More speed confusion
Peter Riedt
The formula for the addition of velocities is given as (v1+v2)/
(1+v1*v2/c^2) #1. Why is it necessary to include the square of the
speed of LIGHT into the formula? Well, the gurus who invented
relativity also postulated that the speed of light is the maximum
speed that is possible and therefore any composite speed had to be
adjusted accordingly. This they achieved with the formula given above
but it is only a conjecture even though it is claimed that particle
physics prove it correct. Now anything can be proved by particle
physics especially that pigs fly. Particle physics is the ultimate
cash cow. The billions spent on it must be justified to prevent the
cash cow and her guardians from being subject to starvation or
poverty.
If the above formula is correct, it is only half the story. The
missing other half is the formula (v1-v2)/(1+v1*v2/c^2) #2. Let me
illustrate it by two scenarios.
In the first scenario, Glaucus chases the nymph Scylla. Glaucus runs
at 20km/hr (v1) and Scylla runs away at 12km/hr (v2). Their combined
speed (negative closing speed) is v1-v2 or 8km/hr. Using formula #2,
their combined speed is 7.999999968km/sec.
In the second scenario, Glaucus and Scylla (after changing her mind)
run towards each other to embrace at 20km/sec and 12km/sec
respectively. Their combined speed (positive closing speed) is v1+v2
or 32km/sec. Using formula #1, their combined speed is 31.99999997.
Formula #1 was obtained by Fizeau (1819-1896) in his experiments about
the propagation of light through flowing water in 1851. AE, citing
Fizeau, proposed this formula for the addition of velocities of all
physical objects (AE, ‘Relativity: The Special and General Theory’,
Chapter 13). He however missed formula #2 and was not aware that if an
object with a velocity of 300000km/sec chases an object running away
at 300000km/sec, their combined speed is -300000km/sec. As in particle
physics.
Peter Riedt
Tom Roberts
> The formula for the addition of velocities is given as (v1+v2)/
> (1+v1*v2/c^2) #1. Why is it necessary to include the square of the
> speed of LIGHT into the formula?
That isn't really the speed of light, it is the symmetry speed of the
spacetime manifold. While Einstein originally based SR on the speed of
light, today we know better and base it on a symmetry of spacetime.
Experimentally, the velocity of that symmetry is equal to the vacuum
speed of light, with very small errorbars.
> but it is only a conjecture even though it is claimed that particle
> physics prove it correct.
It is far more solid than a mere conjecture. It is part and parcel of
modern physics, and has extensive experimental support.
> [...]
You obviously know nothing about particle physics.
You also fail to understand how the addition of velocities formula is
applied to practical problems (hint: velocities have signs in the mode
you use this formula).
You also don't know much about the history of SR (AE did not cite Fizeau
for the composition of velocities formula, he derived it from the
transform equations, as did Lorentz before him).
Rather than spouting nonsense and conspiracy theories, you should STUDY
modern physics before attempting to write about it.
Tom Roberts
Androcles
"Peter Riedt" <rie...@yahoo.co.uk> wrote in message
news:d122c984-5128-4f1b...@i28g2000prd.googlegroups.com...
More speed confusion
The formula for the addition of velocities is given as (v1+v2)/
(1+v1*v2/c^2) #1. Why is it necessary to include the square of the
speed of LIGHT into the formula? Well, the gurus who invented
relativity
============================================
Whoa! Only one crank moron is responsible, all the rest are sheep!
One doesn't call sheep "gurus".
(1+x)/(1+ (1*x)/1^2) = 1 for all x.
Albertito
Haven't you still realized that SR is bullshit and
all supporters of it are bullshitters?
http://en.wikipedia.org/wiki/Bullshit#.22Bullshit.22_in_philosophy
See also Labov's Test:
"According to Jim Holt's discussion in the New Yorker
of Gerald Cohen's paper "Deeper into Bullshit", Cohen
independently discovered what I have always privately
called Labov's Test:
...how could one prove ... that a given statement
is hopelessly unclear, and hence bullshit? One
proposed test is to add a 'not' to the statement
and see if that makes any difference to its plausibility.
If it doesn't, that statement is bullshit."
http://itre.cis.upenn.edu/~myl/languagelog/archives/002402.html
How to apply a Labov's Test to SR?
Take, for instance, the twin paradox and add a 'not' to it, such that
now you have "the stay-at-home twin aged slower than the travelling
twin". If by means of the SR's postulates, it resulted a time
contraction and a length dilation, then SR would be as self-
consistent
as it is. Any SR's supporter would be defending that a moving clock
ticks faster than a clock at rest, and that its length is larger.
Peter Riedt
Androcles, I wrote gurus but meant 'gurus'.
What about (1-x)/(1+(1*x)/1^2)?
Peter Riedt
Androcles
What about (1-x)/(1+(1*x)/1^2)?
===========================
i) (1-x)/(1+(1*x)/1^2) = (1-x)/(1+(1*x)/1) because 1^2 = 1
ii) (1-x)/(1+(1*x)/1) = (1-x)/(1+(1*x)) because x/1 = x
iii) (1-x)/(1+(1*x)) = (1-x)/(1+(x)) because x*1 = x
iv) (1-x)/(1+(x)) = (1-x)/(1+x) because (x) = x
Hence
v) (1-x)/(1+(1*x)/1^2) = (1-x)/(1+x)
What you probably meant was
"What about (1-x)/(1+(1*-x)/1^2)?"
Here we use dummy variable, y = -x.
vi) (1+x)/(1+ (1*x)/1^2) = 1 for all x.
let y = -x
vii) (1+(-y))/(1+ (1*(-y))/1^2) = 1
viii) (1+(-y))/(1+ (1*(-y))/1) = 1 because 1^2 = 1
ix) (1+(-y))/(1+ (1*(-y))) = 1 because anything divided by 1 = the same
anything
x) (1+-y)/(1+ (1*-y)) = 1 remove redundant parenthesis pairs.
xi) (1-y)/(1+ (1*-y)) = 1 because 1+-y = 1-y
xii) (1-y)/(1+ (-y)) = 1 because 1* anything = the same anything.
xiii) (1-y)/(1+ -y) = 1 remove redundant parenthesis pairs.
xiv) (1-y)/(1-y) = 1 because 1+-y = 1-y
Let a = (1-y)
xv) a/a = 1 by substituting a in eq. (xiv).
but 'a' divided by 'a' DOES equal 1.
Quod erat demonstrandum.
Hence (1-x)/(1+(1*-x)/1^2) = 1
Einstein is viewed as a genius only by those with no mathematical
ability whatosover. Only a complete idiot would write a mathematical
paper with fourteen "assumes" in it.
1) But it is not possible without further ASSUMPTION to compare, in respect
of time, an event at A with an event at B.
2) We ASSUME that this definition of synchronism is free from
contradictions, and possible for any number of points;
3) In agreement with experience we further ASSUME the quantity 2AB/(t'A-tA)
= c
4) Current kinematics tacitly ASSUMES that the lengths determined by these
two operations are precisely equal
5) and where for brevity it is ASSUMED that at the origin of k, tau = 0,
when t=0.
6) If no ASSUMPTION whatever be made as to the initial position of the
moving system and as to the zero point of tau
7) We now have to prove that any ray of light, measured in the moving
system, is propagated with the velocity c, if, as we have ASSUMED, this is
the case in the stationary system
8) If we ASSUME that the result proved for a polygonal line is also valid
for a continuously curved line,
9) and our equations ASSUME the form
10) When phi = 0 the equation ASSUMES the perspicuous form
11) the equation for phi' ASSUMES the form
12) for the law of motion of which we ASSUME as follows
13) we may and will ASSUME that the electron, at the moment when we give it
our attention
14) From the above ASSUMPTION, in combination with the principle of
relativity
9) 10) and 11) are excusable, the rest are not.
Einstein was no guru, Einstein was a slick confidence trickster.
Many criminals are clever.
http://tinyurl.com/raf4jp
Juan R.
> Peter Riedt wrote:
>> The formula for the addition of velocities is given as (v1+v2)/
>> (1+v1*v2/c^2) #1. Why is it necessary to include the square of the
>> speed of LIGHT into the formula?
>
> That isn't really the speed of light, it is the symmetry speed of the
> spacetime manifold. While Einstein originally based SR on the speed of
> light, today we know better and base it on a symmetry of spacetime.
> Experimentally, the velocity of that symmetry is equal to the vacuum
> speed of light, with very small errorbars.
I really love your explanation:
1) c is not the speed of ligh in vacuum but the symmetry speed of the
spacetime manifold.
2) Experimentally, the velocity of that symmetry is equal to the vaccumm
speed of light, with very small errorbars.
Tom Roberts: "adding still more speed confusion" :-D)
--
http://www.canonicalscience.org/
Usenet Guidelines:
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Sue...
> More speed confusion
>
> The formula for the addition of velocities is given as (v1+v2)/
> (1+v1*v2/c^2) #1. Why is it necessary to include the square of the
> speed of LIGHT into the formula?
It isn't necessary.
<< The main types of particle combinations used at RHIC
are p + p, d + Au, Cu + Cu and Au + Au. The projectiles
typically travel at a speed of 99.995% of the speed of
light in vacuum. For Au + Au collision, the center-of-mass
energy \sqrt{s_{NN}} is typically 200 GeV (or 100 GeV per nucleus); >>
http://en.wikipedia.org/wiki/Relativistic_Heavy_Ion_Collider
<< Einstein's relativity principle states that:
All inertial frames are totally equivalent
for the performance of all physical experiments.>>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
Sue...
