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Is it true that when you take the covariant derivative of a covariant tensor, do you always have to do with a subscript? What if you do it wrt a superscript?Does the first term (with the partial derivative) take a minus sign? More specifically, is this true?

$$\nabla^{\mu}R_{\mu\nu} = -\frac{\partial R_{\mu\nu}}{{\partial x_{\mu}}} + \text{(Christoffels)}$$

Where does the minus sign come from? Is there a proof for this, or is it just a definition? Also, is there a change in the signs for Christoffel symbols(not the change if the the tensor's indices change position, but the change when the index of the covariant differential changes)?

I want to know the PROOF/REASON behind the minus sign.

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  • $\begingroup$ Remember that $x^\mu x_\mu = x_\mu x^\mu$. In other words, if the subscript of the covariant derivative is contracted, then it doesn't really matter if it is an upper index provided that the other (contracted) index is lowered (and vice versa). Also, the minus sign you are referring to only happens when you raise/lower the time component (or spatial component, depending on the metric signature you are using). $\endgroup$
    – Hunter
    Commented Jun 9, 2014 at 15:30
  • $\begingroup$ The metric signature is (-+++), and I read this thing on the internet. He didnt explain how did he get it, so I got confused.. $\endgroup$
    – GRrocks
    Commented Jun 9, 2014 at 16:09
  • $\begingroup$ "I read this thing on the internet"... What exactly did you read on the internet, i.e. what exactly confuses you? $\endgroup$
    – Hunter
    Commented Jun 9, 2014 at 16:12
  • $\begingroup$ The reason you have a minus sign there.....And here is the link sites.google.com/site/generalrelativity101/… $\endgroup$
    – GRrocks
    Commented Jun 9, 2014 at 16:14

2 Answers 2

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No. The subscript is the defined thing. If you have the superscript, you just assume raising with the metric tensor, so:

$$\nabla^{\mu}R_{\mu\nu} \equiv g^{\mu\alpha}\nabla_{\alpha}R_{\mu\nu}$$

which you expand normally with partial derivatives and Christoffels. Of course, since we know that $\nabla^{a}\left(R_{ab} - \frac{1}{2}Rg_{ab} \right)= 0$, we know right away that we can simplify $\nabla^{\mu}R_{\mu\nu}$ to $\frac{1}{2}\nabla_{\nu}R$

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  • $\begingroup$ thanx for the answer @jerry Schirmer...however, Can you explain where the minus sign comes from(I know the Bianchi identities, but was trying to derive the LHS of the field equations in the manner einstein might have, as he didnt have those identities....) $\endgroup$
    – GRrocks
    Commented Jun 9, 2014 at 16:12
  • $\begingroup$ @GRrocks: why do you dismissively say that the Bianchi identities wheren't around? I can't say for sure that they weren't, but most of Luigi Bianchi's work predates 1916: en.wikipedia.org/wiki/Luigi_Bianchi $\endgroup$
    – Zo the Relativist
    Commented Jun 4, 2015 at 18:57
  • $\begingroup$ @Jerruy Schirmer I really don't think Einstein would have used the Bianchi identities. I am not really into biographies, but I read one and it said that Einstein first gave the field equations only in terms of the Ricci tensor, and later realised the mistake and changed the equations to what we now study. He would'nt have done it in the first place if he knew the identities, would he? I was just toying around with the idea that how would he have done it...... $\endgroup$
    – GRrocks
    Commented Jun 6, 2015 at 10:53
  • $\begingroup$ @GRrocks: if he didn't use the general theorem, my guess would be that he brute force derived the identity $\nabla^{b}R_{ab} = \frac{1}{2}\nabla_{a}R$ $\endgroup$
    – Zo the Relativist
    Commented Jun 6, 2015 at 14:56
  • $\begingroup$ hehe....maybe....wouldn't be surprised if he did it though....... $\endgroup$
    – GRrocks
    Commented Jun 9, 2015 at 9:10
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From your comments, I will try to answer what confuses you. Let us take a metric signature: $$ \eta_{\mu \nu} = \mathrm{diag}(-1,1,1,1) $$ and let us consider some general $x^\mu$. We will denote the time component of $x^\mu$ by $x^0$. If we want to lower the index of $x^0$, we get: $$x^0 = \eta^{0 \mu} x_\mu = \eta^{0 0} x_0 + \eta^{0 1} x_1 + \eta^{0 2} x_2 + \eta^{0 3} x_3 \tag{1}$$ Since $\eta^{0 0} = -1$ and $\eta^{0 1} = \eta^{0 2} = \eta^{0 3} = 0$, equation $(1)$ becomes: $$x^0 = - x_0$$ and so we get the minus sign.

Note that if we only consider the spatial component $x^i$ (where $i$ is either the $1$st, the $2$nd or the $3$rd component), then we lower the index again as: $$ x^i = \eta^{i \mu} x_\mu = \eta^{i0} x_0 + \eta^{i1} x_1 + \eta^{i2} x_2 + \eta^{i3} x_3 = x_i$$ and so we don't get a minus sign.

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  • $\begingroup$ OKKK, so the minus sign is really because we are lowering the indices of the covariant derivative wrt the MINKOWSKI metric..But would this be valid if we did it wrt the METRIC TENSOR for ANY line element?? $\endgroup$
    – GRrocks
    Commented Jun 9, 2014 at 16:24
  • $\begingroup$ Yeah, for any Lorentzian metric in its canonical form this will be valid (remember it always possible to put a metric into its canonical form for some point on the manifold). What kind of metric do you have in mind? $\endgroup$
    – Hunter
    Commented Jun 9, 2014 at 16:41
  • $\begingroup$ Thanx @Hunter, I was just thinking of something arbitrary(e.g swarzchild) $\endgroup$
    – GRrocks
    Commented Jun 9, 2014 at 17:01
  • $\begingroup$ I'm not sure, but you'll probably have to look into curvilinear coordinates; remember that the basis of the tangent and its cotangent space must satisfy $e^\mu e_\nu = g_{\mu \nu} e^\mu e^\nu = \delta^\mu_\nu$. $\endgroup$
    – Hunter
    Commented Jun 9, 2014 at 17:35
  • $\begingroup$ @GRrocks: you can always diagonalize the metric at a point by just defining new coordinates $z^{a} = \lambda^{a}{}_{b}x^{b}$ for some set of constant $\lambda$. $\endgroup$
    – Zo the Relativist
    Commented Jun 9, 2014 at 18:29

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