Fluid Dynamics: Dyadic product

In summary, the conversation discusses a problem with moving on with a proof and whether a dot product can be taken between a vector and a larger expression. The expert suggests taking a step back and evaluating the dot product term-by-term, and also provides a test for the individual to try. The summary concludes with the individual admitting that they do not know how to take the dot product of a vector with a tensor.
  • #1
Feodalherren
605
6

Homework Statement


Hi, I wasn't sure whether to post this here or in the engineering forums. Since it's mainly math/theory I figured here would be more appropriate. Feel free to move it if it doesn't belong here. All relevant info etc. is in the picture, thanks.

Homework Equations


asd.png


The Attempt at a Solution


Ok so my problem is, how do I move on with the proof from here?
Can you take the dot product between just the vector V=<Vx, Vy, Vz> and this whole big mess?
 
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  • #2
Feodalherren said:

Homework Statement


Hi, I wasn't sure whether to post this here or in the engineering forums. Since it's mainly math/theory I figured here would be more appropriate. Feel free to move it if it doesn't belong here. All relevant info etc. is in the picture, thanks.

Homework Equations


asd.png


The Attempt at a Solution


Ok so my problem is, how do I move on with the proof from here?
Can you take the dot product between just the vector V=<Vx, Vy, Vz> and this whole big mess?
Yes. It shouldn't be hard since dot products between two same unit vectors is 1 and dot products between two non-same orthogonal unit vectors is 0.

Chet
 
  • #3
I get the wrong result. My result becomes the whole thing added together. Since you multiply row by column, all of the i, j and k's in my dyad would be dotted by themselves.
 
  • #4
Let's take a step backwards. Here is a test for you.

$$\vec{i}\centerdot \vec{i}\vec{i}=$$
$$\vec{i}\centerdot \vec{j}\vec{i}=$$
$$\vec{i}\centerdot \vec{k}\vec{i}=$$
$$\vec{i}\centerdot \vec{∇}\vec{v}=$$
In the last problem, evaluate the summation term-by-term.

Chet
 
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  • #5
[itex]
i * ii = (1)i = i
[/itex]

[itex]
i * ji = (0)i = 0
[/itex]

[itex]
i*ki = (0)i=0
[/itex]

[itex]
i * ∇V =>
[/itex]

[itex]
∇V =[/itex]

[itex](∂/∂x)Vx (ii) , (∂/∂x)Vy (ij) , (∂/∂x)Vz (ik)[/itex]

[itex](∂/∂y)Vx (ji) , (∂/∂y)Vy (jj) , (∂/∂y)Vz (jk)[/itex]

[itex](∂/∂z)Vx (ki) , (∂/∂z)Vy (kj) , (∂/∂z)Vz (kk)[/itex]

[itex]
i * ∇V = (1i + 0j + 0k) * ∇V =
[/itex]

[itex]
(∂/∂x)Vx (i*i)i + 0((∂/∂y)Vx (j*j)i + 0(∂/∂z)Vx (k*k)i + (∂/∂x)Vy (i*i)j + 0(∂/∂y)Vy (j*j)j + (0)(∂/∂z)Vy (k*k)j ... etc.
[/itex]

[itex]
= (∂/∂x)Vx i + 0(1)i +0(1)i + (∂/∂x)Vy(1) j + 0(1)j + 0(1)j ... etc[/itex]

So it seems that the middle terms cancel in this case only because it is i dot delV, and not V dot delV, since V would have i j and k components, meaning no terms would be zero. Clearly I'm going completely off the rails somewhere but I can't seem to figure it out.
 
