# Question about contraction with metric tensor

I just starting to study GR and I could not prove the following: if I have to tensors $T_{\mu\nu}$ and $Q_{\mu\nu}$ such that $T_{\mu\nu}=Q_{\mu\nu}$, why can I multiply both sides of the equation by $g^{\mu\nu}$, i.e., why is valid $g^{\mu\nu}T_{\mu\nu}=g^{\mu\nu}Q_{\mu\nu}$? Why some authors call it "tracing the equation with $g^{\mu\nu}$"?

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Repeated indices are summed over. What is your problem with the formalism specifically? –  user18764 Mar 10 '13 at 18:55
If $x=5$, why is it valid that I can do $2x = 10$? –  Muphrid Mar 10 '13 at 19:10
If I multiply both sides of $T_{\mu\nu}=Q_{\mu\nu}$ by a scalar, than it is ok! But I cant see why I can multiply by $g^{\mu\nu}$ since there is sum implicit! –  robert Mar 10 '13 at 19:15
Write the sum out in components (don't need all 4 dimensions, just try two). It should become obvious then. –  twistor59 Mar 10 '13 at 19:20
thank you very much! –  robert Mar 10 '13 at 20:09

Note that $g^{\mu\nu}T_{\mu\nu}$ is just another way of writing $$\sum_{\mu,\nu = 0}^3 g^{\mu\nu}T_{\mu\nu}$$ Now, if we know that $T_{\mu\nu} = Q_{\mu\nu}$ for every $\mu,\nu = 0,\dots 3$, then we can simply substitute $Q_{\mu\nu}$ in for $T_{\mu\nu}$ in the sum. In other words $$\sum_{\mu,\nu = 0}^3 g^{\mu\nu}T_{\mu\nu} = \sum_{\mu,\nu = 0}^3 g^{\mu\nu}Q_{\mu\nu}$$ but the sum on the right can be written using the summation convention as $g^{\mu\nu}Q_{\mu\nu}$. So, putting this all together, we have shown that $$g^{\mu\nu}T_{\mu\nu} = g^{\mu\nu}Q_{\mu\nu}$$ Authors call it "tracing" because they are making an analogy with taking the trace of a matrix. Note, in particular, that if $g^{\mu\nu} = \delta^{\mu\nu}$ then we would have $$g^{\mu\nu}T_{\mu\nu} = \delta^{\mu\nu}T_{\mu\nu} = T_{\mu\mu} = \sum_{\mu=0}^3T_{\mu\mu} =\mathrm{tr}(T)$$ where $T$ is the matrix with components $T_{\mu\nu}$.
Hello, it should be $T_{\mu\mu}$ in the last equation after summing over $\nu$ with the delta function. –  gns-ank Mar 10 '13 at 20:16