How Do You Continue the Proof for the Dyadic Product in Fluid Dynamics?

[Feodalherren]

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

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?

 

[Chestermiller]

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

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

 

[Feodalherren]

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.

 

[Chestermiller]

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

 

[Feodalherren]

<br /> i * ii = (1)i = i<br />

<br /> i * ji = (0)i = 0<br />

<br /> i*ki = (0)i=0<br />

<br /> i * ∇V =&gt;<br />

<br /> ∇V =

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

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

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

<br /> i * ∇V = (1i + 0j + 0k) * ∇V =<br />

<br /> (∂/∂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.<br />

<br /> = (∂/∂x)Vx i + 0(1)i +0(1)i + (∂/∂x)Vy(1) j + 0(1)j + 0(1)j ... etc

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.

 

[Chestermiller]

<br /> i * ii = (1)i = i<br />

<br /> i * ji = (0)i = 0<br />

<br /> i*ki = (0)i=0<br />

<br /> i * ∇V =&gt;<br />

<br /> ∇V =

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

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

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

No. This is supposed to be a summation:

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

<br /> i * ∇V = (1i + 0j + 0k) * ∇V =<br />

<br /> (∂/∂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.<br />

<br /> = (∂/∂x)Vx i + 0(1)i +0(1)i + (∂/∂x)Vy(1) j + 0(1)j + 0(1)j ... etc

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

 

[Feodalherren]

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

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!

 

[Chestermiller]

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

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

 

[Feodalherren]

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

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

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

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.

 

[Feodalherren]

Oh wait I see I factored it wrong. Give me a minute

 

[Feodalherren]

Okay I solved it. Thanks! :)

 

[Chestermiller]

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

 

[Feodalherren]

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 :)

 

[Chestermiller]

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