“Vector add to scalar” in left part of Navier-Stokes equation












2












$begingroup$


The left part of Navier-Stokes equation is:
$dfrac{Dvec{v}}{D t}= dfrac{partialvec{v}}{partial t}+ vec{v} cdotnabla vec{v}$



Let's take $vec{v}$ as a two dimentional vector: $(u,v)$. Then: $dfrac{partialvec{v}}{partial t}$ is $(dfrac{partial u}{partial t}, dfrac{partial v}{partial t})$, which is a vector. However, $vec{v} cdotnabla vec{v}$ will be $(u,v)cdot(dfrac{partial u}{partial x},dfrac{partial v}{partial y}) = udfrac{partial u}{partial x}+vdfrac{partial v}{partial y}$, which is a scalar. How a "vector" can be summed with a "scalar"?



Of course my thought is wrong somewhere. Please help me.






pde







share|cite|improve this question









$endgroup$












  • $begingroup$
    directional derivative
    $endgroup$
    – timur
    Dec 10 '18 at 7:40






  • 3




    $begingroup$
    $nablavec{v}$ is not $(dfrac{partial u}{partial x},dfrac{partial v}{partial y})$, it is the rank-2 tensor with entries $dfrac{partial u}{partial x},dfrac{partial u}{partial y},dfrac{partial v}{partial x},dfrac{partial v}{partial y}$
    $endgroup$
    – user10354138
    Dec 10 '18 at 7:46






  • 1




    $begingroup$
    @user10354138 -- Then the dot must be interpreted as tensor contraction. I guess that produces the same results, but it's probably not what was meant by the equation.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 7:51








  • 2




    $begingroup$
    @mr_e_man Yes it is what is meant by the equation, in the definition of $frac{D}{Dt}=frac{partial}{partial t}+mathbf{v}cdotnabla$ as the material derivative.
    $endgroup$
    – user10354138
    Dec 10 '18 at 8:16










  • $begingroup$
    @user10354138 -- I mean it's probably supposed to be $(vcdotnabla)v$ instead of $vcdot(nabla v)$.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 19:18
















2












$begingroup$


The left part of Navier-Stokes equation is:
$dfrac{Dvec{v}}{D t}= dfrac{partialvec{v}}{partial t}+ vec{v} cdotnabla vec{v}$



Let's take $vec{v}$ as a two dimentional vector: $(u,v)$. Then: $dfrac{partialvec{v}}{partial t}$ is $(dfrac{partial u}{partial t}, dfrac{partial v}{partial t})$, which is a vector. However, $vec{v} cdotnabla vec{v}$ will be $(u,v)cdot(dfrac{partial u}{partial x},dfrac{partial v}{partial y}) = udfrac{partial u}{partial x}+vdfrac{partial v}{partial y}$, which is a scalar. How a "vector" can be summed with a "scalar"?



Of course my thought is wrong somewhere. Please help me.






pde







share|cite|improve this question









$endgroup$












  • $begingroup$
    directional derivative
    $endgroup$
    – timur
    Dec 10 '18 at 7:40






  • 3




    $begingroup$
    $nablavec{v}$ is not $(dfrac{partial u}{partial x},dfrac{partial v}{partial y})$, it is the rank-2 tensor with entries $dfrac{partial u}{partial x},dfrac{partial u}{partial y},dfrac{partial v}{partial x},dfrac{partial v}{partial y}$
    $endgroup$
    – user10354138
    Dec 10 '18 at 7:46






  • 1




    $begingroup$
    @user10354138 -- Then the dot must be interpreted as tensor contraction. I guess that produces the same results, but it's probably not what was meant by the equation.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 7:51








  • 2




    $begingroup$
    @mr_e_man Yes it is what is meant by the equation, in the definition of $frac{D}{Dt}=frac{partial}{partial t}+mathbf{v}cdotnabla$ as the material derivative.
    $endgroup$
    – user10354138
    Dec 10 '18 at 8:16










  • $begingroup$
    @user10354138 -- I mean it's probably supposed to be $(vcdotnabla)v$ instead of $vcdot(nabla v)$.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 19:18














2












2








2


1



$begingroup$


The left part of Navier-Stokes equation is:
$dfrac{Dvec{v}}{D t}= dfrac{partialvec{v}}{partial t}+ vec{v} cdotnabla vec{v}$



Let's take $vec{v}$ as a two dimentional vector: $(u,v)$. Then: $dfrac{partialvec{v}}{partial t}$ is $(dfrac{partial u}{partial t}, dfrac{partial v}{partial t})$, which is a vector. However, $vec{v} cdotnabla vec{v}$ will be $(u,v)cdot(dfrac{partial u}{partial x},dfrac{partial v}{partial y}) = udfrac{partial u}{partial x}+vdfrac{partial v}{partial y}$, which is a scalar. How a "vector" can be summed with a "scalar"?



