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dyadic product between scalar and vector

Anonymous Tue 10 Nov 13:53:31 2020 No.12326148 View Reply Original Report

Quoted By: >>12326165 >>12326171 >>12326215 >>12326963

hello folks,

my fluiddynamics prof showed us that term (pic related)

is a vector times (scalar product) a vector, resulting in a scalar that then gets computed with another vector via dyadic product. how is this possible? i cant seen to find a definition of dyadic product between scalar and vector?

what am i missing?

Anonymous

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Anonymous Tue 10 Nov 2020 14:01:57 No.12326165 Report

Quoted By: >>12326202

>>12326148
dyads are essentially outer products. i feel like the language of dyadics isn't really necessary. anyway, if you are trying to understand the equation you posted, then expand it like so:
$ (\vec{u} \cdot \nabla) \vec{q} = \left(u_x \frac{\partial}{\partial x} + u_y \frac{\partial}{\partial y} + u_z \frac{\partial}{\partial z} \right) \vec{q} = \begin{bmatrix} u_x \frac{\partial q_x}{\partial x} + u_y \frac{\partial q_x}{\partial y} + u_z \frac{\partial q_x}{\partial z} \\ u_x \frac{\partial q_y}{\partial x} + u_y \frac{\partial q_y}{\partial y} + u_z \frac{\partial q_y}{\partial z} \\ u_x \frac{\partial q_z}{\partial x} + u_y \frac{\partial q_z}{\partial y} + u_z \frac{\partial q_z}{\partial z} \\ \end{bmatrix} $

Anonymous

Anonymous Tue 10 Nov 2020 14:02:53 No.12326171 Report

Quoted By: >>12326202

>>12326148
So what you're talking about isn't the dyadic product, that's more associated with the tensor product (I believe the tensor product was actually referred to as the dyadic product in older texts. What is in the parenthesis is called the DOT or SCALAR product, which will always return a scalar. And then hence, you multiply q by a scalar quantity, THIS here can be called a "dyadic product" because of how you distribute the scalar to the terms of q, but your professor is over-complicating things by not just referring to it as scalar multiplication.

It's just scalar multiplication anon, if your professor refers to scalar multiplication as a dyadic product just to obfuscate things for students who dont know what the fuck that is, he should not be teaching. It's scalar multiplication between a vector and a scalar.

Anonymous

Anonymous Tue 10 Nov 2020 14:15:29 No.12326202 Report

Quoted By: >>12326236

>>12326165
noice that clears it up, so it is basically scalar multiplication.
also neat latex, how do i use it here?

>>12326171
thank you! well i especially asked that in the lecture aswell but it seemed so clear to everyone else, so she didnt spend much more time elaborating.
I told her that due to mathematical definitions, it should be

$\nabla \vec{q} \cdot \vec{u}$

so you get

matrix times vector which will result in a vector again. that way the dyadic products definition is used while the term remains the same.

thank both of you for telling me that it is simple scalar multiplication, idk how i couldnt see it on my own

Anonymous

Anonymous Tue 10 Nov 2020 14:20:36 No.12326215 Report

Quoted By:

>>12326148
Fluid dynamics is a bit of a notational clusterfuck.
There's no dyadic product in your picture.
However, you can show that $(\vec{u}\cdot\nabla)\vec{u} = \nabla \cdot (\vec u \otimes \vec{u})$ for incompressible fluids.

Anonymous

Anonymous Tue 10 Nov 2020 14:23:00 No.12326224 Report

Quoted By: >>12326238 >>12326255

any good books for starting fluid dynamics?

Anonymous

Anonymous Tue 10 Nov 2020 14:25:18 No.12326236 Report

Quoted By: >>12326271

>>12326202
>idk how i couldnt see it on my own
it's understandable. del acts like a vector, but is an operator, so it can get a little confusing. the term in question is strange looking because you treat del like a vector without it having operated on anything yet. you can do this is some special cases, but care should be taken in general to make sure you know what you're doing.

