Universe Expansion

[David Levy]

You really need to understand these equations before you go plugging numbers in to make a point - but even before that you need to have a feeling from the maths of whereabouts your answers must lie.

 

 

I hope that I have understood Mordred explanation correctly.

If so, my mathematical calculation should be perfectly O.K.

In any case;

Mordred - I would mostly appreciate to get your feedback.

Do you see any mathematical error in my calculation?

Edited February 10, 2016 by David Levy

[Strange]

 

Do you see any mathematical error in my calculation?

 

I can't understand what you have done, but it looks wrong and the result seems to be wrong.

 

If I stick "(3 * (13249 km/sec/Megaparsec)^2 / (8 * \pi * G) ) * c^2 in joules / m^3" into Wolfram Alpha, I get 2.964 x 10^-5 J/m³

http://www.wolframalpha.com/input/?i=%283+*+%2813249+km%2Fsec%2FMegaparsec%29^2+%2F+%288+*+\pi+*+G%29+%29+*+c^2+in+joules+%2F+m^3

Please also be aware that I have only used the total Mass/energy of the visible universe, So the real density of the whole Universe should be even higher than 1.57 Joules/ m³.

 

If you have used the mass-energy of the visible universe in the volume of the visible universe then the density should be the same for the whole universe.

[Mordred]

Yes those pointed out by Imaatsfal for starters. However you also are still misunderstanding.

 

H compared the H today....

 

This means H then is a multiple of H compared to H today.

 

Today's H is 67.9 km/sec/Mpc

H then is 67.9 km/s/Mpc* 23257.149

 

Now stop and think H is the "rate of expansion."

 

In other words the universe is expanding at (67.9 km/s/Mpc*23257.149) at z=1100.

 

So ask yourself why are you calculating the rate of expansion when you already have the rate of expansion?

 

(67.9 km/s/Mpc*23257.149) at z=1100.

[Strange]

Today's H is 67.9 km/sec/Mpc

H then is 67.9 km/s/Mpc* 23257.149

 

Doh. Missed that. So now Wolfram Alpha gives me 0.42 J m^-3

Is that more like it?

 

 

David, I recommend Wolfram Alpha when your math skills are not great (like me!) - it understands all sorts of symbolic representations which saves having to look things up. In fact, I could probably have put H₀ in there...

 

Yep: "(3 * (23257.149 * H_0)^2 / (8 * \pi * G) ) * c^2 in joules / m^3" works. Cool.

http://www.wolframalpha.com/input/?i=%283+*+%2823257.149+*+67.9+km%2Fsec%2FMpc%29^2+%2F+%288+*+\pi+*+G%29+%29+*+c^2+in+joules+%2F+m^3

[imatfaal]

 

 

I hope that I have understood Mordred explanation correctly.

If so, my mathematical calculation should be perfectly O.K.

In any case;

Mordred - I would mostly appreciate to get your feedback.

Do you see any mathematical error in my calculation?

 

 

FFS - you lecture us on complex physics and cannot even do sums!

 

You put this

 

[latex] \rho_c=\frac{3.H^2}{8.\pi.G} [/latex]

 

with H as 2.3 x10^4 (more on this later)

 

- the other values are well known.

 

But you got 3.2 x10^-1

 

With your figures the answer is from Wolfram Alpha - is 9.5 x 10^17 - You are about a billion billion time out.

 

I made this about as plain as I could with an order of magnitude approximation and you still have the brass neck to claim you are correct.

 

What's worse is your formulation of the sum was completely wrong because you didnt have the physics knowhow to realise you were using a completely erroneous figure of H - which is why even the "correct" mathematical answer is way way out.

 

=====

 

Moving on to the physics - your figure of H is taken from a table which clearly says it is the ratio of H to H0; that is to say the ratio of the H then to the H now. That is why it is unitless - it is a ratio - it is also why the current figure is 1.

