Observing galaxies that recess faster than c

[virgil1612]

Hello,

I've read the article at

http://pages.erau.edu/~reynodb2/LineweaverDavis_BigBang_SciAm_March05p36.pdf

that was recommended several times on this forum. At the question "Can we see galaxies receding faster than light?" they answer:
"Sure we can, because the expansion rate changes over time. The photon initially is unable to approach us. But the Hubble distance is not constant; it is increasing and can grow to encompass the photon. Once that happens, the photon approaches us and eventually reaches us"

I know that we can see these galaxies. I just don't understand their explanation. Could someone help me understand?

Thanks, Virgil.

 

[rootone]

Due to expansion they may be moving away from you 'here and now' at a rate exceeding c.
However the light you see 'now' would have been emitted billions of years ago when they we not receding as fast, so those photons DO reach you eventually

 

[marcus]

Their explanation is pretty good.
The galaxy is not moving in the usual sense. It is not approaching any destination, should not be thought of as going through space. The distance between stationary objects can increase (that's what curved geometry is about, distances can change, don't have to obey ordinary Euclidean expectations.

Most of the galaxies we can see with telescopes WERE ALREADY RECEDING FASTER THAN LIGHT WHEN THEY EMITTED THE LIGHT WHICH WE ARE NOW GETTING FROM THEM.

That is true for any galaxy with redshift greater than 1.6 and most visible galaxies have redshifts z > 1.6

 

[virgil1612]

Their explanation is pretty good.
The galaxy is not moving in the usual sense. It is not approaching any destination, should not be thought of as going through space. The distance between stationary objects can increase (that's what curved geometry is about, distances can change, don't have to obey ordinary Euclidean expectations.

Most of the galaxies we can see with telescopes WERE ALREADY RECEDING FASTER THAN LIGHT WHEN THEY EMITTED THE LIGHT WHICH WE ARE NOW GETTING FROM THEM.

That is true for any galaxy with redshift greater than 1.6 and most visible galaxies have redshifts z > 1.6

When they emitted the light that we see NOW, shouldn't they have receded slower than c, so that THAT light eventually reaches us?
When talking about the moment the galaxies emit the light that we see now, are we doing some kind of velocity addition between velocity of light and the recessing velocity at that point?

 

[virgil1612]

Due to expansion they may be moving away from you 'here and now' at a rate exceeding c.
However the light you see 'now' would have been emitted billions of years ago when they we not receding as fast, so those photons DO reach you eventually

This makes sense. Thanks.

 

[marcus]

What Rootone said actually does not make sense to me because it makes it seem that when the galaxy emitted the light we are getting the distance to it was actually increasing SLOWER than c. But the whole point is that for most galaxies we can see, we are getting light that was emitted by them when the distance to the source was increasing FASTER than c.

 

[virgil1612]

What Rootone said actually does not make sense to me because it makes it seem that when the galaxy emitted the light we are getting the distance to it was actually increasing SLOWER than c. But the whole point is that for most galaxies we can see, we are getting light that was emitted by them when the distance to the source was increasing FASTER than c.

I just begin reading cosmology (the descriptive way), but for me it's just the opposite. For a photon emitted by a distant galaxy to reach us, that galaxy, at the moment when it emitted the photon, should have moved, with respect to us, slower than light.

That's just the way I understand it and I could be wrong, but I'm sure that I've read things along these lines, I could try to find those references if it would be helpful.

 

[marcus]

When they emitted the light that we see NOW, shouldn't they have receded slower than c, so that THAT light eventually reaches us?...

No, and it is very simple. There is no tricky "velocity addition" involved. Take a simple example of a galaxy with redshift z = 1.6 which emitted a photon we are now receiving. The distance to the galaxy and the space around the galaxy was increasing at c at the moment of emission.
So the photon made no progress. It was heading towards us but the distance between us did not decrease. It stayed essentially the same distance from us for a long time.

That distance, that is increasing exactly as fast as light travels is called the HUBBLE DISTANCE.

If a photon is trying to get to us and it is that distance from us, it makes no progress.
If it is WITHIN that distance it makes progress
If it is farther than that it gets swept back, loses ground, distance it has to go is getting bigger.

Lineweaver and Davis point out that the Hubble distance, or Hubble radius, INCREASES so it gradually reaches out and TAKES STRUGGLING PHOTONS IN.

If, today, a photon is emitted towards us at the Hubble radius (which is currently 14.4 billion LY) it will at first make zero progress, it will stay at that distance. But the Hubble radius is constantly increasing (because as a percentage rate, the expansion rate is decreasing) so after a while the Hubble radius will be 14.5 billion LY. thank goodness! says the photon, I am now within the Hubble radius! I am making progress towards my goal.

 

[virgil1612]

No, and it is very simple. There is no tricky "velocity addition" involved. Take a simple example of a galaxy with redshift z = 1.6 which emitted a photon we are now receiving. The distance to the galaxy and the space around the galaxy was increasing at c at the moment of emission.
So the photon made no progress. It was heading towards us but the distance between us did not decrease. It stayed essentially the same distance from us for a long time.

That distance, that is increasing exactly as fast as light travels is called the HUBBLE DISTANCE.

If a photon is trying to get to us and it is that distance from us, it makes no progress.
If it is WITHIN that distance it makes progress
If it is farther than that it gets swept back, loses ground, distance it has to go is getting bigger.

