Is the Hubble Shift a relativistic illusion?

[captcass]

Sorry, the graph was not a response to your post above it, I put it up before I saw your post. Then a tree hit a pole and I lost power last night. So....

"Really not sure how you drew this conclusion from the time dilation formula. Yes time runs slower in a stronger gravitational potential. However time dilation has nothing to do with seperation distance, except length of time for a signal between events to reach each other"

 

Using the time dilation formula, the difference in rates decreases with distance. This means time is slowing over distance. until it again = 1 s/s of our inertial frame at infinity. Going slower than what? From where? The top of our heads? It can't slow from 1 s/s to become 1 s/s.

 

"I will always see my own time as the fastest. I will also see my own length as the longest."

 

This is for motion. At the center of a gravity well we see our own time as slowest. If you are at the bottom of Mt. Everest and a buddy is at the top, both will agree his clock is going faster. If you double the height, time will still appear to be going faster, but the difference in the rates halves, so time is going slower.

 

This is the conundrum. How can time go faster slower with altitude? Time does not go faster and faster with altitude (distance), it goes slower and slower until 1 s/s = 1 s/s.

 

If time is slower, equaling 1 s/s at infinity, then time is going faster as the distance to us shortens. How can this be in a gravity well, where time is slowing as the well deepens?

Likewise, from infinity time slows until 1 s/s = 1 s/s in our inertial frame. Slows from what?

 

This is the head twister. The answer is that there are 2 mirror gradients that equalize the rate of time to 1 s/s in any inertial frame, so we are always experiencing 1 s/s.

 

I fully realize how difficult this is to visualize. That is because we are always limited to our 1 s/s perception and our position at the bottom of the gravity well that always makes time to appear to be going faster with altitude. But if we stand outside the gradients, an observer placed equilaterally at infinity, we would see the interplay of both gradients. This is what the graph shows.

 

The 2 s/s apparent rate within the time vortices of our quanta decreases to 1 s/s at infinity as the rate of time from infinity decreases in the gravity well until our 1 s/s perceived rate in our inertial frame. This has to be a reciprocal relationship, so if the perceived rate within our quanta is 2 s/s, then it also has to be 2 s/s at infinity.

 

The proof is in the derivations of the dRt's to and from the midpoint (dRt = .5), which mirror themselves. We do not get .5 s/s at that point using the time dilation formula from us, we get a much smaller dRt. But we get that same dRt halving .5 from the midpoint to our inertial frame.

 

"m/s is a unit of velocity. Not a unit of time. Unit for time is seconds."

 

If a body has a velocity of so many m/s, are not so many seconds passing per meter of motion? s/m? But I am not talking about the evolution of a body moving through preexisting space. I am talking about space itself, and the apparent events within that space, evolving forward over time. How the spatial aspect evolves forward over time.

 

"That graph is meaningless. Go ahead devide by infinity. Why do you think infinity has 1 s/s in that graph? as you approach infinite redshift time gets slower and slower. It doesn't jump back to being in the same frame as frame zero"

 

Again, there are two mirror gradients. If time is slowing in a gravity well to 1 s/s, what rate is it slowing from?

 

Wine, anyone?

 

Sorry, got to go to work now....

[Mordred]

Your obviously using the time dilation formula incorrect if you have this conclusion...

 

 

"Using the time dilation formula, the difference in rates decreases with distance. This means time is slowing over distance. until it again = 1 s/s of our inertial frame at infinity. Going slower than what? From where? The top of our heads? It can't slow from 1 s/s to become 1 s/s."

 

The formula does not give 1 second/second compared by the at rest clock at infinity.

 

This is where you are wrong. See the graph I posted. When v equals c time essentailly stops. This is what the formula shows.

 

[latex]\acute{t}=\frac {t}{\sqrt {1-v^2/c^2}}[/latex]

 

set v=c. You will get 1-1 under the denominator. Which is equal to zero. This will give a divide by zero. Time becomes infinitely slowed down. Divide any number by zero. You won't get a value of one.

 

Secondly where do you see distance in that formula???? Distance doesn't affect time. Yes the gravitational potential decreases with distance but you only account for the difference in gravitational potential between observers.

