Contraction of a rank 4 tensor

Warren2007
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I'm trying to contract a rank 4 tensor with covariant rank 2 and contravariant rank 2 with four different indices

[T[ab][cd]]

to get a scalar value T and I have no idea how to do it as I'm sure a or b does not equal c or d.

Any help would be much appreciated.
 
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You contract a tensor by putting the same index in one up slot and one down slot and summing. So one way to contract T^{ab}{}_{cd} down to a scalar would be as T^{ab}{}_{ab} = \sum_{i, j} T^{ij}{}_{ij}. Note that this is not necessarily equal to T^{ab}{}_{ba} = \sum_{i, j} T^{ij}{}_{ji}! (Here i, j are indices; if you are dealing with tensors over an n-dimensional space, then 1 \leq i, j \leq n is the range of the sums.)
 
But if a does not equal c and b does not equal d then how do you convert T[ab/cd] to T[ab/ab].
 
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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