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Modulus of a tensor

by Jhenrique
Tags: modulus, tensor
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Jhenrique
#1
Mar10-14, 02:53 AM
P: 686
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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DrewD
#2
Mar10-14, 06:10 PM
P: 445
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.


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