Merged: Why the James Webb Telescope rewrites/doesn't the laws of Physics/Redshifts

[Mike Helland]

Originally Posted by jonesdave116

What is your equation, and what does it give as the energy at d = 1 in the first part of that graphic? 6eV or 8eV? Either way, your photon needs to remember how far it has travelled, as the two parts of its journey see it losing different percentages of its energy.
That is what you need to solve, and a linear relationship fails from the get-go.

The equation I'm using is this:

d = -(1 / (1 + z) - 1) c / H₀

As you will note, d and z no longer have a linear relationship.

As per this:

d = 1 >>>>>>>>>>>>>>> d = 0
8eV >>>>>>>>>>>>>>> 4eV

d=1 is actually 7 billion light years, that's where z=1 and you will get 50% of the energy.

At d=2, that's 14 billion light years, z will be approaching infinity, and the photon energy will be approaching 0.

d = 2 >>>>>>>>>>>>>>>> d = 1 >>>>>>>>>>>>>>> d = 0
12eV >>>>>>>>>>>>>>>> 6eV >>>>>>>>>>>>>>> 0eV

[zooterkin]

Originally Posted by jonesdave116

Nothing has a blueshift that high. Not even close. It is either redshifted or blueshifted. A redshifted galaxy cannot have a blueshift!

He’s treating a redshift as a negative blueshift.

[jonesdave116]

Originally Posted by zooterkin

He’s treating a redshift as a negative blueshift.

Which is like saying that black is white.

[zooterkin]

Originally Posted by jonesdave116

Which is like saying that black is white.

I didn't say it made any sense. I suspect it's something so absurd that anyone who does understand the topic would assume he meant something else.

[jonesdave116]

Originally Posted by Mike Helland

The equation I'm using is this:

d = -(1 / (1 + z) - 1) c / H₀

As you will note, d and z no longer have a linear relationship.

As per this:

d = 1 >>>>>>>>>>>>>>> d = 0
8eV >>>>>>>>>>>>>>> 4eV

d=1 is actually 7 billion light years, that's where z=1 and you will get 50% of the energy.

At d=2, that's 14 billion light years, z will be approaching infinity, and the photon energy will be approaching 0.

d = 2 >>>>>>>>>>>>>>>> d = 1 >>>>>>>>>>>>>>> d = 0
12eV >>>>>>>>>>>>>>>> 6eV >>>>>>>>>>>>>>> 0eV

What is that nonsense? Seriously? It loses 50% of its energy travelling from d = 2 to d = 1, and then loses 100% of its remaining energy going precisely the same distance from d = 1 to d = 0? That is as silly as Lerner's nonsense.

Your 'equation' makes zero sense.

And z = 1 means the light has been travelling for ~ 7 Ga. Z= 2 means it has been travelling for ~ 10 Ga. It is not infinity, or even close. Otherwise we would not see photons from those distances. We see the CMB photons at ~ 13.8 Ga. How close do you think the epoch of recombination actually is? Somewhere near Andromeda?

Sorry, but you are not explaining anything. We can independently measure photons from sources at z = 2 and z = 1. They agree with cosmological redshift. Put the z = 1 source between us and the z = 2 source. What do we see in emission and absorption from the z = 1 source? Trust me, it isn't zero as you would have it.

This is going nowhere. Just as Lerner's nonsense did. Face it - there is no tired light mechanism to explain cosmological redshift, and there is no equation to explain it.

[Mike Helland]

Originally Posted by zooterkin

I didn't say it made any sense. I suspect it's something so absurd that anyone who does understand the topic would assume he meant something else.

Background

Light is made of particles called photons, and photons have a wavelength, and that wavelength determines what color the photon is.

Red light has a higher wavelength than blue light. If a photon's wavelength were to increase for some reason, that is called redshift. And if the wavelength decreases, it is called blueshift. Whether the colors are red or blue before or after the shift, or whether they are in the visible range of light at all, doesn't matter. Just whether the wavelength increases (redshift) or decreases (blueshift).

As the universe expands, the wavelength of photons gets stretched, causing redshift.

Redshift and wavelength

Based on the original wavelength (w_emit) of the photon, and wavelength it was observed to have after traveling through the expanding universe (w_obs), the amount of redshift (z) can be described as:

1 + z = w_obs / w_emit

w_obs = w_emit(z + 1)

This shows the effect of redshift z (shown on the x-axis) on the wavelength observed (shown on the y-axis) of a photon emitted with a wavelength of 1 nm.



The red and blue shading indicates where values of z are redshifted or blueshifted. When z > 0, that's redshift. Notice that when z = -1, the wavelength becomes zero. So the range of blueshift is finite and actually very small, -1 < z < 0.

