Is Carlos Barcelo's Research Redefining Cosmology and Gravity?

[Mike2]

This may be a matter of definitions. As SpaceTiger wrote, with accelerated expansion it is usually meant $\ddot a > 0$.

You may define "accelerated expansion" to be $\dot H > 0$, but then you are not using the standard definition for it, and furthermore, you must be aware that both are not equivalent.

You can understand this also with the deceleration parameter $q = - a \ddot a / \dot a^2$. Accelerated expansion means $q < 0$ whereas $\dot H > 0$ means $q < -1$.

Accelerated expansion requires dark energy and $\dot H > 0$ requires, at least in flat space, a phantom energy.

You might be right. I'd have to see the math worked out in full. But doesn't the supernovae data say that it is the Hubble parameter, H= 72km/(sec*Mpc), that is increasing in more recent times?

 

[SpaceTiger]

If the expansion rate does not change and stays at 72km/(sec*Mpc), then the proper distance to the Hubble sphere (where the expansion rate is the speed of light) does not change either.

That's correct. In a pure $\Lambda$ phase, the Hubble sphere has a constant physical size, but a decreasing comoving size. This means that it will enclose fewer objects as time goes on, but not recede in proper distance. Again, acceleration generally refers to the behavior of objects at a constant comoving distance, not a constant physical distance.

Well, I don't mean to be a pest, but you just got through saying that $$H^2=\frac{\Lambda}{3}$$.

I said that expression would be valid in a pure $\Lambda$ phase. Surely you know that $\Lambda CDM$ is not pure $\Lambda$? Prior to the SNe results, many assumed that the universe was matter-dominated (and flat), which would give the following for the Friedmann equation:

$$H^2=\frac{8\pi G\rho}{3}$$

 

[Mike2]

That's correct. In a pure $\Lambda$ phase, the Hubble sphere has a constant physical size, but a decreasing comoving size. This means that it will enclose fewer objects as time goes on, but not recede in proper distance. Again, acceleration generally refers to the behavior of objects at a constant comoving distance, not a constant physical distance.

Then what's this talk about a possible big rip if the universe keeps on accelerating in its expansion? In such a scenario, even galaxies are torn apart by the space between them accelerating away faster than gravity can hold it together. I'm not aware of the mention of phantom energy as a cause of the big rip, they only presume a cosmological constant. In such a case, space would be accelerating ever faster so that the distance at which it recedes at at the speed of light gets closer.[/QUOTE]

 

[SpaceTiger]

Then what's this talk about a possible big rip if the universe keeps on accelerating in its expansion? In such a scenario, even galaxies are torn apart by the space between them accelerating away faster than gravity can hold it together. I'm not aware of the mention of phantom energy as a cause of the big rip, they only presume a cosmological constant. In such a case, space would be accelerating ever faster so that the distance at which it recedes at at the speed of light gets closer.

Nah, the big rip won't occur with a cosmological constant, it only occurs with phantom energy, as hellfire said. Basically, it happens because the dark energy density is increasing with time (rather than being constant, as with $\Lambda$), eventually becoming larger than the binding energy of the universe's constituents (galaxies, atoms, etc.) and tearing them apart.

 

[George Jones]

Nah, the big rip won't occur with a cosmological constant, it only occurs with phantom energy, as hellfire said. Basically, it happens because the dark energy density is increasing with time (rather than being constant, as with $\Lambda$), eventually becoming larger than the binding energy of the universe's constituents (galaxies, atoms, etc.) and tearing them apart.

According to current ideas in physics, John Baez writes in an http://math.ucr.edu/home/baez/end.html" that galaxies and atoms will eventually come apart even without a big rip. Of course, future discoveries could change this scenario.

John Baez[/URL]]Indeed, this whole discussion should be taken with a grain of salt: future discoveries in physics could drastically change the end of this story.

 

[Mike2]

Nah, the big rip won't occur with a cosmological constant, it only occurs with phantom energy, as hellfire said. Basically, it happens because the dark energy density is increasing with time (rather than being constant, as with $\Lambda$), eventually becoming larger than the binding energy of the universe's constituents (galaxies, atoms, etc.) and tearing them apart.

This issue is a stumbbling block for me. I must be missing something fundamental, and I can't proceed until it's resolved. I don't seem to be making my question very clear. So let me try again.