Juan R.
> On May 24, 4:20 am, Peter Riedt <rie...@yahoo.co.uk> wrote:
> Haven't you still realized that SR is bullshit and all supporters of it
> are bullshitters?
Haven't you still realized that your 'theory' is bullshit and all
supporters of it are bullshitters? :-D
Sue...
> Peter Riedt wrote:
> > The formula for the addition of velocities is given as (v1+v2)/
> > (1+v1*v2/c^2) #1. Why is it necessary to include the square of the
> > speed of LIGHT into the formula?
>
> That isn't really the speed of light, it is the symmetry speed of the
> spacetime manifold.
<<
Pseudoscientists invent their own vocabulary in
which many terms lack precise or unambiguous
definitions, and some have no definition at all.
Listeners are often forced to interpret the
statements according to their own preconceptions.
What, for for example, is "biocosmic energy?" Or
a "psychotronic amplification system?" Pseudoscientists
often attempt to imitate the jargon of scientific
and technical fields by spouting gibberish that sounds
scientific and technical. Quack "healers" would be
lost without the term "energy," but their use of the
term has nothing whatsoever to do with the concept
of energy used by physicists. >>
Distinguishing Science and Pseudoscience
Rory Coker, Ph.D.
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html
Sue...
Peter Riedt
Androcles, excellent and comprehensive derivation resulting in the
answer 1=1.
To which I like to add, if we add the speed of light to any other
speed up to 300000km/sec, the answer is always 300000km/sec but if we
subtract the speed of an object from the speed of light using the
formula(c-v)/(1+c*v/c^2)[negative closing speed as in source and
target approaching each other] the result is less than c-v e.g.
for 300000km/sec-30km/sec the composite speed is not 299970km/sec but
299940.006km/sec. However if both c and v are 300000km/sec, they
approach each other at a whopping c-v = 0km/sec!
Peter Riedt
Albertito
<juanREM...@canonicalscience.com> wrote:
> Albertito wrote on Sun, 24 May 2009 04:38:20 -0700:
>
> > On May 24, 4:20 am, Peter Riedt <rie...@yahoo.co.uk> wrote:
> > Haven't you still realized that SR is bullshit and all supporters of it
> > are bullshitters?
>
> Haven't you still realized that your 'theory' is bullshit and all
> supporters of it are bullshitters? :-D
>
>
> Usenet Guidelines:http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
No.
Androcles
Peter Riedt
=============================================
Not by my computer's spreadsheet it doesn't.
c v c+v (c+v)/(1+v/c)
1 2 3 1
1 1.75 2.75 1
1 1.5 2.5 1
1 1.25 2.25 1
1 1 2 1
1 0.75 1.75 1
1 0.5 1.5 1
1 0.25 1.25 1
1 0 1 1
1 -0.25 0.75 1
1 -0.5 0.5 1
1 -0.75 0.25 1
1 -1 0 #DIV/0!
1 -1.25 -0.25 1
1 -1.5 -0.5 1
1 -1.75 -0.75 1
1 -2 -1 1
1 -2.25 -1.25 1
1 -2.5 -1.5 1
1 -2.75 -1.75 1
1 -3 -2 1
c v c+v (c+v)/(1+v/c)
300000 600000 900000 300000
300000 500000 800000 300000
300000 400000 700000 300000
300000 300000 600000 300000
300000 200000 500000 300000
300000 100000 400000 300000
300000 0 300000 300000
300000 -100000 200000 300000
300000 -200000 100000 300000
300000 -300000 0 #DIV/0!
300000 -400000 -100000 300000
300000 -500000 -200000 300000
300000 -600000 -300000 300000
300000 -700000 -400000 300000
300000 -800000 -500000 300000
300000 -900000 -600000 300000
300000 -1000000 -700000 300000
300000 -1100000 -800000 300000
300000 -1200000 -900000 300000
300000 -1300000 -1000000 300000
300000 -1400000 -1100000 300000
Of course computers are only logical and perform
arithmetic perfectly, the relativistic computer has yet
to be invented.
Tom Roberts
> Tom Roberts wrote on Sat, 23 May 2009 22:35:44 -0500:
>
>> Peter Riedt wrote:
>>> The formula for the addition of velocities is given as (v1+v2)/
>>> (1+v1*v2/c^2) #1. Why is it necessary to include the square of the
>>> speed of LIGHT into the formula?
>> That isn't really the speed of light, it is the symmetry speed of the
>> spacetime manifold. While Einstein originally based SR on the speed of
>> light, today we know better and base it on a symmetry of spacetime.
>> Experimentally, the velocity of that symmetry is equal to the vacuum
>> speed of light, with very small errorbars.
>
> I really love your explanation:
> 1) c is not the speed of ligh in vacuum but the symmetry speed of the
> spacetime manifold.
> 2) Experimentally, the velocity of that symmetry is equal to the vaccumm
> speed of light, with very small errorbars.
> Tom Roberts: "adding still more speed confusion" :-D)
I did not add any more confusion.
The original question was essentially: Why is light so special that its
speed is involved in the velocity-composition formula for every object?
I merely pointed out that light is not special in that sense, but rather
the geometry of the manifold is special. Indeed, light is affected the
same as any other object by this geometry; it "just happens" to travel
with the speed of the geometrical symmetry (which permitted the
historical confusion I pointed out, and the use of the same symbol for
two VERY DIFFERENT theoretical quantities that have the same value).
Tom Roberts
PD
The confusion is completely abated if you use sensible units for space
and time (common, natural units). Then the magnitude of speed is
0<v<1, and the composition formula is simple:
(v1 + v2) / (1 + v1*v2).
PD
Paul B. Andersen
> Androcles, excellent and comprehensive derivation resulting in the
> answer 1=1.
>
> To which I like to add, if we add the speed of light to any other
> speed up to 300000km/sec, the answer is always 300000km/sec but if we
> subtract the speed of an object from the speed of light using the
> formula(c-v)/(1+c*v/c^2)[negative closing speed as in source and
> target approaching each other] the result is less than c-v e.g.
> for 300000km/sec-30km/sec the composite speed is not 299970km/sec but
> 299940.006km/sec. However if both c and v are 300000km/sec, they
> approach each other at a whopping c-v = 0km/sec!
>
> Peter Riedt
The correct formula is: (c-v)/(1-c*v/c^2) = c
--
Paul
PD
Sign error in the above formula. Do you know where it is?
Androcles
"Paul B. Andersen" <paul.b....@guesswhatuia.no> wrote in message
news:4A1C4340...@guesswhatuia.no...
Bwahahahahaha!
http://focus.aps.org/story/v5/st19
"In the last decade, physicists have discovered a way to slow photons down
to the speed of a bicycle. "
http://www.rsc.org/publishing/journals/JM/article.asp?doi=b708474a
"In this paper we will look at slow photons mainly through the eye of
chemistry and highlight some recent developments in this exciting and
emerging field"
Peter Riedt
Paul, agreed.
(300000-300000)/(1-300000*300000/300000*300000)=0
Peter Riedt
PD
Oh, Peter, please. Do you really have this much difficulty with
working number problems?
Peter Riedt
>
> - Show quoted text -
Androcles, perhaps Paul's formula works better.
Peter Riedt
Androcles
Peter Riedt
======================================
That IS the dickhead's formula, you incompetent lunatic!
Sheesh, you are as stooopid as he is.
Androcles
Paul, agreed.
(300000-300000)/(1-300000*300000/300000*300000)=0
========================================
So you agree Tusseladd's c = 0.
You are both raving mad and YOU, Riedt, left out the
parentheses!
(300000-300000)/(1- [300000*300000]/[300000*300000])
=(300000-300000)/(1-90000000000/90000000000)
=(300000-300000)/(1-1)
= 0/0
= Divide by zero.
FAILED!
Riedt's gaffe:
(300000-300000)/(1-300000*300000/300000*300000)=0
Parsing left to right,
(300000-300000)/(1-90000000000/300000*300000)
(300000-300000)/(1-30000*300000)
(300000-300000)/(1-90000000000)
0/ (-89999999999) = 0
But c-v = 0 when v = c anyway, you pair of idiotic cretins.
For Riedt's education:
http://en.wikipedia.org/wiki/Parsing
(Follow-ups set to alt.morons )
Paul B. Andersen
No.
0/0 is undetermined, it's not zero
But lim[(c-v)/(1-c*v/c^2)] = c when v -> c
Note that the "speed addition formula" is a misnomer,
it is a speed tranformation formula.
If the speed of an object is u in an inertial frame K,
then this speed transforms to w in a frame K' which is
moving with the speed v relative to K, where
w = (u+v)/(1+u*v/c^2)
The 'object' can be light (or a photon) and thus u = c,
but the speed of frame K' relative to frame K cannot be c,
so both u and v can never both be c.
v can however be arbitrary close to c, so the speed c
will _always_ transform to c for all v.
--
Paul
Androcles
So much fun to see you squirming with simple algebra, Tusseladd,
and how you hope to take a limit!
"lim[(c-v)/(1-c*v/c^2)] = c when v -> c ".. ahahahahahahaha!
Pity a spreadsheet doesn't agree with you!