Redshift and frequency

A photon also has a frequency, which is related to wavelength by the speed of light (c):

c = wavelength * frequency

There are also formulas for z using frequency:

1 + z = f_emit / f_obs

f_obs = f_emit / (1 + z)

This shows the effect of redshift z (shown on the x-axis) on the frequency observed (y-axis) of a photon emitted with a frequency of 1 Thz.




The first thing you notice is that increasing redshift has the opposite effect on frequency as it does on wavelength. A higher z means a lower observed frequency.

The other thing to notice is that the frequency approaches zero as redshift approaches infinity, and that the frequency approaches infinity as redshift approaches z = -1.

In the case of both wavelength and frequency, the blueshift region is small, -1 < z < 0, and the redshift region is infinite, z > 0.

Redshift and energy

A photon also has energy, and that too is affected by redshifting. The energy of a photon and its frequency are directly related by Planck's constant (h):

E = h f

That being the case, the formulas for redshift using energy are similar to those for redshift using frequency:

1 + z = E_emit / E_obs

E_obs = E_emit / (1 + z)

And the graph looks the same as it does for frequency. The more redshift, the less energy.

Redshift z goes in the direction of wavelength (more redshift, higher wavelength), but in the opposite direction of energy and frequency (more redshift, less energy and frequency).

Blueshift

That redshift is going "with" wavelength, or "against" energy and frequency shouldn't make a difference. That can be investigated by inverting the redshift z formulas into blueshift b formulas:

1 + b = w_emit / w_obs
1 + b = f_obs / f_emit
1 + b = E_obs / E_emit

Or:

w_obs = w_emit / (b + 1)
f_obs = f_emit(b + 1)
E_obs = E_emit(b + 1)

Now let's compare the blueshift b to the observed wavelength:



And observed frequency:



We see now the situation has reversed. Not only does energy and frequency rise when blueshift increases, but all the redshifting happens when -1 < b < 0.

All of z > 0 now fits into -1 < b < 0.

Does that affect anything?

Distance

The z-distance relation is:

d = zc / H0
z << 1

This is only considered valid for very small values of z. At z = 10, the distance would be 140 billion light years. And since z can go up to infinity, the sky is the limit.

How about when redshifts are represented as negative blueshifts (-b) instead of z? Perhaps:

d = -bc / H0

Since b has to be greater than -1, the largest distance this relationship can produce is d = c / H₀, or Hubble's length.

To compare z and -b, assume a photon is emitted at 1 MeV. When the photon is observed at the following values, we can see a big difference.

Observed (MeV)	Redshift z	Blueshift b	Distance(z)	Distance(-b)
1	0	0	0 Bly	0 Bly
0.5	1	-0.5	14 Bly	7.0 Bly
0.33	2	-0.66	28 Bly	9.3 Bly
0.25	3	-0.75	42 Bly	10.5 Bly
0.2	4	-0.8	56 Bly	11.2 Bly

We know from the redshift equations that 1 / (1 + z) will give you the percentage of a frequency or energy that will be observed. To get the blueshift b, from that percentage you just subtract 1, so:

b = 1 / (1 + z) - 1

Therefore, we could say:

d = -(1 / (1 + z) - 1) c / H0

Let's compare the original z-distance relationship with this new one.



The new z-distance relationship seems to be closely mimicking the lookback times from the current standard cosmological model, LCDM. LCDM has more parameters, and so leads to a wider variety of predictions. Let's compare two of those sets of parameters with the new z-distance relationship. Consider the distance d to be divided by c the speed of light, converting 1 billion light years (Gly) into 1 billion years (Gyr):

[Mike Helland]

Originally Posted by jonesdave116

What is that nonsense? Seriously? It loses 50% of its energy travelling from d = 2 to d = 1, and then loses 100% of its remaining energy going precisely the same distance from d = 1 to d = 0? That is as silly as Lerner's nonsense.

From d = 2 to d = 1 it loses 50%.

From d = 2 to d = 0 it loses 50% x 2.

That's if by "d" you mean distance. You seem to be using it interchangeably with "z" in the next sentence.

Quote:

And z = 1 means the light has been travelling for ~ 7 Ga. Z= 2 means it has been travelling for ~ 10 Ga. It is not infinity, or even close.

I said at d=2, z=infinity. I didn't say at z=2, z=infinity.