Lineweaver and Davis point out that the Hubble distance, or Hubble radius, INCREASES so it gradually reaches out and TAKES STRUGGLING PHOTONS IN.

If, today, a photon is emitted towards us at the Hubble radius (which is currently 14.4 billion LY) it will at first make zero progress, it will stay at that distance. But the Hubble radius is constantly increasing (because as a percentage rate, the expansion rate is decreasing) so after a while the Hubble radius will be 14.5 billion LY. thank goodness! says the photon, I am now within the Hubble radius! I am making progress towards my goal.

Extremely interesting! Please give me some time to think about it, and I'll post back tomorrow.
Virgil.

 

[Chronos]

Maybe this will help http://arxiv.org/abs/astro-ph/0011070, Superluminal Recession Velocities.

 

[Drakkith]

But the Hubble radius is constantly increasing (because as a percentage rate, the expansion rate is decreasing)

Can you elaborate on this, Marcus? Is there a difference between the rate of expansion (which is increasing) and the rate of expansion as a percentage (which is decreasing)?

 

[marcus]

There's a problem in the English language around the word "rate". People confuse it with "speed".

If a distance grows at a constant rate (say 1% per year) then it lengthens at an increasing speed. Of course! It is growing exponentially at 1% per year.
Exponential growth at a constant rate has this kind of "acceleration" because the principal is growing.

A distance can even grow at a gradually DECLINING rate and if the decline is slow enough it will still lengthen at an increasing speed. It can exhibit NEAR exponential growth, even if the rate slowly declines from 1% to .99% to .98% leveling out say at .97%.
It can "accelerate" even though the growth rate is declining.

In the standard cosmic model, based on the Friedmann equation, the distance growth rate has declined since very early times, and it is expected to continue declining.
What you hear a lot about is that people discovered in 1998 that the decline had slowed enough so that if you focus on a particular distance between two stationary objects that distance would grow at increasing speed. The transition from deceleration to acceleration we think happened around year 8 billion.

The distance growth RATE continues to decline, but gradually enough that this "near exponential growth" behavior has begun to be detectable.

If you calculate what the Hubble growth rate actually is (cut through the confusing units) it comes out to a percentage rate of 1/144 percent per million years and its decline seems destined to continue more and more gradually so that it levels off at a rate of 1/173 percent per million years.

this corresponds to a Hubble radius now of 14.4 billion LY and an eventual Hubble radius of 17.3 billion LY. You can see the past and future behavior of the Hubble radius (and thus its reciprocal, the Hubble rate) by clicking on Lightcone---a table making calculator that embodies the standard cosmic model:
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html
the R column gives the Hubble radius (aka the Hubble distance) in billions of LY.
It shows how it is currently 14.4 and expected to converge to 17.3 in future (as the Hubble rate declines)

 

[Drakkith]

I'm having trouble converting what you've said into something I know about. I've rarely dealt with expansions and scale factors. Let me try to understand this using my new calculus 1 knowledge.

If a distance grows at a constant rate (say 1% per year) then it lengthens at an increasing speed. Of course! It is growing exponentially at 1% per year.
Exponential growth at a constant rate has this kind of "acceleration" because the principal is growing.

If I graph the distance to a galaxy undergoing this kind of motion over time, I get a function which is increasing (because it is moving away from us). Exponential growth is an exponential function, like F(t) = A^t, right?
The 1st derivative is the velocity, which is also increasing.
The 2nd derivative is the acceleration. In this type of motion is the acceleration increasing? I want to say yes but I can't work out the mathematical details at the moment. I just know that if the velocity is increasing, the acceleration is positive.

 

[marcus]

I'm having trouble converting what you've said into something I know about. I've rarely dealt with expansions and scale factors. Let me try to understand this using my new calculus 1 knowledge.
If I graph the distance to a galaxy undergoing this kind of motion over time, I get a function which is increasing (because it is moving away from us). Exponential growth is an exponential function, like F(t) = A^t, right?
The 1st derivative is the velocity, which is also increasing.
The 2nd derivative is the acceleration. In this type of motion is the acceleration increasing? I want to say yes but I can't work out the mathematical details at the moment. I just know that if the velocity is increasing, the acceleration is positive.

That's right! the derivative of e^x is e^x
so the derivative of THAT is e^x
all the derivatives are the same and they are all increasing.

If you look at a different timescale, or put some constant k in front then you have e^kx and the derivatives involve powers of that constant term but that doesn't change the qualitative behavior . The derivatives all look like e^x with different constant terms in front and they are all increasing.

That is growth at a constant rate . But you can have nearly the same shaped curves with growth at a
gradually declining rate if the decline is gradual enough. Think of it as e^kx with k nearly constant but very slowly dwindling down to some k_∞

 

[Drakkith]

Thanks.

A distance can even grow at a gradually DECLINING rate and if the decline is slow enough it will still lengthen at an increasing speed. It can exhibit NEAR exponential growth, even if the rate slowly declines from 1% to .99% to .98% leveling out say at .97%.
It can "accelerate" even though the growth rate is declining.

Is 'declining rate' talking about acceleration while 'lengthen at an increasing speed' talking about velocity?
That would mean velocity is increasing, and that the acceleration is positive, but that it is decreasing, right?

