Derivative of a tensor with respect to a tensor. If this gradient is zero, then there are no tidal forces.

Derivative of a tensor with respect to a tensor The metric component g 12 here measures the alignment of the basis vector e 1 with To do functional derivative of some actions, we need to know a functional differential of metrics $ is a tensor because actions must be scalars, but i don't have any proof. σ = λtr(ϵ) + 2μϵ σ = λ t r (ϵ) + 2 μ ϵ. The same goes with your case, you have to put a Dirac delta in $$ \frac{\delta g^{\mu\nu Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site root of a tensor. 4) that if ZßZ −XÐQÑ 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. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Partial derivative of tensor with respect to tensor. The directional derivative provides a systematic way of finding th My question is related to continuum mechanics, taking partial derivative of tensor with respect to tensor. Hot Network Questions Is partial correctness decidable? tensor is always a subset of the union of the index sets of the multiplication’s arguments, that is, s 3 (s 1 [s 2). Visit Stack Exchange Abstract The object of the paper is the derivative with respect to a second-order tensor. Commented Apr 13, 2020 at 13:52 On varying a tensor with respect to the metric. In fact, the derivatives of the square root of a symmetric tensor, as well as those of the stretch and rotation tensor are explicitly I've intentionally avoided any of the trappings of Riemannian geometry or relativity in general because it isn't necessary here and I think the calculus is instructive, but it is also possible to derive this from the property that the covariant derivative of the inverse metric tensor is identically 0, which can be derived from the covariant Stack Exchange Network. Consider that “taking the derivative with respect to a matrix” means de facto taking the derivatives of that expression with respect to each element of the matrix. But when you are working with total derivative, you should note that It doesn't matter whether your function depends on your affine parameter explicitly or not. constant For an element-wise calculation, the gradient of the sum gives the derivative of each element with respect to its Can someone please show me step by step how to do this derivative? I understand the tensor notation itself but I have problems to understand such a derivative, especially when I have a term like $\frac{\partial(\partial^{\mu}A^{\nu})}{\partial(\partial_{\sigma} A_{\lambda})}$. t a vector. GradientTape() as t: t. Now, this process involves calculating the derivative of the eigenvectors of the right Cauchy-Green deformation tensor with respect to the tensor. The second variational derivative of Ricci tensor with respect to metric. One approach is to use index notation $$\eqalign{ f_i &= W_{ij} x_j \cr\cr \frac{\partial f_i}{\partial W_{mn}} &= \frac{\partial W_{ij}} Derivative of a row vector with respect to a column vector; derivative with respect to a row vector assuming numerator layout. cell 2,1 in the grid). 0 votes. I have now specified that in the question. The partial derivative of A tensor product A with respect to A. On varying a tensor with respect to the metric. Viewed 92 times 1 $\begingroup$ In Sean Caroll's Spacetime and Geometry Textbook, at page 183 (discussing scalar-tensor theories) Carroll defines a conformal metric by performing a conformal transformation as: rates of change (with respect to time, position) 123 3 2 1 w t B B t w G. An obvious interpretation is that you are taking the component-wise logarithm of the tensor. float32 or tf. the naive derivative expressed in polar or spherical According the first line on page $2$ of this paper, A smooth vector field $\\xi$ on a Riemannian manifold $(M, g)$ is said to be a conformal vector field if its flow consists of conformal transform Presumably some kind of tensor with three indices? $\endgroup$ – anon. Perhaps it will be interesting to other users. Derivative of the metric with respect to inverse metric. I haven't been able to find a simple recipe for writing down a solution. The tf. In general, taking the partial derivative of the tensor component-wise with respect to time will not result in another tensor. Matrix derivative of a matrix with constraints. Then : = [(+ )] = = =: Therefore, = Here is the fourth order identity tensor. Modified 1 year ago. Here, we have the tensor product of basis vectors. gradient of force vector). Its components are given by the matrix $\mathbf{G}$: The sign $\otimes$ denotes the tensor product. Tensor, you need to call GradientTape. M. Variable() # Must be a tf. About variation of Ricci tensor. is a function of x) and is a vector function of , the vector-by-vector chain rule states: The Second Piola-Kirchhoff Stress $\boldsymbol{S}$ is related to the Material Tangent Tensor $\mathbb{C}$ by $\mathbb{C} = 2\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{C}}$. For example, a tensor with contravariant rank 2 and covariant rank 1, written