[catsmate]

Originally Posted by jonesdave116

Yeah, right. That's Kroupa. I take him about as seriously as Lerner. He's a MONDist, last I heard. How's that going, Pavel? On its death bed, last time I looked. Fails miserably at large scales. When he can explain the colliding cluster lensing observations without invoking dark matter, I might listen. McGaugh and the rest of the MONDists can't.

He is still, moderately interesting stuff back in '12

[Mike Helland]

Originally Posted by Mike Helland

The new z-distance relationship seems to be closely mimicking the lookback times from the current standard cosmological model, LCDM. LCDM has more parameters, and so leads to a wider variety of predictions. Let's compare two of those sets of parameters with the new z-distance relationship. Consider the distance d to be divided by c the speed of light, converting 1 billion light years (Gly) into 1 billion years (Gyr):

http://www.internationalskeptics.com...282fae4fca.png

Here are some more sets of parameters:

[jonesdave116]

Originally Posted by Mike Helland

From d = 2 to d = 1 it loses 50%.

From d = 2 to d = 0 it loses 50% x 2.

That's if by "d" you mean distance. You seem to be using it interchangeably with "z" in the next sentence.



I said at d=2, z=infinity. I didn't say at z=2, z=infinity.

Errrr, nope! Care to plug the numbers into your equation and show how that impossibility is possible?

Here it is;

d = -(1 / (1 + z) - 1) c / H0

Let's plug in z = 1;

d = -(1/ (1 + 1) -1) = -0.50 assuming you are dividing by (1 + z) before you subtract the -1?;

-(1 / 2 = 0.5 - 1) = -(-0.5).

And now z = 2;

d = -(1/ (1 + 2) -1) = -(1 / 3 = 0.333) - 1 = -(-0.667).

What are you doing with the minus sign before the parentheses? Anything > z = 1 is going to give you a minus number in the parentheses anyway???

So, from that perspective, we still have a linear equation. Which doesn't work.

Fancy plugging in z = 0.5? You end up with -(1 (1 + 0.5) -1) = -(-0.333).

What does that say about a photon travelling from z = 1 to z = 0? And what is its energy at z = 0.5?

z = 1 >>>>>>>>>>>>>>>> z = 0.5 >>>>>>>>>>>>>>> z = 0 (Earth)
12eV >>>>>>>>>>>>>>>> ??eV >>>>>>>>>>>>>>> 6eV

Your equation tells us that it should indeed lose 50% of its energy from z = 1 to z = 0.

It also tells us that a photon from z = 0.5 should lose 1/3rd of its energy. So, if it arrives at Earth with the 6eV calculated above then it must have had an energy of 9eV at z = 0.5.

However, if it started its trip at z = 1, and arrived at z = 0.5 with 9eV, that means it only lost 1/4 of its energy from z = 1 to z = 0.5 (3 / 12)! It should lose 1/3rd, as it did on the second part of its journey.

By your reckoning, the photon from z = 1 should arrive at z = 0.5 with 8eV of energy.
If it then arrives at Earth (d = 0), with 6eV, it has only lost 1/4 of its energy
from z = 0.5 to z = 0.

Sorry, but your equation makes no more sense than Lerner's.


You have ended up with the exact same problem as he did. Because there are zero tired light mechanisms, nor equations, that make any sense. As Ben m said, you need some sort of exponent. And none of them work.

[Mike Helland]

Originally Posted by jonesdave116

d = -(1 / (1 + z) - 1) c / H0

Let's plug in z = 1;

d = -(1/ (1 + 1) -1) = -0.50

You got two things wrong here.

Try a calculator.

[Mike Helland]

Originally Posted by jonesdave116

So, from that perspective, we still have a linear equation.

How many linear equations are in the graph below:

[jonesdave116]

Originally Posted by Mike Helland

You got two things wrong here.

Try a calculator.

No I didn't. If you think I did, do the equation yourself. And show us what the eV values are in the example I gave. Get on with it.

EDIT:

How hard can this be? How much energy, percentage-wise, is lost by a photon travelling from z = 1 to z = 0?
How much energy is lost, percentage-wise by a photon travelling from z = 0.5 to z = 0?
If a photon is travelling from z = 1 to z = 0, what is its value as it passes z = 0.5 on its way to z = 0? Is it the same percentage loss on both halves of its journey? It had better be.

What happens if the object at z = 0.5 happens to be a galaxy that absorbs and re-emits photons from the z = 1 source? What about the emission from the z = 0.5 galaxy itself? Are the two values the same? They had better be, because otherwise you are violating actual observation, not to mention basic physics.

