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

[Mike Helland]

Originally Posted by zooterkin

At d=1, there are two photons, each has an energy of 8eV. What property of the second one means that it loses all its energy while the other loses only half?

It traveled twice as far.

d=1 is the starting point of the first photon, and the midway point of second photon. Agreed?

[erwinl]

Originally Posted by Mike Helland

It traveled twice as far.

d=1 is the starting point of the first photon, and the midway point of second photon. Agreed?

Does that mean that a photon contains the information about when/where it originated?

[zooterkin]

Originally Posted by Mike Helland

It traveled twice as far.

Not from where they both had the same energy. From there, they both travelled the same distance.

Quote:

d=1 is the starting point of the first photon, and the midway point of second photon. Agreed?

Yes.

[Mike Helland]

Originally Posted by erwinl

Does that mean that a photon contains the information about when/where it originated?

At the very least, the model would.

[Mike Helland]

Originally Posted by zooterkin

Not from where they both had the same energy. From there, they both travelled the same distance.

Right.

If the d=2 photon was received at d=1 and then re-emitted then both photons should be received at d=0 identically.

[zooterkin]

Originally Posted by Mike Helland

Right.

If the d=2 photon was received at d=1 and then re-emitted then both photons should be received at d=0 identically.

Why?

ETA: How is the 8eV that the first photon has different from the 8eV the second one has?

[erwinl]

Originally Posted by Mike Helland

At the very least, the model would.

A model should try to describe reality.
Would the photon contain that information in reality?

[Ziggurat]

Originally Posted by Mike Helland

Because it matters where the signal started.

Not to the energy of the photon, it doesn't. Any process that takes one 8 meV photon and turns it into a 4 meV photon is going to take any 8 meV photon and turn it into a 4 meV photon.

Quote:

Do you know where electrical relays come from?

Telegraphs. You can send a signal over a wire for so long that it degrades. So they had batteries in places that could receive a signal, and trigger a fresh one.

You're thinking about it as if there was a relay at d=1 in the bottom scenario.

No, he isn't. If you want to use the telegraph relay analogy, if the voltage of the signal drops by half in the first 100 km, then a second 100 km will drop it to 1/4 (half of half), not to zero. You need relays to keep the signal well above the noise floor, not because the signal actually dropped to zero.

[Mike Helland]

Originally Posted by erwinl

A model should try to describe reality.
Would the photon contain that information in reality?

Does the photon contain the fine structure constant?

[Mike Helland]

Originally Posted by zooterkin

Why?

ETA: How is the 8eV that the first photon has different from the 8eV the second one has?

What is observed is a redshift-distance relationship.

I suppose my thinking is that the distance a photon has traveled may be relevant.

[Mike Helland]

Originally Posted by Ziggurat

Not to the energy of the photon, it doesn't. Any process that takes one 8 meV photon and turns it into a 4 meV photon is going to take any 8 meV photon and turn it into a 4 meV photon.

I can see why that would seem to be a reasonable assumption.

Quote:

No, he isn't. If you want to use the telegraph relay analogy, if the voltage of the signal drops by half in the first 100 km, then a second 100 km will drop it to 1/4 (half of half), not to zero. You need relays to keep the signal well above the noise floor, not because the signal actually dropped to zero.

True enough. It was just an analogy.

[Ziggurat]

Originally Posted by Mike Helland

Does the photon contain the fine structure constant?

Not on its own.

[Mike Helland]

Originally Posted by Ziggurat

Not on its own.

So what contains that?

The only way a non-expansion theory could work is if the energy of a photon is a function of the distance or time it traveled. Which would mean the fine structure constant is an initial value for a photon that, over cosmological distances, diminishes.

So what contains the fine structure constant in reality?

[Ziggurat]

Originally Posted by Mike Helland

So what contains that?

Correlations between multiple photons.