 

=====

 

Have a bit of humility and learn some basic maths and physics before attempting to lecture the members on the finer points of astrophysics

 

@Strange

 

I think you need to have Hubble's parameter in its base SI form - ie second^-1 to give it as a density; but not entirely sure how well Wolfie A handles mad unit transformations. edit It turns out brilliantly - I hadn't realised you could do what you did

 

The critical density is currently calculated at about 10^-26 kg/m3 - an increase of H by a factor of 23,000 would lead to an increase in the critical density of about 500 million

o/t

Chapeau to Wolfram Alpha for its amazing ability with units

[Strange]

@Strange

 

I think you need to have Hubble's parameter in its base SI form - ie second^-1 to give it as a density; but not entirely sure how well Wolfie A handles mad unit transformations

 

The critical density is currently calculated at about 10^-26 kg/m3 - an increase of H by a factor of 23,000 would lead to an increase in the critical density of about 500 million

 

If I get it to display the result in kg/m³, it shows it is spot on:

"(3 * (23257.149 * 67.9 km/sec/Mpc)^2 / (8 * \pi * G) ) in kg / m^3" = 4.684x10^-18 kg/m³

 

Brilliant!

o/t

Chapeau to Wolfram Alpha for its amazing ability with units

 

I am constantly amazed at the things you can "get way with". I have used things like "mass of Earth" or "distance to moon" in equations before now.

Edited February 10, 2016 by Strange

[imatfaal]

 

If I get it to display the result in kg/m³, it shows it is spot on:

"(3 * (23257.149 * 67.9 km/sec/Mpc)^2 / (8 * \pi * G) ) in kg / m^3" = 4.684x10¹⁸ kg/m³

 

Brilliant!

 

I am constantly amazed at the things you can "get way with". I have used things like "mass of Earth" or "distance to moon" in equations before now.

 

I will take a tiniest tiniest bit of credit for some that. I filled in a survey many years ago as an early user of Wolfram Alpha and my big complaint was that whilst it understood pi it didnt parse any physical constants - my two bugbears were G and h - I got a lovely chatty message back saying that so many respondents had asked for that implementation and it was in the works.

 

My other grumble was that if you used a major European city in a query it always defaulted to the American version - and it still does! Ask about Westminster and it asked you Westminster Colorado (been there Phi?) or Westminster California - and I every time I want to scream at the screen

Edited February 10, 2016 by imatfaal
and that will be a ten to the minus eighteen :-)

[Mordred]

Nice if you guys think that tables neat the formula used to calculate that tables presents some other neat details.

 

 

[latex]H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}[/latex]

 

Take a closer look at the terms under the square root.

 

Notice that the density of matter and the density of radiation change at a different rate this is due to having different equations of state.

 

Side note the Cosmological constant doesn't change.

David take note as that should indicate how matter and radiation evolve to calculate total density

[Strange]

I like the fact that a really short and simple post highlights how inadequate David's naive extrapolations are.

[Phi for All]

My other grumble was that if you used a major European city in a query it always defaulted to the American version - and it still does! Ask about Westminster and it asked you Westminster Colorado (been there Phi?) or Westminster California - and I every time I want to scream at the screen

 

I've done a lot of business in Westminster, CO. Fifteen minutes from me. And no way does it deserve a preferential query over a district in London. Our Westminster doesn't even have their own palace (The Royale Palace is actually an events center for quinceanera parties, sorry).

 

Shame on Wolfram. We're awesome, but Earth shouldn't default to America.

[David Levy]

Yes those pointed out by Imaatsfal for starters. However you also are still misunderstanding.

H compared the H today....

This means H then is a multiple of H compared to H today.

Today's H is 67.9 km/sec/Mpc
H then is 67.9 km/s/Mpc* 23257.149

Now stop and think H is the "rate of expansion."

In other words the universe is expanding at (67.9 km/s/Mpc*23257.149) at z=1100.

So ask yourself why are you calculating the rate of expansion when you already have the rate of expansion?

(67.9 km/s/Mpc*23257.149) at z=1100.

 

Thanks for the update.

So, if I understand you correctly:

 

The critical density formula is:

So, lets set the following:

 

 

Pc = K * H²

 

K = 3 / (8 * 3.14 * G)

So, K could be considered as constant.

 

Now you claim that:

 

Today's H is 67.9 km/sec/Mpc

 

Based on that H, the value of the TODAY critical density is

"If you calculate this out it will work out to roughly to "

 

 

Therefore,

 

Pc (for today) = K * H² * = K * (67.9)² = 6 * 10^-10 J/m³

 

H then is 67.9 km/s/Mpc* 23257.149

 

Pc (when z=1100) = K * H² * = K * (67.9 * 23257)² = K * (67.9)² * (23257)² J/m³

 

However:

 

I have proved that:

 

K * (67.9)² = 6 * 10^-10 J/m³

 

 

Therefore:

 

Pc (when z=1100) = K * H² * = K * (67.9 * 23257)² = ( K * (67.9)² ) * (23257)² = 6 * 10^-10 * (23257)² J/m³

 

Hence: the updated maximal critical density is:

Pc (when z=1100) = 6 * 10^-10 * (23257)² = 6 * 10^-10 * 5.3 * 10⁸ = 3.18 * 10^-1 Joules/ m³

 

Do you agree???

