Given an expression

$$g_{\mu\nu}v^{\nu} v^{\rho} k_{\nu} \delta^{\mu}_{\alpha} k_{\rho} \delta^{\nu}_{\beta}$$

if we contract $v^{\rho} k_{\rho}$, we get $v \cdot k$, then the rest of it reads $g_{\mu\nu}v^{\nu} k_{\nu} \delta^{\mu}_{\alpha} \delta^{\nu}_{\beta}$.

At this point, we can further contract $v^{\nu} k_{\nu}$ to get another factor of $v \cdot k$, and the rest contracts to $g_{\alpha \beta}$, so that the whole expression contracts to $(v \cdot k)^2 g_{\alpha \beta}$.

However, it seems that we can also contract $g_{\mu\nu}v^{\nu} \delta^{\mu}_{\alpha}$ and $k_{\nu} \delta^{\nu}_{\beta}$ separately to get $(v \cdot k) v_{\alpha} k_{\beta}$ as the final result.

In general, $(v \cdot k) v_{\alpha} k_{\beta}$ and $(v \cdot k)^2 g_{\alpha \beta}$ have different tensor structures. Where exactly does my derivation go wrong? I feel I might miss something obvious, but a hint is most appreciated.


1 Answer 1


The expression given makes no sense, since in Einstein notation each index can appear at most twice. In your expression the index $\nu$ shows up four times, and therefore one runs into ambiuities of what to contract with what. Where did this object come from? If it came from an earlier computation of yours, do it again taking care to never use the same name for two pairs of dummy indices.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.