Relativity COTD, thursday
josX
\ | /
\|/
--*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
/|\ |
/ | \ | -a- a--->
-b- |
|
Relativity says that light's speed is the same in every reference-frame,
however when we attach a reference-frame to a moving lightfront, it's
speed can obviously not be lightspeed, but should instead be 0.
This simple problem is unsolvable in modern science. What is the speed
of a lightbeam from a reference-point that moves with the light? It cannot
be answered in relativity:
The law for relativity speed addition is:
V' = (V + v0)/(1 + V * v0/c^2)
Where
v0 is the speed of the secondary reference frame relative to 'us'
V is the speed of the object in question relative to the secondary
reference-frame
V' is the speed of the object relative to 'us'.
When v0 = c, and V'= c as well, what is V ?
V' = (V + v0)/(1 + V * v0/c^2)
c = (V + c )/(1 + V * c /c^2)
c = V + c
c * (1 + V * /c ) = V + c
c + V * c/c = V + c
c + V = V + c
c + V = V + c
This is always true for whatever value of V, V can be anything.
The answer we were looking for ofcourse is 0, easily arrived at by the
Galilean speed addition formula:
V' = V + v0
c = V + c
c-c= V
0= V
--
jos
Robert Kolker
josX wrote:
> Reference-point contradiction in relativity:
>
> \ | /
> \|/
> --*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> /|\ |
> / | \ | -a- a--->
> -b- |
> |
>
> Relativity says that light's speed is the same in every reference-frame,
> however when we attach a reference-frame to a moving lightfront, it's
> speed can obviously not be lightspeed, but should instead be 0.
A reference frame cannot be so attached. It is physically meaningless to
do so.
Bob Kolker
Eric Gisse
>Reference-point contradiction in relativity:
>
>\ | /
> \|/
>--*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> /|\ |
>/ | \ | -a- a--->
> -b- |
> |
>
>Relativity says that light's speed is the same in every reference-frame,
>however when we attach a reference-frame to a moving lightfront, it's
>speed can obviously not be lightspeed, but should instead be 0.
How do you give a photon a reference frame when it has no mass?
>This simple problem is unsolvable in modern science. What is the speed
>of a lightbeam from a reference-point that moves with the light? It cannot
>be answered in relativity:
No shit.
"The laws of physics are the same in any inertial frame, regardless of
position or velocity".
Do you know what inertia is?
Does a photon have inertia?
In my Introduction to Philosophy class, it is explained to me that in
order to truly argue against something, you have to understand it
first. Be it religion....or a scientific theory.
You demonstrate again and again that you DO NOT UNDERSTAND.
Mitchell Jones
***{Forget the fact that he used the word "attached." His point is that in
the reference frame of the moving photon, its speed is obviously zero.
Thus, unarguably, it is false to claim that the speed of light is the same
in all reference frames. Hence Einstein's claim that the speed of light is
the same in all reference frames is obviously wrong. Do you have a
response to the actual point being made, or not? --MJ}***
===============================================
Killfile inmates: Charles Cagle, Stephen Speicher, Mati Meron.
Mitchell Jones
<kseggR...@uas.alaska.edu> wrote:
> On 17 Oct 2002 03:29:27 GMT, jo...@mraha.kitenet.net (josX) wrote:
>
> >Reference-point contradiction in relativity:
> >
> >\ | /
> > \|/
> >--*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> > /|\ |
> >/ | \ | -a- a--->
> > -b- |
> > |
> >
> >Relativity says that light's speed is the same in every reference-frame,
> >however when we attach a reference-frame to a moving lightfront, it's
> >speed can obviously not be lightspeed, but should instead be 0.
>
> How do you give a photon a reference frame when it has no mass?
***{Your argument is circular. It is a precept of the Einstein theory,
which he is denying and you are defending, that the photon has no mass.
Hence, in effect, you are saying that the Einstein theory can't be wrong
about the speed of light, because that would mean it is also wrong about
the mass of the photon. Well, since you are the proponent of the Einstein
theory, that's your problem, not his.
In addition to your argument being circular, it is based on a premise that
is not merely false, but ludicrous: there is no basis whatsoever for your
assumption that the origin of a coordinate system must have a mass-bearing
particle at its center. The center of mass of a binary star, for example,
is typically out in space somewhere between the two stars, rather than
within one of them, and yet, despite that, astronomers routinely analyze
the celestial mechanics of such systems by placing the origin of their
coordinates at the center of mass of the system.
--Mitchell Jones}***
[snarling and gnashing of teeth deleted]
m4r...@xs4a11.nl
> Reference-point contradiction in relativity:
>
> \ | /
> \|/
> --*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> /|\ |
> / | \ | -a- a--->
> -b- |
> |
>
> Relativity says that light's speed is the same in every reference-frame,
> however when we attach a reference-frame to a moving lightfront, it's
> speed can obviously not be lightspeed, but should instead be 0.
Relativity says you cannot attach a frame of reference to a photon,
so how does your going OUTSIDE of relativity prove that there is a
contradiction WITHIN relativity? Answer: it doesn't, you're just being
a moron, as always: http://www.xs4all.nl/~marcone/josboersema.html
(if people would simply reply to josX's postings by posting the
above URL, he would be gone within a week...)
