20061207, 22:37  #1 
May 2005
Argentina
2×3×31 Posts 
Tensor Analysis books
Wich book is the best to initiate to study tensor analysis? I have LeviCivita "The absolute differential calculus" but is to abstract to me. I would prefer one with more numerical examples and graphics if possible.
Thanks in advance, Damian. 
20061211, 19:23  #2 
∂^{2}ω=0
Sep 2002
Repรบblica de California
5×2,351 Posts 
We used this one in my graduatelevel differential geometry class, but I suspect you might also find it somewhat "abstract" for your taste, even though Fomenko and Novikov are wellknown relativity physicists and try to keep things "physically grounded" whenever possible.
If differential geometry were easy, general relativity would be a high school subject. 
20061212, 08:57  #3  
Bamboozled!
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May 2003
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11647_{10} Posts 
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It does have a lot of pretty pictures and sound physical interpretations of objects which are often treated as very abstract mathematical constructs. Paul 

20061212, 17:03  #4 
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
Tensor Analysis
Damian, I'm not in the league with Ewmayer or Xilman nor I can I ever measure up to them but I would advise the Schaum's outline series of Theory and problems of 'Vector Analysis and an introduction to Tensor Analysis' There are about 60 pages at the end devoted to Tensor Analysis and I think sufficient to move on to perhaps those recommended by our colleagues. It has 480 solved problems so you can get a better idea on the subject. It is by Murray R. Spiegel, PhD. This is widely available in the U.S. libraries and in New York I can assure you as I picked my copy on sale from them for a throwaway price. My copy is collecting dust on my shelves . I do not profess to have gone thru or even understood it, but I know a good book when I see one . Mally Last fiddled with by mfgoode on 20061212 at 17:04 
20061214, 00:16  #5 
May 2005
Argentina
272_{8} Posts 
thanks for the replys. I'm downloading Thorne and Wheeler Gravitation book.
A newbie question: Does covariant and contravariant concept has anything to do with the transpose of a vector? I ask because I see that a_i*b^i gives the dot product (that is the same as the product of a vector matrix with the traspose of the other vector matrix), and it also the same result as the contraction of tensors (the summation convention) Another question: how can I use tex tags in these posts? Thanks in advance Damian. 
20061214, 09:42  #6 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
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I may be terribly wrong Damian. I think you are jumping the gun, but it all depends on your level. I knew a buddy of mine who was studying this book on gravitation for his PhD thesis. As xilman says it is more than just Tensor calculus and as ewmayer says about differential geometry. The climb to Tensors is long and tedious and requires a good foundation in modern geometry. This sounds simple but I dont want to discourage you though. The covariant curvature tensor is of fundamental importance in Einstein's general theory of Relativity. The contravariance has to do with curvilinear coordinate systems. They are both related with the latter coming before the former. All the best, Mally 
20061214, 15:46  #7  
Aug 2002
2^{3}×1,069 Posts 
Quote:


20061214, 16:08  #8 
May 2005
Argentina
2·3·31 Posts 
Thanks,
what I ment was: if I have two tensors and , then the tensor contraction equals the dot product of two vectors, wich itself equals the product of column vector matrix with file vector Is this casual, or there is a connection between covariance/contravariance and transpose of matrices. I guess the answer is that is casual, because I can have a rank 3 tensor, and how would I define its transpose since it is "similar" to a three dimensional matrix. Thanks, Damian. 
20061214, 17:35  #9  
Bamboozled!
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May 2003
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Paul 

20061214, 19:57  #10  
May 2005
Argentina
2×3×31 Posts 
Quote:
The vector related formula would be Because it would be inconsistent to write: Why does that happen? (have to transpose only one variable and put it before the matrix?) 

20061214, 20:19  #11  
∂^{2}ω=0
Sep 2002
Repรบblica de California
5·2,351 Posts 
Quote:
The tensor index notation replaces this row/columnbased convention with a different one, based on implied summation over a repeated index. This leads to a less visually intuitive procedure than above, but again, it is unambiguous and (at least for vectors and matrices) completely equivalent to conventional matrix multiply. In A_{ij}x^iy^j, the fact that you can either do the index sum over i (equivalent to x^t A) or j (== A y) first simply reflects the associativity of matrix multiply. 

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