The photons in both cases are arriving from z = 0.5. Understand? One of them started at z = 1, before it got to the galaxy at z = 0.5, but the galaxy doesn't know that. As far as it is concerned, it absorbed a photon from somewhere, at a certain value, and this caused it to emit another photon towards us in emission. Using real physics, we can measure the distance to the z = 0.5 galaxy using either its own emissions, or those it has absorbed and re-emitted from a more distant object. Because they will be the same.

Give us a value, using your 'equation', for the energy of the photon at z = 0.5 after it is emitted from z = 1. Then we can use that value, and send the same energy of photon right back at it from Earth. Will it arrive at z = 0.5 with the same value as it arrived with at Earth when travelling in the other direction?

You are in the same boat as Lerner. Your photons need to know how far they have travelled. What is the IQ of a photon?

[Mike Helland]

Originally Posted by jonesdave116

d = -(1 / (1 + z) - 1) c / H0

Let's plug in z = 1;

d = -(1/ (1 + 1) -1) = -0.50

d = 0.5 c / H0, or about 7 billion light light years. Not -0.50. You did the math wrong and got the equation wrong.

Quote:

How much energy, percentage-wise, is lost by a photon travelling from z = 1 to z = 0?

Photons don't travel from z=1 to z=0.

z is not distance.

You can't keep that straight for some reason.

[jonesdave116]

Originally Posted by Mike Helland

How many linear equations are in the graph below:

http://www.internationalskeptics.com...282df3335f.png

You said;

Quote:

From d = 2 to d = 1 it loses 50%.

From d = 2 to d = 0 it loses 50% x 2.

Lol. So, you are using your constant to multiply a negative number, that isn't negative in the real world, to make and negative percentage become zero (i.e. -100%) for any distance greater than d = 2? Brilliant! No wonder I couldn't grok it! It is even crazier and more unscientific than I had first given it credit for!

So, if z = 1, your constant c also equals 1? And if z = 2, c = 2?
So, if your 'equation', for want of a better word, says that a photon loses 50% of its energy from z = 1 to z = 0, you just multiply 50% x 1? Got it. So, you end up with 50%. Fantastic.

If it comes from z = 0.5, we end up with -( 1/(1 + 0.5) -1) x c (0.5, in this case) = -0.333 (-33.3%) x 0.5 = -0.1667 (-16.67%)! It only loses 1/6th of its energy from z = 0.5 to z = 0!

Holy Hell! That means a 10eV photon emitted a z = 1, should arrive at Earth (z = 0) with a energy of 5eV. But it had to pass z = 0.5 on the way! If we have a photon arriving at Earth with an energy of 5eV, that means it had to have had 6eV of energy at z = 0.5 (6 x 5/6 = 5).
But it has already travelled from z = 1 to z = 0.5, and managed to lose 4eV on the way! 40% of its original energy. It should only have lost 1/6th.

You say that photons from z = 2 lose 100% of their energy

So, the photon from z = 1 still has energy when it arrives at Earth. 50% of it, in fact. So, let's stick to the example above.
A 10eV photon leaves z = 1, and gets to us at z = 0 with 5eV of energy. So far so good (kind of). OK. It doesn't stop at Earth. Let's say that it interacts with out atmosphere. It gets absorbed, and re-emitted by some atom or other. It carries on for another z = 1 in the opposite direction from which it arrived. With what energy does it arrive at its new destination? It is a brand new photon, created by emission when the Earthly atom de-excited. It should get to its new destination with 2.5eVof energy. There are alien astronomers there, wanting to study our atmosphere, for clues of a possible intelligent civilisation. I fear that they may well be disappointed, if some posts on this thread are anything to go by!
So, do the aliens see the 2.5eV photon, or has it disappeared for reasons unknown? After all, it has only travelled from Earth to z= 1, even though the photon that brought it into existence originated at z = 2 from the aliens.
Do you see where this is going?

If a photon is not absorbed and re-emitted from z = 2, we can never see it. If it is absorbed and re-emitted along the way, we can see it! Christ alone knows what energy it is supposed to arrive at, be absorbed at, and re-emitted at!

It doesn't really matter, because the whole 'equation' is a complete train wreck, and is totally unphysical.

[jonesdave116]

Originally Posted by Mike Helland

Photons don't travel from z=1 to z=0.

z is not distance.

You can't keep that straight for some reason.

Yes they do. We measure them all the time. What you can't get straight is that your equation fails laughably. And they are the time that the light has been travelling. Which tells us the current distance, not the distance they were emitted at. Because the universe is expanding.
So, when real scientists say that a galaxy is at a redshift of 2, how far was it in your fantasy universe when it emitted the light we detect?

Quote:

d = 0.5 c / H0, or about 7 billion light light years. Not -0.50. You did the math wrong and got the equation wrong.