as a (2,1) tensor, transforms I apologize in advance if this question is too vague, but I have no idea where to start. The proof follows a wholly different approach from those in the works cited above. Calling Sequences. I want to define tensor which is the derivative of the former tensor with respect to the coordinate axis: $$ X = (x_1, x_2, x_3) \\ TD_{\alpha \beta \gamma} = \frac{\partial T_{\alpha \beta}}{\partial x_\gamma} $$ However, I want to get the result in terms of general expressions of tensors, for example: $$ As you can see, each tensor has been assigned with a particular set of attributes. This notation is basically identical to the tensor multiplication I have the tensor which is expressed in terms of coordinate vector. backward (gradient = custom I have trouble figuring out derivatives in tensor notation in SR. placeholder(tf. r. 2005 . This is the general equation of Christoffel symbo After that, in order to calculate the constitutive material tensor, I need to calculate the derivative of the stress tensor with respect to the Lagrangian strain tensor again. Jacobian(Z,X) is a bit closer to what I'm looking for but not exactly what I am looking for, because it computes the gradient of every element in Z with respect to every element in X, whereas I only want to compute the gradient of z_{b,t,j} with respect to x_{j,i} for Before I can tell you what a tensor is, I must tell you what a vector really is; in fact, you will later see that a vector is a type of tensor. Define the function . $$ g_{mn} g^{nr} = \delta^r_m \\ \delta g_{mn} g^{nr} + g_{mn The variation of the Riemann curvature tensor. To understand the indices, consider a simpler example: $$\frac{\partial}{\partial x^i}(x^j x_j) = \frac{\partial}{\partial x^i}(x^j x^k g_{jk})= \delta^j_i x_j + \delta^k_i x_k = 2x_i$$. 3) together with (2. Here's the reason: Usually we compute the Green-Lagrange strain tensor from the deformation gradient with its definition $$ \mathbf{E}(\mathbf{F})=\frac{1}{2}(\mathbf{F}^T\mathbf{F}-\mathbf{I}) \tag{1} $$ It is easy to I have been doing some calculation on variation of Ricci's tensor with respect to the metric, that, according with S. It is a function of the metric tensor and varies with respect to changes in the metric, allowing us to make predictions about the behavior of space-time in the presence of matter and energy. autograd. In this context, we introduce a Partial derivative symbol with a spatial subscript, rather than an index, are used to denote partial di erentiation with respect to that spatial variable. It does not depend on either a metric tensor or a connection: it requires only the structure of a differentiable manifold. This covariant derivative does commute with the metric - in the jargon, it is "metric compatible". 6. The Kronecker Delta is related to the derivatives of coordinate axis variables with respect to themselves. grad) # outputs tensor([7. sin(x)*torch. In classical field models, Hilbert’s energy–momentum tensor for a matter field is defined by a variational derivative of a Lagrangian with respect to the metric tensor. On covariant derivative. Tensor - divergence I. This is because the metric can be defined as the dot product of basis vectors, which are naturally covariant objects. It is conceptually related to the idea of tidal forces though. It is used to calculate the Lie derivative of a tensor field along a given direction. Derivative of Riemann tensor respect to Riemann tensor. Start with an x,y,z coordinate system and ask, "What are Concept Question 2. About the variation, or better the derivative, of a tensor it My question is addressed to the reputable community of physicists in connection with the ignorance of some of the subtleties of mechanics. $\begingroup$ I am just trying to figure out what is the correct way of defining the correct functional derivative of gμνgμν with respect to the components of g itself. they correspond to the same state of deformation. In our ordinary formalism, the covariant derivative of a tensor is given by its partial derivative plus correction terms, one for each index, involving the tensor and the Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site The determinant of a metric makes perfectly good sense, but it is not a function, rather a $2$-density. watch(x_tensor) output = model(x_tensor) result Very briefly, the covariant derivative consists of taking the partial derivative, then for each index throw in a correction term using the Christoffel symbols (which have three indices: $\Gamma^\gamma_{\mu\nu}$, and is thus a $4\times 4\times 4$ table of numbers at each point) to ensure that what you get out in the end is indeed a tensor. (b) Under the conditions of Theorem 3, we define the tensor the (— m)th intrinsic derivative, of W with respec 2 t to z by The exterior derivative of a totally antisymmetric type (0, s) tensor field with components A α 1 ⋅⋅⋅α s (also called a differential form) is a derivative that is covariant under basis transformations. Tensor([4]),requires_grad=True) y = torch. For example what would be the solution to the following Derivative with respect to the spacetime derivative of a field $\phi$ 0. Then create a tensor input value with requires_grad = True. However, there are cases when the output function is an arbitrary tensor. Modified 4 years, 1 month ago. In index notation with respect to an orthonormal basis The first claim comes from contracting the metric with its inverse and differentiating both sides. Ogden, Non-Linear Elastic Deformations, 1997, p. Tensor(6. $\begingroup$ I am assuming that the divergence is taken with respect to the last tensor leg (i. How is this multiaxial relationship derived in this paper? 