> > > Nope - m_0 remains the same. Your assumptions that mass SHRINKS when > > > accelerated is contradicted by nearly a century of real world > > > observations.
I don't give a shit who has observed what. The *rest mass* decreases.
> > m_0 DOESNOT remain the same.
> In reality-based physics, it does.
> > m_0 **WAS** 1Kg, before the accelleration.
No comment hey? I wonder why?
> > A hydrogen atom's m_0 is *LESS* after emmission because of the > > emmitted photon & because it has momentum energy too.
> In order for a hydrogen atom to EMIT anything, it MUST be above ground > state - ie, have energy to spare.
The *spare energy* is the in the *Rest Mass* of the unbound electron & proton you twit.
It emmits energy, gains momentum, Their *REST MASS* has *DECREASED*.
> > 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.
> > 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.
> YOU HAVE TWO DIFFERENT ROCKET MASSES SIMPLETON !
That only matters to *YOU* because of the way I set up the initial conditions.
2&3 are equivelant setups,
I placed a different mass in 3 to first ask;
1) Why according to **YOUR** gibbermath, has the angle changed? 2) What is the energy of the deflected photons in 3)? 3) If there are no photons deflected in 3), why do you have them deflected in 2), when they are identical setups? 4) How (now that you no no other perameters have you determined the energy of the deflected photons? 5) Ask you to explain what is so unique about *MY* setup, that *YOU* coincidentally has E_y * 0.14142 and mass is 0.14142.
I know, I wonder why you don't?
> For the first two cases, m = 0.1414; for the third, it is 0.5707. And > NOTHING HAS REFLECTED YET !
So deflect from 3, and tell me what & why you get.
> > So in essence, there is something *RATHER SPECIAL* to my initial idea.
> > In short, *it's perfection*.
> Translation - you made sh*t up, and are quite proud of your idiocy.
> > 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.
> Where did you get the 'idea' that gravity is a THRUST effect ? You > seem to be whining that gravity INDUCES matter to magically radiate > enough thrust to accelerate it.
Much the same way accellerated charged particles emmit.
The rocket shows that any mass that accellerates (BY ITS OWN ENERGY) radiates.
> > > 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.
> There was no reflection yet BECAUSE YOU STILL HAVE PHOTONS * COMING > FROM THE RIGHT * - they haven't hit anything yet !
Oish, Lets pretend they have, now what?
> Once they hit the rocket, THEY WOULD BE * GOING TO THE RIGHT *.
With what energies?
> > Especially since 2 & 3 are identical setups, just different mass- > > energies.
> Nope - just different DISTRIBUTION of masses and energies. All you > did was add mass to the rocket and hoped no one would notice.
Quite the opposite. I was hoping you would notice the difference, and then scratch your head because you *NOW* cannot work out the energy of the deflected photon.
I wonder why, since 2) & 3) are the same setups?
(Because your so stupid, by "setup" I mean a mass being hit by photons from the left, about to be reflected)
> > > 2) If mass reduced when accelerated, particle accelerators wouldn't > > > work.
> > If you had **READ** for comprehension. It depends *HOW* it was > > accellerated.
> How convenient ! And just WHY should it matter ?
> > They are accellerated 3 ways.
> > 1) Hit by photons/energy. (Their mass increases)
> > 2) Move away from a negative field. But remember, their mass is > > increased placing them in the field.
> Since they are MOVING, they have a gamma >1. You 'think' that just > might have something to do with it ?
It takes an increase in Potential energy, to place it in a negative field, If it was stationary in that field, It's *Rest Mass* would be greater to start with.
> > If they were statically held in a negative field, their mass would be > > heavier.
> You 'determined' that how ?
In 3), the electron is stationary, yet has the energy it was given in 2)
For that brief moment, your electron has a greater *Rest Mass*.
1)
Field ... electron at rest ... () ... ******************************* 2)
Field ... Give energy here ... ()<------ ... Electron moves toward the field ********************************** 3)
Field ... () For an instance, the electron ... is stationary. ********************************* 4)
Field ... electron accellerates away from field ... ---->() ... ***********************************
> > When you let go, and it moves into a neutral field, (LIKE OUR ROCKET) > > it emmits a photon, gains KE, and you can work out its rest mass as > > m_0 = 9.1093897*10^-31Kg, (As you would expect).
> So - it gains mass because it is FED ENERGY first in order to radiate > it later.
Yes.
> Are rockets single electron emitters, suspectible to being held in > charged electromagnetic fields ?
Your missing the point.
The rocket represents the electron at 3).
> By your gibberings, electrons should SHRINK when they are accelerated;
It did. Go back and look again.
> decades of high energy particle physics shows they get heavier in > accordance to relativity (ie, m' = m/sqrt(1-v^2/c^2)).
> Oh, right - by spintwittian decree, THAT KIND OF ACCELERATION DOESN'T > MAGICALLY SHRINK MASS !!!1!1!1
From 3) to 4) above the electrons mass decreased due to the accelleration.
> > 3) Move it towards a positive field. Exactly the same, the rest mass > > of 9.1093897*10^-31 emmits a photon (like the rocket), it > > accellerates, it's *REST MASS* is now lower. Bringing it back into the > > field uses the equation I mentioned *MUCH EARLIER*- Hide quoted text -
> Ah, so you have the DELUSION that gravity is exactly identical to > ELECTROMAGNETIC fields ?
Whoever said that?
But they follow similar rules GMm/r^2, kQQ/r^2
> Once again - since it is MOVING, its relativistic mass is greater. It > emits a photon to RETURN TO its standard rest mass.
In which scenario?
> You seem to get things backwards spintwitty - you seem to 'think' that > the fields INDUCE the electron to emit a photon to move, when it is > usually the other way around (the fields MOVE the electron, which > releases a photon to return to its standard rest mass).
