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

[Acitnoids]

Thanks OnlyMe, you posted the lion's share of my post.

Originally Posted by [/b]origin[/b]
For r=2
H = 142 km/s/Mpc (oops it is going bad this 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.

This is where you are wrong. In the first example we used a distance for 1Mpc so the answer would be:
H=71.3 km/s/Mpc.
This means that for every 1Mpc distance the recession velocity increases by 71.3 km/s. In the second example you used a distance of 2Mpc so the answer would be:
H=142 km/s/2Mpc.
This means that for every 2Mpc distance the recession velocity increases by 142 km/s. This goes for a distance of 1Gly too, the answer would be H=21,882 km/s/Gly and the same reason applies. So yes, 2X the HC.
.
Mpc is the traditional unit used and so there was no reason to change it as the standard. Nothing says that H has to be in units of Mpc.
.
If you're so inclined you could rearange this equation to find the age of the universe.
(r*c)/H=a
The most accurate measurement of H has it at 73.8 +/- 2.4 km/s/Mpc. That would put the actual age of the universe somewhere between 13.7Gyr, 13.2Gyr and 12.8Gyr old.
http://blogs.discovermagazine.com/b...panding-at-73-8-2-4-kmsecmegaparsec-so-there/
I warn you, this article is total pop-sci. It's like reading a transcrip of Science Friday's on NPR.

 

[OnlyMe]

If your read my post as having a different intent than origin's either you are mistaken or I explained my intent badly. I agree with what I understand origin's intent to be.

 

[Acitnoids]

Originally Posted by [/b]OnlyMe[/b]
If your read my post as having a different intent than origin's either you are mistaken or I explained my intent badly. I agree with what I understand origin's intent to be.

In that case, I don't think you fully understand what the Hubble constant represents. Do the math yourself. What is the recession velocity of space at 4Mpc distance using H=71 km/s/Mpc (that would be H*4). Same question but now H=142 km/s/2Mpc (that would be H*2). I'm making this as simple as I can. Why do you think H can only be expressed in units of Mpc?

 

[origin]

This is where you are wrong. In the first example we used a distance for 1Mpc so the answer would be:
H=71.3 km/s/Mpc.
This means that for every 1Mpc distance the recession velocity increases by 71.3 km/s. In the second example you used a distance of 2Mpc so the answer would be:
H=142 km/s/2Mpc.

Oh come on that is just silly.

Your equation for 2 Mpc yields the following:
c/(a/r) = 299,792 km/s/(4202Mpc/2Mpc) = 142 km/s/Mpc!

This is simple division. You are so concerned about admitting that your equation does not give the Hubble Constant that you are making weird math errors.

Everyone makes mistakes - refusing to admit that is not going to help your case, it will do the opposite!



edit to add: Here is how to admit a mistake.

I screwed up! The result of 142 km/s/Mpc is not correct; the correct answer is :

142 km/s I made the silly mistake of not canceling out the units of Mpc. Geeze! The velocity result makes perfect sense as a result for your equation, I will show you later.

 

[origin]

Here is your equation:

c/(a/r) = H

Which is actually,

c/(a/r) = v

c = 299792 km/s
a = 13.7 Gly or 4202 Mpc
r = some distance in Mpc
v = recession velocity at distance r

The equation is simply based on the ratio of your assumed values and the linear relationship between velocity and distance.

The ratio is:

c/a = v/r which gives you: cr/a = v Or c/(a/r) = v

Lets choose 2 points to determine the slope of this linear relationship.

x = 0, 4202Mpc
y = 0, 299792 km/s

$$\frac{y_2 - y_1}{x_2 - x_1}$$ = slope

=$$\frac{299792 km/s- 0}{4204 Mpc- 0}$$ = 71.3 km/s/Mpc.

This is the Hubble Constant - the slope of the line from the recession velocity VS distance.

You took values from the this relationship to make your equations so you are implicitly using the Hubble Constant to find the rescession velocities.

Hope this helps.

 

[Acitnoids]

Originally Posted by origin
Everone makes mistakes - refusing to admit that is not going to help your case, it will do the opposite!

I have no problem admitting when I'm wrong but, I have to be wrong in order for that to happen.

