Unraveling Tensor Notation: Extracting Equations

[junglebeast]

I can't seem to wrap my mind around it. I understand the concept of it, but I can't figure out how to translate that concept into actually extracting the individual equations from tensor notation.

For example,

<br /> a^i \: b^j \: c^k \: \epsilon_{jqs} \: \epsilon_{krt} \: \tau_i^{qr} = 0_{3 \times 3}<br />

note that a,b,c are 3 \times 1 and \tau is 3\times 3 \times 3.

This represents 9 equations. I understand how to calculate the value of the http://mathworld.wolfram.com/PermutationSymbol.html" , but this is complicated by having both superscripts and subscripts, and I'm also not sure if the subscripts of \epsilon count in Einstein summation. My biggest problem is that I don't understand the "method" that can be used to extract the actual equations out of this!

If someone could show me how to extract just 1 of the equations that would help a lot

 

[tiny-tim]

<br /> a^i \: b^j \: c^k \: \epsilon_{jqs} \: \epsilon_{krt} \: \tau_i^{qr} = 0_{3 \times 3}<br />

note that a,b,c are 3 \times 1 and \tau is 3\times 3 \times 3.

This represents 9 equations.

… this is complicated by having both superscripts and subscripts, and I'm also not sure if the subscripts of \epsilon count in Einstein summation. My biggest problem is that I don't understand the "method" that can be used to extract the actual equations out of this!

If someone could show me how to extract just 1 of the equations that would help a lot

Hi junglebeast!

(btw, you needn't say a,b,c are 3 \times 1 and \tau is 3\times 3 \times 3 … it's obvious from the number of indices )

Yes, it's 9 equations, for each of the 3 values of s and t.

And yes, all subscripts and superscripts count, even in deltas and epsilons.

Extracting one of the 9 equations simply involves fixing say s = 2, t = 3, and summing over all the rest … doesn't that .pdf (which I haven't looked at) give any examples?

 

[junglebeast]

Hi junglebeast!

(btw, you needn't say a,b,c are 3 \times 1 and \tau is 3\times 3 \times 3 … it's obvious from the number of indices )

Yes, it's 9 equations, for each of the 3 values of s and t.

And yes, all subscripts and superscripts count, even in deltas and epsilons.

Extracting one of the 9 equations simply involves fixing say s = 2, t = 3, and summing over all the rest … doesn't that .pdf (which I haven't looked at) give any examples?

Hi tim!

Let me start with something simpler...I'll try to show my process

<br /> a^i b^j = R^{ij}<br />

I understand how I can convert the above into linear algebra,

<br /> \mathbf{a} \mathbf{b} ^\mathsf{T} = \mathbf{R}<br />

...it's just the outer product. I could write this out elementwise as

<br /> \mathbf{R}(i,j) = a(i) b(j)<br />

(where parenthesis are used to indicate the indices)

Ok, so now I try to do this with 3 vectors. Written in tensor notation, it is

<br /> a^i b^j c^k = R^{ijk}<br />

If I'm not mistaken, this is a 3x3x3 cube which can be written out elementwise as

<br /> \mathbf{R}(i,j,k) = a(i) b(j) c(k)<br />

Now, \tau is also a cube...and these somehow combine together to make a 3x3 matrix (ignoring the \epsilon which just control the sign or cancellation).

To extract one of the equations it makes more sense for me to think about holding q and r constant. So let's say I choose q = r = 1. Ignoring the $\epsilon$ factors, I will try to sum over the rest...

<br /> \sum_i \sum_j \sum_k \sum_s \sum_t a(i) \: b(j) \: c(k) \: \tau(i,q,r) = 0<br />

Well that seems to contradict what I was doing earlier which did not have summations

 

[tiny-tim]

Hi junglebeast!

Now, \tau is also a cube...and these somehow combine together to make a 3x3 matrix (ignoring the \epsilon which just control the sign or cancellation).

No, you've lost me here.

epsilon doesn't "just control the sign or cancellation" … it's an integral part of the summation, and all three of its indices have to be summed over.

Yes, R_ijk is a 3x3x3 cube, and so are Rⁱ_jk and so on …

in Cartesian coordinates, the "cubes" have the same entries except that some entries are multiplied by minus-one (though in other coordinate systems, it's more complicated).

(and I haven't followed your final question)