"George Dishman" wrote in message
ups.com...
Max Keon wrote:
"George Dishman" wrote in message
...
Max Keon wrote:
You are still in the wrong universe. My equation doesn't design
the zero origin universe, the universe designs the equation.
Sure, and then the equation tells us what will happen to
the planets in your universe - they spiral into the Sun.
Theories often have consequences that weren't anticipated
by their originators. Planck could never have known the
consequences of his original lecture on quantisation as a
solution to black body radiation.
And that sent you all off on another wild goose chase. _Blackbody
radiation has never had anything to do with quantitization_. The
value for the photon has just been plucked out of the air. It
could have been anything at all. But you must surely know that?
Take this part of the formula for example: (2*pi*h*c^2).
Since every component is a constant, the whole thing can be
treated as a simple multiplier. That's all it is, isn't it! So
why isn't it stated as 3.747e-16. But 3.747e-16 what? It's
nothing more than a nominated surface area per unit radius from
the blackbody source on which to establish an energy quantity
according to wavelength. If the unit radius is 1 meter, the
designated surface area is 3.747e-16 square meters.
According to Planck, the energy received from a 4000 K radiator
at 7.25e-7 meter wavelength is
(2 * pi * h * c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
= 1.315e+13 (# units of energy, whatever they are)
Here's an equation that gives the emissive power per wavelength
received at unit radius from the same blackbody source for a 1
square meter surface area. w is wavelength, m is surface area.
m^2 / w^5 / (1.0145^(m / (w * t)) - 1) = 3.509e+28 (# units)
3.509e+28 * 3.747e-16 = 1.315e+13, as the Planck "photon" would
specify it for the tiny surface area.
Try any temperature-wavelength combination you like and you will
always get the right answer.
Planck's photon is entirely dependent on "e" which has no claim
to fame other than it never repeats. That is crap George.
Nice rant Max, but it doesn't even touch on the reason
why quantisation is required, the shape of the spectrum.
It's a little off topic, but the easiest way to demonstrate my
point (to anyone) is to set up a simple program. Each equation
set can be extracted and compared outside the program. This
program of course runs as it is in Qbasic.
Notice the absence of Planck's constant in the alternative
equations. All of the magic in Planck's formula is performed by
the EXP function, which is the same as 2.71828^(h*f) ect.
CLS
pi = 3.14159
c = 3E+08
k = 1.3807E-23
h = 6.6262E-34
m = 1 'Meter unit.
INPUT " Radiator temperature (degrees K)"; t
IF t = 0 THEN END
PRINT " Note: Peak emission wave length is";(4000 / t) * 725; "nm"
PRINT "For spectral energy density the peak is";(4000/t)*1285;"nm"
ma: INPUT " Wave length in nanometers."; w
IF w = 0 THEN END
w = w / 1E+09 'wavelength is reset to meters.
cm = .01 / w
f = c / w
PRINT
'---------------------------------------
PRINT " Planck Alternative"
PRINT
PRINT " Emissive power per wavelength."
plnka = (2*pi*h* c ^ 2) / (w ^ 5 * ((EXP((h * f) / (k * t))) - 1))
alta = m ^ 2 / w ^ 5 / (1.0145016# ^ (m / (w * t)) - 1)
PRINT plnka, alta * 3.747E-16; "per meter units"
'The multiplier is constant.
plnkb = (2*pi*h*(c*10)^2)/((w*10)^5*((EXP((h* f) / (k * t))) - 1))
altb = (m*10)^2/(w*10)^5/(1.0145016#^((m* 10) / (w * 10 * t)) - 1)
PRINT plnkb, altb * 3.747E-16; "per cm units."
PRINT
PRINT " Spectral energy density."
PRINT " Frequency ="; f; "~/sec. 0r"; cm; "~/cm."
plnkc = (8*pi* h * f ^ 3) / (c ^ 2 * (EXP((h * f) / (k * t)) - 1))
altc = f ^ 3 / (c ^ 2 * (1.0145016# ^ (m / ((c / f) * t)) - 1))
PRINT plnkc, altc * 1.6654E-32; "per meter units"
'The multiplier is again constant.
plnkd = (8*pi*h*f^3)/((c * 10) ^ 2 * (EXP((h * f) / (k * t)) - 1))
altd = f^3/((c*10)^2*(1.0145016#^(m*10/(((c * 10) / f) * t)) - 1))
PRINT plnkd, altd * 1.6654E-32; "per cm units"
GOTO ma
------
These are two sets of randomly chosen results for a 4000 k
radiator.
