# A question about static spacetime in GR

• I
• andyl01
andyl01
Hello,
I'm studying General Relativity using Ray D'Inverno's book. [Moderator's note: link deleted due to possible copyright issues.]

. I don't understand what the author writes in paragraph 14.3 ("Static solutions") where he demonstrates that for a static spacetime there are no cross-terms (dx)^0 (dx)^\alpha in (ds)^2 , where x^0 is a timelike coordinate e x^\alpha is a spacelike coordinate. He says that the "assumption that the solution is static means that (ds)^2 is invariant under a time reversal about any origin of time"... but a time reversal is a coordinate transformation and (ds)^2 is a scalar, so is automatically invariant under a time reversal (for it is a coordinate transformation), and not only specifically for a static metric.
Soon after the author demonstrates (Eq. 14.22) that a metric does not contain cross terms of the type indicated above and so he concludes that the metric is static. I don't understand how to prove this, is there someone who can give me a detailed demonstration of this? Thank you.

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andyl01 said:
a time reversal is a coordinate transformation and (ds)^2 is a scalar, so is automatically invariant under a time reversal (for it is a coordinate transformation)
No, that's not correct. A "time reversal" is not a coordinate transformation in the sense in which a metric in GR is invariant under coordinate transformations, because it is not continuous, it is discrete. So not all metrics in GR are invariant under time reversal; only the static ones are, as d'Inverno says.

To see a counterexample, consider Kerr spacetime, which is stationary (see d'Inverno's definition of that term in section 14.1), but not static. Time reversal flips the sign of the cross term in the metric (in Boyer-Lindquist coordinates, which are the simplest ones to use for this purpose since there is only one cross term), so it is not invariant under time reversal.

andyl01 said:
Soon after the author demonstrates (Eq. 14.22) that a metric does not contain cross terms of the type indicated above and so he concludes that the metric is static. I don't understand how to prove this
You have this backwards. He says that if the metric is static, the cross terms will vanish. He defines "static" earlier, in section 14.2 (though I agree he should be clearer about the logical sequence and give the definition of "static" in that section, instead of after equation 14.22), and then uses that definition (that the timelike Killing vector field in the spacetime is hypersurface orthogonal) to prove that one can always find coordinates in which the ##dt dx^\alpha## cross terms vanish. Equation 14.22 is the last step of that proof, which starts from the assumption that the metric is static.

vanhees71 and PeroK
PeterDonis said:
You have this backwards. He says that if the metric is static, the cross terms will vanish. He defines "static" earlier, in section 14.2 (though I agree he should be clearer about the logical sequence and give the definition of "static" in that section, instead of after equation 14.22), and then uses that definition (that the timelike Killing vector field in the spacetime is hypersurface orthogonal) to prove that one can always find coordinates in which the ##dt dx^\alpha## cross terms vanish. Equation 14.22 is the last step of that proof, which starts from the assumption that the metric is static
But that proof does not start from the assumption that the metric is static... I try to summarize what I think D'Inverno says in the chapter up to sec. 14.3 (it's not easy since the topic is presented in a chaotic way). He defines a static metric as a stationary metric which admits a coordinate frame in which mixed terms (of the type indicated above) vanish, and he says that this means that the line element ## (ds)^2 ## is invariant under a time reversal.
Then he assumes that we have a  stationary metric (i.e. there exists a special coordinate system in which the metric is time-independent), together with a hypersurface-orthogonal timelike Killing vector field, and he proves that such a metric is static by showing the existence of a coordinate system in which the mixed metric elements vanish. But soon after he states that we have proved that "a spacetime is static ##\Leftrightarrow## it admits a hypersurface-orthogonal timelike Killing vector field"... He has actually proved only the ##\Leftarrow## implication!

So:
1) How to prove in detail the other one, namely that the existence of a coordinate system in which the mixed terms of a metric vanish (i.e. the metric is static) implies the existence of a timelike hypersurface-orthogonal Killing vector field?

2) I still don't understand why the fact that a time reversal is discrete may cause ##(ds)^2## not to be invariant under such a transformation. Why "discreteness" implies that we can't regard it as a valid coordinate transformation?