0. According to what I've read, the derivative of a tensor is not in general a tensor (according to Steven Weinberg). Is a derivative with respect to the particular coordinate A connection is metric compatible if the covariant derivative of the metric with respect to that connection is everywhere zero. To record gradients with respect to a tf. Before you write me off as a nut, take a look at the directional derivative of some scalar function f (xj): v·∇ f I see that the need to apply the product rule when using curvilinear coordinates results in a term for the derivative of the coefficients that still follows tensor transformation rules, and that it is the derivative of the bases vectors that is the problem. Could someone please help explain why the partial derivative of the strain energy with respect to strain components gives the stress components? My attempt at deriving ($\star$) The second derivative is just the stiffness tensor $$\frac{\partial^2 \Pi}{\partial \boldsymbol{\varepsilon} \partial \boldsymbol{\varepsilon}} = \boldsymbol{M}, Partial derivative of the partial derivative of a function, with respect to the function itself Hot Network Questions Why did the US Congress ban TikTok and not the other Chinese social network apps? consider $\boldsymbol{F}$ is a second order tensor, and $\boldsymbol{F}^{-T}$ the invers traspos of the same tensor. Inverse of The object of the paper is the derivative with respect to a second-order tensor. the usual vector derivative constructs (∇, ∇·, ∇×) in terms of tensor differentiation, to put dyads (e. Hot Network Questions The first side consists of the sum of the squares of the two derivatives of the metric tensor with respect to the four indices in the brackets, and the second side consists of the sum of the squares of the two derivatives of the metric tensor with respect to Let $ X $ be a smooth vector field, $ X _ {p} \neq 0 $, $ p \in M $, and let $ U $ be a tensor field of type $ ( r, s) $, that is, $ r $ times contravariant and $ s $ times covariant; by the covariant derivative (with respect to the given connection) of $ U $ at $ p \in M $ along $ X $ one means the tensor (of the same type $ ( r, s) $) And how exactly do you define said functional derivative? Is it supposed to be the Gateaux or the Frechet derivative evaluated at the metric or something else? The way the Ricci tensor is normally defined it does not make sense to "derive by the metric". # Manually define the custom derivative for the function def custom_derivative (x): return 1 / x # Compute gradients with respect to x using the custom derivative y. How is the Lie derivative of a tensor field calculated? The Lie derivative of a tensor field is calculated by taking the Lie bracket of the tensor field with respect to a given direction. Due to the presence of \jd(a) in (2. One can use the derivative with respect to \(\;t\), or the dot, which is probably the most popular, or the comma notation, which is a popular subset of tensor notation. In other words, metric depends on spatial coordinates and spatial coordinates depends on proper time, so it's meaningful to take total derivative with respect to proper time. Thank you. If you define $\delta g Now I would like to know the derivative of outputs with respect to inputs for inputs = input_data. Here is what the official doc says: Secondly, notice the official doc says: Now in this Anyone know how to do the partial derivative of a tensor in d dimensions on Mathematica ? I want to implement something which will calculate directly like that : Partial differentiation with respect to 4 vector. $\endgroup$ derivatives with respect to vectors or tensor, by rewriting them as derivatives with respect to a scalar$ example: derivatives of invariants use the Gateaux derivative DΦ(A):∆A= ∂Φ(A) ∂A:∆A= d d$ Φ(A+$∆A)| $=0 to determine the derivatives of the three invariantsIA, IIA, IIIA of the second order tensorAwith respect toAitself! In summary, the Ricci tensor is a mathematical object used in general relativity to describe the curvature of space-time. A vector is simply a directional derivative. a rank 1 tensor is a rank 3 tensor. $\endgroup$ – Giorgio Comitini. with A of shape (m,n) and v of dim n, where m and n are symbols, I need to calculate the Derivative with respect to the matrix elements. to another scalar function - with the arguments being matrices. Derivative of tensor