Rank vs Order of Tensors | Tensor Basics

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I am confused about the difference between the rank and order of a tensor.

On p 71 of Mathematical Physics 2nd Ed (Kusse and Westwig, 2006 Wiley-VCH), the rank of a tensor is described as identifying the number of basis vectors of the tensor but in some other books, this seems to be described as the order of a tensor.

Online, I saw both terms used as though they refer to different things, but I can't rely on that information because I saw it in Wikipedia. In Wikipedia, I also saw a statement that the dyadic product between vectors creates a tensor of rank one, but that is not what I read in the book mentioned above.


thanks
 
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The phrase "rank of a tensor" generally just refers to how many and what type of indices it has. "rank of a matrix" usually means the dimension of it's range. They are two completely different usages of the word "rank". The second meaning is what the wikipedia entry is referring to.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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