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  • #6
Feodalherren said:
[itex]
i * ii = (1)i = i
[/itex]

[itex]
i * ji = (0)i = 0
[/itex]

[itex]
i*ki = (0)i=0
[/itex]

[itex]
i * ∇V =>
[/itex]

[itex]
∇V =[/itex]

[itex](∂/∂x)Vx (ii) , (∂/∂x)Vy (ij) , (∂/∂x)Vz (ik)[/itex]

[itex](∂/∂y)Vx (ji) , (∂/∂y)Vy (jj) , (∂/∂y)Vz (jk)[/itex]

[itex](∂/∂z)Vx (ki) , (∂/∂z)Vy (kj) , (∂/∂z)Vz (kk)[/itex]

No. This is supposed to be a summation:

$$\vec{∇}\vec{V}=\frac{\partial V_x}{\partial x}\vec{i}\vec{i}+\frac{\partial V_y}{\partial x}\vec{i}\vec{j}+\frac{\partial V_z}{\partial x}\vec{i}\vec{k}+\frac{\partial V_x}{\partial y}\vec{j}\vec{i}+\frac{\partial V_y}{\partial y}\vec{j}\vec{j}+\frac{\partial V_z}{\partial y}\vec{j}\vec{k}+\frac{\partial V_x}{\partial z}\vec{k}\vec{i}+\frac{\partial V_y}{\partial z}\vec{k}\vec{j}+\frac{\partial V_z}{\partial z}\vec{k}\vec{k}$$
[itex]
i * ∇V = (1i + 0j + 0k) * ∇V =
[/itex]

[itex]
(∂/∂x)Vx (i*i)i + 0((∂/∂y)Vx (j*j)i + 0(∂/∂z)Vx (k*k)i + (∂/∂x)Vy (i*i)j + 0(∂/∂y)Vy (j*j)j + (0)(∂/∂z)Vy (k*k)j ... etc.
[/itex]

[itex]
= (∂/∂x)Vx i + 0(1)i +0(1)i + (∂/∂x)Vy(1) j + 0(1)j + 0(1)j ... etc[/itex]

So it seems that the middle terms cancel in this case only because it is i dot delV, and not V dot delV, since V would have i j and k components, meaning no terms would be zero. Clearly I'm going completely off the rails somewhere but I can't seem to figure it out.
Don't be so hard on yourself. This is all correct. To summarize:

$$\vec{i}\centerdot \vec{∇}\vec{V}=\frac{\partial V_x}{\partial x}\vec{i}+\frac{\partial V_y}{\partial x}\vec{j}+\frac{\partial V_z}{\partial x}\vec{k}$$
Now for the next part of your test:
$$(v_x\vec{i})\centerdot \vec{∇}\vec{V}=$$
$$(v_y\vec{j})\centerdot \vec{∇}\vec{V}=$$
$$(v_z\vec{k})\centerdot \vec{∇}\vec{V}=$$
$$(v_x\vec{i}+v_y\vec{j}+v_z\vec{k})\centerdot \vec{∇}\vec{V}=$$
$$\vec{V}\centerdot \vec{∇}\vec{V}=$$
Make sure you collect coefficients of i, j, and k.

Chet
 
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  • #7
$$\vec{∇}\vec{V}=\frac{\partial V_x}{\partial x}\vec{i}\vec{i}+\frac{\partial V_y}{\partial x}\vec{i}\vec{j}+\frac{\partial V_z}{\partial x}\vec{i}\vec{k}+\frac{\partial V_x}{\partial y}\vec{j}\vec{i}+\frac{\partial V_y}{\partial y}\vec{j}\vec{j}+\frac{\partial V_z}{\partial y}\vec{j}\vec{k}+\frac{\partial V_x}{\partial z}\vec{k}\vec{i}+\frac{\partial V_y}{\partial z}\vec{k}\vec{j}+\frac{\partial V_z}{\partial z}\vec{k}\vec{k}$$

then

$$(v_x\vec{i})\centerdot \vec{∇}\vec{V}=$$

$$V_x\frac{\partial V_x}{\partial x}(1)\vec{i}+V_x\frac{\partial V_y}{\partial x}(1)\vec{j}+V_x\frac{\partial V_z}{\partial x}(1)\vec{k}+V_x\frac{\partial V_x}{\partial y}(0)\vec{i}+V_x\frac{\partial V_y}{\partial y}(0)\vec{j}+V_x\frac{\partial V_z}{\partial y}(0)\vec{k}+V_x\frac{\partial V_x}{\partial z}(0)\vec{i}+\frac{\partial V_y}{\partial z}(0)\vec{j}+V_x\frac{\partial V_z}{\partial z}(0)\vec{k}$$

$$=V_x\frac{\partial V_x}{\partial x}(1)\vec{i}+V_x\frac{\partial V_y}{\partial x}(1)\vec{j}+V_x\frac{\partial V_z}{\partial x}(1)\vec{k}$$

