Einstein's Cat
Galaxies that are greater than a distance of c/H metres from Earth have recessional velocities exceeding the speed of light and begin to fade. Thus, theoretical astronomers 3 trillion years in the future will see only the Milky Way in the night sky. What is the reason for this fading of galaxies?

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You just said the reason: they retreat faster than the speed of light.

Einstein's Cat
You just said the reason: they retreat faster than the speed of light.
However why would that fact mean a photon emitted from that galaxy would never reach Earth?

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However why would that fact mean a photon emitted from that galaxy would never reach Earth?

As the photon moves towards Earth at ##c##, the distance between Earth and the photon is increasing at greater than ##c##, so (as long as space keeps expanding at its current rate or faster) the photon will simply get further and further away.

If the expansion of space was suddenly or gradually to stop, then that would be a different matter!

Einstein's Cat
andrew s 1905
Two points. As the universe expands all galaxies that move away from us tend to fade just because of the inverse square rule for intensity (all else being constant). As for the expansion itself have a look at this http://arxiv.org/abs/astro-ph/0310808 and you will see we can still and do observe galaxies who's recession velocity is greater than c.

Regards Andrew

mfb and Einstein's Cat
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A more accessible description in discussion here:
http://physics.stackexchange.com/qu...ceding-faster-than-light-visible-to-observers

... so amend the above observation: the galaxies are receeding from faster than the speed of light and not all the photons are able to reach a slower expansion part of the Universe. But you get the idea that farther away a galaxy the harder it is for it's light to reach us... at some point it won't be able to. The recession speed being equal to the speed of light is just not the cutoff point.
https://en.wikipedia.org/wiki/Event_horizon#Cosmic_event_horizon

andrew s 1905
I would emphasis that it is a cut-off, as Simon points out, rather than a fading. Give or take interstellar and inter galactic absorption all photons from a galaxy (treated as a point source) emitted towards us at the same time either all make it or they all don't.

Regards Andrew

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Galaxies that are greater than a distance of c/H metres from Earth have recessional velocities exceeding the speed of light and begin to fade. Thus, theoretical astronomers 3 trillion years in the future will see only the Milky Way in the night sky. What is the reason for this fading of galaxies?

There is a fading (and not just at and beyond the Hubble radius). The expansion of the universe has, in two ways, diminished the energy flux that we receive beyond just a distance effect. The energy of light is inversely proportional to its wavelength (energy of a photon is ##E=hc/\lambda)##. As the light travels to us, the expansion of the universe expands the wavelength of the light by a factor of 1+z, where z is redshift. Also, the expansion of the universe decreases the rate at which we receive photons, as compared to the rate at which photons left a source, by another factor of 1+z (gravitational time dilation). Consequently, as redshift goes continuously towards infinity (cosmological horizon, not Hubble radius), intensity continuously decreases towards zero.

andrew s 1905
andrew s 1905
Also, the expansion of the universe decreases the rate at which we receive photons, as compared to the rate at which photons left a source, by another factor of 1+z (gravitational time dilation).

Can you point me at an explanation of this effect as I would like to understand it. I have come across gravitational time dilation in association with mass but not the expansion of the universe before.

Also do you agree it fades towards a cut-off or are you saying a cut-off does not exist ?The paper I referenced implies there is one.

Thanks Andrew

George, I have managed to track down some papers on cosmological time dilation that fits your formula and some Sn Ia results that seem to confirm it. (e.g. http://www.ppd.stfc.ac.uk/ppd/resources/pdf/ppd_seminar_100609_talk_1.pdf [Broken] and my original link!!) I assume this is what you intended. Thanks

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Einstein's Cat
There is a fading (and not just at and beyond the Hubble radius). The expansion of the universe has, in two ways, diminished the energy flux that we receive beyond just a distance effect. The energy of light is inversely proportional to its wavelength (energy of a photon is ##E=hc/\lambda)##. As the light travels to us, the expansion of the universe expands the wavelength of the light by a factor of 1+z, where z is redshift. Also, the expansion of the universe decreases the rate at which we receive photons, as compared to the rate at which photons left a source, by another factor of 1+z (gravitational time dilation). Consequently, as redshift goes continuously towards infinity (cosmological horizon, not Hubble radius), intensity continuously decreases towards zero.
What exactly is the cosmological horizon? And also could one just apply the typical equation for the Doppler effect to a ray of light emitted from a retreating galaxy to calculate the redshift as well as the calculation you suggested?

