Merged Relativity+ / Farsight

Here's something similar to what Farsight has been doing here: the Koide formulaWP for the leptons' masses. It works remarkably well, but is it a lucky accident?

I took a look at what's in ctamblyn's spoiler tags:

I looked at the five "identities" on the bottom, and the first and fourth ones essentially state that pi is an algebraic number. Since that is not the case, those "identities" are thus incorrect, even if they are rather close numerically. The third and fifth ones are also close numerically, but like the first and fourth ones, their errors are well above Mathematica's default floating-point precision.

The second identity is, however, correct, and it's fairly easy to prove. It's equivalent to
cos(pi/7) + cos(3pi/7) + cos(5pi/7) + cos(7pi/7) + cos(9pi/7) + cos(11pi/7) + cos(13pi/7) = 0

= Re( sum of exp(k*pi*i/7) for k from 1 to 13 by 2 )
= Re( exp(pi*i/7) * sum of exp(2k*pi*i/7) for k from 0 to 6 )
= Re( exp(pi*i/7) * (1-exp(14pi*i/7))/(1 - exp(2pi*i/7)) )
= 0

Mathematica doesn't handle trig functions as well as I might want, so I worked with Chebyshev polynomials in x = cos(pi/7).
1st kind: Tn(cos(a)) = cos(n*a)
2nd kind: Un(cos(a)) = sin((n+1)*a)/sin(a)
 
(about a Wikipedia article on two-photon processes)
And that a photon of light, a single wave, spends its time constantly morphing into an electron and a positron, which then magically morph back into a single photon, which nevertheless manages to keep on going at the speed of light! It just isn’t true.
It may seem totally silly, but that's indeed what happens.

Propagators are an important part of both classical and quantum field theories. You insert a particle at position x and remove it from position y and calculate a strength function for going from x to y: P(x,y).

The inverse-square law of gravity and electrostatics is a simple example of a propagator in action, and there are fancier ones that I could point out.

When one tries to calculate a photon's propagator with quantum field theory, one gets the bare-photon one + photon-electrons-photon + photon-electrons-photon-electrons-photon + ...

But by redefining the bare-photon one and various other quantities, it's possible to get the appropriate zero-four-momentum limit for the photon, the one that we commonly experience. That's renormalization, and it got Richard Feynman his Nobel Prize.

So Farsight's attempts to laugh away this effect are a miserable failure.

The upshot is that the standard model doesn't offer any explanation of how pair production works,
Yes it does, and you can predict its rate from the Standard Model.

and it doesn't tell you what the electron is.
Yes, a Dirac spinor field.

(John Archibald Wheeler...)
But as we know, he was confused about curved space v curved spacetime, and didn’t follow it through using displacement current.
How was that supposed to be the case?

He understood the equations of general relativity, and he even wrote a big fat book about GR with two of his colleagues.

Have a read of the Watt balance section of the wikipedia kilogram article, wherein the proposal is to define the kilogram using h and c and not much else. If you can do that for the kilogram, you should be able to do something similar for electron mass too.
Talk about mixing up two different sorts of things.

Measurement units are essentially arbitrary, while something like the electron's mass isn't.
 
There's a lot of issues with the Standard Model, more than people appreciate.
I agree - the successes and failures of the Standard Model are not discussed enough.
So what, Farsight?

What is worse is people displaying ignorance, e.g. your example of mentioning the (not Standard Model and not that good) two-photon physics article on Wikipedia.

You need to think about it, Farsight.
What you quoted was standard quantum electrodynamics. It states that
“A photon can, within the bounds of the uncertainty principle, fluctuate into a charged fermion-antifermion pair, to either of which the other photon can couple”.
This is not a definition of pair production :jaw-dropp!

That article links to Pair production
Pair production refers to the creation of an elementary particle and its antiparticle, usually when a photon (or another neutral boson) interacts with a nucleus or another boson. For example an electron and its antiparticle, the positron, may be created. This is allowed, provided there is enough energy available to create the pair – at least the total rest mass energy of the two particles – and that the situation allows both energy and momentum to be conserved.

The upshot is that you are making up fairy tales about what you read.
Two-photon physics is about real photons.
Pair production is about real photons.
A real photon does really turn into virtual electron-positron pairs all on its own all the time.
 
