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I was unsure whether to ask this here or at a physics SE.

Wald's "General Relativity" defines parallel transport as follows:

$\nabla$ is a derivative operator (is linear, obeys Leibniz rule, commutative with contraction, torsion free and is consistent with the notion of vectors as directional derivatives). A vector $v^b$ given at each point of a curve C is parallel transported is said to be parallelly transported as one moves along the curve if the equation

$t^a \nabla _a v^b =0$

is satisfied along the curve, where $t^a$ are vectors tangent to the curve.

How does this definition reproduce what we intuitively understand as parallel transport? Also, other references use different terminology, with a $\nabla _v$ meaning the derivative along a vector $v$, (the same role $t^a$ plays in the definition above) which is easier to understand but (1) seems ill defined, unlike wald's definition, and (2) both definitions should be related in some way.

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up vote 3 down vote accepted

When a thing is parallel transported, doesn't it remain in a sense unchanged?

Intuitively the equation $$ t^a \nabla _a v^b =0 $$ means that vector $v$ is constant (the derivative is zero!) with respect to differentiation along the velocity vector of curve $C$. This velocity vector $$ t=\dot{C} $$ provides a natural way of expressing the direction in which vector $v$ is being transported.

Which vector $v$ is meant in your last paragraph? It looks to me that there is a confusion. Wald's definition can be written in terms of derivatives along vectors as well, just slightly different notation is used. Namely, $$ \nabla_{t} v^b:= t^a \nabla_ a v^b $$ Notice that Wald uses so called abstract index notation, while other sources that you assume may use "invariant" or "coordinate" ways of writing down that same things.

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Thanks for the answer! it's a lot clear now :) the $v$ in $\nabla _v$ on the last paragraph is the same tangent vector $t^a$. Sorry for the mix up, I should have said that. –  Diego Aug 23 '11 at 9:09
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