Homework Helper
What exactly is the cosmological horizon? And also could one just apply the typical equation for the Doppler effect to a ray of light emitted from a retreating galaxy to calculate the redshift as well as the calculation you suggested?
Not good enough? You can also google "cosmological horizon" and get a range of articles explaining it at a variety of different levels ... you can, then, pick the one most suited to your understanding. Basically, it is the radius at which the galaxies disappear due to cosmological expansion.
The same articles will likely explain why we don't just interpret the redshift observed as a Doppler effect ... one of the side effects of doing this, for instance, would be that distant galaxies do not retreat faster than light: creating some um geometry problems.

quarkstar
galaxy fading sounds like the discrepancy between intensity-based distance and redshift-based distance in dark energy (as an explaination). is there anybody with hard data on intentsities and redshifts or are the galaxy's general output too fuzzy for redshift lines? There must be a general output from a galaxy that has useable data lines, or Hubble et al (Silpher) couldn't use it for the original work on the expanding universe.

quarkstar
the question being does the fading data coincide with the intensity problems in dimness of Sn1a information?

quarkstar
even further would there be an inflection point that coincides with the Sn1a data inflection point in the divergence of intensity vs redshift distance conundrum? (aka dark energy) does it agree with estimates- perlmutter reiss - z~.7 ? or is the fading of galaxies being noticed observationally but without a good analytic technique to categorize it's qualities?

andrew s 1905
galaxy fading sounds like the discrepancy between intensity-based distance and redshift-based distance in dark energy (as an explaination). is there anybody with hard data on intentsities and redshifts or are the galaxy's general output too fuzzy for redshift lines? There must be a general output from a galaxy that has useable data lines, or Hubble et al (Silpher) couldn't use it for the original work on the expanding universe.

There is real data discussed in both the links I made above. Regards Andrew

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Can you point me at an explanation of this effect as I would like to understand it. I have come across gravitational time dilation in association with mass but not the expansion of the universe before.

It is an effect due to redshift, independent of the cause of the redshift, i.e., it is present for cosmological redshift, for redshift caused by a massive object, and even for redshift caused by relative motion between source and receiver in special relativity.

Imagine that observers A and B have identical watches. A sends a light signal to B that B sees redshifted, so that each photon B receives has lower energy (by the redshift factor) than each photon that A sends out.This also means that there is an observed frequency shift for all frequencies, including the rotational frequencies of the second hands of the watches for A and B. B uses one eye to watch the A's second hand and one eye to watch his own second hand. B observes the rotational period of A's second hand to be larger (again, by the redshift factor) than the period (1 minute) of his own second hand. Suppose that A's experimental set up sends out one photon for per revolution of his second hand, i.e., at the rate of 1 photon per minute according to A. B sees the A's rotational period to greater than 1 minute, so B receives photons at a rate of less (by a redshift factor) than one a minute. Putting stuff together, the energy flux received by B is reduced by two factors of redshift compared to the energy flux sent out by A.

Also do you agree it fades towards a cut-off or are you saying a cut-off does not exist ?The paper I referenced implies there is one.

Yes, this illustrated well by panel 3 in figure 1 of the paper. I hope to get back to this, and to some other points.

As the photon moves towards Earth at ##c##, the distance between Earth and the photon is increasing at greater than ##c##, so (as long as space keeps expanding at its current rate or faster) the photon will simply get further and further away.

If the expansion of space was suddenly or gradually to stop, then that would be a different matter!
This is incorrect. Photons emitted from galaxies with superluminal recession velocities will indeed reach earth. See the section "Superluminal recession and the Hubble sphere" here https://www.physicsforums.com/insights/inflationary-misconceptions-basics-cosmological-horizons/. The OP might find the full article of interest.

Einstein's Cat
Einstein's Cat
See George Jones' response #16 above regarding redshift.

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As the photon moves towards Earth at ##c##, the distance between Earth and the photon is increasing at greater than ##c##, so (as long as space keeps expanding at its current rate or faster) the photon will simply get further and further away.

If the expansion of space was suddenly or gradually to stop, then that would be a different matter!

This is incorrect. Photons emitted from galaxies with superluminal recession velocities will indeed reach earth. See the section "Superluminal recession and the Hubble sphere" here https://www.physicsforums.com/insights/inflationary-misconceptions-basics-cosmological-horizons/. The OP might find the full article of interest.

Having read your excellent Insight, I can now see that I was ... correct!

andrew s 1905
Having read your excellent Insight, I can now see that I was ... correct!
As the photon moves towards Earth at ##c##, the distance between Earth and the photon is increasing at greater than ##c##, so (as long as space keeps expanding at its current rate or faster) the photon will simply get further and further away.
/QUOTE]

Sorry this is just not correct.

Regards Andrew

Having read your excellent Insight, I can now see that I was ... correct!
Care to elaborate?

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Care to elaborate?

"Consider a galaxy located beyond the Hubble radius at a distance dH that emits a photon towards Earth. Of course, locally this photon is traveling at c in accordance with special relativity. But, on account of the expansion, the photon is initially moving away from Earth with a speed vtot=vrec−c>0 (where positive velocities point away from Earth, in the direction of expansion.) ..."