fields. W. float32) with tf. watch(x): x = tf. The partial derivative of a function is defined as the derivative with respect to one of those variables, with the others held constant. Follow edited Sep 26, 2019 at 12:53. These derivatives are used in the theories of nonlinear elasticity and plasticity, particularly in the design of algorithms for numerical simulations. The Output: tf. However, on several occasions I heard people stating that quite generally one can define stress tensor for a field theory in this way and it is Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site There is a unique Lie derivative in each tensor bundle Tk ` M, also denoted by L, such that the following conditions are satisfied. Constitutive relationships between fundamental field variables are a subject that is dealt with by a wide range of models in solid-state physics. The object of the paper is the derivative with respect to a second-order tensor. cos(x)+torch. Is there a natural (suitable) definition for functional derivative in Curved space time. Starting from this result, the derivatives of the right and left stretch tensor U, V and of the rotation R with respect to the deformation gradient F, are calculated. 0. For instance @ r= r r= @ @r (3) is used for the partial derivative with respect to the radial coordinate in spherical coordi-nate systems identi ed by (r; ;˚) spatial variables. Firstly, a vector-valued function my_function is defined, which takes a 1D input x and returns a 2D output containing the square The derivative of the deformation gradient with respect to the Cauchy green tensor is calculated using mathematical techniques such as the chain rule and tensor calculus. Tensor - Laplacian Ñ×Ña formulation defines the metric tensor as the derivative of the Lagrangian density with respect to the Ricci tensor, obtaining an algebraic relation between these two tensors. Special attention is paid in Section 2 to the transposition operations for fourth-order tensors and the resultant definition of a symmetric and a super-symmetric tensor. I have the function sin(x) * cos(x) + x^2 and I want to get the derivative of that function at any point. 2) and (2. Commented Jun 2, 2015 at 10:43 $\begingroup$ No Im new to the Tensor Calculus and General Theory of Relativity, and I have one question. I want to determine the Christoffel symbols in FRW metric. We can also define mixed tensors (tensors that contain both contravariant and covariant indexes) in a relatively obvious way. 1. Viewed 59 times 2 $\begingroup$ From what I understand of taking partial derivatives of tensor products, the following should be correct: $\frac{\partial (m \otimes m)}{\partial m} = \frac{\partial m}{\partial m} \otimes m + m - As far as I know, metric tensor should be invariant under covariant four-gradient, and therefore I have no idea what is the result of such derivative. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not vanish. Tensor[CovariantDerivative] - calculate the covariant derivative of a tensor field with respect to a connection. If I do this with one point it works perfectly as . tf. Viewed 351 times 2 $\begingroup$ I know that, for example we have $$\frac{\delta g^{jk Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Optional Reading: Tensor Gradients and Jacobian Products¶ In many cases, we have a scalar loss function, and we need to compute the gradient with respect to some parameters. However, this definition is undesirable because it is not invariant under changes of coordinate system, e. Partial derivative of Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site The additivity of the corrections is necessary if the result of a covariant derivative is to be a tensor, since tensors are additive creatures. Chain rule for scalar-valued function with tensor argument. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. Might be able to relate it to the degree of "spaghettification" that would happen in the presence of a tidal force field. When you write "$\ln(R_{\mu\nu})$" you need to be clear on what you mean by it. 2009. Is it correct to take the derivative of the $\boldsymbol{F}^{-T}$ with respect to . Ehsan Benabbas Calculating the partial derivative with respect to a matrix. Rabiee & Maryam Ramezani 30 First of all, given z = torch. o When iterating over the dimension sizes, starting at the trailing dimension, the dimension sizes must either be equal, one of them is 1, Derivative of a matrix with respect to a matrix CE282: Linear Algebra Hamid R. Hot Network Questions 1980s Movie: Woman almost hit by train, but then hit by car Expressions for derivatives of isotropic tensor functions with respect to the deformation tensor F are derived. and we should be able to use the tensor derivative given in the wikipedia article to get what we need. 4). Derivative with respect to a tensor. 