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Fluid stress tensor in cylindrical coordinates?

  1. Feb 28, 2012 #1
    For fluid with viscosity [itex]\mu[/itex] our stress strain relationship takes the form

    [tex]\sigma_{ij} = -p \delta_{ij} + 2 \mu u_{ij}.[/tex]

    I was wondering how to express this in cylindrical coordinates. The strain tensor I can calculate in cylindrical coordinates (what I get matches eq 1.8 in [1]). But how would the [itex]\delta_{ij}[/itex] portion of the stress strain relationship be expressed in cylindrical coordinates?

    For example, if we considered a non-viscous fluid, the very simplest stress tensor, we have in rectangular coordinates

    [tex]\sigma_{ij} = -p \delta_{ij}.[/tex]

    It's not obvious to me how this would be expressed in cylindrical form. I wanted to try some calculations with the traction vector [itex]T_i = \sigma_{ij} n_j[/itex] in a cylindrical coordinate system, but I'm not sure how to express it. I figured the place to start was with the stress tensor.


    [1] L.D. Landau, EM Lifgarbagez, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.
  2. jcsd
  3. Feb 29, 2012 #2
    After some thought, I think the answer is that this would be exactly like the rectangular case, where for a non-viscious stress [itex]\sigma_{ij} = -p \delta_{ij}[/itex], the cylindrical representation would be exactly the same, with the only difference being that i,j have to index over [itex]\{r, \theta, z\}[/itex]. So, we'd have a -p component for [itex]\sigma_{rr}, \sigma_{\theta\theta}, \sigma_{zz}[/itex].
    Last edited: Feb 29, 2012
  4. Feb 29, 2012 #3
    I know nothing about fluid mechanics but comparing this tensor with Maxwell stress tensor in electromagnetism, you seem to be right. However in Maxwell stress tensor, only two components of matter, the radial and the tangential. Hence I expect an expression like the following:


    r: radial
    t: tangential

    see http://en.wikipedia.org/wiki/Maxwell_stress_tensor
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