So we agree.
Now lets say a slightly positive field was the *standard universal field strength*.
Electrons would have less mass, and protons would have more.
Now lets say, we lived in a universally stronger gravitational field.
Everything in that field would have *LESS* mass.
This is why you perceive the outer galaxy stars to orbit faster than they should.
When you look at a star in the inner galaxy, you say "it has x brightness" "it must be y mass".
But it doesn't, every atom that makes that star has lost mass, because of all the other atoms in the inner galaxy.
In short, your inner galaxy has less mass per star, than your outer galaxy.
> On Nov 3, 10:39 pm, Prof Weird <pol...@msx.dept-med.pitt.edu> wrote:> On Nov 3, 4:48 pm, spintronic <spintro...@hotmail.com> wrote:
> . > .
> > > > > 1) I gave a thought experiment proving mass reduces in a gravitational > > > > > field, and certain equivelant accellerations.
> > > > Whole bunch of silly right there :
> > > No, someone stole & or duplicated my ideas, it's been published.
> > RiiIIiiIIiight !
> Go and look it up.
> > > > 2) If mass reduced when accelerated, particle accelerators wouldn't > > > > work.
> > > If you had **READ** for comprehension. It depends *HOW* it was > > > accellerated.
> > How convenient ! And just WHY should it matter ?
> > > They are accellerated 3 ways.
> > > 1) Hit by photons/energy. (Their mass increases)
> > > 2) Move away from a negative field. But remember, their mass is > > > increased placing them in the field.
> > Since they are MOVING, they have a gamma >1. You 'think' that just > > might have something to do with it ?
> It takes an increase in Potential energy, to place it in a negative > field, > If it was stationary in that field, It's *Rest Mass* would be greater > to start with.
> > > If they were statically held in a negative field, their mass would be > > > heavier.
> > You 'determined' that how ?
> In 3), the electron is stationary, yet has the energy it was given in > 2)
> For that brief moment, your electron has a greater *Rest Mass*.
> 1)
> Field > ... electron at rest > ... () > ... > ******************************* > 2)
> Field > ... Give energy here > ... ()<------ > ... Electron moves toward the field > ********************************** > 3)
> Field > ... > () For an instance, the electron > ... is stationary. > ********************************* > 4)
> Field > ... electron accellerates away from field > ... ---->() > ... > ***********************************
> > > When you let go, and it moves into a neutral field, (LIKE OUR ROCKET) > > > it emmits a photon, gains KE, and you can work out its rest mass as > > > m_0 = 9.1093897*10^-31Kg, (As you would expect).
> > So - it gains mass because it is FED ENERGY first in order to radiate > > it later.
> Yes.
> > Are rockets single electron emitters, suspectible to being held in > > charged electromagnetic fields ?
> Your missing the point.
> The rocket represents the electron at 3).
> > By your gibberings, electrons should SHRINK when they are accelerated;
> It did. Go back and look again.
> > decades of high energy particle physics shows they get heavier in > > accordance to relativity (ie, m' = m/sqrt(1-v^2/c^2)).
> > Oh, right - by spintwittian decree, THAT KIND OF ACCELERATION DOESN'T > > MAGICALLY SHRINK MASS !!!1!1!1
> From 3) to 4) above the electrons mass decreased due to the > accelleration.
> > > 3) Move it towards a positive field. Exactly the same, the rest mass > > > of 9.1093897*10^-31 emmits a photon (like the rocket), it > > > accellerates, it's *REST MASS* is now lower. Bringing it back into the > > > field uses the equation I mentioned *MUCH EARLIER*- Hide quoted text -
> > Ah, so you have the DELUSION that gravity is exactly identical to > > ELECTROMAGNETIC fields ?
> Whoever said that?
> But they follow similar rules GMm/r^2, kQQ/r^2
> > Once again - since it is MOVING, its relativistic mass is greater. It > > emits a photon to RETURN TO its standard rest mass.
> In which scenario?
> > You seem to get things backwards spintwitty - you seem to 'think' that > > the fields INDUCE the electron to emit a photon to move, when it is > > usually the other way around (the fields MOVE the electron, which > > releases a photon to return to its standard rest mass).
> So we agree.
> Now lets say a slightly positive field was the *standard universal > field strength*.
> Electrons would have less mass, and protons would have more.
> Now lets say, we lived in a universally stronger gravitational field.
> Everything in that field would have *LESS* mass.
> This is why you perceive the outer galaxy stars to orbit faster than > they should.
> When you look at a star in the inner galaxy, you say "it has x > brightness" "it must be y mass".
> But it doesn't, every atom that makes that star has lost mass, because > of all the other atoms in the inner galaxy.
> In short, your inner galaxy has less mass per star, than your outer > galaxy.- Hide quoted text -
On Nov 4, 6:58 am, spintronic <spintro...@hotmail.com> wrote:
> On Nov 3, 10:23 pm, Prof Weird <pol...@msx.dept-med.pitt.edu> wrote:> On Nov 3, 4:38 pm, spintronic <spintro...@hotmail.com> wrote:
> . > .
> > > > 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.
> > For this level of calculation they do.
> For *YOUR* calculations, rockets have wavelengths on the quantum > scale?????
spintwitty looks at his reflection and sneers :
> LMFAO!!!!!!!!!!!!!!!!!!!!!!! YOU ARE BY FAR THE BIGGEST TWIT I HAVE > EVER KNOWN.
The equations are actually :
hc/E' - hc/E = hc/E(particle) (given that E = hc/W, W = hc/E)
Divide both sides of the equality by hc, and you get :
1/E' - 1/E = 1/E(particle).