Really? So, what is the energy loss between an object at z = 1 by the time we detect it? What is it at z = 0.5, on its way to us?

Show me the maths. I am not interested in your wonky equation, which makes zero sense. We know z because we know the rest wavelength and the measured wavelength. If it doubles, so does z.

What happens in the fantasy universe you inhabit?

[Mike Helland]

Originally Posted by jonesdave116

So, when real scientists say that a galaxy is at a redshift of 2, how far was it in your fantasy universe when it emitted the light we detect?

9.33 Gly.

Quote:

Really? So, what is the energy loss between an object at z = 1 by the time we detect it? What is it at z = 0.5, on its way to us?

1 / (1 + z)
z = 1, energy remaining is 50%, so energy loss is 50%
z = 0.5, energy remaining is 66%, so energy loss is 33%

[jonesdave116]

Originally Posted by Mike Helland

9.33 Gly.


1 / (1 + z)
z = 1, energy remaining is 50%, so energy loss is 50%
z = 0.5, energy remaining is 66%, so energy loss is 33%

So, from your own words, a 12eV photon emitted at z = 1 gets to Earth (z = 0) with 6eV of energy remaining, and 6eV lost. As the photon had to pass z = 0.5 on its way here, where it may or may not have been absorbed and re-emitted, its energy is ???????????????

You see, if we emit a 9eV photon from z = 0.5, then it will lose one third of its energy, and reach Earth at the measured 6eV. But if it had 9eV of energy at z = 0.5, that means it has only lost 25% of its energy on the trip from z = 1 to z = 0.5.

Conversely, if it obeyed your rules, and lost 33% of its energy from z = 1 to z = 0,5, it would arrive at z = 0.5 with 8eV of energy. And then only lose a further 25% between z= 0.5 and Earth. How is this possible?

It is important that we know this, because if our z measurements and equations were all wrong, as you suggest, we would see an obvious difference in absorption and emission wavelengths from intermediate galaxies and distant quasars, for instance. We don't. How come?

As I keep saying - just like Lerner, your photons need to remember how far they have travelled. Any galaxy they encounter along the way, will have to know where they are going to, in order to know how much it needs to fiddle around with electron excitation levels from photons absorbed from more distant sources, to make them match the galaxy's own emission wavelengths. All this just to fool us poor saps on Earth into believing that our equations are correct after all. Which is not very scientific.

Far more parsimonious to believe that the equations are correct, and that there is neither a mechanism, nor an equation, that explains tired light, and a static universe. Which is why pretty much nobody bothers with such things anymore.

[Mike Helland]

Originally Posted by jonesdave116

So, from your own words, a 12eV photon emitted at z = 1 gets to Earth (z = 0) with 6eV of energy remaining, and 6eV lost.

Let's start with a handy chart.

z	E left (%)	d (Gly)
0	100	0
0.5	66.6	4.66
1	50	7

A photon with a z=1, has half its energy, and has traveled 7 billion light years. Ok.

Quote:

As the photon had to pass z = 0.5 on its way here, where it may or may not have been absorbed and re-emitted, its energy is ???????????????

Let's check the chart. A photon from z=0.5 is from 4.66 Gly.

So where does a z=1 photon actually meet up with z=0.5 photon?

7 - 4.66 = 2.33 Gly

A galaxy that is 2.33 Gly has a z=0.2.

|-----------------7--------------|
A          B                     C
|---2.33---|---------4.66--------|

A and C will see each others photons as z=1 (with 50% of its energy)
A and B will see each others photons as z=0.2 (with 83%)
B and C will see each others photons as z-0.5 (with 66%)

A photon leaves A (d=0,z=0) and arrives at B (d=2.33 Gly, z=0.2).

It will have 83.33% of its energy. If it continues on to C, which is twice as far as it has already gone, (d=7 Gly, z=1) it will have 50% of it it started with, losing an additional 33.33% from point B.

If a photon is emitted from B (d=0, z=0) and arrives at C (d=4.66 Gly, z=0.5) it will have 66.66% of its energy. It too will have lost 33% from point B.

*edit* You can see it adds up clearly by this:

z	E lost (%)	d (Gly)
0	0	0
0.2	16.66	2.33
0.5	33.33	4.66
1	50	7

16.66 + 33.33 = 50%
2.33 + 4.66 = 7 Gly

*edit 2*

A 12 eV photon leaving A will pass B with 10 eV and arrive at C with 6 eV.

12 - (12 * 0.1666) - (12 * 0.3333)

If a photon leaves B with 9 eV, it will arrive at c with 6 eV.

