The Compton Experiment Detected Absolute Motion of the Graphite Target @ 9,302,480 m/Sec

[kenseto]

The Compton Effect Experiment
The experimental set up for the Compton Effect Experiment is simple. It consists of an incident x-ray source that aims at a graphite target. The wavelength of the scattered x-rays are measured at the various deflection angles. The results of this experiment showed that the scattered x-rays have intensities peaked at two wavelengths. One peaked at the same wavelength as the incident x-ray and the other peaked at a longer wavelength (red shifted) than the incident x-ray. The difference between the two wavelengths is called the Compton shift. The Compton shift increases as the scattering angle is increased.

Current Interpretation of the Compton Experiment
The peak that has the same wavelength as the incident x-ray is the result of photons colliding with the combined electrons of the carbon atom. Each of these combined electrons has an effective mass of 22,000 electron mass. Therefore, a photon colliding with it will retain almost all of its energy after the collision and thus the wavelength shift would be immeasurably small. The other peak is the result of photons losing some of their energy by colliding with the free electrons. These photons would have lost some energy and they would appear as being red-shifted.

New Interpretation of the Compton Experiment
The red-shifted peak is the reflection of the incident x-ray by the carbon nuclei that are in a state of receding absolute motion with respect to the incident beam. This process is the same as bouncing a radio beam off a receding object. The return beam will be red-shifted. The other peak is the result of absorption and re-emission of the incident photons. There is no energy lost by these re-emitted photons and therefore no change in wavelength was observed.

With this new interpretation, the absolute motion of the graphite target can be calculated as follows:

Data from the Compton experiment:
Incident wavelength=λ=0.0709E-9 meters
Wavelength at 90 degree after emerge from Graphite Target =λ’=0.0731E-9
Delta(λ)= λ’-λ = 0.0022E-9 meters
Frequency of incident beam = 4.22842E18 Cycles/s

The absolute motion of the graphite target Va can be calculated according to the following equation as follows:
Va= Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree)
= (0.0022E-9)(4.22842E18)(1)
= 9,302,524 m/s
= 9,302.524 km/s

The detection of the absolute motion of the graphite target invalidate all pass conclusions that there is no ether and no absolute motion. Also it invalidated the SR postulate that the speed of light is at constant c as measured by all observers.

[John Heath]

A secondary Compton gamma scatting off protons would have greater back scattering than forwards as the proton is stuck in place. Graphite is a solid.

In a real Compton experiment there are greater forward secondary Compton gammas than back scattered so it should be electron hits not proton.

I have a Compton experiment video to see my point.

https://www.youtube.com/watch?v=oY7EFL4AN9A

[Steve BH]

As explained to you before, even using your own formula, you'll get a DIFFERENT velocity for the same carbon block if you use a different incident frequency, since all other factors are THE SAME (and delta lambda will be the same-- which is the whole point of the Compton formula). And since you can use two incident beams at the same time on the same block, you will get the hilarious and nonsensical result that the one block is moving at two different velocities away from you, at the same time! (All while appearing at rest, which another bit of madness)

Use three different incident beams of three different wavelengths, and you'll get THREE different velocities.

How dare you promulgate results of this type and make fun on Einstein?

[Koobee Wublee]

On Sunday, December 3, 2017 at 8:59:51 PM UTC-8, Steve BH wrote:

> Why don't you try using the actual formula for the Doppler effect:

Uh... How do you derive the relativistic Doppler effect? You can find textbooks showing this, but they are all somewhat different. So, which source would you like to present? And we will talk about it. <shrug>

[Steve BH]

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

It's a simple derivation.

For small velocities v/c << 1 the redshift z = (Δλ)/λ = v/c.

But since Δλ is a constant for any angle in Compton, that means for any input λ you choose, you get out a different (v/c), because that redshift Δλ/λ will be a different number with Δλ fixed.

[David (Kronos Prime) Fuller]

Steve BH
- show quoted text -

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

It's a simple derivation.

For small velocities v/c << 1 the redshift z = (Δλ)/λ = v/c.

But since Δλ is a constant for any angle in Compton, that means for any input λ you choose, you get out a different (v/c), because that redshift Δλ/λ will be a different number with Δλ fixed.

That’s just Bulk Modulus & Wave Speed with Velocity Relative To Wave Speed as redshift

[Steve BH]

On Sunday, December 3, 2017 at 2:14:05 PM UTC-8, kenseto wrote:

Come on. Your number is 3% of the speed of light, so v/c = 0.031. That means the redshift z = delta(lambda)/lambda = 0.031.

Okay, you used 90degrees so delta-lambda is the Compton wavelength 0.0022 E-9.

So to get lambda in, you just multiply 0.0022 E-9 by 1/0.031 and obtain 0.0733 E-9 m input. No need to go through the frequency.

So I check your math, but now hear mine. Use an input frequency twice as much (or an input wavelength half as much), and your redshift doubles to .061. and now v/c is 6% of the speed of light, which is twice as fast. Wups.

[Odd Bodkin]

It is often the case that there is more than one good way to derive a
result. I realize that there are some autistic people for whom this is
unacceptable, who believe there can only be one correct way to make a
derivation.

--
Odd Bodkin -- maker of fine toys, tools, tables

[kenseto]

On Monday, December 4, 2017 at 12:07:31 AM UTC-5, Steve BH wrote:
> On Sunday, December 3, 2017 at 2:14:05 PM UTC-8, kenseto wrote:
> > The Compton Effect Experiment
> > The experimental set up for the Compton Effect Experiment is simple. It consists of an incident x-ray source that aims at a graphite target. The wavelength of the scattered x-rays are measured at the various deflection angles. The results of this experiment showed that the scattered x-rays have intensities peaked at two wavelengths. One peaked at the same wavelength as the incident x-ray and the other peaked at a longer wavelength (red shifted) than the incident x-ray. The difference between the two wavelengths is called the Compton shift. The Compton shift increases as the scattering angle is increased.
> >
> > Current Interpretation of the Compton Experiment
> > The peak that has the same wavelength as the incident x-ray is the result of photons colliding with the combined electrons of the carbon atom. Each of these combined electrons has an effective mass of 22,000 electron mass. Therefore, a photon colliding with it will retain almost all of its energy after the collision and thus the wavelength shift would be immeasurably small. The other peak is the result of photons losing some of their energy by colliding with the free electrons. These photons would have lost some energy and they would appear as being red-shifted.
> >
> > New Interpretation of the Compton Experiment
> > The red-shifted peak is the reflection of the incident x-ray by the carbon nuclei that are in a state of receding absolute motion with respect to the incident beam. This process is the same as bouncing a radio beam off a receding object. The return beam will be red-shifted. The other peak is the result of absorption and re-emission of the incident photons. There is no energy lost by these re-emitted photons and therefore no change in wavelength was observed.
> >
> > With this new interpretation, the absolute motion of the graphite target can be calculated as follows:
> >
> > Data from the Compton experiment:
> > Incident wavelength=λ=0.0709E-9 meters
> > Wavelength at 90 degree after emerge from Graphite Target =λ’=0.0731E-9
> > Delta(λ)= λ’-λ = 0.0022E-9 meters
> > Frequency of incident beam = 4.22842E18 Cycles/s
> >
> > The absolute motion of the graphite target Va can be calculated according to the following equation as follows:
> > Va= Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree)
> > = (0.0022E-9)(4.22842E18)(1)
> > = 9,302,524 m/s
> > = 9,302.524 km/s
> >
> > The detection of the absolute motion of the graphite target invalidate all pass conclusions that there is no ether and no absolute motion. Also it invalidated the SR postulate that the speed of light is at constant c as measured by all observers.
>
>
> As explained to you before, even using your own formula, you’ll get a DIFFERENT velocity for the same carbon block if you use a different incident frequency, since all other factors are THE SAME (and delta lambda will be the same--

Your explanation is based on bogus assertions. My calculations are based on actual experimental data. Are you saying that you get different Compton shifts when different incident frequency is used? In that case doesn’t this invalidate your previous bogus claim that: (λ1’-λ1)=(λ2’-λ2)=(λ3’-λ3)...... ?????

[kenseto]

So your calculation of 0.0733E-9 m incident wavelength disagreed with the actual wavelength 0.0709 E-9 Compton used. Why is your calculated wavelength is superior? I don’t think so. I think you are just making bogus assertions. BTW frequency is used to determine the absolute motion of the target.

[Steve BH]

Compton got .0731 OUT. You posted it yourself. Take his input o.0709 plus the Compton shift of .00243 and you get .07333 as the output. That is the answer. This is not complicated. You get the same shift no matter what the input is. If your input is 1, your output is 1.00243 at the same angle.

That shift of .00243 is the same whatever you START with. You are really being obtuse here.

[Steve BH]

No, you always get the same Compton wavelength shift at the same angle. But you add that to the input wavelength to get the output wavelength, so yes, if you use three different inputs, you will get three different outputs with the same shift. God, your math is horrid.