 

[Chalnoth]

Extremely interesting! Please give me some time to think about it, and I'll post back tomorrow.
Virgil.

Another way to look at it is through the rate of expansion rather than distance. In the very early universe, the rate of expansion was much, much higher. But that rate has decreased over time as our universe has become less dense. Early-on, the high expansion rate caused many photons to fail to make progress towards us. But as that expansion rate dropped, a number of those photons were still close enough that they did start making progress.

To put some numbers on this, the photons from the cosmic microwave background that we see today were, at the time they were emitted, only about 46 million light years away. But that was in the very early universe, when the rate of expansion was much, much faster (as in something like 17,000 times faster). That really rapid rate of expansion easily increased the distance between us and those photons. That rapid expansion was such that it made it so that those photons actually had to traverse some 13.8 billion light years before they could actually reach us.

Because the rate of expansion isn't slowing down as much any longer (and the expansion rate will probably eventually become a constant), there are a great many photons that were emitted from further away that can never reach us.

 

[Monsterboy]

The distance growth RATE continues to decline, but gradually enough that this "near exponential growth" behavior has begun to be detectable.

If you calculate what the Hubble growth rate actually is (cut through the confusing units) it comes out to a percentage rate of 1/144 percent per million years and its decline seems destined to continue more and more gradually so that it levels off at a rate of 1/173 percent per million years.

Does this mean that the magnitude of acceleration of expansion of the universe is decreasing?

 

[Chalnoth]

Does this mean that the magnitude of acceleration of expansion of the universe is decreasing?

I'm not sure that's a good way to look at it.

First of all, "acceleration of expansion" is a confusing phrase. The expansion rate itself is going down and approaching a constant value. What an accelerated expansion means is that individual objects in the universe are accelerating away from one another.

I'm not so sure that the rate of acceleration between objects is decreasing, however. I'd have to do some fairly specific math to figure out whether the acceleration between objects is currently increasing or decreasing, however we do know that in the distant past, objects in our universe were decelerating with respect to one another, and quite rapidly. The fact that they're accelerating recently indicates that the rate of acceleration has increased a lot: it's gone from a negative value to a positive one. If the change in the rate of acceleration were simple enough, then that would imply that the current rate of acceleration between objects may still be increasing.

 

[Monsterboy]

What an accelerated expansion means is that individual objects in the universe are accelerating away from one another.

No , I didn't say objects , I meant accelerated expansion of space between the objects without the objects moving through space.

 

[Chalnoth]

My point is that the distances between individual objects increase at an accelerating pace, but the rate of expansion is still dropping and approaching a constant value.

 

[marcus]

==quote==
Take a simple example of a galaxy with redshift z = 1.6 which emitted a photon we are now receiving. The distance to the galaxy and the space around the galaxy was increasing at c at the moment of emission.
So the photon made no progress. It was heading towards us but the distance between us did not decrease. It stayed essentially the same distance from us for a long time.

That distance, that is increasing exactly as fast as light travels is called the HUBBLE DISTANCE.

If a photon is trying to get to us and it is that distance from us, it makes no progress.
If it is WITHIN that distance it makes progress
If it is farther than that it gets swept back, loses ground, distance it has to go is getting bigger.

Lineweaver and Davis point out that the Hubble distance, or Hubble radius, INCREASES so it gradually reaches out and TAKES STRUGGLING PHOTONS IN.

If, today, a photon is emitted towards us at the Hubble radius (which is currently 14.4 billion LY) it will at first make zero progress, it will stay at that distance. But the Hubble radius is constantly increasing (because as a percentage rate, the expansion rate is decreasing) so after a while the Hubble radius will be 14.5 billion LY. thank goodness! says the photon, I am now within the Hubble radius! I am making progress towards my goal.
==endquote==
Extremely interesting! Please give me some time to think about it, and I'll post back tomorrow.
Virgil.

Hi Virgil, glad you found that interesting, and I hope you will contribute some more to the discussion. Here is a simple graph of how that story. The red curve shows the past experience, collectively, of photons we are now receiving. The ones that were emitted recently had an easy straightforward job getting here. The ones emitted a long time ago, like in year 1 billion, at first got farther away because the space they were traveling thru got farther away (canceling their forward progress).

The blue curve is the Hubble distance which grows and in a sense reaches out to photons so that some of them eventually make it even though they may have, at first, gotten farther away (because the distance to the space they were traveling thru was increasing.)

You can see that some of the photons we are receiving today started out in year 1 billion at a distance of 4 billion LY (if you could have paused geometric expansion at that moment to have time to measure, with radar or string or however)
As long as they were outside the Hubble radius they made no progress, got farther way, their motion thru space canceled by the distance growth.
As long as the red curve is above the blue, things get worse for them.
But the blue curve is always rising, the Hubble radius is extending, and at a certain moment, in year 4 billion, the photon finds itself making ZERO progress but at least not being dragged back. It's forward motion thru space is exactly canceled by the space getting farther away.

The curves cross there, and after that, after year 4 billion, the photon makes progress and gradually reduces the distance to us.
It is always traveling thru space at the standard speed c. but now because it is within the Hubble radius the space it has to travel is not not growing so much.

You can see that the turnaround distance is about 6 billion LY, half way between the 4 and the 8. More precisely it is 5.8 billion LY.