But then, you are proven liar as well as a idiot, anyway.
http://www.androcles01.pwp.blueyonder.co.uk/E%5E2/DeriveMC2.htm
Peter Riedt
>
> - Show quoted text -
Androcles, if you insert additional parentheses as you did, the answer
is division by zero but if you leave it as Paul suggested i.e. (c-v)/
(1-c*v/c^2, the answer is 0. But this 0 is not = c as Paul indicated;
it is the composite speed of two objects (including two rays of light)
approaching each other at c (negative closing speed). It means that if
two particles at 300000km/sec each try to collide, they never can
because their combined speed is zero, nada, zilch, nix.
Peter Riedt
Androcles
=============================================
That's what the spreadsheet says!
but if you leave it as Paul suggested i.e. (c-v)/
(1-c*v/c^2, the answer is 0.
============================================
Tusseladd can no more pull equations out of his arse than Einstein can.
Einstein's idiotic equation is given in
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img76.gif
(LOOK IT UP!)
"It follows, further, that the velocity of light c cannot be altered by
composition with a velocity less than that of light. For this case we obtain
V = (c+w) / (1+ w/c) = c
It doesn't matter if you use u, v, or w for the velocity, it is still the
velocity.
If we set w = -v (i.e. travel backwards), then we can directly substitute:
V = (c+ -v) / (1+ -v/c) = c
If I have a spreadsheet and set cell A1 to 1, A2 to 1 and A3 to A1+A2
then A3 will display 2 .
If I then set A2 to -1 then A3 will display 0.
That's VERY basic algebra, which Tusseladd cannot manage.
==============================================
But this 0 is not = c as Paul indicated;
==============================================
So why did you agree with the moron? It's people like you that
spread confusion.
And don't call him "Paul", he's a Norwegian troll (tusseladd).
==============================================
it is the composite speed of two objects (including two rays of light)
approaching each other at c (negative closing speed). It means that if
two particles at 300000km/sec each try to collide, they never can
because their combined speed is zero, nada, zilch, nix.
Peter Riedt
===============================================
Absolute nonsense. I see cars coming toward me at -140 mph
every day and they are only travelling at -70 mph with respect to
the road.
The car directly in front of me that is doing 70 mph w.r.t. the road isn't
getting any further from me, it is doing 0 mph relative to me.
How you can ever hope to understand Einstein's con when you
can't even manage to add numbers for an opening speed is astounding.
Paul B. Andersen
Good grief. :-)
I see any further comment from my part would be futile.
--
Paul
PD
No, sorry, you've made a mistake. The composition of speeds is a
transformation between two inertial frames. Light is not tied to any
inertial reference frame; put more carefully, there is no inertial
reference frame in which a ray of light is at rest.
In a sense, there IS no way to find a relative speed between two
photons in any frame, though you can certain calculate a closing
speed, which is something different -- an unmeasurable quantity.
>
> Peter Riedt
Albertito
A proton and anti-proton are accelerated in
a ring with opposite directions, achieving
tangential velocities of 0.99994c, such that
they will collide at certain point in the tring.
What will be the speed of the proton in the
moment of the collision as measured in the
rest frame of the antiproton?
PD
0.999999982c.
A similar calculation for heavy ions at RHIC tells you the Lorentz
contraction of one nucleus as seen by the other, and the accordingly
higher quark-gluon density. This leads to certain predictions about
what will be seen in surrounding detectors in terms of secondary
particle production. This seems to be well matched by actual
measurements.
"Sue" has been asking the same question in a number of other threads,
wherein he asks "what's twice 0.999994c?". The answer to that question
is obvious, but it doesn't have anything to do with the velocity of
the proton relative to the antiproton.
PD
Dono
>
>
> A proton and anti-proton are accelerated in
> a ring with opposite directions, achieving
> tangential velocities of 0.99994c, such that
> they will collide at certain point in the tring.
> What will be the speed of the proton in the
> moment of the collision as measured in the
> rest frame of the antiproton?
2v/(1+(v/c)^2)=0.999999982c.
Can you show why?
Albertito
iF the ring has a radius of 1 m, then, at such
speed of 0.999999982c, as measured in the rest
frame of the anti-proton, the proton would be
cycling around the ring at a frequency of
f = 1.000000018 Hz/s. As measured in the rest
frame of the ring, the proton is cycling at
f' = 1.00006 Hz/s. Since both particles travel
always in opposite directions with opposite
tangential velocities of equal magnitude.
Why is the proton seen by the anti-proton
cycling at less frequency than that seen in
the rest frame of the ring?
Dono
>
> Why is the proton seen by the anti-proton
> cycling at less frequency than that seen in
> the rest frame of the ring?
Is it, Albertshito?
PD
Because of time dilation, silly. You are changing reference frames
from one in which the ring is at rest to one in which the antiproton
is at rest. When you do that, then the time between the same two
events (proton and antiproton subsequent passing each other at a
particular location) changes. Special relativity is completely self-
consistent.
But while we're on the subject, note your calculation of f and f' is
likely wrong, though you didn't specify the radius of the ring. As it
is, it appears you have a ring that reaches a good fraction of the
distance to the Moon. And furthermore, the units of frequency are Hz
or 1/s, but not Hz/s.
PD
Albertito
Sorry, that was a typo, I meant cycles/s
which of course are Hz. But, you read wrongly,
I did specify the radius of the ring, I said
it is 1 m. You say that each particle is cycling
at different frequency as measured in different
rest frames, and that's because of time dilation,
ok? Then, there is no agreement as to where the point
of collision will be located in the ring, but we
know those two particles will collide in a point
of the rest frame of the ring. This resembles a
fucking twin paradox. Can you please predict
unambiguously a point of collision under SR's
assumptions?
Dono
Albertshito,
Try calculating the value for the induction B for a radius of 1m, a
frequency f=1Hz and a speed of .9994c. What value did you get?
> Then, there is no agreement as to where the point
> of collision will be located in the ring,
Prove it, Albertshito.
PD
Ah, so you did, my error.
In that case the numbers you have for f are completely wrong. You
perhaps forgot to include the value of c in m/s. Or perhaps you'd like
to recast the radius of the ring to be 1 second (in natural units), in
which case we're back to a really big ring.
> You say that each particle is cycling
> at different frequency as measured in different
> rest frames, and that's because of time dilation,
> ok?
Yes.
> Then, there is no agreement as to where the point
> of collision will be located in the ring,
Why would you think that? There is agreement about the events, just
not about the time between them.
Now, if you want follow this through to the bottom, we're quickly
going to run into a minor problem that the proton and antiproton are
not at rest in any *inertial* frame, because it is a ring. But we can
get around that by straightening out the ring to be linearly and just
ask about the time and distance between an antiproton passing a proton
and the same antiproton passing another proton spaced appropriately in
the lab frame.
> but we
> know those two particles will collide in a point
> of the rest frame of the ring. This resembles a
> fucking twin paradox. Can you please predict
> unambiguously a point of collision under SR's
> assumptions?
Of course you can. But notice that the *distance* between the
crossings is different in the lab frame than it is in the antiproton
frame, and the *time* between crossings is different in the lab frame
than it is in the antiproton frame. This is all consistent with
relativity and with the velocity transformation law I gave you. It's
also true that there IS an invariant quantity pertaining to subsequent
crossings (same between lab and antiproton frames) -- and that's the
quantity [del(t)]^2 - [del(x)]^2 in natural units.
If it helps, the circumference of the ring is only 2pi*(1m) in only
one reference frame.
PD
Albertito
If there is agreement between events and there is no
agreement about times between them, then there is no
agreement about distances, either. But, the collision
is an important event in any particle accelerator. A lot
of efforts are put in predicting where and when a
collision will take place. You can't rely on SR for
that prediction at ultra-high energies, unless you want
a great failure. SR is becoming obsolete very quickly
as we reach new high energy records in accelerators.
PD
Yes. That's what I just said.
> But, the collision
> is an important event in any particle accelerator. A lot
> of efforts are put in predicting where and when a
> collision will take place.
In the lab frame, yes. The fact that the distance is different in a
different frame doesn't mean the antiproton thinks the collision takes
place at a different point on the ring.
My, my. Don't you think it might do you good to learn what relativity
really does say rather than just assume that it's gotta be wrong?
> You can't rely on SR for
> that prediction at ultra-high energies, unless you want
> a great failure.
Seems to be working so far.
> SR is becoming obsolete very quickly
> as we reach new high energy records in accelerators.
Well, let's see, we have a small bet on whether the LHC will be able
to circulate protons. You say no, and if the answer is yes, then
you've promised to shut up about it.
Peter Riedt
>
> - Show quoted text -
Androcles et al, this is the end to the confusion: If two particles
are in direct collision say at a speed of 300000km/sec each, their
closing speed is 600000km/sec e.g. the distance between them shrinks
at that speed.
Peter Riedt
Androcles
Peter Riedt
============================================
Right, Riedt.
Right, Galileo.
Right, Newton.
Wrong, Tusseladd.
Wrong, Einstein.
Wrong, Minkowski.
And it does happen if they ever make LHC work, so you can throw
special relativity in the trash can because theoretical physicists are
miserable failed mathematicians.
Here's another one:
http://www.ivorcatt.com/2804.htm
Cephalobu...@comcast.net
> "Paul B. Andersen" <paul.b.ander...@guesswhatuia.no> wrote in messagenews:4A1D9434...@guesswhatuia.no...