[Koobee Wublee]

On Sunday, December 3, 2017 at 10:39:37 PM UTC-8, Steve BH wrote:
> On Sunday, December 3, 2017 at 10:22:49 PM UTC-8, Koobee Wublee wrote:

> > How do you derive the relativistic Doppler effect? You can find
> > textbooks showing this, but they are all somewhat different. So,
> > which source would you like to present? And we will talk about it.
> > <shrug>
>
> https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

This derivation is one of the worst Koobee Wublee has ever since. It is full of handwavig, fudging, and fabrication --- work of an alchemist who knows nothing about science, physics, and relativity. <shrug>

For instance, the author started with the Lorentz transform written in the special case where only inertial frames of reference apply:

** x = (x’ + v t’) / √(1 – v²/c²)
** y = y’
** z = z’
** t = (t’ + v x’/c²) / √(1 – v²/c²)

Where very importantly,

** v = speed of primed observed by unprimed

And then, the longitudinal velocity transform equation:

** dx/dt = (v + dx’/dt’) / (1 + (v/c²) dx’/dt’)

The velocity transform can only be derived from the more general case of the Lorentz transform:

** dx = (dx’ + v dt’) / √(1 – v²/c²)
** dy = dy’
** dz = dz’
** dt = (t’ + v dx’/c²) / √(1 – v²/c²)

The latter form would not allow the author to fudge coordinates. <shrug>

Koobee Wublee stops at where the fudging becomes more creative. The author claimed erroneusly:

“The period... is not, however, t₂ – t₁ because...”

In any transform including the Lorentz transform, (dt = t₂ – t₁) always. <shrug>

Given past demonstrations to your intellectual level, Koobee Wublee is certain that you do not understand what Koobee Wublee is talking about. With that said, Koobee Wublee is obliged to show the following energy transform where both longitudinal and transverse Doppler effect can be calculated. This is called professionalism, embraced by most of the engineers, instead of alchemy which is a prefer way of the Einstein dingleberries. <shrug>

** E = (E’ + v\ * p’\) / √(1 – v²/c²)

Where

** E = energy observed in unprimed
** E’ = energy observed in primed
** p’ = momentum vector observed in primed
** v\ = velocity observed by unprimed
** * = dot product of 2 vectors

How do you derive the equation above? Hint: not from the very special case of the Lorentz transform. <shrug>

[Koobee Wublee]

On Monday, December 4, 2017 at 12:08:27 PM UTC-8, Steve BH wrote:

> You are really being obtuse here.

X-ray can strike an atom too. The scattering in the case is very much the same with the classical Compton effect. Since the mass becomes so much larger, the scattering becomes less and thus less detectable. <shrug>

Do you suppose when x-ray strikes an electron, that electron is stationary? If not, how would you modify the Compton effect to accommodate for this more general situation? <shrug>

[David (Kronos Prime) Fuller]

Koobee Wublee wrote

Koobee Wublee Is my Hero ...

I think I have a “Man Crush”

[David (Kronos Prime) Fuller]

Koobee Wublee wrote

On Sunday, December 3, 2017 at 10:39:37 PM UTC-8, Steve BH wrote:
> On Sunday, December 3, 2017 at 10:22:49 PM UTC-8, Koobee Wublee wrote:

> > How do you derive the relativistic Doppler effect? You can find
> > textbooks showing this, but they are all somewhat different. So,
> > which source would you like to present? And we will talk about it.
> > <shrug>
>
> https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

This derivation is one of the worst Koobee Wublee has ever since. It is full of handwavig, fudging, and fabrication --- work of an alchemist who knows nothing about science, physics, and relativity. <shrug>

For instance, the author started with the Lorentz transform written in the special case where only inertial frames of reference apply:

** x = (x’ + v t’) / √(1 – v²/c²)
** y = y’
** z = z’
** t = (t’ + v x’/c²) / √(1 – v²/c²)

Where very importantly,

** v = speed of primed observed by unprimed

See .... what Mr Koobee Wublee is doing different is using a Transformation for each of the primed and unprimed and the rest is “JUST HIDDEN BULK MODULUS & WAVE SPEED & DENSITY” of space time and the velocity relative to wave speed.

But you guy believe space time is nothing, no Aether Medium.
So this is all moot.

[Odd Bodkin]

You’re talking with someone who cannot simplify -6/-2.

[mlwo...@wp.pl]

Said an idiot rejecting Pythagorean theorem.

[Odd Bodkin]

Heck no. A lie, as expected from an incompetent modeler.

[mlwo...@wp.pl]

A sad truth.

> as expected from an incompetent modeler.

You're a woodworker, I'm an engineer of modelling
reality (because that's exactly what we, programmers,
are paid for). Don't fool yourself more than you've
already done, boy.

[Python]

mlwo...@wp.pl wrote:
> You're a woodworker, I'm an engineer of modelling

An engineer who rejects 90% of Physics and Mechanics,
including classical Newtonian/Galilean principles
established four centuries ago...

> reality (because that's exactly what we, programmers,
> are paid for). Don't fool yourself more than you've
> already done, boy.

You're the one fooling yourself, Mr Wozniak.

[rotchm]

But the facts are here...YOU failed to solve the simple math questions
posed in this NG; Odd did solve them.

Wanna shows us your skills again at some highschool logic problems?

[kenseto]

The actual shift was 0.0022E-9.

>
> That shift of .00243 is the same whatever you START with. You are really being obtuse here.

So the same shift for all incident x-rays? Do you have any data to show that? Does that mean that the same frequency output? Do you realize that in order to maintain a shift of 0.00243E-9 for all the sources the incident wavelength, the output wavelength and the frequency must shifted to specific values--I look forward for you to give us data to support your claim.

[kenseto]

So how does that invalidate my calculations that the same shift gives the same state of absolute motion of the graphite target?

[mlwo...@wp.pl]

On Tuesday, 5 December 2017 14:43:39 UTC+1, Python wrote:
> mlwo...@wp.pl wrote:
> > You're a woodworker, I'm an engineer of modelling
>
> An engineer who rejects 90% of Physics and Mechanics,
> including classical Newtonian/Galilean principles
> established four centuries ago...

Yet it moves - Galileo was rejecting them himself,
poor idiot.

[mlwo...@wp.pl]

On Tuesday, 5 December 2017 16:28:44 UTC+1, rotchm wrote:
> On Tuesday, December 5, 2017 at 8:01:21 AM UTC-5, mlwo...@wp.pl wrote:
> > On Tuesday, 5 December 2017 13:25:51 UTC+1, Odd Bodkin wrote:
>
> > > as expected from an incompetent modeler.
> >
> > You're a woodworker, I'm an engineer of modelling
> > reality (because that's exactly what we, programmers,
> > are paid for). Don't fool yourself more than you've
> > already done, boy.
>
> But the facts are here...YOU failed to solve the simple math questions

A lie, as expected from well known lying shit.
And, speaking of math, it's always good to remind
that your idiot guru had to reject it's oldest, very
important part - as it didn't want to fit his moronic
mumble he had to invent another, more "cooperative".

[rotchm]

On Tuesday, December 5, 2017 at 11:43:13 AM UTC-5, mlwo...@wp.pl wrote:
> On Tuesday, 5 December 2017 16:28:44 UTC+1, rotchm wrote:

> > But the facts are here...YOU failed to solve the simple math questions
>
> A lie, as expected from well known lying shit.

So, you refuse to try again!?
Want me to give out another simple highschool math problem to see that you will fail it once more? Why dont you want to try? Are you a coward too?

[mlwo...@wp.pl]

Speaking of math, it's always good to remind

that your idiot guru had to reject it's oldest, very
important part - as it didn't want to fit his moronic

mumble he invented another, more "cooperative".

[John Heath]

You are over thinking. Slow down as it is easy. Think of a pool table with a cue ball , gamma source and a red ball , electron. If you hit the red ball bang on all the energy went into the red ball electron. If you just nick the side of the of the red ball , electron , 99 percent of the energy stays with the cue ball , Compton secondary emission and 1 percent of the energy to the red ball , electron. It is just a matter of making sure energy is always conserved.

[Steve BH]

For this experiment where you are getting v/c about 0.031 you don't even NEED the relativistic Doppler equation. The ordinary Doppler equation (which is the same for v/c << 1) gives the same number you calculated for speed.

For Δλ = 2.43 E-12 m (a constant in all Compton experiments for theta = 180)
λ = incident initial wavelength

Δλ =~ λ * [v/c]

v = c Δλ/λ

You can plug this into the values you gave from the original Compton experiment and get the same v you got.

Unfortunately, since c and Δλ are constants, if you use any other λ for your incident radiation source you will get a different v.



Since c and Δλ are constants, this gives a different v for every

[Steve BH]

When an X-ray strikes an atom or electron bound to an atom, the equation is the same Compton equation but you use the mass of the atom, not the mass of one electron. So Δλ is m/M times the classical Compton value, or 1/24,000 of it for a carbon atom. This has since been measured, but it was too small for Compton to measure. It accounted for his beam being reflected with the wavelength "unchanged". It actually had been changed, but the change was Δλ = Δλc/24,000.

[Steve BH]

Yes, the shift (which is Δλ = 2.43 x 10^-12 m for 90 degrees and 4.86 x 10^-12 m for 180 degrees) is the same for all incident X-rays. Compton did not himself get exactly our modern value, which you can look up in any physics handbook.

Do you see an expression for a dependence of Δλ on the λ of the incident X-rays? No, you don't. It doesn't matter what λ for the incident X-rays is, the Δλ is the same. That's the whole point of Compton equation, which you are looking at but do not understand. Incredible.

http://www.spectrumtechniques.com/PDF/Compton%20Scattering%20Experiment%20by%20Prutchi.pdf

This is a student experiment. It has data for a number of incident lambdas. One is from Cs-187 where λ = 1.87 x 10^-12 m. So the energy of the back-scattered peak is going to be Eo λ/[λ+Δλ].

