I don't think space is expanding.

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hecd2 said:
The wavelength is constant on reflection and the speed changes and therefore...?
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The wavelength is not constant on reflection.

Code: 
if (!photon.flipped && photon.y > boundary) {
    photon.flipped = true
    photon.wavelength = photon.wavelength * photon.dv

dv here is 1+z, works for any z.
 
Mike Helland said:
The wavelength is not constant on reflection.

Code: 
if (!photon.flipped && photon.y > boundary) {
    photon.flipped = true
    photon.wavelength = photon.wavelength * photon.dv

dv here is 1+z, works for any z.
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I don’t understand your code. Your output demonstrates that wavelength is constant on reflection. I have no idea where your bug is or why what you are modelling is different from what you think you are modelling.
 
hecd2 said:
I don’t understand your code. Your output demonstrates that wavelength is constant on reflection. I have no idea where your bug is or why what you are modelling is different from what you think you are modelling.
Click to expand...

It's not different from what he thinks he's modeling. It's different from what he says he's modeling.
 
hecd2 said:
I don’t understand your code. Your output demonstrates that wavelength is constant on reflection.
Click to expand...

How so?

That's not what the code is doing.

Are you watching the speed of the dials at the bottom of each sim?

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

That final wave that ripples through them is unique to the decelerated photon. It comes up with the same answer as the normal photon, but the dials move different.
 
Mike Helland said:
How so?

That's not what the code is doing.
Click to expand...
I don’t give a tinker’s arse for what your code is doing. I look at the output and I see that the result of your coding, whatever your intention, is constant wavelength on reflection.

Are you watching the speed of the dials at the bottom of each sim?

That final wave that ripples through them is unique to the decelerated photon. It comes up with the same answer as the normal photon, but the dials move different.
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That’s how I know the wavelength is constant on reflection. Your speed is less before reflection and so is the rate of rotation of the phase. The distance travelled per rotation is the wavelength. The distance travelled per rotation is the same before and after reflection. Your wavelength is constant so your frequency must go up on reflection (in your scenario).

It’s a poor show when you don’t even know what your model says, and it puts your arrogance from an hour ago about having silenced all your critics into perspective.
 
Mike Helland said:
It's 250 nm before, and 500 nm after.

The dial's actually rotate 4 times faster (I think) after they hit the mirror.
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How is the mirror conveying 4x rotation speed to the dials, in your theory? And why do you think, not know? Shouldn't that 4x increase fall out of your calculations quite naturally and incontrovertibly? Maybe you should ask Pixie to go over your math.
 
theprestige said:
How is the mirror conveying 4x rotation speed to the dials, in your theory? And why do you think, not know? Shouldn't that 4x increase fall out of your calculations quite naturally and incontrovertibly? Maybe you should ask Pixie to go over your math.
Click to expand...

It's because the wavelength doubles and so does dt.
 
Mike Helland said:
It's 250 nm before, and 500 nm after.

The dial's actually rotate 4 times faster (I think) after they hit the mirror.
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Four? Why? Anyway, there is something wrong because the output you have requires no change of wavelength. I don't think your light is travelling in straight lines from point A to B via the mirror (that might be a optical illusion though), but in any case, you cannot get the output you got with different wavlengths before and after reflection. So find your bug.
 
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Mike Helland said:
http://www.internationalskeptics.com/forums/imagehosting/thum_76218605556a3541be.png
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This is not good for my blood pressure, so I'm going to bugger off in a minute. 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. The convention is to put the dt at the end of the integrand , by the way.

Look, all your phases in the two cases are identical. That means that the paths as an integer plus fraction of a wavelength are the same in both cases. The wavelength is constant in the first case, so it must be constant in the second - the output tells me that. Find your bug.
 
theprestige said:
No. These are your inventions to justify your pretty pictures. Show us the math. Feynman has math, behind his pretty pictures
Where's your math?
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Right, one of the terms in Feynman's sum is dt.

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

Code: 
photon.dial +=  Math.sqrt(Math.pow(photon.dx * photon.dt , 2) + Math.pow(photon.dy * photon.dt , 2))  / photon.wavelength  * photon.dt
photon.x += photon.dx * photon.dt
photon.y += photon.dy * photon.dt 

if (!photon.flipped && photon.y > boundary) {
    photon.flipped = true
    photon.wavelength = photon.wavelength / photon.dt
    photon.dt = 1

The first line does most of the work. Take your speed, apply dt, add that to the dial, divide it by wavelength, and then you again, per the formula, multiple dt.

After adding this to the dial, it actually moves the photon by those amounts. If it hasn't reflected yet, and if it's made it to the boundary, set flipped to true, and divide wavelength by dt, which is 0.5, so multiple by two. Last set the clock to full speed, dt=1.
 
That integral represents the action of a massive particle by the way. You haven't been introduced to actions and Lagrangians (that's way down the line at the rate you are going) so why are you pretending you understand that integral. And it has nothing to do with photons. You're just posing.
 
hecd2 said:
This is not good for my blood pressure, so I'm going to bugger off in a minute. How in the name of Beelzebub does the infinitesimal time in an integral double?
Click to expand...

I don't know about that, but to make the photon move at less than c, I use a smaller dt.
 
This is the important thing:
Look, all your phases in the two cases are identical. That means that the paths as an integer plus fraction of a wavelength are the same in both cases. The wavelength is constant in the first case, so it must be constant in the second - the output tells me that. Find your bug.
See you tomorrow, by which point you will hopefully have found out what's wrong.
 
Mike Helland said:
They are only identical at the very last moment.

It's a whole other story up until that point.

It's something else.
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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.
 
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