In GR, coordinate are just a tool for us to describe the physics, they should be equivalent. However, in standard form of FLRW metric, it can be inferred that the universe is expanding, but we can do a coordinate transformation to make the spatial part static or changing in a different way with respect to time. Is there a notion of expanding universe which does not depend on coordinates?
Asked
Active
Viewed 2,027 times
6
-
1I'm afraid it isn't clear to me what you mean. Can you rephrase your question in a more mathematical way to clarify exactly what you're asking? Are you comparing the different interpretations of the expansion given by comoving coordinates and "everyday" coordinates? – John Rennie Jan 25 '15 at 16:37
-
What do you mean by "why we on Earth only confirm the standard coordinates used in FLRW metric through our observation"? Which observations are you talking about? In both special and general relativity, all coordinate systems predict the same things about local physical observations, like the proper time on an observer's clock at the moment they receive light from various distant events. – Hypnosifl Jan 25 '15 at 16:39
-
what I mean is the fact that universe is expanding as we observed is only predicted by the FLRW metric in certain special coordinates. If I do a coordinate transformation, the space could be static. Does that mean we are in a special frame? – Shadumu Jan 25 '15 at 17:16
-
1@user3229471: You should edit your comment into the post. – Kyle Kanos Jan 25 '15 at 17:25
-
But what do you mean that the "universe is expanding as we observed"? What specific physical observations are you talking about? If you phrase things in terms of local observations like the relation between the redshift of a galaxy and the apparent brightness of standard candles in it (which can be used to deduce distance in the standard cosmological coordinate system), then all coordinate systems predict the same results for these local observations, even if the observations don't have the same meaning in terms of coordinate distances etc. – Hypnosifl Jan 25 '15 at 17:43
-
@Hypnosifl I see your point. But why this particular coordinate is used to describe the metric but not any other coordinates – Shadumu Jan 25 '15 at 21:29
-
I'm sure you could use a different set of coordinates to express the metric, but it'd probably be more complicated mathematically. If you specifically want to see an example of the FLRW metric expressed in a different coordinate system (with a different definition of simultaneity do that each spatial hypersurface would have a different curvature than the hypersurfaces of simultaneity of the standard cosmological coordinate system), you might consider asking a new question about this, maybe someone here would know of some published example, or could construct one themselves. – Hypnosifl Jan 25 '15 at 22:40
-
2Am I the only one who finds this question perfectly clear? The answer I was going (actually started) to write would have shown how cosmological time and proper distance along associated spatial slicings are special and why FLRW spacetime is not Minkowski space, even though conformal time and comoving coordinates make it look that way – Christoph Jan 25 '15 at 23:22
-
@Hypnosifl what about length contraction in SR, do you think that is physical or just a matter of different coordinate distances? What if the FRW metric was first proposed in some other coordinates, we might change the way we think about the evolution of universe, space might be shrinking or static in some coordinates, though the metric/physics is still the same. – Shadumu Jan 26 '15 at 15:10
-
I'm confused, the expansion of the universe is a conclusion that can be drawn in a coordinate-independent way. We can see the expansion of the metric without making a choice of coordinates. $ds^2=-dt^2+a(t)^2d\Sigma^2$. This shows that the spatial dimensions expand with the scale factor and $d\Sigma$ is coordinate independent. So perhaps I misunderstand the question – Jim Jan 26 '15 at 15:33
-
@Jim you can do a general coordinate transformation in which the t and space coordinates mixed up, the new set of coordinates is equally good but the interpretation can be different, though physics being the same. – Shadumu Jan 26 '15 at 15:40
-
Wouldn't that no longer be the FLRW-metric? Perhaps a demonstration is in order. Could you give an example of such a transformed FLRW-metric? or at least just the $g_{\mu\nu}$? – Jim Jan 26 '15 at 15:56
-
@Jim Sorry, I can't give you the metric, I am just thinking that in principle you can do it. You are free to choose a coordinate to describe the same metric. It is just your interpretations could be different, just as the length of a rod will have different values in different frames in SR. – Shadumu Jan 26 '15 at 16:04
-
Ah, I understand now. I think that would be more a problem with the coordinate transformation. The system transformed to could be unphysical or in a complicated, non-inertial, and unsymmetric motion. But I think that in all cases over long time spans, the FLRW-metric always looks like it s expanding. I haven't got a math proof in mind to back up that claim, however – Jim Jan 26 '15 at 16:13
-
