Re: Weirdo's square of negative mass.

On Oct 23, 6:18 am, spintronic <spintro...@xxxxxxxxxxx> wrote:
On 22 Oct, 19:57, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:

On Oct 21, 10:11 am, spintronic <spintro...@xxxxxxxxxxx> wrote:

On 19 Oct, 16:41, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:> On Oct 19, 7:58 am, spintronic <spintro...@xxxxxxxxxxx> wrote:

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Spintwitty demonstrates his idiocy by making up unrealistic numbers,
CLAIM that is what reality expects, then calls everyone else stupid

YOU are STOOPID!

Screamed spintwitty at his reflection in the mirror ...

YOU claim a

0.14Kg mass hit by 0.43Kg of photons and deflecting 0.06Kg has a
momentum of 0.49KgC

Which is physically IMPOSSIBLE!

No - I stated a 0.14 kg mass is hit by 0.43 kg C of photons.  Some of
these reflect back.

Therefore, the 0.14 kg mass has momentum 0.43 kg c PLUS the momentum
of the photons reflected back - in this case, 0.06 kg C of momentum.
That is the ONLY way to conserve energy and momentum - something that
tends to happen in reality-based physics.

Not in your reality.

Question:

Why are you using 0.14000000000Kg above and 0.141421356 kg ?

Oh, ya, that's it, because your *supposed*  "ONLY way to conserve
energy and momentum" is **FLAWED**!

No, simpleton, it's because there was no point in typing out
'0.141421345 kg' each and every fracking time for a hypothetical
example trying to explain something to a blithering idiot.

Once again, simpleton : momentum before : - 0.43 kg C
Momentum after : - 0.43 kg C + (- 0.06 kg c) + ( 0.06 kg c), or - 0.43
kg c.

No RETARD!

p = sqrt((0.51Kg^2C^4 - 0.14000^2C^4)/C^2)

In MY UNIVERSE, that's

p = 0.490407993KgC

p = sqrt((0.51^2 kg^2 C^4 - 0.141421356^2 kg^2 c^4)/c^4) = 0.49 kg C
of momentum.

Still pointlessly fixated by significant digits ?

************NOT**************

p = 0.49KgC.

Kapiche?

Momentum of the 0.14 kg object is indeed - 0.49 kg C.  Even your math
shows that.

No it ***DOESN'T*** you RETARDED GIBBERTARD!!!!!!!!!

If every single attoJoule of energy is conserved, the momentum of a
0.141421356 kg object moving
at 0.960784313 c is indeed 0.49 kg C (to within as many significant
digits your calculator supports).

Adding the **MAXIMUM** allowed momentum, 0.43 + 0.06 is 0.49.

But ***YOUR*** setup has

p = sqrt(((0.51Kg^2C^4)-(0.14KgC^2C^4))/C^2) = 0.490407993KgC

That is certainly NOT my setup - that is yours.  It DEMONSTRATES why
there was NEVER 0.51 kg of mass.

IT IS YOURS YOU GIBBERTARD

p_f^2 = (p_x' + p_x)^2 + (p_y' + p_y)^2 + (p_z' + p_z)^2

you have y',y,z', & z = 0

which SIMPLIFIES to

p_f^2 = (0.43 + 0.06)^2 = MAXIMUM of ***0.49KgC****

YET YOU ALSO HAVE ****** 0.490407993KgC*****

You TWIT!

Screamed spintwitty at the face in the mirror once he realized no one
was falling for his posturing.

You never mentioned ANYTHING about other spatial dimensions before;
why now ?

Oh, right - you've been shown to be stupid many times over, and are
frantically farting out a smokescreen.

Once again, buffoon : what is the point of carrying out all operations
to 7 or more decimal points ?

Starting with 1 kg c^2 of energy (the 1 kg mass), and 0.49 kg c^2 goes
right (as you have repeated screamed and bellowed), then the remaining
mass has energy 0.51 kg c^2 - composed of REST MASS and KINETIC
ENERGY.

0.51 kg c^2 J = sqrt(p^2 c^2 + m_0^2 c^4)

Square both sides to get :

0.51^2 kg^2 c^4 = p^2 c^2 + m_0^2 c^4

Divide both sides by c^4 to get :

0.51^2 kg^2 = p^2/c^2 + m_0^2

Given there is p = 0.49 kg c going RIGHT (as you have repeated
screamed and bellowed), the momentum of the mass going LEFT must equal
that, so p = 0.49 kg c

Meaning 0.51^2 kg^2 = 0.49^2 kg^2 + m_0^2

Therefore, m_0 MUST equal sqrt (0.51^2 kg^2 - 0.49^2 kg^2), or
0.141421356 kg.

