum world.
>>We should make sure we're talking about the 
>>same thing. It is *invariant* energy which can
>>only vary discretely and NOT continuously.
>>A simple atom emits and absorbs energy
>>by photons. 
>>>>Is true generally, and applies to LF, so 
>>>>f^u=DP^u/ds =0.
>>>>>>>Moreover,
>>>>>>>geodesic acceleration should be independent of mass, which applies to
>>>>>>>forces such as centrifugal, coriolis, and of course, gravitation.
>>>>>>True for acceleration of point masses, (but not true
>>>>>>for relatively large masses, that's why acceleration
>>>>>>is a better term than force in GR, otherwise you'll
>>>>>>need to include the term k*t_uv >0 we figuring G_uv).
>>>>>>>Acceleration due to Lorentz force will be inversely dependent on mass,
>>>>>>>making it a no go.
>>>>>>But I wrote the Lorentz force vanishes. Charged particle
>>>>>>motion must respect Quantum Theory, Lorentz force does
>>>>>>not, so the stated objection evaporates.
>>>>>But why are you making that assumption?  I can make that statement
>>>>>certainly, but I'm excluding a large variety of systems by doing it.
>>>>>So I don't think it is very realistic.
>>>>Consider, for example f_0 = q*F_0i*U^i,
>>>>which in vectors is approximately 
>>>>f_0 = q*E_i (dot) V_i.
>>>>If q moves in direction of E_i then it's
>>>>potential energy will vary continuously and
>>>>this is prohibited by QT, therefore 
>>>>E_i (dot) V_i =0   and thus f_0 =0.
>>>>Simply stated, charge q's  V_i  is always 
>>>>perpendicular to E_i. 
>>>But potential still varies continuously in QT and its changes are very
>>>easily tracked in the position representation.  
>>We're talking about Lorentz force not potential.
>>>I don't know how you came to that conclusion.

>>Fair enough. Just to make clear definitions, define 
>>invariant energy P by P*P = P_u*P^u  ok?
>>An invariant force would be f = dP/ds, and 
>>dP = f*ds   ok?
>>The Lorentz force f_u *dx^u = f*ds,  ok?
>>f_u*dx^u = q*F_uv *(dx^v/ds)* dx^u    ok?
>>Since dx^u*dx^v is symmetrical and F_uv is 
>>antisymmetrical F_uv*dx^u*dx^v =0 when
>>summed    ok?
>>Then f*ds = 0 = dP      ok?
>>That's it! 
>>To summarize: The Lorentz Force Equation
>>predicts dP=0. But we know P is not a constant,
>>therefore P varies discretely, ie by quanta. 
>>Best Regards Ken S. Tucker
>
>Yeah, I think I see what you're trying to say.  The quantity f ds is
>actually invariant work.  

Right (and thanks). Perhaps I might check out the 
following with you to ascertain co-understanding,
($ = integral)

f = dP/ds =0    ok?

$ f ds = constant   (= invariant energy)        ok?

>And your result threw me for a few minutes
>since magnetic fields do no work, but electric fields do.  Then I
>realized that it could be broken down as  f_i * d x^i  + f _0 * d x^0
>= 0.  The first term is the work done by the electric field and the
>second term is non-zero so they will balance out.  Note that f_0 is
>essentially E dot v and will be related to power.  

Ah, but f_0 is an enigma, lets study it together.

1) If f_0 is non-zero then E dot v is non-zero. 
If E dot v is non-zero then *spiral* orbits become 
possible, for example, the charge q can move slowly 
into the electric field E. (this is the old classical idea of
how electrons move in the nuclear  field, and was 
replaced by Plancks Quantum hypothesis (imho)).
So a non-zero f_0 violates QT.

2)You pointed out E dot v is "power", well consider 
electrical current moving threw a resistor and this
produces power by the product Volts x Current 
and in turn produces heat. The production of heat
is by quanta, ie. infared photons, hence power is
quantized. This can be solved and explained by,

$ f_0 ds = q*$ F_0i dx^i =  constant.

For simplicity, work the RHS and get 
q*E*x = q*Q/r = energy.=constant =quanta.

>I'm still not sure where you're going with this, but I think I
>understand your argument a little better.  Thanks.

You're very welcome, thank YOU.
I'm sorry my understanding is not simple, blame QT,
I (we) are working in the system. I think we agree
f=0, so the discussion is on the LF components f_0
and f_i being zero or not. (I'm arguing they are zero).

Best Regards
Ken S. Tucker