You then go on to explain that if the expansion is decelerating, then the recession velocity will decrease and the photon will start moving towards the Earth.

Someone then asked the following question:

"So once the rate of expansion of a distant galaxy exceeds c, it will never slow to a recession velocity of less than c. So I am struggling to see how light emitted from a galaxy that is receding from us >c can ever reach us?"

To which you replied:

"When the universe is accelerating, there is an event horizon. In this case, there are indeed events (like the emission of a photon from a distant galaxy) that will never be observable by us. The misconception that snares many people is that this is also true during even decelerated expansion as long as the galaxy is receding at superluminal speeds."

I take it from this that if the rate of expansion continues to accelerate, then the photon will always be in a region of space that is receding superluminally (hence will never reach us); but if the expansion decelerates, then it may reach us, as above, subject to a more complicated calculation.

Given that I separated those two cases in my post (as underlined in post #20), I don't see my error.

As long as what you mean by "as long as space keeps expanding at its current rate or faster" is that $\ddot{a}<0$, then, yes, we are in agreement.

andrew s 1905
Sorry I just don't see this In the link I posted it says:

"We have seen that the speed of photons propagating towards us (the slope of our past light cone in the upper panel of Fig. 1) is not constant, but is rather vrec −c. Therefore light that is beyond the Hubble sphere has a total velocity away from us. How is it then that we can ever see this light? Although the photons are in the superluminal region and therefore recede from us (in proper distance), the Hubble sphere also recedes. In decelerating universes H decreases as ˙ a decreases (causing the Hubble sphere to recede). In accelerating universes H also tends to decrease since ˙ a increases more slowly than a. As long as the Hubble sphere recedes faster than the photons immediately outside it, ˙ DH > vrec −c, the photons end up in a subluminal region and approach us. Thus photons near the Hubble sphere that are receding slowly are overtaken by the more rapidly receding Hubble sphere."

At current conditions i.e. space expanding at it current rate and acceleration we can see galaxies beyond the Hubble sphere.

What am I missing here.

Regards Andrew

Einstein's Cat
farther away a galaxy the harder it is for it's light to reach us... at some point it won't be able to.
Surely however, despite the continuous diminishing of light, light will reach us, so is the fading a product of the inability for human technology to detect it or a physical phenomenon?

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After the cosmological horizon, there is no known mechanism for light to reach us.
The "fading", as explained above, is the real fact that fewer photons reach us ... no amount of detector sensitivity can change that.

Einstein's Cat
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As long as what you mean by "as long as space keeps expanding at its current rate or faster" is that $\ddot{a}<0$, then, yes, we are in agreement.

Not quite. I was thinking of the case where ##H## is constant. Near the beginning of your Insight you say:

"... for a given rate of expansion, ##H## ..."

But, as the article progresses it seems to focus on ##q## as the rate of change of expansion: "... for the case accelerated expansion, ##q<0##...".

So, when cosmologists say that the expansion is accelerating, do you mean ##q < 0## rather than ##H' > 0##?

Also, by my calculations ##q < 0## is equivalent to ##\ddot{a} > 0##. Is that correct?

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At current conditions i.e. space expanding at it current rate and acceleration we can see galaxies beyond the Hubble sphere.

Consider a photon seen now by us that was emitted by a galaxy at the event on our past lightcone withe approximate coordinates ##t=2## and ##D=5##. From the top panel of Figure 3, this was below (earlier) than the maximum bulge (proper distance) of our past lightcone.

According to equation (20) of the paper (and using the terminology and notation of the paper), the velocity of the photon is

$$v_\mathrm{tot} = v_\mathrm{rec} + v_\mathrm{pec} .$$

Here, ##v_\mathrm{rec}## is the recessional velocity of the galaxy at the emission event, and ##v_\mathrm{pec} = -c## (negative, since the photon was emitted in our direction).

At the emission event, our past light cone is "getting bigger", so ##v_\mathrm{tot} > 0##, which means ##v_\mathrm{rec}>c##. As the photon proceeds up our past lightcone, it passes galaxies with differing ##v_\mathrm{rec}##, but the peculiar velocity of the photon remains constant at ##v_\mathrm{pec} = -c##. In order for the photon to start moving towards us, it must have ##v_\mathrm{tot} < 0##, i.e., ##v_\mathrm{tot}## must decrease, so, at some times in the past (but not necessarily now) the photon must have had ##\dot{v}_\mathrm{tot} < 0##. From the constancy of ##v_\mathrm{pec}##, this means that ##\dot{v}_\mathrm{rec} < 0##, and equation (19) gives ##\ddot{R}<0##.