 

If we place two observers onto a uniform homogeneous and isotropic mass distribution. It doesn't matter how close or far those two observers are. There is no time dilation in this case.

 

If one of the observers is in a stronger gravitational potential. While the other is in the GP of the background space. You only calculate the change between the two reference frames. The GP from the mass source will become essentially identical to the background GP at a certain distance. This background gravitational potential will stay constant as distance increases or until it hits another anistropy.

 

I think you better study the basic formulas again. Your making serious math errors.

Edited September 21, 2016 by Mordred

[captcass]

Thanks, I'll think about this at work and get back later. Off the top of my head, though, I would again say that you are referring to motion induced effects, not static effects.

[Mordred]

Thanks, I'll think about this at work and get back later. Off the top of my head, though, I would again say that you are referring to motion induced effects, not static effects.

No I'm discussing basic math. Which you evidentally got wrong on the time dilation formula. Unfortunately you wrote the entire 88 page article based on this math error....

Edited September 21, 2016 by Mordred

[captcass]

I am using the formula for a stationary, non-rotating body, where To = Tr*sq.rt.(1 - (Ro/r)), which is the distance aspect.This allows us to determine time vs distance. These derivations have nothing to do with velocity. Ro is where v has to = c in order to escape the gravity well. It is the event horizon of a black hole or a quanta.

 

Both black holes and quanta are time vortices. Just on a different scale. Approaching a quanta is the same as nearing a velocity of c or the event horizon of a black hole. In a quanta, the update is nearly instantaneously, every 4 - 16*10^-65 s, repeating updating temporal time in the local frame. It is moving the local frame forward in time at nearly c^2, thus creating an apparent rate of 2 s/s and an apparent dRt relative to infinity of 1 s/s. If there were no difference in the rates of time, there would just be the eternal moment and time would appear to stop. There would be no update. The screen would go dark. Eternity in an absolute void. Meet the Creator.

 

Traveling at c has the same effect, you get to the eternal moment in totally dimensionless ("flat") space.

 

Getting back to the graph:

 

2 s/s 2 s/s

l l

l l

l .5 s/s l

l l

l_1 s/s__________________________1 s/s l

Frame 0 (Frame 1) (near) oo

 

I mislabeled the Frames. Frame 0 is actually Frame 1, which is our inertial frame, and the other end is not infinity, but near infinity. Frame 0 is the center of the energy density, which is in each of the the quanta that make us up, where the difference in rates is 4-16*10^-65 s/s. Thus, even though time appears to be slowing towards the bottom of the gravity well, the rate of update increases as the dRt between the update and temporal evolution decreases, and that creates an impression of quickening time as regards the forward evolution of the continuum. These are the mirroring effects.

 

This is why I really appreciate your feedback, it makes me find better ways to express what I am seeing..

 

Anyway, it is not possible for time to stop. It is time's nature to move forward. Even in the eternal moment, the observer's rate of time would apparently be 1 s/s. This is why we cannot travel at C, nor reduce space to 0.

 

Back to the graph:

As distance always stretches out to infinity from us in all directions, we can be visualized as being at the midpoint where the dRt = .5. This equates to Frame 1, our perceptual frame, where 1 s/s = 1 s/s, not Frame 0 in our quanta.

If we are at a .5 dRt and we experience a 1 s/s rate, which we do, then at the extreme ends of the gradient, time must nearly be at a 2 s/s rate. The extreme ends are infinity and within the quanta within us.

 

Even though we can visualize this scenario, we are still at a 1 s/s rate and therefore experience the external universe as is portrayed by Frame 1 in the graph.

[Strange]

I am using the formula for a stationary, non-rotating body, where To = Tr*sq.rt.(1 - (Ro/r)), which is the distance aspect.

 

 

How is this formula derived?

 

And what evidence do you have that it is correct?

