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

Interpretation of vorticity with non-vanishing strain tensor

  1. Aug 14, 2014 #1

    WannabeNewton

    User Avatar
    Science Advisor

    Hi guys. Let me just say at the outset that I know very little fluid mechanics but I keep coming back to the same issue over and over in a general relativity related problem so I figured I'd just ask the fluid mechanics question here.

    In countless places the interpretation of the vorticity vector of a fluid is given as follows. If we consider any fluid element and an infinitesimal displacement vector between this and a neighboring fluid element, such that the displacement vector remains fixed to this neighboring fluid element, then the vorticity vector measures the instantaneous angular velocity of the displacement vector, say relative to inertial guidance gyroscopes comoving with the considered fluid element. Of course this makes perfect sense when the fluid has a vanishing strain tensor so that it describes a rigid motion and indeed the claim is very easy to prove mathematically in the case of rigid motion.

    But does it still hold if the motion isn't rigid i.e. if the strain tensor is non-vanishing? If the displacement vector is ##\delta x^i## then the relative velocity is ## \delta\dot x^i = \sigma^{i}{}{}_j \delta x^j## in the absence of vorticity and compression, so that there is only shear. But if ##\sigma_{ij}## has off-diagonal components then certainly ##\delta \dot x^i## will have a part that represents rotational velocity, right? Only if ##\delta x^i## is aligned with the principal axes of ##\sigma_{ij}## will there be no such rotational velocity. And indeed I have searched as much as I can and I can't find any mathematical proof of the statement that vorticity measures the entirety of the instantaneous angular velocity of the infinitesimal displacement vector between neighboring fluid elements when the strain tensor is non-zero.

    Thanks in advance.
     
  2. jcsd
  3. Aug 14, 2014 #2
    I can only say that fluid/aerodynamics vorticity is calculated via Biot-Savart law, and it works for magnetic fields as well. So maybe you can get some information about it from there, or from Maxwell's equations and Gauss's law of magnetism.

    Where in GR do you encounter vorticity?
     
  4. Aug 14, 2014 #3
    Hi WBN,

    In what you've written, σ is the velocity gradient tensor. The velocity gradient tensor can be resolved into a symmetric part and an antisymmetric part:

    [tex]σ=\frac{σ+σ^T}{2}+\frac{σ-σ^T}{2}[/tex]

    where T represents the transpose. The antisymmetric part is called the vorticity tensor, and corresponds to the average rate of rotation of the fluid parcel, so it can be viewed as a "rigid rotation." The symmetric part is called the rate of deformation tensor, or the rate of strain tensor. This is the part that gives rise to viscous stresses.

    Hope this helps.

    Chet
     
  5. Aug 14, 2014 #4

    WannabeNewton

    User Avatar
    Science Advisor

    Hi Chet thanks for the reply. Sorry if my notation is nonstandard in fluid mechanics but by ##\sigma## I actually meant purely the shear part of the strain tensor i. e. when I wrote ##\delta \dot x = \sigma \cdot \delta x ## I was already assuming zero vorticity and compression. But if ##\sigma \cdot \delta x## has a component orthogonal to ##\delta \dot x##, which I think it certainly can in general when ##\sigma## has off-diagonal components, then this represents a relative rotational velocity between fluid elements not accounted for by the vorticity so how does one interpret vorticity in the general case of non-vanishing strain, as opposed to its clear meaning in the case of rigid rotation (vanishing strain) wherein it fully and entirely represents the angular velocity of rotation of ##\delta x##?
     
  6. Aug 14, 2014 #5
    The vorticity tensor represents the average rate of rotation of the fluid lines (averaged circumferentially), and the rate of strain tensor represents deformations relative to the average rotation rate. If you average the rotation rates of the lines operated upon by the rate of strain tensor, the average will be zero. A deformation in which the vorticity tensor is zero would be referred to as a pure strain deformation. For shear between parallel plates, vx=γy, and vy=0, where γ is the so-called "shear rate". This can be written as vx=γy/2+γy/2, vy=γy/2-γy/2. The average rotation rate of the fluid elements would be γ/2. The principal components of the rate of deformation tensor would involve pure extension along x = y, and pure compression along x = -y.