So some of the photons that are coming to the HST (hubble space telescope) at this moment started out towards us in year 1 billion at a distance of 4 billion LY and got "dragged back" for the first 3 billion years so that they were actually 5.8 Gly away, and then were rescued by the slowing rate of distance growth (the Hubble distance is ONE OVER THE RATE, the reciprocal of the rate, so as the growth rate declines the Hubble distance increases, as the blue curve shows.)

 

[marcus]

My point is that the distances between individual objects increase at an accelerating pace, but the rate of expansion is still dropping and approaching a constant value.

Good point! Here is a plot of the distance growth rate (the socalled Hubble parameter) over time.
The present is at 0.8 on the x axis.
I don't happen to have a plot of it with the x-axis labeled in billions of years and the present labeled "13.8"
The time scale here is 17.3 billion years = 1.
So 13.8 billion years = 0.8.
But aside from the time scale it gives a good idea of how the expansion rate has changed over time:

The growth rate is expressed as the fractional amount of growth per billion years if expansion would stay constant at that level for the whole billion year time interval.
Readers can see that at x = 0.1 the rate is 0.4 per billion years
The rate is such that if it stayed steady all that time then in a billion years the distance would be 0.4 or 40% longer.
Of course the rate does not stay steady, it declines. So maybe we should imagine a briefer time interval.
That same rate would be 0.0004 per million years. Or 0.04 percent per million years, if it's easier to think about it that way rather than as 0.0004.

Is it clear that at the present time (x = 0.8) the growth rate is about 0.07 per billion years? If you go out to x = 0.8 then the curve is a little more than one square up. One graph square is 0.05, two squares is 0.1. The curve is about 7/10 of two squares up, so it is at 0.07.

0.07 is about 1/14. So at the present rate, if it would remain steady that long, in a billion years a distance would increase by 1/14 of its size.

That is the same as saying the Hubble time is about 14 billion years, or expressing the same thing as a distance, the Hubble distance is 14 billion LY.
That is what is meant by saying the Hubble distance is the RECIPROCAL of the rate. It is a convenient handle on the growth rate which is easier to grasp than these very tiny numbers and which lengthens inversely as the rate declines.

As Chalnoth just said, the rate is dropping and approaching a constant value. You can see that in the plot.
The curve is leveling out around 0.06 or 1/17 per billion years.
By the same token, the Hubble time is INCREASING AND LEVELING OUT AT 17 BILLION YEARS or more precisely at 17.3 billion years. And another way of saying that is the Hubble distance is increasing towards a longterm level of 17.3 billion LY. Hubble time and distance are both convenient inverse handles on the growth rate.

That is what the longterm growth rate level of 0.06 shown by the curve means.

 

[Chalnoth]

By the way, I did some math, and as long as I calculated this correctly, the acceleration of the scale factor is:

{d^2 a \over dt^2} = H^2\left(1 - {3 \over 2}{\rho_m \over \rho_c}\right)

This is assuming no spatial curvature and a cosmological constant. At late times, when this equation is valid, the matter density fraction drops monotonically, which means that the distance between objects accelerates at an increasing pace into the future (in the presence of a cosmological constant).

 

[Jorrie]

By the way, I did some math, and as long as I calculated this correctly, the acceleration of the scale factor is:

{d^2 a \over dt^2} = H^2\left(1 - {3 \over 2}{\rho_m \over \rho_c}\right)

This is assuming no spatial curvature and a cosmological constant. At late times, when this equation is valid, the matter density fraction drops monotonically, which means that the distance between objects accelerates at an increasing pace into the future (in the presence of a cosmological constant).

Is this equivalent to the classical deceleration parameter:

${d^2 a / dt^2} = aH_0^2 (\Omega_\Lambda - \Omega_m/(2a^3))$

as per prof. Peebles' 1992 (textbook)?

The latter is expressed in present values and I presume that your equation uses the values at time t.

 

[tkjtkj]

No, and it is very simple. There is no tricky "velocity addition" involved. Take a simple example of a galaxy with redshift z = 1.6 which emitted a photon we are now receiving. The distance to the galaxy and the space around the galaxy was increasing at c at the moment of emission.
So the photon made no progress. It was heading towards us but the distance between us did not decrease. It stayed essentially the same distance from us for a long time.

That distance, that is increasing exactly as fast as light travels is called the HUBBLE DISTANCE.

If a photon is trying to get to us and it is that distance from us, it makes no progress.
If it is WITHIN that distance it makes progress
If it is farther than that it gets swept back, loses ground, distance it has to go is getting bigger.

Lineweaver and Davis point out that the Hubble distance, or Hubble radius, INCREASES so it gradually reaches out and TAKES STRUGGLING PHOTONS IN.

If, today, a photon is emitted towards us at the Hubble radius (which is currently 14.4 billion LY) it will at first make zero progress, it will stay at that distance. But the Hubble radius is constantly increasing (because as a percentage rate, the expansion rate is decreasing) so after a while the Hubble radius will be 14.5 billion LY. thank goodness! says the photon, I am now within the Hubble radius! I am making progress towards my goal.

You've presented for cosmology an extremely easy-to-imagine model ... But as we all know here, being imaginably logically clear does not mean it is correct.
It all goes back to the very logical (at the time!) voyage of the sun around our Earth ;)

But i do like sitting here finally able to imagine !

