Expanding confusion: Time integral of the reciprocal of the scaling factor from Hubble parameter equation

[andrew s]

It is always normalised about 1. In the region about now. It is continuous on 0 to infinity.

To make matters concrete let's fix now to epoch 2000 and assume t = 13.6 Gyr then. Then a0 = a(13.6Gyr) =1.

Now consider light from andromeda arriving in the year 2000 tr = 13.6 Gyr a(tr) = 1

It was emitted ~2.5 Myr prior to 2000 so te = 13.6 Gyr - 2.5 Myr  when a(tr) = 1 -  every small amount say 0.999999 (made up number)

So (1+ z) = 1/0.999999 ~ 1 i.e we would not detect any red shift due to expansion but only peculiar velocity in fact it's blue shifted as its moving towards us.

So to answer your question yes these points have physical meaning but the effect are very small until you get to large distances. Indeed Hubble' 1929 data showed significant scatter due to the relatively close proximity of the galaxies he measured and their peculiar motion.

Regards Andrew 

Edited August 12, 2023 by andrew s

[kurdewiusz]

24 minutes ago, andrew s said:

It is always normalised about 1. In the region about now. It is continuous on 0 to infinity.

Normalized value is always <=1. It's not normalized for > 1. Conclusion - it's not always normalized. a0 is fixed. My previous objections remain exactly the same.

"In the region about now" - what you say here, is that 1 = 1.

Edited August 12, 2023 by kurdewiusz

[andrew s]

2 minutes ago, kurdewiusz said:

Normalized value is always <=1. It's not normalized for > 1. Conclusion - it's not always normalized. a0 is fixed.

Sorry that totally wrong. 

Let's take a simple example my hight. 

When born I was about 1.7 ft tall at 12yrs I was 5 ft and at 25yrs 6 ft.

Now I decide to scale myself about age 12 so the scale the a0(12) = 5/5 = 1. At birth my scaled hight was a(0.001) = 7/5 = 0.34 and at 25 yrs a(25) = 6/5 = 1.2.

Yes a0 is fixed but "a" varies continuously  about a0 = 1

As I age I will srunk but the Universe is not expected to.

Regards Andrew 

[kurdewiusz]

@andrew s I get it, really, but in any case, that doesn't solve my problem. Let me repeat:
I am asking you about the point (Now, 1) and its proximity, where normalized a0/1 becomes unnormalized a(tr)/a(te). Consider the point that lies on the curve and its coordinates are (Now+dt, 1+da), where  dt and da are infinitesimal and positive. Normalized and unnormalized values need to be equal for a=1. How would you explain the fact, that for a=1.00000001 the values of a(tr) and a(te) are practically equal, since we're already in unnormalized range for a > 1.

[George Jones]

@kurdewiusz, in your opinion, is the Friedmann (differential) equation of standard cosmology correct?

[kurdewiusz]

YES
@George Jones In my opinion, the source of inconsistency between general relativity (which is supposedly a basis of calculations giving 46.5 GLy and 3.2 c values) and the Doppler is in calling the integration-based calculated recession velocity (as well as the proper size of the universe) a part of GR. Friedmann equations, as a solution of Einstein's equations, are a part of GR. Friedmann–Lemaitre–Robertson–Walker metric is a part of GR. Explicit form of the scale factor, derived from the Friedmann equations, is a part of GR. In my opinion, what is not a part of GR, is the integration of the general metric that gives the current, proper distance. This metric equation uses the explicit form of the scale factor derived from GR and may be valid for every single spacetime frame, but if you integrate it, you get this: https://physics.stackexchange.com/questions/769308/ - DELETED / BANNED.

Edited August 12, 2023 by kurdewiusz

[George Jones]

Okay, I have much more important things to do, like watch the first Saturday of Premier League football.

[kurdewiusz]

@George Jones you asked, not my fault you don't like it

[George Jones]

@kurdewiusz, I have largely stopped arguing with people with very non-standard physics views. I had too many bad experiences at Physic Forums, where I was a moderator for years. I will let others here carry the torch.

[andrew s]

Ok one very last attempt.  Note a(tr) and a(te) are the normalised scale factors at the time the light is received and emitted  respectively. 

Looking at my height about age 12 for a small positive absolute increase in time my hight would be 5 + dh ft say and my scaled hight would be  1 + (dh/5)

Looking at a change in hight about age 25 for a similar absolute change my hight would change to 6 + dh and my scaled hight would change to 1.2 + (dh/5)

I have done my best not sure I can make it any clearer. 

Regards Andrew 

[kurdewiusz]

@George Jones i get it. Thank you for your neutral attitude and your choice of words - my respect.

[kurdewiusz]

@andrew s a is normalized for a<=1, a(tr) may be normalized, a(te) may be normalized, their quotient a(tr)/a(te) is > 1, therefore not normalized. And the scale factor function (on the plot given in this post) ranges from 0 to infinity for time as well as for value, therefore it's not normalized.

Normalized and unnormalized values need to be equal for a=a0=1. You didn't explain the fact, that for a=1.00000001 the values of a(tr) and a(te) are practically equal, since we're already in unnormalized range for a > 1.

z+1 = a0/a = a(tr)/a(te) = 1 for a=1

It just ocurred to me, that i explained it, but as a mistake, in the description of my drawing: "then the photons reaching us would have the original wavelength form the time of their emission".

Edited August 12, 2023 by kurdewiusz

[andrew s]

A final try. I have tried to show normalised values can be greater than one e.g. my age at 25.

I think you are hung up on the term normalised.  A better term would be scaled.