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andyl01 said:
that proof does not start from the assumption that the metric is static
Yes, it does, although d'Inverno does not state it that way at the start of the proof. See below.

andyl01 said:
He defines a static metric as a stationary metric which admits a coordinate frame in which mixed terms (of the type indicated above) vanish
No, he doesn't, but I agree the presentation does not make it easy to see the actual logic he is using. Section 14.3 contains two arguments. The first, which is the first few paragraphs of the section (up to the end of the paragraph after equation 14.16), is stated backwards: he states the conclusion first (that in a static metric, we would expect the cross terms to vanish--note the word expect, which means he is not stating a premise, he is stating a conclusion he is going to establish), and only states the premise afterwards, after equation 14.16 (the premise is that a static metric is invariant under time reversal).

The second argument goes from the next paragraph up to equation 14.22, and its premise is at least stated at the outset: that the timelike Killing vector field is hypersurface orthogonal. He then shows that that condition also leads to the conclusion that the cross terms in the metric vanish (equation 14.22). Only after that does he actually state the definition of "static", which is the same as the premise he started with in the second argument; and then he states that the metric being static (i.e., the timelike Killing vector field being hypersurface orthogonal) implies that the cross terms vanish, which has already been proven (the argument leading up to equation 14.22).

IMO it would have been better to state the definition of "static" at the start of section 14.3 and then show what it implies, but the fact that d'Inverno didn't do that but stated things backwards does not change the logic. The logic is what I described above. Also, it would have been better if d'Inverno hadn't stated the first argument at all until after giving the definition of static and showing what it implies, since the first argument doesn't make use of the definition of static he gives later as a premise, and he doesn't actually prove that his definition of static implies the premise he is using (although that can be proven), so giving that argument at that point just confuses the issue. But again, that doesn't change the logic.

andyl01 said:
Then he assumes that we have a  stationary metric (i.e. there exists a special coordinate system in which the metric is time-independent)
No, stationary is not defined by that condition. Stationary is defined, as d'Inverno states in section 14.1, as the spacetime having a timelike Killing vector field. The fact that a coordinate chart exist in which the metric is independent of the timelike coordinate ##t## is a consequence of that definition. Unfortunately section 14.1, like section 14.3 (see my previous post), presents the argument backwards, stating the conclusion first and the premise only at the end.

andyl01 said:
, together with a hypersurface-orthogonal timelike Killing vector field, and he proves that such a metric is static
No, again, you have this backwards. Having a timelike Killing vector field that is hypersurface orthogonal is the definition of static--d'Inverno explicitly states that after equation 14.22.

@andyl01 if you want to see a presentation of this same material in a more logical order, see Wald, section 6.1. There the terms "stationary" and "static" are defined first using the definitions I described above (which d'Inverno gives only after discussing their consequences), and then the implications for coordinate charts are derived from those definitions. (Note that Wald also shows why in a static spacetime ##ds^2## is invariant under time reversal, in correct logical order.)

PeterDonis said:
No, again, you have this backwards. Having a timelike Killing vector field that is hypersurface orthogonal is the definition of static--d'Inverno explicitly states that after equation 14.22.
D'Inverno introduces in sect. 14.1 the concept of stationary spacetime and static spacetime, without (at the beginning) mentioning Killing vector fields. Then he proves, for example for the concept of stationary spacetime, that the initial definition is equivalent for the metric to admit a timelike Killing vector field. This is proved as a theorem (see from Eq. 14.2 until the end of sect. 14.1) and since there exists such an equivalence, we can take as the definition of stationary spacetime that one mentioning a timelike killing vector field. I think he is trying to do the same with static spacetime. He defines a static spacetime at first without mentioning killing vector fields, and then he proves as a theorem that a metric is static if and only if it admits a hypersurface-orthogonal timelike Killing vector field. Since there is such an equivalence, we can take this new one as new definition of static spacetime.
Notice that in the new edition of the book, "Introducing Einstein's relativity. A deeper understanding", in sect. 15.3 (the corresponding section of sect. 14.3 in the old edition) D'Inverno realizes that he forgot the ##\Rightarrow## implication and adds : "
Conversely, if there exists a coordinate system in which the metric takes the form (15.19), one can show (exercise) that $$X^a = \delta^a_0$$
is a hypersurface-orthogonal timelike Killing vector".