You would have to reach the same conclusion in the expanding universe. If the photons are redshifted by the expansion of space, how can it affect different photons differently?

I think you might have found an interesting flaw in the expanding universe.

[Mike Helland]

Originally Posted by Mike Helland

I think you might have found an interesting flaw in the expanding universe.

I should mention that I started a new thread a bit ago, that wasn't about JWST or tired light, just redshift quantification.

But it got merged into this thread, so here we are. This all applies to the expanding universe, and nothing to do with tired light:

----

A 12 eV photon leaves point A (d=0, z=0, E=12 eV) it passes by point B (d=2.33 Gly, z=0.2, E=10 eV), and arrives at point C (d=7 Gly, z=1, E=6 eV).

It has 50% its original energy, after losing 16.66% between A and B, and another 33.33% between B and C.

As the first photon passed point B, say a second photon is emitted at 9 eV from point B (d=0, z=0, E=9 eV) and arrives at point C (d=4.66 Gly, z=0.5, E=6 eV).

It arrives after losing 33.33% of its energy between B and C, just like the first, and arrived with 6 eV just like the first. But it left point B with less energy than the first photon had at the same point and time.

|-----------------7-------------|
A          B                    C
|---2.33---|---------4.66-------|

First photon
A (d=0,        z=0,    E=12 eV)
B (d=2.33 Gly, z=0.2,  E=10 eV)
C (d=7 Gly,    z=1,    E=6 eV)

Second photon
B (d=0,        z=0,    E=9 eV)
C (d=4.66 Gly, z=0.5,  E=6 eV)

I don't see the issue here, except that if the redshifts were caused by expanding space, how could that be?

[Mike Helland]

Originally Posted by Mike Helland

z	E lost (%)	d (Gly)
0	0	0
0.2	16.66	2.33
0.5	33.33	4.66
1	50	7

16.66 + 33.33 = 50%
2.33 + 4.66 = 7 Gly

Since 1 + b = 1 / (1 + z):

z	b	E lost (%)	d (Gly)
0	0	0	0
0.2	-0.166	16.66	2.33
0.5	-0.33	33.33	4.66
1	-0.5	50	7

All the columns add up, except for z.

[Pixel42]

Originally Posted by Mike Helland

I should mention that I started a new thread a bit ago, that wasn't about JWST or tired light, just redshift quantification.

But it got merged into this thread, so here we are.

One thread is more than sufficient for each brand of physics crackpottery.

[Mike Helland]

It turns out that:



Is exactly equal to LCDM in a default state.

What are the odds of that?

Check it out, I made an interactive version, so even if you don't care about the hypothesis, you can still see how the parameters effect the graph.

https://mikehelland.github.io/hubble...other/lcdm.htm

[W.D.Clinger]

Originally Posted by Mike Helland

The equation I'm using is this:

d = -(1 / (1 + z) - 1) c / H₀

Originally Posted by jonesdave116

Errrr, nope! Care to plug the numbers into your equation and show how that impossibility is possible?

Here it is;

d = -(1 / (1 + z) - 1) c / H0

Let's plug in z = 1;

d = -(1/ (1 + 1) -1) = -0.50 assuming you are dividing by (1 + z) before you subtract the -1?;

-(1 / 2 = 0.5 - 1) = -(-0.5).

And now z = 2;

d = -(1/ (1 + 2) -1) = -(1 / 3 = 0.333) - 1 = -(-0.667).

The calculation I highlighted in blue is incorrect, for two reasons: the sign of -0.5 is wrong and the factor c / H0 has been omitted. The calculation I highlighted in yellow is incorrect for only the second of those two reasons. The calculation I highighted in green is correct because, although it omits the factor c/H0, it does not claim to calculate d.

Originally Posted by Mike Helland

You got two things wrong here.

Try a calculator.

Originally Posted by jonesdave116

No I didn't. If you think I did, do the equation yourself.

I'm with Mike Helland on this one point.

It is conceivable that Mike Helland's recent posts have been right about other things as well, but I haven't looked hard enough to find examples of such.

[Mike Helland]

Originally Posted by W.D.Clinger

It is conceivable that Mike Helland's recent posts have been right about other things as well, but I haven't looked hard enough to find examples of such.

Inconceivable!

So, you're pretty well up to speed on much of this. Does this look familiar at all to you?

[W.D.Clinger]

How is that different from

d = (z/(z+1)) c/H₀

In other words, why are you writing z/(z+1) as -(1/(1+z)-1) ?

[fertilizerspike]

"Redshift" bears no relationship to distance

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[Mike Helland]

Originally Posted by W.D.Clinger

How is that different from

d = (z/(z+1)) c/H₀

In other words, why are you writing z/(z+1) as -(1/(1+z)-1) ?