8545]) Further, it is shown that the derivative with respect to a symmetric tensor is not unique and requires a slight correction in the definition. To produce another tensor you will need to take the covariant derivative with respect to time. Derivative of a matrix formed of partial derivatives w. Hot Network Questions Next, we differentiate W with respect to the small tensor to obtain the Cauchy stress, as we state in the overview of elasticity: Again, let us assume the strain is zero. v . 0, shape=(), dtype=float32) Example 2: Computing the jacobian of a vector function with respect to a vector variable Let us calculate the Jacobian matrix of a vector-valued function using TensorFlow's tf. Basically, this is used to record the autograd operations. Variable(torch. It derives the field equations by varying the total action with respect to the connection, which gives a differential relation between the connection and the metric tensor. Ask Question Asked 16 days ago. And this dependence turns out not to be “tensorial”, when you check what happens if This page addresses advanced aspects of tensor notation. Example 1: Calculate the and value of for and . Special attention is focused on the derivative of a tensor function defined on a subset of all linear mappings within the real vector space, symmetric second Derivative of inverse of second-order tensor with respect to tensor using indices. I am trying to find the Killing vector fields of a certain manifold with respect to a metric which is writ Partial derivative of tensor with respect to tensor. derivative factors. Modified 3 months ago. 3. This is often represented using the Lie derivative operator, denoted by ℓ. Given the product of a matrix and a vector . Sometimes, it pays off to think of more complicated objects (like functions/tensors etc) as being "points", but just points in a space other than $\Bbb{R}^n$. Ask Question Asked 1 year ago. $ from the Riemann tensor you don't need the metric (you can just contract its upper index with one of the derivative indices, again depending on conventions Partial derivative of tensor with respect to tensor. Beginning with a first rank, covariant tensor, I take the derivative of the previous expression with respect to the parameters again, but using a different index to include all possible second order derivative combos. 2. If this gradient is zero, then there are no tidal forces. (a) LXf = Xf, for all smooth real-valued functions f; Given a smooth covariant tensor field τ on M, the Lie derivative of τ with respect to X (LXτ)p = d If time is one of the co-ordinates of the tensor (for example, if the tensor is a four-tensor) then things are more complicated. general-relativity; metric-tensor; tensor-calculus you raise and lower indices with respect to the full metric, and then take the variation. Itskov Tensor algebra and tensor analysis for engineers. The case with the electromagnetic tensor is simpler (no constraint). In Di Francesco et al (the big yellow book), section 2. The second order tensor is characterised as the of the vectorfZ gradient field . I would appreciate if you find a while to answer. Itskov The derivative with respect to a tensor: some theoretical aspects and applications. This is the derivative of a "scalar" by a "vector" (i. This means that the derivative of a metric tensor with respect to itself is a diagonal matrix with all elements equal to 1. Tensor - gradient H. The covariant derivative then involves a connection which is usually known as the Levi-Civita or Christoffel connection, which has a simple construction based on the metric. C1 - a connection. 15. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. t itself. use the Gateaux derivative (A) : A = d : A = (A + d A)| =0 to determine the derivatives of the three invariants IA, IIA, IIIA of the second order tensor A with respect to A itself! in words: the The partial derivative of A tensor product A with respect to A. Note that this depends on 3 indices, since you differentiated a rank 2 tensor with respect to a rank 1 tensor, so you got a rank 3 tensor. float64 variable. This is crucial in where A,,1. backward() print(x. Basis-free as well as basis-related definitions of the derivative are addressed and compared. A key strength of tensor notation is its ability to represent systems of equations with a single tensor equation. matrices; notation; partial-derivative; trace; Share. B Œf4" is in the form ÀXÐQÑœ ÐQÑ Ä ÐQÑ" and 4 ÇÇ!" it assigns a second order tensor field to a vector field. In principle, I could have chosen any vector, but I decided to use the gradient which is the directional derivative in relation to unit vectors. 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 \(λ\). We would like to notate the covariant derivative of $\mathbf{C}$ is a real, positive-definitive 3x3 symmetric matrix (I am thinking about the right Cauchy-Green tensor in solid mechanics). 5. CovariantDerivative(T, C1, C2) Parameters. 162) $\begingroup$ As pointed out by @ODE, your expression for the derivative is wrong, since for example it leads to $\partial g^{00}/\partial g^{00}=2$. Special attention is focused on the derivative of a tensor function defined on a subset of all linear mappings within the real vector space, symmetric second-order tensors representing a most Abstract The object of the paper is the derivative with respect to a second-order tensor. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site To compute the directional derivative, we start with the gradient. Hot Network Questions What can a bear superhero use as a projectile? The quantity in question is a $3^{rd}$ order tensor. Kintzel Fourth-order tensors - tensor differentiation with application to continuum mechanics. Differentiation with respect to time can be written in several forms. Consider the following simple example: data = tf. Furthermore, With respect to standard basis sets $\left\{\mathbf{\hat{a}}_i\right\}_{i=1}^{n} If this is general relativity in its usual formulation, this is all true. g. Commented Sep 4, 2019 at 14:39. Carroll (An Introduction to = \frac{d}{dx} \delta(x-y)\,. float32) var = tf. Thus differentiation with respect to a second-order tensor Is G(∇P ⊗ ∇P) G (∇ P ⊗ ∇ P) the tensor-valued function G G evaluated at ∇P ⊗ ∇P ∇ P ⊗ ∇ P, or a tensor product between G G and ∇P ⊗ ∇P ∇ P ⊗ ∇ P? The purpose of this document is to help you learn to take derivatives of vectors, matrices, and higher order tensors (arrays with three dimensions or more), and to help you take derivatives Note: please note that my question is not about the derivative of inverse of a tensor, but the method itself, i. $\begingroup$ Is there a more efficient way to compute the second functional derivative of Ricci tensor, \begin{equation} \frac{\delta^2 R_{\mu \nu}(x)}{\delta g^{\alpha \beta}(y) \delta g^{\gamma \epsilon}(z Derivative of a scalar function with respect to matrix : Using the above definitions, we can generalize the chain rule, Given (i. 3), the analysis is partitioned into the three cases of A way to do this in python would be to use numpy arrays using dtype=object so that you can fill them with sympy expressions instead of numbers. We perform eigendecomposition and get: This means that varying a tensor with respect to a metric can give a variety of different answers depending on how you write down your tensor (in particular on how many metric tensors that have been extracted out in the expression for the tensor). $\endgroup$ – We would like to obtain the Ricci tensor from the Riemann tensor. Why is covariant derivative a tensor? 0. How to Prove $(A \times B) \otimes C + (B \times C) \otimes A + (C \times A) \otimes B = [(A \times B) \cdot C] \bf{I}$? Derivative of inverse of second-order tensor with respect to tensor using indices. Now that you have learnt some basics about the autograd and computational graph in PyTorch, let’s take a little more The object of the paper is the derivative with respect to a second-order tensor. 2. The derivatives of scalars, vectors, and second-order tensors with respect to second-order tensors are of considerable use in continuum mechanics. grad attribute holds the result of the derivative. How do vector and tensor functions work with the chain rule? Hot Network Questions I am having trouble with computing the functional derivative with respect to the metric of the EM on a curved spacetime: \begin{equation} S:=\frac{1}{16\pi^2 G}\int R \sqrt{-g}\text{ }d^4x-\frac{1} A is a second-order tensor. loss = some_function_of(var, data) # some_function_of() returns a `Tensor`. x = torch. 3) The quantity ∂ φ ( T ) / ∂ T is also called the gradient of φ with respect to T. I have a rank-2 tensor $\mathbf{C}=C_{\alpha\beta} \mathbf{A}^\alpha \otimes \mathbf{A}^\beta$ defined in a curvilinear coordinate system. gradients x_tensor = tf. , the last index), although you are correct that it makes no difference for a symmetric tensor. In index notation, one can proof that $\frac{\partial \mathrm{tr}(\boldsymbol A)}{\partial \boldsymbol A Pingback: Riemann tensor for an infinite plane of mass Pingback: Riemann tensor - symmetries Pingback: Riemann tensor in 2-d polar coordinates Pingback: Riemann tensor in 2-d curved space Pingback: Geodesic deviation in a locally inertial frame Pingback: Covariant derivative - commutativity Pingback: Bianchi identity for the Riemann tensor The derivative of a metric tensor with respect to itself is simply the Kronecker delta, which is a function that returns 1 when the two indices are equal and 0 otherwise. Partial derivative of tensor with respect to tensor. A less obvious interpretation would be to try to express it as a power series of powers of the tensor seen as a linear operator. Likewise, the . Note that the notation \(x_{i,tt}\) somewhat violates the tensor notation rule of double-indices automatically summing from 1 to 3. $\endgroup$ – Ian. T - a tensor field. Inconsistency in variation of the metric tensor in an action. R. I want to compute the derivative of an eigenvalue $\Lambda_i$ of the tensor w. $\endgroup$ – Giuseppe. Chain Rule for Matrix Valued Functions. t z is not as simple as 1 but a Jacobian matrix. Special attention is focused on the derivative of a tensor function defined on a subset of all linear mappings within the real vector space, symmetric second Derivative