# Tensor notation

1. Jun 15, 2009

### 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,

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

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" [Broken], 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

Last edited by a moderator: May 4, 2017
2. Jun 15, 2009

### tiny-tim

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?

3. Jun 15, 2009

### junglebeast

Hi tim!

$$a^i b^j = R^{ij}$$

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

$$\mathbf{a} \mathbf{b} ^\mathsf{T} = \mathbf{R}$$

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

$$\mathbf{R}(i,j) = a(i) b(j)$$

(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

$$a^i b^j c^k = R^{ijk}$$

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

$$\mathbf{R}(i,j,k) = a(i) b(j) c(k)$$

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...

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

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

4. Jun 15, 2009

### tiny-tim

Hi junglebeast!
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, Rijk is a 3x3x3 cube, and so are Rijk 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)