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I don't know much about tensor calculus and here is something I'm trying to figure out.

$$T=\mu({\nabla}\vec{V}+{\nabla}\vec{V}^T)$$

T is viscous stress tensor and $\vec{V}$ is the velocity vector. How do we get a tensor of rank 2 by adding a vector (gradient of velocity) to its transpose? My best guess is that the gradient of a vector is a 2nd rank tensor (though we are told in engineering schools that the gradient is only defined for scalar fields). Am I right? Am I missing something here?

Thanks.

Eman
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  • See: http://math.stackexchange.com/questions/156880/gradient-of-a-vector-field – Emilio Novati Feb 28 '16 at 20:33
  • $\nabla V $ gives you a tensor field. It is certainly not true that the gradient is only defined for scalar fields. – user_of_math Feb 28 '16 at 20:34
  • @Emilio Novati Thanks a lot for your responses. – Eman Feb 28 '16 at 20:39
  • @user_of_math Thanks for clearing that up. – Eman Feb 28 '16 at 20:39
  • In components with respect to an orthonormal basis of $\mathbb{R}^3$ the expression you wrote reads $T_{ij}=\mu(\partial_iV_j+\partial_jV_i)$. – Andrea Feb 28 '16 at 20:53
  • @Andrea Thanks a lot. This is exactly the term we write in each component of Navier-Stokes equation. Unfortunately, most of the engineering textbooks do not use precise mathematical notations and they just jump into the physical details without using exact mathematics and it always makes lots of confusion for me until I find out what the mathematical ideas behind them are. – Eman Feb 28 '16 at 21:02

2 Answers2

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If $\mathbf{v}=(v_1,v_2v_3)$ is a vector, its gradient is the matrix:

$$ \nabla \mathbf{v}= \begin{bmatrix} \frac{\partial v_1}{\partial x_1}&\frac{\partial v_1}{\partial x_2}&\frac{\partial v_1}{\partial x_3}\\ \frac{\partial v_2}{\partial x_1}&\frac{\partial v_2}{\partial x_2}&\frac{\partial v_2}{\partial x_3}\\ \frac{\partial v_3}{\partial x_1}&\frac{\partial v_3}{\partial x_2}&\frac{\partial v_3}{\partial x_3}\\ \end{bmatrix} $$ that is the so called Jacobian matrix.

Note that the elements in the $i-$row of this matrix are the components of the gradient of the $i-$componet if the vector $\mathbf{v}$.

This matrix contains all the informations about the first order changes of the vector $\mathbf{v}$ in any direction, and it transforms as a covariant tensor for linear transformation of the coordinates. But note that it is not covariant for curvilinear changes of coordinates, so it is not properly a tensor.

For a simple introduction to tensor and covariant derivative you can see here.

Emilio Novati
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  • Thanks again. If it's not a tensor, then the stress tensor itself wouldn't be a tensor. So I need to study a little more about tensors and find a concrete definition to understand what you said properly. – Eman Mar 01 '16 at 22:25
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To take it a bit further than Emilio and to answer your question: $\vec{\nabla}\vec{v}$ and its transpose are already rank 2 tensors. Scalars are rank 0, vectors are rank 1 and each subsequent addition of a 'component' increases the rank by one.

The result of the addition of two rank 2 tensors is still a rank 2 tensor: $$\vec{\nabla}\vec{v} + \left(\vec{\nabla}\vec{v}\right)^T = \left[\begin{array}{ccc} \partial_{x}u & \partial_{y}u & \partial_{z}u\\ \partial_{x}v & \partial_{y}v & \partial_{z}v\\ \partial_{x}w & \partial_{y}w & \partial_{z}w \end{array}\right] + \left[\begin{array}{ccc} \partial_{x}u & \partial_{x}v & \partial_{x}w\\ \partial_{y}u & \partial_{y}v & \partial_{y}w\\ \partial_{z}u & \partial_{z}v & \partial_{z}w \end{array}\right] = \left[\begin{array}{ccc} 2\partial_{x}u & \partial_{y}u+\partial_{x}v & \partial_{z}u+\partial_{x}w\\ \partial_{x}v+\partial_{y}u & 2\partial_{y}v & \partial_{z}v+\partial_{y}w\\ \partial_{x}w+\partial_{z}u & \partial_{y}w+\partial_{z}v & 2\partial_{z}w \end{array}\right] $$

Clearly the number of 'components' are preserved and the result is a rank 2 tensor.

nluigi
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