Hmm yeah I can already tell that's wrong. I guess I at least identified my problem now. I don't know how to do the dot product if I don't have two vectors of the same size. I don't know how to take the dot product of a vector with a tensor.

I really appreciate your help by the way, thank you Sir!
 
  • #8
Feodalherren said:
$$\vec{∇}\vec{V}=\frac{\partial V_x}{\partial x}\vec{i}\vec{i}+\frac{\partial V_y}{\partial x}\vec{i}\vec{j}+\frac{\partial V_z}{\partial x}\vec{i}\vec{k}+\frac{\partial V_x}{\partial y}\vec{j}\vec{i}+\frac{\partial V_y}{\partial y}\vec{j}\vec{j}+\frac{\partial V_z}{\partial y}\vec{j}\vec{k}+\frac{\partial V_x}{\partial z}\vec{k}\vec{i}+\frac{\partial V_y}{\partial z}\vec{k}\vec{j}+\frac{\partial V_z}{\partial z}\vec{k}\vec{k}$$

then

$$(v_x\vec{i})\centerdot \vec{∇}\vec{V}=$$

$$V_x\frac{\partial V_x}{\partial x}(1)\vec{i}+V_x\frac{\partial V_y}{\partial x}(1)\vec{j}+V_x\frac{\partial V_z}{\partial x}(1)\vec{k}+V_x\frac{\partial V_x}{\partial y}(0)\vec{i}+V_x\frac{\partial V_y}{\partial y}(0)\vec{j}+V_x\frac{\partial V_z}{\partial y}(0)\vec{k}+V_x\frac{\partial V_x}{\partial z}(0)\vec{i}+\frac{\partial V_y}{\partial z}(0)\vec{j}+V_x\frac{\partial V_z}{\partial z}(0)\vec{k}$$

$$=V_x\frac{\partial V_x}{\partial x}(1)\vec{i}+V_x\frac{\partial V_y}{\partial x}(1)\vec{j}+V_x\frac{\partial V_z}{\partial x}(1)\vec{k}$$

Hmm yeah I can already tell that's wrong. I guess I at least identified my problem now. I don't know how to do the dot product if I don't have two vectors of the same size. I don't know how to take the dot product of a vector with a tensor.

I really appreciate your help by the way, thank you Sir!
I don't know why you think you got it wrong. You got it right. Just get rid of those (1)'s. 1 times (anything) is equal to (anything).

As far as the dot product of a vector with a tensor is concerned, that is a new vector. Dotting a vector with a tensor just maps the vector into a new vector.

With this information, you should be able to do the other problems I posed in post #6.

Chet
 
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  • #9
$$(v_x\vec{i})\centerdot \vec{∇}\vec{V}=$$

$$V_x\frac{\partial V_x}{\partial x}(1)\vec{i}+V_x\frac{\partial V_y}{\partial x}(1)\vec{j}+V_x\frac{\partial V_z}{\partial x}(1)\vec{k}+V_x\frac{\partial V_x}{\partial y}(0)\vec{i}+V_x\frac{\partial V_y}{\partial y}(0)\vec{j}+V_x\frac{\partial V_z}{\partial y}(0)\vec{k}+V_x\frac{\partial V_x}{\partial z}(0)\vec{i}+\frac{\partial V_y}{\partial z}(0)\vec{j}+V_x\frac{\partial V_z}{\partial z}(0)\vec{k}$$

