Definition of time-independent scalar field in GR

In summary, a time-independent scalar field in General Relativity (GR) is a scalar function that does not vary with time and is defined over a spacetime manifold. Such fields can influence the geometry of spacetime and are often used to model static distributions of mass or energy. Unlike dynamic fields that change with time, time-independent scalar fields maintain constant values at each point in space, contributing to the overall gravitational field without temporal evolution.
  • #1
Clvrhammer
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TL;DR Summary
The author asks about the definition of time-independence of a scalar field in the context of General Relativity as therein the notion of time is dependent on the coordinate chart and the causal structure of the spacetime.
I was wondering how the notion of a time-independent field translates into the context of General Relativity. In order to specify my confusion, consider a scalar field ##\phi## in Schwarzschild spacetime with usual coordinates ##(t,r,\theta,\phi)##. Its metric is
$$g = - f(r) \, dt^2 + f(r)^{-1} \, dr^2 + r^2 \, d\Omega^2 \quad , \qquad f(r) = 1 - \frac{2m}{r}$$
with ##d\Omega^2## being the round metric one the unit 2-sphere. Naively, time-independence of ##\phi## means ##\nabla_t \phi = 0##. However, in the region ##\{ r < 2m \}##, the coordinate ##t## is no longer a time-function. Instead, the coordinate ##r## is. Does this now mean that in the interior region time-independence of ##\phi## means ##\nabla_r \phi = 0##?

Moreover, consider the same problem in Eddington-Finkelstein coordinates.
$$g = - f(r) \, dv^2 + 2 \, dv \, dr + r^2 \, d \Omega^2$$
In the exterior region, neither ##v## nor ##r## are time-functions. What does time-independence mean there? On any timelike path, the coordinate ##v## is strictly increasing, so one may view ##v## as some pseudo-time-function and define time-independence as ##\nabla_v \phi = 0## there. However, in the interior region, ##r## is a time-function but ##v## remains strictly increasing on timelike paths. Does time-independence there then mean both ##\nabla_v \phi = \nabla_r \phi = 0## or only one of these conditions?

So my question may be summarised as: Given a set of coordinates, how does one define time-independence of a scalar field? Does it mean the scalar field is unchanged with respect to some time-function? How about in coordinate charts where none of the coordinates are time-functions?

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ADDENDUM:
Def.: A function ##u## is called time-function, if its gradient ##\nabla u## is everywhere timelike past-pointing.
 
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  • #2
As you say, "time independence" as such is not very well defined as you may define whatever coordinate system you will (it doesn't even need to have a timelike coordinate). However, to have some more firm definition of this would mean in the context of a stationary spacetime, you would globally select "time" as given by the corresponding Killing field.
 
  • #3
Orodruin said:
However, to have some more firm definition of this would mean in the context of a stationary spacetime, you would globally select "time" as given by the corresponding Killing field.
Thank you for your answer. Unfortunately, I do not quite understand your last statement. What do you mean by "corresponding Killing field"? Corresponding to what?
 
  • #4
Clvrhammer said:
Thank you for your answer. Unfortunately, I do not quite understand your last statement. What do you mean by "corresponding Killing field"? Corresponding to what?
A stationary spacetime by definition admits a timelike Killing field. So, that Killing field.
 
  • #5
Clvrhammer said:
A function ##u## is called time-function, if its gradient ##\nabla u## is everywhere timelike past-pointing.
In pretty much any spacetime of physical interest, there will be such a function (in fact there will in general be many of them, none of which are uniquely picked out by any feature of the spacetime geometry), so you can always define "time independence" with respect to it. But such a definition might not correspond to any quantity of physical interest and so might not be very useful.

In the case of a stationary spacetime, the timelike Killing vector field defines a time function, and the metric is constant along its integral curves, so this is a time function that can be useful. However, it might only be useful on a portion of the spacetime (in the case of Schwarzschild spacetime, it's only useful outside the horizon since the Killing vector field is only timelike outside the horizon).

Note also that if a time function exists in a spacetime, its existence is an invariant, independent of any choice of coordinates. But of course some choices of coordinates will be better adapted to that time function than others.

In short, the general answer to your question is that the concept of "time independence" does not have a unique, well-defined meaning in GR. In some special cases you can pick a useful concept of "time independence", but it will still be limited.
 

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