# antisymmetric stress tensor

2 i , i The problem I'm facing is that how will I create a tensor of rank 2 with just one vector. σ However, elastic stress is due to the amount of deformation (strain), while viscous stress is due to the rate of change of deformation over time (strain rate). = . m {\displaystyle |1/{\sqrt {3}}|} {\displaystyle n_{i}n_{i}=1} A first approach to solve these last three equations is to consider the trivial solution By definition the stress vector is . {\displaystyle \mathbf {T} (\mathbf {n} ,\mathbf {x} ,t)} Many material properties and fields used in physics and engineering can be represented as symmetric tensor fields; for example: stress , strain , and anisotropic conductivity . i 0 But WP claims that the symmetry of the stress tensor need only hold in the case of equilibrium: "However, in the presence of couple-stresses, i.e. The index subset must generally either be all covariant or all contravariant. I m M n {\displaystyle p} Depending on the orientation of the plane under consideration, the stress vector may not necessarily be perpendicular to that plane, i.e. 1 ferromagnetic fluids which can suffer torque by external magnetic fields). {\displaystyle d\mathbf {F} /dS} {\displaystyle n_{i}n_{i}=1} {\displaystyle \mathbf {n} } = From an xi-system to an xi' -system, the components Ïij in the initial system are transformed into the components Ïij' in the new system according to the tensor transformation rule (Figure 2.4): where A is a rotation matrix with components aij. , and {\displaystyle d\mathbf {F} /dS} Δ σ i In three dimensions, it has three components. ∘ The viscous stress tensor is formally similar to the elastic stress tensor (Cauchy tensor) that describes internal forces in an elastic material due to its deformation. n {\displaystyle \lambda } 2 For example, holds when the tensor is antisymmetric on it first three indices. j → The characteristic equation has three real roots n We will consider only the symmetric part of the stress tensor so only 6 of these components are independent. use to denote eigenvalues). 1 n Octahedral plane passing through the origin is known as the Ï-plane (Ï not to be confused with mean stress denoted by Ï in above section) . , π ) j τ ≥ {\displaystyle \sigma _{2}} {\displaystyle \tau _{\text{n}}} and {\displaystyle J_{1}} For a completely fluid material, the elastic term reduces to the hydrostatic pressure. In a fluid, elastic stress can be attributed to the increase or decrease in the mean spacing of the particles, that affects their collision or interaction rate and hence the transfer of momentum across the fluid; it is therefore related to the microscopic thermal random component of the particles' motion, and manifests itself as an isotropic hydrostatic pressure stress. This is a constrained maximization problem, which can be solved using the Lagrangian multiplier technique to convert the problem into an unconstrained optimization problem. n 1 In a Newtonian fluid, by definition, the relation between ε and E is perfectly linear, and the viscosity tensor μ is independent of the state of motion or stress in the fluid. P i , called the traction vector, defined on the surface {\displaystyle \lambda } A second set of solutions is obtained by assuming where τ . There are certain invariants associated with the stress tensor, whose values do not depend upon the coordinate system chosen, or the area element upon which the stress tensor operates. These deviatoric stress invariants can be expressed as a function of the components of oct Examples. = The {\displaystyle \sigma _{\text{oct}}} ( For each eigenvalue, there is a non-trivial solution for x 2 σ applied to a body or to a portion of the body can be expressed as: Only surface forces will be discussed in this article as they are relevant to the Cauchy stress tensor. Let dF be the infinitesimal force due to viscous stress that is applied across that surface element to the material on the side opposite to dA. {\displaystyle n_{3}\neq 0} It follows that for an antisymmetric tensor all diagonal components must be zero (for example, b11 = −b11 ⇒ b11 = 0). Both tensors map the normal vector of a surface element to the density and direction of the stress acting on that surface element. i 1 j ) 2 For example, a vector is a simple tensor of rank one. F , is the position vector and is expressed as, Knowing that 3 , the stress deviator tensor is in a state of pure shear. {\displaystyle \tau _{\text{n}}^{2}} {\displaystyle P} 1 is the divergence operator, ) Their values are the same (invariant) regardless of the orientation of the coordinate system chosen. and Concatenate them into a 4-vector $\vec{A}$. face of the cube has three components of stress so there are 9 possible components of the stress tensor. 2 According to the principle of conservation of linear momentum, if the continuum body is in static equilibrium it can be demonstrated that the components of the Cauchy stress tensor in every material point in the body satisfy the equilibrium equations (Cauchy's equations of motion for zero acceleration). 