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## Parallel Transport in Curvilinear Space

The concept of parallel transport was first introduced by the famous german mathematician C.F. Gauss. The idea was that it would be a way to measure the curvature, as oppose to the generalization of curvature along a line as the modulus of the derivative of the unit tangent vector with respect to arc length.*

k=|dT/ds|

Parallel transport as oppose to ordinary curvature along a line, is a uniquely simple idea, separate from the concept of a rate of change. Basically as one moves through a curvilinear coordinate system, the orientation of a vector in a parallel vector field changes with respect to the direction of the axes, or basis vectors at that point.

Supposing we lived in a curvilinear space, then we would treat any vector as a line if it maintained constant direction with respect to our axes. This, however would not show the true change of our coordinate system. An excellent example is the curved surface we live on, the earth. You start on the north pole facing down the prime meridian, and walk down to the equator. After this you then continue west along the equator till you reach ninety degrees east all while still facing south. The next step is to walk north (while facing south) until you get to the north pole again. You should now be facing at a right angle to your original starting direction.

Another way of viewing this is by placing your arm straight up with your palm facing inwards. Then bring it down ninety degrees so it is in front of you. After this move it ninety degrees so it is to your side. Lastly bring it back up. Your palm should now be facing forwards. This is because you have parallel transported your hand by traversing an imaginary (not numerically) sphere.

Now onto the mathematical formulation of the problem. To understand this you will need to understand advanced calculus (although you don't need a thorough understanding of analysis), tensor calculus, and coordinate systems. Unfortunately E2 has the weakness that it doesn't allow pictures, and this math benefits greatly from the use of pretty diagrams (no ASCII will only confuse you).

Let us start with our generalized coordinates, xi, which has basis vectors, Ei. In this space we suppose the existence of a constant cartesian vector field, Vi, that is one with constant magnitude and sense. The vector in the field can be represented by the sum of its components times the basis vectors, ViEi. As usual with tensor algebra the summation convention holds between two like contravariant and covariant indices. If we want to measure the change in Vi with respect to the basis vectors when traveling along some line in the coordinate system, we must take the difference in changed vector and original vector. We can change the basis vectors by moving along the line and then take the the scalar product of the vector with respect to the modified basis vectors, to find the components of the old vector in the new basis. Then we can subtract the dot product of the old basis with respect to the vector to get the difference.

ΔVi=Ei(xkxk).V-Ei(xk).V

Then using a simple rule of analysis, known as the mean value theorem of multiple variables (I think that's it) we can get to a profound result:

ΔVi=∂Ei/∂xk.VΔxk

Okay I lied it's not profound yet. We have to mess around with V a little. Well recalling that is in fact the sum of the components and the basis vectors: ViEi, we can arrive at our startling conclusion:

ΔVi=∂Ei/∂xk.VjEjΔxk

ΔVi=∂Ei/∂xk.EjVjΔxk

ijk=∂Ei/∂xk.Ej

ΔVi=-ΓijkVjΔxk

Now one may ask why this is a profound result. Well, this is a profound result, because ijk is of course the familiar Cristoffel Symbol of the second kind, also known as a metric connection or affine connection. So now the above result is quite simple (if you're down with tensor analysis). The subject of parallel transport of a closed curve can then be taken as parallel transport along two vectors, Δai and Δbi

ΔViTot=-ΓijkVjΔakijk(x+Δa)Vj(x+Δa)Δbkijk(x+Δb)Vj(x+Δb)ΔakijkVjΔbk

Once again, the mean value theorem comes in handy:

ΔViTot=(∂(ΓijkVj)/∂xlakΔbl-(∂(ΓijkVj)/∂xlbkΔal

ΔViTot=(∂Γijk/∂xl)VjΔakΔblijk(∂Vj/∂xlakΔbl-(∂Γijk/∂xl)VjΔalΔbkijk(∂Vj/∂xlalΔbk

Because this is a parallel transport, a condition is met that the covariant derivative of the vector Vi is zero and hence:

Vi/∂xj=-ΓijkVk

ΔViTot=(∂Γijk/∂xl)VjΔakΔblimkΓmjlVjΔakΔbl-(∂Γijk/∂xl)VjΔalΔbkimlΓmjkVjΔakΔbl

Most of that works out because of dummy indices. Now if you're a fan of my write ups, you'll remember the Riemann Curvature tensor,Rijkl, (gotta love them shameless plugs). This gives a final succinct result of:

ΔViTot=RijklVjΔakΔbl

As discussed in the aforementioned wu this is an excellent method of describing curvature as it can be generalized to higher dimensions (such as 4-dimensional space-time), and describes curvature of an anisotropic coordinate system.

*Of course other types of curvature exist as an extension of this, such as gaussian or total curvature which is the product of curvature along the two degrees of freedom of a surface, but this is a wu on parallel transport, not curvature.

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