I don't think space is expanding.

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I've told you - I don't give two hoots what your process is. The output at the end can ONLY result from wavelengths being the same before and after reflection. You have an almighty cock-up somewhere. Find it. For example, are your starting and ending points in the same location in both cases? Are your trajectories straight lines? I don't know what the problem is with your model but I know there is one. Find it.

If I take out the * dt, it works like you would think, producing Snell's law.

This would mimic the light changing mediums, it's clock doing nothing out of the ordinary, but it's wavelength changing.

When I include the time step in the sum, to mimic slow light in a vacuum, it's works just like regular moving light in a vacuum.

The reason it does this, is that Fermat's least time principle would normally prefer the photon that hits the mirror the soonest, ditches its low v for a high v, getting to the observer before photons that reflect closer to the observer point.

In this situation, when the photon experiences slower time (in the relativistic view, it would be a non-zero time) then the least time principle no longer favors the photon that reaches the mirror first. It may get faster the soonest, but now it has the longest length to the observer point post-reflection, which is when its clock runs the fastest.
 
I've told you - I don't give two hoots what your process is. The output at the end can ONLY result from wavelengths being the same before and after reflection. You have an almighty cock-up somewhere. Find it. For example, are your starting and ending points in the same location in both cases? Are your trajectories straight lines? I don't know what the problem is with your model but I know there is one. Find it.

I can get the same thing in the classical version too:

https://mikehelland.github.io/hubbles-law/other/reflection3.htm

I just make one change:

photon.clock += 1/(1+z)

That's it.

When z=0, clock ticks 1, 2, 3.

When z=1, clock ticks 0.5, 1, 1.5, 2...
 
I can get the same thing in the classical version too:

https://mikehelland.github.io/hubbles-law/other/reflection3.htm

I just make one change:

photon.clock += 1/(1+z)

That's it.

When z=0, clock ticks 1, 2, 3.

When z=1, clock ticks 0.5, 1, 1.5, 2...
It's hard to know exactly what you are doing without the maths (quoting chunks of incomprehensible code is not presenting the maths). Your description is incoherent. But, it seems like you are introducing a fudge factor so that when the path lengths are different before and after reflection you fudge them to be the same. This is worthless and equivalent to:

Mike: I think F = m(a^2)

Us: No it's not. The experiments prove that's not correct.

Mike: It is true if I set m proportional to 1/a - then it matches the experiments.

Your "photon clock" is just a fudge factor to get the answer you want. Such a thing doesn't and can't exist. The only clock that matters is the one in the rest frame of the mirror.

I can tell by looking at your output that the phase for each path is exactly the same as in the base case. I don't care how you got there but in the real world the only way that can happen is if the wavelength is the same before and after reflection. There's nothing more to be said.
 
If I take out the * dt, it works like you would think, producing Snell's law.

This would mimic the light changing mediums, it's clock doing nothing out of the ordinary, but it's wavelength changing.

When I include the time step in the sum, to mimic slow light in a vacuum, it's works just like regular moving light in a vacuum.

The reason it does this, is that Fermat's least time principle would normally prefer the photon that hits the mirror the soonest, ditches its low v for a high v, getting to the observer before photons that reflect closer to the observer point.

In this situation, when the photon experiences slower time (in the relativistic view, it would be a non-zero time) then the least time principle no longer favors the photon that reaches the mirror first. It may get faster the soonest, but now it has the longest length to the observer point post-reflection, which is when its clock runs the fastest.

You really don’t understand how to do integrals, because you are doing them completely wrong. One of the obvious clues is that the Feynman path integral should produce the same answer as the classical method, but it doesn’t. That’s because, again, you are doing it wrong.

I was trying to figure out how to describe your approach to physics, and I’ve got the answer now. You aren’t doing science, you are playing Calvinball.
 
You really don’t understand how to do integrals, because you are doing them completely wrong. One of the obvious clues is that the Feynman path integral should produce the same answer as the classical method, but it doesn’t. That’s because, again, you are doing it wrong.

The classical and quantum methods give the same result. They both produce Snell's law, until you add my hypothesis, then they produce regular angles.


https://mikehelland.github.io/hubbles-law/other/reflection_nm.htm

https://mikehelland.github.io/hubbles-law/other/reflection3.htm

I don't expect you to accept that.

But as far as I'm concerned, the reflection problem is solved.
 
I can tell by looking at your output that the phase for each path is exactly the same as in the base case. I don't care how you got there but in the real world the only way that can happen is if the wavelength is the same before and after reflection. There's nothing more to be said.

Do the math.

Use dt=0.5 before the reflection, and dt=1 after.

You'll seee.
 
The classical and quantum methods give the same result. They both produce Snell's law, until you add my hypothesis, then they produce regular angles.


https://mikehelland.github.io/hubbles-law/other/reflection_nm.htm

https://mikehelland.github.io/hubbles-law/other/reflection3.htm

I don't expect you to accept that.

But as far as I'm concerned, the reflection problem is solved.

Calvinball it is. Have fun with your delusions, you clearly aren’t interested in learning anything.
 