Here is your equation:
c/(a/r)=H
Which is actually,
c/(a/r)=v

If you were making this point about the other equation in the O.P. that also equals H:
1-(1/(Z+2/2)^2)c=H
(Hope this expression makes it clearer)
Then I'd say ; "yes, you are right and I was wrong. That equation does not equal H, it actually equals v. Thank you for pointing that out." But, here's the difference between the two.
.
The only units in play for the other equation are those found in "c" (be it km/s, mi/s or whatever units you want to use) so there is no way to determine a ratio for the recession velocity of space. However, the equation we are currently discussing has such a unit, "r". This unit is what determines a ratio for the recession velocity of space. If r=1Gly then H will be expressed as 21,882 km/s/1Gly meaning that for every 1Gly distance the recession velocity of space will increase by 22,882 km/s. At a distance of 2Gly the recession velocity would be 2*H (43,764 km/s) and so on. This is the very definition of the Hubble constant.
.
Nothing says that H has to be in units of Mpc! That expression is nothing more than a standard unit like Astronomical Units or Solar Masses.

 

[origin]

If you were making this point about the other equation in the O.P. that also equals H:
1-(1/(Z+2/2)^2)c=H
(Hope this expression makes it clearer)

I frankly do not know how you got this equation, but it looks interesting. But what I can say about this equation is that the solution cannot possibly be the Hubble Constant, for 2 reasons:
1. The results you will obtain with that equation is simply a velocity. The Hubble Constant is the slope of the line for recession velocity VS distance. So the units must be velocity/distance.
2. The other more fundamental reason is the results are not a constant!

Then I'd say ; "yes, you are right and I was wrong. That equation does not equal H, it actually equals v. Thank you for pointing that out." But, here's the difference between the two.

.

No problem.

Nothing says that H has to be in units of Mpc! That expression is nothing more than a standard unit like Astronomical Units or Solar Masses.

True. I only used Mpc because those are the traditional units of the Hubble Constant. However as I stated before the Hubble Constant absolutely must be in the form of velocity/distance. It can be furlongs/fortnight/cubits!

 

[Acitnoids]

Originally Posted by origin
I frankly do not know how you got this equation, but it does look interesting.

That is one of the six equations found in the opening post (O.P.). As I said in my last post ; 1-(1/(Z+2/2)^2)c=H does not equal H, it equals the recession velocity of space (v). Through the course of this conversation I was able to notice that. As the title says, ...

But what I can say about this equation is that the solution cannot possibly be the Hubble Constant, ...

I'm glad to see you agree with what I said in my last post.

No problem.

I'm sorry, I don't know what you mean by this.

However as I stated before the Hubble Constant absolutely must be in the form of velocity/distance.

That's exactly what I showed in my last post using c/(a/r)=H. Good to see we're on the same page with that.
.
.
By the way, I never said that c/(a/r)=H was "my" equation. I said:

Originally Posted in the Opening Post
What (my system) does is give a ratio that shows a relationship between like units but, like I said before, these numbers also change with time.
.
If you were looking at an object 1 Gly away then you would be observing our universe as it was one billion years ago. The ratio between the uAs of that timeframe (180=12.7) and our current timeframe (180=13.7) can be used to determine the recession velocity and cosmological red-shift of that object.

 

[origin]

That's exactly what I showed in my last post using c/(a/r)=H. Good to see we're on the same page with that.
.
.
By the way, I never said that c/(a/r)=H was "my" equation. I said:

What we have here is a failure to communicate...

I thought you were agreeing that c/(a/r) = v. I did not realize you were talking about your other equation. Which I am glad to see we agree is the recession velocity!

I went through an entire post clearly showing you that c/(a/r) = recession velocity.

For crying out loud what do you think the units are for:

km/s/(Mpc/Mpc) or km/s/(miles/miles) or km/s/(furlong/furlong)????

They all equal km/s!

 

[Acitnoids]

Originally Posted by origin
I went through an entire post clearly showing that c/(a/r) = recession velocity.
For crying out loud what do you think the units are for:
Km/s/(Mpc/Mpc) or Km/s/(miles/miles) or Km/s/(furlong/furlong)????
They all equal km/s!

