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I'm looking at Navier-Stokes equation in index notation and how to get them in vector notation:

$$ {\partial u_i \over \partial t}+ u_j {\partial u_i \over \partial x_j}= -\frac{1}{\rho}{\partial p \over \partial x_i}+ \nu {\partial^2 u_i \over \partial x_j \partial x_j}+g_i $$

1st equation: The local acceleration, pressure, and body forcing terms seem simple. Is it correct to say as they have 1 free index (i) that they are tensors of rank 1, thus vectors where i = 1,2, and 3 for 3D?

2nd Question: For the convection acceleration term (I assume the diffusive term would be similar): $$ u_j {\partial u_i \over \partial x_j}, $$ is it correct to say the $u_j {\partial \over \partial x_j}$ component, as it has no free indices and 1 dummy index (j), is a rank 0 tensor and thus a scalar (after summation)?

3rd Question: If Question 2 is correct can $u_i$ in the same term, a vector, simply then be multiplied by $u_j {\partial \over \partial x_j}$, a scalar, to then get a vector?

This is probably a softball question for anyone that does a lot of tensor calculus. I'm trying to teach some things about the Navier-Stokes equations and I'm trying to make sure all of my definitions regarding tensor rank and index notation are correct.

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  • $\begingroup$ Your second equation is indeed a scalar operator hitting the vector, it is the velocity dotted into grad then acting on the velocity. $\endgroup$ – JamalS Jul 12 '20 at 16:17
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Is it correct to say as they have 1 free index (i) that they are tensors of rank 1, thus vectors where i = 1,2, and 3 for 3D?

Yes but... not always. In general, an expression with a free index $i$ might be interpreted as the $i$-th component of a $1 \times n$ array of numbers. If it has 2 free indices $i,j$ as the (i,j)-th component of a matrix, and so on. Yet, arrays and tensors are not the same thing. Tensors must obey certain transformation rules under rotations for example, and we 'represent' these tensors with arrays of numbers. For your problem, these quantities are vectors but you can always avoid getting into the details of the transformation properties and just think about them as arrays. Actually, whenever a dummy index arises from non-relativistic physics, it is almost always certain that the quantity is also the component of a vector.

is it correct to say the $𝑒_𝑗\frac{βˆ‚}{βˆ‚π‘₯_𝑗}$ component, as it has no free indices and 1 dummy index (j), is a rank 0 tensor and thus a scalar (after summation)?

Yes, it is a scalar operator (no free indices) but take into account that this might change depending on the object on which it acts, e.g: if it acts on a function (scalar) it will be scalar, if it acts on a vector (1 free index) it will be a vector, and so on. Also, take into account that you cannot add up quantities with different ranks (i.e: a number + a vector is not a valid operation, therefore, if you identify that an equation is made up of a sum of vectors like Navier Stokes, then... the other terms must also be vectors).

If Question 2 is correct can $𝑒_𝑖$ in the same term, a vector, simply then be multiplied by $𝑒_𝑗\frac{βˆ‚}{βˆ‚π‘₯_𝑗}$, a scalar, to then get a vector

Yes, read the previous comments.

Also, you might find useful to interpret things like $\frac{βˆ‚}{βˆ‚π‘₯_𝑗}$ as the components of $\vec{\nabla}$ (a gradient), contractions as dot products, $𝑒_𝑗\frac{βˆ‚}{βˆ‚π‘₯_𝑗}$ as a directional derivative $\vec{u}\cdot\vec{\nabla}$, $\frac{βˆ‚}{βˆ‚π‘₯_π‘—βˆ‚π‘₯_𝑗}$ as a Laplacian $\Delta^2$, etc. Further, take into account that $𝑒_𝑗\frac{βˆ‚}{βˆ‚π‘₯_𝑗}\neq \frac{βˆ‚π‘’_𝑗}{βˆ‚π‘₯_𝑗}$ (the first one is a directional derivative, and the latter a divergence).

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  • $\begingroup$ Thanks, this helps a lot. For $u_j {\partial \over \partial x_j}$ are you saying it is always defined as a scalar operator as it has no free indices? But if it is applied to a scalar the term is overall a scalar and if applied to a vector the result is overall a vector? Or is it referred to as a vector operator if applied to a vector? $\endgroup$ – acollins Jul 12 '20 at 16:53
  • $\begingroup$ it is always a scalar operator, but if it is applied to a scalar/vector the term is overall a scalar/vector. $\endgroup$ – JuanC97 Jul 12 '20 at 17:37
  • $\begingroup$ Please check the update I made to my last answer. I originally had a lapsus and ended up saying just the opposite to what I meant to, sorry but now it's fixed. $\endgroup$ – JuanC97 Jul 12 '20 at 17:48

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