My twin galaxy puzzle Please help

[Fredrik]

Yes, r == psi. I simply use A=1, which means dimensionless r and a scalefactor with units of time or length.
Proceed with the total derivative
$$dt' = \frac{\partial t'}{\partial t}dt + \frac{\partial t'}{\partial r} dr = cosh(r)dt + t\, sinh(r) dr$$
and so on, insert in the metric, sort it out and find the minkowski metric.

OK, I'm going to have another look at it. (I haven't done it yet).

Edit: I still haven't looked at it, but I realized that I forgot to mention why I would be very surprised if this works. If a change of variables can bring the metric to "Minkowski form", then the spacetime we're dealing with is Minkowski space. That spacetime is the solution with k=0, ρ=0, Λ=0, so I don't expect to find another copy of it among the k=-1 solutions.

Spatial geodesics generally have not much physical meaning. In this case, it's clear that the curved spatial geodesics can't be spacetime geodesics.

That's not a FRW universe, since you have constant $$\dot a$$. I don't know if this visualization is helpful,

I'm aware of this flaw in the visualization I described, but it's a minor flaw, and you could easily have corrected it rather than dismiss the whole thing. I thought about making a post earlier to say that time is represented by a function of the distance from the center instead, but I was too lazy. What that function is depends on the density and pressure.

I think this way to visualize a FLRW solution (with positive curvature) is very useful. It has helped me to understand many weird things about cosmology in the past.

Edit: The "flaw" discussed above isn't really a flaw. It's a just convenient way to represent the positive-curvature solution. It should be adequate for most purposes. It's true that if a(t)=constant*t for all t, then it isn't a FLRW solution, but we really don't have to think of the radius of a sphere as a(t). The radius R can be a function of a(t): R(t)=r(a(t)), and if we choose the r to be the inverse of a, we'll have R(t)=t. Yes, I know that a isn't invertible, but the restriction of a to the open interval from the big bang to the moment where the expansion reverses is invertible, so we can at least represent the era during which the universe is expanding this way.

The cause is inflation (in the standard model)

That's definitely incorrect. Inflation only explains why the distance is currently so large. The effect we're talking about (distant objects moving away from us at a speed that grows with distance) is present even in a FLRW solution without inflation.

What we have now is an almost empty spacetime, where we can start with flat/Newtonian/postNewtonian approximations at each event, and try to look at it from this point of view.

Yes, but that's not what you're doing. We're talking about a region of spacetime that we have selected specifically because it's so large that its curvature can't be neglected. I will make this point more clear below.

No. Examine the empty universe example, where normal coordinates apply to all of spacetime. Neither is c!=1, nor are the time coordinates the same.

I will look more closely at this claim. I'm not 100% sure that the visualization I suggested is adequate for this, but if it is, it implies that you're wrong about this.

You have the metric, the worldline, and you can read off the time coordinates and integrate to get proper time.

That doesn't answer my question. How do you know that the result of that procedure is that "events separated by a certain proper time on [the world line] are separated by a larger coordinate time."? Suppose that P and Q are two events on the world line of galaxy A, with P being "earlier" than Q. Suppose that we're considering the normal coordinate system associated with galaxy B's motion at an event R on its world line, where R has been chosen such that the time coordinate of P is =0 in this coordinate system. Do you know how to calculate the time coordinate of Q? I don't.

I also expect the difference between the time coordinates of P and Q to depend on the choice of R (the origin of the normal coordinate system). What if we e.g. choose R such that the time coordinate of Q is =0 and calculate the time coordinate of P? Will the difference be the same? It would be in SR, but this isn't SR.

Normal coordinates define a set of observers that are at rest wrt the origin. FRW comoving particles have relative velocity to these observers, therefore, at any point, there is time dilation wrt the respective "static" observer.

I don't know why you're saying "therefore" as if the conclusion is an immediate consequence of what you said before. It makes me think that you keep making the (big) mistake to think that SR results can be immediately transferred to GR. Galaxy B is at rest in the normal coordinates of galaxy B, and in FLRW coordinates. Galaxy A is at rest in FLRW coordinates, and in the normal coordinates of galaxy A, but not in the normal coordinates of galaxy B. The only conclusion you can make immediately (due to the equivalence principle) is that if an object X is stationary in the normal coordinates of galaxy B and its world line intersects the world line of galaxy A, then there's time dilation between the normal coordinate systems of object X and galaxy A. You can not immediately conclude anything about what the time dilation is between the normal coordinate systems of the galaxies. You would have to calculate it, and as I said before, the result may depend on which point on galaxy B's world line you take to be the origin of its normal coordinates.