>how do i use it here?
see sticky. use html-like [?][/?] tags in your post where "?" is "math" or "eqn" for inline or full line latex environments

Anonymous

Anonymous Tue 10 Nov 2020 14:26:09 No.12326238 Report

Quoted By:

>>12326224
Fluid Mechanics by Joseph H. Spurk and Nuri Aksel is what I am using for studying beside the Lecture im visiting

Anonymous

Anonymous Tue 10 Nov 2020 14:30:20 No.12326255 Report

Quoted By:

>>12326224
Yes, but the answer depends on what you want to do with it.
Landau-Lifschitz, or Bachelor are good for general purpose.
Schlichting for Aerodynamics.
Ferzinger or Moukalled for numerics/finite volumes
Otherwise there's a lot of acoustics, magnetohydrodynamics, atmospheric fluid dynamics etc.

Anonymous

Anonymous Tue 10 Nov 2020 14:38:44 No.12326271 Report

Quoted By:

>>12326236
>the term in question is strange looking because you treat del like a vector without it having operated on anything yet

thanks a bunch for this addition

Anonymous

Anonymous Tue 10 Nov 2020 14:48:21 No.12326300 Report

Quoted By: >>12326320

The J programming language has monadic and dyadic operators. It's the only programming language I know of that has dyads. The monadic and dyadic properties of operators enable every operator to be "rank polymorphic" in respect to the given arguments. What this means is that any function implemented using rank polymorphic operators will never make a specific reference to its parameter, and will never define what it is. Then the function you defined will automatically be callable with an array of any number of dimensions, ie of any rank. A scalar is then not even treated as a scalar, but as a vector of rank 0 (zero dimensions).

Hope this helps in some way, I know the language was carefully designed to be useful to mathematicians so I assume this concept is useful in some way. A dyadic operator is just an operator with two operands, left and right, which is agnostic to the shape or dimensionality of the operands. This is in contrast to a monadic operator, which is the same, except it only operates on a single operand at a time. The "at a time" part is important here, because dyads and monads may be chained for any number of given arguments, repeating its operation compositionally. This is how it achieves rank polymorphism.

Anonymous

Anonymous Tue 10 Nov 2020 14:58:40 No.12326320 Report

Quoted By: >>12326369

>>12326300
yes I understand what you are saying, but it is rather complicated compared to the first answers.

Anonymous

Anonymous Tue 10 Nov 2020 15:16:28 No.12326369 Report

Quoted By: >>12326432

>>12326320
Yeah, sorry, it's an odd language.

Say you want to define a function that multiplies each value in a vector pairwise, reducing the length of each dimension by a factor of 2, then adds each element resulting from the multiplication pairwise, again reducing dimension length by 2. You might define it as `+ *`. The operators go right to left. The rightmost operator, dyadic multiplication, is applied to the first pair of elements in the given array. The result of this operator is then passed to the plus. The plus needs two elements since it's also dyadic. The multiplication dyad will multiply the next pair of values and pass them to the plus, which then applies to this pair dyadically. This pairwise operation is applied across each dimension separately. If there are an odd number of elements in the vector, the final value is applied monadically across the operators I believe. The same would happen with a rank 0, or scalar, argument.

This might be more confusing, sorry if it is. To make things worse, in J you can even pass operators or functions as a vector, which would be composed by the given operators. You would likely need to check the documentation to see what this means in terms of addition and multiplication.

There is no other language that does anything like this btw. So it's kind of strange to even describe.

Anonymous

Anonymous Tue 10 Nov 2020 15:37:51 No.12326432 Report

Quoted By:

>>12326369
What would happen with + * would be that the operators themselves would be composed into a single monadic operator (dyadically). This would apply the operation of each operator from right to left on each value of the given input, the * would be monadic, and the plus would by dyadic with respect to the given element's initial value, and the result of the monadic * applied to it. The monadic definition of * is "sign", giving -1 if negative, 0 for 0 and 1 for positive. So an array of positive numbers would result in each value being incremented by 1 essentially.

This is probably not helpful at all, sorry. It has to be designed like this in order for these concepts to work universally, unfortunately. You could define a dyadic multiplication for the multiplication as /*, which would collapse each dimension to a scalar, then pass the result to the +, which would be monadic for rank 0 input, and dyadic across dimensions (of size 1) for rank > 0

Anonymous

Anonymous Tue 10 Nov 2020 18:00:33 No.12326963 Report

Quoted By:

>>12326148
You can also drop the parentheses and interpret it as vector times velocity gradient(matrix)

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