# Error in Carroll?

1. Feb 18, 2014

### dEdt

On page 161 of Carroll's Spacetime and Geometry, Carroll writes that
$$\delta g_{\mu\nu}=-g_{\mu\rho}g_{\nu\sigma}\delta g^{\rho\sigma}.$$
$\delta g_{\alpha \beta}$ denotes an arbitrary, infinitesimal variation of the metric.

Why is there a minus sign? By the regular rules of raising and lowering indices, shouldn't it just be
$$\delta g_{\mu\nu}=g_{\mu\rho}g_{\nu\sigma}\delta g^{\rho\sigma}?$$

2. Feb 18, 2014

### Bill_K

Nope, he's correct. The easiest way to see it is to vary the Kronecker delta, δμσ = gμνgνσ. You get

0 = (δgμν)gνσ + gμν(δgνσ)

from which the minus sign.

3. Feb 18, 2014

### George Jones

Staff Emeritus
What happens when both sides of $\delta^\alpha_\beta = g^{\alpha \mu} g_{\mu \beta}$ are varied?

4. Feb 18, 2014

### bcrowell

Staff Emeritus
As a concrete example, say the metric starts out as $g_{\mu\nu}=\operatorname{diag}(1,-1)$, and then we change it to $g'_{\mu\nu}=\operatorname{diag}(1+\epsilon,-1)$. Inverting the matrix to find the upper-index versions, we have $g^{\mu\nu}=\operatorname{diag}(1,-1)$ and $g'^{\mu\nu}=\operatorname{diag}(1-\epsilon,-1)$ (neglecting terms of order ε2). So $\delta g_{\mu\nu}=\operatorname{diag}(\epsilon,0)$ and $\delta g^{\mu\nu}=\operatorname{diag}(-\epsilon,0)$. Clearly the minus sign is right. Raising an index would have given $\delta g^{\mu\nu}=\operatorname{diag}(+\epsilon,0)$, which is wrong.

What's less obvious to me is where the red flag should have gone up to warn me that I couldn't raise the index on $\delta g$. I suppose this is because the metric has special status in index gymnastics notation. When we change the metric, we're also, as a side-effect, changing the piece of apparatus that we use for raising and lowering indices. If it had not been for this fact, then it would have seemed logical to me to argue that the difference of two tensors should be a tensor, and that we should therefore be able to raise and lower its indices as usual.

Am I right in thinking that the difference between two metrics is *not* a tensor, for these reasons?

5. Feb 18, 2014

### Bill_K

You get what I wrote. The variation of δμσ is zero, and the variation of gμνgνσ is (δgμν)gνσ + gμν(δgνσ)

6. Feb 18, 2014

### dEdt

I'm thinking the problem is this: If we have a tensor $A_{\alpha\beta}$, we can define the tensor $A^{\alpha\beta}=g^{\alpha\gamma}g^{\beta\delta}A_{\gamma\delta}$. But when we're dealing with the tensor $\delta g_{\alpha\beta}$, we can't just define $\delta g^{\alpha\beta}=g^{\alpha\gamma}g^{\beta\delta}\delta g_{\gamma\delta}$ because we have an independent definition for $\delta g^{\alpha \beta}$ in terms of the variation of the inverse metric, and these two definitions may not be identical (indeed they're not).

Maybe another way of putting it is that the inverse of the infinitesimal variation of the metric does not equal the infinitesimal variation of the inverse metric.

7. Feb 18, 2014

### Bill_K

That's what I would say. δgμν and δgμν are both tensors, but they're not the same tensor, they differ by a sign.

8. Feb 18, 2014

### George Jones

Staff Emeritus
Note the almost simultaneous timings of our two posts. When I started typing, you had not posted. I phrased my post as a question in order to try and get dEdt to work things out.

9. Feb 18, 2014

### bcrowell

Staff Emeritus
Which I guess means that it's a case where index gymnastics notation doesn't quite work as nicely as it usually does.

10. Feb 18, 2014

### Bill_K

It's a notational conflict, trying to use the same name for two different objects. You can define Aμν = δgμν and raise its indices the usual way to get Aμν. And you can define Bμν = δgμν, and even lower its indices if you like, but Aμν and Bμν aren't the same object, they need to be called by different names.