 

[Chalnoth]

You've presented for cosmology an extremely easy-to-imagine model ... But as we all know here, being imaginably logically clear does not mean it is correct.

Sure, but this is correct. Well, as correct as a non-mathematical description can be.

 

[Chalnoth]

Is this equivalent to the classical deceleration parameter:

${d^2 a / dt^2} = aH_0^2 (\Omega_\Lambda - \Omega_m/(2a^3))$

as per prof. Peebles' 1992 (textbook)?

The latter is expressed in present values and I presume that your equation uses the values at time t.

It should be virtually identical, which means I made a mistake. But fortunately it looks like the general structure is unchanged.

 

[GeorgeDishman]

For a more qualitative explanation, there is a similarity between the model of space "stretching" and the classic puzzle of the ant on a rubber rope:

http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

 

[Chalnoth]

For a more qualitative explanation, there is a similarity between the model of space "stretching" and the classic puzzle of the ant on a rubber rope:

http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

The only issue with this particular model is that no matter the starting parameters, the ant always reaches the end of the rope eventually.

In our universe, the rate of expansion is a bit different. Instead of a linear increase in the length of the rope per unit time, our universe is approaching a proportional increase in its scale factor. The difference between the linear increase and the proportional increase is astronomical: the proportional increase leads to exponential growth. There is matter out there today that light rays emitted from Earth can never reach because of this proportional expansion.

 

[virgil1612]

The blue curve is the Hubble distance which grows and in a sense reaches out to photons so that some of them eventually make it even though they may have, at first, gotten farther away (because the distance to the space they were traveling thru was increasing.)
View attachment 82596

You can see that the turnaround distance is about 6 billion LY, half way between the 4 and the 8. More precisely it is 5.8 billion LY.

So some of the photons that are coming to the HST (hubble space telescope) at this moment started out towards us in year 1 billion at a distance of 4 billion LY and got "dragged back" for the first 3 billion years so that they were actually 5.8 Gly away, and then were rescued by the slowing rate of distance growth (the Hubble distance is ONE OVER THE RATE, the reciprocal of the rate, so as the growth rate declines the Hubble distance increases, as the blue curve shows.)

Some great points being made here, and that graph is very clear, thanks.

A question: I'm using Hubble's law with v = c, to find the distance where an object should be to recess at the speed of light and I find approx. 14 x 10⁹ l.y. (depending on the value of H). But then I look at the cosmological redshift, z = v/c, and I know we observed quasars with z = 6, would it mean recessing at 6c? But then it doesn't fit with those objects at 14 x 10⁹ l.y. that recess with c. What am I doing wrong here?

 

[marcus]

Some great points being made here, and that graph is very clear, thanks.

A question: I'm using Hubble's law with v = c, to find the distance where an object should be to recess at the speed of light and I find approx. 14 x 10⁹ l.y. (depending on the value of H).

That's right! I usually use 14.4 billion Ly as the value of the Hubble distance, but it does depend on the value for H.
==quote==
But then I look at the cosmological redshift, z = v/c, and I know we observed quasars with z = 6, would it mean recessing at 6c? But then it doesn't fit with those objects at 14 x 10⁹ l.y. that recess with c.
==endquote==
There's no substitute for checking out Jorrie's "Lightcone" calculator. It makes tables of universe history (past and future) where you choose the limits and the number of steps. It gives recession speeds both back then (when light was emitted) and now (when light is received).
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html

Check it out (ask questions here about any columns in the table you don't understand).
you will see that there is no simple exact correspondence between z and any speed v over c. That z=v/c thing is only an approximation and only works for comparatively near thing. Distance expansion is different from motion thru space. cosmo redshift is not doppler due to motion at some given time. cosmo redshift indicates the total amount of stretching that happened while the light was in transit.

since the rate of stretching constantly changes thru out history, the amount of stretch (I use the letter s = z+1 for the enlargement factor) does not depend in any simple way on the speed of recession at one or another time.
Lightcone also uses the symbol S = z+1 for "stretch factor". If some light comes in and its wavelengths are 3 times what they were when it started out then the stretch S = 3
that corresponds to a redshift of z = S-1 = 2.
I find S more intuitive, easier to use, because it corresponds to the actual enlargement factor by which both distances and wavelengths have expanded while the light is in transit.
Using z instead of S is just an historical accident, a quirk custom that astronomers got into in the early days and became traditional.

 

[marcus]

If you go to the LightCone calculator
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html
and want to use it to print a table IN A POST, so we can discuss it here at PF, say.
Then look at the yellow stripe of output options and tick the button that says "Tex script"
and then press "calculate"