>
> > 0/0 is undetermined, it's not zero
>
> > But lim[(c-v)/(1-c*v/c^2)] = c when v -> c
>
> > Note that the "speed addition formula" is a misnomer,
> > it is a speed tranformation formula.
> > If the speed of an object is u in an inertial frame K,
> > then this speed transforms to w in a frame K' which is
> > moving with the speed v relative to K, where
> > w = (u+v)/(1+u*v/c^2)
>
> > The 'object' can be light (or a photon) and thus u = c,
> > but the speed of frame K' relative to frame K cannot be c,
> > so both u and v can never both be c.
> > v can however be arbitrary close to c, so the speed c
> > will _always_ transform to c for all v.
>
> > --
> > Paul
>
> Bwahahahaha!
> So much fun to see you squirming with simple algebra, Tusseladd,
> and how you hope to take a limit!
>
> "lim[(c-v)/(1-c*v/c^2)] = c when v -> c ".. ahahahahahahaha!
> Pity a spreadsheet doesn't agree with you!
Good grief, Androcles!
(c-v)/(1-c*v/c^2) = (c-v)/(1-v/c) = c*(c-v)/(c-v)
The above equals c everywhere except at v = c where it is
undefined. So Paul's limit is easily seen to be correct without
using a spreadsheet.
What in heck are you using for a "spreadsheet", Androcles?
Why do you constantly insist of making a fool of yourself?
Not that you will answer, of course. Even though you are reading
this, you are pretending to have "plonked" me, so you will seeth
in silence.
Jerry
Peter Riedt
>
> - Show quoted text -
Jerry, there is an infinite number of cases where v1=v2. Are their
combined speeds in every case undefined?
Peter Riedt
Peter Riedt
> On May 23, 11:20 pm, Peter Riedt <rie...@yahoo.co.uk> wrote:
>
> > More speed confusion
>
> > The formula for the addition of velocities is given as (v1+v2)/
> > (1+v1*v2/c^2) #1. Why is it necessary to include the square of the
> > speed of LIGHT into the formula?
>
>
> << The main types of particle combinations used at RHIC
> are p + p, d + Au, Cu + Cu and Au + Au. The projectiles
> typically travel at a speed of 99.995% of the speed of
> light in vacuum. For Au + Au collision, the center-of-mass
> energy \sqrt{s_{NN}} is typically 200 GeV (or 100 GeV per nucleus); >>http://en.wikipedia.org/wiki/Relativistic_Heavy_Ion_Collider
>
> << Einstein's relativity principle states that:
>
> All inertial frames are totally equivalent
> for the performance of all physical experiments.>>
>
> http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
>
> Sue...
>
>
>
> > Well, the gurus who invented
> > speed that is possible and therefore any composite speed had to be
> > adjusted accordingly. This they achieved with the formula given above
> > but it is only a conjecture even though it is claimed that particle
> > physics prove it correct. Now anything can be proved by particle
> > physics especially that pigs fly. Particle physics is the ultimate
> > cash cow. The billions spent on it must be justified to prevent the
> > cash cow and her guardians from being subject to starvation or
> > poverty.
>
> > If the above formula is correct, it is only half the story. The
> > missing other half is the formula (v1-v2)/(1+v1*v2/c^2) #2. Let me
> > illustrate it by two scenarios.
>
> > In the first scenario, Glaucus chases the nymph Scylla. Glaucus runs
> > at 20km/hr (v1) and Scylla runs away at 12km/hr (v2). Their combined
> > speed (negative closing speed) is v1-v2 or 8km/hr. Using formula #2,
> > their combined speed is 7.999999968km/sec.
>
> > In the second scenario, Glaucus and Scylla (after changing her mind)
> > run towards each other to embrace at 20km/sec and 12km/sec
> > respectively. Their combined speed (positive closing speed) is v1+v2
> > or 32km/sec. Using formula #1, their combined speed is 31.99999997.
>
> > Formula #1 was obtained by Fizeau (1819-1896) in his experiments about
> > the propagation of light through flowing water in 1851. AE, citing
> > Fizeau, proposed this formula for the addition of velocities of all
> > physical objects (AE, ‘Relativity: The Special and General Theory’,
> > Chapter 13). He however missed formula #2 and was not aware that if an
> > object with a velocity of 300000km/sec chases an object running away
> > at 300000km/sec, their combined speed is -300000km/sec. As in particle
> > physics.
>
> > Peter Riedt- Hide quoted text - >
> - Show quoted text -
Sue, the relativistic heavy ion collider in Upton, New York cost by
2005 1.2 billion dollars.
Peter Riedt
Cephalobu...@comcast.net
No. Paul was very specifically referring to the case in your
originally posted scenario where Glaucus is a pulse of light
approaching at speed c while Scylla is a material object running
away from Glaucus, i.e.
v1 = c while v2 is any speed below c.
The situation is rather like the old Warner Brothers cartoons
with Pepe Le Pew the skunk taking the part of Glaucus, while
the black cat with a white stripe painted down her back
(usually anonymous, but sometimes identified as "Penelope" or
"La Belle") takes the part of Scylla. No matter how fast
Pelelope runs, Pepe Le Pew approaches inexorably at the same
speed.
Jerry
Dirk Van de moortel
KYiTl.98498$0V4....@newsfe25.ams2
Brilliant - as always:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SpreadsheetLimit.html
Dirk Vdm
G
"
So the only information about electromagnetism contained in the
apparently sophisticated equations (9) and (10) is about the two
constants in electromagnetism: the fixed velocity c, and that E, H at
every point are in fixed proportion Z0. The remaining content of
Maxwell’s Equations is hogwash.
"
Thanks for the link. Funilly enough, this is what I suspected, however
I cannot prove it.
Is the world really so messed up?
yua...@gmail.com
> Juan R. González-Álvarez wrote:
> I merely pointed out that light is not special in that sense, but rather
> the geometry of the manifold is special. Indeed, light is affected the
> same as any other object by this geometry; it "just happens" to travel
> with the speed of the geometrical symmetry (which permitted the
> historical confusion I pointed out, and the use of the same symbol for
> two VERY DIFFERENT theoretical quantities that have the same value).
It "just happens" is an interesting choice of phrase.
Do you think it's just a coincidence?
Love,
Jenny
Androcles
Peter Riedt
========================================
Tom&Jeery can't help constantly making a fool of himself,
fools never can.
Ref:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
What kind of lunacy prompted Einstein to say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
Not that Tom&Jeery will answer, of course. Even though he is reading
this, he will bluster and pretend to change bedpans thinking he can fool
others.
Tom Roberts
No. Of course not. That's why I put it in quotes.
As I said, light is affected the same as any other object by this
geometry. A light pulse has zero mass, and massless objects must move
with the symmetry speed of the manifold. That's why we use the same
symbol for these two very different concepts -- it can be confusing to
newbies and dilettantes, but not to people who understand the basics of
modern physics.
Tom Roberts
Androcles
"G" <gehan.am...@gmail.com> wrote in message
news:8090bb94-6528-477e...@21g2000vbk.googlegroups.com...
==========================================
Dunno about the world, engineers work around the problems.
Are the academics so fucked up? Yes they are, always have been.
Galileo was threatened with excommunication (which then was
to be socially ostracised as well) for supporting Copernicus's
view of the sun being at the centre of the Universe, and by the
very same bigots as Jeery and Tusseladd. Human nature doesn't
change in a few hundred years.
The only academics that are not fooled are the mathematicians,
the rest are opinionated idiots. Academics persuade politicians
who hold the purse strings, but LHC and ISS were built by engineers
for academics. LHC is actually a white elephant but it kept the Swiss
and the French busy for a while. The engineers don't care as long as
they get paid, the politician's don't care as long as they are in charge
and go down in history and the academics don't care as long as they
can run their mouth with their heads up their arses persuading each
other how clever they are.
Those that can, do. Those that can't, teach.
Newton was a mathematician, Galileo was a mathematician, Kepler
was a mathematician. Einstein was a failure.
http://en.wikipedia.org/wiki/How_many_angels_could_sit_on_a_needle
Dirk Van de moortel
> "G" <gehan.am...@gmail.com> wrote in message
> news:8090bb94-6528-477e...@21g2000vbk.googlegroups.com...
[snip snot]
> Is the world really so messed up?
>
> ==========================================
> Dunno about the world, engineers work around the problems.
> Are the academics so fucked up? Yes they are, always have been.
> Galileo was threatened with excommunication (which then was
> to be socially ostracised as well) for supporting Copernicus's
> view of the sun being at the centre of the Universe, and by the
> very same bigots as Jeery and Tusseladd. Human nature doesn't
> change in a few hundred years.