You know from atomic orbitals that "electrons exist as standing waves".
We know that you are quote mining that article, Farsight :jaw-dropp!
atomic orbitals
In this sense, the electrons have the following properties:
Wave-like properties:
  1. The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves. The lowest possible energy an electron can take is therefore analogous to the fundamental frequency of a wave on a string. Higher energy states are then similar to harmonics of the fundamental frequency.
  2. The electrons are never in a single point location, although the probability of interacting with the electron at a single point can be found from the wave function of the electron.
Particle-like properties:
  1. There is always an integer number of electrons orbiting the nucleus.
  2. Electrons jump between orbitals in a particle-like fashion. For example, if a single photon strikes the electrons, only a single electron changes states in response to the photon.
  3. The electrons retain particle like-properties such as: each wave state has the same electrical charge as the electron particle. Each wave state has a single discrete spin (spin up or spin down).
 
That's an understatement.

No, don't do that. It won't work.

I will work out 4πn/√(c³) for you though. It's 2.418x10-12 s1.5/m1.5. Does that look like a wavelength to you?
It certainly doesn't to me. The Compton wavelength is 2.426x10-12 m. Do you see the difference? Hint: it's not 8x10-15.
You've done it wrong. Planck length is l=√(ћG/c³). When you replace √(ћG) with 4πn where n is a suitable value, l=4πn/√c³) still yields a length.

edd said:
This is because the formulae for them aren't dimensionally incorrect, like Worsley's are.
You made the mistake. The expression is dimensionally correct, ctamblyn knows it, only he isn't honest enough to correct you. What you've done edd is create a straw man, and now you're clutching at straws. You can diffract an electron. Atomic orbitals do employ spherical harmonics. And despite what lpetrich says, a photon does not spontaneously morph into an electron and positron, which manage to morph back into a SINGLE photon which nevertheless manages to propagate at c.

All: I see my standard-model post has been moved out of the standard-model thread. Did somebody complain about it?
 
You've done it wrong. Planck length is l=√(ћG/c³). When you replace √(ћG) with 4πn where n is a suitable value, l=4πn/√c³) still yields a length.
Um, edd did exactly what you said:
"Replace √(ћG) with 4πn where n is a suitable value. Now set n to 1, and work out 4πn/√(c³)."

How exactly does that work out to a length??
 
...A real photon does really turn into virtual electron-positron pairs all on its own all the time.
No it doesn't. You must know it doesn't, because the electron-positron pair has to morph back into a single photon, which is in breach of conservation of momentum. And if it did, that photon couldn't be propagating at the speed of light, because electrons and positrons can't travel at the speed of light. And of course if the photon energy is less than 1022keV, there isn't enough energy to create an electron and a positron.

Even a child can see the problems.
 
You've done it wrong. Planck length is l=√(ћG/c³). When you replace √(ћG) with 4πn where n is a suitable value, l=4πn/√c³) still yields a length.

You said "set n to 1". 1 banana? 1 fruit-bat? Edd was right, and I was too generous. Even though I was, and even if it hadn't given the wrong answer, all you (or Worsley) had done at best is add a new parameter to the model (namely n) in order to pretend to explain another one. It's pointless.

ETA: What Roboramma said.

All: I see my standard-model post has been moved out of the standard-model thread. Did somebody complain about it?

Ask in Forum Management.
 
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Mr. Duffield, the lack of response of the curvature issue - something you promised to address on the previous page of this very thread, but never did - is just as pertinent to the severe shortcomings of Relativity+ as the failure to recognise Andrew Worsley's numerology for what it is. And the curvature issue is but the most recent line of discussion, embarrassing for your Relativity+ theory, which has been abandoned without you properly addressing the issue. There are many more.
 
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ctamblyn: I've been busy, remember? OK, where were we? Looking at your post 1172:

1. I claimed that spatial sections through the Schwarzschild geometry have negative curvature.

No problem, you can claim anything you like.

You claimed that it wasn't actually spatial curvature.

Yes I did, I said it was a curvature of the metric. It's a curvature of your measurements, made using things moving through space over time. It isn't a curvature of space.

3. I showed how GR predicts a departure from Euclidean-ness in the spatial part of metric, using a measurement which doesn't depend on clocks.

But it does depend on the motion of light. You said "measured locally using standard metre sticks". The metre is the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second. And it's a measurement. It's a curvature in your metric.

4. You tried to claim that my rods were dependent on the motion of light through space.

They are. See above.

5. I pointed out that many choices of rod are possible, not just light-based ones, and all will agree because they are all responding to the same curvature of the same space.

You said you can also use mean travel distance of beta-decaying particles (with no e/m involved). Sorry, but beta decay involves the weak interaction, and e/m is still involved because what we're dealing with is the electroweak interaction. Search on beta decay. You also said this:

For a ruler, just to avoid confusion, I will use a near-monochromatic source of neutral delta baryons, and measure the distance they need to travel before half the the particles have decayed via the strong force. That way we get a perfectly good ruler without needing to appeal to electromagnetic interactions.