@user3229471 - I would say that length contraction is dependent on one's choice of simultaneity convention which is coordinate-dependent, although given a spacelike curve between points on the worldlines of either end of an object (a curve that would be confined to a single simultaneity surface if it was intended as a measure of instantaneous "length" in some coordinate system), the proper distance along that spacelike curve is coordinate-independent (just like proper time along timelike curves is coordinate-independent). – Hypnosifl Jan 26 '15 at 16:17
-
@Jim think about the positive cosmological constant Lambdavacuum solution. In some coordinates, the universe seems static, whereas in some other, the universe is expanding. I beg similar things can happen to FRW – Shadumu Jan 26 '15 at 16:21
-
By the way, I do know of a special case that might interest you--in the standard coordinate system for FLRW universes the curvature of space in each simultaneity surface is hyperbolic whenever the mass density is below a critical value, and this is true even when the density goes to zero. This is known as the Milne model, but it is really just a different coordinate system on flat Minkowski spacetime. The 2nd large diagram here shows how lines of simultaneity look when plotted in an inertial frame. – Hypnosifl Jan 26 '15 at 16:24
-
@Hypnosifl so indeed when we use another frame and talk about whether stars are getting further away from us, we need to make sure simultaneity when we measure, then we may obtain different conclusions depending on what frame we choose, just like the length contraction. What we observe is redshift and it should be coordinate independent, but when we propose the FRW model to fit the data, we are free to choose a coordinate to describe it. In some of them, we may say the universe is not expanding. – Shadumu Jan 26 '15 at 16:26
3 Answers
7
The standard coordinate system is the mathematically simplest, but I don't think it's actually the most physically intuitive. This is because we live on objects that are gravitationally bound, and admist objects that are electromagnetically defined. This means that our local length scales are not affected by cosmological expansion. But, if you look at the FRLW metric, in its standard form (I choose the flat cosmology for simplicity):
$$ds^{2} = - dt^{2} + a(t)^{2}\left(dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}\right)$$
you can tell that, for some constant-t observer, the ruler actually expands with time by a factor $a$. For this reason, when describing cosmological observations, I actually like to use a different coordinate system, where you replace $r$ with $R = a(t)r$. Then, you have $dR = {\dot a}r\,dt + a\,dr \rightarrow dr =dR- H (R/a)\,dt$, and the metric becomes (note that I used the relation $H = \frac{\dot a}{a}$, to replace $a$ with Hubble's "constant"):
$$ds^{2} = -\left(1-H^{2}R^{2}\right)dt^{2} + 2dR\,dt\left(-HR\right) + dR^{2} + R^{2}d\theta^{2} + R^{2}\sin^{2}\theta d\phi^{2}$$
In terms of direct physics, this coordinate system is a lot clearer. You see that, for a constant-t observer, there is a coordinate singularity at $R = \frac{1}{H}$, corresponding to the cosmological horizon. Furthermore, this coordinate system has a $g_{tr}$ coordinate, which, it can be shown, corresponds to the frame dragging of the system -- so space naturally expands away at a velocity proportional to $HR$, which gives you Hubble's law.
Zo the Relativist
- 41,373
- 2
- 74
- 143
-
But note that all coordinate systems are equal, at base. This is just a different way to look at the physics of this coordinate system. – Zo the Relativist Jan 26 '15 at 18:18
-
-
It could be that in some coordinates the universe appears not to be expanding as a(t). Though all the physics being unchanged, how we interpret them will depend on coordinates. Is there a notion of expanding universe which does not depend on coordinates? – Shadumu Jan 26 '15 at 19:45
-
@user3229471: you have to talk about what measurement you're making, at the end of the day. What I did here was take the dynamics out of the $r$ coordinate, and put them in the $t$ coordinate. In the ADM language, all of the physics in this version of the FRLW metric lives in the lapse and shift. – Zo the Relativist Jan 26 '15 at 19:58
-
-
Also, @user3229471, note that the scale factor $a(t)$ is not an observable of the theory -- it is set by a choice of the time coordinate and by an initial scale for the universe. The Hubble parameter, though, is an observable. – Zo the Relativist Jan 26 '15 at 20:00
-
@JerrySchirmer say I measure the distance between two galaxy using the FRW metric in some coordinate. Certainly I will not get the same value and the value could be changing differently in different coordinates, not necessarily expanding. – Shadumu Jan 26 '15 at 20:07
-
If you have two galaxies, and you're measuring the distance between them in some reference frame, the result will be coordinate-independent. – Zo the Relativist Jan 26 '15 at 20:10
-
Explicitly, if you're measuring the distance along some path $\gamma$, then the distance will be:
$I = \int \sqrt{-g}\epsilon_{abcd}v^{a}w^{b}z^{c}dx^{d}$
where $v,w,z$ are three unit vectors normal to $\gamma$.
