Re: derivative question

Am Mon, 12 May 2008 18:34:19 -0700 schrieb Mike
<michael.h.williamson@xxxxxxxxx> in
6d597313-e544-4547-b878-cff2eb53e106@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

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?

Sometimes it helps to think of a practical example. So let's consider a
particle that is moving inertially near a black hole. Since it is moving
inertially, the 4-force is zero (going to use MACSYMA itensor notation):

f([],[mu])=0

Note that this tensor is contravariant. Now, in Relativity the force is
the absolute derivative of momentum with respect to proper time. This is
the relativistic extension of Newton's second law. So:

0=f([],[mu])=Dp([],[mu])/dtau
=dp([],[mu])/dtau + p([],[alpha])*chris([mu],[alpha,beta])*dx([],[beta])/
dtau

In this example "tau" is the parameter, equivalent to your "s". The first
term in the sum is what is actually going to be observed, ie a change in
momentum if chris([mu],[alpha,beta]) is not zero in all terms, ie there
is curvature of space-time.

So near a black hole the momentum of a photon can be observed to change
even though it is moving inertially.

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.

Oh, I forgot to mention that if there is inertial motion of the observer
then you can just convert the tau to observed time like

dp([],[mu])/dt=( dp([],[mu])/dtau )/gamma
where gamma=contains velocity with respect to the black hole
=1/sqrt(1-v^2/c^2)

Take care.
.

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