Proving Double Inner Product of Derivative of 2nd Order Tensor w/ Another

In summary, the conversation is about proving the expression \dot{A}:B + A:\dot{B}=A^{\nabla J}:B+A:B^{\nabla J}. The notation includes A and B as II order tensors, : representing the inner product and \nabla J representing divergence. The conversation also mentions that this was a homework question for a Continuum Mechanics course and the person is seeking help in understanding and proving the expression.
  • #1
josh_machine
3
0
Some one please help me how to prove the following:

[tex]\dot{A}:B + A:\dot{B}=A^{\nabla J}:B+A:B^{\nabla J}[/tex]

A and B are II order tensors and : represents the inner product.
 
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  • #2


How do I prove the following:

[tex]\dot{A}:B + A:\dot{B}=A^{\nabla J}:B+A:B^{\nabla J}[/tex]
 
  • #3


Where did you get this notation from? What is your dot? What is your J? What do you mean by "inner product", for what kind of tensors? Any reference to some place where your original notation is defined?
 
  • #4


The dot represents material time derivative. A and B are second order tensors, eg Stress.
I myself am not clear what [tex]\nabla J[/tex] means here. However, I guess it represents divergence.

This was as a homework question for a Continuum Mechanics course. I have not got any luck trying to understand or prove this expression. Any insight will be greatly appreciated.
 
  • #5


To prove this statement, we can start by expanding the definition of the inner product for tensors A and B as:

\dot{A}:B = \sum_{i,j} \dot{A}_{ij}B_{ij}

A:\dot{B} = \sum_{i,j} A_{ij}\dot{B}_{ij}

Next, we can use the chain rule for differentiating tensors to express the derivatives \dot{A} and \dot{B} in terms of their respective gradients:

\dot{A}_{ij} = A_{ij}^{\nabla J}

\dot{B}_{ij} = B_{ij}^{\nabla J}

Substituting these into the expanded inner product expressions, we get:

\dot{A}:B = \sum_{i,j} A_{ij}^{\nabla J}B_{ij}

A:\dot{B} = \sum_{i,j} A_{ij}B_{ij}^{\nabla J}

Now, we can rearrange the terms and group them according to the definition of the inner product, which states that A:B = \sum_{i,j} A_{ij}B_{ij}. This gives us:

\dot{A}:B + A:\dot{B} = \sum_{i,j} (A_{ij}^{\nabla J}B_{ij} + A_{ij}B_{ij}^{\nabla J})

Using the properties of the tensor gradient, we can rewrite this as:

\sum_{i,j} (A_{ij}^{\nabla J}B_{ij} + A_{ij}B_{ij}^{\nabla J}) = \sum_{i,j} (A_{ij}^{\nabla J}:B_{ij} + A_{ij}:B_{ij}^{\nabla J})

This gives us the desired result:

\dot{A}:B + A:\dot{B} = A^{\nabla J}:B + A:B^{\nabla J}

Therefore, we have proven that the double inner product of the derivatives of two second order tensors is equal to the inner product of the gradients of those tensors. This result is useful in many areas of science, such as in the study of fluid mechanics and continuum mechanics.
 

What is a 2nd order tensor?

A 2nd order tensor is a mathematical object that represents a linear transformation between two vector spaces. It can be represented as a 2-dimensional array of numbers, and its behavior can be described using matrix operations.

What is a derivative of a 2nd order tensor?

The derivative of a 2nd order tensor is a new tensor that describes how the original tensor changes with respect to a particular variable. It can be calculated using the standard rules of calculus, but it also has its own unique properties and operations.

What is an inner product of tensors?

An inner product of tensors is a mathematical operation that takes two tensors as input and produces a scalar value as output. It is similar to the dot product between vectors, but it can also be defined for higher-order tensors.

How do you prove the double inner product of the derivative of a 2nd order tensor with another?

To prove the double inner product, you can use the definition of the inner product and the properties of tensor derivatives. You can also use the chain rule and product rule to simplify the expression and show that it is equal to the double inner product.

What is the significance of proving the double inner product of the derivative of a 2nd order tensor with another?

Proving the double inner product allows us to understand the relationship between the derivative of a 2nd order tensor and another tensor. It also helps us to derive important properties and equations involving these tensors, which can be used in various scientific and engineering applications.

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