And that is exactly what it is. Eo from Cs-137 is 662 keV. The shift of 4.86 x 10^-12 m puts the reflected beam at 180 degrees at 1.87 + 4.86 = 6.73 x 10^-12 m. And what is that energy? Why, it's 662 keV [1.87/6.73] = 184 keV. They measure 180 keV. Similar shifts for other gammas for other elements are given in the table. Why don't you calculate them for yourself?



Give you data?? The Compton experiment has been reduced to the level of a college

[Steve BH]

No, you're UNDERthinking it. These are not billiard balls of the same mass. Forget the angle and just look at reflections straight back. You'll see that different fractions of the energy are transferred because the same amount of photon momentum must be transferred, no matter what its energy is. The result is the relativistic Compton equation for 180 degree backscatter:

Δλ = 2h/mc = 4.86 x 10^-12 m.

It's so very simple that apparently nobody on this list understands it. The change in wavelength of the 180 degree backward reflected photon is the same. No matter what energy and wavelength it starts with. If it starts at λ1, it will come back at wavelength λ1+Δλ. And if it goes in at λ2 it will still come back at λ2+Δλ, shifted by the absolute same amount.

Of course this represents very different energies depending on what λ (the incoming photon wavelength) is. If it's already 4.86 x 10^-12 m (an energy of exactly half the electron rest energy), its energy will be cut in half. If it's less than this, it energy will be MORE than cut in half. Photons of less energy than 4.86 x 10^-12 m lose less. A photon of 1/10th of this energy will lose only 10% of what energy it has. And so on.

Come on, folks, this is high school physics. I'm going to keep bonking you with it till you get it.

[Steve BH]

On Sunday, December 3, 2017 at 10:22:49 PM UTC-8, Koobee Wublee wrote:
> On Sunday, December 3, 2017 at 8:59:51 PM UTC-8, Steve BH wrote:
>
> > Why don't you try using the actual formula for the Doppler effect:
>
> Uh... How do you derive the relativistic Doppler effect? You can find textbooks showing this, but they are all somewhat different. So, which source would you like to present? And we will talk about it. <shrug>


Forget the relativistic Doppler effect. You can see that at the speed you calculate, just 3% of c, you can use the regular Doppler effect and the answer is the same.

For small velocities v/c << 1 the redshift z = (Δλ)/λ = v/c

Alas for you, c and Δλ are constants for the same scattering angle. That means you get a different v for every λ.

You are pwned.

[Steve BH]

http://www.spectrumtechniques.com/PDF/Compton%20Scattering%20Experiment%20by%20Prutchi.pdf

Read it and weep.

[John Heath]

Now that I think of it you are right. Why over simplify , whats the big hurry? Let us do the real Compton experiment with correct ratios energy wise. You will forgive me if I stick to my original pool table in 2 dimensions only , it help me think. You have taken the cue ball away and left me with a fluffy little ping pong ball for the gamma photon and the electron will be a tiny but heavy ball bearing. So there we have it a ping pong ball for the photon and a tiny but heavy ball ball bearing for the electron. You will note how Mr photon is larger then the electron so a 100 percent reflection back is not possible as the photon will go through and around the electron with only a partial reflection back for a head on target hit. This is about as close as I can get it . If wish to tweak the ping pong gamma photon , ball bearing electron , pool table model for anything you feel is missing please do.

Lets smash some electrons.

Quote

look at reflections straight back. You'll see that different fractions of the energy are transferred because the same amount of photon momentum must be transferred, no matter what its energy is. The result is the relativistic Compton equation for 180 degree backscatter:

Δλ = 2h/mc = 4.86 x 10^-12 m.

End quote

As I said there is no hurry. This thread could go on for months and still not completely cover the Compton experiment. I am a science nerd so if it takes months then so be it. I would like to touch on microwave ovens before proceeding to Δλ = 2h/mc = 4.86 x 10^-12 m.

I am a bacteria living in your microwave oven. You turn on the microwave oven. I do not care as it will have little effect on me. The reason is a microwave GHz photon with a wave length in the range of inches would have little effect on me at 1 / 100 of an inch. I would have a voltage gradient of 10 m volts tops. If I were a cheese slice it would be a different story but a bacteria on problem.

With this in mind the max energy a photon can convey to anything it runs into will be limited to the ratio of the photon wavelength to the physical size of what ever it runs into. photon wave length / size of target = the max amount of energy that can be exchange. This is not the amount of energy exchange. It only sets a limit to what the max energy exchange can be for a given event.

This is a good start to a detailed look at the Compton experiment with one formula that can be trusted.

photon wave length / target physical size = max possible energy exchange.

Time for bed. Feel free to fiddle with the pool table , ping pong and ball bearing so that you are comfortable with it. Also if the wave length / size formula for max energy exchange rubs you the wrong way please state so now as I will be using it to set the physical size of the electron based on hard data from a real Compton test.

[Koobee Wublee]

On Tuesday, December 5, 2017 at 6:41:53 PM UTC-8, Steve BH wrote:

> On December 4, 2017 at 1:35:31 PM UTC-8, Koobee Wublee wrote:

> > X-ray can strike an atom too. The scattering in the case is very
> > much the same with the classical Compton effect. Since the mass
> > becomes so much larger, the scattering becomes less and thus less
> > detectable. <shrug>
>

> When an X-ray strikes an atom or electron bound to an atom, the
> equation is the same Compton equation but you use the mass of the
> atom, not the mass of one electron.

Can you please stop echoing what Koobee Wublee has said? Instead, please address Koobee Wublee’s concern below. Maybe you will understand where Mr. Seto is coming from. <shrug>

[kenseto]

NO, speed of the photon is not c after it collide with the electron--how can it maintains to be c when you acknowledged that it looses momentum after collision?

The correct equation is:
Va=c'Δλ/λ
The equation says that the absolute motion of the target (Va) is constant and that the speed of the photon is changed after the collision.

> Since c and Δλ are constants, this gives a different v for every

NO.....c becomes c’ after collision.

[Odd Bodkin]

Because the momentum of a photon is not related to its speed. Why do you
think it is?

>
> The correct equation is:
> Va=c'Δλ/λ
> The equation says that the absolute motion of the target (Va) is
> constant and that the speed of the photon is changed after the collision.
>
>> Since c and Δλ are constants, this gives a different v for every
>
> NO.....c becomes c’ after collision.
>

[Steve BH]

For light p = E/c and E = pc. Even Maxwell knew this much. c = E/p where c is constant.

Einstein notes that E = pc = hf = hc/λ. Thus p = h/λ for light. There is only one c to factor out. De Broglie suggests p = h/λ for other particles as well, and that is our entry to quantum mechanics.

You want many different velocities of light and that's dishonest. You said your equation was Compton's, just derived by other methods. Compton has one c.

It's also wrong. There is just one speed of light, c, with one value, 3e8 m/sec. There is no ballistic light or slow light in vacuum, because it bounced off some electron. That would completely mess up the derivation, since λf = c' .

Also the photoelectric effect equations disappear since E = hf can no longer be true if f = c/λ and you have different c's to choose from to give you different photon E's for different c's at the same frequency.

Why don't you just think this over. You're wrong.

[David (Kronos Prime) Fuller]

kenseto wrote

NO, speed of the photon is not c after it collide with the electron--how can it maintains to be c when you acknowledged that it looses momentum after collision?

Bulk Modulus & Wave Speed of the Medium is c

((6.6740800122e-11 Pascals) / (3.71295775e-28 * 2 * (kg / (meter^3))))^0.5 = 299792458 m/s
https://en.wikipedia.org/wiki/Friedmann_equations#Density_parameter

c is the speed of “CAUSALITY”


http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html

http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/souspe2.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/imgsou/soundwater.gif

[kenseto]

ROTFLOL.....so you have this magic photon that can maintain its speed in space after loosing some of its momentum. Since momentum is: p=mv, how does your magic photon decreases it momentum without reducing its speed? Do you just making up stuff to suite your theory?

My formula to determine the absolute motion of the target is:

Va= Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree)

In this case Va is constant and Delta(λ) and frequency (f) are measured quantities.
There is no different Va with different incident beam.

[rotchm]

Correct. Its an empirical fact. Then again, if "it* loss some of its momentum, then "it" is no longer the same particle (the previous "it") that is was. So you cant say "it" has the same speed c; its the new particle, the new entity, the new "it" that has speed c (it is still a photon). But all this is way above your head. You are still struggling to understand how to turn on your microphone!

> Since momentum is: p=mv,

No it is not. Have you ever read any physics books (more importantly, have you understood them?). Obviously you havent. Start by reading the booklet that can with your microphone so that you may learn to turn it on...


<rest of idiot ken's nonsense & confusions snipped>

[Michael Moroney]

kenseto <set...@att.net> writes:

>ROTFLOL.....so you have this magic photon that can maintain its speed in
>space after loosing some of its momentum. Since momentum is: p=mv,

No, Stupid Ken. Momentum of a photon is *not* mv. How could it be? m=0
for a photon! The momentum of a photon is h/lambda.

> how does your magic photon decreases it momentum without reducing its
>speed?

It gets a larger lambda, Stupid Ken.