Then you will get a Latex version of the output, already highlighted, so you just say "copy" and paste it into your post.
When I say "Tex script" and then "calculate", and then say "Command-C" to copy it copies the code on my scratchpad
then I go to Physicsforums, to this post and I say "Command-V" to paste and this is what I get---just a sample, we could change the columns and the upper and lower limits and the number of rows etc. This is just the default you get if you don't do anything:
{\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}} {\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&D_{hor}(Gly)&V_{now} (c)&V_{then} (c) \\ \hline 0.001&1090.000&0.0004&0.0006&45.332&0.042&0.057&3.15&66.18\\ \hline 0.003&339.773&0.0025&0.0040&44.184&0.130&0.179&3.07&32.87\\ \hline 0.009&105.913&0.0153&0.0235&42.012&0.397&0.552&2.92&16.90\\ \hline 0.030&33.015&0.0902&0.1363&38.052&1.153&1.652&2.64&8.45\\ \hline 0.097&10.291&0.5223&0.7851&30.918&3.004&4.606&2.15&3.83\\ \hline 0.312&3.208&2.9777&4.3736&18.248&5.688&10.827&1.27&1.30\\ \hline 1.000&1.000&13.7872&14.3999&0.000&0.000&16.472&0.00&0.00\\ \hline 3.208&0.312&32.8849&17.1849&11.118&35.666&17.225&0.77&2.08\\ \hline 7.580&0.132&47.7251&17.2911&14.219&107.786&17.291&0.99&6.23\\ \hline 17.911&0.056&62.5981&17.2993&15.536&278.256&17.299&1.08&16.08\\ \hline 42.321&0.024&77.4737&17.2998&16.093&681.061&17.300&1.12&39.37\\ \hline 100.000&0.010&92.3494&17.2999&16.328&1632.838&17.300&1.13&94.38\\ \hline \end{array}}

 

[marcus]

If you want to see what the block of Latex code looks like that makes that table, that you need to copy and paste, just press "reply" at the preceding post with the table. You don't have to be able to read Latex code to use this feature, you only need to copy and paste the block of code. The PF system will automatically make the table for you.
Here is a sample of the code block, without the "tex" tags that make it get interpreted and turn into a table

...\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}...
...\hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&D_{hor}(Gly)&V_{now} (c)&V_{then} (c) \\ \hline 0.001&1090.000&0.0004&0.0006&45.332&0.042&0.057&3.15&66.18\\ \hline 0.003&339.773&0.0025&0.0040&44.184&0.130&0.179&3.07&32.87\\ \hline 0.009&105.913&0.0153&0.0235&42.012&0.397&0.552&2.92&16.90\\ \hline 0.030&33.015&0.0902&0.1363&38.052&1.153&1.652&2.64&8.45\\ \hline 0.097&10.291&0.5223&0.7851&30.918&3.004&4.606&2.15&3.83\\ \hline 0.312&3.208&2.9777&4.3736&18.248&5.688&10.827&1.27&1.30\\ \hline 1.000&1.000&13.7872&14.3999&0.000&0.000&16.472&0.00&0.00\\ \hline 3.208&0.312&32.8849&17.1849&11.118&35.666&17.225&0.77&2.08\\ \hline 7.580&0.132&47.7251&17.2911&14.219&107.786&17.291&0.99&6.23\\ \hline 17.911&0.056&62.5981&17.2993&15.536&278.256&17.299&1.08&16.08\\ \hline 42.321&0.024&77.4737&17.2998&16.093&681.061&17.300&1.12&39.37\\ \hline 100.000&0.010&92.3494&17.2999&16.328&1632.838&17.300&1.13&94.38\\ \hline

One of the many good things about LightCone calculator is that it knows LaTex so it has this copy paste option. If you don't do anything it just shows you the table on your computer screen. But if you tick the "Tex script" button and then say calculate it gives you a block of stuff that will create the same table in a post here.

So you can call attention to something and ask about it.

 

[virgil1612]

If you want to see what the block of Latex code looks like that makes that table, that you need to copy and paste, just press "reply" at the preceding post with the table. You don't have to be able to read Latex code to use this feature, you only need to copy and paste the block of code. The PF system will automatically make the table for you.
Here is a sample of the code block, without the "tex" tags that make it get interpreted and turn into a table

...\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}...
...\hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&D_{hor}(Gly)&V_{now} (c)&V_{then} (c) \\ \hline 0.001&1090.000&0.0004&0.0006&45.332&0.042&0.057&3.15&66.18\\ \hline 0.003&339.773&0.0025&0.0040&44.184&0.130&0.179&3.07&32.87\\ \hline 0.009&105.913&0.0153&0.0235&42.012&0.397&0.552&2.92&16.90\\ \hline 0.030&33.015&0.0902&0.1363&38.052&1.153&1.652&2.64&8.45\\ \hline 0.097&10.291&0.5223&0.7851&30.918&3.004&4.606&2.15&3.83\\ \hline 0.312&3.208&2.9777&4.3736&18.248&5.688&10.827&1.27&1.30\\ \hline 1.000&1.000&13.7872&14.3999&0.000&0.000&16.472&0.00&0.00\\ \hline 3.208&0.312&32.8849&17.1849&11.118&35.666&17.225&0.77&2.08\\ \hline 7.580&0.132&47.7251&17.2911&14.219&107.786&17.291&0.99&6.23\\ \hline 17.911&0.056&62.5981&17.2993&15.536&278.256&17.299&1.08&16.08\\ \hline 42.321&0.024&77.4737&17.2998&16.093&681.061&17.300&1.12&39.37\\ \hline 100.000&0.010&92.3494&17.2999&16.328&1632.838&17.300&1.13&94.38\\ \hline

One of the many good things about LightCone calculator is that it knows LaTex so it has this copy paste option. If you don't do anything it just shows you the table on your computer screen. But if you tick the "Tex script" button and then say calculate it gives you a block of stuff that will create the same table in a post here.

So you can call attention to something and ask about it.

Thanks a lot Marcus, definitely lots of things to look at. I will try that calculator and I will post questions.
Virgil.

