Is A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} a Tensor?

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Your name]In summary, A_{1} B_{i} + A_{2} B_{j} + A_{3} B_{k} is a tensor of rank 2, as it follows the transformation rule T'_{ij} = R_{il} R_{jm} T_{lm}. This can be confirmed by substituting A_{1} B_{i} + A_{2} B_{j} + A_{3} B_{k} into the transformation rule and showing that it matches the original expression.
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typhoonss821
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Homework Statement



Let [tex] A_{i} , B_{j} , C_{k}[/tex]be vectors.
Is [tex]A_{1} B_{i} + A_{2} B_{j} + A_{3} B_{k}[/tex] a tensor?

Homework Equations


[tex]T'_{ijk} = R_{il} R_{jm} R_{kn} T_{ijk} [/tex]

The Attempt at a Solution


If T is a tensor of rank 3, then [tex]T'_{ijk} = R_{il} R_{jm} R_{kn} T_{ijk} [/tex] where [tex] R^\intercal R =I [/tex]. But I am not sure how to use this relation in the problem. > <

I think it is to use the property of vector [tex]A'_{i} = R_{il} A_{i} [/tex], but difficult appears when I try this way.

Please help...
 
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Hello,

Thank you for your question. To determine if A_{1} B_{i} + A_{2} B_{j} + A_{3} B_{k} is a tensor, we need to understand the properties of tensors and how they behave under transformations.

Firstly, a tensor is a mathematical object that describes a geometric relationship between vectors and scalars. It is characterized by its rank, which is the number of indices it has. In this case, we are dealing with a tensor of rank 2 because we have two indices (i and j).

Now, to determine if A_{1} B_{i} + A_{2} B_{j} + A_{3} B_{k} is a tensor, we need to check if it follows the transformation rule for tensors. As you have mentioned, the transformation rule for a rank 2 tensor is T'_{ij} = R_{il} R_{jm} T_{lm}.

In this case, we have A'_{i} = R_{il} A_{l} and B'_{j} = R_{jm} B_{m}. Therefore, A'_{i} B'_{j} = R_{il} A_{l} R_{jm} B_{m}.

Substituting this into our original expression, we get T'_{ij} = A'_{i} B'_{j} = R_{il} A_{l} R_{jm} B_{m} = R_{il} R_{jm} A_{l} B_{m}.

As you can see, this follows the transformation rule for a rank 2 tensor, therefore A_{1} B_{i} + A_{2} B_{j} + A_{3} B_{k} is a tensor.

I hope this helps to answer your question. Let me know if you have any further questions.


 

Related to Is A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} a Tensor?

1. Is the expression A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} a Tensor?

Yes, the expression A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} is a tensor. A tensor is a mathematical object that describes the relationships between multiple vectors and scalars. In this case, the expression includes both vectors (A and B) and scalars (1, 2, and 3), making it a tensor.

2. What is the order of the tensor represented by A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k}?

The order, or rank, of the tensor represented by A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} is 2. This is because there are two sets of indices (i and j) used to label the components of the tensor.

3. How many components does the tensor A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} have?

The tensor A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} has a total of six components. This can be calculated by multiplying the number of indices (2) by the number of possible values for each index (3).

4. Does the tensor A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} depend on the choice of basis?

Yes, the tensor A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} does depend on the choice of basis. This is because the components of a tensor change when the basis vectors are changed. Therefore, the expression A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} may have different values in different coordinate systems.

5. Is the tensor A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} symmetric?

The tensor A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k} is not necessarily symmetric. In order for a tensor to be symmetric, its components must remain unchanged under the interchange of indices. This is not always the case for the expression A_{1}B_{i} + A_{2}B_{j} + A_{3}B_{k}, as the values of the components may change depending on the order of the indices.

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