<p>What you have taken as data are the redshifts, <span class="math-container">$z$</span>. <em>You</em> have then converted them to a velocity <span class="math-container">$v$</span> using a formula based on the Special Relativistic Doppler effect which tends towards <span class="math-container">$c$</span> at large redshifts and distances. Therefore the general shape of your graph is inevitable,  whether the expansion is accelerating or not.</p>
<p>Leaving aside that trivial point, if the expansion is accelerating, then as we look to distant galaxies, at high redshift, <em>in the past</em> then, for a given distance, we might expect them to be receding from us <em>slower</em> than would be the case if the expansion rate were constant.</p>

CC BY-SA 4.0
2024-12-15T16:09:04.320Z
If the general shape of the graph is "logarithm" how did we conclude the expansion is accelerated ?
Fabien
CC BY-SA 4.0
2024-12-15T17:04:06.463Z
@Fabien this is NOT the graph plotted in peer-reviewed journal papers (which you could look at). There isn't a big difference in the shape of the graph between constant, decelerated or accelerated expansion. Why do you think there should be?
ProfRob
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2024-12-16T09:49:54.273Z
I added in the question the reason why I think it should be exponential (or at least "above linear")
Fabien
CC BY-SA 4.0
2024-12-16T13:57:31.323Z
I don't understand : "we might expect them to be receding from us slower than would be the case if the expansion rate were constant"... On the contrary it should be : "for distant galaxies, for a given distance, we might expect them to be receding from us FASTER than would be the case if the expansion rate were constant" no ?
Fabien
CC BY-SA 4.0
2024-12-16T14:03:38.913Z
@Fabien Huh? When things are accelerating, they are slower in the past.
ProfRob
CC BY-SA 4.0
2024-12-16T14:15:42.300Z
Oo... Ok sure yes, I understand... So you mean IT IS normal that the curve has logarithm form as it is accelerating expansion ? (I made a mistake on the data source and corrected in the question)
Fabien
CC BY-SA 4.0
2024-12-16T14:21:39.243Z
@Fabien No, Constant expansion and decelerating expansion would also have the same general shape, as I have explained several times now, because you have chosen to plot velocity calculated from the relativistic Doppler effect, which cannot be $>c$. They would be different *in detail*.
ProfRob
CC BY-SA 4.0
2024-12-16T14:48:49.097Z
Thank you for your answer in https://chat.stackexchange.com/rooms/156692/discussion-between-fabien-and-profrob
Fabien
CC BY-SA 4.0
2024-12-15T15:42:27.990Z
2024-12-16T09:58:55.590Z
2024-12-16T09:58:55.590Z
ProfRob
<p>I try to understand the accelerated expansion of the universe, and my first idea to do so was to draw a graph of the speed of objects (in km/s) relative to their distance (in MParsecs).</p>
<p>My data source is <a href="https://supernova.lbl.gov/Union/figures/SCPUnion2.1_mu_vs_z.txt" rel="nofollow noreferrer">Union2.1 Compilation Magnitude vs. Redshift Table</a> from <a href="https://supernova.lbl.gov/Union/" rel="nofollow noreferrer">https://supernova.lbl.gov/Union/</a></p>
<p>This is how I converted <a href="https://en.wikipedia.org/wiki/Distance_modulus" rel="nofollow noreferrer">distance modulus</a>  into parsecs :</p>
<p><a href="https://i.sstatic.net/igcFixj8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/igcFixj8.png" alt="enter image description here" /></a></p>
<p>And Z into speed :</p>
<p><a href="https://i.sstatic.net/2a2L7xM6.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2a2L7xM6.png" alt="enter image description here" /></a></p>
<p>The result is :</p>
<p><a href="https://i.sstatic.net/iPiBxlj8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iPiBxlj8.png" alt="enter image description here" /></a></p>
<p>Which seems to be a curve of &quot;logarithmic&quot; form, and NOT exponential as I would expect for &quot;accelerated&quot; expansion...</p>
<p>I would like to point out that in my opinion an &quot;acceleration of the expansion&quot; should give increasingly greater speeds depending on the distance, or at least constant, linear, whereas here we have lower and lower speeds as objects are further and further away...</p>
<p>Can someone help me to understand where is my mistake, please ?</p>