Okay I start to understand the logic... but I still don't understand why the fact that a time reversal is discrete may cause ##(ds)^2## not to be in general invariant under such a transformation. Why "discreteness" implies that we can't regard it as a valid coordinate transformation?

andyl01 said:
D'Inverno introduces in sect. 14.1 the concept of stationary spacetime and static spacetime, without (at the beginning) mentioning Killing vector fields.
Yes, which, as I have already said a couple of times now, is backwards. The definitions of those concepts involve Killing vector fields. As you will see if you consult other references, such as the Wald reference I gave you. The fact that d'Inverno chooses what is IMO a confusing way of presenting the topic does not change the definitions of those concepts.

andyl01 said:
Then he proves, for example for the concept of stationary spacetime, that the initial definition is equivalent for the metric to admit a timelike Killing vector field. This is proved as a theorem
andyl01 said:
He defines a static spacetime at first without mentioning killing vector fields, and then he proves as a theorem that a metric is static if and only if it admits a hypersurface-orthogonal timelike Killing vector field.
You are relying too much on the order in which d'Inverno presents things. His presentation, as I've already said, is backwards. A better description of what he is doing is that he gives, first, a description of what "stationary" and "static" imply as regards a coordinate chart, since he thinks that will be easier for the reader to grasp at the outset. Then he shows that the coordinate-independent definitions of "stationary" and "static", which are the actual rigorous definitions of those terms, do in fact imply the coordinate properties he has described.

cianfa72
andyl01 said:
Why "discreteness" implies that we can't regard it as a valid coordinate transformation?
Because a coordinate transformation, in the sense you are using the term (a transformation that leaves the line element ##ds^2## unchanged), is a continuous transformation.

andyl01 said:
Why "discreteness" implies that we can't regard it as a valid coordinate transformation?
Perhaps another way of looking at it will help: "time reversal" means not just the coordinate change ##t \to - t##, but also interpreting the new ##t## coordinate after the change as going forward in time.

For example, an object rotating clockwise about a given axis is stationary; the "time reversal" of this is the object rotating counterclockwise about the same axis. These are not the same, and the spacetime geometry associated with them will not be the same either (the "twist" or "frame dragging" due to the object's rotation will be in the opposite direction).

But an object just sitting still is static, and the "time reversal" of this is the object just sitting still, i.e., the same thing. And the same goes for its spacetime geometry.

## What is static spacetime in General Relativity?

Static spacetime in General Relativity refers to a spacetime that is both time-independent and has no rotation. This means that the metric components do not change with time, and there is no cross-term involving time and space coordinates in the metric tensor. Essentially, the spacetime geometry remains constant over time, and the gravitational field is unchanging.

## How is static spacetime different from stationary spacetime?

Static spacetime is a special case of stationary spacetime. While both have metrics that do not change with time, static spacetime has the additional property that there are no off-diagonal terms involving time and space coordinates in the metric tensor. In stationary spacetime, there can be such terms, which often correspond to rotational effects. For example, the spacetime around a rotating black hole (described by the Kerr metric) is stationary but not static.

## Can you give an example of a static spacetime solution in General Relativity?

A classic example of a static spacetime solution in General Relativity is the Schwarzschild solution. This solution describes the spacetime outside a spherically symmetric, non-rotating, uncharged massive object. The Schwarzschild metric is time-independent and has no rotation, making it a static spacetime.

## What are the conditions for a spacetime to be considered static?

For a spacetime to be considered static, it must satisfy two main conditions: (1) The metric must be time-independent, meaning that the components of the metric tensor do not change with time. (2) The spacetime must have a timelike Killing vector field that is orthogonal to a family of hypersurfaces, indicating that there is no rotation or cross-term involving time and space coordinates in the metric tensor.

## Why is the concept of static spacetime important in General Relativity?

The concept of static spacetime is important in General Relativity because it simplifies the analysis of gravitational fields and spacetime geometry. Static spacetimes often serve as idealized models for understanding more complex scenarios. They provide insights into the properties of gravitational fields around non-rotating bodies and help in studying phenomena such as black holes, gravitational time dilation, and the behavior of test particles in a gravitational field.

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