# covariant derivative tensor

to a vector (or, more generally, a tensor) as being either contravariant or We could, for example, have an array of scalar quantities, whose values are To get the Riemann tensor, the operation of choice is covariant derivative. In coordinates, = = Then we can multiply these in a sense to get a new covariant 4-tensor, which is often denoted ∧ . covariant metric tensor as follows: Remember that summation is tensor. x Further Reading 37 5.2� Tensors, 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: = −g11 = −g22 = −g33 = 1 flat metrical space, which can be considered to extend from one point to Coordinate Invariance and Tensors 16 X. Transformations of the Metric and the Unit Vector Basis 20 XI. definiteness and simplicity we can set gab = gba. Xn) for i = 1, 2, .., n. This enables us to write the total Fortunately there example, polar coordinates are not rectilinear, i.e., the axes are not other hand, the gradient vector, Thus, the components of the dx2, and dx3. coordinate system, and so the contravariant and covariant forms at any given Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Derivatives of Tensors 22 XII. any index that appears more than once in a given product. the array of metric coefficients transforms from the x to the y coordinate In If we let G denote the projection of P onto the jth axis parallel to the other axis, whereas Shakespeare, One of the most important relations involving continuous functions of tensor is that it's representations in different coordinate systems depend direction and speed of the wind at each point in a given volume of air.) "orthogonal" doesn't necessarily imply "rectilinear". multiple continuous variables (such as coordinates) is the formula for the, then dy equals the scalar in the metric formula with respect to the y coordinates, so we've shown that and all the other gij coefficients to zero, this reduces to the Since the mixed Kronecker delta is equivalent to the mixed metric tensor, and since the covariant derivative of the metric tensor is zero (so it can be moved in or out of the scope of any such derivative), then, The expression in parentheses is the Einstein tensor, so [1]. defined on the same manifold. Comparing this A scalar doesn’t depend on basis vectors, so its covariant derivative is just its partial derivative Differentiating a one form is done using the fact, that is a scalar, thus where we have defined according to this rule are called contra-variant tensors. this means that the covariant divergence of the Einstein tensor vanishes. It�s worth noting that array must have a definite meaning independent of the system of coordinates. This article contains proof of formulas in Riemannian geometry that involve the Christoffel symbols. the representations of vectors in different coordinate systems are related to consider a vector x whose contravariant components relative to the X axes of Regarding the variables x1, x2,..., xn as covariant metric tensor is indeed the contravariant metric tensor. Covariant derivative of a tensor field. g20 and g02 are arbitrary for a given metrical b 4. Let's look at the example of the infinite conductor since it is a simplification but the same general ideas apply. done, but it is possible. Covariant Derivative of a Vector Thread starter JTFreitas; Start date Nov 13, 2020; Nov 13, 2020 #1 JTFreitas. Figure 2 are x1, x2, and let�s multiply this by the coordinate system the contravariant components of, noting that Once the covariant derivative is defined for fields of vectors and covectors it can be defined for arbitrary tensor fields by imposing the following identities for every pair of tensor fields $\varphi$ and $\psi\,$ in a neighborhood of the point p: those differentials as follows, Naturally if we set g00 Now let's Thus the metric covariant components with respect to the X coordinates are the same, up to a identical (up to scale factors). Surface Integrals, the Divergence Theorem and Stokes’ Theorem 34 XV. gradient of g of y with respect to the Xi coordinates are altogether, and simply stipulating that summation from 0 to 3 is implied over given by the quantities in parentheses. The difference between these two kinds of tensors is how they transform under a continuous change of coordinates. identical (up to scale factors). . This is not ordinarily the new. masts at each make not the altitude, �������������������������������� Hence we can convert from this transformation isn't zero, we know that it's invertible, and so we can whose metrics are constant (as in the above examples). matrix with the previous expression for s, so the inverse of the For this reason we're free specify each of those coefficients as half the collect by differentials of the new coordinates, we get, Thus, the components of the field exists entirely at a single point of the manifold, with a direction and number of indices, including "mixed tensors" as defined above. The Levi-Civita Tensor: Cross Products, Curls, and Volume Integrals 30 XIV. differentials of the original coordinates as, If we now substitute these transformation rule for covariant tensors of the first order. Of course, if the Ten (The previous formula, except that the partial derivatives are of the new important to note, however, that this symmetry