Since 1/E is NOT a wavelength, the equality is still valid - if you want the wavelength E', you can just do the math then multiply (1/E (particle) - 1/E) by hc. Removing common factors from both sides makes the math a bit easier AND still gets the correct answer.
> > > > > 4) You are *duped* by a quirk of the setup.
> > > > Nope - m_0 remains the same. Your assumptions that mass SHRINKS when > > > > accelerated is contradicted by nearly a century of real world > > > > observations.
> I don't give a shit who has observed what. The *rest mass* decreases.
So you keep bellowing and screaming; too bad that observations of the REAL WORLD show you are wrong.
> > > m_0 DOESNOT remain the same.
> > In reality-based physics, it does.
> > > m_0 **WAS** 1Kg, before the accelleration.
> No comment hey? I wonder why?
And m_0 REMAINS 1 kg, no matter how hard you stomp your foot and scream otherwise.
> > > A hydrogen atom's m_0 is *LESS* after emmission because of the > > > emmitted photon & because it has momentum energy too.
> > In order for a hydrogen atom to EMIT anything, it MUST be above ground > > state - ie, have energy to spare.
> The *spare energy* is the in the *Rest Mass* of the unbound electron & > proton you twit.
Ah - you seem to have forgotten about binding energy. It has been known for quite some time that the energy of a hydrogen atom is a bit less than a free electron + a proton. IT IS THE REASON YOU HAVE TO SUPPLY ENERGY TO GET THE ELECTRON LOOSE !
A hydrogen atom IN ITS GROUND STATE has no energy to spare; in order to eject a photon or an electron, energy MUST be supplied (and no, sneering at it and DEMANDING it conform to your delusions will not do the job).
> It emmits energy, gains momentum, Their *REST MASS* has *DECREASED*.
In reality, a hydrogen atom NOT IN GROUND STATE emits a photon, gains momentum, and the REST MASS REMAINS THE SAME.
> > > 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.
> > > 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.
> > YOU HAVE TWO DIFFERENT ROCKET MASSES SIMPLETON !
> That only matters to *YOU* because of the way I set up the initial > conditions.
It matters to anyone with a functional brain - in two cases, you have a rocket with mass 0.14142; in another example, its mass is now 0.5707. And you SERIOUSLY expect them to behave exactly the same way ?
> 2&3 are equivelant setups,
No, they're not - you removed the energy going right and added it to the mass !
In example 1 and 2, the momentum = 0
In example 3, momentum = 0.429 ! For someone that is supposedly inerrant and smarter than every physicist on the planet, you sure do make some simple-minded errors (like adding kg to kg m/sec to get kg !)
> I placed a different mass in 3 to first ask;
> 1) Why according to **YOUR** gibbermath, has the angle changed?
What angle simpleton ? The photons are a beam in all cases, so the angle of the photons going left, hitting the mass and coming back right is 180 degrees.
> 2) What is the energy of the deflected photons in 3)?
x = mcp/(2p + m) = (0.5707 x 0.429 x c)/(2x0.429 + 0.5707) = 0.171365787 kg c^2 J.
> 3) If there are no photons deflected in 3), why do you have them > deflected in 2), when they are identical setups?
The setups are NOT identical - setup 2 has mass 0.14142 to be hit with 0.429 kg C momentum of photons (total momentum of the system = 0); in setup 3, you have mass 0.5707 to be hit with 0.429 kg C momentum of photons (total momentum of the system = 0.429).
You'd have to have your head shoved 3.1415 feet up your own arse to 'think' the setups are identical !
> 4) How (now that you no no other perameters have you determined the > energy of the deflected photons?
If they come in from the right and go back to the right, deflection is 180 degrees.
> 5) Ask you to explain what is so unique about *MY* setup, that *YOU* > coincidentally has E_y * 0.14142 > and mass is 0.14142. > I know, I wonder why you don't?
Your setup is not unique; doing the maths correctly tends to limit answers.
> > For the first two cases, m = 0.1414; for the third, it is 0.5707. And > > NOTHING HAS REFLECTED YET !
> So deflect from 3, and tell me what & why you get.
I did - you get 0.1714 kg C momentum of photons going right after they deflect.
Mass of 0.5707 has v = 0.724637 c.
> > > So in essence, there is something *RATHER SPECIAL* to my initial idea.
> > > In short, *it's perfection*.
> > Translation - you made sh*t up, and are quite proud of your idiocy.
> Yes.
Glad to see you admit that you make sh*t up, and are PROUD to be a gibbering moron there spintwitty !
> > > 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.
> > Where did you get the 'idea' that gravity is a THRUST effect ? You > > seem to be whining that gravity INDUCES matter to magically radiate > > enough thrust to accelerate it.
> Much the same way accellerated charged particles emmit.
> The rocket shows that any mass that accellerates (BY ITS OWN ENERGY) > radiates.
RiiIIiiIIiiight ! And gravity doesn't add energy so the mass DOES NOT HAVE TO RADIATE why ?
Oh, right - because spintwitty the bellicose wishes it so !!1!
> > > > 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.
> > There was no reflection yet BECAUSE YOU STILL HAVE PHOTONS * COMING > > FROM THE RIGHT * - they haven't hit anything yet !
> Oish, Lets pretend they have, now what?
> > Once they hit the rocket, THEY WOULD BE * GOING TO THE RIGHT *.
> With what energies?
> > > Especially since 2 & 3 are identical setups, just different mass- > > > energies.
> > Nope - just different DISTRIBUTION of masses and energies. All you > > did was add mass to the rocket and hoped no one would notice.
> Quite the opposite. I was hoping you would notice the difference, and > then scratch your head because you *NOW* cannot work out the energy of > the deflected photon.
> I wonder why, since 2) & 3) are the same setups?
Once again, simpleton : 2 and 3 are NOT the same setup. You have two different masses and the momentums are different (in example 2, momentum = 0 ; in example 3, momentum = 0.429 to the left).
