- #1
da_willem
- 594
- 1
I'm reading a book (intro to, by Davidson) about MHD now, but found I'm a bit rusty on tensors and curvilinear coordinates. It is written that for a circular flow the azimuthal component of the NS equations in the steady state gives (with F some body force)
\tau _{r \theta} r^2 =-\int _0 ^r r^2 F_{\theta} dr
Shouldn't this read, for axisymmetric flow, without the square on both r's? I would argue that the remaining terms in the NS equations
\sigma _{ij,i}+F_j=0
would yield for the azimuthal (\theta) component (any suggestions welcome if the notation is obscure):
F_{\theta} = -\nabla \cdot \overline{\overline{\sigma}}_{\theta}=-\frac{1}{r} \frac{d}{dr}(r\sigma_{r \theta})
Now the author continues,
\tau_{r \theta}=\mu r \frac{d}{dr}(\frac{u_{\theta}}{r})
In which he says he used Newtons law of viscosity, which I think one can write
\tau_{ij}=\mu u_{i,j}
(Is it by the way ok to write this as \overline{\overline{\tau}}=\mu \nabla \vec{u}?)
But how does one come from that to the (r, theta) component?
\tau _{r \theta} r^2 =-\int _0 ^r r^2 F_{\theta} dr
Shouldn't this read, for axisymmetric flow, without the square on both r's? I would argue that the remaining terms in the NS equations
\sigma _{ij,i}+F_j=0
would yield for the azimuthal (\theta) component (any suggestions welcome if the notation is obscure):
F_{\theta} = -\nabla \cdot \overline{\overline{\sigma}}_{\theta}=-\frac{1}{r} \frac{d}{dr}(r\sigma_{r \theta})
Now the author continues,
\tau_{r \theta}=\mu r \frac{d}{dr}(\frac{u_{\theta}}{r})
In which he says he used Newtons law of viscosity, which I think one can write
\tau_{ij}=\mu u_{i,j}
(Is it by the way ok to write this as \overline{\overline{\tau}}=\mu \nabla \vec{u}?)
But how does one come from that to the (r, theta) component?
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