[captcass]

At https://en.wikipedia.org/wiki/Gravitational_time_dilation, look under "Outside a non-rotating sphere"

[imatfaal]

At https://en.wikipedia.org/wiki/Gravitational_time_dilation, look under "Outside a non-rotating sphere"

 

ok the formula there is gravitational time dilation under schwartzchild metric

[latex]

 

\frac{t_0}{t_f}=\sqrt{1-\frac{r_s}{r}}

[/latex]

 

t_f being time of fast-ticking at infinite distance

t_0 being proper time of slow-ticking observer

 

r_s being the schwartchild radius

r being distance (schwartchild coordinate) of the slow-ticking observer

 

what you have is To/Tr = √((1-(Ro/R))

 

[latex]

 

\frac{T_o}{T_r}=\sqrt{1-\frac{R_o}{R}}

[/latex]

 

For a non-rotating stationary Earth, To / Tr = √((1-(Ro/R)), where To / Tr is the ratio between times on the surface, To, and at distance R, Tr, and Ro is the Schwarzschild radius of the Earth...

 

So the Right Hand sides of both equations are the same - but the Left Hand sides are very different . The correct version in wikipedia is the ratio of time for a slow observer in a gravity well over that of fast-ticker at infinite distance (ie zero gravitational potential); whereas yours is the ratio of time on the earth's surface and time at the arbitrary distance r

[captcass]

In my last post I have it as to = Tr*sq. rt ((1-(Ro/r). Same difference.

Tr is the time at radius r, not infinity unless r is infinity. As in all equations things can be moved from one side to the other.

No wonder you think I'm talking gibberish, wrong formula.

Since we are talking gravitational dilation, I assumed you knew the formula, which I use in my paper.

This is why I have urged people to read it from the beginning so they understand the basic precepts and postulates I am using..

[imatfaal]

In my last post I have it as to = Tr*sq. rt ((1-(Ro/r). Same difference.

Tr is the time at radius r, not infinity unless r is infinity. As in all equations things can be moved from one side to the other.

No wonder you think I'm talking gibberish, wrong formula.

Since we are talking gravitational dilation, I assumed you knew the formula, which I use in my paper.

This is why I have urged people to read it from the beginning so they understand the basic precepts and postulates I am using..

 

Please point out in my post where I have mistaken a formula from your work. I have copied a formula and an explanation - and as far as I am concerned it does not match up with any I know

 

For a non-rotating stationary Earth, To / Tr = √((1-(Ro/R)), where To / Tr is the ratio between times on the surface, To, and at distance R, Tr, and Ro is the Schwarzschild radius of the Earth...

 

It is this bit - please provide a derivation or at least an explanation as to why it looks so like the standard gravitational time dilation formula with a couple of crucial changes. I will remind you that when Strange asked where the above formula came from you tersely directed him at wikipedia. Your formula does not appear in wikipedia. So what is it?

[captcass]

Sorry, didn't mean to appear terse. I was at work and didn't have time for a longer post and see no reason to go through derivations of formulas that are readily available online.

The formulas used by others here are correct for time dilation effects due to motion, but not gravitational dilation. The formulas for gravitational time dilation and their derivations can be found at the Wikipedia site I provided a link to above.

I'm confused as to why you think they are not there. Mordred has the formula in his last post.

I am simply using Ro for Rs (Schwarzschild radius). Is that the point of confusion? Ro is often used to denote the Schwarzchild radius.

This is not a formula used just for infinity, the "r" in the denominator is a variable used to determine the time difference at different distances. If the formula was just for infinity, "r" would always be infinity.

 

Does this help? I'm confused as to why you folk don't understand this.

[imatfaal]

I have engrossed the subscripts

 

[latex]\frac{t_{observer}}{t_{fastticker}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}}[/latex]

 

This is canonical version from Wikipedia based on the schwartzchild solutions of the EFE

 

 

[latex]\frac{T_{on the surface}}{T_{Observer_R}}=\sqrt{1-\frac{R_{schwartzchild}}{R_{observer}}}[/latex]

 

This is what you have written

 

 

 

Surely you can see they are NOT THE SAME

[mp][/mp]

 

Your problem seems to be that you think that the gravitational time dilation equation you have posted gives the time dilation between two arbitrary points at different levels of gravitational potential. It does not. It gives the amount that an observer's clock [proper time] is slower than the clock of someone at zero gravitational potential ie infinite distance [coordinate time]

[Mordred]

The other problem is neither the OPs equation. Nor the Schwartzchild equation can be used when the recessive velocities are greater than c. One should be able to calculate velocity of said object with the time dilation formulas.