Those static observers are additionally up or down a gravitational well, therefore their clocks are generally ticking at a different rate than the one at the origin.

I think we should ignore the fact that spacetime geometry isn't really FLRW near a galaxy for now. We shouldn't introduce additional complications until we have solved the "simple" problem.

 

[Ich]

I still haven't looked at it, but I realized that I forgot to mention why I would be very surprised if this works. If a change of variables can bring the metric to "Minkowski form", then the spacetime we're dealing with is Minkowski space. That spacetime is the solution with k=0, ρ=0, Λ=0, so I don't expect to find another copy of it among the k=-1 solutions.

Reduced-circumference polar coordinates (as Wiki calls them), where k is -1,0,1, are ill-defined in this case. Try hyperspherical ones, where
$$(\frac{\dot a}{a})^2 = \frac{k}{a^2}$$
You see that k=0 is valid only for $$\dot a=0$$. These are Minkowski coordinates.
However, you may choose any constant $$\dot a$$ you like; that only means that you choose to measure space with moving (comoving) rods. Lorentz contraction means that radial distances are measured shorter than tangential distances (no typo: shorter), which means negative curvature of space.
Spacetime is still empty, so I would be really surprised if you could not change your coordinates to Minkowski ones. Just do it, it's not too tedious.

I'm aware of this flaw in the visualization I described, but it's a minor flaw, and you could easily have corrected it rather than dismiss the whole thing.

I don't dismiss it. To the contrary, my increased interest in cosmology started with calculating the trajectories of particles that are not glued to the usual Balloon model, but float freely (look https://www.physicsforums.com/showthread.php?t=294690"). The model is extremely poweful, but arbitrarily complicated and counter-intuitive if you try to use a different time than cosmological time.

The cause is inflation (in the standard model)

That's definitely incorrect. Inflation only explains why the distance is currently so large. The effect we're talking about (distant objects moving away from us at a speed that grows with distance) is present even in a FLRW solution without inflation.

The effect is certainly present, but the cause lies mainly in the past, like a initial condition.

What we have now is an almost empty spacetime, where we can start with flat/Newtonian/postNewtonian approximations at each event, and try to look at it from this point of view.

Yes, but that's not what you're doing. We're talking about a region of spacetime that we have selected specifically because it's so large that its curvature can't be neglected.

Not necessarily. I'd help a lot if you'd convince yourself that the difference I'm talking about is also present in flat spacetime, therefore a coordinate effect. We can add curvature in succesive steps as indicated.

Examine the empty universe example, where normal coordinates apply to all of spacetime. Neither is c!=1, nor are the time coordinates the same.

I'm not 100% sure that the visualization I suggested is adequate for this, but if it is, it implies that you're wrong about this.

If I'm wrong about the possibility of Minkowski coordinates in an empty spacetime, I'll withdraw immediately from this forum and my job, and live happily to the end of my days as a gardener.
You realize what you're claiming here?

Do you know how to calculate the time coordinate of Q? I don't.

In an empty model, I do know. In a de Sitter universe, I know also. In general spacetimes, I don't know, except numerically.

I don't know why you're saying "therefore" as if the conclusion is an immediate consequence of what you said before. It makes me think that you keep making the (big) mistake to think that SR results can be immediately transferred to GR.

If you read carefully, I refined my statement such that I'm talking about a local comparison - galaxy moving past an observer who is at rest with the origin. Of course I can make the calculation in SR.

You can not immediately conclude anything about what the time dilation is between the normal coordinate systems of the galaxies.

I don't - not any more. I tried to explain in my last post that there are other second order corrections which have to be included. These go by the name of "gravitational time dilation". In this (second order - ) picture, time dilation due to velocity additionally applies.

I think we should ignore the fact that spacetime geometry isn't really FLRW near a galaxy for now. We shouldn't introduce additional complications until we have solved the "simple" problem.

