| |
 |
|
|
Science Forum Index » Physics - Relativity Forum » derivative question...
Page 1 of 1
|
| Author |
Message |
| Mike... |
 Posted: Mon May 12, 2008 3:34 pm |
|
|
|
Guest
|
Hi, I have a question.
If I have a path that is parameterized by a variable 's',
the derivative of the coordinates with respect to 's' is a
contravariant tensor, yes? I have a textbook that uses
the covariant derivative of this tensor (to derive the
absolute derivative). The first term of the covariant derivative
is not sensible to me (not the term involving the
Christoffel symbol). What does it mean to take the
partial derivative of a function of 's' with respect to
one of the coordinates?
-Mike |
|
|
| Back to top |
|
| Mike... |
| Posted: Wed May 14, 2008 5:02 pm |
|
|
|
Guest
|
On May 14, 12:04 pm, The TimeLord <math-n-physics-... at (no spam) att.com> wrote:
Quote: Also, for a covariant tensor, the expression above would be a difference
instead of a sum, but the same conclusions would apply; the first term is
what you actually observe.
I hope this helps.
Thanks, I understand your explanation. My question is just a
mathematical one,
though.
Given a path parameterized by 's' with functions of the coordinates
x0(s), x1(s),
x2(s), x3(s), the derivatives of these with respect to 's' is a
contravariant tensor,
the four-velocity vector. This is not a vector field though, right?
That is,
the vector is not defined at all points, only points along the path.
So in taking
the covariant derivative of this vector, as I have seen, it is not
clear to me
what the term (partial derivative) d[dx0/ds]/dx0 means, for example.
The textbook I have contracts this tensor with the tangent vector dX/
ds to get
the absolute derivative, and continues without further consideration.
Does taking the partial here with respect to x0 mean to take the
derivative
of [dx0/ds] while keeping x1, x2, and x3 fixed?
Thanks,
-Mike |
|
|
| Back to top |
|
| |
|
Page 1 of 1
All times are GMT - 5 Hours
The time now is Sun Oct 12, 2008 6:44 pm
|
|