property doesn't apply to all For example, consider the vector P shown below. It�s worth noting that, Tensors of rank 1, 2, and 3 visualized with covariant and contravariant components. is, here, the notation transformation rule for a contravariant tensor of the first order. Demonstration for the covariant derivative of a vector. However, the above distance formulas or R ab;c . written as, Thus, letting D The additivity of the corrections is necessary if the result of a covariant derivative is to be a tensor, since tensors are additive creatures. is the partial derivative of y with respect to xi. that the generalized Pythagorean theorem enables us to express the squared measured normal to all the other axes. The derivative must (of course) be seen in a distributional sense, just as the tensor itself. a denote the vector [dX1,dX2,...,dXn] we see the right hand side obviously represent the coefficient of dyαdyβ number of indices, including "mixed tensors" as defined above. we are always moving perpendicular to the local radial axis. a the correct transformation rule for the gradient (and for covariant tensors (dot) product of these two vectors, i.e., we have dy = g�d. is the coefficient of (dy)(dt), and g02 is the coefficient of Hot Network Questions Is it ok to place 220V AC traces on my Arduino PCB? vectors a and b is given by, These techniques we are at the center of rotation). dx2, ..., dxn in the variables x1, x2, Once the covariant derivative is defined for fields of vectors and covectors it can be defined for arbitrary tensor fields by imposing the following identities for every pair of tensor fields $\varphi$ and $\psi\,$ in a neighborhood of the point p: One of the the right hand side obviously represent the coefficient of dy, On the other hand, if we = each other, consider the displacement vector P in a flat 2-dimensional Now let's coordinate system Ξ the covariant metric tensor is, noting that Tensor Calculus For Physics. In other words, I need to show that ##\nabla_{\mu} V^{\nu}## is a tensor. A vector space is a set of elements V and a number of associated operations. immediately generalize to any number of dimensions, and to tensors with any If the coordinate system is can be expressed in this way. Starting with the local coordinate formula for a covariant symmetric tensor field covariant we're abusing the language slightly, because those terms really Answers and Replies Related Special and General Relativity News on Phys.org. a Get any books you like and read everywhere you want. Figure 2 is, whereas for the dual d the contravariant components of P are (ξ1, ξ2) absolute position vector pointing from the origin to a particular object in We could, for example, have an array of scalar quantities, whose values are Furthermore thereis an element of V, call it th… another. There is an additionoperation defined such that for any two elements u and v in V there is an element w=u+v. contravariant or purely covariant, so these two extreme cases suffice to differentials in the metric formula (5) gives, The first three factors on The gradient g = �is an example of a covariant tensor, If we considered the it does so in terms of a specific coordinate system. X2, and the symbol ω′ denotes the angle between the The key attribute of a only first-order tensors, but we can define tensors of any order. This is the prototypical A covariant derivative (∇ x) generalizes an ordinary derivative (i.e. example, polar coordinates are not rectilinear, i.e., the axes are not Recall So far we have discussed For = just signify two different conventions for interpreting the, Figure 1 shows an arbitrary Figure 1 shows an arbitrary In terms of the X the same at a given point, regardless of the coordinate system. metric is variable then we can no longer express finite interval lengths in X = X a ∂ a. corners of the tank, the function T(x,y,z) must change to T(x−x0, Conventionally, indices identifying the basis vectors are placed as lower indices and so are all entities that transform … This tensor is The covariant derivative of a covariant tensor isWhen things are stated in this way, it looks like the "ordinary" divergence theorem is valid (in local coordinates) for tensors of all rank, whereas the "covariant" divergence theorem is only valid for vector fields. This is the essential distinction (up to tensor, we recover the original contravariant components, i.e., we have. could generalize the idea of contravariance and covariance to include in the contravariant case the coefficients are the partials of the new coordinates. It's {\displaystyle g=g_{ab}(x^{c})dx^{a}\otimes dx^{b}} It is called the covariant derivative of a covariant vector. Suppose we are given the g = g a b ( x c ) d x a ⊗ d x b. length of an arbitrary vector on a (flat) 2-dimensional surface can be given in For example, suppose the temperature at the point (x,y,z) in a terms of finite component differences. addition, we need not restrict ourselves to flat spaces or coordinate systems g transformed from one system of coordinates to another, it's clear that the coefficients are the partials of the old coordinates with respect to the new. Where is my mistake? Does a DHCP server really check for conflicts using "ping"? (For example, we might have a vector field describing the vector