Covariant derivatives We understand a covariant derivative as a covariant derivative operator (see Wald). If the covariant derivative is 0, it means that the vector field is parallel transported along the curve. ... by using abstract index notation. Even if a vector field is constant, Ar;q∫0. This correction term is easy to find if we consider what the result ought to be when differentiating the metric itself. 103-106, 1972. In the case where the whole curve lies within a plane of simultaneity for some observer, \(σ\) is the curve’s Euclidean length as measured by that observer. The covariant derivative is a generalization of the directional derivative from vector calculus. However when we prove that the covariant derivative of a $(0,2)$ tensor is the above, we use the fact that the covariant derivative satisfies a Leibniz rule on $(0,1)$ tensors: $\nabla_a(w_b v_c) = v_c\nabla_a(w_b) + w_b\nabla_a(v_c)$. Does it make sense to ask how the covariant derivative act on the partial derivative $\nabla_\mu ( \partial_\sigma)$? Let's consider what this means for the covariant derivative of a vector V. It means that, for each direction, the covariant derivative will be given by the partial derivative plus a correction specified by a matrix () (an n × n matrix, where n is the dimensionality of the manifold, for each). Unlimited random practice problems and answers with built-in Step-by-step solutions. Self-check: Does the above argument depend on the use of space for one coordinate and time for the other? While I could simply respond with a “no”, I think this question deserves a more nuanced answer. The following equations give equivalent notations for the same derivatives: \[\partial _\mu = \frac{\partial }{\partial x^\mu }\]. At \(P\), the plane’s velocity vector points directly west. To connect the two types of derivatives, we can use a total derivative. The other meaning (which is the correct one, in this case) is that it is the composition of two operators: The covariant derivative… The covariant derivative of a contravariant tensor (also called the "semicolon derivative" since its symbol is a semicolon) is given by (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. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Derivatives of Tensors 22 XII. This is something that is overlooked a lot. The most general form for the Christoffel symbol would be, \[\Gamma ^b\: _{ac} = \frac{1}{2}g^{db}(L\partial _c g_{ab} + M\partial _a g_{cb} + N\partial _b g_{ca})\]. III. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. a Christoffel symbol, Einstein Because we construct the displacement as the product \(h\), its derivative is also guaranteed to shrink in proportion to for small . Index Notation (Index Placement is Important!) Then if is small compared to the radius of the earth, we can clearly define what it means to perturb \(γ\) by \(h\), producing another curve \(γ∗\) similar to, but not the same as, \(γ\). Geodesics play the same role in relativity that straight lines play in Euclidean geometry. Example \(\PageIndex{1}\): Christoffel symbols on the globe, As a qualitative example, consider the airplane trajectory shown in figure \(\PageIndex{2}\), from London to Mexico City. However, one can have nongeodesic curves of zero length, such as a lightlike helical curve about the \(t\)-axis. By symmetry, we can infer that \(Γ^θ\: _{φφ}\) must have a positive value in the southern hemisphere, and must vanish at the equator. Now suppose we transform into a new coordinate system \(X\), and the metric \(G\), expressed in this coordinate system, is not constant. Deforming the geodesic in the \(xy\) plane does what we expect according to Euclidean geometry: it increases the length. In Sec.IV, we switch to using full tensor notation, a curvilinear metric and covariant derivatives to derive the 3D vector analysis traditional formulas in spherical coordinates for the Divergence, Curl, Gradient and Laplacian. [ "article:topic", "authorname:crowellb", "Covariant Derivative", "license:ccbysa", "showtoc:no" ], constant vector function, or for any tensor of higher rank changes when expressed in a new coordinate system, 9.5: Congruences, Expansion, and Rigidity, Comma, semicolon, and birdtracks notation, Finding the Christoffel symbol from the metric, Covariant derivative with respect to a parameter, Not characterizable as curves of stationary length, it could change for the trivial reason that the metric is changing, so that its components changed when expressed in the new metric, it could change its components perpendicular to the curve; or. The logarithmic derivative of \(e^{cx}\) is \(c\). The only nonvanishing term in the expression for \(Γ^θ\: _{φφ}\) is the one involving \(∂_θ g_{φφ} = 2R^2 sinθcosθ\). Contravariant and covariant derivatives are then defined as: ∂ = ∂ ∂x = ∂ ∂x0;∇ and ∂ = ∂ ∂x = ∂ ∂x0;−∇ Lorentz Transformations Our definition of a contravariant 4-vector in (1) whist easy to understand is not the whole story. Coordinate Invariance and Tensors 16 X. Transformations of the Metric and the Unit Vector Basis 20 XI. Covariant derivative with respect to a parameter The notation of in the above section is not quite adapted to our present purposes, since it allows us to express a covariant derivative with respect to one of the coordinates, but not with respect to a parameter such as \ (λ\). To make the idea clear, here is how we calculate a total derivative for a scalar function \(f(x,y)\), without tensor notation: \[\frac{\mathrm{d} f}{\mathrm{d} \lambda } = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \lambda } + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \lambda }\], This is just the generalization of the chain rule to a function of two variables. The Metric Generalizes the Dot Product 9 VII. Our \(σ\) is neither a maximum nor a minimum for a spacelike geodesic connecting two events. If so, what is the answer? since its symbol is a semicolon) is given by. Here we would have to define what “length” was. One thing that the two paths have in common is that they are both stationary. In Euclidean geometry, we can specify two points and ask for the curve connecting them that has minimal length. The resulting general expression for the Christoffel symbol in terms of the metric is, \[\Gamma ^c\: _{ab} = \frac{1}{2}g^{cd}(\partial _a g_{bd} + \partial _b g_{ad} - \partial _d g_{ab})\]. Example \(\PageIndex{2}\): Christoffel symbols on the globe, quantitatively. Maximizing or minimizing the proper length is a strong requirement. Symmetry also requires that this Christoffel symbol be independent of \(φ\), and it must also be independent of the radius of the sphere. This trajectory is the shortest one between these two points; such a minimum-length trajectory is called a geodesic. A vector lying tangent to the curve can then be calculated using partial derivatives, \(T^i = ∂x^i/∂λ\). Self-check: In the case of \(1\) dimension, show that this reduces to the earlier result of \(-\frac{1}{2}\frac{\mathrm{d} G}{\mathrm{d} X}\). It … The equations also have solutions that are spacelike or lightlike, and we consider these to be geodesics as well. For this reason, we will assume for the remainder of this section that the parametrization of the curve has this property. In this case it is useful to define the covariant derivative along a smooth parametrized curve \({C(t)}\) by using the tangent to the curve as the direction, i.e. Covariant Derivative of Basis Vector along another basis vector? Covariant derivative - different notations. This is described by the derivative \(∂_t g_{xx} < 1\), which affects the \(M\) term. Covariant and Lie Derivatives Notation. New York: McGraw-Hill, pp. Let G be a Lie group and P → M be a principal G-bundle on a smooth manifold M. Suppose there is a connection on P; this yields a natural direct sum decomposition $${\displaystyle T_{u}P=H_{u}\oplus V_{u}}$$ of each tangent space into the horizontal and vertical subspaces. \(G\) is a second-rank tensor with two lower indices. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In general, if a tensor appears to vary, it could vary either because it really does vary or because the metric varies. Recall that affine parameters are only defined along geodesics, not along arbitrary curves. Physically, the ones we consider straight are those that could be the worldline of a test particle not acted on by any non-gravitational forces (section 5.1). Figure 5.6.5 shows two examples of the corresponding birdtracks notation. The covariant derivative of the r component in the q direction is the regular derivative plus another term. A geodesic can be defined as a world-line that preserves tangency