 

the limit is the Hubble horizon for those formulas. Yet we measure recessive velocity up to 3.2 c at redshift 1100.

 

You can't derive that redshift via the Schwartzchild metric.

 

Even the cosmological redshift formula which includes an expanding volume requires corrections above Hubble horizon.

Edited September 22, 2016 by Mordred

[Strange]

I have engrossed the subscripts

 

[latex]\frac{t_{observer}}{t_{fastticker}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}}[/latex]

 

Just to make that really explicit:

 

[latex]\frac{t_{observer-at-r_{observer}}}{t_{observer-at-infinity}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}}[/latex]

[imatfaal]

 

Just to make that really explicit:

 

[latex]\frac{t_{observer-at-r_{observer}}}{t_{observer-at-infinity}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}}[/latex]

 

Yeah - I don't know how to do spaces in a subscript so I tried to keep it to one word t_fasttickeratinfinitedistance wasn't very intelligible

[Mordred]

 

Yeah - I don't know how to do spaces in a subscript so I tried to keep it to one word t_fasttickeratinfinitedistance wasn't very intelligible

Neither do I lol

 

Just to make that really explicit:

 

[latex]\frac{t<br>{observer-at-r_{observer}}}{t_{observer-at-infinity}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}}[/latex]

br doesn't work lol

Edited September 22, 2016 by Mordred

[imatfaal]

[latex] \frac{t_{observer \ at \ r}}{t_{fast \ ticker \ at \ infinite \ distance}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}} [/latex]


[latex] \frac{T_{on \ the \ surface \ of \ earth}}{T_{observer \ at \ R}}=\sqrt{1-\frac{R_{schwartzchild}}{R_{observer}}}[/latex]

Tada!

space backslash space does it for me

---

Mordred

On a more serious note - is the OP right to be using Schwartzchild Solution for the EFE in this circumstance. We are not talking about a single heavy mass and the vacuum solution of the space around

[Mordred]

[latex] \frac{t_{observer \ at \ r}}{t_{fast \ ticker \ at \ infinite \ distance}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}} [/latex]

 

 

[latex] \frac{T_{on \ the \ surface \ of \ earth}}{T_{observer \ at \ R}}=\sqrt{1-\frac{R_{schwartzchild}}{R_{observer}}}[/latex]

 

Tada!

 

space backslash space does it for me

 

---

Mordred

 

On a more serious note - is the OP right to be using Schwartzchild Solution for the EFE in this circumstance. We are not talking about a single heavy mass and the vacuum solution of the space around

No he should be using the Newton approximation for observers on Earth. Though that won't help him solve beyond Hubble limit without including an expanding volume. The Newton solution and cosmological redshift corrections are previously shown on this thread.

 

Lol very few people are aware the cosmological redshift formula isn't accurate beyond Hubble horizon.

Weins Displacement law has more to do with the original emitter frequency. Which correlates back to hotter temperatures in the past.

 

Anyways lets look at your questions above in more detail after.

 

lets look at the corrections to the redshift formula.

 

First we define a commoving field. This formula though it includes curvature (global) you can set for flat spacetime. A static universe is perfectly flat.

 

[latex]ds^2=c^2dt^2 [\frac {dr^2}{1-kr^2}+r^2 (d\theta^2+sin^2\theta d\phi^2)][/latex]

 

we write [latex](x^0,x^1,x^2,x^3)=(ct,r,\theta,\phi)[/latex]

 

we set the above as

[latex]g_{00}=1,g_{11}=-\frac{R^2(t)}{(1-kr^2)},g_{22}=-R^2 (t)r^2, g_{33}=-R^2 (t)r^2sin^2\theta [/latex]

 

the geodesic equation of the above is

 

[latex]\frac {du^\mu}{d\lambda}+\Gamma^\mu_{\alpha\beta}\mu^\alpha\mu^\beta=0 [/latex]

 

if the particle is massive [latex]\lambda[/latex] can be taken as the proper time s. If it is a photon lambda becomes an affine parameter.

 

So lets look at k=0.

 

we set [latex]d\theta=d\phi=0 [/latex]

 

this leads to

 

[latex]ds^2=c^2t^2-R^2 (t)dr^2=c^2dt^2-dl^2=dt^2 (c^2-v^2)[/latex]

 

where dl is the spatial distance and v=dl/dt is the particle velocity in this commoving frame.

