Partial derivative of metric tensor. Let's look at the partial derivative first.

Partial derivative of metric tensor The Levi-Civita Tensor: Cross Products, Curls and Volume Integrals 12. This is analogous to the ordinary metric tensor, except the inverse metric operates on covectors instead of vectors. 1. I cannot simply raise indices, since the partial derivative is in general not a tensor. Mar 12, 2008 · Could anybody help to spot the inconsistency in the following reasoning? When calculating the normal derivative of the metric tensor I get: [tex] \partial_\mu g^{\rho \sigma} = g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + 2 \partial_\mu g^{\rho \sigma}, [/tex] (1) Mar 23, 2021 · $\begingroup$ it being a constant tells us that the partial derivative will be zero, Deriving the Covariant Derivative of the Metric Tensor. Yet at the same time $$\\delta^\\mu_\\nu = The covariant derivative of a scalar is just the partial derivative at least second derivatives of the metric. Third, the metric in a coordinate basis gives spacetime length and time through dx = dxµe May 3, 2015 · Well, the partial derivative operator in your question is indeed (locally) a tensor, being a section of the tangent bundle. ∂φ(T)/∂T is also called the gradient of . During the deriving of the metric tensor, he is talking about how the vector transforms in a change of basis. Oct 7, 2024 · In differential geometry we may define as basis vectors the partial derivatives $\frac{\partial}{\partial x^i}$. In general, partial derivatives of the components of a vector or a tensor are not components of a tensor. Intrinsic, coordinate-free interpretation of "inverse" of a metric tensor. Indeed, given a vector eld V , under a coordinate transformation, the partial derivatives of its components transform as @V 0 @x 0 = @x Mar 13, 2023 · Is the addition of a Christoffel symbol and the partial derivative of a vector a tensor? 4. Let's look at the partial derivative first. Modified 4 years, {\mu\nu}$ are the components of the inverse metric tensor. Apr 18, 2020 · It's because the determinate of the metric isn't a function - it's a tensor density. 18. g. Apr 9, 2021 · It must be the partial derivative of ##A^n## if you understand ##\nabla_i g_{kn}## with ##n## fixed (a 1-covariant tensor). Viewed 376 times Sep 21, 2018 · More generally, for a tensor of arbitrary rank, the covariant derivative is the partial derivative plus a connection for each upper index, minus a connection for each lower index. 15, v, j, with j i i v, j v | g. Einstein field equations from covariant derivative of a general linear gauge transformation. 15. Jun 28, 2012 · I mean, prove that covariant derivative of the metric tensor is zero by using metric tensors for Gammas in the equation. If the connection is metric compatible, it does commute with the metric tensor and there is no ambiguity. All of this continues to be true in the more general situation we would now like to consider, but the map provided by the Apr 14, 2019 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Oct 26, 2014 · Partial derivative with respect to metric tensor Derivative Electromagnetism Metric Metric tensor Partial Partial derivative Tensor Oct 26, 2014 #1 so in a geometric, basis-independent way. 1973, Arfken 1985). Confusion about Lie derivative. $$ \partial_\lambda g_{\mu\nu By default, ResourceFunction ["ExtrinsicCurvatureTensor"] evaluates all partial derivatives of the spatial metric tensor and gauge variables automatically. The problem comes from the fact that you cannot calculate a functional derivative the way you tried. Jun 2, 2015 · There is no problem, however, since the derivative is identically zero. Ask Question Asked 4 years, 5 months ago. Using the identity $$\text{ln(det g)=Tr(ln g)}$$ and taking the partial derivative, one can show Nov 19, 2022 · Partial derivative of metric. Just as the covariant derivative of a contraction can be seen just as the partial derivative of it or as the contraction of the covariant derivative of the whole tensor. Since the metric in these coordinates is given by $\eta_{\mu\nu}$ (or $\delta_{\mu\nu}$), and these objects have constant coefficients, we have $\partial_\sigma\eta_{\mu\nu}=0$, and the Covariant Derivatives and Curvature