# Speed > c

1. Jun 1, 2015

### Omega0

Hi,

Say c is a reliable constant in nature. From what point in time there was a possible velocity between point A and B > c?

2. Jun 1, 2015

### Chalnoth

In curved space-time, there is no speed limit between objects separated by some distance. The speed of light limitation is only a limitation at a single point. That is, the speed limit in General Relativity is the statement that nothing can outrun a beam of light.

The reason for this is that in General Relativity, there is no single definition of relative speed at different points. There are many possible choices and no way to say which is better. What this means is that you can write down your equations one way and say that points A and B are moving away from one another at 0.5c, and then write them down another way and say they're moving away at 2.0c. So there's just no way to answer your question.

3. Jun 1, 2015

### Omega0

No, I would call it a reference frame, this might be big, depending from the curvature.
If you measure and you measure from your point of view 0.5c than you measure 0.5c. I expect out there the same equations. I can see the coordinate system. I can transform. GRT doesn't mean that everything is relative, it just means that it makes it more complicated to transform (compared with SRT).
If I go to the limit c then my measurement ends. After the limit c I still have a universe which I can't measure, right? When the object would move avay < c I could measure something...
Sorry but I see a logical break in "comoving coordinates"....

4. Jun 1, 2015

### Staff: Mentor

Measure what? Local measurements are not a problem, and the limit of c applies to those. But how would you measure how fast an object a billion light years away is moving relative to you?

Measurement of what?

I think you have a misunderstanding of them. They're perfectly logical, but they certainly don't work the same as inertial coordinates in SR do. It seems to me that you are expecting them to work like SR inertial coordinates; that is bound to lead to confusion.

5. Jun 1, 2015

### Chalnoth

There is no unambiguous way to make a global reference frame. And even if you decide upon a specific global reference frame, you still have multiple options for defining speed.

The only situation in which it makes sense to compare speeds far away is in flat space-time. In practical situations, this means distances much shorter than the space-time curvature.

It is fundamentally impossible to directly measure the speed of a faraway object when spacetime is curved. The speed has to be interpreted, and that interpretation varies depending upon how you write down your equations.

For example, one way to think of speed is to change in distance over time. There is a way to define a distance specifically: you can use the proper distance. But to do that, you have to make a choice of time: you have multiple choices of which time at point A corresponds to a time at point B. One way to do this would be to use the CMB: we can call the time the same if the CMB temperature is the same. Using this definition, the change in distance over time is and always has been greater than c for most of the visible galaxies (because there are more galaxies visible further away than nearby).

6. Jun 3, 2015

### Omega0

That's the point, how we measure this? How me measure which speed have current objects relative to us?
The history of the universe has to be taken into account, right?
I like to understand how I can have an accelerating unviverse and dark energy if I can't measure the spead? What measurement have we?
Maybe I gett something wrong but are not that comoving coordinates the same as SR inertial coordinates from a point of say a galaxy?
It is said (Liddle, An Introduction to Modern Cosmology) that effects of the expansion having effect only to objects millions of lightyears away.
But for me it makes no sense to say that this effect is not aware to us - to our galaxy our the solar system or us - the point is just, the effect is so very very small - it's a joke.
It would be a logical contradiction to separate the comoving coordinates from the phyisical coordintes in a way that you say: "Well, little things like solar systems or galaxies feel nothing - but on the long distances it is felt, space is evolving".
This doesn't work logically since then you would need to define from when the influence ends to work or begins.
Do I think right?
PS: It something like in QM, the proton doesn't care about the electron mass since the electon charge is so dominant - but the mass is there, if so very very small

Last edited: Jun 3, 2015
7. Jun 3, 2015

### Staff: Mentor

And the answer is that there is no way to measure the "speed" of an object a billion light years away from us, relative to us, because that concept is not well-defined.

Because the "acceleration" of the universe's expansion is not defined in terms of the "speed" of distant objects. It's defined in terms of the universe's scale factor and the fractional rate at which it increases. We measure the increase of the scale factor over time by making various measurements of the light we receive from distant objects: the redshift of the light, the angular size of the objects on the sky, and their brightness. The observed relationships between these measurements tell us how the scale factor of the universe has behaved over time.

No. Objects at rest in comoving coordinates will be moving away from us in SR inertial coordinates centered on us.

What this means is that objects close enough to use are not "comoving" in the global sense of the comoving coordinates used in cosmology; their motion relative to us is not determined by the expansion of the universe. For example, our "Local Group" of galaxies (our galaxy, the Andromeda galaxy, and a few other smaller ones like the Magellanic Clouds) are gravitationally bound to each other, so their relative motion is not due to the universe's expansion. To see objects whose motion relative to us is due to the universe's expansion, you have to look far enough away.

Not really. The expansion of the universe, in and of itself, is not a "force"; it's just the inertia of objects that were flying apart in the past. So objects which are gravitationally bound don't feel any small "force" pushing them apart.