 

Assuming it to be a massive particle of mass "m" [latex]q=m (\frac {dl}{ds})c=(1-\frac {v^2}{c^2})^{\frac{1}{2}}[/latex]

 

from the above a photon emitted at time [latex]t_1[/latex] with frequency [latex]v_1 [/latex] which is observed at point P at time [latex]t_0 [/latex] with frequency [latex]v_0[/latex]

 

with the above equation we get

 

[latex]1+z=\frac {R (t_0)}{R (t_1)}[/latex]

 

Please note were still in commoving coordinates with a static background metric.

 

[latex]z=\frac {v}{c}[/latex] is only true if v is small compared to c.

 

from this we get the Linear portion of Hubbles law

 

[latex]v=cz=c\frac{(t_0-t_1)\dot{R}t_1}{R(t_1)}[/latex]

 

now the above correlation only holds true if v is small. When v is high we depart from the linear relation to Hubbles law.

 

We start hitting the concave curved portion.

 

The departures from the linear relation requires a taylor series expansion of R (t) with the present epoch for this we will also need H_0.

 

note the above line element in the first equation does not use the cosmological constant aka dark energy. This above worked prior to the cosmological constant

 

Now for the departure from the linear portion of Hubbles law.

 

[latex] v=H_Od, v=cz [/latex] when v is small.

 

To this end we expand R (t) about the present epoch t_0.

 

[latex]R (t)=R[(t_0-t)]=R(t_0)-(t_0)-(t_0)\dot {R}(t_0)+\frac {1}{2}(t_0-t)^2\ddot{R}(t_0)...=R (t_0)[1-(t_0-t)H_o-\frac {1}{2}(t_0-t)q_0H^2_0...[/latex]

 

with [latex]q_0=-\frac{\ddot{R}(t_0)R(t_0)}{\dot{R}^2(t_0)}[/latex]

 

q_0 is the deceleration parameter. Sometimes called the acceleration parameter.

 

now in the first circumstances when v is small. A light ray follows

 

[latex]\int_{t_1}^{t^0} c (dt/R (t)=\int_0^{r_1}dr=r_1 [/latex]

 

with the use of this equation and the previous equation we get

 

 

[latex]r=\int^{t_0}_t=\int^{t_0}_t cdt/{(1-R (t_0)[1-(t_0-t)H_0-...]}[/latex]

 

[latex]=cR^{-1}(t_0)[t_0-t+1/2 (t_0-t)^2H_0+...][/latex]

 

here r is the coordinate radius of the galaxy under consideration.

 

Solving the above gives..

 

[latex]t_0-t=\frac {1}{c}-\frac {1}{2}H_0l^2/c^2 [/latex]

 

which leads to the new redshift equation

 

[latex]z=\frac {H_0l^2}{c+\frac {1}{2}(1+q_0)H^2_0l^2/c^2+O (H^3_0l^3)}[/latex]

Here is the workup starting with the FLRW ds^2 line element.

 

The last equation is the corrected redshift formula when recessive velocity exceeds c

 

The FLRW metric derives from the Newton approximation then adds volume change.

Edited September 22, 2016 by Mordred

[Strange]

On a more serious note - is the OP right to be using Schwartzchild Solution for the EFE in this circumstance. We are not talking about a single heavy mass and the vacuum solution of the space around

 

Indeed. It would only make sense if there were no other masses in the universe. In which case we would not see any receding galaxies!

[Mordred]

 

Indeed. It would only make sense if there were no other masses in the universe. In which case we would not see any receding galaxies!

side note on Hubbles law. The formula v=Hd is a linear formula. It can only be used below Hubble horizon. Once you get beyond Hubble horizon the redshift scale is no longer linear with distance. It now follows an inverted curve.

 

same problem with the cosmological redshift.

 

Gravitational redshift formula is also limited in allowable range determined by the corresponding recessive velocity values.

 

To put it into perspective. This is the equation he needs to compare against beyond Hubble horizon.

 

[latex]z=\frac {H_0l^2}{c+\frac {1}{2}(1+q_0)H^2_0l^2/c^2+O (H^3_0l^3)}[/latex]

 

but he needs to match this below Hubble horizon.