Even before the Friedmann-Robertson-Walker metric or the application of GR, the expansion rate was measured by the red shift of distant galaxies. Eventually, this recession rate was pinned down to H = 72km/(sec*Mpc) - the result of direct measurement. This means that objects that are 1Megaparsec away are receding at 72km/sec. Objects that are 2Mpc away are receding at 144km/sec. Ultimately there is a distance that is receding at the speed of light - the Hubble sphere. Am I right so far?

Then came the observation that the rate of expansion is not constant, that more distance supernovae were dimmer than expected because they had receded faster than expected of a linear Hubble law. This means that points of space had a rate of recesion that changed. Is this right so far?

OK, then doesn't that mean the Hubble rate of 72km/(sec*Mpc) is not constant but has changed?

 

[hellfire]

The Hubble law relating proper distance (on a simultaneity hypersurface) to recession speed is always linear. For the current time:

$$d_p = \frac{1}{H_0} v$$

For small redshifts this can be approximated to:

$$d_p = \frac{1}{H_0} z$$

However, what you have probably read is about the recession speed vs. luminosity distance (instead of proper distance). As you can see in Ned Wright's tutorial on Supernova Cosmology the relation depends on the cosmological model.

For small redshifts, the luminosity distance can be written as:

$$d_L = \frac{1}{H_0} \left(z + \frac{1}{2} (1 - q_0) z^2 + ...\right)$$

where $q_0$ is the current deceleration parameter. You can see that the more $q_0 < 0$ the more deviates this from the linear relation.

 

[SpaceTiger]

According to current ideas in physics, John Baez writes in an http://math.ucr.edu/home/baez/end.html" that galaxies and atoms will eventually come apart even without a big rip. Of course, future discoveries could change this scenario.

The heat death is really quite different from the big rip. In the scenario Baez describes, atoms are gradually ionized as their density decreases and galaxies slowly boil away. In the big rip, on the other hand, both are suddenly torn apart by the increasing energy density of the phantom energy.

 

[SpaceTiger]

OK, then doesn't that mean the Hubble rate of 72km/(sec*Mpc) is not constant but has changed?

It has indeed changed, but I don't think that explains the confusion you're having. Remember that we're looking back in time when we look at high redshifts, so they weren't expecting a linear relation (nor did they get one). In both $\Lambda CDM$ and in the matter-dominated universe, the Hubble constant will decrease with time, but it will decrease much more quickly in the latter.

 

[George Jones]

The heat death is really quite different from the big rip.

I didn't mean to imply that it wasn't, I just meant to highlight that a big rip isn't the only way that galaxies and atoms can come apart.

 

[Mike2]

It has indeed changed, but I don't think that explains the confusion you're having. Remember that we're looking back in time when we look at high redshifts, so they weren't expecting a linear relation (nor did they get one). In both $\Lambda CDM$ and in the matter-dominated universe, the Hubble constant will decrease with time, but it will decrease much more quickly in the latter.

Then what can an accelerating universe means except that distances between points are receding away from each other with ever greater velocity. I suppose it does complicate things to have to guess what the simultaneous points are doing when we can only look down our light cone to the past. But I would think the we would eventually witness what is now simultaneous. Or are you saying that the distance to the Hubble sphere will appear to be getting farther away because we are accelerating away from it? That the light emitted there is still within our past light cone, but since we are accelerating, it will take longer to get here, and so it appears further away? That the path length to reach an accelerating object is longer than for a inertial body?

Perhaps if we were to rescale these numbers to distances where the time lag was negligable and assume we could accurately measure the recession velocity at these close distances, then the situation might become more obvious. 75km/sec/Mpc = 2.4311183144246E-12m/sec/m. Assume we can measure such small distances as accurately as we do a mile. On this scale what does it mean that the universe is accelerating in its expansion?

 

[SpaceTiger]

Then what can an accelerating universe means except that distances between points are receding away from each other with ever greater velocity.

Take an object that's expanding away from us at a constant comoving distance. As time goes on, how does its proper distance from us change? If it accelerates away, then we say the universe is accelerating. Note that this is not the same as asking what will happen to a set of hypothetical objects at the same proper distance at a variety of different times. In a pure $\Lambda$ universe, the Hubble constant remains constant and this set of objects all recedes from us at the same rate, but a single object will accelerate away from us if followed through time. This is because its proper distance is increasing with time, and a constant Hubble constant means that its speed should increase as it moves away from us.