– Zo the Relativist Jan 26 '15 at 20:12 -
Your metric doesn't have a singularity at $R=1/H$. The $t$ coordinate becomes lightlike there, and spacelike beyond, but it's never singular. The cosmological horizon generally isn't at $R=1/H$ anyway, except when $Ω=Ω_Λ$. Also, the $dR,dt$ term isn't evidence of physical frame dragging, it's just a coordinate artifact. You can tell by the fact that it's linear in $R$ at the origin. Any first-order deviation from the Minkowski metric at the origin can always be removed by a change of coordinates. For best behavior at the origin you'd need a non-FLRW time coordinate, which starts to get messy. – benrg Aug 13 '20 at 00:14
-
I have a small doubt. dt will be equal to the proper time measured by which observer. Like in the Schwarzschild coordinates dt will be equal to the proper time of the observer at infinity or in the gullstrand coordinates dt will be equal to the proper time of a radially falling observer. Similarly here dt would be equal to the proper time measured by which observer – Shashaank Feb 23 '21 at 16:43
2
Comments to the question (v3):
It is true that there exists a huge freedom to choose local coordinates in GR, but it is not possible to alter the metric tensor $g_{\mu\nu} dx^{\mu}dx^{\nu}$ (when we include the basis elements $dx^{\mu}$ and $dx^{\nu}$).
Given an arbitrary but single fixed spacetime point $p$, there exist Riemann normal coordinates.
We cannot get the metric components $g_{\mu\nu}$ on an arbitrary prescribed symmetric form (with Minkowski signature) in an open neighborhood, no matter how small. It is not a free lunch!
Qmechanic
- 201,751
0
All coordinate systems are equal, but some systems are more equal than others ;)
In case of Friedmann universes, there's a distinguished set of coordinates that corresponds to a family of freely-falling observers which see the universe as isotropic and chosen so that matter is distributed homogeneously within a spatial slice at constant time.
Furthermore, we could choose our coordinates so that the time-like coordinate agrees with proper time of our observers and the space-like coordinates agree with proper distance within a spatial slice.
This is just one possible choice among many: For example, observers in relative motion would not see the universe as isotropic, and their description of the matter distribution would be as valid as the one we chose - just less convenient.
Even if we keep our set of observers, we're free to scale our coordinates as we see fit. For example, using conformal time and comoving coordinates makes FLRW spacetime look deceptively like Minkoswki space with a static matter distribution:
(source)
Note how the yellow light rays are given by straight lines and that galaxies would sit at a fixed comoving distance.
This, however, is misleading, the same way that doing a logarithmic plot does not change the underlying function. Note that whatever the coordinates, the math will of course still work out thanks to the magic of tensor calculus: spacetime will remain curved, proper distance within spatial slices will increase and light will experience redshift.
-
i think the distance within spatial slice at some constant time is not the same in different coordinates, just like length contraction. Hence, in some coordinates, the universe could be not expanding as a(t), is there a notion of expanding universe independent of coordinates. – Shadumu Jan 26 '15 at 19:42
-
@user3229471: the distance within a spatial slice is well-defined; length-contraction is only possible because observers in relative motion do not agree on spatial slicing... – Christoph Jan 26 '15 at 19:56
-
the distance within a spatial slice is not the proper distance. it clearly varies in different coordinates – Shadumu Jan 26 '15 at 19:59
-
@user3229471: it will not vary between coordinates that agree on spatial slicing – Christoph Jan 26 '15 at 20:09
-
spatial slicing means the spatial part of the metric at some constant time right? then if in two coordinates the time part in the metric is different, when we calculate the distances between two stars (set dt=0), we will get different results. – Shadumu Jan 26 '15 at 20:19