------
Radiator temperature (degrees K)? 4000
Note: Peak emission wave length is 725 nm
For spectral energy density the peak is 1285 nm
(which can obviously be converted to frequency).
Wave length in nanometers.? 400
Planck Alternative
Emissive power per wavelength.
4.523494E+12 4.523528E+12 per meter units
4.523494E+09 4.523528E+09 per cm units.
Spectral energy density.
Frequency = 7.5E+14 ~/sec. 0r 25000 ~/cm.
9.650121E-09 9.650578E-09 per meter units
9.650121E-11 9.650578E-11 per cm units
Wave length in nanometers.? 800
Planck Alternative
Emissive power per wavelength.
1.285609E+13 1.28561E+13 per meter units
1.285609E+10 1.28561E+10 per cm units.
Spectral energy density.
Frequency = 3.75E+14 ~/sec. 0r 12500 ~/cm.
1.097053E-07 1.097097E-07 per meter units
1.097053E-09 1.097097E-09 per cm units
And two results for a 2.73 K radiator.
Radiator temperature (degrees K)? 2.73
Note: Peak emission wave length is 1062271 nm
For spectral energy density the peak is 1882784 nm
Wave length in nanometers.? 1e6
Planck Alternative
Emissive power per wavelength.
1.929899E-03 1.929902E-03 per meter units
1.929899E-06 1.929902E-06 per cm units.
Spectral energy density.
Frequency = 3E+11 ~/sec. 0r 10 ~/cm.
2.573199E-17 2.573306E-17 per meter units
2.573199E-19 2.573306E-19 per cm units
Wave length in nanometers.? 2e6
Planck Alternative
Emissive power per wavelength.
9.028213E-04 9.028193E-04 per meter units
9.028213E-07 9.028193E-07 per cm units.
Spectral energy density.
Frequency = 1.5E+11 ~/sec. 0r 5 ~/cm.
4.815047E-17 4.815229E-17 per meter units
4.815047E-19 4.815229E-19 per cm units
------
Frequency is also given in cycles per cm, which is how the CMBR
monopole is normally depicted. A common multiplier is of course
required to align the power spectrum peak with 384 MJy/sr.
Power * 7.94e+18 does the trick.
How hard would it be
to detect the anomalous acceleration on the outward and inward
legs? How hard would it be to notice that one is pointing in the
opposite direction to the other, relative to the Sun? Whatever is
the cause, there should be some sort of anisotropy evident in the
result
Sure, but that gets you no farther forward in identifying
the _cause_ of the effect.
It would certainly prove that the effect is real though. Then we
can move on.
We already know the anomaly is real without a doubt.
Yes, but you still think it could be caused by systematic errors.
So you really have no idea what the actual cause might be.
But GR's amazing feat is quite irrelevant anyway because
instantaneous action at a distance was never a requirement in
anybody's gravity.
It was in Newton's.
Only because he didn't understand gravity at all. The fact is
that something was missing in the way he analyzed planetary
motion. And that is clearly still the case today.
No, GR resolved that problem 90 years ago. Science
has moved on a long way since then.
And how much farther it would have gone if it was on the right
path. Mankind might not be in the mess it's now in. The clock is
certainly ticking.
Go and learn vectors, then read up on the
two-body problem
The following accompanies an animation at this address.
http://www.optusnet.com.au/~maxkeon/binstar.html
The animation depicts a binary star pair (blue) in a concentric
orbit about each other. The star masses are identical. This an
impossible scenario of course because the orbit velocity of each
star is faster than light speed.
I assume you mean your illustration shows impossible
speeds.
The geometry can obviously be
altered to overcome that problem, but my point is still made.
The geometry is not a problem. There are many such
binary systems in reality but obviously the orbital speed
is much slower. Something around 300 km/s is the top
speed for a "grazing" binary IIRC but binaries of nearly
equal mass are not unusual.
The action of gravity at one point around the orbit only is
shown, and that is obviously common throughout the entire orbit
cycle. The nearest red dot in the direction of motion is the
retarded position of each companion star from the viewpoint of
each star.
The '\' line travels the straight line distance to the intersect
point with its companion at the speed of light. The action of
gravity on each star is necessarily shifted to point toward the
retarded image of its companion, which is 90 degrees offset from
its true position around the orbit path. The apparent gravity
link with each companion star can obviously never be generated
where the companion currently resides.
Sure, but only because you have assumed there is abberation
which isn't the case in Newtonian gravity or GR, you are only
describing a strawman so I don't see the point.
No. I'm merely describing how gravity works in the real universe.
The focal point of the orbit radius is perpendicular to the
natural tangent,
No, an ellipse has two focii and for the simpler case
of a circle, they coincide at the centre.