> The only academics that are not fooled are the mathematicians,
Androcles, the mathematician, and limits:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SpreadsheetLimit.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Limit.html
Androcles, the mathematician, and complex numbers:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroclesOperators.html
Androcles, the mathematician, and Boolean algebra:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XOROnceMore.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORrevisited.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORContinued.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORpersistence.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORWildStab.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LooksBoolean.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/XORforever.html
Androcles, the mathematician, and calculating 2*5:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/HumanBeing.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/MaliciousImbecile.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/StupidQuestion.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/TheProblem.html
Androcles, the mathematician, and differentials:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DiffConst.html
Androcles, the mathematician, and integrals:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Integral.html
Androcles, the mathematician, and geometry:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SimpleEnough.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FullyAware.html
Androcles, the mathematician, and transformations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroTransform.html
Androcles, the mathematician, and calculations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Percentages.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FALSE.html
Androcles, the mathematician, and groups:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroGroups.html
Androcles, the mathematician, and logs:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LogsHuh.html
Androcles, the mathematician, and vectors:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/IdiotVectors.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroVec.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/VectorLength.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/VectorSpaces.html
Androcles, the mathematician, and polar coordinates:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PolarManager.html
Androcles, the mathematician, and equations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GOGI-GIGO.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Doofus.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SetSolve2.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AndroDistri.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Pythagoras.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ToothlessBite.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Competent.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/UseTrans.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Sheesh.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DivZero.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Think.html
Androcles, the mathematician, and square roots:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GoodTeachers.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/TwoTurds.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/STILL.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CanSpecify.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Nearly.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Quadratic.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GrowUp.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Tautology.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Material.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GIVEN.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PythagoRescue.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Humour.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
Androcles, the mathematician, and partial differential equations:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff2.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff3.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff4.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NotFxy.html
Dirk Vdm
Peter Riedt
> On May 30, 1:01 am, Peter Riedt <rie...@yahoo.co.uk> wrote:
>
> > Jerry, there is an infinite number of cases where v1=v2. Are their
> > combined speeds in every case undefined?
>
> No.
Jerry, correct. In every case the combined speed is well defined. It
is the sum of their individual speeds i.e. the distance between them
shrinks by their combined speed.
Peter Riedt
G
> "G" <gehan.ameresek...@gmail.com> wrote in message
>
> news:8090bb94-6528-477e...@21g2000vbk.googlegroups.com...
> On May 29, 1:59 pm, "Androcles" <Headmas...@Hogwarts.physics> wrote:
>
>
>
> > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
>
> >news:8019d7ca-8f18-4104...@q14g2000vbn.googlegroups.com...
> > On May 28, 11:37 am, "Androcles" <Headmas...@Hogwarts.physics> wrote:
>
> > > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
>
> > >news:8a726271-3b2e-4921...@r31g2000prh.googlegroups.com...
> > > On May 28, 2:57 am, "Androcles" <Headmas...@Hogwarts.physics> wrote:
>
> > > > "Peter Riedt" <rie...@yahoo.co.uk> wrote in message
>
> > > >news:5c5ba431-1bc6-4b43...@j9g2000prh.googlegroups.com...
> > > > On May 27, 3:30 am, "Paul B. Andersen"
>
> > > > <paul.b.ander...@guesswhatuia.no> wrote:
> > > > > Peter Riedt wrote:
> > > > > > Androcles, excellent and comprehensive derivation resulting in the
> > > > > > answer 1=1.
>
> > > > > > To which I like to add, if we add thespeedof light to any other
> > > > > > we
> > > > > > subtract thespeedof an object from thespeedof light using the
> > > > > > formula(c-v)/(1+c*v/c^2)[negative closingspeedas in source and
> > > > > > target approaching each other] the result is less than c-v e.g.
> > > approaching each other at c (negative closingspeed). It means that if
> > > two particles at 300000km/sec each try to collide, they never can
>
> > > Peter Riedt
> > > ===============================================
> > > Absolute nonsense. I see cars coming toward me at -140 mph
> > > every day and they are only travelling at -70 mph with respect to
> > > the road.
> > > The car directly in front of me that is doing 70 mph w.r.t. the road
> > > isn't
> > > getting any further from me, it is doing 0 mph relative to me.
>
> > > How you can ever hope to understand Einstein's con when you
> > > Hide
> > > quoted text -
>
> > > - Show quoted text -
>
> > Androcles et al, this is the end to the confusion: If two particles
> > are in direct collision say at aspeedof 300000km/sec each, their
Is the LHC the supercollider that accelerates particles to .99 c in
one direction and .99 c in the other direction so they can slow down
enough to collide at .99c? Why bother ?
Androcles
===============================================
The LHC is the big relativistic machine that doesn't work.
http://www.androcles01.pwp.blueyonder.co.uk/MC2.htm
Newton: 2+2 = 4
Einstein: 2+ 1/sqrt[(1-0.865965^2/1^2)] ~= 4 within the limits
of experimental error.
"Everything should be as simple as possible, but not simpler."- Einstein.
koobee...@gmail.com
> As I said, light is affected the same as any other object by this
> geometry.
Your model of your so-called geometry minimizes on the accumulated
spacetime, proper time, or whatever you want to call it. Any particle
massless or not moves from one fixed point in spacetime to another
also fixed in spacetime has its path uniquely identified by the
minimum amount of spacetime (than any other paths) it accumulates in
doing so. That is great until you include the photons themselves.
The photons always exist with a spacetime of exactly zero. Thus, it
is absolutely impossible to identify a unique path of trajectory for a
photon because any path you can trace between these two fixed points
in spacetime exactly accumulates a spacetime of very exactly zero.
Your model of geodesics must call out for uniqueness in light.
<shrug>
> A light pulse has zero mass, and massless objects must move
> with the symmetry speed of the manifold.
This is not true. A particle with zero rest mass (or
indistinguishable from zero) can simply travel at any speed under the
speed of light. In doing so, it becomes very difficult to detect.
One example is the neutrino. <shrug>
> That's why we use the same
> symbol for these two very different concepts
Do you mean ‘c’? If so, so what?
> -- it can be confusing to
> newbies and dilettantes, but not to people who understand the basics of
> modern physics.
It looks like the geodesics are confusing the heck out of professors,
and that is no understatement. <shrug>
Eric Gisse
> On May 30, 7:10 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > As I said, light is affected the same as any other object by this
> > geometry.
>
> Your model of your so-called geometry minimizes on the accumulated
> spacetime, proper time, or whatever you want to call it.
Maximizes. Get it right.
[snip rest]
Cephalobu...@comcast.net
You are confusing relative speed with closing speed.
Jerry
Tom Roberts
> On May 30, 7:10 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> As I said, light is affected the same as any other object by this
>> geometry.
>
> Your model of your so-called geometry minimizes on the accumulated
> spacetime, proper time, or whatever you want to call it.
That is one way of defining geodesics, which as you point out does not
work for massless objects.
The better, more natural way to define geodesics is that they
parallel-transport their own tangent vector. This works for all
geodesics, and is expressed via the usual geodesic equation. This is
more natural, because it resides at a lower level of structure than the
metric -- one needs only a connection; if a metric is present (and in
physics there always is one) the two definitions are equivalent when
they are both well defined (as long as the connection is consistent with
the metric).
> Your model of geodesics must call out for uniqueness in light.
And it does. The geodesic equation applies equally well to timelike,
spacelike, or null paths. A geodesic is UNIQUELY specified by giving a
point and a tangent vector at that point; the character of the initial
tangent vector (timelike, spacelike, or null) determines the character
of the entire geodesic path. (Other types of boundary conditions can be
used to solve that differential equation.)
>> A light pulse has zero mass, and massless objects must move
>> with the symmetry speed of the manifold.
>
> This is not true.
Yes, it is true in both SR and GR.
> A particle with zero rest mass (or
> indistinguishable from zero) can simply travel at any speed under the
> speed of light.
Not true. In SR and GR a particle with zero mass must move at the
symmetry speed of the manifold, c. A particle with a small mass that is
indistinguishable from zero will move with a speed that is
indistinguishable from c.
> In doing so, it becomes very difficult to detect.
> One example is the neutrino.
Neutrinos are very difficult to detect because they interact only
weakly, not because they have such small masses. Proof: photons have
even less mass (zero) but are comparatively easy to detect -- because
they interact electromagnetically, not weakly. Electromagnetic
interactions are MUCH stronger than weak interactions at the energies
accessible to us.
>> That's why we use the same
>> symbol for these two very different concepts
>
> Do you mean �c�?
Yes, of course.
>> -- it can be confusing to
>> newbies and dilettantes, but not to people who understand the basics of
>> modern physics.
>
> It looks like the geodesics are confusing the heck out of professors,
> and that is no understatement.
No. As always, YOU are confused, because you refuse to study modern physics.
Tom Roberts
Steak
>
> Yes, of course.
>
> >> -- it can be confusing to
> >> newbies and dilettantes, but not to people who understand the basics of
> >> modern physics.
>
> > It looks like the geodesics are confusing the heck out of professors,
> > and that is no understatement.
>
> No. As always, YOU are confused, because you refuse to study modern physics.
>
> Tom Roberts
amazing to realise, that since i told you that todays
physics is all about modelling, your each 2nd sentence
is about models
and speaking the witch, you say you do so much modelling,
tell us what kinda models do you use, domains, boundary conditions,
numerical schemes, algorithms, constraints, error analysis,
validation, optimization, software and else
Tom Roberts
Actually, this depends on one's choice of metric signature and the
character of the geodesic. For the signature -+++, a spacelike geodesic
minimizes it and a timelike geodesic maximizes it; roles are reversed
for the signature +---. A null geodesic does neither, but is a
stationary point of the integral (this does not uniquely determine the
path).
One can say "extremizes" for all cases.
Tom Roberts
Steak
>
> Yes, of course.
>
> >> -- it can be confusing to
> >> newbies and dilettantes, but not to people who understand the basics of
> >> modern physics.
>
> > It looks like the geodesics are confusing the heck out of professors,
> > and that is no understatement.