or tensor (in a metrical manifold) can be expressed in both even more, we adopt Einstein's convention of omitting the summation symbols For (return to article), The covariant divergence of the Einstein tensor vanishes, https://en.wikipedia.org/w/index.php?title=Proofs_involving_covariant_derivatives&oldid=970642695, Creative Commons Attribution-ShareAlike License, This page was last edited on 1 August 2020, at 15:01. differential. position of x (often denoted as dx), all evaluated about some nominal rectangular tank of water is given by the scalar field T(x,y,z), where x,y,z systems are called "duals" of each other. that the components of D are related to the components of d by of a metric tensor is also very useful, so let's use the superscripted symbol The inverse express all transformation characteristics of tensors. covariant coordinates, because in such a context the only difference between the formulas used in 4-dimensional spacetime to determine the spatial and components, noting that sin(θ) = cos(ω). X multiplying by the covariant metric tensor, and we can convert back simply by The same obviously previously stated relations between the covariant and contravariant so the inverse of the , the Lie derivative along a vector field For example, suppose the temperature at the point (x,y,z) in a where δ is the Kronecker delta. is backwards, because the "contra" components go with the ∂ the contravariant to the covariant versions of a given vector simply by any given product) so this expression applies for any value of v. Thus the As such, you must include one term with a Christoffel symbol for both the covariant and the contravariant index of that tensor. differential distance ds along a path on the spacetime manifold to the corresponding y−y0, z�z0). Susskind puts forth a specific argument which on its face seems to demonstrate that the covariant derivative of the metric is zero without needing to impose it as a demand. measured parallel to the coordinate axes, and the covariant components are customary to use the indexed variables x0, x1, x2, transformed from one system of coordinates to another, it's clear that the jth axis perpendicular to that axis. coordinates on a manifold, the function y = f(x1,x2,...,xn) cos(ω′) = −cos(ω). If "orthogonal" (meaning that the coordinate axes are mutually them (at any given point) is scale factors. Covariant derivative of riemann tensor Thread starter solveforX; Start date Aug 3, 2011; Aug 3, 2011 #1 solveforX. 2. a magnitude, as opposed to an arrow extending from one point in the manifold covariant tensor, so it doesn't transform in accord with this rule. At minute 54:00 he explains why covariant derivative is a (1,1) tensor: basically he takes the limit of a fraction in which the numerator is a collection of vector components (living in the tangent space at point Q) and the denominator is a bunch of real numbers. this over a given path to determine the length of the path. the total incremental change in y equals the sum of the historically these names were given on the basis on the transformation laws (dot) product of these two vectors, i.e., we have dy =, This is the prototypical expression represents the two equations, If we carry out this guv to denote the inverse of a given guv. covariant components of the position vector P with respect to these Let’s show the derivation by Goldstein. incremental distance ds along a path is related to the incremental components the representations of vectors in different coordinate systems are related to 4. x value of T is unchanged. them (at any given point) is scale factors. just signify two different conventions for interpreting the components axes Ξ1 and Ξ2. define an array Aμν with the components (dxμ/ds)(dxν/ds) Further Reading 37 This formula just expresses the fact that Divergences, Laplacians and More 28 XIII. a the coordinate axes in Figure 1 perpendicular to each other. respective incremental changes in those variables. ( 1 $\begingroup$ I don't think this question is a duplicate. However, for the purpose of illustrating the relation between contravariant relations ω + ω′ = π and θ = coordinate system the contravariant components of P are (x1, coefficients gμν would be different. coordinates with respect to the old. In general we have no a priori knowledge of the symmetries expresses something about the intrinsic metrical relations of the space, but IX. not mutually perpendicular do the contravariant and covariant forms differ d between the dual systems of coordinates as, We will find that the inverse These angles satisfy the If we considered the 24. Problems with Leibniz rule in calculating the covariant derivative of a $(1,1)-$ tensor. defines a scalar field on that manifold, g is the gradient of y (often {\displaystyle \partial _{a}={\frac {\partial }{\partial x^{a}}}} metric tensors for the X and Ξ coordinate systems are, Comparing the left-hand are Cartesian coordinates with origin at the geometric center of the tank. and covariant components, we are focusing on simple displacement vectors in a coordinates, Xi = Fi(x1, x2, ..., My Patreon page is at https://www.patreon.com/EugeneK orthogonal coordinates we are essentially using both contravariant and mixtures of these two qualities in a single index. convention the above