> Do you just making up stuff to suite your theory?

No, that's what you are doing. The momentum of a photon being h/lambda
has been part of QM/SR since the photon was discovered, it was not "made
up".

>My formula to determine the absolute motion of the target is:

Doesn't matter what your formula is, since IRT remains a total, complete
failure.

[David (Kronos Prime) Fuller]

We’re inside a black hole, c will always be measured to be c because the pressure is also Constant ...... you dishonest Obfuscating Fools

[Odd Bodkin]

Yes!

> Since momentum is: p=mv,

Wrong! That is NOT momentum. It applies to no object in the real world.
Measurements have already confirmed that. The expression p=mv is a handy
APPROXIMATION that only applies to slow moving objects. A photon does not
fit that restriction and this approximate formula is therefore completely
wrong for it.

> how does your magic photon decreases it momentum without reducing its
> speed? Do you just making up stuff to suite your theory?
>
> My formula to determine the absolute motion of the target is:
> Va= Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree)
> In this case Va is constant and Delta(λ) and frequency (f) are measured quantities.
> There is no different Va with different incident beam.
>

[Steve BH]

No, that would be you. Nobody who knew any physics would go around saying that the momentum of a photon was mv. Or that photons bounced off things and lost momentum by losing speed. Do we get slow photons, there, Booblie? Maybe they stop and stand still, like billiard balls. What's the slowest a photon can go, in vacuum?


=====

Booblie:

My formula to determine the absolute motion of the target is:
Va= Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree)
In this case Va is constant and Delta(λ) and frequency (f) are measured quantities. There is no different Va with different incident beam.

==============


Talk about making things up! I was promised the Compton equation derived by other means. It's not above. You have the Delta(λ) and (1-cos(theta)) on the same side now, so we can't even divide by the Compton equation and see what's left.

Let's help a bit. Set theta = 90 degrees, so 1-cos90 = 1. Then the Compton equation reads:

Delta(λ) = h/mc

Pretty simply, eh?

The problem is it has no variables! Theta was the last one. All that is left is h, m and c, which are all constants of nature:

Delta(λ) = h/mc = 0.00243 Angstrom.

Delta(λ) is a constant also in all Compton experiments. You asked me to provide experimental proof of this using different gamma radiation sources, and I did, above. And you ignored this, because you had to. At some point the crackpot must ignore the data and claim it is all a vast conspiracy against them. And now we have you down to the grade school physics (momenta of photons = mv) and to the point to the point you must ignore well known and simple equations, and even simpler experiments demonstrating them.

You retrodicted a velocity from ONE experiment (Compton's) but it doesn't work for any OTHER experiment of that type, because I posted results of some of those. Delta(λ) is the same even if the radiation comes from Cs-137. Oops. Explain that for us all.

[kenseto]

Momentum of a photon is:
p=h/λ=kg*m(meters)/s=m(mass)*v
That’s high school math.

[kenseto]

Momentum of a photon is:
p=h/λ=kg*m(meters)/s=m(mass)*v

That’s high school math--dimensional analysis

>
>
> =====
>
> Booblie:
>
>
> My formula to determine the absolute motion of the target is:

> Va= Momentum of a photon is:

> In this case Va is constant and Delta(λ) and frequency (f) are measured quantities. There is no different Va with different incident beam.
>
> ==============
>
>
> Talk about making things up! I was promised the Compton equation derived by other means. It's not above. You have the Delta(λ) and (1-cos(theta)) on the same side now, so we can’t even divide by the >Compton equation and see what's left.

I did not promise anything. Why do we need to derive the Compton equation by other mean?
My equation is:
Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
Va = (h/mc)*f(1- cos Φ)
It is designed to calculate the value of absolute motion Va of the graphite target using the Compton equation.

[Odd Bodkin]

Sorry, Ken, but no. Of course any momentum is going to have those
dimensions. This does not mean this is how you calculate it. The m for a
photon is zero. And yet the momentum of light is not zero. So how are you
going to resolve that problem, insisting that momentum is mv.

Look in your textbook. It will tell you that p=mv is not right and is only
an approximation.

[Steve BH]

> > > > > The equation says that the absolute motion of the target (Va) is constant and that the speed of the photon is changed after the collision.
> > > > >
> > > > > > Since c and Δλ are constants, this gives a different v for every
> > > > >
> > > > > NO.....c becomes c’ after collision.
> > > >
> > > >
> > > >
> > > > For light p = E/c and E = pc. Even Maxwell knew this much. c = E/p where c is constant.
> > > >
> > > > Einstein notes that E = pc = hf = hc/λ. Thus p = h/λ for light. There is only one c to factor out. De Broglie suggests p = h/λ for other particles as well, and that is our entry to quantum mechanics.
> > > >
> > > > You want many different velocities of light and that's dishonest. You said your equation was Compton's, just derived by other methods. Compton has one c.
> > > >
> > > > It's also wrong. There is just one speed of light, c, with one value, 3e8 m/sec. There is no ballistic light or slow light in vacuum, because it bounced off some electron. That would completely mess up the derivation, since λf = c' .
> > > >
> > > > Also the photoelectric effect equations disappear since E = hf can no longer be true if f = c/λ and you have different c's to choose from to give you different photon E's for different c's at the same frequency.
> > > >
> > > > Why don't you just think this over. You're wrong.
> > >
> > > ROTFLOL.....so you have this magic photon that can maintain its speed in space after loosing some of its momentum. Since momentum is: p=mv, how does your magic photon decreases it momentum without reducing its speed? Do you just making up stuff to suite your theory?
> >
> >
> > No, that would be you. Nobody who knew any physics would go around saying that the momentum of a photon was mv. Or that photons bounced off things and lost momentum by losing speed. Do we get slow photons, there, Booblie? Maybe they stop and stand still, like billiard balls. What’s the slowest a photon >can go, in vacuum?
>
> Momentum of a photon is:
> p=h/λ=kg*m(meters)/s=m(mass)*v
> That’s high school math--dimensional analysis
>
> >
> >
> > =====
> >
> > Booblie:
> >
> >
> > My formula to determine the absolute motion of the target is:
> > Va= Momentum of a photon is:
>
> > In this case Va is constant and Delta(λ) and frequency (f) are measured quantities. There is no different Va with different incident beam.
> >
> > ==============
> >
> >
> > Talk about making things up! I was promised the Compton equation derived by other means. It's not above. You have the Delta(λ) and (1-cos(theta)) on the same side now, so we can’t even divide by the >Compton equation and see what's left.
>


> I did not promise anything. Why do we need to derive the Compton equation by other mean?

Because you did say you had another way to the Compton equation:

> > > > > > > And let's see you derive the correct Compton equation for the frequency-shift from your assumptions, so YOU can get the Nobel Prize. Remember, your theory has nothing to do with free electrons, so you can't put the mass of an electron in it anywhere. Good luck.
> > > > >
> > > > > I got no answer. Here is the very simple formula (due to Compton) that fits the data:
> > > > >
> > > > > Delta lambda = [h/mc] [1-cos(theta)]
> > > > >
> > > > > Delta lambda is the change in wavelength of the X-ray beam. Also h is Planck's constant, m the mass of one electron, and c the speed of light. Theta is the scattering angle, which is zero if the beam continues it its initial direction like a pool ball that hits a smaller ball and keeps going in the same direction.
> > > > >
> > > > > You can see that if the beam hits the electron (or misses) and continues in the same direction (theta = 0, cos0 = 1) the change in momentum is minimal, and the wavelength changes by zero.
> > > > >
> > > > > If theta = 90 degrees (right angle scattering), cos(theta) = zero, and the change in wavelength is h/mc [1-0] = one Compton wavelength of the electron, which is 0.0243 Angstroms. The most it can change is twice this, because 1-cos(180) = -1, so this coefficient comes out 2.
> > > > >
> > > > > So never mind the angle. Please explain why the beam changes wavelength by zero Compton electron-wavelength at minimum, and two Compton wavelengths at maximum. This does fit the experimental data.
> > > > >
> > > > > Since you say this is a Doppler effect, you must explain why it only goes from ZERO up to 2h/mc for pure backscatter. You can't use h, since you deny this is a quantum process. You can't use m, the electron mass, as you deny one electron is involved.
> > > > >
> > > > > Show your work.
> > >
> > >
> > > Come on. If you have a better physical theory, what quantitative answers does it give for the Compton experiment that we can compare with the experimental results? Compton got the Nobel because he had to assume light was made of photons to get his result that agreed with reality. What about you?
> >
> > My theory provides a better physical explanation for Compton’s math. That’s all.
>
>
> Where is Planck's constant h in your theory? Where it the mass of the electron m in your theory?
>
> Show how your theory leads to Compton's math.

Booblie: My theory does not claim that it has a different way of deriving h or m of the electron. It merely provides a different physical explanation for the Compton math than the current explanation.


Can you read the above? You wrote it.

> My equation is:
> Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
> Va = (h/mc)*f(1- cos Φ)

> It is designed to calculate the value of absolute motion Va of the graphite target using the Compton equation.
>
> >
> > Let's help a bit. Set theta = 90 degrees, so 1-cos90 = 1. Then the Compton equation reads:
> >
> > Delta(λ) = h/mc
> >

> > Pretty simple, eh?