 

[Mordred]

Ok the 6.0*10-10 value was a calculation done when it was believed H was 61 km/s/Mpc. This wouldn't have mattered if you simply plugged the numbers directly into the formula as Strange showed you.

 

If you want to do the method your using (why I don't know lol but I'll play along.) You would get less risk of rounding errors by using the formula ah well.

 

Using 67.9 into the formula gives you 7.78*10^-10

 

Now if you use 23257.149 instead of 23257 you get 540894979.6.

 

By using the full values (or had you just plugged the numbers in to Wolfram as Strange did)

 

You will end up at 0.421 j/m^3

 

At least your answer is in the right ballpark.

 

Quite frankly the minor difference is negligible. Datasets vary depending on the latest results in particular fine tuning Hubbles constant. Most textbooks typically just use 100 km/s/Mpc.

 

WMAP has a different value. 2013 WMAP H is 69.8 but if you use WMAP values

H/H_0 is 22289.979 ( has different densities for radiation and matter) using the formula you get 0.4087 j/m^3

Edited February 11, 2016 by Mordred

[David Levy]

At least your answer is in the right ballpark.

 

Thanks for the support. I do appreciate it.

 

By using the full values (or had you just plugged the numbers in to Wolfram as Strange did)

You will end up at 0.421 j/m^3

 

WMAP has a different value. 2013 WMAP H is 69.8 but if you use WMAP values

H/H_0 is 22289.979 ( has different densities for radiation and matter) using the formula you get 0.4087 j/m^3

 

O.K.

Let's agree that the maximal value of the critical density of the early universe ((when its temp was 3000K) is 0.421 j/m^3

However I have proved that the density of that early Universe is:

4 × 10⁶⁹ / 2.54 * 10⁶⁹ m³ = 1.57 Joules/ m³

 

This is based on the total mass/energy of the visible Universe.

Never the less, it is clear that the total mass mass/energy of the whole Universe should be higher.

Based on the following article the current size of the Universe is 46 BLY.

http://www.space.com/24073-how-big-is-the-universe.html

"Thus, while scientists might see a spot that lay 13.8 billion light-years from Earth at the time of the Big Bang, the universe has continued to expand over its lifetime. Today, that same spot is 46 billion light-years away."

Therefore, the volume of the current universe is bigger by 37.037 times from the visible Universe:

X = R³ (radius of whole Universe) / R³ (radius of the Visible Universe) = ( R(radius of whole Universe) / R(radius of the Visible Universe) ³

= (46 / 13.8) ³ = 37.037

 

Hence, the density of the early universe based on the total mass/energy of the whole universe is:

1.57 x 37.037 Joules/ m³ = 58.148 Joules/ m³

That value is higher by 138 times than the calculated critical density for that time – which is: 0.421 Joules/ m^3.

As follow: 58.138 / 0.421 = 138

Please also be aware that the information about the mass/energy of the visible universe had been taken from the following site:

 

http://www.physicsoftheuniverse.com/numbers.html

 

I'm not sure if it is fully updated.

Just few years ago, the science have considered that the total galaxies in the universe were only 100 or 200 Billion. Now they already claim that it is about 500 Billion. They also had no clue that 50% of the stars are located outside the galaxies.

 

So, there is good chance that the total mass/ energy of the visible universe is higher than what it is stated at that article.

In any case, even without this help, the density of the early universe is higher by at least 138 times than the critical density.

So, do you agree that it is quite clear that the Expansion can't work at the early Universe?

Edited February 12, 2016 by David Levy

[Mordred]

I don't know why you keeping saying the early universe can't expand especially since you and I both know it did.

 

Especially since the equation we used above shows the expansion rate

 

H/H_O

 

Seriously you really need to realize your doing something wrong especially since the equation used the expansion rate at the time of the CMB.

Seriously David...

 

The early universe was 40Mly. The universe today is how big?

 

Does it make sense to say the early universe can't expand. When the universe today is bigger than the universe then??????

 

Come on use a little common sense.

