Distance Comparison: Converting 48 Distance Modulus to Meters

In summary, the authors of this paper could not have known that there are Type Iax SNe, which would make the distance modulus for any supernova detected at z > ~1 very small.
  • #1
Kairos
177
14
hello

I have problems with the distance units used in cosmology:
I wonder where is located a distant supernova at 48 "distance modulus", compared to the Hubble radius.
Far much less?

please use meters (x10^25 m)

thank you in advance
rachid
 
Astronomy news on Phys.org
  • #2
Kairos said:
hello

I have problems with the distance units used in cosmology:
I wonder where is located a distant supernova at 48 "distance modulus", compared to the Hubble radius.
Far much less?

please use meters (x10^25 m)

thank you in advance
rachid
I am not certain what you are asking. The distance modulus (μ) is merely the difference between apparent magnitude and absolute magnitude. You are also not going to see a distance modulus of 48 because that would translate to ~130 billion light years. The Hubble Radius is essentially our observable universe, 13.78 billion light years. Which means the distance modulus cannot be greater than 43.129 because we cannot see beyond the Hubble Radius.

μ = m - M
10(μ/5)+1 = d (in parsecs)
10(43.129/5)+1 = 4.224 Gpc = 13.78 billion light years = 1.3037 × 1026 meters
 
Last edited:
  • #3
thank you for the length of the radius of Hubble!

but I saw many distance-redshift plots for supernovae (such as this one : http://inspirehep.net/record/824150/plots) with µ largely exceeding 43..
This would mean that these supernovae are very close to the limit of observability?
 
  • #4
|Glitch| said:
The Hubble Radius is essentially our observable universe, 13.78 billion light years.
No, |Glitch|! It's not. That's only where the recession rate reaches c. The observable universe radius in terms of proper distance stretches to about 45 Gly.I think the issue here is that the distance modulus is based on the luminosity distance, which follows the inverse square law only in our nearest neighbourhood, unaffected by the expansion of the universe. In the expanding universe faraway objects, where expansion cannot be neglected, look dimmer as the light gets stretched, so the luminosity distance is no longer the true distance and will always show values much higher than actual.

I'm sorry I don't know how to calculate proper distance from DM in cosmology. I think you need to assume some expansion model first, which makes it not so trivial.
What you could do, is get the distances from redshifts (z), which is what most of the DM graphs are plotted against. Go here:
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html
click on the 'set sample chart range' and then 'calculate'. The columns of the table that should interest you are the two labelled 'S' and 'Dnow'. Deduct 1 from the values in S to get redshift (e.g., z=6.2 means S=7.2). Look for the corresponding nearby value of proper distance (displayed in millions of light years, so you can easily convert these to metres).
 
  • Like
Likes mfb
  • #5
Kairos said:
thank you for the length of the radius of Hubble!

but I saw many distance-redshift plots for supernovae (such as this one : http://inspirehep.net/record/824150/plots) with µ largely exceeding 43..
This would mean that these supernovae are very close to the limit of observability?
The Hubble Radius is defined as the ratio of the velocity of light (c), to the value of the Hubble constant (H0). This gives the distance from the observer at which the recession velocity of a galaxy would equal the speed of light. Therefore, any object with a µ > 43.129 must not have the absolute magnitude they have previously determined.

The problem, I suspect, is their assumption that only Type Ia SNe are visible at distances of z > ~1, and they are using Type Ia SNe as their "standard candle." In 2008, when the paper you posted was written, that was the correct assumption. We knew that super-Chandrasekhar Type Ia SNe have existed since 2003, and they have a greater absolute magnitude than Chandrasekhar Type Ia SNe, but incorporating super-Chandrasekhar Type Ia SNe would appear closer than they are actually. What the authors of the paper could not have known is that we have since discovered sub-Chandrasekhar Type Ia SNe which vary in absolute magnitude between -14.2 and -18.9.

Therefore, if they detect a SN with an apparent magnitude of +28.7 and a redshift z > ~1, it would have been natural form them to assume it was a standard Chandrasekhar Type Ia SNe with an absolute magnitude of -19.3. That would give them a µ of 48. However, if that supernova was a sub-Chandrasekhar Type Ia SNe (a.k.a. Type Iax SNe), then the absolute magnitude would be substantially less, and therefore have a smaller distance modulus. Since the new classification of Type Iax SNe was not established before 2013, there is no way the authors of the paper could have filtered out the "contaminated" data.

See:
Type Iax Supernovae: A New Class of Stellar Explosion - The Astrophysical Journal, Volume 767, Number 1 (2013) free issue
 
Last edited:
  • #6
Thank you very much... I have to read more on this (very complex) subject !

This point is however very confusing because there is a lot of published data (for example by the Nobel prize recipients who discovered the accelerating universe) with µ beyond 43. I don’t understand why these authors did not compare their distances to the Hubble radius
 
  • #7
Again, the distances you get from using the DM calculations in post #2 do not correspond to actual distances where the expansion of the universe needs to be taken into account!
And again, the Hubble radius does not represent the extent of the observable universe.
 