> >
> > The problem is it has no variables! Theta was the last one. All that is left is h, m and c, which are all constants of nature:
> >
> > Delta(λ) = h/mc = 0.00243 Angstrom.

> > > > > The correct equation is:
> > > > > Va=c' Δλ/λ

> My equation is:
> Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
> Va = (h/mc)*f(1- cos Φ)

Sure, sure, sure. Let's just put it all together.

Va = (h/mc)*f(1- cos Φ)

Va = c' Δλ/λ

THEREFORE

(h/mc)*f(1- cos Φ) = c' Δλ/λ

f = c' /λ

fλ = c'

Yeah the starting frequency f and the starting wavelength λ can be anything you decide, because you have inserted a totally bogus slow speed of light c' that you intend to measure after the experiment.

This is not the Compton equation or physics as we know it. It's your screwed up fantasy land.

[kenseto]

This is your equation. My equation has no c’ in it.

[Steve BH]

I simply combined your two equations algebraically. None are mine. If you don’t like what I did, then YOU combine the two equations you have already given and/or show the derivation of your velocity equation. The Compton equation has no c’. You apparently think the Doppler equation (feel free to use any form of this you like) does.

You are not getting out of this. You claim Compton shifted photons come off at some momentum mc’. Is that p= E/c’ perhaps? The Compton equation doesn’t work if that happens for then delta-lambda is not constant for a given angle (please assume that all experiments are done at 90 degrees so we can eliminate the cos term).

[kenseto]

On Saturday, December 9, 2017 at 2:13:31 PM UTC-5, Steve BH wrote:
> I simply combined your two equations algebraically. None are mine. If you don’t like what I did, then YOU combine the two equations you have already given and/or show the derivation of your velocity equation. The Compton equation has no c’. You apparently think the Doppler equation (feel free to use any form of this you like) does.
>
> You are not getting out of this. You claim Compton shifted photons come off at some momentum mc’. Is that p= E/c’ perhaps? The Compton equation doesn’t work if that happens for then delta-lambda is not constant for a given angle (please assume that all experiments are done at 90 degrees so we can eliminate the cos term).

My equations are:

Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
Va = (h/mc)*f(1- cos Φ)

At 90 degree:
Va = (h/mc)*f
Since (h/mc) and f are constant then Va is also constant.

[Steve BH]

I'm sorry, but you start with an equation that is out of nowhere.

Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).

Where did you get THIS? What is frequency f? If it is the input frequency, it is not true that f =c/lambda so Va = [delta-lambda]/lambda?

The problem with that is delta-lambda is a constant (it is h/mc, see Compton), but you can pick any input lambda you want.

I posted several experiments with the same delta lambda (which they must have-- again it is h/mc, not some retarded c' of yours) but different lambdas depending on input radioactive source. Indeed, Cs-137 gives a different lambda than in Compton's original experiment, since he didn't use that isotope.

[kenseto]

On Saturday, December 9, 2017 at 9:07:20 PM UTC-5, Steve BH wrote:
> On Saturday, December 9, 2017 at 1:56:41 PM UTC-8, kenseto wrote:
> > On Saturday, December 9, 2017 at 2:13:31 PM UTC-5, Steve BH wrote:
> > > I simply combined your two equations algebraically. None are mine. If you don’t like what I did, then YOU combine the two equations you have already given and/or show the derivation of your velocity equation. The Compton equation has no c’. You apparently think the Doppler equation (feel free to use any form of this you like) does.
> > >
> > > You are not getting out of this. You claim Compton shifted photons come off at some momentum mc’. Is that p= E/c’ perhaps? The Compton equation doesn’t work if that happens for then delta-lambda is not constant for a given angle (please assume that all experiments are done at 90 degrees so we can eliminate the cos term).
> >
> > My equations are:
> > Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
> > Va = (h/mc)*f(1- cos Φ)
> > At 90 degree:
> > Va = (h/mc)*f
> > Since (h/mc) and f are constant then Va is also constant.
>
>
> I’m sorry, but you start with an equation that is out of nowhere.

Moron it is an equation for the speed of any wave:
Velocity=(wavelength)(Frequency)

>
> Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
>
> Where did you get THIS? What is frequency f? If it is the input frequency, it is not true that f =c/lambda so Va = [delta-lambda]/lambda?

Moron.....your equation: Va = [delta-lambda]/lambda is dimensionless. Gee you are stupid. f is dependent on the angle of detection. At 90 degree, cos(phi) is (1) and thus f at 90 degree is the incoming frequency..

[kenseto]

[kenseto]

You are a moron woodworker. You don’t know any real physics.

[Michael Moroney]

kenseto <set...@att.net> writes:

>> Sorry, Ken, but no. Of course any momentum is going to have those
>> dimensions. This does not mean this is how you calculate it. The m for a
>> photon is zero. And yet the momentum of light is not zero. So how are you

>> going to resolve that problem, insisting that momentum is mv.=20

>You are a moron woodworker.

And you are just a moron.

>You don't know any real physics.

Yet Odd has shown repeatedly he knows tons more real physics than you do.

[Steve BH]

On Sunday, December 10, 2017 at 5:41:03 AM UTC-8, kenseto wrote:
> On Saturday, December 9, 2017 at 9:07:20 PM UTC-5, Steve BH wrote:
> > On Saturday, December 9, 2017 at 1:56:41 PM UTC-8, kenseto wrote:
> > > On Saturday, December 9, 2017 at 2:13:31 PM UTC-5, Steve BH wrote:
> > > > I simply combined your two equations algebraically. None are mine. If you don’t like what I did, then YOU combine the two equations you have already given and/or show the derivation of your velocity equation. The Compton equation has no c’. You apparently think the Doppler equation (feel free to use any form of this you like) does.
> > > >
> > > > You are not getting out of this. You claim Compton shifted photons come off at some momentum mc’. Is that p= E/c’ perhaps? The Compton equation doesn’t work if that happens for then delta-lambda is not constant for a given angle (please assume that all experiments are done at 90 degrees so we can eliminate the cos term).
> > >
> > > My equations are:
> > > Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
> > > Va = (h/mc)*f(1- cos Φ)
> > > At 90 degree:
> > > Va = (h/mc)*f
> > > Since (h/mc) and f are constant then Va is also constant.
> >
> >
> > I’m sorry, but you start with an equation that is out of nowhere.
>
> Moron it is an equation for the speed of any wave:
> Velocity=(wavelength)(Frequency)
>
> >
> > Va = Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree).
> >
> > Where did you get THIS? What is frequency f? If it is the input frequency, it is not true that f =c/lambda so Va = [delta-lambda]/lambda?
>
> Moron.....your equation: Va = [delta-lambda]/lambda is dimensionless. Gee you are stupid. f is dependent on the angle of detection. At 90 degree, cos(phi) is (1) and thus f at 90 degree is the incoming frequency..

Fine, I'll put the c back in for you, if you can't infer it:

Va/c = [delta-lambda]/lambda = (Δλ)/λ

That is the low velocity v/c <<1 Doppler red-shift formula I posted at the beginning of this thread on Dec 3. You think I don't know it? It's you who has no idea how to use it.

This is the equation you decided showed that Va was "9,302,524 m/sec" (or see header of this thread, which differs in the last three sig digits).

[Need I comment on the physics acumen of somebody who calculates v to 6 or 7 decimal places when they only have 3 sig decimal input data?]

In this case we have Compton's 0.0709 nm input = Δλ for input beam. And his Δλ of 0.0022 nm. [The modern measurement is closer to 0.0024 nm]

So v = 3.00e8 m/sec (0.0022/0.0709) = 9.31 x 10^6 m/sec = 9,310,000 m/sec.

But the problem (again) is that at the same angle of 90 degrees you could use any incident λ. You don't need to use Compton's gamma source of 0.0709 nm.

But even if you use a different λ, your Δλ will be the same Compton wavelength of 0.0024 nm (look it up-- it's .0024263102367(11) nm). This is a constant of nature, since it is h/c m_e.

That means Δλ/λ will be different, because the numerator Δλ is the SAME (I'm using the difficult math concept for you) while the bottom divisor HAS CHANGED. If you multiply that new quotient Δλ/λ by c you will get a different v. And yet you have not changed anything in the experiment, but the radiation source. Why isn't the object traveling just as fast through the aether? Δλ/λ cannot change if it does that, but if that is so, the Compton equation is WRONG. Which way do you argue?

So, who's the moron? I think it would be you, here. If you think there is something special about your v of 9300 km/sec, think again. Use an input wavelength of half as much and this "v" will be twice as large (if Compton is right). So your "Va" is not a real thing. It exists only in your mind.

[Odd Bodkin]

Then look in your textbook. It will tell you the same. That wasn’t written
by a woodworker, it was written by a physicist.

But I know you’re terrified of reading what physicists write.

[kenseto]

The Compton Edge for Cs 137, Co-60, Na-22 and Ba-133 are as follows:
Predicted Measured

Cs-137 482 Kv 469 Kv
Co-60 966 Kv 965 Kv
Na-22 344 Kv 333 Kv
Ba-133 210 Kv 213 Kv

I don’t see how these different sources have the same Delta(λ).