You'd have to be pretty slack-witted, inattentive, or willfully stupid to claim they are the same.
On Nov 4, 7:22 am, spintronic <spintro...@hotmail.com> wrote:
> On Nov 3, 10:39 pm, Prof Weird <pol...@msx.dept-med.pitt.edu> wrote:> On Nov 3, 4:48 pm, spintronic <spintro...@hotmail.com> wrote:
> . > .
> > > > > 1) I gave a thought experiment proving mass reduces in a gravitational > > > > > field, and certain equivelant accellerations.
> > > > Whole bunch of silly right there :
> > > No, someone stole & or duplicated my ideas, it's been published.
> > RiiIIiiIIiight !
> Go and look it up.
Why should I do your work for you - you CLAIM others have stolen or duplicated your ideas, so it should be quite simple for you to BACK UP YOUR ASSERTIONS.
> > > > 2) If mass reduced when accelerated, particle accelerators wouldn't > > > > work.
> > > If you had **READ** for comprehension. It depends *HOW* it was > > > accellerated.
> > How convenient ! And just WHY should it matter ?
> > > They are accellerated 3 ways.
> > > 1) Hit by photons/energy. (Their mass increases)
> > > 2) Move away from a negative field. But remember, their mass is > > > increased placing them in the field.
> > Since they are MOVING, they have a gamma >1. You 'think' that just > > might have something to do with it ?
> It takes an increase in Potential energy, to place it in a negative > field, > If it was stationary in that field, It's *Rest Mass* would be greater > to start with.
Or the available energy has increased (rest mass + energy).
> > > If they were statically held in a negative field, their mass would be > > > heavier.
> > You 'determined' that how ?
> In 3), the electron is stationary, yet has the energy it was given in > 2)
> For that brief moment, your electron has a greater *Rest Mass*.
> 1)
> Field > ... electron at rest > ... () > ... > ******************************* > 2)
> Field > ... Give energy here > ... ()<------ > ... Electron moves toward the field > ********************************** > 3)
> Field > ... > () For an instance, the electron > ... is stationary. > ********************************* > 4)
> Field > ... electron accellerates away from field > ... ---->() > ... > ***********************************
> > > When you let go, and it moves into a neutral field, (LIKE OUR ROCKET) > > > it emmits a photon, gains KE, and you can work out its rest mass as > > > m_0 = 9.1093897*10^-31Kg, (As you would expect).
> > So - it gains mass because it is FED ENERGY first in order to radiate > > it later.
> Yes.
And gravity can't feed energy to mass so the mass can radiate gibbertwitons away to accelerate why ?
The question you are adept at evading :
Given that a planet in a circular orbit experiences CONSTANT GRAVITATIONAL ACCELERATION TOWARDS THE SUN, why isn't the Earth blasted by radiation as Mercury and Venus transit the sun ?
The energy released should be HUGE (acceleration is small, but mass is tremendous); and WHY are there even planets LEFT, given that in your gibbertwittian physics they should be radiating away their mass at a furious rate ?
> > Are rockets single electron emitters, suspectible to being held in > > charged electromagnetic fields ?
> Your missing the point.
> The rocket represents the electron at 3).
And matter behaves EXACTLY like an electron in every conceivable case. Wood responds to magnetic fields EXACTLY the same way a single electron would, and only a god-hating piece of atheist filth would DARE say otherwise !!1!!1
> > By your gibberings, electrons should SHRINK when they are accelerated;
> It did. Go back and look again.
They returned to their rest mass; by your 'physics', they should shrink BEYOND THEIR INITIAL REST MASS and become ever smaller.
> > decades of high energy particle physics shows they get heavier in > > accordance to relativity (ie, m' = m/sqrt(1-v^2/c^2)).
> > Oh, right - by spintwittian decree, THAT KIND OF ACCELERATION DOESN'T > > MAGICALLY SHRINK MASS !!!1!1!1
> From 3) to 4) above the electrons mass decreased due to the > accelleration.
After given more energy/mass, they return to their INITIAL REST MASS. In your gibberf*ckian physics, the electrons should shrink beyond their initial rest mass (as you ASSERT matter does in the gravitational fields of a black hole).
> > > 3) Move it towards a positive field. Exactly the same, the rest mass > > > of 9.1093897*10^-31 emmits a photon (like the rocket), it > > > accellerates, it's *REST MASS* is now lower. Bringing it back into the > > > field uses the equation I mentioned *MUCH EARLIER*- Hide quoted text -
> > Ah, so you have the DELUSION that gravity is exactly identical to > > ELECTROMAGNETIC fields ?
> Whoever said that?
> But they follow similar rules GMm/r^2, kQQ/r^2
So of COURSE matter simply MUST radiate because spintwitty noted the formula are similar ! (too bad the constants are different, different orders of magnitude of forces, gravity never repellent as electromagnetism can be, etc)
> > Once again - since it is MOVING, its relativistic mass is greater. It > > emits a photon to RETURN TO its standard rest mass.
> In which scenario?
Yours, simpleton.
> > You seem to get things backwards spintwitty - you seem to 'think' that > > the fields INDUCE the electron to emit a photon to move, when it is > > usually the other way around (the fields MOVE the electron, which > > releases a photon to return to its standard rest mass).
> So we agree.
> Now lets say a slightly positive field was the *standard universal > field strength*.
> Electrons would have less mass, and protons would have more.
> Now lets say, we lived in a universally stronger gravitational field.
> Everything in that field would have *LESS* mass.
Based on what ? Your assertion ? Have you confused mass with weight, or just being a gibbering twit ?
> This is why you perceive the outer galaxy stars to orbit faster than > they should.