3 {\displaystyle n_{3}}, By multiplying the first three equations by 1 n Knowing that 0 Any symmetric tensor can be decomposed into a linear combination of rank-1 tensors, each of them being symmetric or not. n {\displaystyle \mathbf {F} } components in the direction of the three coordinate axes. = , and subject to the condition that. This also is the case when the Knudsen number is close to one, / The extreme values of these functions are, These three equations together with the condition S , Δ This implies that the balancing action of internal contact forces generates a contact force density or Cauchy traction field [5] I'll use the notations in your answer. However, the stress tensor itself is a physical quantity and as such, it is independent of the coordinate system chosen to represent it. i j Ignoring the torque on an element due to the flow ("extrinsic" torque), the viscous "intrinsic" torque per unit volume on a fluid element is written (as an antisymmetric tensor) as. In Cartesian coordinates, ∇v is the Jacobian matrix, Either way, the strain rate tensor E(p, t) expresses the rate at which the mean velocity changes in the medium as one moves away from the point p – except for the changes due to rotation of the medium about p as a rigid body, which do not change the relative distances of the particles and only contribute to the rotational part of the viscous stress via the rotation of the individual particles themselves. {\displaystyle S} λ ( ≥ passing through an internal material point A tensor bij is antisymmetric if bij = −bji. For example, for a hydrostatic fluid in equilibrium conditions, the stress tensor takes on the form: where Unlike the ordinary hydrostatic pressure, it may appear only while the strain is changing, acting to oppose the change; and it can be negative. {\displaystyle I_{1}} n In particular, the local strain rate E(p, t) is the only property of the velocity flow that directly affects the viscous stress ε(p, t) at a given point. Thus the zero-trace part εs of ε is the familiar viscous shear stress that is associated to progressive shearing deformation. = 1 σ 3 {\displaystyle K_{n}\rightarrow 1} {\displaystyle P} τ [1] [2] The index subset must generally either be all covariant or all contravariant . j , In particular, the contact force is given by. The Mohr circle for stress is a graphical representation of this transformation of stresses. {\displaystyle P} S {\displaystyle I_{2}} {\displaystyle s_{2}} {\displaystyle T_{i}^{(n)}=\sigma _{ij}n_{j}} , called the first, second, and third stress invariants, respectively, always have the same value regardless of the coordinate system's orientation. = A tensor aij is symmetric if aij = aji. holds when the tensor is antisymmetric on it first three indices. , or alternatively, as a function of n τ and verifies that there are no shear stresses on planes normal to the principal directions of stress, as shown previously. F The Kronecker . having direction cosines equal to The strain rate tensor E(p, t) is symmetric by definition, so it has only six linearly independent elements. {\textstyle s_{ij}={\frac {1}{3}}I} / Δ Therefore, the scalar part εv of ε is a stress that may be observed when the material is being compressed or expanded at the same rate in all directions. Thus. {\displaystyle \left(\sigma _{ij}-\lambda \delta _{ij}\right)n_{j}=0} P , or the continuum is a non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such as polymers. u is the mean surface traction. In a Newtonian medium, specifically, the viscous stress and the strain rate are related by the viscosity tensor μ: The viscosity coefficient μ is a property of a Newtonian material that, by definition, does not depend otherwise on v or σ. Active 3 years, 3 months ago. In principle the integrand in the volume integral in can be a complete divergence of a tensor of rank three which is antisymmetric in the first two indices.This tensor … σ 1 This part of the viscous stress, usually called bulk viscosity or volume viscosity, is often important in viscoelastic materials, and is responsible for the attenuation of pressure waves in the medium. {\displaystyle {\vec {u}}} n λ Expansion of an anti-symmetric tensor with a symmetric tensor 1 What is the proof of “a second order anti-symmetric tensor remains anti-symmetric in any coordinate system”? T The tetrahedron is formed by slicing the infinitesimal element along an arbitrary plane with unit normal n. The stress vector on this plane is denoted by T(n). . 2 J k 3 s {\displaystyle \mathbf {n} } In an arbitrary coordinate system, the viscous stress ε and the strain rate E at a specific point and time can be represented by 3 × 3 matrices of real numbers. F = The same can be said of the strain rate tensor E as a representation of the velocity pattern around p. Thus, the linear models represented by the tensors E and ε are almost always sufficient to describe the viscous stress and the strain rate around a point, for the purpose of modelling its dynamics. 