Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stress tensor basics

  1. Feb 6, 2010 #1
    I'm trying to understand the meaning of the components of the stress(-energy) tensor. Considering just the space-space components, is this right?

    [tex]T^{ij} = \frac{\mathrm{d} F^i}{\mathrm{d} A^j},[/tex]

    where [itex]F^i[/itex] is the [itex]x^i[/tex]-component of force, [itex]i \neq 0[/itex], and [itex]\mathrm{d} A^j[/itex] is an area element perpendicular to the basis vector [itex]\textbf{e}_j, j \neq 0[/itex]. And does this mean that shear stress only exists in a non-Cartesian (non-Lorentzian) frame, i.e. that in a Cartesian (Lorentzian) frame, all elements off the main diagonal of a matrix representation of this tensor, in the space-space part, will be zero?
  2. jcsd
  3. Feb 7, 2010 #2
    I don't see why the nondiagonal components of this quantity should be zero in an intertial frame with a orthonormal coordinate system. The force acting on any of the infinitesimal surface elements can point in any direction.

  4. Feb 7, 2010 #3
    If [itex]T^{ij}[/itex] is the "i-component of force per unit area across a surface perpendicular to [itex]\textbf{e}_j[/itex]" or, equivalently, "component of momentum that crosses a unit area which is perpendicular to [itex]\textbf{e}_k[/itex], per unit time, with the crossing being from -[itex]x^j[/itex] to +[itex]x^j[/itex]" (Blandford and Thorne: Applications of Classical Physics, 1.89), where [itex]\textbf{e}_j[/itex] indicates the jth basis vector, in what sense does a force perpendicular to a surface "cross" that surface?

    Am I taking "cross" too literally? It suggests to me an arrow drawn from one side of the surface to the other. If I picture an arrow lying parallel to the surface, the arrow doesn't cross the surface at all.
    Last edited: Feb 7, 2010
  5. Feb 7, 2010 #4

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    For me, a convenient model to represent a tensor is the surface of a cube. Each face has three directions associated with it: the surface normal (which is a diagonal component) and two directions which lie within the face: the off-diagonal components.

    The tensor is independent of coordinate frame.
  6. Feb 7, 2010 #5
    Yeah, that formulation seems a bit suspect to me... I would have rather said something like the amount of force in direction i that acts on a surface element with a normal in direction j. Check out [tex]\sigma[/tex] and the figure on the right here:

  7. Feb 8, 2010 #6
    Thanks for your answers. How's this sound? Given the net force [itex]\textbf{F}[/itex] at a point, and surfaces, [itex]A_i[/itex], normal to each basis vector, [itex]\textbf{e}_i[/itex], we subtract in turn from [itex]\textbf{F}[/itex] all forces caused/exerted by anything on the positive side of [itex]A_i[/itex], and take the limit of force wrt an area of that orientation as the area goes to zero, [itex]\frac{\mathrm{d}\textbf{F}}{\mathrm{d}A_i}[/itex]. The resulting vectors are the stress vectors, also called the traction vectors, each acting on the corresponding [itex]\mathrm{d}A_i[/itex], represented by the blue arrows on the cube in this section:


    These are the rows of the matrix representation of the stress tensor discussed in this section of the Wikipedia article, and the columns of Blandford and Thorne's stress tensor, the tensor being symmetric so that its matrix representation is its own transpose. Whereas Blandford and Thorne right-multiply their stress tensor matrix by a unit column vector representing an arbitrary surface, the Wikipedia version has to be left-multiplied by a unit row vector representing the surface. The resulting column (row) vector is the stress vector on the specified surface, i.e. force per unit area acting on the surface, exerted by whatever is on the other side of the surface.
  8. Feb 14, 2010 #7
    Schutz: "[itex]T^{ij}[/itex] = flux of i momentum across j surface" (A First Course in General Relativity, 1st ed.. p. 98, 4.18).

    So, given the definition of flux across a surface in terms of the dot product, why not

    [tex]T^{ij} \hat{\textbf{i}}_i \otimes \hat{\textbf{i}}_j = \lim_{A\rightarrow 0}\frac{F^i}{A}\hat{\mathbf{i}}_i \cdot A^j\hat{\textbf{i}}_j = \lim_{A\rightarrow 0}\frac{F^i}{A}\,A^j\,\delta_{ij}?[/tex]

    Where [itex]A^j = 1[/itex], a unit surface, and the indices i and j (Roman letters) stand for spatial values only: x,y,z, not t. Einstein summation convention in play. The indices on the left stand for one arbitrary particular value of i and j, rather than being abstract (slot-naming) indices indicating a collection of numbers.
  9. Feb 14, 2010 #8
    I believe that there is a common misconception here. Consider two neighbouring cubic fluid elements, contacting each other on a face of constant value of z. Ie. the contact surface is parallel to the x and y directions.