under parallel transport, figure \(\PageIndex{4}\). Spacelike geodesics in special relativity are stationary by the above definition. In special relativity, a timelike geodesic maximizes the proper time (section 2.4) between two events. If we apply the same correction to the derivatives of other tensors of this type, we will get nonzero results, and they will be the right nonzero results. Deforming it in the \(xt\) plane, however, reduces the length (as becomes obvious when you consider the case of a large deformation that turns the geodesic into a curve of length zero, consisting of two lightlike line segments). Applying this to the present problem, we express the total covariant derivative as, \[\begin{align*} \nabla _{\lambda } T^i &= (\nabla _b T^i)\frac{\mathrm{d} x^b}{\mathrm{d} \lambda }\\ &= (\partial _b T^i + \Gamma ^i \: _{bc}T^c)\frac{\mathrm{d} x^b}{\mathrm{d} \lambda } \end{align*}\], Recognizing \(\partial _b T^i \frac{\mathrm{d} x^b}{\mathrm{d} \lambda }\) as a total non-covariant derivative, we find, \[\nabla _{\lambda } T^i = \frac{\mathrm{d} T^i}{\mathrm{d} \lambda } + \Gamma ^i\: _{bc} T^c \frac{\mathrm{d} x^b}{\mathrm{d} \lambda }\], Substituting \(\frac{\partial x^i}{\partial\lambda }\) for \(T^i\), and setting the covariant derivative equal to zero, we obtain, \[\frac{\mathrm{d}^2 x^i}{\mathrm{d} \lambda ^2} + \Gamma ^i\: _{bc} \frac{\mathrm{d} x^c}{\mathrm{d} \lambda }\frac{\mathrm{d} x^b}{\mathrm{d} \lambda } = 0\]. Mathematically, the form of the derivative is \(\frac{1}{y}\; \frac{\mathrm{d} y}{\mathrm{d} x}\), which is known as a logarithmic derivative, since it equals \(\frac{\mathrm{d} (\ln y)}{\mathrm{d} x}\). Because birdtracks are meant to be manifestly coordinate-independent, they do not have a way of expressing non-covariant derivatives. In special relativity, geodesics are given by linear equations when expressed in Minkowski coordinates, and the velocity vector of a test particle has constant components when expressed in Minkowski coordinates. Likewise, we can’t do the geodesic first and then the affine parameter, because if we already had a geodesic in hand, we wouldn’t need the differential equation in order to find a geodesic. With the partial derivative \(∂_µ\), it does not make sense to use the metric to raise the index and form \(∂_µ\). If we do take the absolute value, then for the geodesic curve, the length is zero, which is the shortest possible. One can go back and check that this gives \(\nabla _c g_{ab} = 0\). The additivity of the corrections is necessary if the result of a covariant derivative is to be a tensor, since tensors are additive creatures. In our example on the surface of the earth, the two geodesics connecting \(A\) and \(B\) are both stationary. 12. . In other words, there is no sensible way to assign a nonzero covariant derivative to the metric itself, so we must have \(∇_X G = 0\). Hints help you try the next step on your own. is a generalization of the symbol commonly used to denote the divergence In a Newtonian context, we could imagine the \(x^i\) to be purely spatial coordinates, and \(λ\) to be a universal time coordinate. Applying this to \(G\) gives zero. But if it isn’t obvious, neither is it surprising – the goal of the above derivation was to get results that would be coordinate-independent. Regardless of whether we take the absolute value, we have \(L = 0\) for a lightlike geodesic, but the square root function doesn’t have differentiable behavior when its argument is zero, so we don’t have stationarity. The result is that the geodesic is neither a minimizer nor a maximizer of \(σ\). In that case, the change in a vector's components is simply due to the fact that the basis vectors themselves are not parallel trasnported along that curve. The quantity \(σ\) can be thought of as the result we would get by approximating the curve with a chain of short line segments, and adding their proper lengths. The covariant derivative of the r component in the r direction is the regular derivative. Some Basic Index Gymnastics 13 IX. It does make sense to do so with covariant derivatives, so \(\nabla ^a = g^{ab} \nabla _b\) is a correct identity. The metric on a sphere is \(ds^2 = R^2 dθ^2 + R^2 sin^2 θdφ^2\). However $\nabla_a$ on it's own is not a tensor so how do we have the above formula for it's covariant derivative? In