See hyperphysics. They decay in about 10^-23 seconds. They don't get very far do they? Not enough for any practical use. That's not a "perfectly good ruler". Before you know it you've got a neutron and a neutral pion. You should look up neutral pion decay on wikipedia:

The π0 meson has a slightly smaller mass of 135.0 MeV/c2 and a much shorter mean lifetime of 8.4×10^−17 s. This pion decays in an electromagnetic force process. The main decay mode, with probability 0.98798, is into two photons (two gamma ray photons in this case):

10^−17s? Electromagnetic force process? Two photons? My oh my. Where did that electromagnetism come from? Magic?

ETA:

As for the "many more", you're playing the naysayer, and you know it. See for example On the Wave Character of the Electron by G. Poelz on arXiv:

"A classical model of the electron based on Maxwell's equations is presented in which the wave character is described by classical physics. It uses a circulating massless electric charge field which moves in the spherical background of its own synchrotron radiation. A finite bound system exists which explains the features of the electron, and the wave character of this system yields a tight connection between the classical and the quantum mechanic world. The size of the object follows from the magnetic moment, and the finite synchrotron radiation requests a quantum mechanic core with a size determined via the angular momentum. The power of the synchrotron radiation yields the elementary charge i.e. the fine structure constant alpha. Two circulations are required to obtain a stable circular orbit which is consistent with the description with the Dirac equation. The mass of the particle follows from the internal movement with speed of light. The system appears in the external world as a standing wave with an amplitude propagating like the de Broglie wave".

Don't get in the way of scientific progress, ctamblyn. Or you will get steamrollered.
 
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Robo/edd: I do declare that somebody is trying to change the subject!

They teach you this sort of thing in school, since it's such a powerful way of spotting one's own mistakes. It's not a very powerful way of spotting other people's mistakes because they've usually caught them this way before publishing them.

It's a damning indictment of your competence in physics that you can't see the glaring flaw in this formula. Even a child can see the problems (it's in the higher parts of GCSE maths syllabuses from a quick dig around).

It's therefore very relevant to the discussion of the basic idea that you have any idea what you're talking about.
 
Edd, you made a mistake, get used to it. Now if you'll excuse me, I've got something else I need to do.
 
ctamblyn: I've been busy, remember?

OK, let's accept that it took you two weeks to write two short blog entries.

...
ctamblyn said:

Yes I did, I said it was a curvature of the metric. It's a curvature of your measurements, made using things moving through space over time. It isn't a curvature of space.

The distinction you are trying to draw is not there. In GR (which is based on Riemannian geometry), "curvature" means something very specific - it refers to something which has an effect on lengths and angles (i.e. the metric). You, here, are using the word "curvature" to mean something private that you have invented. Something, I suspect, that you are incapable of defining precisely and which is certainly irrelevant to GR.

Anyway, the point remains that if the spatial sections I described were ordinary, flat, Euclidean 3-space you would not detect the phenomenon I described, therefore the spatial slices are not ordinary, flat, Euclidean 3-space. More precisely, they have negative Gaussian curvature.

ctamblyn said:
But it does depend on the motion of light. You said "measured locally using standard metre sticks". The metre is the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second. And it's a measurement. It's a curvature in your metric.
...

I'll address your responses to 3, 4, and 5 together by noting that I defined a ruler which is not dependent on light, using strong decays:

ctamblyn said:
For a ruler, just to avoid confusion, I will use a near-monochromatic source of neutral delta baryons, and measure the distance they need to travel before half the the particles have decayed via the strong force. That way we get a perfectly good ruler without needing to appeal to electromagnetic interactions.

To this, you responded:

See hyperphysics. They decay in about 10^-23 seconds. They don't get very far do they? Not enough for any practical use. That's not a "perfectly good ruler".

As anyone with the most basic education in physics and mathematics knows:

(a) How far they go depends on how fast they are moving.
(b) Even if they travel only a femtometre, it doesn't prevent me using that as a standard of length. I can use an old trick called "multiplication" to generate longer lengths.

So, yes, it is a perfectly acceptable length standard. The fact that it is embarrassing for your theory because it doesn't depend on electromagnetism is not my problem.
 
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Edd, you made a mistake, get used to it. Now if you'll excuse me, I've got something else I need to do.

Even if you had got your dimensions correct in the post edd replied to - and you didn't, you said "set n to 1" - the rest of the argument is useless, for the reasons I mentioned (most recently) here, which several other people have also pointed out to you.
 

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