# Modulus of a tensor

1. Mar 10, 2014

### Jhenrique

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 $\sqrt{\mathbf{T}:\mathbf{T}}$ ?

2. Mar 10, 2014

### DrewD

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.