general relativity, Minkowski coordinates don’t exist, and geodesics don’t have the properties we expect based on Euclidean intuition; for example, initially parallel geodesics may later converge or diverge. of a vector function in three dimensions, is sometimes also used. Since we have \(v_θ = 0\) at \(P\), the only way to explain the nonzero and positive value of \(∂_φ v^θ\) is that we have a nonzero and negative value of \(Γ^θ\: _{φφ}\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. We want to add a correction term onto the derivative operator \(d/ dX\), forming a new derivative operator \(∇_X\) that gives the right answer. Often a notation is used in which the covariant derivative is given with a semicolon, while a normal partial derivative is indicated by a comma. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. In the math branches of differential geometry and vector calculus, the second covariant derivative, or the second order covariant derivative, of a vector field is the derivative of its derivative with respect to another two tangent vector fields. The trouble is that this doesn’t generalize nicely to curves that are not timelike. For example, if we use the multiindex notation for the covariant derivative above, we would get the multiindex $(2,1)$, which would equally correspond to the operator $$\frac{D}{dx^2}\frac{D}{dx^1}\frac{d}{dx^1}f$$ which is different from the original covariant derivative … It measures the multiplicative rate of change of \(y\). However, this assertion may be misleading. A covariant derivative is a (Koszul) connection on the tangent bundle and other tensor bundles. We could either take an absolute value, \(L = \int \sqrt{|g{ij} dx^i dx^j|}\), or not, \(L = \int \sqrt{g{ij} dx^i dx^j}\). At \(Q\), over New England, its velocity has a large component to the south. it could change its component parallel to the curve. For example, if \(λ\) represents time and \(f\) temperature, then this would tell us the rate of change of the temperature as a thermometer was carried through space. 2 cannot apply to \(T^i\), which is tangent by construction. The required correction therefore consists of replacing \(d/ dX\) with, \[\nabla _X = \frac{\mathrm{d} }{\mathrm{d} X} - G^{-1}\frac{\mathrm{d} G}{\mathrm{d} X}\]. Applying the tensor transformation law, we have \(V = v\frac{\mathrm{d} X}{\mathrm{d} x}\), and differentiation with respect to \(X\) will not give zero, because the factor \(dX/ dx\) isn’t constant. The logarithmic nature of the correction term to \(∇_X\) is a good thing, because it lets us take changes of scale, which are multiplicative changes, and convert them to additive corrections to the derivative operator. Einstein Summation Convention 5 V. Vectors 6 VI. The covariant derivative of a contravariant tensor (also called the "semicolon derivative" We find \(L = M = -N = 1\). If a vector field is constant, then Ar;r =0. To see this, pick a frame in which the two events are simultaneous, and adopt Minkowski coordinates such that the points both lie on the \(x\)-axis. Figure \(\PageIndex{3}\) shows two examples of the corresponding birdtracks notation. 6.1. The easiest way to convince oneself of this is to consider a path that goes directly over the pole, at \(θ = 0\).). https://mathworld.wolfram.com/CovariantDerivative.html. We noted there that in non-Minkowski coordinates, one cannot naively use changes in the components of a vector as a measure of a change in the vector itself. For the spacelike case, we would want to define the proper metric length \(σ\) of a curve as \(\sigma = \int \sqrt{-g{ij} dx^i dx^j}\), the minus sign being necessary because we are using a metric with signature \(+---\), and we want the result to be real. As a special case, some such curves are actually not curved but straight. New York: Wiley, pp. We’ve already found the Christoffel symbol in terms of the metric in one dimension. The notation of in the above section is not quite adapted to our present purposes, since it allows us to express a covariant derivative with respect to one of the coordinates, but not with respect to a parameter such as \(λ\). Of differentiating vectors relative to vectors t transform according to the south out our status page at https:.! The metric on a sphere is \ ( G\ ) is a of. 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