LOOOOOOOOOOOOOOOOOOOOOOOOOOOOOLLLLLLL

Translation : sh*t !! Prof Weird showed spintwitty is a howling
buffoon once again !!! spintwitty must posture and scream in a flaccid
attempt to distract everyone !!1!11!

And ****NOW*** he has ***CHANGED*** back to

"0.141421356 kg."

LOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOLLLL

Translation : sh*t !! spintwitty MUST scream louder and fart more
odiously to distract people from his psychotic fixation on significant
decimal places !!

NOTE FOR THE STOOPID.

0.490407993KgC

is GREATER than

0.49KgC!

True, but not relevant in this example,

YOU UTTER RETARD!!!!!!!!!!!!!!!

Screamed spintwitty at his reflection when he realized no one was
falling for his bluffs

because you're a blithering
idiot that can't tell the difference betwen a mass, a momentum, and
kinetic energy.  And for two significant digits (and given the
rounding off done to keep the math simple enough for you to follow),
is 0.490408 really significantly different from 0.49 ?

YES YOU TWIT.

WHY DONT YOU PUT 0.000407993Kg of matter in your mouth and BLOW IT UP
AND THEN SEE IF IT DOESN'T MATTER.

Focus on the PROBLEMS AT HAND - to 2 decimal places, 0.490408 is
0.49. What, EXACTLY, is the point of carrying out operations to 7 or
more decimal places ?

Oh, right - it gives you an excuse to howl and vomit up insults when
reality does not conform to your delusions ...

Because of your pathological fixation on significant digits, your
example of a 1 kg mass that has 0.49 kg C^2 of light going to the
right boils down to :

m_0 = 0.141421356 kg; 'photonic mass' = 1 kg - m_0 = 0.858578643 kg.

0.429289321 kg C of momentum ( and 0.429289321 kg c^2 J of energy) go
right as photons; the other 0.429289321 kg C of photons go left
(momentum - 0.429289321 kg C), and hit the 0.141421356 kg rocket.

Of the 0.429289321 kg C of momentum hitting, 0.060710677 kg c reflects
back - giving that mass a momentum of 0.49 kg c total (actual value =
0.489999998 kg c).

Velocity of the 0.141421356 kg mass is 0.960784313 c.

Gamma = 1/sqrt(1-v^2/c^2), or 3.606244584
MOMENTUM of the 0.141421356 kg mass = m_0 x v x gamma = 0.4899999 kg c
(close enough to 0.49 kg c)
KINETIC ENERGY of the 0.141421356 kg mass = (gamma-1) x m_0 x c^2 = .
368578644 kg c^2.

If you check, you'll see that energy AND momentum have been conserved.- Hide quoted text -

Now wasn't that easy?

Now all you have to do is work out the angle of deflection,
the 0.06Kg's of photons underwent as they scattered,
add up all 3 spacial vectors.

Why ? You never mentioned them before. The only reason you care
about them now is the fact that reality does not conform to your
delusions, and so you're vomiting up a smokescreen.

If the photons were going LEFT, and are now going RIGHT, the angle of
deflection was 180 degrees.

Then explain why the total momentum 0.489999998KgC is **NOT**
all going in the x-axis (to the right to you).

If there was scattering in all directions, it all wouldn't be. But
since you made no mention of the momentum splitting off in other
directions before this, why care about it now ?

Oh, right - because you've been shown to be wrong, and desperately
need a diversion.

Starting with 1 kg, the only way to get 0.49 kg C^2 J of photons going
RIGHT (as you have repeated screamed, bellowed and howled over the
weeks) is if every single one of them bounces straight back.

There are four scenarios of what happens when a mass is hit by
photons :

1) Transmission - the photons pass straight through without
interaction (like X-rays mostly do for flesh). In your case, ~ 0.43
kgC^2 J go right, 0.43 kg C^2 J go left and straight through the ~
0.14 kg mass. Which doesn't move. Energy and momentum conserved.

2) Total absorbtion - the photons are completely absorbed by the mass,
and NOTHING radiates away. In this case, the rest mass increases (as
an inelastic collision), so momentum is conserved, and the mass moves
left. This scenario only has ~ 0.43 kg C^2 J of photons going right.

3) Absorbtion with re-radiation : the photons are absorbed, momentum
is transferred and the extra energy is radiated away in all
directions. Get almost 0.49 kg C^2 J of photons going right, but the
momentum of the ~0.14 kg mass is ~0.43 kg C.

4) Reflection - the photons hit and bounce straight back. This gives
you 0.49 kg C momentum going right, the mass has momentum 0.49 kg C
going left.

Then explain how your scenario has any meaning at all in my thought
experiment.

It shows that your deranged ideas about mass magically shrinking as it
is accelerated is grossly in error, given the FACT that physics
derived from REAL WORLD OBSERVATIONS shows that mass increases as it
accelerates.

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