    Now, say that element 1 is loosing momentum in the x-direction. It is possible that element 2 is responsible for this. I.e. element 2 is stealing x-momentum from element 1, even though their contact area is parallel to the x-axis. So one says that there is a flow of x-momentum from element 1 to element 2. This flow goes through the contact surface, which is parallel to the x-axis. The actual change in momentum experienced by element 1 is in the x-direction, but the flow of this momentum is actually in the z-direction, ie. into fluid element 2.

    You are not supposed to project the x-momentum onto the normal vector for the contact surface. That gives you zero, and only tells you that the relevant force is acting in the same plane as the contact area between the fluid elements.

    How does this momentum transfer take place? Simply because there is a force acting between them that is parallel to the contact surface, ie. a shearing force.

    So I think it is dangerous to say talk about the stress tensor as a "momentum flux through a certain surface", because that is easy to misunderstand.

    I just looked it up in Landau-Lifschitz "Fluid Mechanics" (ch.1, par.7), and believe their description is consistent with what I said here.

  10. Feb 14, 2010 #9
    Thanks, torquil. From what you say, I think what I might be doing here is confusing x-momentum (the x-component of a momentum vector), [itex]kg \cdot m \cdot s^{-1}[/itex], with the x-component of "momentum flux density" = "stress vector (=traction vector) acting on a surface of constant x", [itex]N \cdot m^{-2}[/itex].

    Are the following definitions right?

    [tex]\text{energy density:- }\lim_{V\rightarrow 0}\sum_{i=1}^{n}\frac{E_i}{V}[/tex]

    [tex]\text{momentum density:- }\lim_{V\rightarrow 0}\sum_{i=1}^{n}\frac{\textbf{p}_i}{V}[/tex]

    where [itex]V[/itex] is volume, [itex]A[/itex] area (do the shapes of the volume are area elements matter when this limit is taken?), [itex]E_i[/itex] is energy of the nth particle, [itex]\mathbf{p}_i[/itex] the momentum of the ith particle.

    [tex]\text{energy flux density:- }\lim_{A,t\rightarrow 0}\frac{\Delta_j \left(\sum_{i=1}^{n}{E_i}\right)}{A\,t}[/tex]

    where [itex]\Delta_j \left(\sum_{i=1}^{n}{E_i}\right)[/itex] is the total energy lost on the negative side of the jth coordinate surface and gained on the positive side, and t is the time over which this change in the location of energy happens.

    [tex]\text{momentum flux density, i.e. (vectorial) stress, i.e. traction:- }[/tex]

    [tex]\lim_{A,t\rightarrow 0}\frac{\Delta_j \left(\sum_{i=1}^{n}{\textbf{p}_i}\right)}{A\,t} = \lim_{A\rightarrow 0}\frac{\textbf{F}}{A}[/tex]

    where [itex]\Delta_j \left(\sum_{i=1}^{n}{\textbf{p}_i}\right)[/itex] is the momentum lost of the negative side of the jth coordinate surface and gained on the positive side over time t, and [itex]\mathbf{F}[/itex] the force exerted on the jth coordinate surface by particles on its negative side, the jth component of momentum flux density = stress = traction being the pressure on that surface, the other components being components of sheer stress.

    Puzzlingly though, of the kinds of things that go by the name of flux--"flux density" (a vector field, not generally dependent on any surface being specified), "flux density across a given oriented surface" (a scalar field defined on the surface), and "flux" (the surface integral of flux density, with respect to an oriented surface)--here we have vector quantities called "flux density" which nevertheless seem to depend on a specified surface, as (ordinarily) do the scalar quantities "flux density across a surface" and "flux".
    Last edited: Feb 14, 2010
  11. Feb 14, 2010 #10
    Hmm, so there's an x-component to the z traction vector, pointing in the -x direction; x-momentum is said to "flow" across the z surface element, even though this "flow" is represented as a decrease in the x-component of the momentum flux density, m.f.d. being another name for the traction vector (also called the stress on that surface).

    If I've got this right, letting the stress tensor act on one unit coordinate surface normal vector gives the relevant traction vector, i.e. the "flux density" across that surface, but this is a vector and plays the role of the kind of quantity that I've been calling "flux density" (which kind of quantity in general doesn't require a surface to be specified for it to be defined, although here it happens that it does). It could be that's what's confusing me. Only when we dot this vector with another unit coordinate surface normal vector do we get the kind of scalar quantity that I've been calling "flux density across a surface".

    (In particular diazona's definitions in #17.)

    Clearly I need better names to tell these entities apart. Perhaps the vector that I've been calling "flux density" could stay as it is, and the scalar be called "flux" (although it is also a kind of density) and the surface integral of flux density "net flux". Or maybe I should call them vector flux density and scalar flux density, which might be less ambiguous. Know any good conventions? Maybe when I'm familiar with them and what they all mean, I'll be able to get by calling them all just "flux" like everyone else, but for now that's too confusing.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook