Forgive Me But I have No One To Discuss This Idea With

[Acitnoids]

I think I've constructed a numerical system that places a value on most every unit of measurement and those values change as the universe ages. I call it "The System of Universal Angles" and it goes like this:
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1) Every number in this system represents the A angle of a triangle, the A point represents a hypothetical "location" for the Big Bang and the a side represents the observed measurement between points B and C.
2) I assumed that the age of the universe (visible horizon) was equal to 180 degrees - a.k.a., a line.
3) I assumed that anything that can be observed has a value greater than (<) 0 degrees - 0 degrees being a point or line.
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If we assume that the age of the universe is percisely 13.7Gyrs old then:
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One billion light years will equal 13.13868613 deg. (uA1Gly)).
One second will equal 4.166250042e-16 deg (uA1(sec)).
One mile will equal 2.236523745e-21 deg (uA1(mi))
One meter will equal 1.389711426e-24 deg (uA1(m)).
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These values can be expressed as:
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180/uA1(Gly)=13.7Gly old/long
180/uA1(sec)=4.320432e+17 seconds old
180/uA1(m)=1.295232928e+26 meters long
uA1(sec)/uA1(m)=299,792,458 m/s
uA1(sec)/uA1(mi)=186,282.397 mi/s
uA1(mi)/uA1(m)=1,609.344 m/mi
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What this does is give a ratio that shows the relationship between like units but, like I said before, these numbers also change with time.
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If you were looking at an object 1Gly away then you would be observing our universe as it was one billion years ago. The ratio between the uAs of that timeframe (180=12.7Gly) and our current timeframe (180=13.7Gly) can be used to determine the recession velocity and cosmological red-shift of that observed object.
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2(a/x)^1/2-2=Z
c/(a/r)=H
1-1/(Z+2/2)^2a=r
1-1/(Z+2/2)^2c=H
a/x=past uA1/equivalent future uA1
a/r=1/1-(x/a)
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Where:
a=distance to observers visible horizon
r=distance to observed object
x=a-r
c=speed of light
Z=Cosmological red-shift
H=Hubble's constant

 

[origin]

These equations do not make any sense if only for the units.

2(a/x)^1/2-2=Z

Z is suppose to be dimensionless.

c/(a/r)=H

this does not give the proper units for the Hubble constant

1-1/(Z+2/2)^2a=r
1-1/(Z+2/2)^2c=H

these units do not make sense either.

a=age of universe
r=distance to observed object
x=a-r

I do not know what time minus distance means.

 

[Acitnoids]

Originally Posted by origin
These equations do not make any sense if only for the units.

I do not know laTeX so if you could post my equations in that format I could tell you if there has been some sort of miscommunication.
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One of the furthest quasars found is called ULAS J1120+0641. It has a redshift of 7.1 and is estimated to be 12.9Gly away from us.
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1-1/(7.1+2/2)^2*13.7Gly=13.0Gly
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This is well within the realm of error for the actual age of the universe.
http://www.sciencedaliy.com/releases/2011/06/110629132527.htm

this does not give the proper units for the Hubble constant

One mega parsec (Mpc) equals 3.26 Mly.
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299,792,458/(13.7e+9/3.26e+6)=7133.7475 m/sec/Mpc
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This is well within the realm of error for the actual age of the universe.

 

[OnlyMe]

I do not know laTeX so if you could post my equations in that format I could tell you if there has been some sort of miscommunication.

There is a pretty good thread in the Physics and Math folder that describes how to use Tex for posting. Check this thread

I did not know anything about LaTex either, they would not even allow calculators in class when I was in school. Heck I still don't know more than the very basics, but that should be enough to get ya going...

 

[origin]

2(a/x)^1/2-2=Z
c/(a/r)=H
1-1/(Z+2/2)^2a=r
1-1/(Z+2/2)^2c=H
a/x=past uA1/equivalent future uA1
a/r=1/1-(x/a)
.
Where:
a=age of universe
r=distance to observed object
x=a-r
c=speed of light
Z=Cosmological red-shift
H=Hubble's constant

It was pretty late when I looked at this and I did miss the carrots- oops. So along those lines I have a question. When you write:

1-1/(Z+2/2)^2c=H

Do you mean:

(1-1/(Z+2/2)^2)*c=H


Also you are calling Z the red shift. I assumed you meant z, which is :

$$z = (\lambda_{obs} - \lambda_{emit{) / \lambda_{emit}$$

Is that correct?