I'm not talking about galaxies. I'm talking about this funny term on the right side of
$$\frac{\ddot a}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^{2}}\right)$$
That's a homogeneous matter distribution, and in normal coordinates you have at any point r a potential proportional to the amount of matter within the sphere with radius r.

 

[michelcolman]

What Ich is saying pretty much agrees with the model I had come up with, but it looks like Frederik is having trouble understanding exactly how it works, so I'll try to explain things the way I see them:

Suppose there is no curvature in space. If there's any matter or energy in it, just assume it's negligible (at least for now, we'll refine this point later). Spacetime is flat.

At some point in the past, the big bang occurred and sent stuff flying apart everywhere with some accelleration proportional to distance (measured locally between nearby points) during a short time, after that the stuff kept flying straight ahead. This is of course a huge oversimplification but never mind that.

Now look at the entire universe from our point of view, applying SR. It is quite obvious that time for distant objects must be running slower, since they are moving relative to us. Don't go all "but they're in a different M4 space" or whatever it was Frederik was babbling about, we are just looking at the entire spacetime from our point of view, assuming a flat and infinite universe in which things just happen to be flying apart but they might as well be flying in different directions, our metric is not tuned to this expansion in any way. The objects are "really" moving relative to us, they are not fixed in some kind of expanding metric (that's a different model which we'll talk about later).

Since we have based our coordinate system on SR, nothing can travel faster than light. Very distant objects will be approaching it, and their time will be grinding to an asymptotic halt. They will never reach our age, but are "stuck" in the big bang.

Locally, of course, time over there is just moving at its normal pace, life will develop, and the aliens there will be convinced that they are in the oldest galaxy and WE will never exist. This is perfectly acceptable as we will never be able to see each other anyway.

Along with time dilation, we obviously have length contraction as well. In fact the entire "infinite" universe actually fits very neatly in a finite sphere with a radius of the age of the universe times the speed of light. At the edge of the sphere, the Big Bang is just starting right now, and will forever be "just starting". Obviously this "edge" is a singularity. Slightly inside the edge, the big bang accelleration is still occurring but time has almost come to a standstill. There's still an infinite amount of space in the tiny shell if you were there locally, it's only shrunk to a finite distance because of length contraction (which asymptotically becomes infinite at the edge). In fact, the location of the aliens is close to the edge from our point of view, but they will actually say they are in the middle and we are near the edge.

Now let's compare this metric with the cosmological model. In that model, the time coordinate is defined as local time experienced by an observer who is moving together with the expansion of the universe. This gets rid of time dilation by definition, the universe is the same age everywhere. Space distances are defined in such a way that the speed of light, measured locally relative to a local, comoving observer, is c. That means that the speed of light in a faraway region of space viewed from our position, has to be vectorially added to the local expansion speed of the universe! Which may be greater than the speed of light itself, by the way, making it impossible for light to get here from there.

In this model, the universe is homogenous, and the very distant alien does exist, right now, but we will never be able to communicate with it because space in between is expanding too rapidly for even light to cross it. So in any case it does not really matter whether it exists or not, and there is no contradiction with the first model. Just like an event can be earlier or later than another event depending on the observer, the first coordinate system moves the aliens off the spacetime map entirely.


OK, so far so good, back to the first model. The only thing to work out now, is what happens when gravity (curvature) is thrown into the mix. For example, if gravity is strong, and there is no dark energy or anything like that, and the universe would somehow be pulled back together in a Big Crunch, there should not be a twin paradox so somehow gravitational time dilation should probably offset the time dilation due to speed (at least in the end). That's the question I asked in the other thread (Twin Paradox in Big Crunch).

This gravitational time dilation would be explained by considering the cosmic gravitational field centered around our position. Obviously the aliens will say the cosmic gravitational field is centered around them, and they are perfectly at liberty to say so.

With any luck, distant clocks will be moving slower initially, then speed up and actually get faster than ours while the universe slows down (no more time dilation due to speed, while gravity still acts), actually getting ahead of our clocks, and then slow down again during the big crunch so they are exactly in sync with ours when we get back together. But someone much smarter than me will have to calculate that some day...