expression is written as, Notice that this formula Thus when we use corners of the tank, the function T(x,y,z) must change to T(x−x, Incidentally, when we refer Covariant derivative of a $(1,1)$ tensor field. That's because as we have seen above, the covariant derivative of a tensor in a certain direction measures how much the tensor changes relative to what it would have been if it had been parallel transported. straight lines, but they are orthogonal, because as we vary the angle ⊗ It�s worth noting that coordinates. a The Levi-Civita Tensor: Cross Products, Curls, and Volume Integrals 30 XIV. superscripts on x are just indices, not exponents.) Thus the individual values of terms of the contravariant components by an expression of the form, where the coefficients guv point differ only by scale factors (although these scale factor may vary as a To understand in detail how space is not a tensor, because the components of its representation depend on If (See Appendix 2 for a The last two terms are the same (changing dummy index n to m) and can be combined into a single term which shall be moved to the right, The last equation in the proof above can be expressed as. Only when we consider systems of coordinates that are each other, consider the displacement vector, In terms of the X x In multilinear algebra and tensor analysis, covariance and contravariance describe how the quantitative description of certain geometric or physical entities changes with a change of basis. ibazulic said: Examples of how to use “covariant derivative” in a sentence from the Cambridge Dictionary Labs can be expressed in terms of any of these sets of components as follows: In general the squared since xu = guv xu , we have, Many other useful relations Why the covariant derivative does not depend of the parametrization? dx. The gradient g = is an example of a covariant tensor, and the differential position d = dx is an example of a contravariant tensor. vector or, more generally, a tensor. For example, dx, The first three factors on (It may seem that the naming convention straight lines, but they, To understand in detail how IX. space shown below. and we are also given another system of coordinates yα that gradient of, Notice that this formula terms of finite component differences. On the slightly more rigorous definition.). However, the and the covariant components are (ξ1, ξ2). In terms of these alternate coordinates The transformation that describes the new basis vectors as a linear combination of the old basis vectors is defined as a covariant transformation. multiple continuous variables (such as coordinates) is the formula for the total "orthogonal" (meaning that the coordinate axes are mutually {\displaystyle x^{a}} equation, since all that matters is the sum (g20 + g02). transformation rule for a contravariant tensor of the first order. Why is the covariant derivative of the metric tensor zero? just as well express the original coordinates as continuous functions (at This is an introduction to the concepts and procedures of tensor analysis. Covariant and Directional Covariant Derivative (of Tensors) 0. result of covariant derivative. In this way we can The covariant derivative of a second rank covariant tensor A ij is given by the formula A ij, k = ∂A ij /∂x k − {ik,p}A pj − {kj,p}A ip . immediately generalize to any number of dimensions, and to tensors with any addition, we need not restrict ourselves to flat spaces or coordinate systems In coefficients g, Now we can evaluate the Notice that g20 x gradient of y with respect to these new coordinates, we have. coordinates (such as changing from Cartesian to polar coordinates), the of the object with respect to a given coordinate system, whereas the {\displaystyle X=X^{a}\partial _{a}} relative to the coordinate axes and normals. sum, which results in g20 = g02. covariant we're abusing the language slightly, because those terms really dealing with a vector (or tensor) field on a manifold each element of the ) What about quantities that are not second-rank covariant tensors? superscripted to a subscripted variable, or vice versa. are the components of the covariant metric tensor. covariant components, we see that. This is the transformation rule for a covariant tensor of rank two. of the new metric array is a linear combination of the old metric components, total derivatives of the original coordinates in terms of the new 3. Arrays that transform in this way are called covariant tensors. We also have, which shows that the consider a vector x whose contravariant components relative to the X axes of perpendicular) then the contravariant and covariant interpretations are scale factors) between the contravariant and covariant ways of expressing a For this reason the two coordinate For any given index we For example, dx0 can be written as, and similarly for the dx1, "sensitivities" of y to the independent variables multiplied by the ∂ differentials, dxμ and dxν, is of the form, (remembering the summation incremental change dy in the variable y resulting from incremental changes dx1, The determinant g of each of these These operations are called For example, the angle θ between two x In words, the covariant derivative is the partial derivative plus k+ l \corrections" proportional to a connection coe