For example, if you look at a photon from the n=1 to n=0 transition of atomic hydrogen, the photon itself does not tell you that it's from that transition. If it hasn't red or blue shifted, the energy will match that transition, but you can't tell it's from that transition because it could be something else that shifted TO that energy. If you observe a whole spectrum, then you can observe a pattern of how photons from different transitions match up with the hydrogen spectrum, even if it's shifted, and you can conclude that it matches hydrogen and assign specific transitions to specific photons you detect. You need multiple photons to make that comparison, though. Any one photon doesn't tell you that.

[Mike Helland]

Originally Posted by Ziggurat

Correlations between multiple photons.

For example, if you look at a photon from the n=1 to n=0 transition of atomic hydrogen, the photon itself does not tell you that it's from that transition. If it hasn't red or blue shifted, the energy will match that transition, but you can't tell it's from that transition because it could be something else that shifted TO that energy. If you observe a whole spectrum, then you can observe a pattern of how photons from different transitions match up with the hydrogen spectrum, even if it's shifted, and you can conclude that it matches hydrogen and assign specific transitions to specific photons you detect. You need multiple photons to make that comparison, though. Any one photon doesn't tell you that.

https://en.wikipedia.org/wiki/Fine-structure_constant

"With the development of quantum electrodynamics (QED) the significance of α has broadened from a spectroscopic phenomenon to a general coupling constant for the electromagnetic field, determining the strength of the interaction between electrons and photons."

I think one thing to keep in mind is that even though redshifted light has traveled billions of light years, ultimately, it's still just two electrons exchanging a photon. A pretty simple Feynman diagram.

(e)/\/\/\/\/\/\/\/\/\/\(e)


If the light is going to be measured in the middle of the journey, and retransmitted, it's not as simple.

(e)/\/\/\/\(e)\/\/\/\/\(e)

If the redshift-distance relationship depends on the distance a photon has traveled, the ultimate receiver of the signal will get photons of different energies depending on how many times and where the light has been measured in the intermediate.

[Ziggurat]

Originally Posted by Mike Helland

If the redshift-distance relationship depends on the distance a photon has traveled

There is no possible way for it to depend on distance like that. And I don't just mean there's no mechanism which can red shift light like that (though that is itself an insurmountable problem, see the thread about tired light), I also mean that the photon itself can't accommodate any such mechanism. Photons have a frequency/wavelength/energy, and they have a polarization, but that's it. They don't have any internal structure or properties in which to store information about how far they have traveled or how much they have red shifted/blue shifted. A 1 meV photon that was emitted at 1 meV is identical to a 1 meV photon that was red shifted from 2 meV. There is no way to distinguish them. The universe cannot treat them differently, because they aren't different.

[Mike Helland]

Originally Posted by Ziggurat

There is no possible way for it to depend on distance like that. And I don't just mean there's no mechanism which can red shift light like that (though that is itself an insurmountable problem, see the thread about tired light), I also mean that the photon itself can't accommodate any such mechanism. Photons have a frequency/wavelength/energy, and they have a polarization, but that's it. They don't have any internal structure or properties in which to store information about how far they have traveled or how much they have red shifted/blue shifted. A 1 meV photon that was emitted at 1 meV is identical to a 1 meV photon that was red shifted from 2 meV. There is no way to distinguish them. The universe cannot treat them differently, because they aren't different.

I can see why that would be a reasonable assumption make.

We could also assume all the electrons in my body just changed places with all the electrons in your body. There's no way of knowing.

[W.D.Clinger]

Originally Posted by Mike Helland

We could also assume all the electrons in my body just changed places with all the electrons in your body. There's no way of knowing.

The two of you could hold hands while one of you grabs the 800 volt output terminal of a 1 amp DC power supply and the other grabs its ground post. We would then have good reason to believe at least some electrons are moving between the two of you, along with a nice visual demonstration of the effect.

[zooterkin]

Originally Posted by Mike Helland

I can see why that would seem to be a reasonable assumption.