CC BY-SA 4.0
2024-12-15T15:29:40.883Z
Maybe you should explain why you think the curve should be an exponential.
ProfRob
CC BY-SA 4.0
2024-12-15T16:28:32.970Z
@ProfRob An "accelerated expansion" should be exponential no ?
Linear would be constant, and logarithm "slowing expansion" ?
Fabien
CC BY-SA 4.0
2024-12-15T17:01:30.943Z
how can it be exponential when you are using a formula that is asymptotic to a constant value? I don't understand your question or why you think the graph should have a different shape.
ProfRob
CC BY-SA 4.0
2024-12-16T09:29:14.047Z
@ProfRob it can be anything, exponential, constant or logarithmic because it's not a "formula" that is plotted it's simple "data", from a table of values in the cited paper, from two columns : Z converted to speed in km/s and DM (distance modulus) converted to parsecs. But it's true that I could have add the label of the axis on the graph....
Fabien
CC BY-SA 4.0
2024-12-16T09:45:01.557Z
Ok just added the label of the axis on the graph.
Fabien
CC BY-SA 4.0
2024-12-17T00:31:56.167Z
A graph of speed over distance should be roughly linear, with the Hubble constant (km/s / Mpc) as a slope. A graph of distance over time should on the other hand be exponential.
controlgroup
CC BY-SA 4.0
2024-12-17T08:11:55.303Z
But the graph of speed over distance is not linear as you see in my question... Or is it what you call "roughly" linear ?
Fabien
CC BY-SA 4.0
2024-12-18T01:39:11.020Z
I think that, for recessional velocities much smaller than $c$, it is roughly linear, yes. Keep in mind that $c\approx300,000\,\text{km}\,\text{s}^{-1}$, though, a relatively-small number. Note that at a little under 1000 Mpc, where the non-linear nature becomes more-apparent, you're already dealing with $v\approx0.16c$, where other relativistic effects come into play. For the very non-linear bits, like those out near 5000 Mpc, you're talking about $v\approx0.5c$.
controlgroup
CC BY-SA 4.0
2024-12-18T08:13:19.480Z
Ok I see. Could you tell me if it's possible to draw, as you said, a "graph of distance over time" with the set of data I mention in my question ? If not, how is it possible to do that or to found a set of data over time ? Does it simply exists ?
Fabien
CC BY-SA 4.0
2024-12-18T13:30:42.427Z
The thing is distances measured in kilometers are functionally nothing in the face of megaparsecs, so if you draw a distance-over-time curve for any one galaxy/quasar it’ll look like a straight line.
controlgroup
<universe><expansion>
Accelerated expansion
Fabien
# Accelerated expansion

(Asked by: username_0 on 2024-12-15)

I try to understand the accelerated expansion of the universe, and my first idea to do so was to draw a graph of the speed of objects (in km/s) relative to their distance (in MParsecs).

My data source is [Union2.1 Compilation Magnitude vs. Redshift Table](https://supernova.lbl.gov/Union/figures/SCPUnion2.1_mu_vs_z.txt) from <https://supernova.lbl.gov/Union/>

This is how I converted [distance modulus](https://en.wikipedia.org/wiki/Distance_modulus) into parsecs :

[![enter image description here](https://i.sstatic.net/igcFixj8.png)](https://i.sstatic.net/igcFixj8.png)

And Z into speed :

[![enter image description here](https://i.sstatic.net/2a2L7xM6.png)](https://i.sstatic.net/2a2L7xM6.png)

The result is :

[![enter image description here](https://i.sstatic.net/iPiBxlj8.png)](https://i.sstatic.net/iPiBxlj8.png)

Which seems to be a curve of "logarithmic" form, and NOT exponential as I would expect for "accelerated" expansion...

I would like to point out that in my opinion an "acceleration of the expansion" should give increasingly greater speeds depending on the distance, or at least constant, linear, whereas here we have lower and lower speeds as objects are further and further away...