I suppose it does complicate things to have to guess what the simultaneous points are doing when we can only look down our light cone to the past.

With the cosmological principle, it's not so difficult, we just assume that on average, the entire universe behaves the same way at a given time in its history.

But I would think the we would eventually witness what is now simultaneous. Or are you saying that the distance to the Hubble sphere will appear to be getting farther away because we are accelerating away from it?

In a pure $\Lambda$ (which would be accelerating) universe the Hubble sphere stays at a constant proper distance. In our current universe it will slowly recede from us until things become completely $\Lambda$-dominated. You shouldn't think of the Hubble sphere as an object, just a scale distance that depends upon the expansion rate.

That the light emitted there is still within our past light cone, but since we are accelerating, it will take longer to get here, and so it appears further away?

We don't directly observe the Hubble sphere, it's just a scale radius.

That the path length to reach an accelerating object is longer than for a inertial body?

It's not acceleration in the Newtonian sense. In a $\Lambda$-dominated universe, distant galaxies will remain on geodesics despite "accelerating" away from us.

Perhaps if we were to rescale these numbers to distances where the time lag was negligable and assume we could accurately measure the recession velocity at these close distances, then the situation might become more obvious.

If the time lag were negligible, then by the cosmological principle, the Hubble constant would be a constant and we would see a linear distance-redshift relationship, regardless of cosmological model.

 

[Mike2]

This is because its proper distance is increasing with time, and a constant Hubble constant means that its speed should increase as it moves away from us.

Well, this much we knew before the supernovae data. What do the researchers of the supernovae data mean by acceleration of expansion? Does it not mean that space at a specified proper distance will receed with greater velocity with time?

 

[SpaceTiger]

Well, this much we knew before the supernovae data. What do the researchers of the supernovae data mean by acceleration of expansion? Does it not mean that space at a specified proper distance will receed with greater velocity with time?

No, it doesn't, read the first paragraph again. That's where I explain what it means.

 

[Mike2]

The Hubble law relating proper distance (on a simultaneity hypersurface) to recession speed is always linear.

We can only get the proper distance by use of a model that we're not quite sure of yet (e.g. what is dark energy)

When I look at the first figure in your reference at:
http://www.astro.ucla.edu/~wright/sne_cosmology.html

I notice how the supernovae line is concave downward. Taking the slope of this curve gives the Hubble parameter for a particular time, right? And I notice how that slope is increasing as we approach the present. I take it this is what is meant by an "accelerating expansion".

But you note how the proper distance recession rate is always the same. I'm not able to understand this yet, and I could use some help. I'm hoping there is a way to get from this to a linear proper distance relation without resorting to a model that we're not sure of yet. For example, is it possible to integrate a curve connecting the dots to get the linear relation? Or is there a non-linear correction factor for say, the interference of dust, that linearized this curve?

If the Hubble diagram were linear, there would be no issue. But since we are dealing with a curved Hubble diagram, this is relatively new (to me), and I don't understand how you get from this to a linear relation with proper distance. Any help you could give would be appreciated.

 

[hellfire]

I don't understand how you get from this to a linear relation with proper distance. Any help you could give would be appreciated.

The linear Hubble law for the proper distance follows from the properties of space-time. Start with the FRW line element with a scale factor $a$ (we do not need the angular coordinates):

$$ds^2 = -c^2 dt^2 + a^2 dr^2$$

Take a hypersurface of constant time $dt = 0$:

$$ds = a dr$$

Note that on a hypersurface of constant time the scale factor is the same everywhere and it is possible to integrate to get the proper distance (which is defined on a hypersurface of constant time), let's call it $D$:

$$d_{proper} = D = a r$$

Now, consider a galaxy located at a proper distance $D$ at a specific time on a specific hypersurface. ¿How does its proper distance vary with time?

$$\frac{dD}{dt} = \frac{da}{dr} r + a \frac{dr}{dt}$$

The second term of the rhs is the peculiar speed which is zero for a comoving galaxy $dr/dt = 0$. It follows then that the proper distance changes according to:

$$\frac{dD}{dt} = \frac{da}{dt} r$$

The definition of the Hubble parameter is:

$$H = \frac{1}{a} \frac{da}{dt}$$

Inserted in the previous equation:

$$\frac{dD}{dt} = H a r$$

Moreover, considering the relation between proper distance and radial coordinate we got above:

$$\frac{dD}{dt} = H D$$

Which is actually the Hubble law:

$$v = H D$$

This is a linear relation on a hypersuface of constant time.