Randy S Abstract Partial derivatives of tensor elds are generally not tensor elds. 13. Jun 16, 2017 · Flat space/spacetime: In flat spacetime, we have a natural connection, one that in a cartesian frame is given by the partial derivatives $\partial_\mu$. I want to determine the Christoffel symbols in FRW metric. Feb 25, 2025 · Determining the partial derivative of a metric tensor. The Lie derivative of a type (r, s) tensor field T along (the flow of) a contravariant vector field X ρ may be expressed using a coordinate basis as [20] Feb 18, 2022 · Often, the semicolon is used to denote covariant derivatives (just like a comma denotes a partial derivative), so: \begin{align} \dot g_{\mu\kappa;}{}^\kappa &\equiv Jan 19, 2021 · Doubt on Cauchy Stress tensor: a partial derivative of metric tensor? Ask Question Asked 4 years, 1 month ago. Aug 14, 2018 · Second, every time an index is raised or lowered, it is done using the metric. The concept of a covariant derivative is a modi cation of the concept of a partial derivative, de ned so that covariant derivatives of tensor elds are still tensor elds. 3. Dec 14, 2022 · I am not quite sure what happens when I take the trace of a partial derivative of a metric tensor. Here, I mention the term ‘covector‘. The components of the metric tensor are functions which eat four spacetime coordinates and spit out real numbers: If partial derivatives of chart coordinates with respect to local coordinates on the manifolds are prescribed at every point, which is done implicitly by assigning a metric tensor to every point, then the partial derivative operators of first order ( \left\{\frac{d}{dr}, \frac{d}{d\theta}\right\} in our concrete example), which form a basis of Feb 6, 2024 · The partial derivative of a tensor does not follow the rule of tensors. The various partial derivatives of a second-order tensor j i i j j j i i i j i j ij There are three important exceptions: partial derivatives, the metric, and the Levi-Civita tensor. He derived a formula using a gradient formula with partial derivatives by simply replacing the gradient by a vector. Why? Because the ordinary derivative of the metric tensor in Gaussian coordinates is zero. Jul 4, 2019 · The local Lorentz frame seems to be interpreted to say that the Space-Time metric can be represented in normal coordinates as diagonal ±1 with first partial derivatives equal to zero at a central point. , wh is the following possible? ([tex]\phi[/tex] is a scalar function) Having discovered that the derivative of a tensor is not a tensor in its own right, we must look for a way to bring differentiation back within the fold of the tensor paradigm. the The partial derivatives are. On varying a tensor with respect to the Apr 8, 2020 · Determining the partial derivative of a metric tensor. In other words, = and thus = = = is the dimension of the covariant derivative of the metric vanishes, Sep 27, 2016 · In the precedent article Covariant differentiation exercise 1: calculation in cylindrical coordinates, we have deduced the expression of the covariant derivative of a tensor of rank 1, i. \(\Gamma^{\mu}_{\alpha\beta}\) denotes Christoffel symbols of the second kind, which is essentially an array of partial derivatives of the metric tensor, computed with respect to the coordinate basis. Examples of at space are the 3D Euclidean space coordinated by a rectangular Cartesian system whose metric tensor is diagonal with all the diagonal elements being +1, and the 4D Minkowski space-time whose metric is diagonal with Jun 17, 2020 · "we know that [the covariant derivative of the metric tensor] is zero. Since the metric tensor is known to be parallel wrt the Levi Civita Connection, this is $=0$. Third, all partial derivatives should be covariant derivatives, which depend on the $\begingroup$ I'd say it's a bit of a stretch to say that you need a metric to define the gradient of a scalar. They are equivalent ways of seeing it. Coordinate Invariance and Tensors 8. Hence a scalar field is a mapping that assigns a scalar to each point of some domain in \(\mathbb R^3\); a vector field assigns a vector to each point of such a domain, etc. The Riemann tensor is of course made from Jan 1, 2020 · A tensor field is a tensor-valued function of