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What is a tensor?by VivaLaFisica
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#1
Jan1413, 04:36 PM

P: 8

I'm a bit confused not just on what exactly a tensor is but also how I should be thinking about a tensor. I figure it's just a vector's bigger brother, but I'm just trying to wrap my head around it.
Any explanations are appreciated. Thanks in advance. 


#2
Jan1413, 04:48 PM

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#3
Jan1413, 04:53 PM

P: 8

But really, this tensor business is boggling me. 


#4
Jan1413, 08:47 PM

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What is a tensor?
If your teacher has told you something like "it's something that transforms under rotations according to the tensor transformation law", it's understandable that you're confused, because what does that statement even mean? It's been almost 20 years since someone tried to explain tensors to me that way, and it still really irritates me when I think about it.
I will explain one way to interpret that statement for the case of "covariant vectors" and "contravariant vectors". (I really hate those terms). Consider a function T that takes coordinate systems to ordered triples of real numbers. We can choose to denote the members of the triple corresponding to a coordinate system S by T(S)^{i}, or by T(S)_{i}. If S' is another coordinate system, that can be obtained from S by a rotation, then it might be useful to have formulas for T(S')^{i} in terms of the T(S)^{i}, and for T(S')_{i} in terms of the T(S)_{i}. Those formulas can be thought of as describing how (T(S)^{1}, T(S)^{2}, T(S)^{3}) and (T(S)_{1}, T(S)_{2}, T(S)_{3}) "transform" when the coordinate system is changed from S to S'. It's possible that neither of those formulas looks anything like the tensor transformation law, but it's also possible that one of them looks exactly like the tensor transformation law. If that's the case with the formula for T(S')^{i}, then some people call the triple (T(S)^{1}, T(S)^{2}, T(S)^{3}) a "contravariant vector". And if it's the case with the other formula, then some people call the triple (T(S)_{1}, T(S)_{2}, T(S)_{3}) a "covariant vector". I think that terminology is absolutely horrendous. The information about how the triple will "transform" isn't present in the triple itself. It's in the function T and the choice about where to put the indices. So if anything here should be called a "contravariant vector", it should be something like the pair (T,"upstairs"). I also don't like that this defines "vector" in a way that has nothing to do with the standard definition in mathematics: A vector is a member of a vector space. If you understand this, then it's not too hard to understand how this generalizes to other tensors. It's just hard to explain it. I consider this entire approach to tensors obsolete. A much better approach is to define tensors as multilinear functions that take tuples of tangent vectors and cotangent vectors to real numbers. This post and the ones linked to at the end explain some of the basics. (If you read it, skip the first two paragraphs. You can also ignore everything that mentions the metric if you want to. What you need is an understanding of the terms "manifold", "tangent space", "cotangent space" and "dual basis"). 


#5
Jan1413, 10:03 PM

P: 109

My explanation is that a tensor is an n dimensional grid of numbers. Since a scalar is a 0d grid, a vector is a 1d grid, a matrix is a 2d grid, we can say that a tensor is any of these, plus 3d grids, 4d grids etc.
But it isn't just a grid of numbers, it also has a few operations on it, such as addition and multiplication, which work like matrix algebra in the 2d case, and real number algebra in the 0d case, etc. 


#6
Jan1413, 10:24 PM

P: 4,573

Also, you might want to consider the analog to the Cartesian product of vector spaces (or linear spaces) to get a "multi"linear object (hence the term multilinear algebra which is another word for tensors).



#7
Jan1413, 10:24 PM

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Thanks
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#8
Jan1513, 12:46 AM

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#9
Jan1513, 12:51 AM

P: 109

Are you suggesting that you need something more for tensors that you don't need for say just matrices? Matrices can be transformed between coordinate systems. 


#10
Jan1513, 01:20 AM

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