Take another look at the animation.
http://www.optusnet.com.au/~maxkeon/binstar.gif There is no
doubt that each star is drawing toward where its companion
appears to be. That's the direction of the pull of gravity at
the star even if that path curves back to the companion. It
doesn't make any difference if the gravity link is curved or
straight, but the distance to where the companion was is
shorter. Dimension was being drawn into the companion's gravity
well at that time, so even if the companion is no longer in the
same place, the full affect of its presence will be felt when
its image coincides with the star's position in space.
which would cause the stars to fly apart. But
the distance from each star to where its companion appears to be
is .707 of the instantaneous distance. According to 1/r^2, the
pull of gravity toward each other is 1 / .707^2 which is double
that required to hold the stars at the instantaneous orbit
radius. The stars would be pulled inward to point in the
direction of the instantaneous orbit tangent.
Even if that's not enough to do the trick, the universe has no
rules which dictate that a binary star pair of specific mass can
only reside at a specific radius for a specific orbit velocity.
A stable orbit exists for every circumstance. Denying that is
absurd.
Crap. For any radius, if the speed is too fast for a circular
orbit the object will move farther away and if it is too slow
it moves closer.
It will find a stable orbit somewhere though, won't it.
Both give stable elliptical orbits because
energy and momentum are conserved, but as sonn as you
start to remove energy the orbit decays. That's what happens
in nature so that's what our maths has to copy.
What energy is removed in a simple orbit? A curved gravity link
isn't going to remove energy.
---
so the force
remains like a spring ..
No it isn't. A spring always pulls in the same direction
like gravity but your anisotropy changes direction. A
spring removes and stores energy while being stretched
and returns it while being relaxed because the force is
in the same direction as the speed.
The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif
Wrong, you have no anisotropy in that.
Of course not. The elastic force applied by the universe has
been counteracted by Mercury's increased centrifugal force as
it orbits the Sun at a faster rate than you would expect. That's
its natural orbit in the prevailing circumstances.
---
The momentum loss immediately converts to potential energy, which
in turn slowly converts to kinetic energy.
Wrong, momentum and energy are not interchangeable, they
are separate quantities, both of which must be separately
conserved (as must the three spatial components of momentum).
In an inelastic, head on collision between two equal masses
the two masses are brought to a halt relative to each other.
Heat energy is obviously generated during the process. So where
does that come from if not from lost momentum?
From lost energy. There is no momentum lost in the crash.
Momentum may not have been lost from the universe, but neither
mass has momentum relative to the other. They are stationary.
The expended energy would need to be put back to restore the
original relationship, and that could point them in any opposing
directions. Their momentum relative to each other, and relative
to the universe, can now be pointing anywhere.
So what are you on about?
Mercury's momentum is transferred to the "elastic sheet" provided
by the mass of the universe. Potential energy is an immediate
consequence of its orbit velocity slowing. That slowly converts
to kinetic energy as it pulls toward the Sun. Mercury's fall
halts when its orbit velocity has increased to the point where
centrifugal forces counteract the fall. It then just cycles
around within the elastic sheet.
---
The need for instantaneous action at a distance to overcome the
problem of the planets spiraling into the Sun should have
sounded alarm bells.
It did and people were trying to resolve it for 200
years until Einstein found the solution.
You know as well as I do that the "solution" fails the Pound and
Rebka test.
What are you talking about Max? Pound-Rebka is one of
the classic tests which _confirm_ GR !
The radioactive iron frequency generator located at the
Pound-Rebka tower base appears to cycle slower than a similar
piece located at the tower top. The logical conclusion is that
the generated frequencies are not the same. But if that was the
case, although GR could probably live with it, it would scuttle
the big bang theory. So the only available option is that the
frequencies generated in each piece must remain constant, that
they only appear to be different through some geometric quirk
of nature.
The problem is that the radioactive iron frequency generator is
much the same as the Caesium atom configuration that drives an
atomic clock. Atomic clocks can run for years, and all the while
the base clock would be losing ticks compared with the top clock.
Those lost ticks can't hide forever in the distance between the
tower base and the top.
When the clocks are brought together, there is no doubt that
the base clock will have lost time compared to the top clock.
_No doubt at all_. The time difference will depend almost
entirely on how long they have been separated over the tower
height.
If the cycle time of an atomic clock oscillator varies with
altitude, so did the radioactive iron frequency source used in
the Pound and Rebka experiment.
There are no tests
whatsoever that GR fails, only known limitations in merging
it with QM, which could as easily indicate a failing of QM.
I don't think that QM is going to be a problem.
-----
Max Keon