  • #8
Ah OK ! that seems more logical. Do you know the extent of the observable universe?
 
  • #9
A little over three times larger than the Hubble radius. See post #3.
 
  • #10
OK thanks,
I will try to understand the theoretical basis of this result.
 
  • #11
Bandersnatch said:
Again, the distances you get from using the DM calculations in post #2 do not correspond to actual distances where the expansion of the universe needs to be taken into account!
And again, the Hubble radius does not represent the extent of the observable universe.
The calculations in post #2 only correspond to the amount of time light has traveled, not distance. By the time that light reaches us, the object is considerably further away. The Hubble Radius is indeed our observable universe, and the distance it took for light to travel 13.8 billion years to reach us equates to ~46 billion light years. Anything beyond the Hubble Radius is moving away from us at a combined speed that faster than the speed of light, which means that it can never be observed.
 
  • #12
|Glitch| said:
The Hubble Radius is indeed our observable universe
I don't see how you can justify saying that.

See here:
Expanding confusion... Davis, Lineweaver.
"Using Hubble’s law (v_rec=HD), the Hubble sphere is defined to be the distance beyond which the recession velocity exceeds the speed of light, D_HS=c/H. As we will see, the Hubble sphere is not an horizon. Redshift does not go to infinity for objects on our Hubble sphere (in general) and for many cosmological models we can see beyond it."

"Light that superluminally receding objects emit propagates towards us with a local peculiar velocity of c, but since the recession velocity at that distance is greater than c, the total velocity of the light is away from us (Eq. 20). However, since the radius of the Hubble sphere increases with time, some photons that were initially in a superluminally receding region later find themselves in a subluminally receding region. They can therefore approach us and eventually reach us. The objects that emitted the photons however, have moved to larger distances and so are still receding superluminally. Thus we can observe objects that are receding faster than the speed of light"
(section 2; 'Hubble sphere' means Hubble radius)

|Glitch| said:
The calculations in post #2 only correspond to the amount of time light has traveled, not distance.
The calculations output units of distance, and use luminosity as their basis. The latter must be adjusted for the effects of expansion on light (like stretching wavelengths).
For equations see this paper:
Distance measures in cosmology. David W. Hogg.
(under 'luminosity distance')
 
  • #13
Bandersnatch said:
I don't see how you can justify saying that.

By your own definition:
Bandersnatch said:
...the Hubble sphere is defined to be the distance beyond which the recession velocity exceeds the speed of light...
Hence, if the galaxy is receding from us at a combined speed that is faster than the speed of light, it can no longer be considered "observable." We can see a galaxy at the very edge of the Hubble Radius as it was 13.8 billion years ago, but in that 13.8 billion years it took the light from that galaxy to reach us, that galaxy is now ~46 billion light years away from us.

Bandersnatch said:
The calculations output units of distance, and use luminosity as their basis. The latter must be adjusted for the effects of expansion on light (like stretching wavelengths).
For equations see this paper:
Distance measures in cosmology. David W. Hogg.
(under 'luminosity distance')
The distance the calculations output is the amount of time it took light to reach us at the time the light was emitted, not the actual distance to the object now. I agree that there are additional adjustments that must be made, not least of which is the VSL in the early universe.
 
  • #14
|Glitch| said:
Anything beyond the Hubble Radius is moving away from us at a combined speed that faster than the speed of light, which means that it can never be observed.
This is wrong. Its initial distance increases, but the Hubble constant is decreasing - the light of objects not too far beyond the Hubble radius will get closer again in the future and reach us in the very distant future.
The particles that emitted the CMB we see today always increased their distance to us faster than the speed of light - they are now at a distance of about 45 Gly.

It would be true in a completely massless universe with exponential expansion from dark energy, but we do not live in such a universe.
 
  • #15
This is likely a stupid question:
If an object just inside the Hubble-sphere is traveling at a speed relative to Earth of 99% of lightspeed away from us, will the light it's sending towards us be at lightspeed or 1% of lightspeed?
 
  • #16
the lightspeed will be c
 
  • #17
mfb said:
This is wrong. Its initial distance increases, but the Hubble constant is decreasing - the light of objects not too far beyond the Hubble radius will get closer again in the future and reach us in the very distant future.
The particles that emitted the CMB we see today always increased their distance to us faster than the speed of light - they are now at a distance of about 45 Gly.

It would be true in a completely massless universe with exponential expansion from dark energy, but we do not live in such a universe.

There is a profound contradiction between the results. The visible supernovae around 50 DM were far beyond the present Hubble distance (c/H0) when they emitted light, this means that the Hubble distance (c/H1) was shorter in the past. Hence, H1> H0. As you explain, the Hubble constant decreased during light flight. This conclusion is exactly opposite to the acceleration deduced from the redshift of these SAME supernovae (Nobel prize 2011)
 
  • #18
Where is the contradiction?
A decreasing Hubble constant means it was larger in the past, which means the Hubble radius was smaller.

With constant expansion, the Hubble constant would still decrease (with 1/time). Accelerated expansion makes it decrease slower than that, but it is always decreasing.
 