I'll don the derivatino tomorrow. I'm goin to fdin my car.

https://mikehelland.github.io/hubble...ue/redblue.htm

https://mikehelland.github.io/hubble...other/lcdm.htm

[Mike Helland]

Originally Posted by W.D.Clinger

How is that different from

d = (z/(z+1)) c/H₀

In other words, why are you writing z/(z+1) as -(1/(1+z)-1) ?

I guess the answer is because that's how I derived it.

Here's an equation for redshift:

1 + z = E_emit / E_obs

When z>0 that's redshift, when z<0, that's blueshift.

But notice this is hyperbolic, and z>0 goes to infinity, while z<0 only goes to -1:



If you invert the formula, the situation reverses:

1 + b = E_obs / E_emit



So 1 + b = 1 / (1 + z)

I think checked this:

d = -bc / H₀

And it produced LCDM lookback times over its domain.

Since 1 + b = 1 / (1 + z), then b = 1 / (1 + z) - 1.

[Mike Helland]

Originally Posted by Mike Helland

A 12 eV photon leaves point A (d=0, z=0, E=12 eV) it passes by point B (d=2.33 Gly, z=0.2, E=10 eV), and arrives at point C (d=7 Gly, z=1, E=6 eV).

It has 50% its original energy, after losing 16.66% between A and B, and another 33.33% between B and C.

As the first photon passed point B, say a second photon is emitted at 9 eV from point B (d=0, z=0, E=9 eV) and arrives at point C (d=4.66 Gly, z=0.5, E=6 eV).

It arrives after losing 33.33% of its energy between B and C, just like the first, and arrived with 6 eV just like the first. But it left point B with less energy than the first photon had at the same point and time.

|-----------------7-------------|
A          B                    C
|---2.33---|---------4.66-------|

First photon
A (d=0,        z=0,    E=12 eV)
B (d=2.33 Gly, z=0.2,  E=10 eV)
C (d=7 Gly,    z=1,    E=6 eV)

Second photon
B (d=0,        z=0,    E=9 eV)
C (d=4.66 Gly, z=0.5,  E=6 eV)

I don't see the issue here, except that if the redshifts were caused by expanding space, how could that be?

Did anyone find the mistake here?

It looks right to me. But obviously can't be, if redshifts are caused by expanding space.

[Ziggurat]

Originally Posted by Mike Helland

A 12 eV photon leaves point A (d=0, z=0, E=12 eV) it passes by point B (d=2.33 Gly, z=0.2, E=10 eV), and arrives at point C (d=7 Gly, z=1, E=6 eV).

It has 50% its original energy, after losing 16.66% between A and B, and another 33.33% between B and C.

33.3% of what? The energy it had at A, not the energy it had at B. It loses 40% of the energy it had at B when traveling from B to C, using your numbers.

Quote:

As the first photon passed point B, say a second photon is emitted at 9 eV from point B (d=0, z=0, E=9 eV) and arrives at point C (d=4.66 Gly, z=0.5, E=6 eV).

It arrives after losing 33.33% of its energy between B and C, just like the first,

No. You are comparing apples to oranges by not taking an equivalent percentage. If your first set of numbers was correct, this second photon should lose 40% of its energy it had at B when traveling from B to C, not 33.33%, and end with 5.4 eV, not 6.

[bruto]

Originally Posted by Mike Helland

From d = 2 to d = 1 it loses 50%.

From d = 2 to d = 0 it loses 50% x 2.

That's if by "d" you mean distance. You seem to be using it interchangeably with "z" in the next sentence.



I said at d=2, z=infinity. I didn't say at z=2, z=infinity.

I defer to others on physics and higher math, about which I am something of an idiot. But the highlighted part seems odd. What causes the percentage to increase?

[Mike Helland]

Originally Posted by bruto

I defer to others on physics and higher math, about which I am something of an idiot. But the highlighted part seems odd. What causes the percentage to increase?

From d=0 to d=1 is some distance without a unit of length.

d=0 to d=2 should be twice that.

[erwinl]

Originally Posted by Mike Helland

From d=0 to d=1 is some distance without a unit of length.

d=0 to d=2 should be twice that.

In your post #140, you say d does have a unit of length (Gly, to be precise).
In post #127 you suggest that d does not have nunits for length and also, if I understand it correctly, does not go higher than 2.

Or did I not understand (which is more than possible)?

[bruto]

Originally Posted by Mike Helland

From d=0 to d=1 is some distance without a unit of length.

d=0 to d=2 should be twice that.