[Steve BH]

Perhaps you don't know what a Compton edge energy is? It's ΔE, or E-E' where E is the photon input energy and E' is the backscatter at the max of 180 degrees. If you got your table from the below source, which it appears you did, it's easier to simply use the values of E and E'. Otherwise you can calculate E' from E-E_comptonedge. And you need a gamma input for E for each radioisotope used.

http://122.physics.ucdavis.edu/course/cosmology/sites/default/files/files/Gamma%20Spec/Compton%20Scattering%20Experiment.pdf

The formula to convert E and E' to Δλ is:

Δλ = hc [1/E' - 1/E]

This formula follows directly from the simple fact that E = hv = hc/λ for both E and E'.

As the paper above points out, the simplest units to use hc in, here, is keV*nm, and the value of hc in those units is 1.24 keV*nm.

Okay, let's do the first one. Cs-137 makes 662 keV gammas (for E), the Compton edge is 482 keV and thus E' is 662-482 = 180 keV. And if you look at scattering spectra for Cs-137 it indeed goes down to 180 keV and quits.

Δλ = 1.24 keV*nm [1/180 keV - 1/662 keV] = 5.02 x 10^-3 nm. Taking half this for the 90 degree Compton electron wavelength gives 2.51 x 10^-3 nm = 2.51 pm. This is reasonably close to the literature 2.43 pm (which predicts Δλ = 4.86 pm for 180 degrees).

Let us do sodium-22 gammas, your third one. Predicted E and E':

Δλ = 1.24 keV*nm [1/167 keV - 1/511 keV] = .000500 nm = 5.00 pm. Again very close to theoretical Δλ = 4.86 pm.

I'll let you do the rest. These are within 3% and are all essentially the same Δλ, independent of source. When you get done you'll find out why Compton won the Nobel prize, and you didn't.

[Steve BH]

Erratum:

Δλ = 1.24 keV*nm [1/167 keV - 1/511 keV] = 0.00500 nm = 5.00 pm. Again very close to theoretical Δλ = 4.86 pm.

One too many zeros in the nm figure. All these Compton wavelengths are really naturally in the 10^-12 m = pm.

[kenseto]

My formula to determine the absolute motion Va of the target is:
Va= Delta(λ output-λ input)*((Frequency (f))(1- cos Φ at 90 degree)
For the detection angle at 90 degree this is simplified to
Va= Delta(measured λ output - measured λ input)*(measured Frequency (f)

I don’t know if Va will be the same with different sources because “measured λ output”, “measured λ input” and “frequency” are different for different sources. That means that we need do experiments to determine Va for different sources to see if Va is constant.

I rejected your equation for Va:
Va=c(Δλ/λ)
Because you automatically made Va to be different for different sources.

[Edwin Huckabee]

kenseto wrote:

>> I'll let you do the rest. These are within 3% and are all essentially
>> the same Δλ, independent of source. When you get done you'll find out
>> why Compton won the Nobel prize, and you didn't.
>
> My formula to determine the absolute motion Va of the target is:
> Va= Delta(λ output-λ input)*((Frequency (f))(1- cos Φ at 90 degree)
> For the detection angle at 90 degree this is simplified to Va=
> Delta(measured λ output - measured λ input)*(measured Frequency (f)

Nice try, but this is only descriptive, like Relativity. Could you pleas
put in it some numbers, numerically, as we say in physics. Your theory is
not a subset in Relativity, is it??

[numbernu...@gmail.com]

The Caltech-MIT LIGO detected stellar black hole gravitational waves using relativity to alter the armature length of Michelson interferometer but the LIGO experiment is based on a constant magnitude of Einstein's relativistic translational velocity but Einstein's translationtal velocity formed by the earth's daily and yearly motions is not constant. At the surface of the earth, for the time of 6:00 pm, the magnitude of the earth's tangential velocity vector that forms Einstein's translation velocity is 462 m/s (fig 7) and increases to 5,077 m/s at 7:00 pm. At midnight, the translational velocity is 30,462 m/s. The magnitude of Einstein's translational velocity increases from 462 m/s to 30,462 m/s (6:00 pm - 12:00 am) yet the LIGO experiment is based on the constant magnitude of Einstein's translational velocity to form the alteration of the armature length that forms the signal used to justify the existence of gravitational waves.

[Carl Susumu]

RE: CERN


In the CERN (2017) particle physics experiment, two hadron beams are collided to form subatomic particles but a proton has a mass 1,888 time larger than an electron and have a charge. The positive charges of the protons of the hadron beam would limit the concentration of protons in the hadron beam since like charge protons would repeal. It is unlikely that a high enough concentration of protons can produce a proton beam that could result in the collision of accelerated proton of the hadron beam. A collision implies a particle beam interacting with a metal structure. There is no physical example of an electron, nuclear or proton beam that interaction produces a collision of the entities within the beams when the beams are combined. When two particle beams are combine at perpendicular angles there is never an interaction that depicts a collision of the particles within the beams is occurring. In addition, it does not appear realistic that high energy protons of the hadron beam can be both accelerated and made to propagate on a exact circular path of the 27 km CERN vacuum tube since as the protons propagate through the vacuum tube the velocity of the protons increase which would require magnets that increasing magnetic fields are pointing towards the center of the circular 27 km vacuum tube. The size of the magnets would increase alone the path of the 27 km CERN vacuum tube to produce the increasing magnetic fields required to kept the protons contain in the 27 km circumference path of the CERN vacuum tube but the magnets along the vacuum tube appear to have the same size. Also, when the protons interact during the collision it is more likely that the protons would repel rather then collide and form subatomic particles. The high energy protons of the proton beam are propagating at the velocity of .999c that energy can cut steel which would make it very difficult to contain the CERN hadron beam within the 27 km circular vacuum tube without the protons contacting the outer surface of the vacuum tube enclosure producing a hole in the vacuum tube and eliminating the vacuum. The high energy protons that are propagating through the circular accelerator vacuum tube would have the energy to cut steel which can be tested by placing a .1 thick steel plate in that path of the accelerated proton beam. The accelerated protons at the end of the path through the vacuum tube would produces a hole in the steel test plate is placed in the path of the proton beam when the hadron beam accelerator is activated for approximately 20 seconds a hole would be formed in the steel test plate. Also, when the two hadron beams are combine a 300 $ (used) Geiger counter could be used to determine if the CERN experiment is accurately producing subatomic particles. The ATLAS experiment and Compact Muon Solenoid are the detectors that are used to detect the subatomic particles of the CERN accelerator collision. Also, the Compact Muon Solenoid is a metal structure that weighs 14,000 lb that resemble the Muon detector used the Soudan mine Muon experiment.

[JanPB]

Why do you laugh at people who understand something you do not and are
trying to explain it to you? Are you an adult?

> so you have this magic photon that can maintain its speed in space after loosing some of its momentum.

Where have you been for the last 100+ years? The root cause here is that
_time_ is not exactly what Newton thought it was. Photon's momentum is
not related to its speed, it's not Newtonian mechanics.

> Since momentum is: p=mv,

In Newtonian mechanics, yes.

> how does your magic photon decreases it momentum without reducing its speed?

How about learning physics? Why do you waste your time daydreaming when
you can have MUCH more fun learning the real deal?

> Do you just making up stuff to suite your theory?

It's called "physics". I know you've probably never heard of it but if that's
what you are interested in, STUDY it. Forget about creating new theories and
forget critiquing it until you get a firm grasp of the FACTS (NOT opinions).

> My formula to determine the absolute motion of the target is:
> Va= Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree)

It's gobbledygook. Junk crank pseudoscience.

> In this case Va is constant and Delta(λ) and frequency (f) are measured quantities.
> There is no different Va with different incident beam.

Yeah yeah, whatever. Just don't complain when you realise you've wasted
your life away on garbage.

--
Jan

[Edwin Huckabee]

JanPB wrote:

>> Do you just making up stuff to suite your theory?
>
> It's called "physics". I know you've probably never heard of it but if
> that's what you are interested in, STUDY it. Forget about creating new
> theories and forget critiquing it until you get a firm grasp of the
> FACTS (NOT opinions).

https://www.youtube.com/watch?v=qxJbQpdYINg

[JanPB]

https://www.youtube.com/watch?v=zotVTi3IPYs

--
Jan

[Steve BH]

You can simplify it further as Δ(measured λ output - measured λ input) = Δλ

Va= Δλ *(measured Frequency (f) input)

And I believe you called a moron because you didn't think I knew that f = c/λ

So now your own equation reads Va = c Δλ/λ

Then you say:

> I don’t know if Va will be the same with different sources because “measured λ output”, “measured λ input” and “frequency” are different for different sources. That means that we need do experiments to determine Va for different sources to see if Va is constant.
>
> I rejected your equation for Va:
> Va=c(Δλ/λ)
> Because you automatically made Va to be different for different sources.

The equation you "reject" is not just mine. You reject your own equation where λ is λ(input). The only way you can save it is to claim the "f" is output frequency, even though you used the input frequency in your demo Compton calculation to get Va above at the beginning of this thread:

kenseto: Frequency of incident beam = 4.22842E18 Cycles/s


Now you want to use the output f, do you? It doesn't matter. Δλ is a constant in all these experiments for the same angle. It is 4.84 picometers at 180 degrees. The speed of light is a constant. Since you pick λ_input and can pick two different ones at the same time for the same sample, but cannot get back two Va's if that number means anything, obviously your original equation is wrong. You just proved it. If you're going to claim you were talking about output λ or f all along:

Va=c Δλ / [λ_ouput] = =c Δλ / [λo]

Then λ_output must always be the same for the same sample (no matter what λ (in) is, because c is the same and Δλ is the same (4.84 picometers at 180 degrees, 2.42 picometers at 90 degrees).