Expansion rate at CMB is 540894979.6 km/s/Mpc. 67.9*H/H_O.

[David Levy]

I don't know why you keeping saying the early universe can't expand especially since you and I both know it did.

Especially since the equation we used above shows the expansion rate

H/H_O

Seriously you really need to realize your doing something wrong especially since the equation used the expansion rate at the time of the CMB.

Come on use a little common sense.

 

 

Well

It's not an issue of "know", "believe" or "little common sense"...

It is a simple mathematics.

The science gave us basic information about the early science and mathematical formulas.

It is quite simple to calculate the density and the critical density.

Somehow, based on those formulas and information it is clear that the expansion can't work.

We have to ask ourselves –

A. Why a simple calculation had proved that the expansion is not feasible?

B. How could it be that the whole science had missed this critical issue?

Therefore, there are three options -

1. There are errors in the basic information. So, for example, the value of H/H_0 which is considered to be 22289.979 (by WAMP) should be quite higher.

2. There are errors in the formulas. So, for example, a different mathematical formula should be used to calculate the critical density.

Seriously David...

 

The early universe was 40Mly. The universe today is how big?

Does it make sense to say the early universe can't expand. When the universe today is bigger than the universe then??????

3. Yes, it makes sense. The expansion can't work because it is an unrealistic hypothetical theory. Therefore, the science should look for better explanation for what we see.

Edited February 12, 2016 by David Levy

[Mordred]

To add details to Strange answer you can calculate where expansion will overcome gravity.

 

Step one calculate the strength of the cosmological constant.

 

For this you use the critical density formula.

 

[latex] \rho_c=\frac{3H^2}{8\pi G}[/latex]

 

 

If you calculate this out it will work out to roughly to [latex] 6.0 *10^{-10} joules/metre^3[/latex]

 

or alternatively [latex]10^{-26} kg/m^3[/latex]

 

Then you calculate the strength of gravity at a given radius from a mass.

 

[latex]F=\frac{GM_1m_2}{r^2}[/latex]

 

Convert newtons to joules.

 

When the critical density becomes greater than the force of gravity. Expansion takes over.

 

This example here is "if you take a probe fly it from a large scale structure in the universe today...

 

The energy of expansion will overcome gravity from that large scale structure.

 

 

The critical density formula uses the rate of expansion. Which is the value H. Hubbles constant.

 

[latex]\Omega=\frac{\Omega_{total}}{\Omega_{critical}}[/latex]

 

Does not determine the expansion rate of the universe but is used to determine the curvature constant

Edited February 12, 2016 by Mordred

[David Levy]

This example here is "if you take a probe fly it from a large scale structure in the universe today...

The energy of expansion will overcome gravity from that large scale structure.

The critical density formula uses the rate of expansion. Which is the value H. Hubbles constant.

[latex]\Omega=\frac{\Omega_{total}}{\Omega_{critical}}[/latex]

Does not determine the expansion rate of the universe but is used to determine the curvature constant

 

Dear Mordred

 

Please try to answer directly.

Do you see any error in my calculations?

 

If so, please highlight the errors.

If you don't see any error - then you have to agree that the expansion can't work!

[swansont]

 

We have to ask ourselves –

A. Why a simple calculation had proved that the expansion is not feasible?

B. How could it be that the whole science had missed this critical issue?

 

I'm putting my money on "the simple calculation is flawed"

[Mordred]

 

 

Well

It's not an issue of "know", "believe" or "little common sense"...

It is a simple mathematics.

The science gave us basic information about the early science and mathematical formulas.

It is quite simple to calculate the density and the critical density.

Somehow, based on those formulas and information it is clear that the expansion can't work.

We have to ask ourselves

A. Why a simple calculation had proved that the expansion is not feasible?

B. How could it be that the whole science had missed this critical issue?

Therefore, there are three options -

1. There are errors in the basic information. So, for example, the value of H/H_0 which is considered to be 22289.979 (by WAMP) should be quite higher.

2. There are errors in the formulas. So, for example, a different mathematical formula should be used to calculate the critical density.

 

 

3. Yes, it makes sense. The expansion can't work because it is an unrealistic hypothetical theory. Therefore, the science should look for better explanation for what we see.

This full statement doesn't work...

 

We have observable proof that expansion has happened, redshift alone tells you expansion occurs. However we don't rely on just redshift. The temperature change from CMB to now tells us expansion has occurred. Interstellar parallax and intergalactic parallax tells us expansion is occurring.

