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I'm a beginner at relativity, I have a question about eq 2 if its true. I know that the interval can be calculated like this in special relativity: $$ ds^2 = \eta_{\mu\nu}dx^\mu dx^\nu \tag{1} $$ where $\eta_{\mu\nu}$ is the metric tensor (for example the conventional $\text{diag}(-1,+1,+1,+1)$ in Cartesian coordinate system). But what about with the upper index version: $$ ds^2 = \eta^{\mu\nu}dx_\mu dx_\nu, \tag{2} $$ of course the units are $c=1$, $\hbar=1$, $G=1$. The covariant components we can get from this equation: $$ dx_\mu=\eta_{\mu\nu}dx^\nu, \tag{3} $$ also between the upper and down index metric tensors we can write the following: $$ \eta_{\mu\alpha}\cdot\eta^{\alpha\nu}=\delta^\nu_\mu, \tag{4} $$ or more simply: $$ \eta^{\mu\nu}=(\eta_{\mu\nu})^{-1}. \tag{5} $$ Eq 2 seems a bit weird for me, I'm not sure about it. If we write everything out in eq 2 using eq 3, we get the following: $$ ds^2 = \eta^{\mu\nu}\eta_{\mu\alpha}dx^\alpha \eta_{\nu\beta}dx^\beta. \tag{6} $$ Where we can notate $\eta_{\mu\alpha}dx^\alpha=dx_\mu$ and $\eta_{\nu\beta}dx^\beta=dx_\nu$. Is this right?

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  • $\begingroup$ Your eq. 4 is incorrect. The left hand side would sum over both indices leaving 4 on the right hand side. I think you want $\eta_{\mu\alpha} \eta^{\alpha\nu} = {\delta_\mu}^\nu$. $\endgroup$
    – Paul T.
    May 30 '20 at 19:09
  • $\begingroup$ yes, I'll correct that one, thank you $\endgroup$
    – waerx
    May 30 '20 at 19:18
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If you look carefully, your equation (6) is exactly the same as equation (1). To see this, use equation (4) by applying it on the first two terms on the RHS of equation (6). Then you will be left with equation (1).

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  • $\begingroup$ Thank you, so as I understand I can write the following: $$ds^2=\eta^{\mu\nu}\eta_{\mu\alpha}dx^\alpha\eta_{\nu\beta}dx^\beta=\delta^\nu_\alpha dx^\alpha\eta_{\nu\beta}dx^\beta=\eta_{\alpha\beta}dx^\alpha dx^\beta$$ where at last step I used the Kronecker delta index property. $\endgroup$
    – waerx
    May 30 '20 at 19:38
  • $\begingroup$ yes, that's correct! $\endgroup$
    – Ayush Raj
    May 30 '20 at 19:53
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It is right, but I probably would have written (5) $$\eta^{\mu\nu}={(\eta^{-1}})_{\mu\nu}$$ In practice, you wouldn't need to write this because it is clearly true and you never have to write $\eta^{-1}$.

It helps to think of $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$ as index raising and index lowering operators. This idea carries forward to $g^{\mu\nu}$ and $g_{\mu\nu}$ in curved spacetimes and makes the relationship between contravariant and covariant tensors quite easy and natural to work with.

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Equation 2 is an abuse of notation to me. Although, technically equation 1 is also an abuse of notation already (just common place and everyone knows what it means). It's just notation after all. However, the way you've defined them in equation six could be OK but I've not seen anyone write it this way and it'd probably just confuse people as there's no point to it

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