I scaled my age to that at age 12. This led to my scaled age at 0.001 being less than 1 and at age 25 greater than one. That's all ther is to it.

Make a cosmic measurement now (say x meters) and its scale factor is defined as 1 i.e. (x/x = 1).

If the same measurement had been made in the past the actual measurement in meters would be smaller (say y m) the scale factor would be such that y/x < 1.

If made in the future you would get the longer distance (say z m) and the scale factor would be z/x> 1

Say your light ray was emitted at the early time te and received at the later time tr then

(1+z) = a(tr)/a(te) = (z/x)/(y/x) = z/y > 1

Regards Andrew 

Edited August 12, 2023 by andrew s

[kurdewiusz]

z+1 = a0/a = a(tr)/a(te) = (z/x)/(y/x) = z/y = 1 for a=1

a(tr) = (z/x) = a(te) = (y/x) for a = 1

z/y = 1 for a = 1

Changes nothings, adds nothing, explains nothing, a(tr) is still equal to a(te) for a = 1.

[andrew s]

16 hours ago, kurdewiusz said:

z+1 = a0/a = a(tr)/a(te) = (z/x)/(y/x) = z/y = 1 for a=1

a(tr) = (z/x) = a(te) = (y/x) for a = 1

z/y = 1 for a = 1

Changes nothings, adds nothing, explains nothing, a(tr) is still equal to a(te) for a = 1.

Absolutely if you emit a light beam now and detect it now + deltat then it will have travelled c×deltat a very short distance in a very short time and to zeroth order a(tr) = a(te) = 1 but to first order a(tr) = a(te)  + deltat.da(te)/dt

I have done my best to explain this to you.

 a(tr) = a(te) if and only if te = tr I.e they are at the same time. If the time are different the they are not equal. 

You seem to be stuck on a basic fact of calculus.  I have tried my best.

Good luck. I am retiring from the discussion as I don't seem able to help you.

Regards Andrew 

[kurdewiusz]

Please...

For a = 0.99999, te = 0 and tr is the age of the universe. For a = 1.00001, te becomes equal to tr, that is the age of the universe, or both te and tr become almost zero, that is the time of the emission. You change the difinitions of te and tr in a = 1 to explain the equality of a(te) and a(tr), but it's still the same function. If it's the same function, you can't change the definitions of time parameters.

Edited August 12, 2023 by kurdewiusz

[kurdewiusz]

5 hours ago, andrew s said:

No.

a(te) is the scale factor for when the light was emitted I.e. in our past. a(tr) is the scale factor at the time it is received. 

That was your answer to my question, how is it possible, that a(tr)=a(te). Back then you had the old definitions of te and tr in mind, but you changed them.

[andrew s]

15 minutes ago, kurdewiusz said:

Please...

For a = 0.99999, te = 0 and tr is the age of the universe. For a = 1.00001, te becomes equal to tr, that is the age of the universe, or both te and tr becomae almost zero, that is the time of the emission. You change the difinitions in a = 1 to explain the equality of a(te) and a(tr), but it's still the same function. If it's the same function, you can't change the definitiona of time parameters.

At te =0 a = 0  Just look at the scale on the right hand scale of the y axis in the Figure 1 that I posted.

The CMB we see today has a red shift of about 1100 so 

(Z+1) = 1101 = a(tr=13.6 Gyr)/a(te) = 1/a(te) 

a(te) = 1/1101 ~ 0.001 and t ~ 0

I don't think I changed anything tr is always the time received and te the time emitted. 

This is my last reply. Good evening. 

Regards Andrew 

[kurdewiusz]

te=0 for both a=0 and a=0.99999, it's not a variable, but constant. What you wanted to teach me is that t=0 for a=0, no need.

[kurdewiusz]

5 minutes ago, andrew s said:

I don't think I changed anything tr is always the time received and te the time emitted. 

If you don't change the definition of te or tr in a=1, you will never get the equality a(te)=a(tr), but you need it in a=1.

[andrew s]

Ok I lied. te is the time light is emitted. We can observer today light from the CMB emitted at te ~ 0 through any value up to now te ~ 13.6 Gyr since the Big Bang.

For light we see today te only depends on how faraway it was when emitted. It certainly is not a constant.

The end.

Regards Andrew 

PS I lied about not responding again nothing else!

PPS "If you don't change the definition of te or tr in a=1, you will never get the equality a(te)=a(tr), but you need it in a=1."

No no no. For observations now a(tr) = 1, a(te) depends on when the light was emitted. a(te) = a(tr) if and only if they occur simultaneously. 

Edited August 12, 2023 by andrew s

[kurdewiusz]

Bye

Edited August 12, 2023 by kurdewiusz

[kurdewiusz]

Indeed, you need tr=te for a(tr)=a(te) for a=1.

[kurdewiusz]

@andrew s You say, that te is variable. In the observable universe size calculation, that is the integration of the reciprocal of the scale factor function a(t), te is the time of the emission of CMB and tr is the variable, that is Now.

[andrew s]

45 minutes ago, kurdewiusz said:

@andrew s You say, that te is variable. In the observable universe size calculation, that is the integration of the reciprocal of the scale factor function a(t), te is the time of the emission of CMB and tr is the variable, that is Now.

Correct. 

The CMB is the furthest back we can see at a red shift of about 1100 now. 

So for an observation now we have a(te) ~ 0.001, te = 0 and a(tr) =1, tr =Now

Where a(te) is as I calculated before.

Regards Andrew 

PS here is a plot of a(t) based on the Plank18 final mission data from here.

 

Edited August 13, 2023 by andrew s