1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fluid Dynamics: Dyadic product

  1. Sep 11, 2015 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    asd.png

    3. 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?
     
  2. jcsd
  3. Sep 11, 2015 #2
    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
     
  4. Sep 11, 2015 #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.
     
  5. Sep 11, 2015 #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
     
  6. Sep 12, 2015 #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.
     
    Last edited: Sep 12, 2015
  7. Sep 12, 2015 #6
    No. This is supposed to be a summation:

    $$\vec{∇}\vec{V}=\frac{\partial V_x}{\partial x}\vec{i}\vec{i}+\frac{\partial V_y}{\partial x}\vec{i}\vec{j}+\frac{\partial V_z}{\partial x}\vec{i}\vec{k}+\frac{\partial V_x}{\partial y}\vec{j}\vec{i}+\frac{\partial V_y}{\partial y}\vec{j}\vec{j}+\frac{\partial V_z}{\partial y}\vec{j}\vec{k}+\frac{\partial V_x}{\partial z}\vec{k}\vec{i}+\frac{\partial V_y}{\partial z}\vec{k}\vec{j}+\frac{\partial V_z}{\partial z}\vec{k}\vec{k}$$
    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
     
  8. Sep 12, 2015 #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!
     
  9. Sep 12, 2015 #8
    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
     
  10. Sep 14, 2015 #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.
     
    Last edited: Sep 14, 2015
  11. Sep 14, 2015 #10
    Oh wait I see I factored it wrong. Give me a minute
     
  12. Sep 14, 2015 #11
    Okay I solved it. Thanks! :)
     
    Last edited: Sep 14, 2015
  13. Sep 14, 2015 #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
     
  14. Sep 15, 2015 #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 :)
     
  15. Sep 15, 2015 #14
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fluid Dynamics: Dyadic product
  1. Fluid Dynamics (Replies: 1)

  2. Fluid Dynamics (Replies: 9)

  3. Fluid dynamics (Replies: 1)

  4. Fluid Dynamics (Replies: 3)

Loading...