position in space. Even if you're working with a smooth manifold that has no defined notion of metric, the partial derivatives $\partial_i$ are well-defined; at each point they form the basis for the tangent space generated by the given system of local coordinates. $ To Feb 27, 2016 · So far, we have defined both the metric tensor and the Christoffel symbols as respectively: Let's begin by rewriting our metric tensor in the slightly different form g αμ : Now, in this second step, we want to calculate the partial derivative of g αμ by x ν : Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Jul 3, 2016 · Remark 1: As we have seen in our articles Local Flatness or Local Inertial Frames and SpaceTime curvature and Local Inertial Frame (LIF), in a inertial frame of reference, the vanishing of the partial derivatives of the metric tensor at any point of M is equivalent to the vanishing of Christoffel symbols, and then we can write this fundamental Nov 11, 2020 · Partial derivative of metric tensor help. Also the Einstein-Hilbert action involves functional derivatives involving the metric tensor. We cannot just recklessly take derivatives of a tensor’s components: partial derivatives of components do not transform as tensors under coordinate transformations. 71) or connection coefficients Feb 15, 2023 · A colleague of mine gave me the hint, that is as crucial as it is simple, on how to calculate $$ \mathcal{L}_\xi \partial_\lambda g_{\mu\nu}. Mar 5, 2025 · Christoffel symbols of the second kind are the second type of tensor-like object derived from a Riemannian metric g which is used to study the geometry of the metric. Torsion-free, metric-compatible covariant derivative { The three axioms we have introduced Nov 5, 2024 · The components of the metric tensor with mixed indices are just equal to the Kroncker delta: Lie derivative of the partial derivative of the metric. To have an expression which does not depend on g uv being constant (Minkowski or otherwise) you need to use covariant derivatives instead of partial derivatives. Nov 26, 2020 · I was able to follow the steps up until this point (33:18 in the video). So, in any coordinate system, we have [that the ordinary partial derivatives of the metric tensor in arbitrary coordinates minus the two Chrisoffel correction terms] = 0. Start with an x,y,z coordinate system and ask, "What are \( {\partial x \over \partial x} \), \( {\partial x \over \partial y} \), and \( {\partial x \over \partial z} \)?" Feb 14, 2022 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Is there any way to prove this rule using only the definition of the Christoffel via the metric tensor? That is, using: The partial derivatives need some The partial derivative of . That is, the partial derivative of a tensor field of any rank behaves like a tensor in transformation from one RCC system to another RCC system. Feb 17, 2021 · Im new to the Tensor Calculus and General Theory of Relativity, and I have one question. In flat space in Cartesian coordinates, the partial derivative operator is a map from (k, l) tensor fields to (k, l + 1) tensor fields, which acts linearly on its arguments and obeys the Leibniz rule on tensor products. Last edited: Jun 28, 2012 Jun 28, 2012 covariant derivative of a vector is defined to be the complete expression in 1. Partial derivative of metric tensor help. Jun 22, 2020 · Simple four-vector partial derivatives. Contraction of Christoffel symbol and metric tensor. If partial derivatives of chart coordinates with respect to local coordinates on the manifolds are prescribed at every point, which is done implicitly by assigning a metric tensor to every point, then the partial derivative operators of first order ( in our concrete example), which form a basis of covariant vectors, also span the partial The reason why the exterior derivative deserves special attention is that it is a tensor, even in curved spacetimes, unlike its cousin the partial derivative. 82) defines an honest tensor no matter what the metric and coordinates are. space with a diagonal metric tensor whose all diagonal elements are 1; the space is called \curved" otherwise. 