Do you disagree with it?

[Ziggurat]

Originally Posted by Mike Helland

I can see why that would be a reasonable assumption make.

We could also assume all the electrons in my body just changed places with all the electrons in your body. There's no way of knowing.

But that's just it: if your speculation was correct, we could know. It would necessarily have consequences, which is the entire point of your speculation. The whole thing about identical particles is that there ARE no consequences of an exchange. It's impossible.

[jonesdave116]

Originally Posted by Mike Helland

Attachment 47323

It should look like this:

http://www.internationalskeptics.com...607de1eecf.png

At z=1, you're halfway to the beginning of the universe, call that d=1, which would be half of Hubble's length.

Over this distance, a photon will lose half its energy.

If you double that distance, so d=2, then z=infinity, the photon will have redshifted away the other half of its energy.

Trivially wrong. Lerner wants a strictly linear relationship between d & z.
And how in hell would a photon lose all its energy in coming from d = 1 to d= 0, having travelled from d = 2????

If a photon only loses 50% of its energy in travelling initially from d = 1 to d= 0 (i.e. 8eV to 4eV), why would a 12eV photon, on it way from d = 2, arrive at d = 1 with 6eV of energy, and then disappear by the time it gets to d = 0????

You have an 8eV photon losing half its energy from d = 1 to d = 0, and a 6eV photon, travelling the same distance, losing all of its energy!

Sorry, but you need to crack open a calculator and try again! Better still, go back and read ben m's post. The photon, in Lerner's nonsense, needs to remember how much energy it started out with, and how far it has travelled. Do you think photons are intelligent?

The diagram I posted, based on ben m's post, is correct.

If you are using 1/(1 +z), that tells you the energy that the photon has remaining after travelling various distances.
If d = 1, then you have 1/(1 + 1) = 0.5. It retains 50%. So, an 8eV photo from d = 1 (equals z = 1 in Lerner's 'model') will arrive with 4eV of energy at d = 0.

If it starts at d = 2, the equation becomes 1/(2 + 1) = 0.33. It retains 1/3rd of its energy at d = 0. So, a 12eV photon also arrives at d = 0 with an energy of 4eV.
What is the energy as it passes d = 1?

Answers on a postcard.

[Mike Helland]

Originally Posted by jonesdave116

Trivially wrong. Lerner wants a strictly linear relationship between d & z.

Why are you obsessed with that guy?

Quote:

And how in hell would a photon lose all its energy in coming from d = 1 to d= 0, having travelled from d = 2????

Because d=1 is where z=1, and d=2 is where z=infinity.

Divide a frequency by 1+infinity, what do you get?

Quote:

If it starts at d = 2, the equation becomes 1/(2 + 1) = 0.33.

That's cartoonishly wrong.

d=2 is not where z=2.

[Mike Helland]

Originally Posted by zooterkin

Do you disagree with it?

I think a photon that has traveled farther may act differently.

The redshift distance relationship being evidence of that.

Wouldn't be hard to implement in a model.

while (true) {
    photon1.distanceTraveled++
    photon1.energy--
}

There you go.

In fact, you don't even need "photon1.distanceTraveled++" because the energy is decreasing independently anyways.

[Mike Helland]

Originally Posted by jonesdave116

If you are using 1/(1 +z)

It seems to me I don't really have to.

For z to be useful, it's always accompanied by a "1+".

Why?

There is another way.

With redshift z=0, this means it has shifted toward the red 0%.

z can be anywhere from -infinity (blueshift) or infinity (redshift).

You add one to this number (1+z) and then divide the frequency/energy of the photon and that's the observed frequency/energy.

When z=1, it's shifted half way red. What about when z=-1? Divide by zero. Hmmmm.

I have an alternative proposal, with no 1+ and no division.

Instead of redshift, we could have a "color fidelity" measurement, I'll just call "Q" arbitrarily.