The accelerated expansion, due to dark energy, can be thought of as exerting a small "force" even on objects that are gravitationally bound, like the solar system or the galaxy; but on those scales it is way, way too small to matter.

8. Jun 4, 2015

### Omega0

Okay.

Okay, this goes into the direction that I may unwrap the nots in my potentential misconception, thanks - but why does angular size play a role in an isotropic homogenieous universe? This should only play a role in a non-isotropic universe?
I feel not well if you insert into the model now the "brightness". How to solve the problem of the Olbers paradoxon?
Isn't it a horror to introduce a dimmer factor?
This seems to be a misconception. Everything in the universe is bound to each other, in my point of view. The link can be broken but this demands to be out of the light cone. In my eyes it makes no sense to seperate between local and global. If you are introducing a new coordinate system which you call "comoving coordinates" you are introducing something pretty interesting: You feel fine as you did with absolute space and time (or SRT). You have that expansion factor but who cares, we are in our "Local Group". Then we have the void. Now the expansion factor begins to work. Not before...
This is a clear conceptual break. If you have something in physics then it is very often smooth, except from symmetry breaking. So, do you break a symmetry here?
So we are back again to the roots: We introduce an absolute background. After we have left Galileo transforms and Newtons absolute time we construct physics on expanding space-time-coordinates?
Back again: Now a suddenly introduced "anti-gravitation influence" plays a role - but it is too small to matter.
Sorry, but this seems to me like a logical misconception.

9. Jun 4, 2015

### Staff: Mentor

Because something that's farther away should have a smaller angular size, so angular size is an indirect measure of distance. Basically what I was describing is comparing various indirect measures of distance to determine the behavior of the scale factor of the universe.

That paradox is already solved by any model of the universe which either (a) gives the universe a finite age, or (b) gives objects further away increasing redshifts. Since our current model of the universe does both of these things, it is not subject to Olbers' paradox.

Then your point of view is wrong, at least with respect to this discussion. It is true that, considered in isolation, two pieces of matter can affect each other gravitationally no matter how far apart they are. But the universe does not consist of two isolated pieces of matter. There is matter everywhere, and from the viewpoint of any isolated system (our solar system or our galaxy or our Local Group--any system which is locally gravitationally bound), the matter in the rest of the universe is, on average, distributed symmetrically in all directions. That means the average gravitational influence of distant matter on local systems is zero, by the shell theorem.

Comoving coordinates do no such thing. They are just convenient to use in cosmology because observers who see the universe as homogeneous and isotropic, on average, are at rest in these coordinates. There is no claim about absolute space and time involved.

I didn't say the universe was not expanding in our local region. I said that the relative motion of galaxies in our Local Group is not due to that expansion, because those galaxies are gravitationally bound to each other. See above for more on how that works.

We aren't constructing the physics using any particular coordinates; that's not how GR works. You can describe the solution of the Einstein Field Equation that we use to model the universe in any coordinates you want; you will get the same predictions for all physical observables. (Note, however, that whatever coordinates you use must be able to cover a large enough region of the universe; local inertial coordinates centered on our galaxy, for example, will not do that, they will only cover our local region.) As I said above, we choose comoving coordinates because they are convenient: they make the description look simple so it's easier to work with.

It plays a role on large scales--large distances and long times. It is too small to matter on local scales and over short times. There's no logical problem.

I would strongly recommend that you take some time to learn what our current cosmological models actually say before posting further; you appear to have some basic misconceptions, and if you continue to push them as you have in your latest post you are likely to receive a warning. I would recommend checking out Ned Wright's cosmology tutorial and FAQ here:

http://www.astro.ucla.edu/~wright/cosmolog.htm

10. Jun 4, 2015

### timmdeeg

If I imagine the perfect fluid, then the accelerated expansion works at any scale, regardless how small. But in contrast to the fluid, our universe is homogeneus only on very large scales. A galaxy is gravitationally bound and its average matter density is much larger than the matter density of the universe. So, I would have expected, that the accelerated expansion exerts zero force, not a very small one on a galaxy. Am I wrong?

11. Jun 4, 2015

### Staff: Mentor

Yes. At least, it does on our current understanding of dark energy, which is that it, unlike the ordinary matter and energy in the universe, is homogeneous even on the smallest scales. In other words, the dark energy density really is constant everywhere, on all distance scales, whereas the ordinary matter and energy density is only constant on very large scales.

12. Jun 5, 2015

### timmdeeg

I see, I didn't ask the right question.
Let me try again. According to Friedmann the dynamics of the the expansion of the universe depends on the amounts of matter density and lambda. Now, let's compare the effect of the expansion on the scale of a galaxy regarding two cases, both with the same matter denisity and lambda: (i) the universe is perfectly homogeneous and (ii) the universe shows a large-scale structure like ours. With (ii) we should distinguish the effect of the expansion on a galaxy and on a void of the same size.
Could you please explain the outcome?