And I think you've missed my point. The Hubble constant is the ratio of the speed of recession. If you want to know what the ratio of recession is for 1Mpc then (13.7Gly/3.26Mly) = 4202.453988. This shows how many times the recession velocity will increase by 71.3 km/s over 13.7Gly. We both agreed that my example was right for this distance but you insisted that this is were the equation stops equaling H. I then showed you how the ratio of recession for increments of 1Gly is (13.7Gly/1Gly)=13.7. This shows how many times the recession velocity will increase by 21,882.66 km/s over 13.7Gly. What I am saying is that (a/r) is the ratio for the recession velocity of space. In order to calculate the recession velocity you need to divide (a/r) into c and this will give you how many times that speed of recession will increase over that distance all the way up to our visible horizon. This is the Hubble constant.

 

[origin]

And I think you've missed my point. The Hubble constant is the ratio of the speed of recession. If you want to know what the ratio of recession is for 1Mpc then (13.7Gly/3.26Mly) = 4202.453988. This shows how many times the recession velocity will increase by 71.3 km/s over 13.7Gly. We both agreed that my example was right for this distance but you insisted that this is were the equation stops equaling H. I then showed you how the ratio of recession for increments of 1Gly is (13.7Gly/1Gly)=13.7. This shows how many times the recession velocity will increase by 21,882.66 km/s over 13.7Gly. What I am saying is that (a/r) is the ratio for the recession velocity of space. In order to calculate the recession velocity you need to divide (a/r) into c and this will give you how many times that speed of recession will increase over that distance all the way up to our visible horizon. This is the Hubble constant.

I am begining to understand why no one wants to discuss this with you.

Lets try this one more time.

The hubble constant is ~71 km/s/Mpc. That is the hubble constant - look it up. That is the magnitude for those units.

If the equation c/(a/r) = H, is correct then H must equal 71 km/s/Mpc (if a and r are in Mpc). This is rather trivially obvious.

So lets see if the equation works for r=10 Mpc!

c/(a/r) = H = 299,792 km/s (4202 Mpc/10 Mpc) = 713 km/s

No, I am afraid this is not the right magnitude nor the right units. The Mpcs cancel out leaving km/s. This is not the Hubble Constant - it is just that simple.

However, 713 km/s is the recession velocity of a galaxy at 10 Mpc.

OK?

 

[Acitnoids]

Originally Posted by origin
I am begining to understand why no one wants to discuss this with you.

You may be onto something here and, unfortunately for you and I this is also a contributing factor for why it hadn't gotten through my thick skull yet. All I've had to go on is one man's word and, especially on a forum like this one, you never know who's on the other end. That being said, I am willing to reconsider my stance.

Lets try this one more time.
The hubble constant is ~ 71 km/s/Mpc. That is the hubble constant - look it up. That is the magnitude for those units.

I never once denied that 71 km/s/Mpc=H. I tried looking if H could be held in any unit other than Mpc and couldn't find an example. Actually, this thread was the first search option to come up and in the discription it said; "This does not equal the Hubble constant." You can't blame me for trying to articulate why I thought (a/r) was a ratio for the recession velocity, can you? I guess both of us can admit to being a pain in the ass sometimes.

No, I am afraid this is not the right magnitude nor the right units. The Mpcs cancel out leaving km/s. This is not the Hubble Constant - it is that simple. However, 713 km/s is the recession velocity of a galaxy at 10 Mpc.
OK?

Seeing that the other equation in the O.P. had the wrong solution I'll concede this one as well so, OK.
.
I can tell that I've neared its limits but I'd like to thank you for your patients and time. It's just nice to be able to talk this through with someone/anyone in a civilized manner.
.
One more thing.
If the units of (a/r) cancel each other out then what does that say about the first equation in the O.P.?
$$Z={2}\sqrt{\frac{a}{x}}-2$$

 

[origin]

Seeing that the other equation in the O.P. had the wrong solution I'll concede this one as well so, OK.

Good for you! I mean it, that is how learning takes place.

One more thing.
If the units of (a/r) cancel each other out then what does that say about the first equation in the O.P.?
$$Z={2}\sqrt{\frac{a}{x}}-2$$

That means the units cancel which is a good thing. Z is dimensionless not to mention that the square root of the unit is meaningless.


edit to add: There are also volumes that I do not know on this subject - I am no expert.

 

[Acitnoids]

In addition to origin I should also thank OnlyMe for chiming in to help point out my misunderstanding between H and v. You obviously understand the technical stuff better than I.
.
Here is another interesting coincidence that I found in my system.

Originally Posted in the Opening Post
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.

