I am following a standard text on GR. In the chapter on linearized gravity, the metric $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ reduces to $\eta_{\mu \nu}$ when the metric act on tensor components which are $O(h)$. That make sense since we ignore terms smaller than $O(h)$.

Suppose now that $a^\mu$ denote a vector from some arbitrary tensor that is not necessarily $O(h)$ and $T^{\mu \nu}$ the stress energy tensor. It seems to me that the author of the textbook I am reading uses the following prescription, but it could be a coincidence:

  • $T^{\mu\nu} = 0 \implies g_{\mu\nu} a^\mu = \eta_{\mu\nu}a^\mu$ and
  • $T^{\mu\nu} \neq 0 \implies g_{\mu\nu} a^\mu = (\eta_{\mu\nu} + h_{\mu \nu})a^\mu$

My questions then:

(1) Is this a coincidence, theorem, or something else?

(2) If the metric acts on a tensor of unknown magnitude, is it then still possible to deduce whether $g$ reduce to $\eta$ or not?

EDIT: the textbook is Carroll, spacetime and geometry. The most relevant sections are 7.1, 7.2, 7.3, 7.4. Notice how Carroll use of the metric changes as the stress energy tensor changes.

  • $\begingroup$ An easy way to see which metric to use is to note that, in principle, one always uses the full metric $g_{\mu \nu}$ while doing GR. It's just that in linearized gravity, we keep terms (in an expression) of only first order in $h$. So for example, if you have some tensor $T$ that is already linear in $h$, then when you act on $T$ with a metric tensor, you will use $\eta$, so that the final expression is also linear in $h$ $\endgroup$ – Avantgarde May 7 '17 at 16:48
  • $\begingroup$ That's a good answer. To add, thetensor T could be zero and the metric not Minkowski, as in gravitational waves. You have to put in a context for a $\endgroup$ – Bob Bee May 7 '17 at 18:09
  • $\begingroup$ I don't think so, but it could be my mistake. I have updated my question to make explicit what I am asking. Thanks $\endgroup$ – Mikkel Rev May 7 '17 at 18:12
  • $\begingroup$ I guess one thought would be why wouldn't you assume a Minkowski background ($\eta_{\mu\nu}$) with a small perturbation (which can be labelled $h_{\mu\nu}$)? $\endgroup$ – Kyle Kanos May 7 '17 at 18:48
  • $\begingroup$ @Kyle Kanos - We can add small perturbations around any general metric ($g_{\mu \nu}$), and not necessarily $\eta_{\mu \nu}$. But anyway, that's not the intent of OP's question. They are confused when to use the full metric $g_{\mu \nu} + \delta g_{\mu \nu}$ (where $\delta g_{\mu \nu} = h_{\mu \nu}$) and when to use just $g_{\mu \nu}$ $\endgroup$ – Avantgarde May 7 '17 at 19:20

The equations are saying if there is NO stress energy in T, the stress energy tensor, then the metric is described by n (i.e. flat space), but when you have T != 0, then you can describe the metric by the flat space one plus some added little bits which are described by h.

Essentially expanding the metric into a flat metric + corrections is always wrong in an exact sense, but for 'weak' gravitational fields its a really nice way to simplify the math.

1 ? (Perhaps: Separating a metric into flat + h can only be done when it makes sense.) You can tell if it made sense when you get an answer and h < 0.001 or so.

2) No. If you don't know the magnitude of the stress energy, it makes no sense to try to use flat + h as a metric.


2) Yes - one could always use a flat + h metric, but the math would become tricky and harder than a full non linear treatment. You can in theory use any coordinate system on any spacetime, its just that the math gets hard.

The metric reduces to the flat metric if and only if there is no energy around. - Or better it reduces to the flat metric if you don't care about the errors that the flat assumption makes.


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