From Special Relativity to Feynman Diagrams.pdf

# 342 parallel transport there is an equivalent way of

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3.4.2 Parallel Transport There is an equivalent way of describing the curvature of a sphere. Consider a cannon at the north pole (which may metaphorically represent a tangent vector) and let us carry it along a meridian till it reaches the equator at a point A , displacing it in such a way as to always keep it parallel to itself , so that the angle it forms with the meridian remains constant (for example, in Fig. 3.6, the vector forms an angle zero with the arcs NA and BN and π/ 2 with the arc AB ). This is what it is meant by parallel transport . Eventually, the same kind of parallel transport is performed along an arc AB of the equator whose length is 1 4 the circumference, and finally we carry the cannon back to the north pole along the meridian BN , See Fig. 3.6 .

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80 3 The Equivalence Principle Fig.3.6 Parallel Transport We easily realize that after this tour the cannon arrives back rotated by an angle of π 2 , which is exactly the angular excess described by the curvature: θ = α + β + γ π = 3 2 π π = π 2 = k A . (3.44) Indeed the ratio between the angular excess θ and the area of the spherical octant, π R 2 2 , gives exactly the value of the curvature, namely 1 R 2 K . Although this result was obtained in a very particular case, it can be shown to hold for a parallel transport along any closed path γ, not necessarily along geodesic triangles, enclosing an area and on any surface. Indeed one can expect and actually show that in the general case the rotation angle is found by first applying ( 3.44 ) to an infinitesimal area dA and then integrating over the whole area : θ = K ( x ) d A . (3.45) where is the area enclosed by the curve γ. This alternative way of defining the curvature can be easily extended to manifolds with any number of dimensions. Since our goal is to define the curvature for the four- dimensional space–time with local Minkowski metric η αβ , we consider the parallel transport of a vector v μ , (μ = 0 , 1 , 2 , 3 ) along a closed path γ in a four-dimensional manifold. After the trip, the vector will get back to the initial point “rotated” with respect to the original direction. However, as we have previously learned, such a “rotation” is a “four-dimensional rotation” in a space endowed with a metric which has the same signature as the Minkowski one η αβ , 22 and therefore is a Lorentz transformation . In particular, if the closed path γ is infinitesimal, we can obtain the analogous of the Gaussian curvature as in the case of a two-dimensional surface. However, while in two dimensions we just have one orientation for the infinitesimal area dA 22 It can be shown that the signature of the metric, that is the number of positive and negative eigenvalues of the matrix g μν ( x ), is the same at each point of a manifold. Since at a given point the metric can be taken to coincide with η αβ this explains the meaning of the statement in the text.
3.4 Curvature 81 Fig.3.7 Parallel transports along infinitesimal contours with opposite orientations enclosed by the path γ,

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• Fall '17
• Chris Odonovan

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