I assumed that every triangle in this system is an isosceles triangle. Whether they have to be isosceles or not is undetermined, that's just the simplest configuration I used. If you were to plot the lengths for side b using different lengths for the a side you will find that the closer points B and C are to each other the closer they will get to point A. The absolute shortest length side b can be for any given moment is equal to (a/pi) however, this length is not physically possible. Small lengths such as the planck length get very close to equaling (a/pi) but you have to have a really good calculator to see the difference between various small A angles. The opposite is also true. You could plot the lengths of side b over one billion year increments and see how everything moves apart. The longest length side b can be for any given moment equals (a/2). You can do this for any age of the universe and what you'll find is that the further back in time you go the closer everything gets to point A.

 

[Acitnoids]

Since I am just scratching the surface to this system I should probably show you how it works. The best way to do that is to use a natural unit system.

Originally Posted in the Opening Post
One second will equal 4.166250042e-16 deg (uA1(sec))
...
One meter will equal 1.389711426e-24 deg (uA1(m))

For this example I'm going to use the planck length and planck time.
.
uA1(m)*1.616252e-35 m = 2.24612(4)e-59 deg.
uA1(sec)*5.39124e-44 sec = 2.24612(5)e-59 deg.
.
Taking the rounding error into consideration - the A angle for both of these units have the same length for their a sides. This is how you determine if two units are equivalent. The fact that these angles differ slightly shows that I need a more precise value for the planck units.
.
For comparison let me do this agin with the atomic units of length and time.
.
uA1(m)*.52917720859e-10 = 7.354036132e-35 deg. (uA(al))
uA1(sec)*2.418884326505e-17 = 1.007767693e-32 deg. (uA(pt))
.
Obviously these angles are not equivalent so then what is the difference between the two?
.
uA(al)/uA(at) = .007297352537
.
That would be the fine structure constant.
.
http://physics.nist.gov/cuu/Constants/Table/allascii.txt - ca., 6/29/2007

 

[Acitnoids]

There is one last thing I can show without having to dive into the "bottom" half of this system. That would be a photons wavelength, frequency and energy. I'm hesitant to describe this because the equations for the "top" half are upside down. The "bottom" half is much more consistent with known theory and it includes temperature. For the record; the "top" half consists of nothing more than space and time. The "bottom" half contains most every other unit including inverse-space and frequency.
.
The "top" half expresses a photon like this:
w * uA1(m) / uA1(sec) = 1/f
(1/f) * uA1(sec) / uA1(m) = w
h * uA1(sec) / (w * uA1(m)) = E
Where:
w = wavelength (in meters)
f = frequency (in hertz)
h = Planck's constant
E = energy
.
The "bottom" half expresses a photon like this:
(1/w) * uA1(1/m) / uA1(Hz) = f
(1/w) * uA1(1/m) / uA1(K) = K
(1/w) * uA1(1/m) / uA1(eV) = E (in eV)
.
f * uA1(Hz) / uA1(1/m) = 1/w
f * uA1(Hz) / uA1(K) = K
f * uA1(Hz) / uA1(eV) = E (in eV)
.
E * uA1(eV) / uA1(1/m) = 1/w
E * uA1(eV) / uA1(Hz) = f
E * uA1(eV) / uA1(K) = K
.
K * uA1(K) / uA1(1/m) = 1/w
K * uA1(K) / uA1(Hz) = f
K * uA1(K) / uA1(eV) = E (in eV)
Where:
K = Kelvin
.
Have you ever seen anything like this before? Do you understand what I'm showing? Can you see any problems with what I've presented so far?

 

[Acitnoids]

I could be taking this the wrong way but am I being marginalized for some undisclosed reason? Am I being seen as some sort of disgruntled crank for suggesting that a photon should actually equal one number instead of a multitude of units even though those units can be derived from that one number using this system?
.
(1/w) * uA1(1/m) = f * uA1(Hz) = E * uA1(eV) = K * uA1(K)
.
I posted in this sub-forum because my system had never been peer reviewed and I thought that was a prerequisite for the hard science forums. I fear that if this conversation does not run a course it may become fodder for hacks. That is my intention. I would rather this be an example of how people can talk through an unsubstantiated concept with the purpose of learning. I am not a physicist. I shape steel for a living and I'm damn proud of it. So, I want to ask again.
.
Have you ever seen anything like this before? Do you understand what I am trying to show? Can you see any problems with what I've presented so far?