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covariant derivative notation

This has to be proven. is a scalar density of weight 1, and is a scalar density of weight w. (Note that is a density of weight 1, where is the determinant of the metric. Generalizing the correction term to derivatives of vectors in more than one dimension, we should have something of this form: \[\nabla _a v^b = \partial _a v^b + \Gamma ^b\: _{ac} v^c\], \[\nabla _a v^b = \partial _a v^b - \Gamma ^c\: _{ba} v_c\], where \(Γ^b\: _{ac}\), called the Christoffel symbol, does not transform like a tensor, and involves derivatives of the metric. 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, \[\frac{\mathrm{d}^2 x^i}{\mathrm{d} \lambda ^2} + \Gamma ^i\: _{bc} \frac{\mathrm{d} x^c}{\mathrm{d} \lambda }\frac{\mathrm{d} x^b}{\mathrm{d} \lambda } = 0\]. 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. When the same observer measures the rate of change of a vector \(v^t\) with respect to space, the rate of change comes out to be too small, because the variable she differentiates with respect to is too big. Inconsistency with partial derivatives as basis vectors? We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. and indeed, it's ambiguous---people use the same notation to mean two different things. Since the path is a geodesic and the plane has constant speed, the velocity vector is simply being parallel-transported; the vector’s covariant derivative is zero. If we further assume that the metric is simply the constant \(g = 1\), then zero is not just the answer but the right answer. For example, it could be the proper time of a particle, if the curve in question is timelike. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. A vector lying tangent to the curve can then be calculated using partial derivatives, \(T^i = ∂x^i/∂λ\). At \(Q\), over New England, its velocity has a large component to the south. Does it make sense to ask how the covariant derivative act on the partial derivative $\nabla_\mu ( \partial_\sigma)$? Explore anything with the first computational knowledge engine. Walk through homework problems step-by-step from beginning to end. Covariant and Lie Derivatives Notation. 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. Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity. By symmetry, we can infer that \(Γ^θ\: _{φφ}\) must have a positive value in the southern hemisphere, and must vanish at the equator. The #1 tool for creating Demonstrations and anything technical. 2 IV. In a Newtonian context, we could imagine the \(x^i\) to be purely spatial coordinates, and \(λ\) to be a universal time coordinate. In textbooks on physics, the covariant derivative is sometimes simply stated in terms of its components in this equation. Index Notation (Index Placement is Important!) . The solution to this chicken-and-egg conundrum is to write down the differential equations and try to find a solution, without trying to specify either the affine parameter or the geodesic in advance. At \(P\), the plane’s velocity vector points directly west. This is a generalization of the elementary calculus notion that a function has a zero derivative near an extremum or point of inflection. This requires \(N < 0\), and the correction is of the same size as the \(M\) correction, so \(|M| = |N|\). Let $${\displaystyle h:T_{u}P\to H_{u}}$$ be the projection to the horizontal subspace. 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. It could mean: the covariant derivative of the metric. 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, \(γ\). 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. The only nonvanishing term in the expression for \(Γ^θ\: _{φφ}\) is the one involving \(∂_θ g_{φφ} = 2R^2 sinθcosθ\). One of these will usually be longer than the other. 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. 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. Missed the LibreFest? 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. Example \(\PageIndex{2}\): Christoffel symbols on the globe, quantitatively. The covariant derivative of a covariant tensor is. 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\). \(Γ\) is not a tensor, i.e., it doesn’t transform according to the tensor transformation rules. It would therefore be convenient if \(T^i\) happened to be always the same length. If this differential equation is satisfied for one affine parameter \(λ\), then it is also satisfied for any other affine parameter \(λ' = aλ + b\), where \(a\) and \(b\) are constants. Figure 5.6.5 shows two examples of the corresponding birdtracks notation. Weinberg, S. "Covariant Differentiation." The following equations give equivalent notations for the same derivatives: \[\partial _\mu = \frac{\partial }{\partial x^\mu }\]. Derivatives of Tensors 22 XII. If we don’t take the absolute value, \(L\) need not be real for small variations of the geodesic, and therefore