- #1
kent davidge
- 933
- 56
Since it's possible to choose a coordinate chart where the Schwarzschild metric components are dependent on time, why that's not done? Would'nt there be a scenario where such a choice would be useful?
kent davidge said:Since it's possible to choose a coordinate chart where the Schwarzschild metric components are dependent on time, why that's not done? Would'nt there be a scenario where such a choice would be useful?
Quite the opposite. I myself have never seen time dependent metric components in the Schwarzschild metric.PeroK said:Is that not what's usually done?
kent davidge said:Quite the opposite. I myself have never seen components time dependent for the metric.
I suspect this has to do with the physical situation involved. That the solution is for a static, etc... spacetime, so the most natural choice should be the one with time independent metric components.
I don't know. I was asking if there's a case where one would want it.PeroK said:Why would you want time-dependent components for a static solution?
kent davidge said:Would'nt there be a scenario where such a choice would be useful?
thank youIbix said:An obvious time-dependent form is one in which the mass is moving. Presumably you could do it - radar coordinates spring to mind as something with practical utility for a passing spaceship. I haven't seen it done. Probably because you'd want a different solution for every possible powered and unpowered trajectory. Easier just to work in regular Schwarzschild coordinates, I should think.
kent davidge said:it's possible to choose a coordinate chart where the Schwarzschild metric components are dependent on time
Because static only means you can find a coordinate system where the metric takes that form taken in the Schwarzschild coordinates. The definition doesn't say that it's the only possibility.PeterDonis said:Why do you think this is the case?
Any metric of the form ##g_{\alpha \beta}(u^0) du^\alpha du^\beta## where the "time" coordinate is ##u^0## and ##g_{0 0} < 0## and ##\ g_{0, \alpha}, \ g_{i,j} > 0##?PeterDonis said:Can you give an example?
kent davidge said:The definition doesn't say that it's the only possibility.
kent davidge said:Any metric of the form...
yes, it would mean that it's not helpful to have such metric.PeterDonis said:I meant a specific example. If you can't find one after searching the literature, that would be relevant to your OP question, would it not?
wow , thanksPAllen said:The metric in Lemaitre coordinates for the Schwarzschild geometry depends on the timelike coordinate. This is a consequence of covering both the horizon interior and exterior in one coordinate patch, and making the outside seem similar to the inside, which is not static. A world line of constant position in these coordinates is the trajectory of a radial free fall from infinity. These coordinates share with Kruskal coordinates the feature that there is one timelike and 3 spacelike coordinates over the whole patch. However, they don’t, and can’t cover the whole Kruskal geometry - they cover one exterior and part of one interior region.
[edit: These coordinates can be quite useful for many questions relating to radial infall because key facts are just read from the coordinates.]
@PeroK , now hoppefully you can see why your earlier commentPAllen said:These coordinates can be quite useful for many questions relating to radial infall because key facts are just read from the coordinates
does not make much of a sense... depending on the situation it's actually more easy to have a second "more difficult" approach.PeroK said:Is GR not hard enough without worrying about questions like that?
PeroK said:Why would you want time-dependent components for a static solution?
but by a change of coordinate chart we can make it static again, correct? (but I guess in this new chart it becomes non-static outside the event horizon)PeterDonis said:Schwarzschild spacetime is only static outside the event horizon.
kent davidge said:but by a change of coordinate chart we can make it static again, correct?
Just to expand on this - there arePeterDonis said:But that Killing vector field might not be timelike; in Schwarzschild coordinates, ##t## is only timelike outside the horizon.
Ibix said:there are three Killing vector fields in Schwarzschild spacetime
Indeed - can't count on a Friday night.PeterDonis said:No, there are four.
PeterDonis said:No, there are four. Three correspond to spatial rotations (heuristically, you can think of them as rotations about three mutually perpendicular spatial axes). The fourth is the one you describe.
PAllen said:@stevendaryl ,
I have been confused about rotational killing vectors in the past. However I think the following makes it crystal clear that ther are three, not two:
http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GRKilling.pdf
I confuse myself like this sometimes. I remind myself by thinking of Minkowski spacetime whose spacelike Killing fields are the translations and rotations, making six. There are only three spacelike directions, so three Killing vectors at a point must be dependent. But the fields aren't. There's always a way to linearly combine the fields to make a fourth at a point, but never so that it works everywhere.stevendaryl said:Okay, getting back to my intuitive picture: My mistake is cavalierly conflating vectors with vector fields.
I believe so. I specified the space-like ones, which I think are six in number. I could have said Euclidean 3-space, I suppose.vanhees71 said:Shouldn't Minkowski space even have 10 killing vectors (or rather Killing fields, to avoid the sloppy use of math language by physicists that again seems to lead to some confusion ;-)) generating the full orthochronous proper Poincare group (space-time translations, Lorentz boosts, and rotations)?
kent davidge said:Two of the Killing fields are easy to know, it follows directly from the independence of the metric on ##t## and ##\varphi## that ##(-(2m - r), 0,0,0)## and ##(0,0,0, (r \sin \vartheta)^{-2})## are the vectors components of the vectors of the two fields at any point ##(t,r,\vartheta,\varphi)##.
But I'm afraid the vector components of the remaining vector fields can only be found by solving the system of PDE resulting from the Killing equations. And this is really hard.
Or perhaps we can split the metric into parts and compute only the necessary (unknown) terms for the Killing equations?
stevendaryl said:Actually, ##\phi## represents the angle for rotation about the z-axis. You can, from symmetry, immediately get the other two: ##\phi_x## representing rotation about the x-axis, and ##\phi_y## representing rotation about the y-axis. So rather than making a question of solving a differential equation, it becomes a geometric problem of how to compute ##\phi_x## and ##\phi_y## in terms of the usual ##\theta, \phi##.
Ibix said:I remind myself by thinking of Minkowski spacetime whose spacelike Killing fields are the translations and rotations, making six.
kent davidge said:But the relations you gave relate the three parameters (angles) of the spherical coordinates with the Cartesian parameters. I was talking about expressing the remaining Killing vector fields in Schwarszschild coordinates. They will have four components. For example, the (dual of the) timelike one will be ##g_{tt} (1,0,0,0) = -(1 / (1 - 2m / r),0,0,0)##.