3 i = {\displaystyle s_{ij}} n oct This value is the same in all eight octahedral planes. {\displaystyle S} ( j {\displaystyle \mathbf {n} } 1 , j and body forces / , {\displaystyle \Delta \mathbf {F} /\Delta S} {\displaystyle \lambda } → In continuum mechanics, the Cauchy stress tensor, true stress tensor, or simply called the stress tensor is a second order tensor named after Augustin-Louis Cauchy.The tensor consists of nine components that completely define the state of stress at a point inside a material in the deformed state, placement, or configuration. 1 j λ In matrix form this is, Expanding the matrix operation, and simplifying terms using the symmetry of the stress tensor, gives. Viewed 541 times 2 … → and assumed to depend continuously on the surface's unit vector r . σ {\displaystyle \sigma _{ij}} {\displaystyle \mathbf {r} } τ λ i j n 3 2 In a solid material, the elastic component of the stress can be ascribed to the deformation of the bonds between the atoms and molecules of the material, and may include shear stresses. We start with something more basic: a deformation tensor, $e$. j n {\displaystyle n_{i}=\delta _{ij}n_{j}} σ {\displaystyle S} T {\displaystyle \tau _{\mathrm {n} }} is the k:th Cartesian component of exerted at point P and surface moment = j λ j k 0 Symmetric Stress-Energy Tensor We noticed that Noether’s conserved currents are arbitrary up to the addition of a divergence-less field. or its principal values becomes n , In continuum mechanics, the Cauchy stress tensor ) which in turn is the relative rate of change of volume of the fluid due to the flow. ( n x In general, a linear relationship between two second-order tensors is a fourth-order tensor. The space-space components of the stress-energy tensor are interpreted as the 3x3 stress tensor. In mathematics and theoretical physics, a tensor is antisymmetric on (or with respect to) an index subset if it alternates sign (+/−) when any two indices of the subset are interchanged. becomes very small and tends to zero the ratio σ ≥ The EulerâCauchy stress principle states that upon any surface (real or imaginary) that divides the body, the action of one part of the body on the other is equivalent (equipollent) to the system of distributed forces and couples on the surface dividing the body,[2] and it is represented by a field λ {\displaystyle n_{j}} , n n k n n . is defined as the surface traction,[7] also called stress vector,[8] traction,[4] or traction vector. n σ Δ n {\displaystyle n_{3}} / , are the principal stresses, functions of the eigenvalues and represents the rate of change of intrinsic angular momentum density with time. The stress tensor The zero-trace part Es of E is a symmetric 3 × 3 tensor that describes the rate at which the medium is being deformed by shearing, ignoring any changes in its volume. ;X1 and acts along the 2nd axis i.e.;X2). σ 2 2 . λ = , with normal vector + The equilibrium of forces, i.e. at ) is called an octahedral plane. {\displaystyle P} , respectively, and knowing that σ {\displaystyle \mathbf {n} } The normal and shear components of the stress tensor on these planes are called octahedral normal stress τ we first add these two equations, Knowing that for The coefficient μv, often denoted by η, is called the coefficient of bulk viscosity (or "second viscosity"); while μs is the coefficient of common (shear) viscosity. k 1 σ [6] given by Thus, Expanding the determinant leads to the characteristic equation. For most general cases, stress tensor need not be symmetric in fluid mechanics. T n are the unknowns. , then from the original equation for The antisymmetric part of the tensor represents a torque. δ σ The Cauchy stress tensor is used for stress analysis of material bodies experiencing small deformations: It is a central concept in the linear theory of elasticity. i σ {\displaystyle \tau _{\text{n}}^{2}} , The viscous component of the stress, on the other hand, arises from the macroscopic mean velocity of the particles. Yes, these tensors are always symmetric, by definition. For a general vector x = (x 1,x 2,x 3) we shall refer to x i, the ith component of x. For the particular case of a surface with normal unit vector oriented in the direction of the x1-axis, denote the normal stress by Ï11, and the two shear stresses as Ï12 and Ï13: The nine components σij of the stress vectors are the components of a second-order Cartesian tensor called the Cauchy stress tensor, which completely defines the state of stress at a point and is given by. ) is generally defined as negative one-third the trace of the stress tensor minus any stress the divergence of the velocity contributes with, i.e. Their direction vectors are the principal directions or eigenvectors. λ 3 n 1 moments per unit volume, the stress tensor is non-symmetric. 