$$=V_x\frac{\partial V_x}{\partial x}(1)\vec{i}+V_x\frac{\partial V_y}{\partial x}(1)\vec{j}+V_x\frac{\partial V_z}{\partial x}(1)\vec{k}$$

then similarly

$$(V_y\vec{j})\centerdot \vec{∇}\vec{V}=$$

$$=V_y\frac{\partial V_x}{\partial y}\vec{i}+V_y\frac{\partial V_y}{\partial y}\vec{j}+V_y\frac{\partial V_z}{\partial y}\vec{k}$$

and$$(V_z\vec{k})\centerdot \vec{∇}\vec{V}=$$

$$=V_z\frac{\partial V_x}{\partial z}\vec{i}+V_z\frac{\partial V_y}{\partial z}\vec{j}+V_z\frac{\partial V_z}{\partial z}\vec{k}$$

Hmm okay. Now if I factor each term I can get

$$V_x[\frac{\partial V_x}{\partial x}\vec{i}+\frac{\partial V_y}{\partial x}\vec{j}+\frac{\partial V_z}{\partial x}\vec{k} ]+ $$
$$V_y[\frac{\partial V_x}{\partial y}\vec{i}+\frac{\partial V_y}{\partial y}\vec{j}+\frac{\partial V_z}{\partial y}\vec{k}] + $$
$$V_z[\frac{\partial V_x}{\partial z}\vec{i}+\frac{\partial V_y}{\partial z}\vec{j}+\frac{\partial V_z}{\partial z}\vec{k}]$$

$$=V_x[\vec{∇}V_x] + V_y[\vec{∇}V_y] + V_z[\vec{∇} V_z]=$$

And I'm stuck again. I don't see how I can get the dot product to the right place and the V vector out of this again.
 
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  • #10
Oh wait I see I factored it wrong. Give me a minute
 
  • #11
Okay I solved it. Thanks! :)
 
Last edited:
  • #12
You did it correctly, but the way you expressed the final result is not what I had in mind. What I was looking for was expressing the final result in the form $$\vec{V}\centerdot ∇ \vec{V}=(\ \ \ \ \ \ \ \ )\vec{i}+(\ \ \ \ \ \ \ \ )\vec{j}+(\ \ \ \ \ \ \ \ )\vec{k}$$
Just fill in what goes in the parentheses.

Chet
 
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  • #13
Yes Sir, I did that on paper when I re-worked it. I was just too lazy to write it all out on here. Thanks :)
 
  • #14
Feodalherren said:
Yes Sir, I did that on paper when I re-worked it. I was just too lazy to write it all out on here. Thanks :)
Well, what this gives you is the acceleration vector of the fluid at each point in the fluid (in cases where the flow is at steady state).

Chet
 
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What is a dyadic product in fluid dynamics?

A dyadic product is a mathematical operation that combines two vectors to create a third vector. In fluid dynamics, it is used to represent the stress tensor, which describes the forces acting on a fluid element.

How is the dyadic product used in fluid dynamics?

The dyadic product is used to calculate the stress tensor, which is essential in understanding the behavior of fluids. It helps in analyzing the forces and stresses that are acting on a fluid element.

What are the components of a dyadic product in fluid dynamics?

The components of a dyadic product in fluid dynamics are the two vectors that are being multiplied together. These vectors can represent different physical properties, such as velocity and pressure, and their combination gives us insight into the behavior of the fluid.

What is the significance of the dyadic product in fluid dynamics?

The dyadic product is significant in fluid dynamics because it allows us to mathematically describe and analyze the complex behavior of fluids. It helps us understand the forces and stresses that are acting on a fluid, which is crucial in many practical applications, such as aerodynamics and hydrodynamics.

How is the dyadic product related to other mathematical operations in fluid dynamics?

The dyadic product is related to other mathematical operations, such as the dot product and the cross product, which are also used in fluid dynamics. These operations allow us to manipulate and analyze the properties of fluids and provide valuable insights into their behavior.

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