That isn’t doing the math. That isn’t how integrals work.

Here's what Feynman says:

So, I thought I was finding out what Dirac meant, but, as a matter of fact, had made the discovery that what Dirac thought was analogous, was, in fact, equal. I had then, at least,
the connection between the Lagrangian and quantum mechanics, but still with wave functions and infinitesimal times.

It must have been a day or so later when I was lying in bed thinking about these things, that I imagined what would happen if I wanted to calculate the wave function at a finite interval later. I would put one of these factors exp( i  L /  ) in here, and that would give me the wave functions the next moment, t   and then I could substitute that back into (2) to get another factor of exp( i  L /  ) and give me the wave function the next moment, t + 2  , and so on and so on. In that way I found myself thinking of a large number of integrals, one after the other in sequence. In the integrand was the product of the exponentials, which, of course, was the exponential of the sum of terms like  L /  . Now, L is the Lagrangian and  is like the time interval dt, so that if you took a sum of such terms, that’s exactly like an integral. That’s like Riemann’s formula for the integral  Ldt , you just take the value at each point and add them together.
 
Yes. It is obviously wrong and worthless, because that dt in the integral is not something you can ever "double". What you're doing is cargo cult physics.

So you're saying, don't include the timestep in the sum?

And then I should get Snell's Law.

And I told you, yeah, that's what happens. That's what I had first.

Then I added the redshifts.

 
So you're saying, don't include the timestep in the sum?

And then I should get Snell's Law.

And I told you, yeah, that's what happens. That's what I had first.

Then I added the redshifts.

http://www.internationalskeptics.com/forums/imagehosting/thum_762186056142463bf5.png
And you don't understand a word of that Feynman quote. And why should I teach you? You are probably the worst student I have ever encountered. You don't even understand that all you're doing is adding a fudge factor.

I have one more thing to say and then I'm done. You have obviously validated your idea to your own satisfaction. That is the easiest thing in the world to do. What's more difficult is persuading others, something you have utterly failed to do. After a gazillion pages, not a single person who has contributed to or read this thread thinks that you have a case. Not one. There are dozens of physics and science forums on the internet - why don't you go and try your luck with them? I guarantee that you will not persuade a single physicist on any of those forums either, but be my guest - spend your life impotently trying. (I also warn you that you'll be rapidly banned on some of those forums - they do not waste time as we do with propositions which are obviously worthless).

What you have done successfully is to display one fundamental misconception after another, culminating in the idea that the dt of an integral can be "doubled". The proposition that someone who doesn't understand the meaning of
ql_c0a866aef306d642643a2fe72334ca89_l3.png
nor what the dt in the integrand actually means (I learned that when I was 15 years old) could overthrow all of physics would be very funny if it wasn't also so sad.

What have we learned? That the combination of profound ignorance and arrogance is distasteful and completely unproductive. You think you can contribute to physics but you don't want to do the years of hard work to learn the maths and physics as the rest of us do. I think you were right when you said your problems were caused by black mould. I'm out of here.
 
Here's what I have so far:

Photon Velocity

The redshifts in this hypothesis are not caused by the expansion of space, or other interactions as in tired light theories (discussed in a later section). The redshifts are interpreted as their own phenomenon that is as fundamental to nature as inertia. Light from distant galaxies is observed to be redshifted because light redshifts with distance.

The decelerating photon hypothesis is that cosmological redshifts indicate new physics for a photon:

A photon loses frequency as it travels cosmological distances, but with no change to wavelength, resulting in a loss of speed, according to v=c/(1+D/H)^2
The energy of a photon absorbed by matter is emitted as a new photon with D=0, and therefore v=c and an elongated wavelength.

In this interpretation of the redshifts, the frequency decreases while the photon is in flight, and the wavelength increases at the beginning of a new photon's journey. This is different from the expanding theory where frequency and wavelength change together without affecting the photon's velocity.

This poses a number of conflicts with existing physics. One of the most illustrative issues involves the Hubble Space Telescope (HST) and Snell's law.

The hypothesis says that the photon travels at less than c, and then after interacting with matter, new photons that travel at c are emitted. If decelerated photons reach a mirror and change speed, the photon's motion can be compared to that of light changing mediums.

This would result in a change to the angle at which decelerated light reflected off a mirror, which would be relevant in the case of the HST observing highly redshifted galaxies.

One would expect then, that if redshifted photons are traveling at less than c, the HST would not be able to resolve them at the expected angles with any clarity. Since the HST does resolve such galaxies, light must be reflecting normally, which means the photons have to be moving at c, invalidating the hypothesis.

To demonstrate this, I built two models of light reflecting off a mirror. The first using Fermat's least time principle. The second using Feynman path integrals. These demonstrated the problem.

Photons and Reflection

For the purpose of solving the reflection problem, imagine a non-relativistic space. In this space there are observers who experience time according to one master clock. In this space there are also photons, and each photon has its own clock. Vaugley similar to relativity, but not quite. Every photon's clock begins synced with the master clock, but falls off with distance, at a rate of 1/(1 + D/H)^2 which is equal to 1/(1 + z). This is derived in a later section.

