Re: Wheres weirdo?

On Oct 21, 10:06 am, spintronic <spintro...@xxxxxxxxxxx> wrote:
On 19 Oct, 16:17, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:





On Oct 17, 9:04 am, spintronic <spintro...@xxxxxxxxxxx> wrote:

On 15 Oct, 16:17, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:> On Oct 14, 8:30 am, spintronic <spintro...@xxxxxxxxxxx> wrote:

.
.
<snip>

GIGO

You UTTER RETARDED SPASDICK!

screamed spintwitty at his reflection

From:http://en.wikipedia.org/wiki/Compton_scattering

hf + m_0C^2 = hf' + sqrt((p_m_0C)^2 + (m_0C^2)^2)

*****************************
hf = 0.43KgC^2
m_0C^2 = 0.1400000KgC^2

From what orifice did you pull 'hf  = 0.43 kgC^2' ? Having trouble
keeping your masses straight ?  Again ?

You sure you got that formula correct ?  AND that it applies to
something other than hydrogen atoms ?

****************************

hf' = 0.06KgC^2

ERGO

p = ( hf - hf' )/C = 0.43KgC - 0.06KgC

Get THAT???????

You GIBBERTARD???

p = *******0.37KgC********

If the 0.14 kg mass WAS COMPOSED OF PHOTONS, you may have come close
to having a point (given that is what the Compton effect deals with -
the change in wavelength of PHOTONS); is a 0.14 kg mass a photon ?

In other words, you are changing the question to conform to your
delusions, rather than accept that you are so confused that you can't
tell the difference between momentum, mass, and kinetic energy !

Again, oaf : the 0.14 kg mass would have MOMENTUM of 0.49 kg c, and a
KINETIC ENERGY of 0.37 kg c^2 J.

Once again buffoon : 0.14 kg rest mass, 0.86 mass to be converted into
photons.

The 0.86 kg converted into 0.86 kg c^2 J of energy.

Ultimately, 0.49 kg c^2 J to the right, meaning a MOMENTUM of 0.49 kg
c goes right.

As momentum and energy MUST BE CONSERVED, there is a MOMENTUM of 0.49
kg c going left - the MOMENTUM of the 0.14 kg mass.

Which corresponds to a velocity of about 0.96078 c, meaning there is
0.51 kg c^2 J of ENERGY going left (the 0.14 kg mass, plus 0.37 kg c^2
J of KINETIC ENERGY). Energy and momentum are conserved.

It is ***YOU*** & **WILLIAM** who are **CONFUSING** this whole
process.
You **CAN'T** comprehend the thought experiment & so change the rules
to argue.

No, the only person trying to change the rules to make reality conform
to his answers is you spintwitty.

You **TRY** to incorperate compton scattering, which **WILL NOT** work
here!!

You 'determined' that HOW ?

Oh, right - because reality MUST bend to the whims of spintwitty the
bellicose !

See this below?

--------------------------------------------------------------

^ That (up above) is called a **STRAIGHT LINE**!.

Forces like **GRAVITY** act in **STRAIGHT LINES**!

YOURS & WILLIAMS setups ***DO NOT** act in straight lines.

You 'determined' that how ?

Oh, right - it MUST be true in order for spintwitty to be correct
about something.

Again, oaf : the 0.14 kg mass would have MOMENTUM of 0.49 kg c, and a
KINETIC ENERGY of 0.37 kg c^2 J.

ABSOLUTELY IN NO WAY SHAPE OR FORM IS THAT CORRECT!

Actually, it IS correct.

When the 0.86 kg of 'photonic fuel' is converted into photons, 0.43 kg
c^2 J go right, 0.43 kg c^2 J go left.

The photons going RIGHT carry a momentum of 0.43 kg c going right.

The photons going LEFT carry momentum - 0.43 kg c, and hit the 0.14 kg
mass.

Some of the photons reflect back right - in this case, they have a
momentum of 0.06 kg c.

The 0.14 kg mass gains - 0.43 kg c momentum from the photons striking
it, then another - 0.06 kg c as the photons reflect. Momentum
conserved : -0.43 kg c = - 0.49 kg c + 0.06 kg c !

