Parallel Transport and Geodesics

[muhammed yasser raso]

TL;DR Summary

can any arbitrary tensor be parallel transported along any direction

So i am confused as to what can be parallel transported , can an arbitrary tensor be transported along any curve that we wish , or do we define a curve and then solve the equation of parallel transport (which is a linear first order differential equation ) and then the solutions we get from there are the only vectors or tensors that can be parallel transported ?

 

[cianfa72]

So i am confused as to what can be parallel transported , can an arbitrary tensor be transported along any curve that we wish

Sure, as long as you have an affine connection defined on your manifold (either metric compatible or not) then you can transport any tensor along any smooth curve using that connection.

 

[muhammed yasser raso]

thank you , what does the parallel transport equation give us ?

 

[muhammed yasser raso]

because the covariant derivative of a tensor wont always give us zero , what makes the parallel transport equation different ?

 

[Ibix]

You can parallel transport anything. $\nabla_\mu T^{\nu\rho}{}_\sigma$ tells you how the tensor changes from a small step in the direction of your $\mu$th basis vector, and $X^\mu\nabla_\mu T^{\nu\rho}{}_\sigma$ tells you how it changes from a small step in the direction of the vector $X$. By chaining small (infinitesimal, in fact) steps together you can see what happens to an arbitrary tensor as it is transported along a path, using the path's tangent vector as $X$ at each step.

When you solve the geodesic equation, though, you are requiring that the change of the vector you are transporting be zero. That defines a path, because at each step there is only one choice of vector $X$ that gives zero change.

Physically, you can define a vector pointing along your nose. Solving the geodesic equation for that vector tells you where you will end up if you just walk and follow your nose. But you can also stick your arm out to the right and define a vector along that and ask how that vector changes as you follow your nose - or any other path.

 

[cianfa72]

Just to be more specific: suppose you have a curve $C$ and $X^{\mu}$ is the tangent vector along it. Then you have a tensor $T^{\nu\rho}{}_\sigma$ defined on the starting point $P$. Solving the parallel transport equation $$X^\mu\nabla_\mu T^{\nu\rho}{}_\sigma = 0$$ does mean calculate the "parallel transported version" of the tensor $T^{\nu\rho}{}_\sigma$ along that curve $C$.

 

[JimWhoKnew]

thank you , what does the parallel transport equation give us ?

Parallel transport and the differential equation are aspects of the same concept. Which one should be used, depends on the problem you're facing. Loosely speaking, it is the generalization to curved spacetime of the constant field on a curve from ordinary calculus in $R^n$.

Consider first calculus in $R^2$, in Cartesian coordinates. Suppose there is an affinely parametrized open curve $\gamma(\lambda)$ and a point $P=\gamma(0)$ on it. Let $f(P)$ (scalar, vector, tensor) be a function value at $P$. Since the tangent spaces at each point of $R^2$ are "the same", we can allow ourselves to duplicate $f(P)$ at all points of $\gamma$, to create a constant function on it. This function will satisfy $\frac{d f}{d \lambda}=0$. The components of $f$ will be the same at each point.

Consider now polar coordinates on $R^2$. You can create a constant field on a curve from the value at $P$, by demanding that they all have the same "length" and "cartesian direction" (just like before). But this time, if $f(P)$ is a tensor, the components will not satisfy $\frac{d f}{d \lambda}=0$ in general. You'll need additional terms to compensate for using polar coordinates (connections).

Generalizing to curved spacetime, the function on $\gamma$ will be considered as "constant" if it satisfies $\nabla_\vec{u} f=0$, where $\vec{u}$ is the tangent vector to the curve. This means that $f$ is parallel transported from the tangent space of $\gamma(\lambda)$ to the tangent space of an infinitely close $\gamma(\lambda+\epsilon)$ by an expression of the form $$f(\gamma(\lambda+\epsilon))= f(\gamma(\lambda))+\epsilon f(\gamma(\lambda)) \Gamma$$
To convince yourself that this generalization is "good", consider the coordinate transformation to a locally inertial frame at the point $\gamma(\lambda)$. In this frame $g_{\mu \nu}=\eta_{\mu \nu}$ and all $\Gamma$s are zero. So the parallel transport amounts to duplicating the components, and the equation reduces to $\frac{d f}{d \lambda}=0$, Just as in ordinary calculus.