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I am self-studying General Relativity, and the course of study I am following has started to introduce me to index notation. The texts I am using (Carroll, Schutz) begin with a geometric slant on Special Relativity, and I am finding the index notation a bit of a challenge. From my textbooks, $\eta_{\mu\nu}$ = diag (-1,1,1,1) and I understand from web searches that:

For SR:

$$ \eta_{\mu\nu} = \eta^{\mu\nu} $$ and

$$ \partial^\mu = \eta^{\mu\nu} \partial_\nu $$

However, a lot of the stuff I have found on the web seems to use $\eta_{\mu\nu}$ = diag (1,-1,-1,-1) (which I am finding a bit confusing tbh) and states that for a scalar field $\phi(t,x,y,z)$,

$\partial_\mu \phi= \frac{\partial\phi}{\partial x^\mu} = (\frac{1}{c} \frac{\partial \phi}{\partial t}, \nabla)$

and

$\partial^\mu \phi = \frac{\partial\phi}{\partial x_\mu} = (\frac{1}{c} \frac{\partial \phi}{\partial t}, -\nabla)$

So I am thinking that with $\eta_{\mu\nu}$ = diag (-1,1,1,1), I'm looking at

$\partial^\mu \phi = \frac{\partial\phi}{\partial x_\mu} = (\frac{-1}{c} \frac{\partial \phi}{\partial t}, \nabla)$

Is this anywhere near the mark? Please be gentle with me, as all of this notation is very new, and I am not a physics student, just an (older) interested amateur who is struggling with new concepts and new notation.

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    $\begingroup$ $\eta_{\mu\nu} = \eta^{\mu\nu}$ It’s true that the covariant and contravariant components of the Minkowski metric are numerically the same in Cartesian components, but physicists usually don’t write equations like this where the free indices are mismatched. Tensor equations should have the same kind of tensor on both sides. $\endgroup$
    – G. Smith
    Nov 2, 2020 at 1:00

2 Answers 2

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Yes, you are correct. The "real" derivative has a downstairs index because that's just the way derivatives transform, so we always have

$$\partial_\mu = \left(\frac{1}{c} \partial_t, \nabla\right)$$

and $\partial^\mu = \eta^{\mu\nu} \partial_\nu$, because that's what index raising means. So with some simple matrix multiplication we see that in the $(+ - -\, -)$ convention we have

$$\partial^\mu = \left(\frac{1}{c} \partial_t, -\nabla\right)$$

while in the $(- + +\, +)$ convention we have

$$\partial^\mu = \left(-\frac{1}{c} \partial_t, \nabla\right).$$

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  • $\begingroup$ Thanks so much for responding! I thought I was on the right track, but with no-one to ask face to face, you always feel a bit unsure when studying new material. $\endgroup$
    – Essenbee
    Nov 2, 2020 at 17:29
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No worries, you're doing great! I'm glad you're self-studying these things, that takes a lot of courage. The signature of the metric is not important, all that matters is that we are consistent. So whether the signature is + or - (Which means whether the trace of the metric is positive or negative), our physics will be the same.

I initially did this wrong, I had a brain slip up and wrote something very incorrect, as some commenters have pointed out this is more accurate (for the $(-1,1,1,1)$ signature): $$ \partial^0 \phi = \eta^{0 \nu} \partial_\nu \phi = \eta^{00} \partial_0 \phi = - \partial_0 \phi \\ \partial^i \phi = \eta^{i j} \partial_j \phi = \delta^{i j} \partial_j \phi = \partial_i \phi $$

Thank you everyone who pointed out my mistake, I hope this is more clear.

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    $\begingroup$ The part of your equation with the sum is incorrect. You are letting $\mu$ change in the sum. $\endgroup$
    – G. Smith
    Nov 2, 2020 at 1:20
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    $\begingroup$ Many thanks for the responses. Glad to know I am on the right track! I'll have the digest the comments of @G.Smith, and thanks to them as well. $\endgroup$
    – Essenbee
    Nov 2, 2020 at 8:28
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    $\begingroup$ @CuriousHegemon It would be better if you just deleted or corrected the wrong statement, don't leave it there with an edit after it. $\endgroup$
    – Javier
    Nov 2, 2020 at 13:16
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    $\begingroup$ @CuriousHegemon, thanks I understand what G.Smith was saying now! $\endgroup$
    – Essenbee
    Nov 2, 2020 at 17:34
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    $\begingroup$ As the comments have pointed out, this is a bit sloppy. No, I’m sorry, but $$ \partial^\mu \phi= \eta^{\mu \nu} \partial_\nu \phi = (\eta^{00} \partial_0 + \eta^{11} \partial_{1} + \eta^{22} \partial_{2} + \eta^{33} \partial_{3} ) \phi = (-\partial_t \phi / c, \nabla \phi) $$ isn’t “a bit sloppy”. It’s just wrong. Frankly the third part doesn’t even make sense because it isn’t a four-vector like the first, second, and fourth parts. I don’t think you’ve understood your mistake yet. $\endgroup$
    – G. Smith
    Nov 2, 2020 at 17:54

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