> When you look at a star in the inner galaxy, you say "it has x > brightness" "it must be y mass".
Brightness is determined by what nuclear reactions the star peforms, and what brightness corresponds to what mass is pretty well worked out by examining many thousands of stars.
> But it doesn't, every atom that makes that star has lost mass, because > of all the other atoms in the inner galaxy.
RiiIIiiIIiiight !
> In short, your inner galaxy has less mass per star, than your outer > galaxy.
So a G0 class star (mass = 1 solar mass) in the inner galaxy has less mass than a G0 star (mass = 1 solar mass) in the outer rim ?!?!!?!? What fresh gibbering lunacy is that ?
> On Nov 4, 12:22 pm, spintronic <spintro...@hotmail.com> wrote:
> > On Nov 3, 10:39 pm, Prof Weird <pol...@msx.dept-med.pitt.edu> wrote:> On Nov 3, 4:48 pm, spintronic <spintro...@hotmail.com> wrote:
> > . > > .
> > > > > > 1) I gave a thought experiment proving mass reduces in a gravitational > > > > > > field, and certain equivelant accellerations.
> > > > > Whole bunch of silly right there :
> > > > No, someone stole & or duplicated my ideas, it's been published.
> > > RiiIIiiIIiight !
> > Go and look it up.
> > > > > 2) If mass reduced when accelerated, particle accelerators wouldn't > > > > > work.
> > > > If you had **READ** for comprehension. It depends *HOW* it was > > > > accellerated.
> > > How convenient ! And just WHY should it matter ?
> > > > They are accellerated 3 ways.
> > > > 1) Hit by photons/energy. (Their mass increases)
> > > > 2) Move away from a negative field. But remember, their mass is > > > > increased placing them in the field.
> > > Since they are MOVING, they have a gamma >1. You 'think' that just > > > might have something to do with it ?
> > It takes an increase in Potential energy, to place it in a negative > > field, > > If it was stationary in that field, It's *Rest Mass* would be greater > > to start with.
> > > > If they were statically held in a negative field, their mass would be > > > > heavier.
> > > You 'determined' that how ?
> > In 3), the electron is stationary, yet has the energy it was given in > > 2)
> > For that brief moment, your electron has a greater *Rest Mass*.
> > 1)
> > Field > > ... electron at rest > > ... () > > ... > > ******************************* > > 2)
> > Field > > ... Give energy here > > ... ()<------ > > ... Electron moves toward the field > > ********************************** > > 3)
> > Field > > ... > > () For an instance, the electron > > ... is stationary. > > ********************************* > > 4)
> > Field > > ... electron accellerates away from field > > ... ---->() > > ... > > ***********************************
> > > > When you let go, and it moves into a neutral field, (LIKE OUR ROCKET) > > > > it emmits a photon, gains KE, and you can work out its rest mass as > > > > m_0 = 9.1093897*10^-31Kg, (As you would expect).
> > > So - it gains mass because it is FED ENERGY first in order to radiate > > > it later.
> > Yes.
> > > Are rockets single electron emitters, suspectible to being held in > > > charged electromagnetic fields ?
> > Your missing the point.
> > The rocket represents the electron at 3).
> > > By your gibberings, electrons should SHRINK when they are accelerated;
> > It did. Go back and look again.
> > > decades of high energy particle physics shows they get heavier in > > > accordance to relativity (ie, m' = m/sqrt(1-v^2/c^2)).
> > > Oh, right - by spintwittian decree, THAT KIND OF ACCELERATION DOESN'T > > > MAGICALLY SHRINK MASS !!!1!1!1
> > From 3) to 4) above the electrons mass decreased due to the > > accelleration.
> > > > 3) Move it towards a positive field. Exactly the same, the rest mass > > > > of 9.1093897*10^-31 emmits a photon (like the rocket), it > > > > accellerates, it's *REST MASS* is now lower. Bringing it back into the > > > > field uses the equation I mentioned *MUCH EARLIER*- Hide quoted text -
> > > Ah, so you have the DELUSION that gravity is exactly identical to > > > ELECTROMAGNETIC fields ?
> > Whoever said that?
> > But they follow similar rules GMm/r^2, kQQ/r^2
> > > Once again - since it is MOVING, its relativistic mass is greater. It > > > emits a photon to RETURN TO its standard rest mass.
> > In which scenario?
> > > You seem to get things backwards spintwitty - you seem to 'think' that > > > the fields INDUCE the electron to emit a photon to move, when it is > > > usually the other way around (the fields MOVE the electron, which > > > releases a photon to return to its standard rest mass).
> > So we agree.
> > Now lets say a slightly positive field was the *standard universal > > field strength*.
> > Electrons would have less mass, and protons would have more.
> > Now lets say, we lived in a universally stronger gravitational field.
> > Everything in that field would have *LESS* mass.
> > This is why you perceive the outer galaxy stars to orbit faster than > > they should.
> > When you look at a star in the inner galaxy, you say "it has x > > brightness" "it must be y mass".
> > But it doesn't, every atom that makes that star has lost mass, because > > of all the other atoms in the inner galaxy.
> > In short, your inner galaxy has less mass per star, than your outer > > galaxy.- Hide quoted text -
> > - Show quoted text -
> Yeah, that shut you up you weirdo.- Hide quoted text -
> - Show quoted text -
You seem to have this ridiculous notion that I have nothing better to do with my life than show everyone what a howling oaf you are spintwitty.
But, you ARE the arrogant simpleton who once used a paper showing convergent evolution as 'proof' that chromosomes teleport across genera, making phylogenies irrelevant ...
On Nov 6, 9:02 pm, Prof Weird <pol...@msx.dept-med.pitt.edu> wrote:
> On Nov 4, 6:58 am, spintronic <spintro...@hotmail.com> wrote:
. .