Another problem is what the universe will look like around the time of reversal, when expansion turns into contraction. Gravitational effects must somehow undo the length contraction as well, because otherwise we're still stuck with the expanding Big Bang Shell while the big crunch will happen at a definite time in the future. This will probably be solved by considering that the speed of light increases as you move higher up in the gravitational field, allowing things to travel faster than our local speed of light. The universe is probably infinite all along after all...

Whew, does all this make any sense at all?

 

[Ich]

Disclaimer:
Neither do I perceive Fredrik as babbling, nor do I think did I establish my viewpoint well enough that it is only a matter of understanding how it works. Some discussion is definitely justified.
Further, I refrain from describing the whole universe via these "SR-like" coordinates. This can be done, at least for an infinite universe up to the horizon. But the SR/gravitational time dilation concepts only make sense if one compares with a suitable flat background metric, i.e. in a finite neighbourhood of some point.

 

[michelcolman]

Disclaimer:
Neither do I perceive Fredrik as babbling, nor do I think did I establish my viewpoint well enough that it is only a matter of understanding how it works. Some discussion is definitely justified.

I agree, the word "babbling" was a poor choice of words, it was just that he was talking about different M4 frames (or whatever it was) while this had nothing to do with your metric. Just two people misunderstanding each other, happens all the time, I did not mean to imply anything else.

 

[Fredrik]

OK, I finally did the variable change you (Ich) suggested, and I'm getting the result you said I would. I did this a few days ago, but I still can't quite wrap my head around it. It seems completely bizarre to me that one of the k=-1 FLRW solutions is the t>0 half of Minkowski space.

$$ds^2=-d\tau^2+\tau^2(d\psi^2+\sinh^2\psi (d\theta^2+\sin^2\theta\ d\phi^2))$$

$$t=\tau\cosh\psi$$
$$r=\tau\sinh\psi$$

$$\begin{pmatrix}dt\\ dr\end{pmatrix}\begin{pmatrix}\cosh\psi & \sinh\psi\\ \sinh\psi & \cosh\psi \end{pmatrix}\begin{pmatrix}d\tau\\ \tau d\psi\end{pmatrix}$$

The matrix is a hyperbolic rotation, so you can invert it just by changing the sign of the "angle".

$$\begin{pmatrix}d\tau\\ \tau d\psi\end{pmatrix}\begin{pmatrix}\cosh\psi & -\sinh\psi\\ -\sinh\psi & \cosh\psi \end{pmatrix}\begin{pmatrix}dt\\ dr\end{pmatrix}$$

Now it's easy to verify that the line element takes the form

$$ds^2=-dt^2+dr^2+r^2(d\theta^2+\sin^2\theta\ d\phi^2)=-dt^2+dx^2+dy^2+dz^2$$

I'm going to have to do some thinking about what this means.

You seem to be right about the geodesics. A geodesic is a straight line in the t,x,y,z coordinates, but the result

$$t^2-r^2=\tau^2(\cosh^2\psi-\sinh^2\psi)=\tau^2$$

implies that a curve of constant $\tau$ is a hyperbola.

 

[Fredrik]

I agree, the word "babbling" was a poor choice of words, it was just that he was talking about different M4 frames (or whatever it was) while this had nothing to do with your metric. Just two people misunderstanding each other, happens all the time, I did not mean to imply anything else.

That was actually RUTA, not me. And there was no misunderstanding in the post where he talked about that. He was just asking Ich what he meant by "time dilation" in this case.

 

[Ich]

It seems completely bizarre to me that one of the k=-1 FLRW solutions is the t>0 half of Minkowski space.

Actually, it's the x<t, t>0 "wedge" of Minkowski space. Or, more appropriately, the future light cone of a specific "Big Bang Event". The Big Bang itself is the boundary of that cone. As I said, it's quite similar to Rindler coordinates.
The "recession velocity" of comoving objects in FRW coordinates is exactly the rapidity of said objects in Minkowski coordinates. You can switch from one object's viewpoint to another's by a Lorentz transformation; that does not alter the appearance of spacetime.
You'll find further reference if you look up "Milne Model" - but avoid the Wikipedia article.

Ok, but the interesting point is that any FRW spacetime looks locally like Minkowski space, with second order deviations due to gravity. You can describe it pretty well by Newtonian dynamics, and express gravitational time dilation by the Newtonian potential, as in michelcolman's https://www.physicsforums.com/showthread.php?p=2289391#post2289391".