Tensor - the indicial notation, beginners problem

Dafe
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1. The problem statement and attempt at solution

Given the matrix S_ij and a_i evaluate a),b),c),d) and e)

2A1.jpg


For a) I think i use Einsteins convention.
b) I just first sum on i, and then on j giving me 9 terms. The answer I get is 24.

d) can i change m with i since they are both dummy indexes?

e) the same problem as d I guess, is this allowed?

I'm trying to learn continuum mechanics by my self, and this is the first step.

Thank you.
 
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Dafe said:
1. The problem statement and attempt at solution

Given the matrix S_ij and a_i evaluate a),b),c),d) and e)

2A1.jpg


For a) I think i use Einsteins convention.
Yes, that is the sum of numbers on the main diagonal, also called the "trace".

b) I just first sum on i, and then on j giving me 9 terms. The answer I get is 24
?? I get 28, the sum of the squares of all terms. And, of course, (c) is exactly the same as (b).

d) can i change m with i since they are both dummy indexes?
Yes, but why would you want to? In any case this is the sum of the squares of the elements of a, 1+ 4+ 9.

e) the same problem as d I guess, is this allowed?
No, this is not at all the same problem as (d)! (d) was amam which is, as I said, the sum of the squares of the elements of a: the same as the 'dot product' of a with itself. In particular, there is no "S" in (d). (e) is Smnaman. Multiply the "matrix" S with the column vector a, then take the dot product of that with a.

I'm trying to learn continuum mechanics by my self, and this is the first step.

Thank you.[/QUOTE]
 
I didn't mean that e) was the same problem as d), I just wondered if I could change the indices from m to i and so on... You've answered that though, thank you very much!
 
I'll post some more questions here so I don't spam the forums, hope that's alright.

1. The problem statement and attempt at solution:

2A4-1.jpg


As I see it E_kk is a first order tensor and E_ij is a second order one. How do I go from E_ij to E_kk?

Thank you.
 
Try to find out what kronecker delta is! Then you will get the answer.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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