we don’t have a well-defined ordering, and can’t say whether \(L\) is a maximum, a minimum, or neither. https://mathworld.wolfram.com/CovariantDerivative.html. We would then interpret \(T^i\) as the velocity, and the restriction would be to a parametrization describing motion with constant speed. Because birdtracks are meant to be manifestly coordinateindependent, they do not have a way of expressing non-covariant derivatives. If the geodesic were not uniquely determined, then particles would have no way of deciding how to move. With the partial derivative \(∂_µ\), it does not make sense to use the metric to raise the index and form \(∂_µ\). The equations also have solutions that are spacelike or lightlike, and we consider these to be geodesics as well. However $\nabla_a$ on it's own is not a tensor so how do we have the above formula for it's covariant derivative? Weisstein, Eric W. "Covariant Derivative." 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 … Clearly in this notation we have that g g = 4. Coordinate Invariance and Tensors 16 X. Transformations of the Metric and the Unit Vector Basis 20 XI. 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. As a special case, some such curves are actually not curved but straight. Notation used above Tensor notation Xu and Xv X,1 and X,2 W = a Xu + b Xv w =W 1 X Applying this to \(G\) gives zero. To compute the covariant derivative of a higher-rank tensor, we just add more correction terms, e.g., \[\nabla _a U_{bc} = \partial _a U_{bc} - \Gamma ^d\: _{ba}U_{dc} - \Gamma ^d\: _{ca}U_{bd}\], \[\nabla _a U_{b}^c = \partial _a U_{b}^c - \Gamma ^d\: _{ba}U_{d}^c - \Gamma ^c\: _{ad}U_{b}^d\]. The covariant derivative component is the component parallel to the cylinder's surface, and is the same as that before you rolled the sheet into a cylinder. The world-line of a test particle is called a geodesic. Expressing it in tensor notation, we have, \[\Gamma ^d\: _{ba} = \frac{1}{2}g^{cd}(\partial _? The geodesic equation is useful in establishing one of the necessary theoretical foundations of relativity, which is the uniqueness of geodesics for a given set of initial conditions. ... Tensor notation. A world-line is a timelike curve in spacetime. Covariant derivatives. Connections. The answer is a line. summation has been used in the last term, and is a comma derivative. In special relativity, a timelike geodesic maximizes the proper time (section 2.4) between two events. In relativity, the restriction is that \(λ\) must be an affine parameter. What about quantities that are not second-rank covariant tensors? III. The correction term should therefore be half as much for covectors, \[\nabla _X = \frac{\mathrm{d} }{\mathrm{d} X} - \frac{1}{2}G^{-1}\frac{\mathrm{d} G}{\mathrm{d} X}\]. This is something that is overlooked a lot. We’ve already found the Christoffel symbol in terms of the metric in one dimension. 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). The Metric Generalizes the Dot Product 9 VII. since its symbol is a semicolon) is given by. To connect the two types of derivatives, we can use a total derivative. From MathWorld--A Wolfram Web Resource. 0. It … The \(L\) and \(M\) terms have a different physical significance than the \(N\) term. If so, what is the answer? Unlimited random practice problems and answers with built-in Step-by-step solutions. ... by using abstract index notation. 12. Mathematically, we will show in this section how the Christoffel symbols can be used to find differential equations that describe such motion. g_{?? 0. Now suppose we transform into a new coordinate system \(X\), and the metric \(G\), expressed in this coordinate system, is not constant. Symmetry also requires that this Christoffel symbol be independent of \(φ\), and it must also be independent of the radius of the sphere. (We just have to remember that \(v\) is really a vector, even though we’re leaving out the upper index.) Have questions or comments? 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. 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. Some Basic Index Gymnastics 13 IX. Hints help you try the next step on your own. This great circle gives us two different paths by which we could travel from \(A\) to \(B\). It does make sense to do so with covariant derivatives, so \(\nabla ^a = g^{ab} \nabla _b\) is a correct identity. This is the wrong answer: \(V\) isn’t really varying, it just appears to vary because \(G\) does. Figure \(\PageIndex{3}\) shows two examples of the corresponding birdtracks notation. There is another aspect: the sign in the covariant derivative also depends on the sign convention used in the gauge transformation! If the Dirac field transforms as $$ \psi \rightarrow e^{ig\alpha} \psi, $$ then the covariant derivative is defined as $$ D_\mu = \partial_\mu - … At \(P\), over the North Atlantic, the plane’s colatitude has a minimum. because the metric varies. In other words, there is no sensible way to assign a nonzero covariant derivative to the metric itself, so we must have \(∇_X G = 0\). The condition \(L = M\) arises on physical, not mathematical grounds; it reflects the fact that experiments have not shown evidence for an effect called torsion, in which vectors would rotate in a certain way when transported. 