>
> No. As always, YOU are confused, because you refuse to study modern physics.
>
> Tom Roberts
or maybe you let your students to do the models
please ask them
in order for me to learn relativity much more !
yua...@gmail.com
> Not true. In SR and GR a particle with zero mass must move at the
> symmetry speed of the manifold, c. A particle with a small mass that is
> indistinguishable from zero will move with a speed that is
> indistinguishable from c.
A particle with a small mass (indistinguishable from zero but not
zero) mving at high velocity (indistinguishable from c but not c) can
(in principle) be chased and caught. At that tme, its velocity
(relative to the chaser) will be zero (which is distinguishable from
c).
Consequently A particle with a small mass that is "indistinguishable
from zero" will be moving with a speed that is *distinguishable* from
c.
> >> -- it can be confusing to
> >> newbies and dilettantes, but not to people who understand the basics of
> >> modern physics.
That's apparently not true.
Love,
Jenny
G
"Thus Einstein's E = mc2 is actually derived from Newtonian
Mechanics."
If that is true, Scandal , I say , Scandal.
Could it be really that simple? Given the amount of energy stored
in Albert Einstein's reputation is greater than any scientist who ever
lived, one has to doubt. Nothing can approach the greatness
of Einstein, Androcles, you can come to within 99.99999% of Einstein
but you can never exceed Einstein.
Also, if you were calculating the energy of released by projectiles
arranged in a sphere around a spherical charge what is the total
energy? You would have to integrate mv^2 around the sphere, am I right?
yua...@gmail.com
> yuan...@gmail.com wrote:
> > It "just happens" is an interesting choice of phrase.
> > Do you think it's just a coincidence?
> No. Of course not. That's why I put it in quotes.
> As I said, light is affected the same as any other object by this
> geometry. A light pulse has zero mass, and massless objects must move
> with the symmetry speed of the manifold. That's why we use the same
> symbol for these two very different concepts -- it can be confusing to
> newbies and dilettantes, but not to people who understand the basics of
> modern physics.
So when you write 'it "just happens" ' you don't mean 'it just
happens' (i.e that it's accidental). You mean that it must happen.
Of course, I knew that - I had noticed the quote marks. But why not
write what you mean?
___________________
it "just happens" to travel
with the speed of the geometrical symmetry (which permitted the
historical confusion I pointed out, and the use of the same symbol for
two VERY DIFFERENT theoretical quantities that have the same value).
__________________
should surely have been:
________________
it necessarily travels with the speed of the geometrical symmetry
(which permitted the historical confusion I pointed out, and the use
of the same symbol for two INTIMATELY CONNECTED theoretical quantities
that have the same value).
________________
Love,
Jenny
Androcles
==========================================
For pity's sake, I PROVED it.
1/2 mv^2 + 1/2 mv^2 = mv^2
A bullet masses a gram, a gun masses a kilogram.
the bullet's speed is 1000 metres/sec, the gun's speed
(recoil) is 1 metre/sec.
That's Newton's third law.
mV = Mv
1000 * 1 = 1 * 1000.
Bullet's energy = 1/2 * 1 * 1000^2 = 500,000
Gun's energy = 1/2 * 1000 * 1^2 = 500
The bullet has 1000 times the energy of the gun and
travels 1000 times faster.
If the gun and bullet have the same mass then they share the
energy in the explosive charge (or the spring in the pop gun)
and get 1/2mv^2 each.
Of course it is that simple!
========================================
Given the amount of energy stored
in Albert Einstein's reputation is greater than any scientist who ever
lived, one has to doubt. Nothing can approach the greatness
of Einstein, Androcles, you can come to within 99.99999% of Einstein
but you can never exceed Einstein.
==========================================
Spare me your drooling hero worship, Einstein was nowhere near as
great as Newton and even Newton's first two laws belong to Galileo.
You've been reading the popular press again. Read Newton's Principia
Mathematica instead, it's been around for over 300 years.
http://members.tripod.com/~gravitee/
(Just hit "cancel" to continue)
Also read Galileo:
http://www.webexhibits.org/calendars/year-text-Galileo.html
==========================================
Also, if you were calculating the energy of released by projectiles
arranged in a sphere around a spherical charge what is the total
energy? You would have to integrate mv^2 around the sphere, am I right?
===========================================
The integral of the velocity is zero, half the masses go the other way.
But to answer your question, begin with one mass. Split it in two,
the halves fly apart, North and South.
1/2 mv^2 + 1/2mv^2 = mv^2
Divide it into four, the quarters fly apart N ,S, East, West.
1/4 mv^2 +1/4 mv^2 +1/4 mv^2 +1/4 mv^2 = mv^2
Divide it by 6, the parts fly N ,S, E, W, Up, Down.
1/6 mv^2 + 1/6 mv^2 + 1/6 mv^2 + 1/6 mv^2 + 1/6 mv^2 + 1/6 mv^2 =mv^2
mv^2 IS the integral.
There is a frame of reference in which the total momentum of the
particles of the Universe is zero, even if they all move relative to
that frame. That frame is Newton's absolute frame.
Androcles' third law:
For every photon there is an equal and opposite rephoton.
http://www.androcles01.pwp.blueyonder.co.uk/rephoton.gif
PD
Well, we'll find out in late September, won't we?
Tom Roberts
> On May 31, 7:32 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> Not true. In SR and GR a particle with zero mass must move at the
>> symmetry speed of the manifold, c. A particle with a small mass that is
>> indistinguishable from zero will move with a speed that is
>> indistinguishable from c.
>
> A particle with a small mass (indistinguishable from zero but not
> zero) mving at high velocity (indistinguishable from c but not c) can
> (in principle) be chased and caught. At that tme, its velocity
> (relative to the chaser) will be zero (which is distinguishable from
> c).
You ave just described a situation for a particle with a mass
DISTINGUISHABLE from zero. If its mass were indistinguishable from zero,
you would not be able to catch up to it -- doing so is merely one way of
distinguishing its mass from zero.
Tom Roberts
G
It isn't proven until the relativists accept it?
G
What is the difference?
Also, truth by calculation does not happen anymore? Computing power
is cheap these days.
doug
G wrote:
You could try studying to find out some physics.
Androcles
=====================================
Huh! By that argument you are not your mother's son unless the
fuckwits accept it! You do come across as completely stooopid
sometimes, what drugs are you taking?
koobee...@gmail.com
> koobee.wub...@gmail.com wrote:
> > Your model of your so-called geometry minimizes on the accumulated
> > spacetime, proper time, or whatever you want to call it.
>
> That is one way of defining geodesics, which as you point out does not
> work for massless objects.
There is actually no other way. Any geodesic models must work for
particles with mass >= 0. If not, it is just wrong. <shrug>
> The better, more natural way to define geodesics is that they
> parallel-transport their own tangent vector.
What makes you think your definition of geodesics is valid without any
mathematics backing it up?
> This works for all
> geodesics, and is expressed via the usual geodesic equation.
So, your model does not satisfy the variational principle, and that is
really, really bad. <shrug>
> This is
> more natural, because it resides at a lower level of structure than the
> metric
This is nonsense if you really have understood the fundamental issues
involved with the geodesics. The variational principle manifests the
parallel, vertical, side-ways, or whatever tangential transports.
<shrug>
> -- one needs only a connection;
You really do not understand the subject well. The variational
principle allows you to identify these connections. <shrug>
> if a metric is present (and in
> physics there always is one) the two definitions are equivalent when
> they are both well defined (as long as the connection is consistent with
> the metric).
The connections are always consistent with the metric under the
variational principle. <shrug> Please discuss physics with
mathematics instead of word salad.
> > Your model of geodesics must call out for uniqueness in light.
>
> And it does. The geodesic equation applies equally well to timelike,
> spacelike, or null paths. A geodesic is UNIQUELY specified by giving a
> point and a tangent vector at that point; the character of the initial
> tangent vector (timelike, spacelike, or null) determines the character
> of the entire geodesic path. (Other types of boundary conditions can be
> used to solve that differential equation.)
Your model does not work, and you are trying to weasel your way out of
it. What you are talking about makes no mathematical sense. <shrug>
> A light pulse has zero mass, and massless objects must move
> with the symmetry speed of the manifold.
>
> > This is not true.
>
> Yes, it is true in both SR and GR.
Well, I am not objecting to the first half of that sentence. It is
the second half that is false. <shrug>
> > A particle with zero rest mass (or
> > indistinguishable from zero) can simply travel at any speed under the
> > speed of light.
>
> Not true. In SR and GR a particle with zero mass must move at the
> symmetry speed of the manifold, c. A particle with a small mass that is
> indistinguishable from zero will move with a speed that is
> indistinguishable from c.
Massless object must moves at the speed of light if it has non-zero
energy or momentum what is a special case. Massless objects can very
well move under the speed of light and offer no momentum and
observable energy. This is all in the mathematics of GR and SR where
SR agrees with the mathematical foundations of the Voigt transform in
this aspect as well. <shrug>
> > In doing so, it becomes very difficult to detect.
> > One example is the neutrino.
>
> Neutrinos are very difficult to detect because they interact only
> weakly, not because they have such small masses. Proof: photons have
> even less mass (zero) but are comparatively easy to detect -- because
> they interact electromagnetically, not weakly. Electromagnetic
> interactions are MUCH stronger than weak interactions at the energies
> accessible to us.
