Tensor product in momentum equation

jinggca · September 8, 2022, 20:21

I am wondering that when we derived the momentum equation using the differential form of the transport equation, $\frac{d\phi}{dt}+\nabla\cdot(\phi \mathbf{u}) + s = 0$ , and substituted $\rho \mathbf{u}$ for $\phi$ , we get a term that looks like $\nabla\cdot(\rho \mathbf{u}\mathbf{u})$ . How do we know we should use tensor product between the two $\mathbf{u}$ , instead of say cross product or dot product? I know that the term will only have the correct shape (i.e. it has to be a vector after taking the divergence so that it can be added to other terms in the equation), but is there a physical meaning behind the tensor product? Thank you!

LuckyTran · September 8, 2022, 23:47

Well yes the tell is that it must be a tensor and not a vector or scalar, which immediately eliminates the possibility of a cross or inner product. Even compilers will know not to compile a sequence that has mismatched operands and throw an error if the classes are incorrect, there's no reason an intelligent human being cannot do better.

Obviously it is better if everything comes with a dictionary and a manual of how to do it. But outer and inner products already have their notations. It would be a little silly not to use them. For example in a polynomial equation ab = c, how do you know ab is the product of a and b and not addition of a and b, a+ b? Is it because you seek a higher authority to decide things for you? Or is it because addition already has a notation a+b and if you wanted it to signify addition, you would write a + b and not ab? Dyadic notation is extension of vector notation into dyadics and specifically here you are use nabla vector notation where those have clear conventions for the dot and cross products (i.e. $\\nabla\cdot$ already implies dot operator being applied so why would you ignore this convention in the very next symbol $\nabla\cdot(\rho \mathbf{u}\mathbf{u})$ ).

$\rho \mathbf{u}\mathbf{u}$ are your transport rates. They are the x,y,z (i,j,k) components of momentum transported by the x,y,z(i,j,k) components of the mass flux.

jinggca · September 9, 2022, 01:31

Apart from your unprofessionalism I think you missed the point of my question, and perhaps I did not make it very clear so my apology.

I guess a better way to phrase my question is this: the dummy property $\phi$ in the transport equation can be both scalar and vector, or even higher order tensors. If it is a scalar quantity, then the operation between $\phi$ and $\mathbf{u}$ is very clear and without ambiguity. However, if $\phi$ is now a vector quantity, the definition of the same operation is ambiguous, because there are more than one type of vector product. My source of the derivation of the transport equation did not make it clear how to interpret such an operation, but went straight ahead using the tensor product, and thus arises my question.

jinggca · September 9, 2022, 01:32

I think after a bit of thought I can answer this question myself now. Basically if $\phi$ is a vector quantity, then the transport equation is applied to every element of the vector individually. So for example if $\phi$ is a 2x1 vector, then there will be 2 transport equations, one for $\phi_1$ and one for $\phi_2$ . In order to keep things compact and not having to write multiple equations, using the tensor product will yield exactly the same system of equations in vector form, eliminating the need for multi-lined, system of equations.

LuckyTran · September 9, 2022, 03:30

Now do it for the triple products and quadruple products and higher order products. If the uu is not a dyadic product then you'll a growing number of notations in order to determine if it is the inner product, double inner product, triple inner product, double outer product, triple outer product, and so on. The point of notational conventions, is actually following them. That's why they're conventions.

Also note that the dyadic product is a subclass of tensor products and it is wrong in general to call it a tensor product. It works for navier-stokes because the velocity vector is rank 1 and uu actually is a dyadic product.

FMDenaro · September 9, 2022, 03:56

Quote:

Originally Posted by jinggca

Apart from your unprofessionalism I think you missed the point of my question, and perhaps I did not make it very clear so my apology.

I guess a better way to phrase my question is this: the dummy property $\phi$ in the transport equation can be both scalar and vector, or even higher order tensors. If it is a scalar quantity, then the operation between $\phi$ and $\mathbf{u}$ is very clear and without ambiguity. However, if $\phi$ is now a vector quantity, the definition of the same operation is ambiguous, because there are more than one type of vector product. My source of the derivation of the transport equation did not make it clear how to interpret such an operation, but went straight ahead using the tensor product, and thus arises my question.

Look at the time derivative. If you have phi as vector you cannot think to have a scalar product in the convective term, it must be a tensor to be congruent with the vectore time derivative.

sbaffini · September 9, 2022, 07:40

What Lucky was probably trying to convey is that you are using your source material wrong or it is wrong the material itself.

If you were using the Einstein notation, you would have 0 doubt about everything. So, in Lucky terms, it is your impression (or a problem in the material you are reading) that the operation is subject to interpretation, but it actually isn't. To be more precise, where would a cross or dot product come from? They are simply not there.