 

Your misguided calculation tells us you can't do math...

 

What part of the formulas your using, uses the expansion rate in those calculation don't you understand?

 

Earlier this thread I provided the formula to calculate the rate of expansion.

 

The acceleration equation is given as

[latex]\frac{\ddot{a}}{a}=-\frac{4\pi G\rho}{3c^2}(\rho c^2+3p)[/latex]

This leads to

[latex]H^2=\frac{\dot{a}}{a}=\frac{8\pi G\rho}{3c^2}-\frac{kc^2p}{R_c^2a^2}[/latex]

 

In other words your not using the right formulas. The rest of the math is moot because your using the wrong procedure and formulas

This formula is a short cut method, it is derived from the last set of formulas (through many steps in particular the equations of state)

 

[latex]H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}[/latex]

 

The advantage of the last derivitive is that were using already previously calculated and measurement confirmed values.

Edited February 12, 2016 by Mordred

[David Levy]

Mordred

Just a simple question -

Please...

Do you see any error in my calculation?

The sun will not fall over our head if my calculations are correct.

Maximum - WE will get some sort of reward from the science community...

Edited February 12, 2016 by David Levy

[Mordred]

I already explained The errors wrong formula usage.

 

This value for one is incorrect your scaling down density without including the equations of state for matter and radiation.

 

1.57 x 37.037 Joules/ m3 = 58.148 Joules/ m3

 

you didn't catch the hint from under the square root in this equation.

[latex]H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}[/latex]

 

Note matter and radiation density changes at different rates.

Edited February 12, 2016 by Mordred

[David Levy]

I already explained The errors wrong formula usage.

This value for one is incorrect your scaling down density without including the equations of state for matter and radiation.

1.57 x 37.037 Joules/ m3 = 58.148 Joules/ m3

 

Thanks

 

Your reply isn't fully clear to me.

 

You claim that the following calculated density of the early Universe (based on the whole Universe) is incorrect:

1.57 x 37.037 Joules/ m³ = 58.148 Joules/ m³

 

So, let me start by asking the following:

Do you agree that the density of that early Universe (based on the Visible Universe) is:

4 × 10⁶⁹ / 2.54 * 10⁶⁹ m³ = 1.57 Joules/ m³

 

you didn't catch the hint from under the square root in this equation.

[latex]H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}[/latex]

Note matter and radiation density changes at different rates.

 

What is the source for this mathematical formula?

In your explanation about the density (earlier) you didn't mention it at all.

Why suddenly it pop up?

Is it just because that the results are different from your expectations?

Can you please explain how shall we use it?

Would you kindly set the calculation, or let me know what should be the expected density based on the whole Universe mass/energy?

Edited February 12, 2016 by David Levy

[Strange]

What is the source for this mathematical formula?[/size] [/size]

 

https://en.wikipedia.org/wiki/Friedmann_equations

 

Why suddenly it pop up?

 

It was first mentioned in this thread three days ago.[/size] [/size]

[David Levy]

Sorry.

This formula doesn't contradict my calculation.

 

https://en.wikipedia.org/wiki/Friedmann_equations

 

"An expression for the critical density is found by assuming Λ to be zero (as it is for all basic Friedmann universes) and setting the normalised spatial curvature, k, equal to zero. When the substitutions are applied to the first of the Friedmann equations we find:

The density parameter (useful for comparing different cosmological models) is then defined as:

Omega = P / Pc

 

As you can see, Omega represents the ratio between the density to critical density. It has no effect on my calculation!!!

 

So, it is clear (to me) that:

A. The density of the early Universe (based on the Visible Universe) is:

4 × 10⁶⁹ / 2.54 * 10⁶⁹ m³ = 1.57 Joules/ m³

B. The density of the early Universe (based on the whole Universe) is:

1.57 x 37.037 Joules/ m³ = 58.148 Joules/ m³

 

Do you agree with A?

If no, Why?

If yes, why you do not agree with B?

 

If you still don't agree, please let me know what should be the density based on your calculations?

Edited February 12, 2016 by David Levy

[Mordred]

David you cannot directly take the mass density of the full universe and directly convert it to Joules then state that it will be the energy density in the past . You must apply the equations of state for each particle species matter changes in energy density at a different rate than radiation.

 

The equations of state is in the link Strange posted.

Shrink the universe enough and all particles become relativistic hence radiation.