Mitchell Jones
> josX wrote:
> > Reference-point contradiction in relativity:
> >
> > \ | /
> > \|/
> > --*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> > /|\ |
> > / | \ | -a- a--->
> > -b- |
> > |
> >
> > Relativity says that light's speed is the same in every reference-frame,
> > however when we attach a reference-frame to a moving lightfront, it's
> > speed can obviously not be lightspeed, but should instead be 0.
***{It would be amusing to let the erroneous, arrogant, and vituperative
replies to your fascinating argument go on and on, but, to be fair, many
relativists do not word things the way you do, above. Instead, some state
that the speed of light will be measured to be the same by all observers,
regardless of their reference frames. Ultimately, therefore, someone is
going to point out that an "observer" can't be pushed up to lightspeed to
move along with the photon, and, once someone says that, your argument
will be rejected by all of the Einsteinian true believers, despite the
fact that, before they encountered your reasoning, many of them were
guilty of the precisely the fallacious thinking that you have exposed!
None will admit that, however. Instead, all will claim to have known it
all along, and will decry your ignorance and shower you with imprecations,
as usual. (It's real sad! :-) --MJ}***
Dirk Van de moortel
"Mitchell Jones" <mjo...@jump.net> wrote in message news:mjones-1710...@66-105-229-12-aus-02.cvx.algx.net...
I don't think that many people, after having seen, understood and
used the factor gamma = 1/sqrt(1-v^2), will get it in their head to
put v=1 and continue calculating, let alone declare that the theory
must be wrong, knowing well that this is outside of the theory by
design.
Dirk Vdm
Aardvark
Yes, and he made it. The concept of a "reference frame of the moving
photon" is physically meaningless.
Relativity is mathematically self-consistent, given its initial
postulates. Setting up a gedanken-experiment that is at variance with
the postulates is not a test of the internal-consistency of the
theory. Is this too difficult for you?
The fact that 3^2 = 9 does not invalidate the statement "x^2 = 4, if x
= 2".
Tris
josX's argument is outside the bounds of SR. He is once again attempting to
prove there is a self-contradiction by _not_ using SR. Every argument of
his is a straw man. Are you saying SR is internally inconsistent? I'm
certain everyone would be interested in a detailed answer in the
affirmative.
The "Einstein theory" _could_ be wrong about the speed of light and the mass
of the photon. If someone shows a photon has mass, or the that speed of
light is _not_ constant, then they have _empirically_ refuted it.
Empiricism is the only way, AFAIK, to prove SR is wrong.
I wish anti-SRists would actually do some experiments instead of constantly
spewing bullshit.
Dirk Van de moortel
"Mitchell Jones" <mjo...@jump.net> wrote in message news:mjones-1710...@66-105-229-19-aus-02.cvx.algx.net...
> In article <3DAE2FCC...@attbi.com>, bobk...@attbi.com wrote:
>
> > josX wrote:
> > > Reference-point contradiction in relativity:
> > >
> > > \ | /
> > > \|/
> > > --*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> > > /|\ |
> > > / | \ | -a- a--->
> > > -b- |
> > > |
> > >
> > > Relativity says that light's speed is the same in every reference-frame,
> > > however when we attach a reference-frame to a moving lightfront, it's
> > > speed can obviously not be lightspeed, but should instead be 0.
> >
> > A reference frame cannot be so attached. It is physically meaningless to
> > do so.
> >
> > Bob Kolker
>
> ***{Forget the fact that he used the word "attached." His point is that in
> the reference frame of the moving photon, its speed is obviously zero.
Bob's point is that according to the theory, the photon cannot
have a reference frame to begin with, so there is no way to
define, let alone measure "speed wrt a photon". Saying that
its own speed is obviously zero, does not make sense.
> Thus, unarguably, it is false to claim that the speed of light is the same
> in all reference frames. Hence Einstein's claim that the speed of light is
> the same in all reference frames is obviously wrong.
It is not wrong if the claim explicitly excludes light from having
a reference frame.
The claim can be labelled wrong if an experiment is carried
out with an observer sitting on a photon.
It can also be labelled wrong if it induces internal inconsistencies.
> Do you have a
> response to the actual point being made, or not? --MJ}***
The response has been made hundreds of times.
The actual point being made by Boersema, is void.
Perhaps you don't, but by now, we assume that Boersema is
perfectly aware of that.
Dirk Vdm
Dirk Van de moortel
"Mitchell Jones" <mjo...@jump.net> wrote in message news:mjones-1710...@66-105-229-26-aus-02.cvx.algx.net...
The difference is that in this example the center of mass of
the binary star is not moving at light speed wrt the observed
phenomena, so according to the theory, and consistently with
the used equations, one can easily put a physical observer at
that place, whereas one can neither put a photon there, nor
can one put a physical observer on a photon, which is what
Boersema is using to prove the theory wrong.
So the no-mass-argument being circular indeed, this
mass-center-example does not refute it and should i.m.o.
not be used as a counter-example. I think it could turn
Boersema even more confused than he already pretends to
be.
Dirk Vdm
BernardZ
mjo...@jump.net says...
> