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Doubt:-

What is the physical meaning of tensor? How can I see this special case of tensor definition as a definition of vectors(quantities with both magnitude and direction)in $\mathbb R^3$? Please help me. I could see $L_0(V)$ as a set of scalars. So, $k-\text{tensors on } V$ is a generalization of scalars. When $k=0$ tensors are scalars. How can I see vectors like this?

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  • $\begingroup$ Why negative vote? Why is the question not eligible for an answer? Please help me to correct the question. $\endgroup$
    – user464147
    Jul 5, 2018 at 2:54
  • $\begingroup$ A $k$-tensor is essentially a matrix represented in $k$ dimensions, so a 'vector is a $1$-tensor and a matrix is a $2$- tensor. A confusing issue is that in physics people say tensor when they mean what a mathematician would call a tensor-valued function or a tensor field. $\endgroup$
    – Ian
    Jul 5, 2018 at 3:26
  • $\begingroup$ You mean 2-tensor means $Row 1-[f(e_1,e_1) f(e_1, e_2)], Row 2-[f(e_2,e_1) f(e_2, e_2)]$ like this? $\endgroup$
    – user464147
    Jul 5, 2018 at 3:37
  • $\begingroup$ How it is the generalisation of ordinary vectors? $\endgroup$
    – user464147
    Jul 5, 2018 at 3:38
  • $\begingroup$ the set of all tensors is also a vector space $\endgroup$
    – janmarqz
    Jul 5, 2018 at 4:51

1 Answer 1

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0-tensors are constant functions, which we identify with scalars.

1-tensors are linear functions, which we identify with vectors. This identification amounts to selecting an inner product: we identify the vector $x$ with the function $y \mapsto \langle x,y \rangle$.

2-tensors are bilinear functions, which we identify with matrices. This identification also amounts to selecting an inner product: we identify the matrix $A$ with the function $(x,y) \mapsto \langle x,Ay \rangle$.

Things become a bit foreign when we go to $k$-tensors with $k>2$. One way to think about it is that a $k$-tensor takes a vector and gives back a $(k-1)$-tensor. Thus for instance a $3$-tensor takes a vector and gives back a matrix.

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  • $\begingroup$ How do you guarantee that $V$ is Inner product space? $\endgroup$
    – user464147
    Jul 5, 2018 at 15:47
  • $\begingroup$ You don't, but selecting an inner product provides such an identification. Without selecting an inner product, tensors remain abstract linear functions. $\endgroup$
    – Ian
    Jul 5, 2018 at 16:01
  • $\begingroup$ Now i have one more doubt, stress tensors are actually bilinear. Right?en.wikipedia.org/wiki/Cauchy_stress_tensor $\endgroup$
    – user464147
    Jul 5, 2018 at 16:21
  • $\begingroup$ That is correct, though as I mentioned physicists often say tensor when they mean tensor field. $\endgroup$
    – Ian
    Jul 5, 2018 at 16:48
  • $\begingroup$ Thank you very much @Ian $\endgroup$
    – user464147
    Jul 5, 2018 at 17:00

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