Note that the definition of luminosity distance is a different one.

I hope this helps. Otherwise keep asking and I will do my best.

 

[hellfire]

Ned Wright has a great page explaining this and the relation to other distances here: http://www.astro.ucla.edu/~wright/cosmo_02.htm

 

[Mike2]

The linear Hubble law for the proper distance follows from the properties of space-time. Start with the FRW line element with a scale factor $a$ (we do not need the angular coordinates):

$$ds^2 = -c^2 dt^2 + a^2 dr^2$$

Take a hypersurface of constant time $dt = 0$:

$$ds = a dr$$...

I hope this helps. Otherwise keep asking and I will do my best.

Thanks, hellfire, for your help. It was gracious of you to take the time to do the math, and I appreciate it.

Yes, the Hubble constant is spatially invariant, the same at every point in space at a specific time. That's implied by the cosmological principle, and now you've proven it from the metric.

However, my concern is with how the Hubble constant may vary with time. Ned Wright's website that I link to shows recession velocity plotted against luminousity distance. The slope of this curve would seem to give the Hubble constant at various distances. This slope increases for closer distance. And since closer distance means more recent time, the curve would seem to indicate an increase of the Hubble parameter with more recent time. And with increased Hubble parameter comes an increase in the rate at which the universe is expanding. I thought this is what was meant by accelerating expansion.

But it would seem that luminosity distance needs to be corrected for things like redshift. So now I'm not sure that the graph does indicate an increase of the Hubble parameter with time.

And why do I keep asking?... I simply took the word of those who said that the universe was accelerating in its expansion. I thought this meant that the recession velocity for a fixed physical distance was increasing with time. This would mean that the distance out to which the recession velocity was the speed of light was getting closer. I had hoped that this might mean that there would be an entropy associated with this horizon that would then be seen as shrinking. And if horizon entropy constrained the entropy inside it, then a shrinking horizon might be a force of creation inside. But now I'm hearing that the GR predicts never an increase in the Hubble parameter, only decreases to a final value, and that even the graph against luminosity distance when corrected for redshift, etc, does not indicate a change in the Hubble constant, that the acceleration in expansion is only appearent, not actual.

 

[George Jones]

As an example of what hellfire and Space Tiger have said, consider a toy universe and galaxies A, B, C, D at three different instants of cosmological times, t = 1, t = 2, and t = 3.

At times t = 1, t = 2, and t = 3, the proper distances to galaxies A, B, C, D are given by the table:

$$\begin{matrix}<br /> & | & A & B & C & D \\<br /> -- & | & - & - & - & - \\<br /> t = 1 & | & 1 & 2 & 3 & 4 \\<br /> t = 2 & | & 4 & 8 & 12 & 16 \\<br /> t = 3 & | & 9 & 18 & 27 & 36<br /> \end{matrix}$$

At times t = 1, t = 2, and t = 3, the recessional speed of galaxies A, B, C, D are given by the table:

$$\begin{matrix}<br /> & | & A & B & C & D \\<br /> -- & | & - & - & - & - \\<br /> t = 1 & | & 2 & 4 & 6 & 8 \\<br /> t = 2 & | & 4 & 8 & 12 & 16 \\<br /> t = 3 & | & 6 & 12 & 18 & 24<br /> \end{matrix}$$

What are the values of the Hubble constant $H$ at the three times? Since $v = H d$, the Hubble constant is given by $H = v/d$. This give that $H$ equals 2, 1, and 2/3 at times 1, 2, and 3.

Note: 1) at each instant in time, the Hubble constant is constant, i.e., independent of the galaxy used to calculate it; 2) the Hubble constant decreases with time.

What about acceleration or deceleration of the expansion of this universe? During the time interval from $t = 1$ to $t = 2,$ Galaxy A "moves" a distance $\Delta d = 4 - 1 = 3$. During the later but equal-length interval from $t = 2$ to $t = 3,$ the same galaxy, Galaxy A, "moves" a greater distance, $\Delta d = 9 - 4 = 5$. This is an indication that the expansion of the universe is accelerating. The fact that this universe is accelerating is independent of which galaxy is used.

