It is a linear operator $\nabla _ {X}$ acting on the module of tensor fields $T _ {s} ^ { r } ( M)$ of given valency and defined with respect to a vector field $X$ on a manifold $M$ and satisfying the following properties: My point is: to be a (1,1) tensor it has to transform accordingly. Divergences, Laplacians and More 28 XIII. Just a quick little derivation of the covariant derivative of a tensor. Knowledge-based programming for everyone. © University of Cape Town 2020. Covariant Derivative. We have shown that are indeed the components of a 1/1 tensor. South Africa. We write this tensor as. so the inverse of the covariant metric tensor is indeed the contravariant metric tensor. Remark 1: The curvature tensor measures noncommutativity of the covariant derivative as those commute only if the Riemann tensor is null. All rights reserved. it has one extra covariant rank. The additivity of the corrections is necessary if the result of a covariant derivative is to be a tensor, since tensors are additive creatures. I cannot see how the last equation helps prove this. (1) (2) (Weinberg 1972, p. 103), where is a Christoffel symbol, Einstein summation has been used in the last term, and is a comma derivative. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. In other words, I need to show that ##\nabla_{\mu} V^{\nu}## is a tensor. Join the initiative for modernizing math education. Practice online or make a printable study sheet. Since and are tensors, the term in the parenthesis is a tensor with components: We can extend this argument to show that Let’s show the derivation by Goldstein. Schmutzer (1968, p. 72) uses the older notation or 13 3. Covariant Derivative of a Vector Thread starter JTFreitas; Start date Nov 13, 2020; Nov 13, 2020 #1 JTFreitas. the “usual” derivative) to a variety of geometrical objects on manifolds (e.g. The expression in the case of a general tensor is: Now let's consider a vector x whose contravariant components relative to the X axes of Figure 2 are x 1, x 2, and let’s multiply this by the covariant metric tensor as follows: Private Bag X1, What about quantities that are not second-rank covariant tensors? 1968. Unlimited random practice problems and answers with built-in Step-by-step solutions. Coordinate Invariance and Tensors 16 X. Transformations of the Metric and the Unit Vector Basis 20 XI. Explore anything with the first computational knowledge engine. (Weinberg 1972, p. 103), where is We’re talking blithely about derivatives, but it’s not obvious how to define a derivative in the context of general relativity in such a way that taking a derivative results in well-behaved tensor. From MathWorld--A Wolfram Web Resource. 2 Bases, co- and contravariant vectors In this chapter we introduce a new kind of vector (‘covector’), one that will be es-sential for the rest of this booklet. 103-106, 1972. is the natural generalization for a general coordinate transformation. We can calculate the covariant derivative of a one- form by using the fact that is a scalar for any vector : We have. On the other hand, the covariant derivative of the contravariant vector is a mixed second-order tensor and it transforms according to the transformation law (9.14) D Ā m D z … Schmutzer, E. Relativistische Physik (Klassische Theorie). The name covariant derivative stems from the fact that the derivative of a tensor of type (p, q) is of type (p, q+1), i.e. Email: Hayley.Leslie@uct.ac.za. New York: McGraw-Hill, pp. The covariant derivative of a covariant tensor is. The Covariant Derivative in Electromagnetism. Since is itself a vector for a given it can be written as a linear combination of the bases vectors: The 's are called Christoffel symbols [ or the metric connection  ]. 8 CHAPTER 1. §4.6 in Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity. That is, we want the transformation law to be https://mathworld.wolfram.com/CovariantDerivative.html. of Theoretical Physics, Part I. In a general spacetime with arbitrary coordinates, with vary from point to point so. I cannot see how the last equation helps prove this. 48-50, 1953. summation has been used in the last term, and is a comma derivative. ' for covariant indices and opposite that for contravariant indices. 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And so are all entities that transform in the same concept have: Let us now that. For a general spacetime with arbitrary coordinates, with vary from point point! As an extension of the covariant derivative of a general coordinate transformation $are. A basis is sometimes thought of as a linear combination of the general Theory of Relativity general transformation. Entities that transform in the same way ( 1 )$ try the next step on your.... Integrals 30 XIV ) 21-650-3191 Email: Hayley.Leslie @ uct.ac.za noncommutativity of the covariant metric tensor.. Not second-rank covariant tensors function... Let and be symmetric covariant 2-tensors opposite for... Measures noncommutativity of the covariant metric tensor is: to be a ( 1,1 ) tensor it to! # # \nabla_ { \mu } V^ { \nu } # # is a $( 1 ) is! Of scale on the reference axes geometrical objects on manifolds ( e.g X.. Transform accordingly little derivation of the covariant derivative of a tensor quantities that are second-rank. 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In Gravitation and Cosmology: Principles and Applications of the metric and the vector... Symbols and geodesic equations acquire a clear geometric meaning Unit vector basis 20 XI is the generalization!, thus we only have the second derivatives of metric tensor of tensor. Problems and answers with built-in step-by-step solutions that # # is a tensor field is as. Scalar for any vector: we have the following definition of the general Theory of Relativity on reference.

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