This means the 0.14 kg mass has momentum - 0.49 kg c going left, while
there is (0.43 + 0.06) kg c momentum going right. Momentum is
conserved.

Of the 0.43 kg c^2 J of photons that went LEFT, only 0.06 kg c^2 J
return - meaning the 0.14 kg mass has ENERGY of 0.37 kg c^2 J. This
is its KINETIC ENERGY.

The KINETIC ENERGY of a 0.14 kg mass moving at 0.96071 c is indeed
0.37 kg c^2 (KE = m_0 x c^2 x gamma-1)

Your 0.06Kg's of deflected PHOTONS ***ARE NOT** deflected back in a
straight line.

Actually, they are. Upon what basis did you 'determine' otherwise ?

Oh, right - your pathological need to argue. Since reality-based
physics isn't giving you the answers you want, you invoke all sorts of
weird stuff to counteract it.

The rocket has

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

NOW 0.49KgC **IS NOT** 0.490407993KgC

You did the math wrong : (0.51 kgc^2 - 0.14 kg c^2)/c^2 = 0.37 kg !

Why you would want the square root of a MASS is beyond me, but it's
your delusion ... !

Using the "law of Parallelogram's" what you propose is
**IMPOSSIBLE**!!

Nope - doing YOUR math is impossible, for what is the SQUARE ROOT OF
0.37 KG ?!

As the **MAXIMUM** allowed momentum is

0.43 + 0.06 = 0.49 and is **LESS** than 0.490407993

SO YOUR PROPOSAL IS IMPOSSIBLE!

Now, THIS ROCKET HAS TO TRAVEL IN A STRAIGHT LINE JUST AS THE LINES OF
THE FIELD DO.

THE ONLY WAY TO DO IT IS THE WAY **** I *** PURPOSED!

                1KgC^2 Rest
                  |
<--0.51KgC^2 = E' |
                  |
-p = 0.49KgC      |  p = 0.49KgC

V = P/m' = 0.960784314C

Rest mass =   0.141421356Kg

Kapiche??

Yes - you're a blithering idiot.

You are using NEWTONIAN physics where RELATIVISITIC physics is more
applicable.

At 0.960784314 c, gamma = 3.606244506

Relativistic mass = m_0 x gamma = 0.141421356 kg x 3.606244506 = 0.51
kg.

Kinetic energy : gamma-1 x m_0 x c^2 = 2.606244506 x 0.141421356 kg x
c^2 = 0.368578632 kg c^2 J (about 0.37 kg c^2 )

Momentum : m_0 x V x gamma = .141421356 kg x 0.960784314 c x
3.606244506 = 0.489999987 kg c (about 0.49 kg c).

Please note that the 0.1414 kg mass has a MOMENTUM of 0.49 kg c, and a
KINETIC ENERGY of 0.37 kg c^2 - just like I've been telling you for
weeks now !

.

Relevant Pages

  • Re: Wheres weirdo?
    ... If the 0.14 kg mass WAS COMPOSED OF PHOTONS, ... tell the difference between momentum, mass, and kinetic energy! ...
    (talk.origins)
  • Re: Weirdos square of negative mass.
    ... CLAIM that is what reality expects, ... of the photons reflected back - in this case, 0.06 kg C of momentum. ... there was NEVER 0.51 kg of mass. ...
    (talk.origins)
  • Re: The Black Hole (Thought experiment.)
    ... that any mass that uses 1/2 of it's PE to ... photonic energy. ... sides of the (Centre of momentum frame) Given by. ... The same as it was before the photons were emitted. ...
    (sci.physics.relativity)
  • Re: black holes and singularity
    ... A single photon doesn't have mass. ... and your basis for flotons vs. photons. ... Yes, if anything is conserved, it's momentum. ... do you account for the moon being deflected, if mass in the moon isn't ...
    (sci.physics.relativity)
  • Re: Work - impulse
    ... another object of different mass at rest, ... Kinetic energy is conserved if the collision is perfectly elastic ... Momentum is a vector. ... No magic, no bogus concepts, no violations. ...
    (sci.physics)