This toy model is a Freidman-Robertson-Walker universe that has its scale factor given by $a(t) = t^2.$

 

[Mike2]

What about acceleration or deceleration of the expansion of this universe? During the time interval from $t = 1$ to $t = 2,$ Galaxy A "moves" a distance $\Delta D = 4 - 1 = 3$. During an equal intreval of time, but now from $t = 2$ to $t = 3,$ Galaxy A "moves" a distance $\Delta D = 9 - 4 = 5$. Thiis is an indication that the universe is accelerating. The fact that this universe is accelerating is independent of which galaxy is used.

Yes, of course. There is no argument there. Even if the Hubble constant were to decrease with time, the simple fact that space is expanding with time means that galaxies pick up speed with time which is an acceleration. This much was know when the Hubble law was first discovered. So am I now to believe that the supernova data just now discovered this effect? I don't think so. I'm sure they were pointing to an apparent increase in the Hubble constant when they say the word "acceleration". For they say things like, "the expansion rate of the universe is increasing", etc. The problem I have is that now I'm not so sure that the "expansion rate of the universe is accelerating" after corrections for redshift are taken into account. I would think that such experts would make the distinction clear as to an appearent acceleration and an actual acceleration. Or did they?

 

[George Jones]

Yes, of course. There is no argument there. Even if the Hubble constant were to decrease with time, the simple fact that space is expanding with time means that galaxies pick up speed with time which is an acceleration.

Not necessarily.

This much was know when the Hubble law was first discovered.

No, before the (1998) supernova data, it was widely believed that the universe was decelerating (near t = now). This is why people found the supernova data to be so surprising.

So am I now to believe that the supernova data just now discovered this effect? I don't think so. I'm sure they were pointing to an apparent increase in the Hubble constant when they say the word "acceleration".

No. I think people participating in this thread have tried repeatedly to tell you that this is not the case.

As an example of a decelerating (but still expanding) universe that has a Hubble constant that decreases with time, Consider a toy universe and galaxies A, B, C, D at three differents instants of cosmological time.

Table of proper distances $D$ from Milky Way:

$$\begin{matrix}<br /> & | & A & B & C & D \\<br /> -- & | & - & - & - & - \\<br /> t = 1 & | & 1 & 2 & 3 & 4 \\<br /> t = 2 & | & 1.4 & 2.8 & 4.2 & 5.7 \\<br /> t = 3 & | & 1.7 & 3.5 & 5.2 & 6.9<br /> \end{matrix}$$

Table of recessional speeds $v$ from Milky way:

$$\begin{matrix}<br /> & | & A & B & C & D \\<br /> -- & | & - & - & - & - \\<br /> t = 1 & | & 0.5 & 1 & 1.5 & 2 \\<br /> t = 2 & | & 0.35 & 0.71 & 1.1 & 1.4 \\<br /> t = 3 & | & 0.29 & 0.58 & 0.87 & 1.2<br /> \end{matrix}$$

The Hubble constant equals 1/2, 1/4, and 1/6 at times 1, 2, and 3.

Also, as can be seen, the universe is decelerating.

This toy model is a Freidman-Robertson-Walker universe that has its scale factor given by $a(t) = t^{1/2}$, and, qualitatively, echos the way we thought the universe behaved (near t = now) before we had the 1998 supernova data.

 

[hellfire]

So now I'm not sure that the graph does indicate an increase of the Hubble parameter with time.

Clearly it is not.

If the Hubble parameter would be currently increasing ($\dot H_0 > 0$) then in our flat universe:

$$q_0 < - 1$$

And therefore the curve relating luminosity distance and redshift:

$$d_L = \frac{1}{H_0} \left(z + \frac{1}{2} (1 - q_0) z^2 + ...\right)$$

would be below the blue one here, which corresponds to the de-Sitter model in which:

$$q_0 = - 1$$

In encourage you to check that an increasing Hubble parameter implies $q_0 < - 1$ in a flat space. It is not a difficult task and I already pointed out how to proceed.

 

[Mike2]

So after the luminosity distance is corrected for redshift, is there not even an appearent increase in the Hubble constant.