The reason is since Δλ = λo-λi then λo= Δλ+λi so

Va = c Δλ / [λo] = c Δλ/[Δλ+λi]

That's no good as λi is the only variable and it's still chosen by the experimenter. Choose your λi and it tells you your Va by the above equation. Chose two different λi's and you must get two different Va's if the Compton equation is correct.

If you don't believe it, you have Compton experiment results for four different experiments and λi sources. I told you to calculate all the Δλ's for each, and you'd find them the same. You were too lazy to do that. So I can't help you.

[Michael Moroney]

Edwin Huckabee <t...@neetws.nd> writes:

>kenseto wrote:

>Nice try, but this is only descriptive, like Relativity. Could you pleas
>put in it some numbers, numerically, as we say in physics. Your theory is
>not a subset in Relativity, is it??

Good luck with that, nymshifter. Ken Seto hasn't been able to understand
math beyond the third grade level, so is unable to perform the necessary
math of substituting real numbers.

[kenseto]

I did in the post that started this thread. Here it is again:
Data from the Compton experiment:
Incident wavelength=λ=0.0709E-9 meters
Wavelength at 90 degree after emerge from Graphite Target =λ’=0.0731E-9
Delta(λ)= λ’-λ = 0.0022E-9 meters

Frequency of incident beam = 4.22842E18 Cycles/s

The absolute motion of the graphite target Va can be calculated according to the following equation as follows:
Va= Delta(λ)*((Frequency (f))(1- cos Φ at 90 degree)
= (0.0022E-9)(4.22842E18)(1)
= 9,302,524 m/s
= 9,302.524 km/s

[mlwo...@wp.pl]

Said an idiot rejecting not matching to his mad
ideology pythagorean theorem.

[kenseto]

> You can simplify it further as Δ(measured λ output - measured λ input) = Δλ_k

My equation for Δλ_k is based on two actual measurements. Your equation for Δλ_s is as follows:
Δλ_s = h/mc
Where h (planck’s constant), m (the electron mass) and c(the speed of light) all are universal constants.
I don’t think that Δλ_k is equal to Δλ_s for all different sources.

>
> Va= Δλ *(measured Frequency (f) input)
>
> And I believe you called a moron because you didn’t think I knew that f = c/λ

f’ in my equation is:
f’= f(1- cos Φ) = c’/λ

>
> So now your own equation reads Va = c Δλ/λ

No that’s your equation.....not mine.
My equation for Va at 90 degree detection angle is as follows:
Va= Delta(measured λ output - measured λ input)*(measured Frequency (f’)
Where f’=f at 90 degree detection angle.

>
> Then you say:
>
>
> > I don’t know if Va will be the same with different sources because “measured λ output”, “measured λ input” and “frequency” are different for different sources. That means that we need do experiments to determine Va for different sources to see if Va is constant.
> >
> > I rejected your equation for Va:
> > Va=c(Δλ/λ)
> > Because you automatically made Va to be different for different sources.
>
>
> The equation you "reject" is not just mine. You reject your own equation where λ is λ(input). The only way you can save it is to claim the “f" is output frequency, even though you used the input frequency in your demo Compton calculation to get Va above at the beginning of this thread:

The input f is equal to the output f’ at 90 degree detection angle.

>
> kenseto: Frequency of incident beam = 4.22842E18 Cycles/s

Yes that’s what Compton used for his experiment.

>
>
> Now you want to use the output f, do you? It doesn't matter.

The output f’= input f(1- cos Φ)

>Δλ is a constant in all these experiments for the same angle. It is 4.84 picometers at 180 degrees. The speed of light is a constant. Since you pick λ_input and can pick two different ones at the same time for the same sample, but cannot get back two Va's if that number means anything, obviously your original equation is wrong. You just proved it. If you're going to claim you were talking about output λ or f all along:
>
> Va=c Δλ / [λ_ouput] = =c Δλ / [λo]

This is your equation for Va.....not mine.

[kenseto]

> not related to its speed, it’s not Newtonian mechanics..

You are a failed mathematician. You have no knowledge of physics.

[Gary Harnagel]

On Thursday, December 14, 2017 at 3:36:13 PM UTC-7, kenseto wrote:
>
> On Wednesday, December 13, 2017 at 5:11:03 PM UTC-5, JanPB wrote:
> >
> > Where have you been for the last 100+ years? The root cause here is that
> > _time_ is not exactly what Newton thought it was. Photon's momentum is
> > not related to its speed, it’s not Newtonian mechanics..
>
> You are a failed mathematician. You have no knowledge of physics.

Stupid Simpleton Seto doesn't even realize that he's lost the argument.

[Michael Moroney]

On Thursday, December 14, 2017 at 3:36:13 PM UTC-7, kenseto wrote:

> You are a failed mathematician. You have no knowledge of physics.

Says the man who still can't solve a 3rd grade math problem, or understand
what SR actually says!

[Steve BH]

But what is your explanation for why carbon ALWAYS gives these numbers with this wavelength of incident radiation? But other numbers with OTHER wavelengths of radiation? Surely it can't have the same "absolute motion" all the time through the aether, year by year, day and night. And even if it does, in some kind of macro-galactic way, why would that change if you bombard with it some other gamma emitter? Which it does.

[Steve BH]

Well, that's your problem. Since you refuse to look at real experiments with different sources and do the calculation, I guess you can keep doubting.

http://122.physics.ucdavis.edu/course/cosmology/sites/default/files/files/Gamma%20Spec/Compton%20Scattering%20Experiment.pdf

> >
> > Va= Δλ *(measured Frequency (f) input)
> >
> > And I believe you called a moron because you didn’t think I knew that f = c/λ
>
> f’ in my equation is:
> f’= f(1- cos Φ) = c’/λ

Oh, you're going to use the slow speed of light c' again. A retarded c. We'll just call the above equation the "retarded c Seto equation."

But since c' is not measured, that's a fifth variable and can be anything we like. So we no longer have the Compton equation.



And why f’= c’/λ which means λ = c’/f’

and not f’= c’/λ’ ?

What is λ’- λ = Δλ in terms of f and f' and c and c’?

λ = c’/f’ = c/f

The world waits for the definition of λ’ in terms of c and f or c' and f'

> The input f is equal to the output f’ at 90 degree detection angle.

That means f = f' at cos 90

f = f' = c’/λ’ At 90 degrees

But f= c/λ

c/λ = c'/λ'

cλ' = c'λ


I'm sorry but this is all crazy.

[kenseto]

That’s not just my problem. If Δλ_s is constant for all sources and all detection angles then the Compton experiments shouldl not detect different Δλ_s at different angles.....but it does.

>
> http://122.physics.ucdavis.edu/course/cosmology/sites/default/files/files/Gamma%20Spec/Compton%20Scattering%20Experiment.pdf
>
>
> > >
> > > Va= Δλ *(measured Frequency (f) input)
> > >
> > > And I believe you called a moron because you didn’t think I knew that f = c/λ
> >
> > f’ in my equation is:
> > f’= f(1- cos Φ) = c’/λ

There is no c’/λ in my equation. Simply: f’= f(1- cos Φ)

>
> Oh, you're going to use the slow speed of light c' again. A retarded c. We’ll just call the above equation the "retarded c Seto equation."
> But since c' is not measured, that's a fifth variable and can be anything we like. So we no longer have the Compton equation.

c or c’ are not measured. They are assumed. In any case there is no c or c’ in my equation. There are actual measurements of λ and λ’ and their difference is Delta(Lambda) and Delta(lambda) is different for different sources.

[Gary Harnagel]

It's right up there where you wrote it, imbecile. So you are a liar as well
as an imbecile.

> > Oh, you're going to use the slow speed of light c' again. A retarded c.
> > We’ll just call the above equation the "retarded c Seto equation."
> > But since c' is not measured, that's a fifth variable and can be anything
> > we like. So we no longer have the Compton equation.
>
> c or c’ are not measured. They are assumed. In any case there is no c or c’
> in my equation.

Liar.

> There are actual measurements of λ and λ’ and their difference is Delta
> (Lambda) and Delta(lambda) is different for different sources.

I doubt very much if Stupid Seto has actually made a Compton effect
measurement, so how does he know that delta-lambda is different for different
sources? He doesn't, of course. He is just plying his usual trade of making
baseless assertions. He is a liar and an imbecile and is too arrogant and
stupid to actually read up on the Compton effect:

https://en.wikipedia.org/wiki/Compton_scattering

and learn that it does NOT provide a way to measure one's "absolute velocity."

[Omar Shabsigh]

I wonder! I have been looking at these blogs about Absolute Motion.

Don't we have any logical person on this subject to realize that motion can never be absolute? The notion of motion itself implies that motion can be only relative to the observer or to whatever other reference adopted.

SENSE & LOGIC Gentlemen!!!!


> The detection of the absolute motion of the graphite target invalidate all pass conclusions that there is no ether and no absolute motion. Also it invalidated the SR postulate that the speed of light is at constant c as measured by all observers.

[kenseto]

On Friday, January 19, 2018 at 7:04:01 PM UTC-5, Omar Shabsigh wrote:
> I wonder! I have been looking at these blogs about Absolute Motion.
>
> Don't we have any logical person on this subject to realize that motion can never be absolute? The notion of motion itself implies that motion can be only relative to the observer or to whatever other reference adopted.
>
> SENSE & LOGIC Gentlemen!!!!