 

[virgil1612]

The blue curve is the Hubble distance which grows and in a sense reaches out to photons so that some of them eventually make it even though they may have, at first, gotten farther away (because the distance to the space they were traveling thru was increasing.)
View attachment 82596

How did you get this graph? I looked at the calculator and didn't find any means of graphing something based on that data.

 

[marcus]

How did you get this graph? I looked at the calculator and didn't find any means of graphing something based on that data.

under the yellow bar with display options there is a white space with two things to click on
"set sample chart range" will get a convenient interval of time for drawing charts (i.e. graphs, curves...)

"column definition and selection" will let you eliminate most columns and just include Time (T) Hubble radius (R) and distance then (D_then)
The columns in your table are what turn into curves of the chart (i.e. the graph) when you select graph output.

there is one more thing to know, how to select Time to be the x-axis if it doesn't happen automatically
but for now take a look at the "column definition" and try to produce a table with only three columns

 

[marcus]

When you click "select sample chart range" it sets the stretch factor S limits at 40 and 0.4
S=40 is in the past (light comes to us stretched by that factor) and S=1 is the present (light is not stretched at all)
S=0.4 is in the future when distances will be 2.5 times present size.
To get rid of the future part and make the table (or the graph) stop at present, change 0.4 to 1 in the S_lower box.

Now if you have opened "column selection" menu and de-selected everything except T, R, and D_then

you can make that graph you mentioned. You simply have to tick "chart" in the yellow stripe of display options, and press calculate.

Did you try this? Any luck, Virgil?

 

[virgil1612]

When you click "select sample chart range" it sets the stretch factor S limits at 40 and 0.4
S=40 is in the past (light comes to us stretched by that factor) and S=1 is the present (light is not stretched at all)
S=0.4 is in the future when distances will be 2.5 times present size.
To get rid of the future part and make the table (or the graph) stop at present, change 0.4 to 1 in the S_lower box.

Now if you have opened "column selection" menu and de-selected everything except T, R, and D_then

you can make that graph you mentioned. You simply have to tick "chart" in the yellow stripe of display options, and press calculate.

Did you try this? Any luck, Virgil?

Thanks

 

[virgil1612]

Thanks. No, not yet.

 

[marcus]

Shucks, that's frustrating. Easier to start out making a table, then make a graph when that's easy.
Have you tried making a table with just 3 columns: T, R, and Dthen?

Have you tried setting S_lower to 1, so the table stops at the present and doesn't go on into future?

Any success with the "table" stage of learning?

Any problems? Don't be reluctant to ask. Several people around here use Lightcone and can reply helpfully

 

[virgil1612]

When you click "select sample chart range" it sets the stretch factor S limits at 40 and 0.4
S=40 is in the past (light comes to us stretched by that factor) and S=1 is the present (light is not stretched at all)
S=0.4 is in the future when distances will be 2.5 times present size.
To get rid of the future part and make the table (or the graph) stop at present, change 0.4 to 1 in the S_lower box.

Now if you have opened "column selection" menu and de-selected everything except T, R, and D_then

you can make that graph you mentioned. You simply have to tick "chart" in the yellow stripe of display options, and press calculate.

Did you try this? Any luck, Virgil?

Got it! I think I will play with this a lot.

 

[virgil1612]

Shucks, that's frustrating. Easier to start out making a table, then make a graph when that's easy.
Have you tried making a table with just 3 columns: T, R, and Dthen?
Have you tried setting S_lower to 1, so the table stops at the present and doesn't go on into future?

Hi,

I reproduced the graph.
- for the part where the photons were outside of the Hubble radius, does that mean they weren't visible?
- I added D_now to the graph. What is the difference between D_then and D_now?

Thanks

 

[Chalnoth]

Hi,

I reproduced the graph.
- for the part where the photons were outside of the Hubble radius, does that mean they weren't visible?
- I added D_now to the graph. What is the difference between D_then and D_now?

Thanks

No photon is visible until it reaches the observer.

 

[marcus]

Got it! I think I will play with this a lot.

Congratulations! Good work!

Hi,
I reproduced the graph.
- for the part where the photons were outside of the Hubble radius, does that mean they weren't visible?
- I added D_now to the graph. What is the difference between D_then and D_now?
Thanks

Let me try to explain, just the red curve, D_then. This tells us the histories of all the photons which we are receiving at our telescope today.
Today is year 13.8 billion. Some of the arriving photons were emitted fairly recently, comparatively near by. Like emitted in year 10 billion at a distance of around 3.3 (does the curve at 10 look to you about 3.3 high?) It approached us on an almost straight line timetable. Sliding down the red curve, so to speak. Getting nearer at an almost constant speed.

All the photons traveled along this curve, so to speak, to get to us.
One was emitted at year 4 billion, at a distance of nearly 6 (do you read it at 4 as a little less than 6 high?)
At first he made hardly any progress. The curve of his distance from us is nearly level. He doesn't get much nearer.
But after 6 billion years have passed it is year 10 billion and he has gotten within 3.3 of us. His distance from us follows the curve down. The curve shows his progress.

In year 10 this old photon could just be passing where the other photon I mentioned is emitted (in year 10 billion). They are both 3.3 from us, traveling together at the same speed, down the red curve, arriving here at the same day.