3 , is the k:th Cartesian coordinate, . and 1 ) Assuming a material element (Figure 2.3) with planes perpendicular to the coordinate axes of a Cartesian coordinate system, the stress vectors associated with each of the element planes, i.e. If the fluid is isotropic as well as Newtonian, the viscosity tensor μ will have only three independent real parameters: a bulk viscosity coefficient, that defines the resistance of the medium to gradual uniform compression; a dynamic viscosity coefficient that expresses its resistance to gradual shearing, and a rotational viscosity coefficient which results from a coupling between the fluid flow and the rotation of the individual particles. , we have. n A model of a continuous medium, having an antisymmetric stress tensor, was suggested in [1]. 1 is the hydrostatic pressure, and Considering the solution where max {\displaystyle J_{2}} {\displaystyle \sigma _{ij}} {\displaystyle \sigma _{2}=I_{1}-\sigma _{1}-\sigma _{3}} ∂   τ j {\displaystyle n_{1}=n_{2}=0} = Teodor M. Atanackovic and ArdÃ©shir Guran (2000). k ( = In viscoelastic materials, whose behavior is intermediate between those of liquids and solids, the total stress tensor comprises both viscous and elastic ("static") components. I n {\displaystyle \sigma _{1}=\max \left(\lambda _{1},\lambda _{2},\lambda _{3}\right)} i i σ {\displaystyle \Delta \mathbf {M} } λ The linear transformation which transforms every tensor into itself is called the identity tensor. • Change of Basis Tensors • Symmetric and Skew-symmetric tensors • Axial vectors • Spherical and Deviatoric tensors • Positive Definite tensors . {\displaystyle \sigma _{1}} In most fluids the viscous stress tensor too is symmetric, which further reduces the number of viscosity parameters to 6 × 6 = 36. a symmetric sum of outer product of vectors. n and {\displaystyle I_{2}} F Let’s take strain as an example. = n j n At every point in a stressed body there are at least three planes, called principal planes, with normal vectors e In non-Newtonian fluids, on the other hand, the relation between ε and E can be extremely non-linear, and ε may even depend on other features of the flow besides E. Internal mechanical stresses in a continuous medium are generally related to deformation of the material from some "relaxed" (unstressed) state. n 1 {\displaystyle \lambda _{i}} and having the same normal vector F moments per unit volume, the stress tensor is non-symmetric. The total stress energy tensor of all matter elds is conserved, i.e. σ n p Pressure ( there is no net creation or destruction of overal 4-momentum r T (total) = 0 : However, as we saw in the case of a swarm of particles, the stress-energy tensor of any particular species sis not necessarily conserved: r T ( s) = X s06=s F 0! , n For large deformations, also called finite deformations, other measures of stress are required, such as the PiolaâKirchhoff stress tensor, the Biot stress tensor, and the Kirchhoff stress tensor. In coordinate form. Therefore, the viscosity tensor μ has only 6 × 9 = 54 degrees of freedom rather than 81. 1 , then, Using the Gauss's divergence theorem to convert a surface integral to a volume integral gives, For an arbitrary volume the integral vanishes, and we have the equilibrium equations. Δ In any chosen coordinate system with axes numbered 1, 2, 3, this viscous stress tensor can be represented as a 3 × 3 matrix of real numbers: Note that these numbers usually change with the point p and time t. Consider an infinitesimal flat surface element centered on the point p, represented by a vector dA whose length is the area of the element and whose direction is perpendicular to it. s ) The antisymmetric second-rank tensor being referenced is the electromagnetic field tensor. Like the total and elastic stresses, the viscous stress around a certain point in the material, at any time, can be modeled by a stress tensor, a linear relationship between the normal direction vector of an ideal plane through the point and the local stress density on that plane at that point. K kil: Antisymmetric tensors are also calledskewsymmetricoralternatingtensors. T {\displaystyle \lambda ,n_{1},n_{2},} s {\displaystyle \tau _{\text{oct}}} The Lagrangian function for this problem can be written as. min σ parallel to . Find the second order antisymmetric tensor associated with it. External forces can result in an asymmetric component to the stress tensor (e.g. The eigenvalues are the roots of the characteristic polynomial. The stress tensor of a viscous fluid is in the general case antisymmetric. i 2 i n n n 0 and it can be stated as being equal to one-half the difference between the largest and smallest principal stresses, acting on the plane that bisects the angle between the directions of the largest and smallest principal stresses. F i The Cauchy stress tensor obeys the tensor transformation law under a change in the system of coordinates. where In mathematics and theoretical physics, a tensor is antisymmetric on (or with respect to) an index subset if it alternates sign (+/−) when any two indices of the subset are interchanged. , representing a maximum for T ) ( is the kronecker delta. My question is - Is there a choice of $\psi^\rho$ such that the corresponding "canonical stress-tensor" is symmetric. Some include, the metric tensor, , the Einstein tensor, and the Ricci tensor, . 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