Another question. There are two versions to explain why galaxies don't expand. Often people mention the gravitationally bound system, but others talk about overdensity. What is correct? I mean, if two stars orbit each other in a distance of galaxy scale (other masses very far away), it's a gravitationally bound system then and one can hardly talk about overdensity. Shouldn't one expext, that such a binary system participates in the expansion on much smaller scales?

13. Jun 5, 2015

### Tanelorn

I thought the effect of expansion occurred even on small scales but it was extremely small.

14. Jun 5, 2015

### Staff: Mentor

Yes, but the matter density is highly variable on small enough distance scales--the Earth is about 30 orders of magnitude denser than the average matter density used in the Friedmann equations. The lambda density, OTOH, is constant everywhere.

In case (i), the matter is evenly distributed everywhere (which is of course a huge idealization; the reason we have stars and galaxies today is that the matter was never exactly evenly distributed to begin with), so eveyr single piece of matter will follow an exact "comoving" worldline. Since "expansion" is defined in terms of comoving worldlines, any two pieces of matter, no matter how close together they are, will show the effects of expansion.

In case (ii), on small enough distance scales, the matter is much denser than the average over the universe, so even if the center of mass of a system like a galaxy is following a "comoving" worldline, the rest of the matter in the galaxy is not; the system as a whole is gravitationally bound. So, for example, even if we assume that the center of mass of the Milky Way galaxy is following an exact "comoving" worldline, individual stars are not; the solar system's center of mass, for example, is not following a "comoving" worldline; it's orbiting the center of the galaxy. So you won't be able to see the effects of expansion on the scale of the Milky Way, because there are no pieces of matter actually moving on "comoving" worldlines, so their relative motion won't tell you anything about the expansion.

In fact, from our observations, case (ii) applies up to the scale of galaxy clusters and superclusters. Our galaxy and all the others in our Local Group are part of a cluster/supercluster, which is gravitationally bound even on a scale of tens of millions of light years. So even if the center of mass of the supercluster is moving on a "comoving" worldline, individual galaxies are not. So to see the effects of expansion, we would need to look on a large enough distance scale to see multiple superclusters, and be able to compare the motion of their centers of mass (which will be the average motion of all their galaxies). If we assume each supercluster's center of mass is moving on a "comoving" worldline, then comparing their motion does tell us about the universe's expansion, because we are comparing different "comoving" worldlines.

Both. The reason there are gravitationally bound systems at all is overdensity; some regions started out slightly denser than others, and gravitational clumping greatly magnified those density differences, so that we now have gravitationally bound systems on multiple distance scales.

If the average density of the system containing the two stars, at the distance they're orbiting each other, is not greater than the average density in the universe a a whole, then they can't be gravitationally bound.

15. Jun 7, 2015

### timmdeeg

Thanks for you answer. If I understand you correctly, this means that the Milky Way doesn't expand at all. And it should also mean that the 'force' which the dark energy exerts on all scales has zero (not marginal) effect on gravitationally bound systems.

I'm interested in this question, because I never understood this paper: http://arxiv.org/pdf/astro-ph/9803097v1.pdf
The authors claim that the cosmological expansion affects the solar system, the milky way ... .

16. Jun 7, 2015

### Staff: Mentor

Correct.

It is not correct to say that dark energy has zero effect. Dark energy does affect which worldlines are "comoving" ones, so it does affect (by a very small amount) how much the worldlines of objects in a gravitationally bound system like the Milky Way deviate from being "comoving". This effect can be thought of as a very small amount of tidal gravity.

I believe we had a thread on PF some time ago about this paper; I'll see if I can find it. IIRC, the basic conclusion was that the paper was only partly right: dark energy can be thought of as creating a very small amount of tidal gravity (as above), but expansion in itself does not.

17. Jun 9, 2015

### timmdeeg

18. Jun 13, 2015

### Omega0

First: I didn't want to attack you or any else. I just don't like things that are because they are.

Correct.
That's what I wanted to say, obviously in the wrong way.

I think I understand that galaxies are bound gravitationally to each other.
Sorry if didn't find the right words - I just wanted to s
But there is an influence which is really really small (http://arxiv.org/pdf/astro-ph/9803097v1.pdf).
To say the effect is very very small, that's acceptable.
Otherwhise the transformotion between real coordinates and comoving coordinates would be pretty anstonshing...

To summarize: The effect locally is peanuts but its' existing.

19. Jun 13, 2015

### Chronos

Liking our description of the local universe is strictly optional. The observational evidence does not change. Only interpretations.

20. Jun 13, 2015

### timmdeeg

It depends on what you mean, if you say "effect". As Peter Donis explained, gravitationally bound systems don't expand (not even tiny), but experience a tiny tidal force due to accelerated expansion (second derivative of the scale factor $> 0$), which makes a difference.

Last edited: Jun 13, 2015