Re: Simple four velocity question

On Mar 12, 4:18 pm, Jonathan Doolin <good4us...@xxxxxxxxx> wrote:
On Mar 11, 9:20 pm, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
On Mar 11, 4:26 pm, "Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote:
On Mar 11, 7:27 am, Jonathan Doolin <good4us...@xxxxxxxxx> wrote:
On Mar 10, 9:48 am, Tom Roberts <tjroberts...@xxxxxxxxxxxxx> wrote:

Jonathan Doolin wrote:
Does it seem awkward to anybody else to take the the derivative of the
"four-position" with respect to tau instead of with respect to t?

No. This is the general geometric way to form the tangent vector to a
path (where tau is an affine parameter of the path).

This pretty much assures that for all objects, you'll get the same
first coordinate (c*gamma*t)/d(tau) = d(c*tau)/d(tau) = c,

Not true. It is the norm of 4-velocity that is c, not the "first
coordinate" (by which I assume you mean the time component).

For an object moving with speed v along the x axis of the
inertial frame used for the components of 4-velocity, those
components are (c*gamma,gamma*v,0,0), gamma=1/sqrt(1-v^2/c^2).

and as Mr.
Green said, you don't have a well-defined four-velocity for light.

Yes (more on this soon...). But the tangent vector to the path is well
defined for light, as is the 4-momentum, and they are equally useful.

Is the wikipedia article error?

It looked OK to me. But it did not imply your claim above.

Tom Roberts

Just to be sure, though, look carefully where the article states:
=======================
"Components of the four-velocity"
The relationship between the time t and the coordinate time x0 is
given by
x0 = c*t = c*gamma*tau
=======================
But you appear to have said x0=c*gamma

Are these definitions agreeable,

http://farside.ph.utexas.edu/teaching/jk1/lectures/node15.html

Ken

IMO, Ken, that link entails an ambiguity as to
what type of derivative is employed in Eq.(115),
to justify defining the L.H.S. of (115) as a
tensor. Would you say it's an absolute derivative?
Apparently this does certainly needs the attention
of expert discussion, especially to evaluate Eq.(118).
Ken S. Tucker

So I guess the question is whether it is physically meaningful to take
the second derivative with respect to tau? (Eq 115.)

This may or may not be a difficulty, but I see that the first
derivative takes into account a reference frame with an origin
(t=0,x=0,y=0,z=0) in the rest frame.

Another way of looking at the problem is this.
In an inertial frame (accelometer reading zero),
at a Point P, (origin) is a lab where Physics
experiments are being conducted.

In SR, a multitude of references may have a uniform
motion (not accelerating) with velocities from v=0
to v<c, and take measurements of the Physics at P,
and all FoR's will agree the Physical Laws are true,
at P.
The Lorentz Transform also allows for the arbitary
observer to be displaced in space from P.

When an acceleration is applied,
the "generalized Lorentz Transform" involves a translation, so the
origin will NOT stay in the same place.

I think "shoe-horning" acceleration into LT was tried
after SR1905, but found unsatisfactory necessitating
the full developement of GR1916.

I think this would be okay as long as the accelerating particle is at
the origin.

Yes, that's about my understanding too. I'm not
sure whether it's a physical requirement or a
limitation imposed by using tensor analysis.
In tensors, the transformation between two CS's
is at the same point.
It's a case of interpreting the meaning of the
mathematics, but that's the *safest* one I know.
Regards
Ken S. Tucker
.

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