 

[latex]1+Z=\frac{\lambda}{\lambda_o}[/latex]

 

Not easy when equation 1 uses an expanding volume and you have a limit of v less than c for the gravitational redshift formula. Quite frankly without an increasing volume its most likely impossible... How can you get a signal beyond Hubble limit without a volume change? Using the basic formulas in Either the Schwartzchild metric or Newton approximation won't cut it.

 

This formula for example won't cut it. It has an inherent limit. v less than c for accuracy.

 

[latex]\acute{t}=\frac {t}{\sqrt {1-v^2/c^2}}[/latex]

 

This includes the ones captcass is using. They can at best have a limit to range.

v less than c...

 

The curve ratio is completely different from the first equation to the last equation. The last equation doesn't even match the ratio of change for recessive velocity. Curved vs linear...

Edited September 22, 2016 by Mordred

[captcass]

WOW, you guys are quick! You've added so much I don't have time to read it all right now, but will as time allows.

However, with just a quick glance, I would note the following:

 

You are just misinterpreting my terms. Ro = Schwarzschild radius. r = distance from observer. To/Tr = difference in rates ratio between observer and distant point.

 

I am using the formula beyond its normally accepted limits because I see a nearly static, non-expanding universe. You folks are looking at it like the universe is expanding and the velocities away from us apply. I am saying they do not and the formula therefore works to infinity.

 

Likewise, in a non-=expanding universe, OmegaM and OmegaVac do not apply.

 

As long as you keep trying to compare this to an expanding universe model you will find fault because the two don't mix or apply to each other.

 

I'll follow up if necessary when I have time to review all this you put up

 

Thanks for the feedback.

[Mordred]

Hopefully you will catch what we are really stating when you read it again...

 

You seem to have a major problem understanding one major concept...

 

If you want to make datasets that use an expanding volume and model such with a non expanding volume. You simply cannot arbitrarily choose to ignore those datasets.

 

You must be able to precisely reproduce those datasets. now you don't need actual datasets. You can generate them via the formulas.

 

If you cannot replicate the same graph results that current well tested formulas have shown. Tested in many ways other than redshift. Then your formulas DO NOT WORK

 

So far you have used formulas that lead to an inverted bell curve. To try and replace a linear relation.

 

and then claim falsely that your formulas work. Impossible wrong ratios of change.

 

( man I wish our schooling spent more time teaching Laplace transforms grrr....) It would be so much easier to teach physics if a student can look at a formula and without a single calculation. Know what type of graph it will lead to...

Edited September 22, 2016 by Mordred

[imatfaal]

WOW, you guys are quick! You've added so much I don't have time to read it all right now, but will as time allows.

However, with just a quick glance, I would note the following:

 

You are just misinterpreting my terms. Ro = Schwarzschild radius. r = distance from observer. To/Tr = difference in rates ratio between observer and distant point.

...

 

For a non-rotating stationary Earth, To / Tr = √((1-(Ro/R)), where To / Tr is the ratio between times on the surface, To, and at distance R, Tr, and Ro is the Schwarzschild radius of the Earth...

You seem to be changing your tune.

[captcass]

I am not changing my tune. This is what I have been saying all along, though not clearly enough here obviously.

 

My only objective is to show the red shift can be explained in terms of a non-expanding universe. The mirror gradients accomplish this.

 

This is the most basic of derivations that doesn't include rotation or motion or the presence of other bodies. It is meant merely to show how the effect can be generated.

 

Don't know what else to say........If you want to understand it, you have to put the BB and the accelerating universe aside and just consider the basics of a non-expanding, most likely eternal, universe.

 

I know this is difficult as everything has been developed over the last hundred years assuming the shift is due to the Hubble effect. I fully agree all of that seems to make nearly perfect sense.......except for the conclusions of a singularity, BB and accelerating expanding universe that eventually goes cold.

 

By-the-by, I am certainly not the only one who does not agree with the BB, et al. Lots of folks are looking for other answers to the effects we see.

[Mordred]

what in the world is a mirror gradient ?

 

Were dealing with the basic math issues here...atm let alone the rest of your paper