> > > > Because *ATOMS* HAVE WAVELENGTHS ON THE QUANTUM SCALE. ROCKETS DON'T.
> > > For this level of calculation they do.
> > For *YOUR* calculations, rockets have wavelengths on the quantum > > scale????? > > LMFAO!!!!!!!!!!!!!!!!!!!!!!! YOU ARE BY FAR THE BIGGEST TWIT I HAVE > > EVER KNOWN.
> The equations are actually :
> hc/E' - hc/E = hc/E(particle)
PARTICLE YOU TWIT!!!!!!!!!!!!!!!!!!!!!!!!!
In the *original* compton experiment he scattered x-rays off a carbon mass.
He used the mass & wavelength of the ***ELECTRONS*** in that carbon mass.
HE DID NOT USE THE MASS & WAVELENGTH OF THE CARBON TARGET ITSELF YOU FUCKING MORON!!!!!!!!!!!!
> > > > > > 4) You are *duped* by a quirk of the setup.
> > > > > Nope - m_0 remains the same. Your assumptions that mass SHRINKS when > > > > > accelerated is contradicted by nearly a century of real world > > > > > observations.
> > I don't give a shit who has observed what. The *rest mass* decreases.
> So you keep bellowing and screaming; too bad that observations of the > REAL WORLD show you are wrong.
OBSERVATIONS IN THE REAL WORLD SHOW I AM RIGHT YOU RETARDED MORONIC PIECE OF PIG BOLLOCK.
> > > > m_0 DOESNOT remain the same.
> > > In reality-based physics, it does.
> > > > m_0 **WAS** 1Kg, before the accelleration.
> > No comment hey? I wonder why?
> And m_0 REMAINS 1 kg, no matter how hard you stomp your foot and > scream otherwise.
So m_0 remains 1Kg, when we *ALL* know it is 0.14142Kg in the final analysis?
You are a RETARDED SPASDICK!!!!!!!!!!
> > > > A hydrogen atom's m_0 is *LESS* after emmission because of the > > > > emmitted photon & because it has momentum energy too.
> > > In order for a hydrogen atom to EMIT anything, it MUST be above ground > > > state - ie, have energy to spare.
> > The *spare energy* is the in the *Rest Mass* of the unbound electron & > > proton you twit.
> Ah - you seem to have forgotten about binding energy.
NO YOU MORON, I DOUBT YOU KNOW WHAT THE TERM MEANS.
> It has been known for quite some time that the energy of a hydrogen atom is a bit > less than a free electron + a proton.
BEHOLD, IT'S A MIRACLE. ITS REST MASS IS REDUCED. EXACTLY AS I KEEP SAYING YOU FUCKING MORON.
AFTER EMMITING A PHOTON AND ACCELLERATING (FROM OUR CENTRE OF MOMENTUM FRAME) IT'S REST MASS IS REDUCED.
YOU JUST AGREED YOU RETARDED FUCKING TWIT!!!!!!!!!!!!!!!!!!
> > It emmits energy, gains momentum, Their *REST MASS* has *DECREASED*.
> In reality, a hydrogen atom NOT IN GROUND STATE emits a photon, gains > momentum, and the REST MASS REMAINS THE SAME.
YOU STUPID STUPID STUPID STUPID STUPID RETARD!!!!!!!!!!!!
WE ARE IN THE "CENTRE OF MOMENTUM FRAME".
That means FOR THE RETARDS NAMED WEIRDO, that the *RESTING* Hydrogen atom, consists of the mass of the proton & electron. (You would call it excited) which is a term of misdirection.
It is at REST, it emmits a photon, it accellerates, and its **REST MASS** is REDUCED!!!!!!!!!!!!
> > > > 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.
> > > > 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.
> > > YOU HAVE TWO DIFFERENT ROCKET MASSES SIMPLETON !
> > That only matters to *YOU* because of the way I set up the initial > > conditions.
> It matters to anyone with a functional brain - in two cases, you have > a rocket with mass 0.14142; in another example, its mass is now > 0.5707. And you SERIOUSLY expect them to behave exactly the same > way ?
You do.
You expect any old amount of photons from the right to deflect at 180 degrees.
They don't.
It's an illusion of the setup I created.
Trust me. Even *IF* you were right, (WHICH YOU AREN'T) you are STILL confirming my original argument that
1) *REST MASS* accellerated by it's own PE decreases. 2) This happens in a gravitational field. 3) A mass 100m lower than it started weighs less. 4) No mass enters a Black Hole. 5) Dark Energy is the radiation produced every time you drop something. 6) Dark matter is a result of the fact that stars of the same type, weigh *LESS* in the galactic centre, due to the stronger field they are in. 7) You are a moron!
> On Nov 7, 8:05 pm, William <wpihug...@hotmail.com> wrote:
> > On Nov 7, 10:07 am, spintronic <spintro...@hotmail.com> wrote:
> > > In the *original* compton experiment he scattered x-rays off a carbon > > > mass.
> > > He used the mass & wavelength of the ***ELECTRONS*** in that carbon > > > mass.
> > Indeed. Please stick your fingers in you ears and go > > "WAH! WAH! WAH! IT IS! IT IS!"
> > The scenario given is closely related to Compton > > scattering but it is not Compton scattering.
> ANY scenario, where a photon is reflected with a different energy is > compton scattering
Strange universe you live in. In your universe one can burn .43 kg of atomic fuel and have photons with 0.43 kg c^2 J or energy leave left. The momentum can be transferred to baryons by virtual photons which do not carry energy, the baryons get the energy they need to accelerate from their rest mass, however, there are no mirrors.
On Nov 7, 9:19 pm, William <wpihug...@hotmail.com> wrote:
> On Nov 7, 4:48 pm, spintronic <spintro...@hotmail.com> wrote:
. .
