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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    $\begingroup$ Repeated indices are summed over. What is your problem with the formalism specifically? $\endgroup$
    – user18764
    Mar 10, 2013 at 18:55
  • $\begingroup$ If $x=5$, why is it valid that I can do $2x = 10$? $\endgroup$
    – Muphrid
    Mar 10, 2013 at 19:10
  • $\begingroup$ 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! $\endgroup$
    – robert
    Mar 10, 2013 at 19:15
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    $\begingroup$ Write the sum out in components (don't need all 4 dimensions, just try two). It should become obvious then. $\endgroup$
    – twistor59
    Mar 10, 2013 at 19:20

1 Answer 1


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 $T_{\mu\nu}$.

  • $\begingroup$ Hello, it should be $T_{\mu\mu}$ in the last equation after summing over $\nu$ with the delta function. $\endgroup$
    – aignas
    Mar 10, 2013 at 20:16
  • $\begingroup$ @robert sure thang brotha. $\endgroup$ Mar 10, 2013 at 21:08

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