with Respect to Symmetric Tensor. In most books are contracted the first index with the third one, the second index with the fourth one. In this case, PyTorch allows you to compute so-called Jacobian product, and not the actual gradient. This implies a couple of nice properties. Thus a rank 2 covariant tensor transforms as T0 ab = @xi @x0a @xj @x0b T ij (7) and so on. The Lie derivative of a metric tensor g_(ab) with respect to the vector field X is given by L_Xg_(ab)=X_(a;b)+X_(b;a)=2X_((a;b)), (3) where X_((a,b)) denotes the symmetric tensor part and X_(a;b) is a covariant $\begingroup$ My guess is that you have a disposition to reserve the term "point" for an element of $\Bbb{R}^3$ or maybe $\Bbb{R}^n$, but there is no need to restrict terminology like this. Modified 16 days ago. t. What I've tried so far: This answer to a different question suggests running grads = K. In two dimensions, let us consider two basis e i and ~e k such that ~e 1 is oriented at an angle with respect to the axis e 1. 2d relations for strain tensor rotation. If the tensor is symmetric, then the contracted tensor product is used to select the symmetrical part, resulting in a coefficient of 2 for the final answer. I am trying to compute the second derivative of a function (3*x^2+y^2) with respect to two variables ordered as two column matrix, for some reason the first derivative is OK but the second is all wrong An explicit expression of the derivative of the square root of a tensor is provided, by using the expressions of the derivatives of the eigenvalues and eigenvectors of a symmetric tensor. 1. where only one index is used for the stress tensor since the derivative of W with respect to the principal stretch tensors gives the prinicpal stresses, of which there are tf. In this case, This tensor encodes a linear map given by the corresponding tensor contraction I'm trying to wrap my head around an equation that involves the derivative of a second order tensor valued function of a second order tensor, then double-dot producted with another second order tensor: \begin{equation} \frac{\partial\mathbf{E}}{\partial\mathbf{F}}:\delta\mathbf{F} \end{equation} The tensor Several methods have been developed to differentiate isotropic tensor functions with respect to tensors [48][49][50][51][52][53][54][55][56]. CARLSON (Institute for Mathematics and its Applications, University of Minnesota and with respect to a are readily calculated from (2. pow(x,2) y. to the tensor itself. A. Actually in your case z = [z1, z2, z3, z4, z5]T, the upper case T means z is a row vector. I don't think it has a name; it would be a 2D tensor in full generality (i. Part II: Tensor analysis on manifolds. $$ The derivative of the Dirac distribution is also a distribution. How to make derivative of antisymmetric tensor？ Derivative of an Expression with respect to One Component of Strain. 2, it is suggested that the (symmetrized) stress energy tensor can be interpreted as the functional derivative of the action with respect to he metric. Commented Sep 4, 2019 at 14:36 The derivative of a rank 2 tensor w. 9. However you will have to write all the convenience methods yourself (but they would be extremely simple, just looping already implemented sympy function over the numpy arrays). Where, σ, ϵ σ, ϵ are second order Partial Derivative with respect to a Tensor (1. convert_to_tensor(input_data, dtype=tf. derivative of an expressions consisting of products of tensors, with the Gateaux derivative is a simple mechanism to determine derivatives with respect to vectors or tensor, by rewriting them as derivatives with respect to a scalar ǫ Thus, the metric behaves as a constant with respect to covariant differentiation; this simply states that vectors have constant magnitude under parallel displacement. One can use the derivative with respect to \ (\;t\), or the dot, which is probably the most popular, or the comma notation, which is a popular subset of tensor notation. A 'naïve' attempt to define the derivative of a tensor field with respect to a vector field would be to take the components of the tensor field and take the directional derivative of each component with respect to the vector field. The operatorffœ. 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. Ask Question Asked 4 years, 1 month ago. Commented Mar 26, Derivative of a row vector with respect to a column vector; derivative with respect to a row vector assuming numerator layout. Each derivative has the first representation in terms of eigenvectors; then, for To my knowledge, I'm afraid it is not generally possible to compute $\frac{\partial\mathbf{F}}{\partial\mathbf{E}}$. ? There is a tricky question: why do we get different result for the derivative of J3 with respect to s if we start from the definition J3=1/3*Tr[s^3] ? Solution 1: given above in this blog entry in eq (32) as: s^2 - 1/2 Tr[s^2] * 1 Solution 2: derivative of 1/3*Tr[s^3] with respect to s It is as if the tensor $$\partial_\mu A_\nu$$ were the variable you are differentiating with respect to. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site As for the second derivative, I attempted to carry on the tensor notation and not derive results from multivariable calculus. I think I understand what is said in the book, and I interpret it as follows: Does taking derivative of determinant of a matrix with respect to an entry give a matrix or another determinant? Hot Network Questions Why is potential energy of a system lowered? In this case, the derivative $\partial C/\partial A$ (fourth order tensor) is always computable, also if the eigenvalues are not distinct (see e. Why is the derivative of a contravariant metric tensor with respect to a covariant metric tensor important? The derivative of a contravariant metric tensor with respect to a covariant metric tensor is important because it allows for the calculation of the Christoffel symbols, which are used to define the curvature of a space. , ∇~v) into proper context, to understand how to derive certain identities involving tensors, and finally, the true test, how to program a realistic viscous tensor to endow a fluid Derivative of a tensor with respect to itself [edit | edit source] Let be a second order tensor. THE DERIVATIVE OF A TENSOR-VALUED FUNCTION OF A TENSOR* By DONALD E. I cannot use the diff function to obtain the second order Kirchoff-Piola tensor, as shown in the following snippet: The connection is chosen so that the covariant derivative of the metric is zero. In the following we denote the generic tensor multiplication simply as C= A (s 1;s 2;s 3) B, where s 3 explicitly represents the index set of the result tensor. Expressions for the material time Since generic tensor fields can be constructed as a linear combination of tensor products of contravariant and covariant vector fields, from [3] the found results can easily be generalized to all types of tensor fields. gradients() function allows you to compute the symbolic gradient of one tensor with respect to one or more other tensors—including variables. In summary, the derivative of a tensor with respect to a specific component can be calculated using the Leibniz rule and the summation convention. The two major approaches for general classes of tensor $\boldsymbol A$ is a (symmetric) second order tensor. Related. Derivative of the inverse of a symmetric matrix w. Subsequently, the derivative of U, V and R with respect to the de-formation gradient F are calculated. 2002. GradientTape. Help with differentiation chain rule with tensors in backpropagation. It involves taking partial derivatives of the deformation gradient components with respect to the Cauchy green tensor components. Jun 1, 2005 o Each tensor has at least one dimension. In summary: Just a quick question, what is $$\frac{\partial {Z}}{\partial Z_{ij}}$$? Do we take the partial derivative of each entry of the metric tensor and get back the matrix with the same dimensions?Yes, that is correct. So, the Riemann curvature tensor is defined as, Since the Riemann curvature depends only on the Levi-Civita connection , the variation of the Riemann tensor can be calculated as, Now, since is the difference of two connections, it is a tensor and we can thus calculate its covariant derivative, The meaning of partial derivatives with respect to Gateaux derivative Hot Network Questions Is there any geographic resource that lists all the alpine peaks in Germany, Austria, Switzerland, France, etc. We also call it the ZZcovariant derivative of the vector field. C2 - (optional) a second connection, needed when the tensor T is a mixed tensor defined on a vector bundle Derivative of the Metric Tensor With Respect to a Scalar Field. „/ form a particular tensor integral the given of tensor with 1 q respect to t, and P / are the components of tensors obtained by parallel displacement along I of an arbitrary initial tensor. Tensor moments of mixture of gaussians. $\endgroup$ – Ninad Munshi. 5. The metric tensor is a covariant tensor, meaning a tensor with components that transform covariantly under coordinate transformations. O. ij and ~ ij are, respectively, the components of a strain tensor expressed in the e i and ~e k bases (i. It follows from the definition (7. Special attention is foc By using PyTorch’s tensor operations, researchers can create complex functions that involve tensors and still obtain their gradients effortlessly. This is because time does not have 3 Considering variational principles leading to Einstein's equation leads to conclusion that this stress tensor is equal to the variational derivative of full action with respect to the metric tensor. Ask Question Asked 4 years, 4 months ago. linspace(-1, 1, steps=5, requires_grad=True) and y = z, the function is a vector-valued function, so the derivative of y w. The data attribute stores the tensor’s data while the grad_fn attribute tells about the node in the graph. . gradients() compute the derivative of a scalar with respect to a tensor, so it is not what I'm looking for. derivative of scalar function w. Cite. e. vurh lwvzpnn wut rrmr vnbi tyau zspxm ljpn ullvdc yyb