Of course my thought is wrong somewhere. Please help me.






pde







share|cite|improve this question









$endgroup$




The left part of Navier-Stokes equation is:
$dfrac{Dvec{v}}{D t}= dfrac{partialvec{v}}{partial t}+ vec{v} cdotnabla vec{v}$



Let's take $vec{v}$ as a two dimentional vector: $(u,v)$. Then: $dfrac{partialvec{v}}{partial t}$ is $(dfrac{partial u}{partial t}, dfrac{partial v}{partial t})$, which is a vector. However, $vec{v} cdotnabla vec{v}$ will be $(u,v)cdot(dfrac{partial u}{partial x},dfrac{partial v}{partial y}) = udfrac{partial u}{partial x}+vdfrac{partial v}{partial y}$, which is a scalar. How a "vector" can be summed with a "scalar"?



Of course my thought is wrong somewhere. Please help me.






pde




pde






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 7:24









T XT X

154




154












  • $begingroup$
    directional derivative
    $endgroup$
    – timur
    Dec 10 '18 at 7:40






  • 3




    $begingroup$
    $nablavec{v}$ is not $(dfrac{partial u}{partial x},dfrac{partial v}{partial y})$, it is the rank-2 tensor with entries $dfrac{partial u}{partial x},dfrac{partial u}{partial y},dfrac{partial v}{partial x},dfrac{partial v}{partial y}$
    $endgroup$
    – user10354138
    Dec 10 '18 at 7:46






  • 1




    $begingroup$
    @user10354138 -- Then the dot must be interpreted as tensor contraction. I guess that produces the same results, but it's probably not what was meant by the equation.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 7:51








  • 2




    $begingroup$
    @mr_e_man Yes it is what is meant by the equation, in the definition of $frac{D}{Dt}=frac{partial}{partial t}+mathbf{v}cdotnabla$ as the material derivative.
    $endgroup$
    – user10354138
    Dec 10 '18 at 8:16










  • $begingroup$
    @user10354138 -- I mean it's probably supposed to be $(vcdotnabla)v$ instead of $vcdot(nabla v)$.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 19:18


















  • $begingroup$
    directional derivative
    $endgroup$
    – timur
    Dec 10 '18 at 7:40






  • 3




    $begingroup$
    $nablavec{v}$ is not $(dfrac{partial u}{partial x},dfrac{partial v}{partial y})$, it is the rank-2 tensor with entries $dfrac{partial u}{partial x},dfrac{partial u}{partial y},dfrac{partial v}{partial x},dfrac{partial v}{partial y}$
    $endgroup$
    – user10354138
    Dec 10 '18 at 7:46






  • 1




    $begingroup$
    @user10354138 -- Then the dot must be interpreted as tensor contraction. I guess that produces the same results, but it's probably not what was meant by the equation.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 7:51








  • 2




    $begingroup$
    @mr_e_man Yes it is what is meant by the equation, in the definition of $frac{D}{Dt}=frac{partial}{partial t}+mathbf{v}cdotnabla$ as the material derivative.
    $endgroup$
    – user10354138
    Dec 10 '18 at 8:16










  • $begingroup$
    @user10354138 -- I mean it's probably supposed to be $(vcdotnabla)v$ instead of $vcdot(nabla v)$.
    $endgroup$
    – mr_e_man
    Dec 10 '18 at 19:18
















$begingroup$
directional derivative
$endgroup$
– timur
Dec 10 '18 at 7:40




$begingroup$
directional derivative
$endgroup$
– timur
Dec 10 '18 at 7:40




3




3




$begingroup$
$nablavec{v}$ is not $(dfrac{partial u}{partial x},dfrac{partial v}{partial y})$, it is the rank-2 tensor with entries $dfrac{partial u}{partial x},dfrac{partial u}{partial y},dfrac{partial v}{partial x},dfrac{partial v}{partial y}$
$endgroup$
– user10354138
Dec 10 '18 at 7:46




$begingroup$
$nablavec{v}$ is not $(dfrac{partial u}{partial x},dfrac{partial v}{partial y})$, it is the rank-2 tensor with entries $dfrac{partial u}{partial x},dfrac{partial u}{partial y},dfrac{partial v}{partial x},dfrac{partial v}{partial y}$
$endgroup$
– user10354138
Dec 10 '18 at 7:46




1




1




$begingroup$
@user10354138 -- Then the dot must be interpreted as tensor contraction. I guess that produces the same results, but it's probably not what was meant by the equation.
$endgroup$
– mr_e_man
Dec 10 '18 at 7:51






