Is the Modulus of a Tensor Calculated Differently Than a Vector?

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The modulus of a vector is calculated using the formula √(v · v), while for a rank 2 tensor, the modulus can be expressed as √(T : T). The Euclidean norm is relevant for vectors as it relates to physical concepts, but the notion of magnitude for tensors lacks a universally accepted interpretation. Common norms for tensors include the trace, which only counts diagonal elements, and the Euclidean norm based on those elements. Overall, the discussion highlights the complexities and variations in defining tensor magnitudes compared to vectors.
Jhenrique
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I was thinking... if the modulus of a vector can be calculated by ##\sqrt{\vec{v} \cdot \vec{v}}##, thus the modulus of a tensor (of rank 2) wouldn't be ##\sqrt{\mathbf{T}:\mathbf{T}}## ?
 
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Using ##\sqrt{\vec{v}\cdot\vec{v}}## for the magnitude of a vector makes sense because the Euclidean norm (which that is) relates to our world. I am definitely not the most well versed in tensors in general, but my understanding is that there is no particular idea of a magnitude that makes similar physical sense.

The most common norms that I recall were the trace (which is only a norm if you take the magnitude of the diagonal elements?) and the Euclidean norm using the diagonal elements. I think that might be the what you wrote, but I'm not very familiar with that notation. I think that there are many others too, but those are the only ones that I ever used.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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