FMDenaro · September 9, 2022, 09:07

The meaning of phi is always related to the transport theorem. Any quantity phi is advected by normal component of the mass flux rho u. Consequently no scalar product can appear since

Int[S] rho un phi dS

is the meaningful physical contribution. For the mass phi=1, for the momentum phi=u.

September 8, 2022, 20:21	Tensor product in momentum equation	#1
jinggca New Member William Join Date: Sep 2022 Posts: 8 Rep Power: 4	I am wondering that when we derived the momentum equation using the differential form of the transport equation, $\frac{d\phi}{dt}+\nabla\cdot(\phi \mathbf{u}) + s = 0$ , and substituted $\rho \mathbf{u}$ for $\phi$ , we get a term that looks like $\nabla\cdot(\rho \mathbf{u}\mathbf{u})$ . How do we know we should use tensor product between the two $\mathbf{u}$ , instead of say cross product or dot product? I know that the term will only have the correct shape (i.e. it has to be a vector after taking the divergence so that it can be added to other terms in the equation), but is there a physical meaning behind the tensor product? Thank you!

September 8, 2022, 23:47		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,815 Rep Power: 68	Well yes the tell is that it must be a tensor and not a vector or scalar, which immediately eliminates the possibility of a cross or inner product. Even compilers will know not to compile a sequence that has mismatched operands and throw an error if the classes are incorrect, there's no reason an intelligent human being cannot do better. Obviously it is better if everything comes with a dictionary and a manual of how to do it. But outer and inner products already have their notations. It would be a little silly not to use them. For example in a polynomial equation ab = c, how do you know ab is the product of a and b and not addition of a and b, a+ b? Is it because you seek a higher authority to decide things for you? Or is it because addition already has a notation a+b and if you wanted it to signify addition, you would write a + b and not ab? Dyadic notation is extension of vector notation into dyadics and specifically here you are use nabla vector notation where those have clear conventions for the dot and cross products (i.e. $\\nabla\cdot$ already implies dot operator being applied so why would you ignore this convention in the very next symbol $\nabla\cdot(\rho \mathbf{u}\mathbf{u})$ ). $\rho \mathbf{u}\mathbf{u}$ are your transport rates. They are the x,y,z (i,j,k) components of momentum transported by the x,y,z(i,j,k) components of the mass flux. Last edited by LuckyTran; September 9, 2022 at 00:55.

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September 9, 2022, 01:31		#3
jinggca New Member William Join Date: Sep 2022 Posts: 8 Rep Power: 4	Apart from your unprofessionalism I think you missed the point of my question, and perhaps I did not make it very clear so my apology. I guess a better way to phrase my question is this: the dummy property $\phi$ in the transport equation can be both scalar and vector, or even higher order tensors. If it is a scalar quantity, then the operation between $\phi$ and $\mathbf{u}$ is very clear and without ambiguity. However, if $\phi$ is now a vector quantity, the definition of the same operation is ambiguous, because there are more than one type of vector product. My source of the derivation of the transport equation did not make it clear how to interpret such an operation, but went straight ahead using the tensor product, and thus arises my question.

September 9, 2022, 01:32		#4
jinggca New Member William Join Date: Sep 2022 Posts: 8 Rep Power: 4	I think after a bit of thought I can answer this question myself now. Basically if $\phi$ is a vector quantity, then the transport equation is applied to every element of the vector individually. So for example if $\phi$ is a 2x1 vector, then there will be 2 transport equations, one for $\phi_1$ and one for $\phi_2$ . In order to keep things compact and not having to write multiple equations, using the tensor product will yield exactly the same system of equations in vector form, eliminating the need for multi-lined, system of equations.

September 9, 2022, 03:30		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,815 Rep Power: 68	Now do it for the triple products and quadruple products and higher order products. If the uu is not a dyadic product then you'll a growing number of notations in order to determine if it is the inner product, double inner product, triple inner product, double outer product, triple outer product, and so on. The point of notational conventions, is actually following them. That's why they're conventions. Also note that the dyadic product is a subclass of tensor products and it is wrong in general to call it a tensor product. It works for navier-stokes because the velocity vector is rank 1 and uu actually is a dyadic product.

September 9, 2022, 07:40		#7
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,209 Blog Entries: 29 Rep Power: 41	What Lucky was probably trying to convey is that you are using your source material wrong or it is wrong the material itself. If you were using the Einstein notation, you would have 0 doubt about everything. So, in Lucky terms, it is your impression (or a problem in the material you are reading) that the operation is subject to interpretation, but it actually isn't. To be more precise, where would a cross or dot product come from? They are simply not there.

September 9, 2022, 09:07		#8
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,982 Rep Power: 73	The meaning of phi is always related to the transport theorem. Any quantity phi is advected by normal component of the mass flux rho u. Consequently no scalar product can appear since Int[S] rho un phi dS is the meaningful physical contribution. For the mass phi=1, for the momentum phi=u.