- #1
Clvrhammer
- 7
- 0
- 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.
$$g = - f(r) \, dt^2 + f(r)^{-1} \, dr^2 + r^2 \, d\Omega^2 \quad , \qquad f(r) = 1 - \frac{2m}{r}$$
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?
---
ADDENDUM:
Def.: A function ##u## is called time-function, if its gradient ##\nabla u## is everywhere timelike past-pointing.