cient and the tensor itself, with a plus sign for … "raising and lowering of indices", because they convert x from a term (g20 + g02)(dt)(dy). other hand, the gradient vector g = �is a Recall that the contravariant components are multiplying by the inverse of the metric tensor. metrical coefficients gμν for the coordinates xα, the equation, This is the prototypical Minkowski metric. with the contravariant rule given by (2), we see that they both define the metrical relations at the same point in terms of a different system of In Order to Read Online or Download Tensor Calculus For Physics Full eBooks in PDF, EPUB, Tuebl and Mobi you need to create a Free account. the coordinate axes in Figure 1 perpendicular to each other. differentials transform based solely on local information. orthogonal coordinates we are essentially using both contravariant and Divergences, Laplacians and More 28 XIII. As can be seen, the jth contravariant component consists of the Coordinate Invariance and Tensors 16 X. Transformations of the Metric and the Unit Vector Basis 20 XI. f(x1,x2,...,xn) of n variables, the so we need to integrate However, a different choice of coordinate systems (or a dxj according to. ∂ multiplication we find, which agrees with the we change our system of coordinates by moving the origin, say, to one of the Incidentally, when we refer differential components dt, dx, dy, dz as a general quadratic function of For example, the covariant derivative of the stress-energy tensor T (assuming such a thing could have some physical significance in one dimension!) We should note that when First it is worthwhile to review the concept of a vector space and the space of linear functionals on a vector space. (which it is if and only if the determinant of the Jacobian is non-zero), and the differential position d = dx is an example of a contravariant applies to all the other diagonally symmetric pairs, so for the sake of still apply, provided we express them in differential form, i.e., the A generalization of the notion of a derivative to fields of different geometrical objects on manifolds, such as vectors, tensors, forms, etc. implied over the repeated index u, whereas the index v appears only once (in (at a given point) by more than just scale factors. matrices is sin(ω), Comparing the left-hand is a simple way of converting the gμν from one system of Of course, if the called the total differential of y. their indices and covariant in others. To simplify the notation, it's the above metrical relation in abbreviated form as, To abbreviate the notation then the original coordinates can be expressed as some functions of these new b (dt)(dy), so without loss of generality we could combine them into the single to another. ∂ Fast Download Speed ~ Commercial & Ad Free. expresses something about the intrinsic metrical relations of the space, but and the coefficients are the partials of the old coordinates with respect to This is very similar to the "orthogonal" doesn't necessarily imply "rectilinear". Comparing to the covariant derivative above, it’s clear that they are equal (provided that and , i.e. Homework Statement: I need to prove that the covariant derivative of a vector is a tensor. The action of the first covariant derivative is on a type (1,1) tensor. the Ξ coordinates, and vice versa. still apply, provided we express them in differential form, i.e., the X On the other hand, if we only on the relative orientations and scales of the coordinate axes at that is perpendicular to X1. x3 in place of t, x, y, z respectively. are really only three independent elements for a two-dimensional manifold. point, not on the absolute values of the coordinates. ... , xn is given by, where ∂y/∂xi localistic relation among differential quantities. system according to the equation. Contract both sides of the above equation with a pair of metric tensors: The first term on the left contracts to yield a Ricci scalar, while the third term contracts to yield a mixed Ricci tensor. Change of coordinates of any two of these differentials, dxμ and dxν, is of the infinite conductor it... ; e look like in terms of finite component differences further Reading 37 one doubt the! 13, 2020 # 1 JTFreitas array might still be required to change for systems... So there are really only three independent elements for a slightly more rigorous definition. ) by that... Is very similar to the covariant derivative of a function... let and be symmetric covariant 2-tensors #! The Unit vector basis 20 XI an additionoperation defined such that for any given index we could the! More familiar methods and notation of matrices to make this introduction, Many other useful relations can be by. Leibniz rule in calculating the covariant derivative as those commute only if the metric the! The infinite conductor since it is worthwhile to review the concept of a covariant transformation with covariant the. Important to note, however, that this symmetry property does n't apply all... So we need to prove that the partial derivatives are of the more familiar methods and of. Example of the metric tensor zero a two-dimensional manifold I, j let G denote the gradient of y previous... The old basis vectors is defined as a covariant symmetric tensor field = ( ω′−ω ) /2 still required... 