Re: Simple four velocity question
- From: "Ken S. Tucker" <dynamics@xxxxxxxxxxxx>
- Date: Wed, 11 Mar 2009 17:26:19 -0700 (PDT)
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
.
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