In my theory I defined that absolute motion as that motion of an object in a stationary ether called the E-Matrix or that motion of an object wrt the light waves moving at constant speed in the E-Matrix.

[Gary Harnagel]

On Saturday, January 20, 2018 at 7:32:56 AM UTC-7, kenseto wrote:
>
> On Friday, January 19, 2018 at 7:04:01 PM UTC-5, Omar Shabsigh wrote:
> >
> > I wonder! I have been looking at these blogs about Absolute Motion.
> >
> > Don't we have any logical person on this subject to realize that motion
> > can never be absolute? The notion of motion itself implies that motion can
> > be only relative to the observer or to whatever other reference adopted.
> >
> > SENSE & LOGIC Gentlemen!!!!

It's NOT a matter of logic. It's a matter of discerning what the experimental
evidence is.

> In my theory I defined that absolute motion as that motion of an object in
> a stationary ether called the E-Matrix or that motion of an object wrt the
> light waves moving at constant speed in the E-Matrix.

"YOU" define? What gices you the right to tell the universe how to behave?
That is SO arrogant of you!

> > > The detection of the absolute motion of the graphite target

There was no such "detection."

> > > invalidate all pass conclusions that there is no ether and no absolute
> > > motion. Also it invalidated the SR postulate that the speed of light
> > > is at constant c as measured by all observers.

Only in your delusional minf.

[Tom Roberts]

On 1/19/18 6:03 PM, Omar Shabsigh wrote:
> Don't we have any logical person on this subject to realize that motion can
> never be absolute? The notion of motion itself implies that motion can be
> only relative to the observer or to whatever other reference adopted.

Hmmm. Newton described both space and time in the Principia. This is usually
translated as

Absolute, true, and mathematical space remains similar and
immovable without relation to anything external.

So "absolute motion" is clearly motion relative to this "absolute space".

Since then we have refined and re-defined these concepts ENORMOUSLY. In
particular, with the perspective afforded by differential geometry, we MODEL the
"space of the world" using a manifold with metric. It is QUITE CLEAR that motion
relative to a manifold is completely undefined; to define motion one must first
apply a coordinate system (i.e. a mapping from the manifold to a region of R^N),
and then define motion relative to the coordinates.

So in this context, "absolute motion" is meaningless.

> SENSE & LOGIC Gentlemen!!!!

Not really. A historical perspective is necessary. As is an understanding of
today's models, how they map to the world we inhabit, and the experimental basis
that establishes their validity.

Tom Roberts

[mlwo...@wp.pl]

The world you inhabit is your moronic delusions.

and the experimental basis
> that establishes their validity.

:)

[Esteban Weddle]

Tom Roberts wrote:

>> SENSE & LOGIC Gentlemen!!!!
>
> Not really. A historical perspective is necessary. As is an
> understanding of today's models, how they map to the world we inhabit,
> and the experimental basis that establishes their validity.

And here is some math

x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

[kenseto]

On Saturday, January 20, 2018 at 3:05:10 PM UTC-5, Gary Harnagel wrote:
> On Saturday, January 20, 2018 at 7:32:56 AM UTC-7, kenseto wrote:
> >
> > On Friday, January 19, 2018 at 7:04:01 PM UTC-5, Omar Shabsigh wrote:
> > >
> > > I wonder! I have been looking at these blogs about Absolute Motion.
> > >
> > > Don't we have any logical person on this subject to realize that motion
> > > can never be absolute? The notion of motion itself implies that motion can
> > > be only relative to the observer or to whatever other reference adopted.
> > >
> > > SENSE & LOGIC Gentlemen!!!!
>
> It's NOT a matter of logic. It's a matter of discerning what the experimental
> evidence is.
>
> > In my theory I defined that absolute motion as that motion of an object in
> > a stationary ether called the E-Matrix or that motion of an object wrt the
> > light waves moving at constant speed in the E-Matrix.
>
> "YOU" define? What gices you the right to tell the universe how to behave?
> That is SO arrogant of you!

Yes idiot, I am free to define any term for my theory. In my the E-Matrix is Newton’s absolute space so any motion of an object in the E-Matrix is absolute motion.

>
> > > > The detection of the absolute motion of the graphite target
>
> There was no such “detection."

Sure there was such detection.

>
> > > > invalidate all pass conclusions that there is no ether and no absolute
> > > > motion. Also it invalidated the SR postulate that the speed of light
> > > > is at constant c as measured by all observers.
>

> Only in your delusional mind.

[Gary Harnagel]

On Monday, January 22, 2018 at 12:49:30 PM UTC-7, kenseto wrote:
>
> On Saturday, January 20, 2018 at 3:05:10 PM UTC-5, Gary Harnagel wrote:
> >
> > It's NOT a matter of logic. It's a matter of discerning what the
> > experimental evidence is.
> >
> > > In my theory I defined that absolute motion as that motion of an object in
> > > a stationary ether called the E-Matrix or that motion of an object wrt the
> > > light waves moving at constant speed in the E-Matrix.
> >
> > "YOU" define? What gices you the right to tell the universe how to behave?
> > That is SO arrogant of you!
>
> Yes idiot, I am free to define any term for my theory.

And then to observe that it does NOT predict what actually occurs. You
keep forgetting about that most important freedom.

> In my the E-Matrix is Newton’s absolute space so any motion of an object
> in the E-Matrix is absolute motion.

Which has never been measured, so is completely irrelevant.

"the whole change in the conception of the ether which the special theory of
relativity brought about, consisted in taking away from the ether its last
mechanical quality, namely, its immobility."

http://www.relativitybook.com/resources/Einstein_aether.html

> > > > > The detection of the absolute motion of the graphite target
> >
> > There was no such “detection."
>
> Sure there was such detection.

You are either a liar, a mental case or an ignoramus.

> > > > > invalidate all pass conclusions that there is no ether and no absolute
> > > > > motion. Also it invalidated the SR postulate that the speed of light
> > > > > is at constant c as measured by all observers.
> >
> > Only in your delusional mind.

And it remains so.

[rotchm]

Idiot ken, you still have an ongoing discussion in the other thread that you seem to have cowardly forgotten. Below are the questions that still await your input...

On Saturday, January 20, 2018 at 7:44:17 PM UTC-5, kenseto wrote:
> On Saturday, January 20, 2018 at 11:06:34 AM UTC-5, rotchm wrote:

> > In our discussion here, it is "gravity", "NM", "F ~ 1/r²" that
> > is discussed.
> > How is "F ~ 1/r²", or more particularly, F = -k²/r² a composite
> > force?
>
> Hey moron the Newton equation is as follows:
> F_newton=GMaMb/r^2

Thats the magnitude. There is a " - " in front to indicate its attractive.
That's why I write for you F = -k²/r², or just F = -1/r² for simplicity.

Now, explain how this F = -1/r² is a composite force?

> The DTG equation is as follows:
> F_dtg=(F_aa/F_ab)G*MaMb/r^2(ja)(+/-jb)

Irrelevant. We are NOT discussing your formulas. We are discussing ONLY
F = -1/r² in the Newtontian sense.



> Both F_newton and F_dtg are composite forces, they include an
> attractive EM force and a repulsive effect force.

Where in F = -1/r² is there an EM repulsive force?

> That’s why F_newton ... will predict that the moon can orbit the earth.

That's NOT what you have been saying. Again, this is what you have been claiming: "If gravity is only an attractive force as physicists think then the moon is not able to maintain a stable orbit."

This means that if F = -1/r² then the moon is not able to maintain a stable orbit.

Now you say it can orbit the earth? Make up your mind!

[kenseto]

On Monday, January 22, 2018 at 5:17:03 PM UTC-5, rotchm wrote:
> Idiot ken, you still have an ongoing discussion in the other thread that you seem to have cowardly forgotten. Below are the questions that still await your input...
>
> On Saturday, January 20, 2018 at 7:44:17 PM UTC-5, kenseto wrote:
> > On Saturday, January 20, 2018 at 11:06:34 AM UTC-5, rotchm wrote:
>
> > > In our discussion here, it is "gravity", "NM", "F ~ 1/r²" that
> > > is discussed.
> > > How is "F ~ 1/r²", or more particularly, F = -k²/r² a composite
> > > force?
> >
> > Hey moron the Newton equation is as follows:
> > F_newton=GMaMb/r^2
>
> Thats the magnitude. There is a " - " in front to indicate its attractive.
> That's why I write for you F = -k²/r², or just F = -1/r² for simplicity.
>
> Now, explain how this F = -1/r² is a composite force?
>
> > The DTG equation is as follows:
> > F_dtg=(F_aa/F_ab)G*MaMb/r^2(ja)(+/-jb)
>
> Irrelevant. We are NOT discussing your formulas. We are discussing ONLY
> F = -1/r² in the Newtontian sense.
>
>
>
> > Both F_newton and F_dtg are composite forces, they include an
> > attractive EM force and a repulsive effect force.
>
> Where in F = -1/r² is there an EM repulsive force?

Moron without the repulsive effect the attractive EM force would be much larger. The reason why gravity is so weak compared to the attractive EM force alone is due to that it is a composite force.
Gee you are so stupid....I suggest that you ask your daddy for clues.