You wrote:

1-1/(7.1+2/2)^2*13.7Gly=13.0Gly

The units do not work of for this because you incorrectily used 13.7 Gly for the age of the universe. The age of the universe is in Gy which is time and not Gly which is distance. So the units still do not make any sense which leads me to believe the equations are not right.

edit to add: Could you liberally use parenthasis in the above equation to show how you got 13.0 Gly, I am at a loss to see how you got that result.

 

[Acitnoids]

Originally Posted by origin
Could you liberally use parenthesis in the above equation to show how you got 13.0 Gly, I am at a loss to see how you got that result.

I'll do better than that and in the process answer your other questions.
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If the observed cosmological redshift is equal to 7.1 you would first add 2:
7.1+2=9.1
You then divide that by 2:
9.1/2=4.55
You then square that number:
4.55^2=20.7025
You then divide that into 1:
1/20.7025=.04830334501
You then subtract that into 1:
1-.04830334501=.951696655
You then multiply that by the distance for the observer's visible horizon (equal to the age of the universe) and that will give you the distance from the observer of the cosmological redshift:
.951696655*13.7Gly=13.03824417Gly
Instead of using 13.7Gly I could use its equivalent value in meters and the answer would be in meters instead of light years:
.951696655*1.295232928e+26m=1.232668845e+26m

 

[origin]

I'll do better than that and in the process answer your other questions.
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If the observed cosmological redshift is equal to 7.1 you would first add 2:
7.1+2=9.1
You then divide that by 2:
9.1/2=4.55
You then square that number:
4.55^2=20.7025
You then divide that into 1:
1/20.7025=.04830334501
You then subtract that into 1:
1-.04830334501=.951696655
You then multiply that by the distance for the observer's visible horizon (equal to the age of the universe) and that will give you the distance from the observer of the cosmological redshift:
.951696655*13.7Gly=13.03824417Gly
Instead of using 13.7Gly I could use its equivalent value in meters and the answer would be in meters instead of light years:
.951696655*1.295232928e+26m=1.232668845e+26m

This still does not help your case because now the age of the universe is in meters. That is the problem. If someone asks me my age and I answer 6ft that is not very helpful! See the problem?

 

[Acitnoids]

It's the distance to the visible horizon that is important. This happens to be tied to the age of the universe. I think you are obsessing to much over my misstatement about describing "a" as being equal to the age. a = THE DISTANCE TO THE OBSERVERS VISIBLE HORIZON!!
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I'll make the change in the O.P. Thank you for pointing that out!

 

[origin]

It's the distance to the visible horizon that is important. This happens to be tied to the age of the universe. I think you are obsessing to much over my misstatement about describing "a" as being equal to the age. a = THE DISTANCE TO THE OBSERVERS VISIBLE HORIZON!!
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I'll make the change in the O.P. Thank you for pointing that out!

I don't think asking for an equation that works is obsessing. And your welcome. With your new definition this equation still does not work:

c/(a/r)=H

H is is suppose to be a constant. But in your equation c is constant and a is constant so the Hubble 'constant' will be different for every object you measure. When the you plug in the edge of the observable universe the hubble 'constant' is equal to the speed of light!

 

[Acitnoids]

Originally Posted by origin
I don't think asking for an equation that works is obsessing.

You are correct but I have never been in a discussion where the other person does not pay attention to a clarification. I did state in the O.P. and in my replies to you that the age of the universe was interchangable to the distance of the observer's visible horizon.

Originally Posted by acitnoids
2) I assumed that the age of the universe (visible horizon) was equal to 180 degrees - a.k.a, a line.

Do you see how I got the distance for an object where Z=7.1? Does that equation work?

Originally Posted by origin
H is is suppose to be constant.