> > > The scenario given is closely related to Compton > > > scattering but it is not Compton scattering.
> > ANY scenario, where a photon is reflected with a different energy is > > compton scattering
> Strange universe you live in.
You are deluded.
> In your universe one can burn .43 kg of atomic fuel and > have photons with 0.43 kg c^2 J or energy > leave left.
Don't ever remember saying that.
I do remember *YOU* saying 0.86Kg of baryonic fuel and 0.43KgC^2 going left.
> The momentum can be transferred to > baryons by virtual photons
Where are you getting this gibberish?
1st) I have never agreed with your scenario. 2nd) I said the momentum comes from real photons, and is transferred through the matter via virtual photons. 3rd) Misdirection by twisting someones words is the sign of a desperate man.
> which do not carry energy,
All photons carry energy. So I can only assume, more desperation on your part.
> the baryons get the energy they need to accelerate from their > rest mass, however, there are no mirrors.
I don't see the need for a connection between reducing rest mass & mirrors.
But then I'm not one of the 2 retards in this conversational 3some.
In *YOUR* scenario, the baryons are the mirrors. *NOT* the rocket as a whole.
By the way. You didn't answer, so I'll ask again.
And quite frankly I'm getting tired of your shit.
You know I'm right, but your just too much of a prick to admit it!
Unanswered Question Repeated:
QUOTE" Are you (crazily) bouncing your photons off the rockets mass of 0.141Kg, in comptons equation?
OR
Are you bouncing your photons off the baryons the rocket is made up of?
On 8 Nov, 05:18, William <wpihug...@hotmail.com> wrote:
> On Nov 7, 11:18 pm, spintronic <spintro...@hotmail.com> wrote:
> > All mirrors do work by compton scattering.
> Rule number one: When you are deep in a hole > stop digging.
The only 3 retards digging holes here, are *YOU*, *WEIRDO*, and now *SHIT4BRAINS AKA Y.O.O AKA P.O.O-Head*. And he's only chirped up, since the math stopped.
*ALL* photons that reflect off mirrors do so with either *LESS ENERGY*, *MORE ENERGY*, or the *SAME ENERGY*.
The reflected *ENERGY* depends on 4 things.
1) The *SUBATOMIC* (GET THAT WORD? WELL HERE'S ANOTHER) *PARTICLE* it reflected off. 2) The *INITIAL* energy of the photon. 3) The reletavistic *MOTION* of the *SUBATOMIC -PARTICLE*. (GET THEM WORD'S?) and 4) The *ANGLE* of deflection.
*IF* Red. light that bounces off a glass mirror. It is *STILL* Red, *ONLY* because of the *MAJOR* difference in the *WAVELENGTH* of the *ELECTRONS* in the *SURFACE* of that mirror, and the *WAVELENGTH* of the Red light itself.
HOWEVER: there is *STILL* an energy difference between the initial & final red photon,
and *COMPTON'S* EQUATION EXPLAIN THE DIFFERENCE.
YOU FUCKING RETARDS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ONE THING IS CERTAIN, PHOTONS ***DO NOT*** BOUNCE OFF *SUBATOMIC* PARTICLES, THAT HAVE THE MASS OF ROCKETS!!!!!!!!
On Nov 8, 1:14 pm, spintronic <spintro...@hotmail.com> wrote:
> The reflected *ENERGY* depends on 4 things. > 3) The reletavistic *MOTION* of the *SUBATOMIC -PARTICLE*.
After you hit .14 kg of baryons with .215 kg c^2 J of photons the motion of the baryons changes and the reflected energy also changes, i.e. the equation you use is no longer valid.
On 8 Nov, 20:01, William <wpihug...@hotmail.com> wrote:
> On Nov 8, 1:14 pm, spintronic <spintro...@hotmail.com> wrote:
> > The reflected *ENERGY* depends on 4 things. > > 3) The reletavistic *MOTION* of the *SUBATOMIC -PARTICLE*.
> After you hit .14 kg of baryons with > .215 kg c^2 J of photons the motion of > the baryons changes and the reflected > energy also changes, i.e. the equation you > use is no longer valid.
> - William Hughes
What are you Jabbing about?
I have been telling YOU that *YOUR* 0.43Kg of photons *CANNOT* deflect at 180degrees.
*DON'T* try to cop out with "Ah well it's 3-1 anyway so compton doesn't work" CRAP.
1) Even at 3-1 compton *ALWAYS* works. It's just harder to work out. 2) *ALL* of this arguing is to show *YOU* that *YOUR* scenario was a non-starter in the first place. Yet *NOW* that you *REALISE* as much, you are *STILL* trying to *DOUBLE-BLUFF* your way out of it. 3) *YOUR* scenario *DOESN'T* work. *END OF*!
4) *MY* scenario works, we only need *ONE* single photon to *PROVE* rest mass reduces, when accellerated by it's *OWN* PE.
5) *MY* scenario works *EXACTLY* as a gravitational field does.
> On 8 Nov, 20:01, William <wpihug...@hotmail.com> wrote:
> > On Nov 8, 1:14 pm, spintronic <spintro...@hotmail.com> wrote:
> > > The reflected *ENERGY* depends on 4 things. > > > 3) The reletavistic *MOTION* of the *SUBATOMIC -PARTICLE*.
> > After you hit .14 kg of baryons with > > .215 kg c^2 J of photons the motion of > > the baryons changes and the reflected > > energy also changes, i.e. the equation you > > use is no longer valid.
> > - William Hughes
> What are you Jabbing about?
Please indicate the first bit you do not understand.
