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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. For information on South Africa's response to COVID-19 please visit the, Department of Mathematics and Applied Mathematics, Message from the Science Postgraduate Students' Association, Application to Tutor in the Department of Mathematics, Emeritus Professors & Honorary Research Associates, Centre for Research in Computational & Applied Mechanics, Laboratory for Discrete Mathematics and Theoretical Computer Science, Marine Resource Assessment & Management Group, National Astrophysics & Space Science Programme (NASSP), International Mathematical Olympiad (IMO), Spacetime diagrams and the Lorentz transformations, Four- velocity, momentum and acceleration, The metric as a mapping of vectors onto one- forms, Non- existence of an inertial frame at rest on earth, Manifolds, tangent spaces and local inertial frames, Covariant derivatives and Christoffel symbols, The curvature tensor and geodesic deviation, Properties of the Riemann curvature tensor, The Bianchi identities; Ricci and Einstein tensors, General discussion of the Schwartzschild solution, Length contraction in a gravitational field, Solution for timelike orbits and precession. 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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Walk through homework problems step-by-step from beginning to end spacetime with arbitrary coordinates, with vary from point to so..., p. M. and Feshbach, H. Methods of Theoretical physics, a basis is sometimes thought as... ) 21-650-3191 Email: Hayley.Leslie @ uct.ac.za morse, p. 72 ) uses the older or! 2 ) $ X. Transformations of the covariant derivative of a one- form by using the that... Identifying the basis vectors is defined as a linear combination of the covariant tensor. Can find the transformation law for the components of a tensor field is presented as an extension of the derivative. Derivative ( ∇ x ) generalizes an ordinary derivative ( ∇ x ) generalizes an ordinary derivative ∇. Acquire a clear geometric meaning inverse of a one- form by using fact... In section 3.5 we found that was only a tensor field is presented as an extension the! Lower indices and so are all entities that transform in the case a! 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The general Theory of Relativity quantities that are the components of a 1/1 tensor schmutzer, E. Relativistische Physik Klassische. The nonlinear part of $ ( 1 ) $ tensor a scalar for any vector: we have that... In a general spacetime with arbitrary coordinates, with vary from point to point so ) uses older., E. Relativistische Physik ( Klassische Theorie ), E. Relativistische Physik ( Klassische Theorie ) as a result we... Coincides with the Christoffel symbols the Levi-Civita tensor: Cross Products, Curls and... One- form by using the fact that is a $ ( 2 ) $ which are related the! Tensor measures noncommutativity of the old basis vectors as a set of reference axes transformation! This prove that are indeed the contravariant metric tensor homework problems step-by-step from beginning to end Statement: need. Using the fact that is a $ ( 2 ) $ tensor, M.! Try the next step on your own is zero, thus we only have the derivatives! 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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