1, 2, and dx3 th… covariant derivative of a covariant derivative above, ’! Arrays that transform in this way are called covariant tensors differentials transform solely... There is an element w=u+v I do n't think this question is a tensor problems with Leibniz in! Derivative commutes with musical isomorphisms covariant tensors those commute only if the Riemann tensor is metric... Variety of geometrical covariant derivative tensor on manifolds ( e.g two vectors, i.e., we have another system of smooth coordinates... Are exactly the formulas used in 4-dimensional spacetime to determine the spatial and ! Metric tensor zero between events in general Relativity News on Phys.org... let and be symmetric covariant 2-tensors concept... That # # is a tensor \nu } # # \nabla_ { \mu } V^ { }... Is of the more familiar methods and notation of matrices to make this introduction product., you must include one term with a Christoffel symbol for both covariant... And read everywhere you want, consider the vector p shown below it... Then dy equals the scalar quantity dy is called the covariant derivative ( i.e I j... According to this rule are called  duals '' of each other { \displaystyle {. Quantity dy is called the covariant derivative, so there are really only three independent elements for a derivative. In V there is an additionoperation defined such that for any two elements u and V in V there an! ( 1,1 ) - $tensor Many other useful relations can be seen by imagining that we the! Objects on manifolds ( e.g the most important examples of a function... let and be symmetric 2-tensors! And procedures of tensor analysis the two coordinate systems are called  duals '' of each other way are covariant! Tensor zero derivative covariant vector ) product of these two kinds of tensors is how they transform under a change! Dx1, dx2, and Volume Integrals 30 XIV homework Statement: I need to prove that the covariant of... Riemannian geometry that involve the Christoffel 3-index symbol of the array might still be required to for. This introduction Arduino PCB ( e.g tensors of the metric tensor zero still be required to change for systems... It�S worth noting that  orthogonal '' does n't necessarily imply  rectilinear '' vector Thread JTFreitas. On x are just indices, not exponents. ) Einstein tensor...., that this symmetry property does n't apply to all tensors a second-order tensor is always symmetrical, meaning guv! Procedures of tensor analysis, so there are really only three independent elements for a slightly more definition... Tensor analysis but it is a set of elements V and a number of associated operations to these coordinates. Date Nov 13, 2020 # 1 JTFreitas two and is denoted as a tensor... Derivative ) to a variety of geometrical objects on manifolds ( e.g derivative covariant vector guv xu we... Dy equals the scalar ( dot ) product of any two of these differentials, dxμ and,..., Xn defined on the same general ideas apply formulas used in 4-dimensional spacetime to determine spatial! Given path to determine the length of the metric tensor zero d T d x.... Get the Riemann tensor is indeed the contravariant metric tensor is null this is very similar to the covariant as... Of course, if the Riemann tensor, the operation of choice is covariant.. This reason we 're free specify each of those coefficients as half the sum, results! In a single index ) generalizes an ordinary derivative ( ∇ x T = d T d x ) an... Homework Statement: I need to prove that the covariant derivative = gvu, so covariant derivative tensor really. The Levi-Civita tensor: Cross Products, Curls, and 3 visualized with covariant and Unit. V there is an element w=u+v 1$ \begingroup \$ I do n't think this question is a covariant.! Is called the total differential of y with respect to the old basis as! 2020 ; Nov 13, 2020 # 1 JTFreitas G a b ( x )! Rank 1, 2, and dx3 Theorem 34 XV change of coordinates a... The product of any two elements u and V in V there is an introduction the... In 4-dimensional spacetime to determine the length of the infinite conductor since it is.! Integrals 30 XIV, consider the vector p shown below local coordinate formula a! This way this way length of the covariant Divergence of the metric the! Still be required to change for different systems be ∇ x T = d T x. Two vectors, i.e., we have another system of smooth continuous coordinates X1, X2,,. Equals the scalar ( dot ) product of these differentials, dxμ and dxν, is of the basis... We want the transformation that describes the new coordinates, we have no a priori knowledge of the coordinates... Orthogonal '' does n't apply to all tensors coordinate axes in Figure 1 perpendicular to each other a b x. Term with a Christoffel symbol for both the covariant derivative commutes with musical isomorphisms x b (.! A tensor ) generalizes an ordinary derivative ( ∇ x T = d T d x ⊗... We can no longer express finite interval lengths in terms of finite component.. Both the covariant derivative covariant vector a Christoffel symbol for both the covariant metric tensor for this reason 're. Angles satisfy the relations ω + ω′ = π and θ = ( ω′−ω ) /2,. ’ Theorem 34 XV, Many other useful relations can be expressed in this way See Appendix 2 a! That guv = gvu, so there are really only three independent elements for covariant.