You can kid kindergarten kids with that one. Photons are easy to
detect because it has a very high observed (relativistic) mass even
though the rest mass is zero. According to the Standard Model,
neutrinos are detectable when they are traveling near the speed of
light and with a lot of energy. Your definition of mass is preventing
you from understanding this. It is much easier to accept any mass as
an observed quantity even though “rest” mass can be zero. <shrug>
> >> That's why we use the same
> >> symbol for these two very different concepts
>
> > Do you mean ‘c’?
>
> Yes, of course.
Hmmm... I have already forgotten what these two very different
concepts. I hope it is not Alzheimer.
> > It looks like the geodesics are confusing the heck out of professors,
> > and that is no understatement.
>
> No. As always, YOU are confused, because you refuse to study modern physics.
I am ever more convinced that the geodesics are confusing the heck out
of professors, and that is no understatement.
As I have said, the geodesic model where the unique path follows the
one with the least accumulated spacetime fails miserably for light.
However, not affecting the result of the field equations under GR, the
geodesic model calling out for the unique path being the one with the
least amount of accumulated local or observed (coordinate) time allows
light to propagate with twice the deflection of Newtonian result.
You just cannot handwave to sow massed particles under one model and
massless particles under the other model together and call that blah,
blah, blah transports. Only Dr. Frankenstein is allowed to do that.
<shrug>
And once again, <CHECKMATE>
koobee...@gmail.com
> Eric Gisse wrote:
> > Maximizes. Get it right.
>
> Actually, this depends on one's choice of metric signature and the
> character of the geodesic.
Not really.
> For the signature -+++, a spacelike geodesic
> minimizes it and a timelike geodesic maximizes it; roles are reversed
> for the signature +---.
This is just not true. For the -+++ signature of spacetime, it is
nonsense because it allows objects to be indirectly identified to
travel beyond the speed of light by an observer. This is forbidden by
SR. For the +--- signature, geodesic model can call out for
minimizing the accumulated spacetime, local time (not to be confused
with proper time), or observed (coordinate) time. <shrug>
> A null geodesic does neither, but is a
> stationary point of the integral (this does not uniquely determine the
> path).
What are you talking about? The integral I take you mean is the
action of accumulating spacetime. It just does not exist because it
is always zero. So, being a stationary point of this integral does
not make any sense. <shrug>
There is no special law of physics regarding the null geodesics. Null
geodesics is a useless term to describe the geodesics of light.
Geodesics with minimum accumulated spacetime fails for light but not
the geodesics with minimum accumulated local or observed time.
<shrug>
> One can say "extremizes" for all cases.
Please don’t try to pull more mathemagic tricks out of the words
“extremize”. Study the calculus of variations thoroughly before you
come back for more. <shrug>
G. L. Bradford
ball-, or point-like particle observationally or measureably multi-sided or
multiplex, but is probably the simplest two-dimensional single-sided
membrane-like tissue only (strictly frontside photo-face only, thus strictly
2-dimensional!). It has no tangible, or even intangible, backside (which
would make it a 3-d object) to ever catch up to, much less surpass. Move
around from front to back to try to observe it, there is nothing there to
it, it does not exist.
GLB
============================
yua...@gmail.com
You were writing about a "small mass that is indistinguishable from
zero".
I was pointing out that, according to SR, there is no such animal.
A small mass cannot move at c, therefore it must move at zero relative
to something.
c is the only speed which is invariant.
You're getting confused.
Lotsa people here think that you understand what you're writing about.
Please be more precise in what you write.
Love,
Jenny
Tom Roberts
> You were writing about a "small mass that is indistinguishable from
> zero".
> I was pointing out that, according to SR, there is no such animal.
Nonsense. There is ALWAYS some measurement resolution. For instance, at
present the upper bound on the mass of a photon is 6*10^-17 eV/c^2 -- a
mass we generally consider to be zero. The upper bound on the masses of
neutrinos is 2 eV/c^2, and there is now a lower bound that is on the
order of a fraction of an eV/c^2 -- this used to be considered to be
zero, but now we know it isn't. These represent the experimental
resolutions of the best experiments that measure these masses.
> A small mass cannot move at c, therefore it must move at zero relative
> to something.
But that frame may well be completely unaccessible to human
experimenters. And even so, the object may be completely unobservable as
it moves (which is the case for both photons and neutrinos, but for
different reasons: the photon can only interact once, and the neutrino
interacts extremely weakly with a cross-section proportional to its energy).
This is about ACTUAL MEASUREMENTS, not abstract concepts and principles.
Tom Roberts
Paul B. Andersen
> On May 29, 7:13 pm, Cephalobus_alie...@comcast.net wrote:
>> On May 27, 4:17 pm, "Androcles" <Headmas...@Hogwarts.physics> wrote:
>>
>>
>>
>>
>>
>>> "Paul B. Andersen" <paul.b.ander...@guesswhatuia.no> wrote in messagenews:4A1D9434...@guesswhatuia.no...
>>>> 0/0 is undetermined, it's not zero
>>>> But lim[(c-v)/(1-c*v/c^2)] = c when v -> c
>>>> Note that the "speed addition formula" is a misnomer,
>>>> it is a speed tranformation formula.
>>>> If the speed of an object is u in an inertial frame K,
>>>> then this speed transforms to w in a frame K' which is
>>>> moving with the speed v relative to K, where
>>>> w = (u+v)/(1+u*v/c^2)
>>>> The 'object' can be light (or a photon) and thus u = c,
>>>> but the speed of frame K' relative to frame K cannot be c,
>>>> so both u and v can never both be c.
>>>> v can however be arbitrary close to c, so the speed c
>>>> will _always_ transform to c for all v.
>>>> --
>>>> Paul
>>> Bwahahahaha!
>>> So much fun to see you squirming with simple algebra, Tusseladd,
>>> and how you hope to take a limit!
>>> "lim[(c-v)/(1-c*v/c^2)] = c when v -> c ".. ahahahahahahaha!
>>> Pity a spreadsheet doesn't agree with you!
>> Good grief, Androcles!
>>
>> (c-v)/(1-c*v/c^2) = (c-v)/(1-v/c) = c*(c-v)/(c-v)
>>
>> The above equals c everywhere except at v = c where it is
>> undefined. So Paul's limit is easily seen to be correct without
>> using a spreadsheet.
>>
>> What in heck are you using for a "spreadsheet", Androcles?
>>
>> Why do you constantly insist of making a fool of yourself?
>>
>> Not that you will answer, of course. Even though you are reading
>> this, you are pretending to have "plonked" me, so you will seeth
>> in silence.
>>
>> Jerry- Hide quoted text -
>>
>> - Show quoted text -
>
> combined speeds in every case undefined?
Of course not.
(v1-v2)/(1-v1*v2/c^2) = 0 if v1 = v2 <> c
v1 = v2 = c is obviously a very special case.
--
Paul
yua...@gmail.com
> yuan...@gmail.com wrote:
> > You were writing about a "small mass that is indistinguishable from
> > zero".
> > I was pointing out that, according to SR, there is no such animal.
> Nonsense.
Nonsense back at you.
As I wrote:
_____________________
A particle with a small mass (indistinguishable from zero but not
zero) moving at high velocity (indistinguishable from c but not c) can
(in principle) be chased and caught. At that tme, its velocity
(relative to the chaser) will be zero (which is distinguishable from
c).
______________________
Note the use of the words "in principle".
In what way does what I wrote conflict with SR?
> But that frame may well be completely unaccessible to human
> experimenters. And even so, the object may be completely unobservable as
> it moves (which is the case for both photons and neutrinos, but for
> different reasons: the photon can only interact once, and the neutrino
> interacts extremely weakly with a cross-section proportional to its energy).
> This is about ACTUAL MEASUREMENTS, not abstract concepts and principles.
I was replying to what you wrote:
________________
A particle with a small mass that is
indistinguishable from zero will move with a speed that is
indistinguishable from c.
________________
What you wrote *does* conflict with SR, which is why I jumped in.
You were making a statement that is wrong *in principle*, as I
claimed.
You did *not* write:
________________
A particle with a small mass that is
indistinguishable from zero will move with a speed that is
very difficult to measure
________________
Please don't try to rewrite history.
How you measure these things was never the issue.
Love,
Jenny
Tom Roberts
> How you measure these things was never the issue.
When I wrote "indistinguishable", I meant experimentally. So how you
measure these things was always an issue, as it always is in physics.
Tom Roberts
yua...@gmail.com
> yuan...@gmail.com wrote:
Distinguishable:
to perceive a difference in : mentally separate
Perceive:
to attain awareness or understanding of
I can distinguish (perceive a world of a difference) between light
speed and not light speed in SR.
If you can't then you don't truly understand SR.
Light speed is invariant and not light speed is not invariant.
The're the 2 kinds of speed we have in SR.
Most physicists (not including you, apparently) consider the above
statements to be experimentally well substantiated.
Massless objects must move with the symmetry speed of the manifold
massive objects mustn't. That hasn't been experimentally confirmed for
infinitesimal masses, but is taken to be true in SR -- it can be
confusing to newbies and dilettantes, but not to people who understand
the basics of modern physics.
The title of this thread was well chosen.
Love,
Jenny
Peter Riedt
wrote:
>
> - Show quoted text -
Paul, correct, it is a special case. SR works only in special cases -
all those which are beyond our experimental facilities.