$begingroup$
@user10354138 -- Then the dot must be interpreted as tensor contraction. I guess that produces the same results, but it's probably not what was meant by the equation.
$endgroup$
– mr_e_man
Dec 10 '18 at 7:51






2




2




$begingroup$
@mr_e_man Yes it is what is meant by the equation, in the definition of $frac{D}{Dt}=frac{partial}{partial t}+mathbf{v}cdotnabla$ as the material derivative.
$endgroup$
– user10354138
Dec 10 '18 at 8:16




$begingroup$
@mr_e_man Yes it is what is meant by the equation, in the definition of $frac{D}{Dt}=frac{partial}{partial t}+mathbf{v}cdotnabla$ as the material derivative.
$endgroup$
– user10354138
Dec 10 '18 at 8:16












$begingroup$
@user10354138 -- I mean it's probably supposed to be $(vcdotnabla)v$ instead of $vcdot(nabla v)$.
$endgroup$
– mr_e_man
Dec 10 '18 at 19:18




$begingroup$
@user10354138 -- I mean it's probably supposed to be $(vcdotnabla)v$ instead of $vcdot(nabla v)$.
$endgroup$
– mr_e_man
Dec 10 '18 at 19:18










2 Answers
2






active

oldest

votes


















4












$begingroup$

As stated in the comments, this is the material derivative. If $mathbf{v} = (v_1,v_2)$ then



$$ (mathbf{v}cdot nabla) mathbf{v} = (mathbf{v}cdot nabla v_1, mathbf{v}cdot nabla v_2) = left(v_1 frac{partial v_1}{partial x} + v_2frac{partial v_1}{partial y}, v_1frac{partial v_2}{partial x} + v_2frac{partial v_2}{partial y}right) $$



You can also intepret it as a matrix product



$$ mathbf{v}cdot nabla mathbf{v} = begin{bmatrix} v_1 & v_2 end{bmatrix}begin{bmatrix} dfrac{partial v_1}{partial x} & dfrac{partial v_2}{partial x} \ dfrac{partial v_1}{partial y} & dfrac{partial v_2}{partial y} end{bmatrix} $$



where $nabla mathbf{v}$ represents a rank-2 tensor






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You're right in that scalars and vectors don't add, but have unfortunately - and understandably - got muddled up with how the operators are put together. If you put brackets around (v dot del) then it will make sense. This will be a scalar operator just like partial_t.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Scalars and vectors can be added in geometric algebra, though that's not happening here.
      $endgroup$
      – mr_e_man
      Dec 10 '18 at 7:46










    • $begingroup$
      That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur.
      $endgroup$
      – Paul Childs
      Dec 10 '18 at 7:51











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    As stated in the comments, this is the material derivative. If $mathbf{v} = (v_1,v_2)$ then



    $$ (mathbf{v}cdot nabla) mathbf{v} = (mathbf{v}cdot nabla v_1, mathbf{v}cdot nabla v_2) = left(v_1 frac{partial v_1}{partial x} + v_2frac{partial v_1}{partial y}, v_1frac{partial v_2}{partial x} + v_2frac{partial v_2}{partial y}right) $$



    You can also intepret it as a matrix product



    $$ mathbf{v}cdot nabla mathbf{v} = begin{bmatrix} v_1 & v_2 end{bmatrix}begin{bmatrix} dfrac{partial v_1}{partial x} & dfrac{partial v_2}{partial x} \ dfrac{partial v_1}{partial y} & dfrac{partial v_2}{partial y} end{bmatrix} $$



    where $nabla mathbf{v}$ represents a rank-2 tensor






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      As stated in the comments, this is the material derivative. If $mathbf{v} = (v_1,v_2)$ then



      $$ (mathbf{v}cdot nabla) mathbf{v} = (mathbf{v}cdot nabla v_1, mathbf{v}cdot nabla v_2) = left(v_1 frac{partial v_1}{partial x} + v_2frac{partial v_1}{partial y}, v_1frac{partial v_2}{partial x} + v_2frac{partial v_2}{partial y}right) $$



      You can also intepret it as a matrix product



      $$ mathbf{v}cdot nabla mathbf{v} = begin{bmatrix} v_1 & v_2 end{bmatrix}begin{bmatrix} dfrac{partial v_1}{partial x} & dfrac{partial v_2}{partial x} \ dfrac{partial v_1}{partial y} & dfrac{partial v_2}{partial y} end{bmatrix} $$



      where $nabla mathbf{v}$ represents a rank-2 tensor






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        As stated in the comments, this is the material derivative. If $mathbf{v} = (v_1,v_2)$ then



        $$ (mathbf{v}cdot nabla) mathbf{v} = (mathbf{v}cdot nabla v_1, mathbf{v}cdot nabla v_2) = left(v_1 frac{partial v_1}{partial x} + v_2frac{partial v_1}{partial y}, v_1frac{partial v_2}{partial x} + v_2frac{partial v_2}{partial y}right) $$