> On Nov 8, 6:00 pm, spintronic <spintro...@hotmail.com> wrote:
> > On 8 Nov, 20:01, William <wpihug...@hotmail.com> wrote:
> > > On Nov 8, 1:14 pm, spintronic <spintro...@hotmail.com> wrote:
> > > > The reflected *ENERGY* depends on 4 things. > > > > 3) The reletavistic *MOTION* of the *SUBATOMIC -PARTICLE*.
> > > After you hit .14 kg of baryons with > > > .215 kg c^2 J of photons the motion of > > > the baryons changes and the reflected > > > energy also changes, i.e. the equation you > > > use is no longer valid.
> > > - William Hughes
> > What are you Jabbing about?
> Please indicate the first > bit you do not understand.
> - William Hughes- Hide quoted text -
> - Show quoted text -
I understand perfectly what you are jabbering about.
Ergo) With 3 photons per baryon - each deflection is different.
But your statement
> i.e. the equation you use is no longer valid.
is rediculous. It just means you have to use it 3 times. But since you cant predict the deflection angles,all you have is a spray of photons.
WHICH IS WHAT IV'E BEEN SAYING FOR DAYS & DAYS NOW.
Your scenario ***DOES NOT*** deflect E_i * 0.14142 by 180 degrees.
So in absolutely no way shape or form does your rocket & spray, act like an accelleration equivelant to gravity.
> On 9 Nov, 00:09, William Hughes <wpihug...@hotmail.com> wrote:
> > On Nov 8, 6:00 pm, spintronic <spintro...@hotmail.com> wrote:
> > > On 8 Nov, 20:01, William <wpihug...@hotmail.com> wrote:
> > > > On Nov 8, 1:14 pm, spintronic <spintro...@hotmail.com> wrote:
> > > > > The reflected *ENERGY* depends on 4 things. > > > > > 3) The reletavistic *MOTION* of the *SUBATOMIC -PARTICLE*.
> > > > After you hit .14 kg of baryons with > > > > .215 kg c^2 J of photons the motion of > > > > the baryons changes and the reflected > > > > energy also changes, i.e. the equation you > > > > use is no longer valid.
> > > > - William Hughes
> > > What are you Jabbing about?
> > Please indicate the first > > bit you do not understand.
> > - William Hughes- Hide quoted text -
> > - Show quoted text -
> I understand perfectly what you are jabbering about.
> Ergo) With 3 photons per baryon - each deflection is different.
------------- Nope. You are quite insistent that the *reflected energy* changes when the motion changes even if the deflection does not change.
> On Nov 8, 8:30 pm, spintronic <spintro...@hotmail.com> wrote:
> > On 9 Nov, 00:09, William Hughes <wpihug...@hotmail.com> wrote:
> > > On Nov 8, 6:00 pm, spintronic <spintro...@hotmail.com> wrote:
> > > > On 8 Nov, 20:01, William <wpihug...@hotmail.com> wrote:
> > > > > On Nov 8, 1:14 pm, spintronic <spintro...@hotmail.com> wrote:
> > > > > > The reflected *ENERGY* depends on 4 things. > > > > > > 3) The reletavistic *MOTION* of the *SUBATOMIC -PARTICLE*.
> > > > > After you hit .14 kg of baryons with > > > > > .215 kg c^2 J of photons the motion of > > > > > the baryons changes and the reflected > > > > > energy also changes, i.e. the equation you > > > > > use is no longer valid.
> > > > > - William Hughes
> > > > What are you Jabbing about?
> > > Please indicate the first > > > bit you do not understand.
> > > - William Hughes- Hide quoted text -
> > > - Show quoted text -
> > I understand perfectly what you are jabbering about.
> > Ergo) With 3 photons per baryon - each deflection is different. . > Nope. You are quite insistent that > the *reflected energy* > changes when the motion changes even if the deflection > does not change.
If you are saying, (That I am saying);
"2 photons with the same initial energy, deflected at exactly the same angle;
can have different deflection energies due to the motion of the subatomic particles they are deflected from"
On Nov 9, 7:56 am, spintronic <spintro...@hotmail.com> wrote:
> "2 photons with the same initial energy, deflected at exactly the same > angle; can have different deflection energies due to the motion of the > subatomic particles they are deflected from"
> Of course they can
And since the equation you use says that the reflected energies are not different the equation you use is not applicable to the case where the motion of the subatomic particles changes.
On 9 Nov, 12:54, William Hughes <wpihug...@hotmail.com> wrote:
> On Nov 9, 7:56 am, spintronic <spintro...@hotmail.com> wrote:
> > "2 photons with the same initial energy, deflected at exactly the same > > angle; can have different deflection energies due to the motion of the > > subatomic particles they are deflected from"
> > Of course they can
> And since the equation you use says that the reflected energies > are not different the equation you use is not applicable to > the case where the motion of the subatomic particles changes.
I never said "W'-W=h/p(1-(cos(A))) applied to photons bouncing off moving particles".
I said the deflected energy depended on 4 things, One of which was the relative motion of the particle". but it is still "compton scattering".
In the case of "inverse compton scattering", p is relativistic.
If the particle is moving away you account for doplershift too.
> On 9 Nov, 12:54, William Hughes <wpihug...@hotmail.com> wrote:
> > On Nov 9, 7:56 am, spintronic <spintro...@hotmail.com> wrote:
> > > "2 photons with the same initial energy, deflected at exactly the same > > > angle; can have different deflection energies due to the motion of the > > > subatomic particles they are deflected from"
> > > Of course they can
> > And since the equation you use says that the reflected energies > > are not different the equation you use is not applicable to > > the case where the motion of the subatomic particles changes.
> I never said "W'-W=h/p(1-(cos(A))) applied to photons bouncing off > moving particles".
And since in my scenario the particles in the mirror R begin stationary, but are moving before all collisions take place. the equation "W'-W=h/p(1-(cos(A)))" does not apply.
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