We should ask however what force in the universe makes objects on a
direct collision course slow down to obey the above formula? If A and
B converge, does A know it is approached by B and it must reduce its
speed and vice versa?
Peter Riedt
Androcles
Peter Riedt
===========================================
Good reply.
Except it really isn't that special, the LHC is designed specifically
for the case where v1 = -v2 = c.
Hydrogen gas has its electrons stripped, nuclei are accelerated
side-by-side so that their relative speed v1-v2 =0, split into two
paths and accelerated so that their closing speed is 2c and Tusseladd
is a babbling moron.
http://hands-on-cern.physto.se/ani/acc_lhc_atlas/lhc_atlas.swf
The thing is, Peter, when Einstein said
the "time" required by light to travel from A to B equals the "time" it
requires to travel from B to A
(from which the cuckoo malformations are derived)
it only applied to a special case, that being when A was at rest relative to
B.
So you are correct, SR only works in special cases, when v = 0.
Juan R.
> yua...@gmail.com wrote:
>> On May 26, 10:43 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>>> I merely pointed out that light is not special in that sense, but
>>> rather the geometry of the manifold is special. Indeed, light is
>>> affected the same as any other object by this geometry; it "just
>>> happens" to travel with the speed of the geometrical symmetry (which
>>> permitted the historical confusion I pointed out, and the use of the
>>> same symbol for two VERY DIFFERENT theoretical quantities that have
>>> the same value).
>>
>> It "just happens" is an interesting choice of phrase. Do you think it's
>> just a coincidence?
>
> No. Of course not. That's why I put it in quotes.
>
> As I said, light is affected the same as any other object by this
> geometry. A light pulse has zero mass, and massless objects must move
> with the symmetry speed of the manifold. That's why we use the same
> symbol for these two very different concepts -- it can be confusing to
> newbies and dilettantes, but not to people who understand the basics of
> modern physics.
The former class includes those stating that c "isn't really the speed of
light". Which curiously also includes those individuals do not
understanding that c is an universal constant.
--
http://www.canonicalscience.org/
Usenet Guidelines:
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Paul B. Andersen
> Paul, correct, it is a special case. SR works only in special cases -
> all those which are beyond our experimental facilities.
The 'speed addition formula' is confirmed by the Fizeau experiment.
It is also confirmed by a fibre optic gyro:
http://home.c2i.net/pb_andersen/pdf/fiber_optic_gyro.pdf
> We should ask however what force in the universe makes objects on a
> direct collision course slow down to obey the above formula? If A and
> B converge, does A know it is approached by B and it must reduce its
> speed and vice versa?
Nothing is "slowing down".
Speed is a frame dependent entity, that's all.
A-> v1 v2 <- B
If two objects A and B are approching each other with constant
speeds v1 and v2 in the 'screen-frame', then
the speed of B in A's rest frame is (v1+v2)/(1+v1*v2/c^2) and
the speed of A in B's rest frame is (v1+v2)/(1+v1*v2/c^2).
All the speeds are constant.
--
Paul
PD
>
> > - Show quoted text -
>
> Paul, correct, it is a special case. SR works only in special cases -
> all those which are beyond our experimental facilities.
> We should ask however what force in the universe makes objects on a
> direct collision course slow down to obey the above formula?
They don't slow down, Peter. This is the velocity they have.
You have the idea somehow that they start out at (v1 + v2) and then
slow down (v1 + v2)/(1 + v1*v2). They never get to (v1 + v2).
Androcles
"Paul B. Andersen" <paul.b....@somewhere.no> wrote in message
news:4A2503A7...@somewhere.no...
Proofs 12 and 16 are countered by proof 18.
Wackypedia
http://en.wikipedia.org/wiki/Mathematical_proof
lists:
1 Direct proof
2 Proof by induction
3 Proof by transposition
4 Proof by contradiction
5 Proof by construction
6 Proof by exhaustion
7 Probabilistic proof
8 Combinatorial proof
9 Nonconstructive proof
10 Elementary proof
Not included:
11 Proof by "everybody knows" (proof by popular opinion).
12 Proof by "because I say so" (proof by assertion).
13 Proof by "it is written" (proof by appeal to authority).
14 Proof by "you prove it isn't!" (proof by simple denial).
15 Proof by "what about the tooth fairy?"(proof by irrelevance).
16 Proof by "I'm smarter than you, so there!" (proof by bluster).
17 Proof by "read a text book" (proof by bluster revision 2).
and the ultimate counter proof:
18 Proof by "You're'n'asshole!" (proof by ad hominem attack).
Proof 18 is my favorite, I use it often. It is very effective when used
against proofs 11-17. Fight fire with fire, I say. Proofs 1-10 have me
defeated, they prevent me from using proofs 11-17 and I have to bite
the bullet and embarrass myself to win the argument (which I must do
at all costs upon pain of death by diarrhea of the verbal kind).
Peter Riedt
>
> > > - Show quoted text -
>
> > Paul, correct, it is a special case. SR works only in special cases -
> > all those which are beyond our experimental facilities.
> > We should ask however what force in the universe makes objects on a
> > direct collision course slow down to obey the above formula?
>
> They don't slow down, Peter. This is the velocity they have.
> You have the idea somehow that they start out at (v1 + v2) and then
> slow down (v1 + v2)/(1 + v1*v2). They never get to (v1 + v2).
>
PD, correct. A's speed remains at v1 and B's speed remains at v2. Only
the distance between them shrinks at v1+v2, not (v1 + v2)/(1 + v1*v2).
Peter Riedt
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>
> > If A and
> > B converge, does A know it is approached by B and it must reduce its
> > speed and vice versa?
>
> > Peter Riedt- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
G
>
> It isn't proven until the relativists accept it?
> =====================================
> Huh! By that argument you are not your mother's son unless the
> sometimes, what drugs are you taking?
What I mean to say, Androcles, is how does one convince a realtivist?
Is there a method?
G
>
> > > > - Show quoted text -
>
> > > Paul, correct, it is a special case. SR works only in special cases -
> > > all those which are beyond our experimental facilities.
> > > We should ask however what force in the universe makes objects on a
> > > direct collision course slow down to obey the above formula?
>
> > They don't slow down, Peter. This is the velocity they have.
> > You have the idea somehow that they start out at (v1 + v2) and then
> > slow down (v1 + v2)/(1 + v1*v2). They never get to (v1 + v2).
>
> +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
> PD, correct. A's speed remains at v1 and B's speed remains at v2. Only
> the distance between them shrinks at v1+v2, not (v1 + v2)/(1 + v1*v2).
>
> Peter Riedt
> +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>
>
>
> > > If A and
> > > B converge, does A know it is approached by B and it must reduce its
> > > speed and vice versa?
>
> > > Peter Riedt- Hide quoted text -
>
> > - Show quoted text -- Hide quoted text -
>
> > - Show quoted text -
Peter, does it mean that the scientist observer watching this in the
control room will
see the particles approach each other at v1+v2?
Are you assuming that all information transfer is limited by the speed
of light?
Also, the particles must be experiencing a heck of a Doppler shift
near the speed of light.
Message Marker #1
G
I have been trying to understand physics, however right at the
beginning,
trying to understand SRT is a problem because what I am being taught
is logically inconsistent
Androcles
The only method is a 2"x 4" laid carefully alongside the ear with a force
of few hundred newtons and see which breaks first, solid bone or pine.
You'd have more fun converting the Pope to Islam. Why bother?
There are lots of interesting and as yet unexplained phenomena in
Nature without troubling the crank Einstein's disciples. Of course if
all you want is to argue with lunatics then waste your time, you'll never
get a conversion. They lack the logic see sense.
Uncle Ben
Jerry, it has been shown here many times that Androcles doesn't know
what a limit is.
Uncle Ben
Uncle Ben
>
> - Show quoted text -
G: Forget about "seeing". That will just confuse you. What it means
if the closing speed of the particles is about 2c, and if they are 10
feet apart, they will collide in about 5 nanoseconds, since c = 1 ft/
nanosecond, approximately. (1 ft = about 0.3 meter.)
And forget about Doppler shift. Particles don't act like waves in
this sense. Photons, yes. Protons, no.
Androcles
"Uncle Ben" <b...@greenba.com> wrote in message
news:6673def1-9e93-4b36...@r13g2000vbr.googlegroups.com...
========================================
Bwahahahahaha!
Who mentioned waves, drunken unk?
What it means is the particle stream will arrive at the opposing particle
with a frequency that is double their frequency when passing a magnet,
and that's a mere 11 kHz in the LHC.
Forget about seeing, you are as blind as a bat, you stooopid old fart.
Paul B. Andersen
So you finally got it.
In the frame of reference where A's speed is v1 and B's speed is v2,
A's speed remains at v1 and B's speed remains at v2. Only
the distance between them shrinks at v1+v2, not (v1 + v2)/(1 + v1*v2).
But you forgot to add:
In A's rest frame, B's speed remains at (v1 + v2)/(1 + v1*v2).
In B's rest frame, A's speed remains at (v1 + v2)/(1 + v1*v2).
All the speeds are constant.
--
Paul
Paul B. Andersen
Quite.
Glenlivet consume -> infinite when Androcles -> bedtime
No limit!
--
Paul
G
wrote:
Is that B's speed or closing speed?
Is that A's speed or closing speed?
Uncle Ben
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
If you can't tell, why not actually read
a textbook on relativity? What are you doing here
anyway without any knowledge of the subject?
Uncle Ben