1
$\begingroup$

I have a really simple question:

Let $$g = \begin{bmatrix} 1&0&0&0\\ 0&-1&0&0\\0&0&-1&0\\0&0&0&-1 \end{bmatrix}$$ be the metric tensor, then it has elements $g_{\kappa \nu}$. In einstein summation notation, what is the inner product between the two tensors $g_{\kappa \nu} g^{\mu \nu}$? Obviously here, we are summing over $\nu$ in both terms, but $\kappa$ and $\mu$ are different, do we just say that $\kappa = \mu$ otherwise the terms are zero, giving us the inner product $g_{\kappa \nu} g^{\mu \nu} = g_{\mu \nu} g^{\mu \nu} = 1 +(-1)^2 + (-1)^2 + (-1)^2$. Or am I missing something fundamental about the tensor relationships / summation notation?

Cheers.

$\endgroup$
1
  • 1
    $\begingroup$ The indices $\kappa$ and $\mu$ are free in the expression $g_{\kappa \nu}g^{\mu \nu}$. Free indices as a usual rule appear on both sides of an equation, the missing free indices in $1+(-1)^2+(-1)^2+(-1)^2$ is a dead give-away that your idea is off. The answer explains further... $\endgroup$
    – James S. Cook
    Commented Jul 25, 2018 at 2:25

1 Answer 1

Reset to default
1
$\begingroup$

The product of $g_{\kappa\nu}g^{\mu\nu}$ will not return a scalar, it will return a mixed tensor, specifically $\delta_{\kappa}^{\mu}$, the Kronecker delta. In fact, $g_{\kappa\nu}g^{\mu\nu}=\delta_{\kappa}^{\mu}$ is often taken as the definition of the inverse metric $g^{\mu\nu}$. The delta has components

$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} .$$

That is, you only perform contraction on the repeated index $\nu$, while the other two stay free. This is true for contracting tensors in general. So $g_{\kappa\nu}g^{\mu\nu}$ is not the same as $g_{\mu\nu}g^{\mu\nu}$.

*EDIT in fact this question has been answered at https://physics.stackexchange.com/questions/66394/kronecker-delta-confusion

$\endgroup$

You must log in to answer this question.