Q = 1/(1+z)

Q goes in the opposite direction of z.

z starts at 0 and goes to infinity (for redshifts).

Q starts at 1 and goes to 0 (for redshifts)

z starts at 0 and goes to -infinity for blueshifts, which gives it higher energy.

If Q grows from 1 to infinity for blueshifts, it gets higher energy.

The energy of a photon and its z are inversely relataed, stretching to infinity in both directions.

The energy of a photon and its Q are not inverse, starting at 1 (100% color fidelity) and going to zero as the photon loses energy, and >1 as it gains energy.

Using Q only, which is as simple as Q=freq_observed/freq_emitted (no 1+ required) simplifies things and avoids a divide by 0 in an arbitrary place.

[Mike Helland]

Originally Posted by Mike Helland

Instead of redshift, we could have a "color fidelity" measurement, I'll just call "Q" arbitrarily.

Q = 1/(1+z)

...

Q=freq_observed/freq_emitted

The first equation is only approximate when z>0 and z<<1, which is where the relation is only considered valid anyways.

When z=1 the frequencies are halved, but where z=-0.5 they are doubled.

When Q=0.5 the frequencies are halved, and when Q=2 they are doubled.

Which makes more sense?

[zooterkin]

Originally Posted by Mike Helland

The first equation is only approximate when z>0 and z<<1, which is where the relation is only considered valid anyways.

When z=1 the frequencies are halved, but where z=-0.5 they are doubled.

When Q=0.5 the frequencies are halved, and when Q=2 they are doubled.

Which makes more sense?

That you don't understand what you're talking about?

[steenkh]

Originally Posted by Mike Helland

I think a photon that has traveled farther may act differently.

Maybe the photons that have traveled far just get tired, take a rest, and will recharge after a rest and start moving again later?

We seem to get back to the tired light hypothesis all the time, even though it is demonstrably wrong. You seem to be impervious to counter arguments: after a while you give up, and then later you just reissue the same tired light idea all over again, as if all the arguments against it never existed.

[Mike Helland]

Originally Posted by steenkh

Maybe the photons that have traveled far just get tired, take a rest, and will recharge after a rest and start moving again later?

What observations support that?

[Ziggurat]

Originally Posted by Mike Helland

I think a photon that has traveled farther may act differently.

How is the even possible? Again, photons have no internal properties that could carry info on how far they have traveled. Hell, how far they have traveled isn't even a single number, it's completely reference frame dependent. So which distance do we use?

[Mike Helland]

Originally Posted by Ziggurat

How is the even possible?

How is it possible some body of mass knows how and how much to warp spacetime?

Quote:

Again, photons have no internal properties that could carry info on how far they have traveled.

It's not like electrons are fortune cookies either, where we can open them up and see their mass written down.

Point is we make observations and we make models of what we observe. I'm not saying the photon knows how far it has traveled, but the universe might.

Quote:

Hell, how far they have traveled isn't even a single number, it's completely reference frame dependent. So which distance do we use?

Hmmm. That's a good question.

Photon's have spin, right?

[Ziggurat]

Originally Posted by Mike Helland

What observations support that?

None. That's rather his point.

[Ziggurat]

Originally Posted by Mike Helland

How is it possible some body of mass knows how and how much to warp spacetime?

Mass is the only required parameter.

Quote:

It's not like electrons are fortune cookies either, where we can open them up and see their mass written down.

You don't have to open them up to find out their mass. It isn't a hidden parameter.

Quote:

Point is we make observations and we make models of what we observe. I'm not saying the photon knows how far it has traveled, but the universe might.

How? Where is this information? It can't be in the photon.

Quote:

Hmmm. That's a good question.

That's why I asked. The problem is, there is no answer.

Quote:

Photon's have spin, right?

Spin is part of polarization, which doesn't change when traveling through a vacuum. And it's quantized, so it can't carry the information you want it to.

