Is ##\eta## Referring to the Minkowski Metric in Tensor Summation?

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sorry solved
 
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By ## \eta ## are you referring to the Minkowski metric? Also, I assume that only indices that appear once raised and once lowered are being summed over, so that ## \eta^{ii} ## is not to be summed, correct?
 
Geofleur said:
By ## \eta ## are you referring to the Minkowski metric? Also, I assume that only indices that appear once raised and once lowered are being summed over, so that ## \eta^{ii} ## is not to be summed, correct?

thanks, the issue was simply that sometimes I forget we are dealing with components and not whole vectors, matrices etc.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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