And then there is the photon who was emitted in year 1 billion, at distance 4.
He arrives on the same day as the other two.
He is aimed towards us but at first he is keeps getting farther because he is traveling thru space which is getting farther.
Eventually in year 4 billion he is almost at distance 6 from us, and he meets the photon which was emitted just then at that same distance, and the two travel together.

At first they make hardly any progress, as I said before, because they are traveling at speed c thru space with is getting farther from us at speed c, but finally by year 10 billion they are only distance 3.3 from us and they meet up with the third photon which was just emitted in year 10 billion. The three travel together and arrive the same day, today.

For the sake of narrative I assumed they were all coming from the same direction, just emitted at different times. they could have been coming from different directions and just arrive simultaneously. The curve only describes the time&distance relation of all the photons arriving today from all directions.
===============
D_then tells the distance to the galaxy that emitted the photon when it emitted it so it tells the distance the photon was from us when it started.
D_now tells what the distance to the galaxy is now today when the photon finally gets here.
If distances and wavelengths have expanded by a factor of 2 while the photon was en route then the distance to the galaxy now will be twice what it was then.

By convention we call a factor of 2 expansion by the name "redshift z = 1"
The redshift number is, by astronomers' tradition, always one less than the actual expansion factor.
If you use Lightcone to make a table and include the expansion or "stretch" factor S they you may notice that the Dnow is exactly S times the Dthen. In the row where S=2, the distance now is twice the distance back then when the light was emitted.

 

[George Jones]

By convention we call a factor of 2 expansion by the name "redshift z = 1"
The redshift number is, by astronomers' tradition, always one less than the actual expansion factor.

$z$ is the fractional redshift or the relative redshift or the fractional change in wavelength or ..., i.e.,

$$z = \frac{\lambda_{ob}}{\lambda_{em}} - 1 = \frac{\lambda_{ob} - \lambda_{em}}{\lambda_{em}} = \frac{\Delta \lambda}{\lambda_{em}} .$$

 

[marcus]

$z$ is the fractional redshift or the relative redshift or the fractional change in wavelength or ..., i.e.,

$$z = \frac{\lambda_{ob}}{\lambda_{em}} - 1 = \frac{\lambda_{ob} - \lambda_{em}}{\lambda_{em}} = \frac{\Delta \lambda}{\lambda_{em}} .$$

Exactly!
And the ratio of wavelengths (observed over emitted) is what some, including myself, have been denoting by small or capital S.
$$\frac{\lambda_{ob}}{\lambda_{em}} = S$$
S is also the ratio by which distances are enlarged while the light is in transit.

Virgil was asking about the relation between distance to source now, and back then when the light was emitted...
$$\frac{D_{now}}{D_{then}} = S$$

So it is written in the light itself, as it comes to us (since we know the wavelengths of light that atoms emit).

 

[virgil1612]

View attachment 82784
===============
D_then tells the distance to the galaxy that emitted the photon when it emitted it so it tells the distance the photon was from us when it started.
D_now tells what the distance to the galaxy is now today when the photon finally gets here.
If distances and wavelengths have expanded by a factor of 2 while the photon was en route then the distance to the galaxy now will be twice what it was then.

Please tell me if I got it correctly.

D_then is the actual distance from the photon to us, from the moment it was emitted until we saw it. It follows the photon and shows the distance taking into account the dimension of the Universe at that point. If I read D_then correctly, not one of the photons we see today, coming from anywhere, has ever been at more than 6 GLy from us.

D_now shows the distance between past positions of the photon and us, calculated now, not when the photon was there. It doesn't show at any point the real distance between a real photon and us.

Virgil.

 

[Chalnoth]

Please tell me if I got it correctly.

D_then is the actual distance from the photon to us, from the moment it was emitted until we saw it. It follows the photon and shows the distance taking into account the dimension of the Universe at that point. If I read D_then correctly, not one of the photons we see today, coming from anywhere, has ever been at more than 6 GLy from us.

D_now shows the distance between past positions of the photon and us, calculated now, not when the photon was there. It doesn't show at any point the real distance between a real photon and us.

Virgil.

No. The actual distance traveled by the photon is given by the light travel time.

D_then is the proper distance measured at the time of emission, which is the distance if you could freeze the expansion at that precise moment, and bounce some light rays back and forth.

D_now is the current distance to the object that emitted the photon measured in the same way.

 

[marcus]

Please tell me if I got it correctly.

D_then is the actual distance from the photon to us, from the moment it was emitted until we saw it. It follows the photon and shows the distance taking into account the dimension of the Universe at that point. If I read D_then correctly, not one of the photons we see today, coming from anywhere, has ever been at more than 6 GLy from us.

D_now shows the distance between past positions of the photon and us, calculated now, not when the photon was there. It doesn't show at any point the real distance between a real photon and us.

Virgil.

Yes! that is good insight! You noticed that not one photon we are seeing has ever been more than 6 Gly from us, at any time (proper distance measured at that time). Actually you can get it down more accurately to 5.8 Gly using Jorrie's calculator. He and I discovered that for ourselves a few years back. It is a nice thing to realize.

I think you got it right. The part about D_now too.
Virgil if I remember right you and I went over the idea of "proper distance" already, didn't we? Defined at some particular instant of universe time.
The distance you would measure with string or radar or any conventional means if you could pause the expansion process at that given moment long enough to make the measurement.