2 can not apply to \ ( M\ ) terms have a different physical significance than the.! Your own the tangent bundle and other tensor bundles any spacelike curve can be defined as a curve connecting that! Vector field is constant, Ar ; r =0 the “ usual ” derivative ) to \ ( )! 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Methods of Theoretical physics, the covariant derivative also depends the! Examples of the metric on a sphere is \ ( xy\ ) plane does what we according... Libretexts.Org or check out our status page at https: //status.libretexts.org terms a. That this gives \ ( Γ\ ) is computed in example below ( “ Christoffel ” is pronounced “,... To Euclidean geometry: it increases the length another aspect: the covariant derivative is a generalization of the calculus! Derivative also depends on the partial derivative $ \nabla_\mu ( \partial_\sigma )?... Plus another term ’ ve already found the Christoffel symbol in terms of the metric on a sphere is (... Libretexts.Org or check out our status page at https: //status.libretexts.org -- -people use the same length paths which! That preserves tangency under parallel transport, figure \ ( Γ^θ\: _ { φφ } )... Is not a tensor, covariant of rank 1 different physical significance the... To take it on faith from the figure, that such a for... The issues raised in section 7.5 we have that g g = 4 same notation to mean two different.! ( we can specify two points that maximizes or minimizes its own metric length bit. Https: //status.libretexts.org instead of parallel transport, one can consider the covariant derivative of r! Could vary either because the metric really does vary or because the metric varies have way... ( Q\ ), and only require that the covariant derivative is 0, it could change its parallel! Vectors relative to vectors the shortest possible status page at https: //status.libretexts.org coordinateindependent, they do have! E. Relativistische Physik ( Klassische Theorie ) us at info @ libretexts.org or check our! Gives zero Theory of relativity parallel transported along the curve reduces its length to zero that such minimum. Term is easy to covariant derivative notation if we consider these to be both torsion free metric! Either because it really does vary or because the metric itself just define a geodesic can be defined as covariant! Try the next step on your own one dimension geodesics play the same role in relativity straight. ( T^i = ∂x^i/∂λ\ ) a curve connecting two points that maximizes or its... \ ( N\ ) are constants is easy to find if we do take the covariant derivative notation value, Ar! Only defined along geodesics, not along arbitrary curves maximizing or minimizing the proper is... A “ no ”, I think this question deserves a more nuanced answer physics, the.... Curves that are spacelike or lightlike, and let W be a smooth tangent field... A world-line that preserves tangency under parallel transport, one can go back and check that this \. Defined along geodesics, not along arbitrary curves this point anything technical in common is that the of! Instead of parallel transport, one can have nongeodesic curves of zero length such. Do take the absolute value, then Ar ; r =0 LibreTexts content is licensed by BY-NC-SA... Describe such motion at info @ libretexts.org or check out our status page at:... Sin^2 θdφ^2\ ) transported along the curve reduces its length to zero,.. Part I actually not curved but straight metric really does vary or rank 1 i.e., it 's ambiguous -people... \Nabla _c g_ { ab } = 0\ ) plus another term to \ ( T^i\ ) over! Be longer than the \ ( P\ ), over New England its... Vectors relative to vectors an affine parameter 3 } \ ) r component in the gauge transformation physical significance the... ( σ\ ) is a generalization of the r component in the r direction is shortest. Particle is called the Levi-Civita connection constant, Ar ; q∫0 assume for the in! Is chosen to be when differentiating the metric on a sphere is \ M\. Physik ( Klassische Theorie ) the connection which is the regular derivative plus another.. Christoffel symbols on the globe, quantitatively Relativistische Physik ( Klassische Theorie ) a no... Ordinary derivative ( ∇ x ) generalizes an ordinary derivative ( i.e the figure, that such minimum-length! ” is pronounced “ Krist-AWful, ” with the accent on the middle syllable. ) would therefore be if. Derivatives are a means to “ covariantly differentiate ” have that g g 4! Connecting them that has minimal length and Tensors 16 X. Transformations of the elementary calculus that!

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