The scale factor ##R## also is often denoted (possibly with different normalization) by ##a##, and, I think, this is what bapowell means by

As long as what you mean by "as long as space keeps expanding at its current rate or faster" is that $\ddot{a}<0$, then, yes, we are in agreement.

For a universe that consists entirely of matter (including dark matter) and dark energy, an exact solution

$$R\left(t\right) = A \sinh^{\frac{2}{3}} \left(Bt\right),$$

where ##A## and ##B## are constants. Except at very early times when the early universe is radiation dominated, this corresponds to our universe. This solution has ##\ddot{R}<0## at early time, and ##\ddot{R}>0## (accelerating expansion) at later times.

So, when cosmologists say that the expansion is accelerating, do you mean ##q < 0## rather than ##H' > 0##?
They mean $\ddot{a}>0$, which is equivalent to $q<0$. It turns out that constant $H$ gives $\ddot{a}>0$. Can you see why?

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They mean $\ddot{a}>0$, which is equivalent to $q<0$. It turns out that constant $H$ gives $\ddot{a}>0$. Can you see why?

Yes, but they are not equivalent. Constant ##H## is ##q = -1## whereas ##q < 0## is equivalent to ##H' > - H^2##. For constant ##q \ne -1## I get:

##H(t) = [(q+1)t + 1/H_0]^{-1}## which goes to ##0## as ##t \rightarrow \infty## (for ##q > -1##)

And:

##d_{H, com} = \frac{C}{H_0}[H_0(q+1)t + 1]^{\frac{q}{q+1}}##

Which goes to ##0## as ##t \rightarrow \infty## for ##-1 < q < 0##.

So, that would be a solution where eventually there is no expansion in terms of ##H## but the comoving Hubble radius has eventually reduced to 0, hence everything not gravitationally bound is gone!

If I've understood your article, then that's the point of ##q##. It determines the long-term behaviour of the comoving Hubble radius, hence whether eventually everything is gone (##q < 0##), or eventually everything comes back into the observable universe (##q > 0##).

Flatland
I'd like to expand on this question a little bit.When another galaxy that is currently receding away from us at greater than c, would there be a point in space where the photons would reach where this other galaxy is receding exactly at c? And if yes would it be possible to be at rest relative to these photons?

I'd like to expand on this question a little bit. What happens when another galaxy that is currently receding away from us at greater than c; would there be a point in space where the photons would reach where this other galaxy is receding exactly at c?
Yes. Since photons from this galaxy eventually reach earth, they must pass a point that is receding from the galaxy at speed c.

Einstein's Cat
After the cosmological horizon, there is no known mechanism for light to reach us.
The "fading", as explained above, is the real fact that fewer photons reach us ... no amount of detector sensitivity can change that.
How long would it take a typical galaxy to fade away so that it is no longer visible where when t is zero the recessional velocity of the galaxy is c from the perspective of an observer on earth? The means of viewing of the observer is irrelevant.

Is there any data on such a subject? I've looked online and created a thread but to no success.

sunrah
How long would it take a typical galaxy to fade away so that it is no longer visible where when t is zero the recessional velocity of the galaxy is c from the perspective of an observer on earth? The means of viewing of the observer is irrelevant.

Is there any data on such a subject? I've looked online and created a thread but to no success.

Any answer to the question you ask depends on your cosmological model. Our universe is accelerating, this means that the Hubble parameter is not constant. That is essentially why we can still see objects outside the Hubble radius
$d_{H} = \frac{c}{H_{0}}$
where $H_{0}$ is the value of the Hubble parameter at the present time. If the Hubble parameter were constant, the Hubble radius (~14 billion light years) would correspond to the size of the visible universe and the time required for light to reach us from this boundary would be the age of the universe (~14 billion years). As it is the size of the visible universe is much greater ~47 billion light years. The reason we see these regions is because how the value of the scale factor, which describes the expansion of our universe, has changed until the present day. If our universe were to stop expanding suddenly, light from the particle horizon region would take about 47 billion years to reach us, but of course galaxies located inside the particle horizon would not "fade", because the universe had ceased expanding.

What you call fading is galaxies leaving the particle horizon, that is crossing the boundary of the visible universe, due to cosmic (accelerated) expansion. The time t required for a galaxy to leave the particle horizon from time t0 where v(t0)=c, depends on the evolution of the scale factor (and therefore Hubble parameter) in the future. This evolution is described by cosmological models like the concordance model. Also, I think standard version of Hubble law needs to be modified as this only applies to low redshift values, i.e. z<<1.

What you really need to understand is that Hubble radius is not a physical boundary - it doesn't really mean much in our particular universe. The particle horizon is the important one. And we can see high redshift galaxies due to the evolution history of our universe. As the universe expands galaxies will leave the particle horizon, eventually all that we will be able to see is our local group of galaxies.

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Einstein's Cat