2): $$ \nabla_\gamma T_{\alpha\beta} = \partial_\gamma T_{\alpha\beta} - \Gamma_{\gamma\alpha}^\mu T_{\mu\beta} - \Gamma_{\gamma\beta}^\mu T_{\alpha\mu} $$ Plugging in the metric, noting that the left hand side Jun 10, 2018 · Yes, the tensor itself is independent of the coordinate system, but the operation of taking a partial derivative is highly dependent on what coordinate system you're using: you vary one of the coordinates while keeping all the other coordinates (in that coordinate system) constant. The Partial Derivative of a Tensor The rules for covariant differentiation of vectors can be extended to higher order tensors. T. The solution will come in the form of a new differential operator, known as the covariant derivative , whose defining characteristic is its tensor property , i. There's a formula that connects derivatives of both cases, so one of them would be enough. Ask Question Asked 2 years, 3 months ago. Dual Vectors 7. Modified 3 years, 6 months ago. May 24, 2023 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have the rst kind with the metric tensor i jk = g ir jkr (5) where gir is the contravariant metric tensor. Covectors are another type of mathematical object you’ll Jun 19, 2022 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Apr 27, 2015 · I understand the two half terms are used to account for the symmetry of the metric tensor. The curvature of spacetime is then given by the Riemann curvature tensor which is defined in terms of the Levi-Civita connection ∇. may denote a tensor of rank (2,0) by T(P,˜ Q˜); one of rank (2,1) by T(P,˜ Q,˜ A~), etc. \end{equati Apr 12, 2020 · Derivative of the metric with respect to inverse metric rho}}=0$$ For a general tensor we have $$\frac{\partial but since the metric tensor is symmetric the Apr 12, 2020 · Derivative of the metric with respect to inverse metric rho}}=0$$ For a general tensor we have $$\frac{\partial but since the metric tensor is symmetric the Apr 22, 2015 · And if they are zero, you can use partials no matter whether the tensor is a true tensor or a tensor density, since although the formula for the covariant derivative of a tensor density is a little different from that for a true tensor, it still involves just the partials and the Christoffel symbols. The metric tensor g de ned by its basis vectors: g = ~e ~e The metric tensor provides the scalar product of a pair of vectors A~and B~by A~B~= g V V The metric tensor for the basis vectors in Figure 1 is g ij= ~e 1~e 1 ~e 1~e 2 ~e 2~e 1 ~e 2~e 2 = 1 0:6 0:6 1 The inverse of g ij is the raised-indices metric tensor for the covector space: gij According to P. 14 of these online notes, this derivative is $$ \frac{\partial(F Write F with upper indices as F with lower indices times the appropriate metric Kronecker Delta and Derivatives of Axis Variables The Kronecker Delta is related to the derivatives of coordinate axis variables with respect to themselves. The unfortunate fact is that the partial derivative of a tensor is not, in general, a new tensor. But some questions which you can write down are not covariantly formulated. Modified 4 years, 1 month ago. It Dec 11, 2022 · Similarly, $\nabla_{\partial_k} g_{ij}$ is the expression for the covariant derivative of the covariant derivative of the metric tensor in a coordinate system. $\endgroup$ – Simply put, the inverse metric tensor is mathematically be defined as a multilinear map that takes two covectors to a number. On varying a tensor with respect to the metric. Hot Network Questions Dec 12, 2021 · The first claim comes from contracting the metric with its inverse and differentiating both sides. May 27, 2016 · It is not clear what raising an index on the partial derivative would imply because it does not commute with the metric tensor. is defined to be a second-order tensor with these partial derivatives as its components: i j T ij e e T ⊗ ∂ ∂ ≡ ∂ ∂φ φ Partial Derivative with respect to a Tensor (1. Transforming the Metric / Unit Vectors as Non-Coordinate Basis Vectors 9. Tensor contraction and covariant derivative. Jul 18, 2021 · $\begingroup$ I still do not know how to mess with mathjax, not necessarily lorentz transformation, a vector with index up is said contravariant, And it also transform contravariant , and I think I can not show image yet, but This appears in several videos of tensors like in the" khan ",That's what it says, but if we use the metric tensor we can lower indexes, and it left me confused, I don't Apr 30, 2020 · Determining the partial derivative of a metric tensor. . 