  • #19
mfb said:
With constant expansion, the Hubble constant would still decrease (with 1/time)

Are you saying that "constant (exponential) expansion" does not mean H = constant?

and calling "acceleration" a slower deceleration is a bit confusing
 
  • #20
Kairos said:
Are you saying that "constant (exponential) expansion" does not mean H = constant?
Constant expansion and exponential expansion are different things.

"Constant" means the distance between two objects increases at a constant rate. That means their relative distance increases slower over time.

"Exponential" means the distance between two objects increases exponentially. That means their relative distance increases the same over time.

That can look a bit confusing, but there is no way to avoid having those two different concepts of relative and absolute.
 
  • #21
mfb said:
"Constant" means the distance between two objects increases at a constant rate. That means their relative distance increases slower over time.

I believe that the debate about the acceleration/deceleration of expansion, always concerns an exponential rate. There are different speeds:

[itex] D(t) = D_0+vt [/itex] (arithmetic progression). The increase is not necessarily slow as v can be illimited!
or
[itex] D(t) = D_0 \ e^{Ht} [/itex] (geometric progression). The Hubble law Vrec=HD proves that this mode of expansion is the good one, but H can be constant or not
 
  • #22
Kairos said:
The Hubble law Vrec=HD proves that this mode of expansion is the good one
It does not, and it is not a good model. You would get this in a completely empty universe with just dark energy, but we do not happen to live in such a universe.
It also does not work at all for the big bang, where distances were zero (or extremely close to zero shortly afterwards).
 
  • #23
This not a model but just maths. V=HD can be rewritten dx/dt=H0*x, whose solution is elementary
 
  • #24
Only if H does not depend on t, but it does.
 
  • #25
OK but precisely the present H0 long remained constant because it has been defined as the slope of the linear Hubble plot (including now for very distant galaxies) and I read that the age of the universe is approximately 1/H0!
I would greatly appreciate knowing the type of function proposed for H(t) and the mathematical type of speed if it is not one of those of post #21
 
  • #26
@Kairos please refer to the paper by Lineweaver and Davis linked to in post #12 (it is not the SciAm popular article by the same authors you may sometimes see linked on PF). It should answer most of your questions much more thoroughly than anyone here could ever do in any single post.

If that's not enough, this thesis paper from Davis:
http://arxiv.org/pdf/astro-ph/0402278v1.pdf
is an extended version of the earlier linked paper (its second half probably of little relevance here, though).

The calculator linked to in post #4 may be of some use. The concise list of equations employed can be accessed from a link at the bottom of that page.
The calculator should make it clear, for example, how the Hubble time ##1/H_0## being approximately close to the age of the universe is coincidental (compare ##T## and ##R##), and how it has never been constant but asymptotically approaches a constant value.
 
Last edited:
  • #27
Kairos said:
OK but precisely the present H0 long remained constant because it has been defined as the slope of the linear Hubble plot (including now for very distant galaxies)
It does not work for very long distances.
Kairos said:
and I read that the age of the universe is approximately 1/H0!
Which is a coincidence today. It was not true 5 or 10 billion years ago, and it won't be true in 5 or 10 billion years.
Kairos said:
I would greatly appreciate knowing the type of function proposed for H(t) and the mathematical type of speed if it is not one of those of post #21
We have a calculatur, but you can also take those formulas and plug in numbers yourself.
 
  • #28
Bandersnatch said:
If that's not enough, this thesis paper from Davis:
mfb said:
We have a calculatur, but you can also take those formulas and plug in numbers yourself.

thank you for these refs
 

1. How is distance modulus related to meters?

Distance modulus is a unit used in astronomy to measure the distance between celestial objects. It is directly related to meters, as it is a logarithmic scale that helps convert the difference in brightness between two objects into a distance in meters.

2. What is the formula for converting distance modulus to meters?

The formula for converting distance modulus to meters is: Distance (in meters) = 10^((m-M+5)/5), where m is the apparent magnitude and M is the absolute magnitude of the object.

3. Can distance modulus be negative?

Yes, distance modulus can be negative. This happens when the absolute magnitude of an object is greater than its apparent magnitude, indicating that the object is relatively closer to Earth.

4. How does a distance modulus of 48 translate to meters?

A distance modulus of 48 would be an extremely large distance in meters, as the formula for converting it to meters involves taking 10 to the power of 48 divided by 5. This would result in a distance of approximately 3.2 x 10^9 meters or 3.2 billion kilometers.

5. Why is distance modulus a useful unit of measurement in astronomy?

Distance modulus is a useful unit of measurement in astronomy because it allows for the comparison of distances between objects that may have vastly different apparent magnitudes. It also provides a more intuitive understanding of distance, as larger values indicate greater distances, similar to how a higher number of kilometers indicates a longer distance on Earth.

Similar threads

  • Astronomy and Astrophysics
Replies
6
Views
2K
Replies
3
Views
980
Replies
50
Views
3K
  • Astronomy and Astrophysics
Replies
1
Views
1K
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
668
Replies
14
Views
1K
Replies
17
Views
2K
  • Astronomy and Astrophysics
2
Replies
35
Views
32K
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top