No it is not.
H is the ratio of the speed of recession of a galaxy (do to the expansion of the universe) to its distance from the observer, the Hubble constant is not actually a constant, but is regarded as measuring the expantion rate today.
http://www.google.com/m?=define+hubble+constant
The recession velocity of an object 1Mpc away is not the same today as it was thirteen billion years ago. The recession velocity of an object 3Mpc away is not the same as an object 1Mpc away.

When you plug in the edge of the observable universe the hubble 'constant' is equal to the speed of light!

That is correct. Beyond that edge light has not had time to reach us yet.

 

[origin]

No it is not.
H is the ratio of the speed of recession of a galaxy (do to the expansion of the universe) to its distance from the observer, the Hubble constant is not actually a constant, but is regarded as measuring the expantion rate today.

I think you need to do a little research on the hubble constant.

Here is how the hubble constant is used.

The value of the Hubble constant is ~ 72 m/s/mpc. By multiplying the distance by the HC you will find the recession speed. The hubble constant does not change with the distance - it would not be very useful if it did.

The recession velocity of an object 1Mpc away is not the same today as it was thirteen billion years ago. The recession velocity of an object 3Mpc away is not the same as an object 1Mpc away.

That is correct and the Hubble Constant gives the relationship between the 2.

 

[Acitnoids]

Originally Posted by origin
The value of the Hubble constant is ~ 72 m/s/mpc.

I think you meant to say 72 km/s/Mpc not 72 m/s/Mpc.
Let's see what my equation gets for a distance of 1Mpc.
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c/(a/r)=H
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1Mpc=3.26Mly
c=299,792.458 km/s
a=13.7Gly
r=3.26Mly
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so:
299,792.458/(13.7e9/3.26e6)=71.33747541 km/s/Mpc
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This is well within the realm of error for the actual age of the universe.
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I can see what you are getting at. My equation describes the recession velocity of space more so than it strickly being applied to the Hubble constant. Is that correct?

 

[origin]

I think you meant to say 72 km/s/Mpc not 72 m/s/Mpc.

I certainly did.

Let's see what my equation gets for a distance of 1Mpc.
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c/(a/r)=H
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1Mpc=3.26Mly
c=299,792.458 km/s
a=13.7Gly
r=3.26Mly.
so:
299,792.458/(13.7e9/3.26e6)=71.33747541 km/s/Mpc

Notice the bolded distance! You changed the distance from Mpc to ly so your answer is actually 71 km/s/ly. So using your equation the hubble constant is off by 6 magnatudes!

Lets look at 2 Mpc and see what happens:

r = 5.52 Mly

299,792.458/(13.7e9/5.52e6)=120.8 km/s/ly

These show a major issue with your equation.

I can see what you are getting at. My equation describes the recession velocity of space more so than it strickly being applied to the Hubble constant. Is that correct?

No, I think that the equations I have looked are just not correct.

 

[Acitnoids]

Originally Posted by origin
Notice the bolded distance! You changed the distance from Mpc to ly so your answer is actually 71 km/s/ly. So using your equation the hubble constant is off by 6 magnatudes!

That is not correct. This is a straw man argument.
Fine, let's put all the values into kilometers:
1Mpc = 3.082087114e19 km
c = 299,792.458 km/sec
a = 1.295232928e23 km
r = 3.082087114e19 km
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so:
299,792.458/(1.295232928e23/3.082087114e19)=71.33747542 km/sec/Mpc
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In case you didn't know, the ratio between "a" and "r" is always the same as long as they are depicted in the same unit. That goes for light years as well as kilometers. I take it Cosmology is not your area of expertise.
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Now, I have answered every one of your arguments so please answer mine.
How did you get a difference of 6 magnitudes?
Do you see how I got the distance for an object where Z=7.1? Does that equation work?
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Every one of my equations is related and consistent. Each one can be used to find the answer to the next. While I appreciate you engaging me in conversation I do not appreciate you making stuff up. Once again, how did you get a difference of 6 magnitudes?

Lets look at 2Mpc and see what happens:
r=5.52Mly

No it does not. 3.26e6*2=6.52e6. Even if Cosmology is not your area of expertise, this is just sloppy.