[Mike Helland]

Originally Posted by Ziggurat

How? Where is this information? It can't be in the photon.

There are inverse distance laws all over physics. Not what's so scary about that.

[Mike Helland]

Originally Posted by zooterkin

That you don't understand what you're talking about?

It's a true fact that to half the energy of photon, you redshift it by z=1.

To double it's energy, you redshift by z=-0.5.

And at z=-1, there is a divide by zero.

Seems pretty simple to understand.

1+z=freq_emit/freq_observed

I suggest:

Q=freq_observed/freq_emit

No 1+ stuff. No dividing, and no divide by zero problem.

[erwinl]

Originally Posted by Mike Helland

It's a true fact that to half the energy of photon, you redshift it by z=1.

To double it's energy, you redshift by z=-0.5.

And at z=-1, there is a divide by zero.

Seems pretty simple to understand.

1+z=freq_emit/freq_observed

I suggest:

Q=freq_observed/freq_emit

No 1+ stuff. No dividing, and no divide by zero problem.

How would you know what the freq_emit of a single photon was?

[Mike Helland]

Originally Posted by erwinl

How would you know what the freq_emit of a single photon was?

We tell the redshift of galaxies by the photons that are missing due to absorption.

[Hellbound]

Originally Posted by Mike Helland

There are inverse distance laws all over physics. Not what's so scary about that.

Sorry, but this is the funniest thing I've read on this forum in a long time.

Inverse square laws don't have anything to do with any values carried by the individual particles. They're and artifact of radial emission and the relationship of distance to area. There's nothing in any individual particle to indicate where it came from, or what point it's at on that inverse square curve.

[Mike Helland]

Originally Posted by Hellbound

They're and artifact of radial emission and the relationship of distance to area.

Distance, you say?

The strong nuclear force does its thing, and it dies out.

The assumption is that EM is infinite, and the observation are that it isn't.

Maybe it dies out too?

[Hellbound]

Originally Posted by Mike Helland

Distance, you say?

The strong nuclear force does its thing, and it dies out.

The assumption is that EM is infinite, and the observation are that it isn't.

Maybe it dies out too?

LOL

You should do a standup routine, really

ETA: If I double the length of a line, it's twice as long. If I double the diameter or a circle, it's four times as large. That's all you really need to know to understand inverse square relationships.

And somehow, you completely missed that.

ETA2: That's also all you need to know to understand that there's a significant difference between things that affect all emissions as a whole (inverse square), and things that apply to individual photons (the absolutely moronic tired light ideas). Of course, if one is completely ignorant of, well, pretty much the majority of modern physics, then I could understand the confusion. I leave it to readers to draw their own conclusions.

ETA3: Not even modern physics, really, unless one takes a loose definition of modern. I'm willing to allow it for the sake of tact.

[Mike Helland]

Originally Posted by Hellbound

LOL

You should do a standup routine, really

ETA: If I double the length of a line, it's twice as long. If I double the diameter or a circle, it's four times as large. That's all you really need to know to understand inverse square relationships.

And somehow, you completely missed that.

ETA2: That's also all you need to know to understand that there's a significant difference between things that affect all emissions as a whole (inverse square), and things that apply to individual photons (the absolutely moronic tired light ideas). Of course, if one is completely ignorant of, well, pretty much the majority of modern physics, then I could understand the confusion. I leave it to readers to draw their own conclusions.

ETA3: Not even modern physics, really, unless one takes a loose definition of modern. I'm willing to allow it for the sake of tact.

That's not lost on me at all.

Inverse square laws could be written more intuitively if their constant's were divided by 4pi, to make the 4*pi*r² more obvious:

https://physics.stackexchange.com/qu...ional-constant

But thinking back to the strong force, it's so much more powerful than the EM force, but it doesn't "work" far from the nucleus.

Maybe that's what redshifted EM is. It makes all atoms stick together. Does it really need to be infinite?