Can someone help me to understand where is my mistake, please ?

**Tags:** universe, expansion
> **Comments:**
> **username_1**: Maybe you should explain why you think the curve should be an exponential.  
> **username_0**: @ProfRob An "accelerated expansion" should be exponential no ? Linear would be constant, and logarithm "slowing expansion" ?  
> **username_1**: how can it be exponential when you are using a formula that is asymptotic to a constant value? I don't understand your question or why you think the graph should have a different shape.  
> **username_0**: @ProfRob it can be anything, exponential, constant or logarithmic because it's not a "formula" that is plotted it's simple "data", from a table of values in the cited paper, from two columns : Z converted to speed in km/s and DM (distance modulus) converted to parsecs. But it's true that I could have add the label of the axis on the graph....  
> **username_0**: Ok just added the label of the axis on the graph.  
> **username_2**: A graph of speed over distance should be roughly linear, with the Hubble constant (km/s / Mpc) as a slope. A graph of distance over time should on the other hand be exponential.  
> **username_0**: But the graph of speed over distance is not linear as you see in my question... Or is it what you call "roughly" linear ?  
> **username_2**: I think that, for recessional velocities much smaller than $c$, it is roughly linear, yes. Keep in mind that $c\approx300,000\,\text{km}\,\text{s}^{-1}$, though, a relatively-small number. Note that at a little under 1000 Mpc, where the non-linear nature becomes more-apparent, you're already dealing with $v\approx0.16c$, where other relativistic effects come into play. For the very non-linear bits, like those out near 5000 Mpc, you're talking about $v\approx0.5c$.  
> **username_0**: Ok I see. Could you tell me if it's possible to draw, as you said, a "graph of distance over time" with the set of data I mention in my question ? If not, how is it possible to do that or to found a set of data over time ? Does it simply exists ?  
> **username_2**: The thing is distances measured in kilometers are functionally nothing in the face of megaparsecs, so if you draw a distance-over-time curve for any one galaxy/quasar it’ll look like a straight line.  

---

## 1 answers

### Answer by username_1 on 2024-12-15. Score: 3 | Accepted answer

What you have taken as data are the redshifts, $z$. *You* have then converted them to a velocity $v$ using a formula based on the Special Relativistic Doppler effect which tends towards $c$ at large redshifts and distances. Therefore the general shape of your graph is inevitable, whether the expansion is accelerating or not.

Leaving aside that trivial point, if the expansion is accelerating, then as we look to distant galaxies, at high redshift, *in the past* then, for a given distance, we might expect them to be receding from us *slower* than would be the case if the expansion rate were constant.

> **Comments:**
> **username_0**: If the general shape of the graph is "logarithm" how did we conclude the expansion is accelerated ?  
> **username_1**: @Fabien this is NOT the graph plotted in peer-reviewed journal papers (which you could look at). There isn't a big difference in the shape of the graph between constant, decelerated or accelerated expansion. Why do you think there should be?  
> **username_0**: I added in the question the reason why I think it should be exponential (or at least "above linear")  
> **username_0**: I don't understand : "we might expect them to be receding from us slower than would be the case if the expansion rate were constant"... On the contrary it should be : "for distant galaxies, for a given distance, we might expect them to be receding from us FASTER than would be the case if the expansion rate were constant" no ?  
> **username_1**: @Fabien Huh? When things are accelerating, they are slower in the past.  
> **username_0**: Oo... Ok sure yes, I understand... So you mean IT IS normal that the curve has logarithm form as it is accelerating expansion ? (I made a mistake on the data source and corrected in the question)  
> **username_1**: @Fabien No, Constant expansion and decelerating expansion would also have the same general shape, as I have explained several times now, because you have chosen to plot velocity calculated from the relativistic Doppler effect, which cannot be $>c$. They would be different *in detail*.  
> **username_0**: Thank you for your answer in https://chat.stackexchange.com/rooms/156692/discussion-between-fabien-and-profrob  

---