I get that. If d0 to d1 is twice the length, it follows that d1 to d2 is the same length as d0 to d1. So what causes the percentage to change in the second half?

I think in most situations, if something loses 50 percent of anything in a given distance, then when it goes twice that given distance, it would lose another 50 percent of that 50 percent unless some other factor made the percentage of loss over distance greater.

[W.D.Clinger]

Originally Posted by erwinl

In your post #140, you say d does have a unit of length (Gly, to be precise).
In post #127 you suggest that d does not have nunits for length and also, if I understand it correctly, does not go higher than 2.

Good catch.

Mike Helland has been consistently sloppy throughout this thread.

With my highlighting:

Originally Posted by Mike Helland

The equation I'm using is this:

d = -(1 / (1 + z) - 1) c / H₀

As you will note, d and z no longer have a linear relationship.

As per this:

d = 1 >>>>>>>>>>>>>>> d = 0
8eV >>>>>>>>>>>>>>> 4eV

d=1 is actually 7 billion light years, that's where z=1 and you will get 50% of the energy.

At d=2, that's 14 billion light years, z will be approaching infinity, and the photon energy will be approaching 0.

The highlighted equation simplifies to

d = (z/(z+1)) c/H₀

from which it is obvious that d cannot exceed c/H₀, which is about 14 billion light-years.

Yet Mike Helland is talking about d=1 or d=2, which is obviously inconsistent with his equation for d (highlighted above). (ETA: According to the equation for d, d=1 and d=2 are both achieved at approximately zero red shift, and it is obvious that Mike Helland thinks he has been talking about measurably positive red shifts.)

I have absolutely no idea of what Mike Helland thinks he means by d=1 or d=2, and have ceased to care. More accurately, I have not cared at any moment since the birth of this thread what Mike Helland thinks he is talking about.

Mike Helland should consider whether the degree of care he has put into describing his equations and numbers is remotely adequate to motivate any care on the part of whatever he believes his audience to be.

[Mike Helland]

Originally Posted by W.D.Clinger

Good catch.

Mike Helland has been consistently sloppy throughout this thread.

With my highlighting:

The highlighted equation simplifies to

d = (z/(z+1)) c/H₀

from which it is obvious that d cannot exceed c/H₀, which is about 14 billion light-years.

Yet Mike Helland is talking about d=1 or d=2, which is obviously inconsistent with his equation for d (highlighted above). (ETA: According to the equation for d, d=1 and d=2 are both achieved at approximately zero red shift, and it is obvious that Mike Helland thinks he has been talking about measurably positive red shifts.)

I have absolutely no idea of what Mike Helland thinks he means by d=1 or d=2, and have ceased to care. More accurately, I have not cared at any moment since the birth of this thread what Mike Helland thinks he is talking about.

Mike Helland should consider whether the degree of care he has put into describing his equations and numbers is remotely adequate to motivate any care on the part of whatever he believes his audience to be.

That's understandable, W.D.Clinger. It was jonesdave116 that was persistent in that I answer a question originally proposed by ben m in this forum far before my time.

Originally it was about a photon from a distance of z=1, and what about a photon z=2, which was proposed to be twice as far away a z=1. Since that's wrong, it may be confusing.

The scenario was eventually modified to be about a z=0.5 photon a z=1 photon.The change in stipulations only adds to the confusion.

I answered the question thoroughly, eventually, and the results are here:

|-----------------7-------------|
A          B                    C
|---2.33---|---------4.66-------|

First photon
A (d=0,        z=0,    E=12 eV)
B (d=2.33 Gly, z=0.2,  E=10 eV)
C (d=7 Gly,    z=1,    E=6 eV)

Second photon
B (d=0,        z=0,    E=9 eV)
C (d=4.66 Gly, z=0.5,  E=6 eV)

This demonstrates that the expanding universe is internally inconsistent.

[Mike Helland]

Originally Posted by bruto

I get that. If d0 to d1 is twice the length, it follows that d1 to d2 is the same length as d0 to d1. So what causes the percentage to change in the second half?

It doesn't. It's the same percent as the photon's original energy.

Say a photon redshifts over 100 million light years. But it doesn't redshift over 1 light second.

If you checked the photon's energy every second, and reset its distance to 0, it would never redshift.

Photons redshift per their original energy, not their energy at any arbitrary point.

[RecoveringYuppy]

Does the phrase "It's the same percent as the photon's original energy" actually mean something in context here?

[Mike Helland]

Originally Posted by RecoveringYuppy

Does the phrase "It's the same percent as the photon's original energy" actually mean something in context here?

Why wouldn't it?