Re: Weirdo's square of negative mass.

On Nov 3, 8:02 pm, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:
On Nov 3, 6:09 am, spintronic <spintro...@xxxxxxxxxxx> wrote:





On Nov 2, 10:41 pm, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:

On Nov 2, 4:49 pm, spintronic <spintro...@xxxxxxxxxxx> wrote:

L = hc/E, so the equation would be more like :

hc (1/0.060710678 kg c^2  - 1/0.429289322 kg c^2 ) = h/c*m_r (1 - cos
A)

h (14.14213566/ kg c) = (1 - cos A ) h/c(0.141421356 kg)

14.14213566/ kg c = (1 - cos A)/(0.141421345 kg c)

Multiply both sides by 0.141421356 kg c to finally yield :

2.000000002 = 1 - cos A; therefore, A = 180 degrees !

(unless, of course,  you'd like to ignore the fact that most
calculators have round-off errors, and wet yourself with rage about
that 2.000000002 not being exactly 2 ?)

This gibberish is just a *QUIRK* OF THE COINCIDENCE i Keep mentioning.

It is not a coincidence when there is only one answer - conservation
of momentum and energy kind of constrain possible results.

You are doing neither.


Williams;  ratio of  final photon / initial photon = rest mass

will always hold.

(m_y_f / m_y_i)  *  m_r_i = m_0.  (ALWAYS)

And that's what's confusing you.

1) Compton *ONLY* deals with quantum particles.

2) You can not have a 0.43Kg single photon.

You can have a single photon with the ENERGY contained within 0.43 kg


This is why you are a retard, and I am not.

h/0.43C = 5.14856*10^-42m
The Plank Length is 1.616252*10^-35m


Your photon is 3139231 shorter, than the units of quantum gravity.
Nice Trick.



- it will have a very short wavelength.

That's like saying, a proton can fit inside an electron because it has
a shorter wavelength.


3) You are saying that if I increase the mass of my mirror, my
laserpen will change colour.

(UTTERLY REDICULOUS)

Yes,  your misrepresentation IS utterly ridiculous, given the FACT
that the mirror far 'outweighs' by several dozens of orders of
magnitude the photons hitting it from the laser pen.

Are you *TOTALLY* retarded? (DONT ANSWER, I ALREADY KNOW YOU ARE)

Are you saying the "m_e" in the equation;

W'-W = (h/m_e C) * (1-(cos(A)))

Is either;

A) The mass of the *CALCITE CRYSTAL* used in the compton experiment?
B) The mass of the electron in the crystal surface?



I'M JUST CURIOUS???????????????????????????????????



The laser light
reflecting off of the mirror would be slightly red-shifted.  So slight
that it would be nigh impossible to detect it (if the mirror is 10^25
x heavier than the photons hitting it, the returning photons would be
redshifted to about 1 + 2/10^25 vs 1.  Good luck finding a shift that
tiny !

No need. red light is not anywheree the ***ELECTRON*** wavelength.

NOTHING TO DO WITH THE MASS OF THE MIRROR YOU TWITTERTARD!!!!!!!!!!!



Compton noticed that some photons came back with nearly the same
energy they left with EVEN WHEN DEFLECTED A FULL 180 DEGREES.

Because *ATOMS* HAVE WAVELENGTHS ON THE QUANTUM SCALE. ROCKETS DON'T.


Reality-based explanation : the photons struck the carbon nucleus.
Given that the nucleus is several thousand times heavier than an
electron, its wavelength shift would be h/m(carbon nuclei)*c.  So the
returning photons would be at a very slightly increased wavelength.

Gibbertwittian 'explanation' - Gawd magically accelerated the photon
by pulling energy out of the luminiferous aether.


You missed a third explanation TWIT.

The x-rays were Inverse Compton scattered by electrons travelling
toward the photon on the lattice surface.

YOU TWIT.


4) You are *duped* by a quirk of the setup.

m_pC/m' = V
sqrt(1-V^2) = gamma
gamma * m' = m_0

Nope - m_0 remains the same.  Your assumptions that mass SHRINKS when
accelerated is contradicted by nearly a century of real world
observations.


m_0 DOESNOT remain the same.


m_0 **WAS** 1Kg, before the accelleration.


A hydrogen atom's m_0 is *LESS* after emmission because of the
emmitted photon & because it has momentum energy too.



m_KE = ((m_0C^2 / gamma) - m_0C^2)/C^2
m_y_f_right = 1/2 (m_p - m_KE)
m_y_i_left = m_p - m_y_f_right.

Now watch.

1/0.429289322 = 2.329431339
1/(0.429289322*0.141421356) = 16.47156696
16.47156696-2.329431339 = 14.14213562
1/ 0.141421356 = 7.071067812

14.14213562 / 7.071067812 = 2 (ALWAYS)

iT'S A QUIRK OF THE SETUP.

What you call a 'quirk' is more likely a side effect of conservation
of momentum and energy.

In *YOUR'S & WILLIAMS* setup. It *DOESNOT* matter wether the 0.8Kg of
fuel, is
a) part of the initial rocket's mass,
b) the annhilation happens some distance away.
c) There are just photons coming to the left.

Try C, and your 14.14213562 / 7.071067812 = 2 (ALWAYS)
Suddenly dissapears.




Watch;

1)

A.m.Bomb is Part Of the Rockat
 ___________________
/ 0.14142Kg<~~~|~~~>
\___________________
Photons weigh 0.858Kg
Total E = 1KgC^2
*******************************
2)

A.M.bomb is not part of the rocket
 ___________
/ 0.14142Kg |<~~~~|~~~~>
\___________| AM-BOMB
Photons weigh 0.858Kg
Total E = 1KgC^2
********************************
3)
No A.M Bomb Yet; Just exactly the same amount of photons
hitting from the right.

 _________
/ 0.5707Kg |<~~~~
\__________|
Photons weigh 0.429Kg
Total E = 1KgC^2 (1-(COS(A)) !=2

Nope - you have a DIFFERENT SETUP.


Same setup.

All 3 have a mass/rocket, and photons coming from the left.

There is (IN YOUR MIND) no difference how much the rockets m_0
actually is.

(To YOU, m_0 is about to grow when hit by photons)

So I choose m_0_r = 0.1Kg and the photons coming from the left to be
0.43Kg.

Now, they are deflected at a different angle.

So in essence, there is something *RATHER SPECIAL* to my initial idea.

In short, *it's perfection*.

At the end of the day, if I concede the m_e to equal the mass of the
rocket (which I don't).

Our scenarios give identical results.


Which means, (LIKE I'VE BEEN SAYING);

"We do not need to look at the internal workings in the rocket, we can
treat it like a particle with "no hair", and this is **EXACTLY** how a
mass acts in a gravitational field".

It's *Rest Mass* changes according to the field strength.




Total E = 0.5707 kg c^2 + 0.429 kg c^2 = 1 KG C^2.  There is no
reflection in above model, so blubbering about (1- cos A) is
premature.


Why is there no reflection? You haven't thouth that through, have you.

Especially since 2 & 3 are identical setups, just different mass-
energies.

.

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