3) The quantity . This is the case for Christo el symbols which are partial derivatives of the metric tensor but are not tensors themselves. This is the general equation of Christoffel symbo The Christoffel symbols of this connection are given in terms of partial derivatives of the metric in local coordinates by the formula = (+) = (, +,,) (where commas indicate partial derivatives). of the corresponding element of the metric tensor Here is the inverse matrix to the metric tensor . For future convenience, define new notation for partial derivatives: Note: Note that the metric tensor may be a function of position in the space. Furthermore I have that \begin{equation} \gamma_{\mu \nu} = \partial_x g_{\mu \nu}(x). Apr 10, 2010 · Hi, why can I raise and lower indices of a partial derivative with the help of the metric tensor? E. In the special case of the partial derivative of a tensor field in the RCC system, the rule of tensors is observed. Obviously since ηµˆνˆ has zero derivatives, it cannot give the connection. You will derive this explicitly for a tensor of rank (0;2) in homework 3. Jun 16, 2015 · So i am studying GR at the moment, and I've been trying to figure out what the derivative (not covarient) of the mixed metric tensor $$\\delta^\\mu_\\nu$$ would be, since this tensor is just the identity matrix surely its derivative should be zero. Our notation will not distinguish a (2,0) tensor T from a (2,1) tensor T, although a notational distinction could be made by placing marrows and ntildes over the symbol, or by appropriate use of dummy indices (Wald 1984). 2. Christoffel symbols and the covariant derivative Pingback: Vectors and the metric tensor Pingback: Metric tensor - inverse and In flat space in Cartesian coordinates, the partial derivative operator is a map from (k, l) tensor fields to (k, l + 1) tensor fields, which acts linearly on its arguments and obeys the Leibniz rule on tensor products. Ask Question Asked 7 years ago. Derivative of the Lagrangian with respect to the metric tensor. Since we haven't studied curved spaces yet, we cannot prove this, but (1. They are also known as affine connections (Weinberg 1972, p. So let's calculate the partial derivatives, without the factor of 1/2: Jan 30, 2023 · My problem is that i don't know how to compute the Lie derivative of the partial derivative of the metric $$ \mathcal{L}_{\xi} \left[\partial_\rho g^{\mu \nu}\right] $$ Either in a covariant or an contravariant form. Second, the metric components in a coordinate basis give the connection through the well-known Christoffel formula involving the partial derivatives of the metric components. Is my proof of the tensor identity true? 0. Further Reading 13. To contradict the above, this point is discussed at 1:20:39 in lecture 5a on Tensor Calculus by Pavel Grinfeld at 1:20:39, where he states that only $ \frac{\partial}{\partial x^{\alpha}}$ exists (as he says, you can only take derivative with respect to coordinates) According to this site, the Lie derivative of a $(0,2)$-tensor is $$ \mathcal{L}_XT_{ab}=\partial_XT_{ab}+T_{cb}\partial_aX^c+T_{ac}\partial_bX^c $$ However, according the same website, the Lie The Lie derivative is another derivative that is covariant under basis transformations. $$ Express the partial derivative using the covariant derivative (with the handy fact that the metric is compatible) plus some terms including the Christoffel symbols, i. Divergences and Laplacians 11. $$ g_{mn} g^{nr} = \delta^r_m \\ \delta g_{mn} g^{nr} + g_{mn The argument for this is that the partial differentiation of the tensor involves evaluating the correctly explains that the partial derivative Metric Tensor. " Here, \(x\) denotes the geodesic and the derivatives are taken with respect to \(s\), an affine parameter that uniquely parameterizes \(x\). As for the second derivatives, they cannot be made to all vanish by any coordinate transformation, because the curvature tensors are tensors, and depend explicitly on second derivatives of the metric, which, if it is Jan 19, 2018 · Once you have defined $\nabla$ on scalars (just the usual differential) and vector fields (via the Levi-Civita axioms), there is a unique extension to all tensors that satisfies the product rule $$\nabla(a \otimes b) = \nabla a \otimes b + a \otimes \nabla b$$ and commutes with contractions; and this extension is by definition the derivative operator we are using when we say $\nabla g = 0. Like the exterior derivative, it does not depend on either a metric tensor or a connection. Formally, this means that it transforms as a section of the bundle associated with the frame bundle and a particular nontrivial character of the general linear group. Action of connection on connection of metric tensor. e. Some Exercises Tensors Sep 23, 2016 · One can show that the associated covariant derivative of an arbitrary 2-tensor satisfies (see, for example, Carroll's GR, section 3. Sep 27, 2020 · Raised index of partial derivative. In the mathematical field of differential geometry, a metric tensor (or simply metric) is an additional structure on a manifold M the partial derivatives ∂ / Jun 29, 2020 · For this derivation, we first need to calculate the partial derivative of the covarinat metric tensor (which can be expressed, as the dot product of two covariant basis vectors). Do I understand metric tensor Nov 29, 2017 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Sep 14, 2022 · By the definition of geodesics, vectors parallel translate along geodesics, and the first derivatives of the metric vanish. All of this continues to be true in the more general situation we would now like to consider, but the map provided by the $\begingroup$ I'd say it's a bit of a stretch to say that you need a metric to define the gradient of a scalar. The Derivatives of Tensors 10. Apr 9, 2020 · May someone confirm or deny that covariant derivative of four-position is just metric tensor? I mean: $\nabla_{\gamma}X_{\alpha} = g_{\gamma \alpha}$ When I try to rewrite it with base vectors i Apr 14, 2019 · The determinant of a metric makes perfectly good sense, but it is not a function, rather a $2$-density. By default, ResourceFunction ["RicciTensor"] evaluates all partial derivatives of the metric tensor automatically. 0. Christoffel symbols of the second kind are variously denoted as {m; i j} (Walton 1967) or Gamma^m_(ij) (Misner et al. Jul 19, 2017 · Clearly, the problem is that partial derivatives are not "metric compatible", which in view of the Leibniz law means they don't annihilate the metric tensor, whereas covariant derivatives are designed to do exactly that. e of a contravariant vector - type (1,0) or of a covariant vector - type (0,1). In certain cases, however, these partial derivatives may be difficult or even impossible to compute, in which case the evaluation may not terminate in a reasonable time. Curvature tensors are de ned in terms of covariant Jul 11, 2019 · Partial derivative is different story though. Viewed 7k times 3 $\begingroup$ This may be a dumb question In the mathematical field of differential geometry, a metric tensor (or simply metric) is an additional structure on a manifold M the partial derivatives ∂ / $$-\\frac{1}{4\\mu_0}F^{pq}F^{jl} \\frac{\\partial}{\\partial{g_{kn}}}(g_{pj}g_{ql})=+\\frac{1}{4\\mu_0} F^{pq} F^{lj} 2 \\delta^k_p \\delta^n_j g_{ql}$$ I need to Determining the partial derivative of a metric tensor. φ with respect to . The Metric as a Generalized Dot Product 6. Apr 1, 2011 · These are not tensor quantities. On varying a tensor with respect to the Apr 16, 2016 · $\begingroup$ Any question that is formulated covariantly can be answered by choosing geodetic coordinates and replacing all covariant derivatives by ordinary partial derivatives of the coefficients. The gradient, which is the partial derivative of a scalar, is an honest (0, 1) tensor, as we have seen. Ask Question be simple results involving the metric tensor and the Kronecker delta, however I am lost as to wether I would Dec 22, 2021 · We know that the cofactor of determinant ##A##, is $$\\frac{\\partial A}{\\partial a^{r}_{i}} = A^{i}_{r} = \\frac{1}{2 !}\\delta^{ijk}_{rst} a^{s}_{j} a^{t}_{k Feb 25, 2025 · Determining the partial derivative of a metric tensor. 7. tbbjvz bmvy azkx aoa vase akafjm iujk ojqqoiboo htrj uqrnqbm rpzlc pgddn gdnne kfggqoz mvtgu