Re: The need of geodesics in physics

Igor <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
>On 28 Oct 2003 11:25:37 -0800, [EMAIL PROTECTED] (Ken S. Tucker)
>wrote:
>
>>Igor <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
>>>On Mon, 27 Oct 2003 21:20:38 +0000 (UTC),
>>>[EMAIL PROTECTED] (Gregory L. Hansen) wrote:
>>>
>>>>In article <[EMAIL PROTECTED]>,
>>>>Ken S. Tucker <[EMAIL PROTECTED]> wrote:
>>>>>Uncle Al <[EMAIL PROTECTED]> wrote in message
>>>>>news:<[EMAIL PROTECTED]>... 
>>>>>>Metric models are geometric and have geodesics.  Affine models are
>>>>>>electromagnetic in form and have no geodesics.  
>>>>>
>>>>>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.

>>>>I'm always tempted to make the connections non-scalar.  If the geodesic 
>>>>equation is
>>>>  a_i = -{i,jk} v_j v_k
>>>>with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the 
>>>>connection, you can multiply through by an m and call -m*{i,jk} the force 
>>>>of gravity.  
>>
>>I have no objection provided m is small relative to 
>>the gravitating body, such as Mercury is to the Sun, 
>>but check out Dover's P of R, A.E.'s GR Eq.(45), 
>>and read geodesic is "the motion of the point". 
>>I figure this is the solution from G_uv =0 (with the
>>energy density being zilch, by the specification of a
>>point) as opposed to the G_uv =k*T_uv, in the case 
>>of a massive particle creating nonlinear *feed-back* 
>>like solutions. Opinions Please.
>>
>>>>So I always want to put some column vectors in there, like
>>>>  a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k
>>>>         (0)                   (1)
>>>>and then multiply through by (m,q).
>>
>>Very interesting how the indexes 1,2 are added
>>within your connections, and may very well be 
>>ok, is a significant departure from conventional
>>mathematics. 
>>I think QT excludes continuous acceleration of
>>charged particles. This is why Lorentz's
>>f_u = q*F_uv *U^v =0
>>may be reasonable.
>>
>>((Knowing Mr. Hansen has R.C. Tolmans, 
>>"Relativity, Thermodynamics and Cosmology", 
>>he may want to glance Eq. (103.1) and (103.2).))
>>
>>>Yeah, I see what you're trying to do (even though your column vectors
>>>got a little skewed).  The first term would be gravitation and the
>>>second the Lorentz, but how you'd ever get a cross product out of a
>>>symmetric connection is a bit of a stretch.
>>
>>The "cross-product" is primarily a magnetic effect,
>>from relative charge motion considered from a rest 
>>frame. If I'm not mistaken, a direct proportionality
>>exists between the Magnetic field and the angular 
>>momentum of the mass the charge creating the current
>>and B-field is attached (at rest) to.
>>  Wouldn't this suggest that the metric describing
>>angular momentum and it's direct analog be equal?
>>Regards
>>Ken S. Tucker
>
>I'm not completely sure I understand what you are trying to say here.
>All I was saying was that we have two completely different types of
>mathematical objects here.  The magnetic force is a cross product of
>two vectors, sometimes called a bi-vector or a second rank
>antisymmetric tensor.  The affine connection, on the other hand, has a
>second rank symmetric structure in it's covariant indices.  

Agreed, what you have stated is rigorously correct
using the concept of the Magnetic field, but the 
magnetic force could be calculated from,

f=qE*(1-v^2),    with  magnetic force basically,

f(m) = -q*E*v^2  = -q*Q/r * v^2/r 

>I don't understand how those two patterns could match up.

Well setting energy (=mass) = q*Q/r  and a
centrifugal acceleration term from the affine 
to be v^2/r  reproduces basic f(m). 
Mr. Hansen is very familiar with centrifugal metrics
including spacetime metrics like g_0i. We discussed
these before, so I have some *very* basic notions,
about these possiblities.

>Your statement about proportionality of B field and angular momentum
>is only true for spin and not angular momentum in general.

Thanks, BTW way I read with interest, awhile ago  
your 2 stage process to generate a photons.
Ken S. Tucker