 

[RealityCheck]

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Relax mate. No need to get stroppy or impatient. Just clarify and leave out the umbrage. It will all come out in the wash. Maybe its all simple/innocent misunderstandings? Be cool.

I was just passing. Cheers!

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[Acitnoids]

Originally Posted by RealityCheck

Relax mate. No need to get stroppy or impatient. Just clarify and leave out the umbrage. It will all come in the wash. Maybe its all simple/innocent misunderstandings? Be cool. I was just passing. Cheers!

I mate, Cheers! :cheers:

 

[origin]

That is not correct. This is a straw man argument.

No it is not a strawman at all. I am simply reporting the results of your equations. I am not attacking you.:shrug:

Lets just work out the problems in the proper units.

c/(a/r)=H

c = 299,792 km/s
a = 4202 Mpc (using your number 13.7 Gly for the edge of the observable universe)

so for r = 1 Mpc

H = 71.3 km/s/Mpc (Ok so far this is very close to the given hubble constant)

for r = 2 Mpc

H = 142 km/s/Mpc (oops it is going bad this is 2X the HC)

And so on so this equation has only one distance that will yield a number that is close to the hubble constant for all other distances it is wrong. Which means this equation does not work.

 

[Acitnoids]

origin,
I want to sincerely apologize for the tone of my last post to you. I must have been hungry or something. I'm usually more civil. I took a queue from RealityCheck and grabbed a beer and relaxed. I very much appreciate your patients and help. Your reply last night was worded differently. I will reply to the edited version later because I want to get all my ducks in a row. Everything I have seen online says the recession velocity of space at 2Mpc does in fact equal 142 km/sec. I mainly wanted to appologize to you.

 

[origin]

origin,
I want to sincerely apologize for the tone of my last post to you. I must have been hungry or something. I'm usually more civil. I took a queue from RealityCheck and grabbed a beer and relaxed. I very much appreciate your patients and help. Your reply last night was worded differently. I will reply to the edited version later because I want to get all my ducks in a row. I mainly wanted to appologize to you.

No problem. I am usually a pain in the ass so I do not mind getting it back so to speak. I made several errors both arithmetic and conceptual in my previous posts especially my last one, which is why I edited it.

Everything I have seen online says the recession velocity of space at 2Mpc does in fact equal 142 km/sec.

That should be correct based on the Hubble constant which is ~ 71 km/sec/Mpc. So at 2 Mpc the recession velocity calculates out to 142 km/sec.

Which means your equation is not giving the Hubble Constant it is actually giving recession velocity at a given distance.

This works because you are implicitly using the Hubble Constant. The edge of the observable universe is where the recession speed is c. So by dividing the speed of light by the ratio of the observable edge of the universe by the distance of a galaxy you will get the recession speed of the galaxy. Which is why the distance of 1 Mpc yields the Hubble Constant.

 

[OnlyMe]

origin,
I want to sincerely apologize for the tone of my last post to you. I must have been hungry or something. I'm usually more civil. I took a queue from RealityCheck and grabbed a beer and relaxed. I very much appreciate your patients and help. Your reply last night was worded differently. I will reply to the edited version later because I want to get all my ducks in a row. Everything I have seen online says the recession velocity of space at 2Mpc does in fact equal 142 km/sec. I mainly wanted to appologize to you.

If between here and 1Mpc from here space expands by 71 km/sec., and from 1Mpc from here and 2Mpc from here space expands by 71 km/sec., then space expands at a rate of 71 km/sec/Mpc. Which would lead to a total recession velocity at 2Mpc of 142 km/sec over 2Mpc. and 213 km/sec over a distance of 3Mpc.

The rate of recession remains the same for any 1Mpc distance measured. This is consitent with the conventional view and does not account for an apparent accelerating expansion.

NOTE: All numbers were from your post and are not meant as anything more than an attempt to point out how as the distance considered is increased a recessional velocity associated with a specific "distance" becomes additive, when applied to greater distances. At 1.5Mpc the recession rate would be 107.5 km/sec, or there abouts...

Though I personally am not sure that the Hubble constant is actually a constant, there is nothing in this discussion that seems to suggest otherwise, as far as I can tell.