        You can also intepret it as a matrix product



        $$ mathbf{v}cdot nabla mathbf{v} = begin{bmatrix} v_1 & v_2 end{bmatrix}begin{bmatrix} dfrac{partial v_1}{partial x} & dfrac{partial v_2}{partial x} \ dfrac{partial v_1}{partial y} & dfrac{partial v_2}{partial y} end{bmatrix} $$



        where $nabla mathbf{v}$ represents a rank-2 tensor






        share|cite|improve this answer









        $endgroup$



        As stated in the comments, this is the material derivative. If $mathbf{v} = (v_1,v_2)$ then



        $$ (mathbf{v}cdot nabla) mathbf{v} = (mathbf{v}cdot nabla v_1, mathbf{v}cdot nabla v_2) = left(v_1 frac{partial v_1}{partial x} + v_2frac{partial v_1}{partial y}, v_1frac{partial v_2}{partial x} + v_2frac{partial v_2}{partial y}right) $$



        You can also intepret it as a matrix product



        $$ mathbf{v}cdot nabla mathbf{v} = begin{bmatrix} v_1 & v_2 end{bmatrix}begin{bmatrix} dfrac{partial v_1}{partial x} & dfrac{partial v_2}{partial x} \ dfrac{partial v_1}{partial y} & dfrac{partial v_2}{partial y} end{bmatrix} $$



        where $nabla mathbf{v}$ represents a rank-2 tensor







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 15:22









        DylanDylan

        12.8k31026




        12.8k31026























            1












            $begingroup$

            You're right in that scalars and vectors don't add, but have unfortunately - and understandably - got muddled up with how the operators are put together. If you put brackets around (v dot del) then it will make sense. This will be a scalar operator just like partial_t.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Scalars and vectors can be added in geometric algebra, though that's not happening here.
              $endgroup$
              – mr_e_man
              Dec 10 '18 at 7:46










            • $begingroup$
              That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur.
              $endgroup$
              – Paul Childs
              Dec 10 '18 at 7:51
















            1












            $begingroup$

            You're right in that scalars and vectors don't add, but have unfortunately - and understandably - got muddled up with how the operators are put together. If you put brackets around (v dot del) then it will make sense. This will be a scalar operator just like partial_t.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Scalars and vectors can be added in geometric algebra, though that's not happening here.
              $endgroup$
              – mr_e_man
              Dec 10 '18 at 7:46










            • $begingroup$
              That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur.
              $endgroup$
              – Paul Childs
              Dec 10 '18 at 7:51














            1












            1








            1





            $begingroup$

            You're right in that scalars and vectors don't add, but have unfortunately - and understandably - got muddled up with how the operators are put together. If you put brackets around (v dot del) then it will make sense. This will be a scalar operator just like partial_t.






            share|cite|improve this answer









            $endgroup$



            You're right in that scalars and vectors don't add, but have unfortunately - and understandably - got muddled up with how the operators are put together. If you put brackets around (v dot del) then it will make sense. This will be a scalar operator just like partial_t.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 7:37









            Paul ChildsPaul Childs

            1757




            1757












            • $begingroup$
              Scalars and vectors can be added in geometric algebra, though that's not happening here.
              $endgroup$
              – mr_e_man
              Dec 10 '18 at 7:46










            • $begingroup$
              That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur.
              $endgroup$
              – Paul Childs
              Dec 10 '18 at 7:51


















            • $begingroup$
              Scalars and vectors can be added in geometric algebra, though that's not happening here.
              $endgroup$
              – mr_e_man
              Dec 10 '18 at 7:46










            • $begingroup$
              That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur.
              $endgroup$
              – Paul Childs
              Dec 10 '18 at 7:51
















            $begingroup$
            Scalars and vectors can be added in geometric algebra, though that's not happening here.
            $endgroup$
            – mr_e_man
            Dec 10 '18 at 7:46




            $begingroup$
            Scalars and vectors can be added in geometric algebra, though that's not happening here.
            $endgroup$
            – mr_e_man
            Dec 10 '18 at 7:46












            $begingroup$
            That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur.
            $endgroup$
            – Paul Childs
            Dec 10 '18 at 7:51




            $begingroup$
            That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur.
            $endgroup$
            – Paul Childs
            Dec 10 '18 at 7:51


















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