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I would like to compute the material derivated of a tensor quantity, in the context of the finite volume method (FVM):

The equation is:

$$ \frac{\mathrm{d} \textbf{T}}{\mathrm{d} t} = \frac{\partial \textbf{T}}{\partial t} + \vec{v} \cdot \nabla \textbf{T} = \underbrace{\frac{\partial \textbf{T}}{\partial t} + \nabla \cdot \left(\textbf{T} \vec{v}\right) - \textbf{T} \left(\nabla \cdot \vec{v}\right)}_{is \: this \: true \: for \: tensors?} $$

I am thinking of solving this by solving each component entry in the tensor. Is this the correct procedure? In python, I used to solve this with solve_ivp but that considers a moving reference frame and in the FVM the frame is stationary.

Any tricks/things to be aware of when solving for a tensor quantity?

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No, the right-hand side of your equation is not valid. I will assume that $\mathbf{T}$ is a second-order tensor. If that's the case we need that the time derivative is second-order. The factor $\nabla \mathbf{T}$ gives you a third-order quantity that is later turned into a second-order one via projection over $v$. Now, the factor $\mathbf{T} v$ is a vector and is turned into a scalar after applying the divergence.

I would say that a trick might be to write the expression in index notation.

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    $\begingroup$ I think whether it's valid or not depends on what the notation is supposed to mean. If $\mathbf{T}v$ is the outer product $\mathbf{T}\otimes v$, which has rank equal to $\text{rank}(\mathbf{T}) + \text{rank}(v)$, then I think it's correct, but your answer assumes that it means the contraction of $\mathbf{T}$ with $v$ along some index. Either way I agree that writing it in index notation will probably help. $\endgroup$ May 27, 2022 at 22:50
  • $\begingroup$ @DanielShapero, you are right. I assumed that the OP meant the contraction. $\endgroup$
    – nicoguaro
    May 28, 2022 at 2:46
  • $\begingroup$ Hi, thanks for the reply. To be honest I do not know if it is a contraction. For a scalar quantity: en.wikipedia.org/wiki/Material_derivative, and from the divergence identity en.wikipedia.org/wiki/Vector_calculus_identities. The expression holds for scalar quantities, and I have also seen it applied to vector quantities. To 2nd order tensor, I do not know ... $\endgroup$
    – user42626
    May 28, 2022 at 8:55
  • $\begingroup$ $ \nabla \cdot (s \vec {v}) = s (\nabla \cdot \vec{v}) + \vec{v} \cdot \nabla s$ $ \Leftrightarrow \nabla \cdot (s \vec {v}) - s (\nabla \cdot \vec{v})= \vec{v} \cdot \nabla s $ $\endgroup$
    – user42626
    May 28, 2022 at 8:58

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