Previously, the code for the model of the decelerating photon looked like this:

Code:
    photon.dx = c / Math.pow(1 + photon.x / H, 2)

The hypothesis was directly changing the velocity based on distance. So solve our problems, we'll approach this slightly differently. The photon needs a clock that stats at zero, and this is what the hypothesis directly changes, like so:

Code:
    // find out the time slice for a photon at this distance
    photon.dt = 1 / Math.pow(1 + photon.x / H, 2)

    // add that time slice to the photon's clock
    photon.clock += dt

    // move the photon at the speed of light for the time slice
    photon.dx = c * dt

The photon will actually always be traveling at c according to its own clock. But since that clock gradually runs slower than the master clock, the photon appears to decelerate according to the observers in the model.

Now when we run the problem demonstration of Fermat's least time principle, rather than choose the photon that arrived first according to the master clock, we choose the photon whose clock has the lowest reading.

Lastly, when the photon hits the mirror, it will be emitted as a new photon with D=0, causing dt=1, causing the v=c, and creating an elongated wavelength.

See: Reflection in classical physics

https://mikehelland.github.io/hubbles-law/other/reflection3.htm

When run, photons reflect at the same angle as photons moving at c in a vacuum, no longer demonstrating a problem with the hypothesis. This works because the least time principle always favored the photons that hit the mirror first when a speed increase was involved. If a photon was the first to reflect and gain a higher speed, it had a considerable advantage over the other photons traveling at the slower speed. In the updated demonstration, the time is measured by each photon's clock, which changes speed when the photon hits the mirror. For the photon that hits the mirror first, the speed boost no longer provides an advantage, because the photon's clock speed will have increased too, canceling out any advantage being first to the region once held.

A similar resolution can be added to Feynman's path integral in quantum electrodynamics. Each photon has a clock, which provides the time slice to be used. Photons that are redshifted have slow clocks dt<1, when they are reflected as fresh photons, their clock speed returns to dt=1.

Accounting for the difference in dt before and after the reflection, the problem demonstration again shows there is no problem at all.

https://mikehelland.github.io/hubbles-law/other/reflection_nm.htm
 
Despite the constantly churning incoherence of his "ideas", Mike Helland has demonstrated one fact beyond all possible doubt: that Mike Helland has absolutely no understanding of calculus. Summarizing for the record:

It's because the wavelength doubles and so does dt.

What is dt?


How in the name of Beelzebub does the infinitesimal time in an integral double? You have no idea what you are even looking at when you look at the integral - as soon as poseurs pretend to understand maths, they are discovered.

I'm using dt=1 for the full speed photon, and dt=0.5 for the half speed photon.

You really don’t understand how to do integrals, because you are doing them completely wrong....

I was trying to figure out how to describe your approach to physics, and I’ve got the answer now. You aren’t doing science, you are playing Calvinball.

Do the math.

Use dt=0.5 before the reflection, and dt=1 after.

Calvinball it is. Have fun with your delusions, you clearly aren’t interested in learning anything.

That isn’t doing the math. That isn’t how integrals work.

I showed you my work.

Yes. It is obviously wrong and worthless, because that dt in the integral is not something you can ever "double". What you're doing is cargo cult physics.

What you have done successfully is to display one fundamental misconception after another, culminating in the idea that the dt of an integral can be "doubled". The proposition that someone who doesn't understand the meaning of
ql_c0a866aef306d642643a2fe72334ca89_l3.png
nor what the dt in the integrand actually means (I learned that when I was 15 years old) could overthrow all of physics would be very funny if it wasn't also so sad.

What have we learned? That the combination of profound ignorance and arrogance is distasteful and completely unproductive. You think you can contribute to physics but you don't want to do the years of hard work to learn the maths and physics as the rest of us do.


The phrase I highlighted is exactly what Mike Helland has taught us. Not much of an accomplishment, really, since we knew it before Mike Helland showed up here, but reminders can be worthwhile, especially when the unproductiveness of folly is demonstrated so clearly as it has been within this thread.
 
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The phrase I highlighted is exactly what Mike Helland has taught us. Not much of an accomplishment, really, since we knew it before Mike Helland showed up here, but reminders can be worthwhile, especially when the unproductiveness of folly is demonstrated so clearly as it has been within this thread.

My computer is better at discrete time steps than spacetime continuums.

There is some evidence that redshifts are quantized.

Make of that what you will.
 
The one solid falsification you had disappeared.

Think about that.

No, Mike, it didn't. You thought it did because you're doing the math wrong. And now you're just proving that in addition to being clueless about physics, you're clueless about calculus.

Here's what Feynman says:

That's nice. It's also irrelevant, because (again) you're still doing the math wrong.

Here's a hint: dt isn't what you think it is. When doing a numerical approximation of the integral, you have to choose a value for your time step, it is not determined by any clock. And if you have chosen a good value for your time step, then making it smaller won't appreciably change the answer. If it does appreciably change the answer, then you done ****** up, boy.
 
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