    Chet
     
  7. Aug 15, 2014 #6
    Here's a brief primer.

    In large deformation mechanics (required for fluids), the starting point is usually the equation:

    [tex]d\vec{x}(t)=\vec{F}(t)\centerdot d\vec{x}(0)[/tex]

    where [itex]d\vec{x}(t)[/itex] is a differential position vector between two neighboring material points at time t, [itex]d\vec{x}(0)[/itex] is the differential position vector between the same two material points at time zero, and [itex]\vec{F}(t)[/itex] is the "deformation gradient tensor," which maps [itex]d\vec{x}(0)[/itex] into [itex]d\vec{x}(t)[/itex] for any arbitrary pair of neighboring material points. If we take the material time derivative of the above equation, we get:

    [tex]\frac{D(d\vec{x}(t))}{Dt}=d\vec{v}(t)=\frac{D\vec{F}(t)}{dt}\centerdot \vec{x}(0)[/tex]
    where [itex]d\vec{v}(t)[/itex] is the difference in velocity between the two material points.
    If we combine the two equations, we obtain:
    [tex]d\vec{v}(t)=\frac{D\vec{F}(t)}{Dt}\centerdot \vec{F}^{-1}(t)\centerdot \vec{dx}(t)[/tex]
    The difference in velocity between the two material points is related to the velocity gradient tensor by:
    [tex]d\vec{v}(t)=\vec{dx}(t)\centerdot ∇\vec{v(t)}=(∇\vec{v(t)})^T\centerdot \vec{dx}(t)[/tex]
    If we combine the previous two equations, we obtain:
    [tex]\frac{D\vec{F}(t)}{Dt}=(∇\vec{v(t)})^T\centerdot \vec{F}(t)[/tex]
    The squared magnitude of the differential position vector [itex]d\vec{x}(t)[/itex] is given by:
    [tex](dx(t))^2=d\vec{x}(0)\centerdot \vec{F}^T(t)\centerdot \vec{F}(t)\centerdot d\vec{x}(0)[/tex]
    If we take the material time derivative of this, we get
    [tex]\frac{D(dx(t))^2}{Dt}=d\vec{x}(t)\centerdot (∇\vec{v(t)}+(∇\vec{v(t)})^T)\centerdot d\vec{x}(t)[/tex]
    The term on the left is the rate of change of the square of the distance between the material points, and relates to the rate of stretching. The term in parenthesis on the right hand side is twice the rate of deformation tensor, and also twice the symmetric part of the velocity gradient tensor. Note that there is no rotational part (vorticity part) involved here. So the rate of stretching of the material lines does not involve the vorticity.

    Chet
     
  8. Aug 15, 2014 #7

    WannabeNewton

    User Avatar
    Science Advisor

    Thanks for the detailed explanations Chet! Actually I'm totally fine with everything you've explained above. I've seen those calculations before and I would like to think that I don't have any troubles with them conceptually. What I was looking for is more along the lines of the statements in the following two excerpts:

    http://books.google.com/books?id=P5...orticity principal axes strain tensor&f=false

    http://books.google.com/books?id=Ok...rotation principal axes strain tensor&f=false

    As you can see, what these both state is that in the case of non-vanishing strain the vorticity measures the rotation of the principal axes of the strain tensor relative to, say, comoving inertial guidance gyroscopes. So I can condense my questions into the following:

    (1) How does one prove more rigorously that indeed the vorticity measures the rotation of the principal axes of strain?

