The stress tensor in continuum mechanics (i.e. the negative of the space-space part of the stress energy tensor in relativity1) takes the form: $$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \sigma_{ij}=-p\delta_{ij}+\eta \l\p{v_i}{x_j}+\p{v_j}{x_i}\r . \tag{1}$$ As the name suggests this is a tensor but I am not clear what type of a tensor and can't seem to find a resource which tells me and it is my assumption that (1) is not written in tensor notation (i.e. correct index position). Thus my question is this: how do we write the stress tensor (1) in actual tensor notation?
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2just raise the indices in $x_j,x_i$ and you're good to go – AccidentalFourierTransform Dec 30 '16 at 15:35
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1@AccidentalFourierTransform are you sure? I am worried that since $\delta_{ij}$ is not necessarily $1$ when $i=j$ this will muck things up – Quantum spaghettification Dec 30 '16 at 15:39
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1what? how can $\delta_{ij}$ not be $1$ when $i=j$? what does $\delta_{ij}$ mean for you? – AccidentalFourierTransform Dec 30 '16 at 15:40
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1@AccidentalFourierTransform See this http://www.physicspages.com/2012/12/31/kronecker-delta-as-a-tensor/ – Quantum spaghettification Dec 30 '16 at 15:42
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Most likely, the reason you can't find the answer is because the engineers who actually use this stuff don't care about labelling things as covariant or contravariant - what matters is doing the correct operations on the quantities, not giving them impressive names. The only way I would ever write your equation (1) is exactly the way you wrote it! – alephzero Dec 30 '16 at 15:44
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@Quantumspaghettification that's terrible notation. It's not what people usually mean when they write $\delta_{ij}$. – AccidentalFourierTransform Dec 30 '16 at 15:45
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1It's a contravarient tensor you are after, (upper indices) and you have the elements of it, rather than the complete tensor, in your post. – Dec 30 '16 at 15:41
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possible duplicate of http://physics.stackexchange.com/q/76048/ – Dec 31 '16 at 01:26
2 Answers
It seems that OP essentially wants to consider an incompressible Newtonian fluid on a Riemannian 3-manifold $(M,g)$ (as opposed to, say, a perfect relativistic fluid).
Next, we should write local coordinates $x^i$ with an upper (as opposed to a lower) index, as user AccidentalFourierTransform suggests in a comment above.
The Kronecker delta should be replaced with the metric tensor, cf. e.g. this Phys.SE post.
Partial derivatives $\partial_i$ should be replaced with (Levi-Civita) covariant derivatives $\nabla_{\!i}$. This point 4 and the previous point 3 were also suggested in the answer by user Chester Miller.
Use the musical isomorphism to raise and lower indices $v_i=g_{ij}v^j$ on the incompressible velocity field, ${\rm div} (v)= \nabla_{\!i}v^i=0$ .
Then both sides of OP's eq. (1) becomes a symmetric (0,2) covariant tensor $$ \sigma_{ij}~=~-pg_{ij}+\eta \left(\nabla_{\!i} v_j+ \nabla_{\!j} v_i\right), \qquad p~=~ -\frac{1}{3} g^{ij}\sigma_{ij}. \tag{1'} $$
The equation you gave is expressed in terms of Cartesian coordinates, where the distinction between covariant and contravariant components does not exist. If you want to express this Newtonian fluid constitutive equation in terms of "actual" tensor notation, the partial derivatives on the right hand side need to be replaced by covariant derivatives, and the Kronecker delta needs to be replaced by the appropriately indexed representation of the metric tensor.
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2Yes, this is indeed correct. +1 [as a small nitpick, one may note that, in Cartesian coordinates, some people will set $x^i=-x_i$, so the distinction might be relevant, depending on conventions] – AccidentalFourierTransform Dec 30 '16 at 19:39