4-Vector Gradient and Contravariant Derivative

Question

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, $eta_{munu}$ = diag (-1,1,1,1) and I understand from web searches that:
For SR:
$$
eta_{munu} = eta^{munu}
$$
and
$$
partial^mu = eta^{munu} partial_nu
$$
However, a lot of the stuff I have found on the web seems to use $eta_{munu}$ = diag (1,-1,-1,-1) (which I am finding a bit confusing tbh) and states that for a scalar field $phi(t,x,y,z)$,
$partial_mu phi= frac{partialphi}{partial x^mu} = (frac{1}{c} frac{partial phi}{partial t}, nabla)$
and
$partial^mu phi = frac{partialphi}{partial x_mu} = (frac{1}{c} frac{partial phi}{partial t}, -nabla)$
So I am thinking that with $eta_{munu}$ = diag (-1,1,1,1), I'm looking at
$partial^mu phi = frac{partialphi}{partial x_mu} = (frac{-1}{c} frac{partial phi}{partial t}, nabla)$
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.

Javier · Accepted Answer

Yes, you are correct. The "real" derivative has a downstairs index because that's just the way derivatives transform, so we always have
$$partial_mu = left(frac{1}{c} partial_t, nablaright)$$
and $partial^mu = eta^{munu} partial_nu$, because that's what index raising means. So with some simple matrix multiplication we see that in the $(+ - -, -)$ convention we have
$$partial^mu = left(frac{1}{c} partial_t, -nablaright)$$
while in the $(- + +, +)$ convention we have
$$partial^mu = left(-frac{1}{c} partial_t, nablaright).$$

CuriousHegemon · Answer

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.

4-Vector Gradient and Contravariant Derivative

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