    (2) Coming back to the equation for the relative velocity with vanishing vorticity and compression, we have ##\delta \dot x^i = \sigma^i{}{}_j \delta x^j## with as usual ##\delta x## the infinitesimal displacement vector between neighboring fluid elements and ##\sigma_{ij}## the strain. If we choose as the basis of a reference frame the principal axes of the strain then in this frame ##\sigma_{ij}## is diagonal and hence ##\delta \dot x^i \propto \delta x^i## relative to this frame meaning in this frame the neighboring fluid elements do not rotate. So is it safe to say that in the case of vanishing vorticity we can always choose our frame to have its axes aligned with the principal axes of strain so as to have a non-rotating frame (since the principal axes don't rotate relative to the gyroscopes) in which the neighboring fluid elements themselves do not rotate? And of course we do this at each point on the flow lines in a smooth manner.
     
  9. Aug 15, 2014 #8
    Hi WBN,

    I think I can answer your questions, but, because I'm an engineer and tend to proceed more intuitively, I'm not sure I can answer them to the degree of mathematical exactness that you may be seeking.

    Regarding your first question, it looks like some of the answer may lie in Section 2.1.

    But, in any event, here goes. Imagine a tiny spherical parcel of fluid at time t, with a position vector [itex]\vec{dx}(t)[/itex] of uniform length dx(t) drawn from the center of the sphere to each point on the surface of the sphere. Since the sphere is very small, the deformation kinematics can be regarded as locally homogeneous (i.e., within the sphere). We are going to consider the kinematics of the sphere strain and rotation during the time interval dt. According to the equations I presented in my previous post, we have

    [tex]\vec{dx}(t+dt)≈\vec{dx}(t)+\vec{Ω}\centerdot \vec{dx}(t)dt+\vec{D}\centerdot
    \vec{dx}(t)dt[/tex]

    where Ω is the vorticity tensor and D is the rate of deformation tensor. This equation shows where the material points of the parcel will be located at time t+dt, given their locations on the sphere at time t. It indicates that it will be comprised of three components: (1) the original location at time t, (2) the displacement resulting from the pure strain Ddt (whose principal axes are fixed over the time interval), and the displacement resulting from a superimposed rigid body rotation associated with the vorticity Ω.

    From the frame of reference of an observer who is rotating along with the fluid parcel, the principal axes of the strain rate tensor would seem to rotate, but in the opposite direction of Ω.

    To answer your second question, the answer is yes.

    Chet
     
  10. Aug 15, 2014 #9

    WannabeNewton

    User Avatar
    Science Advisor

    That makes perfect sense, thank you Chet! You've helped completely clarify something PeterDonis and I have been wondering about for a long while now in GR.
     
  11. Aug 15, 2014 #10
    Wow. I'm very honored to have helped two of my idols.

    Chet
     
  12. Aug 16, 2014 #11

    WannabeNewton

    User Avatar
    Science Advisor

    Haha thanks Chet! Actually I was wondering if you could answer a related question I had; I sort of asked it above but completely muddled the statement of the question.

    So as you've demonstrated the vorticity in general measures the rotation of the principal axes of the strain tensor because it is only these axes which undergo rigid rotation when there is shearing. So when there is vanishing vorticity, the principal axes don't rotate. But if I have a displacement vector that isn't aligned with a principal axes, then can this displacement vector still rotate relative to the principal axes? It would seem that in general it can as is the case e.g. of a sphere being deformed into an ellipse. So does this mean that even when there is vanishing vorticity, displacement vectors that are not aligned with the principal axes can still rotate?

    In such a case what would be the best way of thinking about vorticity as applied to displacement vectors that aren't aligned with the principal axes? That if averaged over infinitesimal circles the rotation of the displacement vectors will vanish when there is vanishing vorticity, because somehow the instantaneous rotation relative to principal axes due to shearing gets averaged out to zero over infinitesimal circles?

    Thanks in advance.
     
  13. Aug 16, 2014 #12
    Sure.
    Yes. In the excerpt you sent me, they showed Stokes Theorem applied to the velocity. Everything you said here follows directly from that.

    Chet
     
  14. Aug 17, 2014 #13

    WannabeNewton

    User Avatar
    Science Advisor

    Awesome, thanks again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Interpretation of vorticity with non-vanishing strain tensor
  1. The Vanishing Energy (Replies: 7)

Loading...