Tensor Calculus and Divergence

In summary, the conversation discusses the dyadic product and the divergence of a tensor. The correct expression for the divergence of the dyadic product is shown, but there is uncertainty about using the product rule and setting it equal to zero. The use of the dyadic product in relation to the momentum flux is also questioned, with the explanation that it is the only way to obtain the correct momentum fluxes.
  • #1
member 428835
Hi PF!

I have a question on the dyadic product and the divergence of a tensor. I've never formally leaned this, although I'm sure it's published somewhere, but this is how I understand the operators. Can someone tell me if this is right or wrong? Let's say I have some vector ##\vec{V} = v_x i + v_y j##. Then ##\vec{V} \otimes \vec{V} = (v_x i + v_y j)^2 = v_x v_x ii + v_x v_y ij+v_yv_xji+v_yv_yjj## (which is a 2 by 2 matrix). Now if I take $$\nabla \cdot (\vec{V} \otimes \vec{V}) = \partial_x (v_x v_x) (i\cdot i)i +\partial_x( v_x v_y)(i\cdot i)j+\partial_x (v_yv_x)(i\cdot j)i+\partial_x(v_yv_y)(i\cdot j)j+\\ \partial_y (v_x v_x) (j\cdot i)i +\partial_y( v_x v_y)(j\cdot i)j+\partial_y (v_yv_x)(j\cdot j)i+\partial_y(v_yv_y)(j\cdot j)j=\\
\partial_x (v_x v_x)i +\partial_x( v_x v_y)j+\partial_y (v_yv_x)i+\partial_y(v_yv_y)j$$ where I could then use the product rule, simplify, factor out ##\nabla \cdot \vec{V}## and set equal to zero by continuity (if ##\vec{V}## was velocity of an incompressible fluid). Do these operations look correct?

Again, I've never been showed this but it looks intuitive and feels correct.
 
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  • #2
joshmccraney said:
Hi PF!

I have a question on the dyadic product and the divergence of a tensor. I've never formally leaned this, although I'm sure it's published somewhere, but this is how I understand the operators. Can someone tell me if this is right or wrong? Let's say I have some vector ##\vec{V} = v_x i + v_y j##. Then ##\vec{V} \otimes \vec{V} = (v_x i + v_y j)^2 = v_x v_x ii + v_x v_y ij+v_yv_xji+v_yv_yjj## (which is a 2 by 2 matrix). Now if I take $$\nabla \cdot (\vec{V} \otimes \vec{V}) = \partial_x (v_x v_x) (i\cdot i)i +\partial_x( v_x v_y)(i\cdot i)j+\partial_x (v_yv_x)(i\cdot j)i+\partial_x(v_yv_y)(i\cdot j)j+\\ \partial_y (v_x v_x) (j\cdot i)i +\partial_y( v_x v_y)(j\cdot i)j+\partial_y (v_yv_x)(j\cdot j)i+\partial_y(v_yv_y)(j\cdot j)j=\\
\partial_x (v_x v_x)i +\partial_x( v_x v_y)j+\partial_y (v_yv_x)i+\partial_y(v_yv_y)j$$
This all looks correct up to here.
where I could then use the product rule, simplify, factor out ##\nabla \cdot \vec{V}## and set equal to zero by continuity (if ##\vec{V}## was velocity of an incompressible fluid).
This part, I'm not so sure about. Why don't you expand it out using the product rule and see what you get?
 
  • #3
Chestermiller said:
This part, I'm not so sure about. Why don't you expand it out using the product rule and see what you get?
I did it on paper and it works! The result is the convective term for the Navier-Stokes Eq! Did you want me to post it?

Since you're here on this thread, I do have a question on Navier-Stokes. If I need to post a different thread I'm happy to, but my question is, why use the dyadic product when considering the momentum flux? Why ##\vec{V}\otimes \vec{V}## rather than, say, the inner/outer product ##\vec{V}:\vec{V}## or some other product that gives a second rank tensor from two vectors?
 
  • #4
Just my 2 cents:

If you take the inner product $V\cdot V$, you lose information about the direction of the momentum flux. If you want to construct a 2-tensor, there is not much other choice then the direct product, since the direct product of a vector with itself is symmetric; the only other choice I can think of is the traceless part of it,

[tex]
V_i V_j - \frac{1}{2} (V \cdot V) \delta_{ij} \,.
[/tex]

Note that this transforms irreducibly under the group of spatial rotations SO(2). Btw, your first expression is correct, because in components it reads (using your incompressibility)

[tex]
\partial_i (V^i V^j) = (\partial_i V^i) V^j + V^i \partial_i V^j = V^i \partial_i V^j
[/tex]
 
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  • #5
joshmccraney said:
I did it on paper and it works! The result is the convective term for the Navier-Stokes Eq! Did you want me to post it?

Since you're here on this thread, I do have a question on Navier-Stokes. If I need to post a different thread I'm happy to, but my question is, why use the dyadic product when considering the momentum flux? Why ##\vec{V}\otimes \vec{V}## rather than, say, the inner/outer product ##\vec{V}:\vec{V}## or some other product that gives a second rank tensor from two vectors?
There is only one way of doing it that gives the momentum fluxes correctly. ##\vec{V}:\vec{V}## is not a 2nd order tensor. In fact, it is not anything.

In elementary treatments, the dyadic form of the momentum flux is typically derived by using a control volume, and evaluating the momentum flux terms in the differential force balance first, and then recognizing that it can also be expressed in terms of the divergence of the dyadic product. Alternately, in relativistic developments, it is obtained by taking the divergence of the stress-energy tensor.
 
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1. What is Tensor Calculus?

Tensor Calculus is a branch of mathematics that deals with the study of tensors, which are mathematical objects that describe linear relations between different sets of vectors, scalars, and other tensors. It is used in various fields such as physics, engineering, and computer science to model and analyze complex systems.

2. What are the applications of Tensor Calculus?

Tensor Calculus has a wide range of applications in fields such as mechanics, electromagnetism, general relativity, computer graphics, and machine learning. It is used to model and analyze physical systems, design control systems, and develop algorithms for machine learning and data analysis.

3. What is the divergence of a vector field?

The divergence of a vector field is a scalar value that represents the magnitude of the outward flux of the vector field at a given point. It is a measure of how much the vector field spreads out or converges at that point. Mathematically, it is defined as the dot product of the vector field and the del operator.

4. What is the relationship between Tensor Calculus and Divergence?

Tensor Calculus is used to express the divergence of a vector field in a general coordinate system, where the vector field and del operator are represented by tensors. This allows for a more flexible and powerful approach to calculating divergence in complex systems. Tensor Calculus is also used to derive and solve equations involving divergence.

5. How is the concept of divergence used in physics?

In physics, divergence is used to describe the flow of a vector field such as velocity or force. It is particularly useful in fluid mechanics, where it is used to characterize the flow of fluids and gases. In electromagnetism, divergence is used to describe the flow of electric and magnetic fields, and in general relativity, it is used to describe the curvature of spacetime.

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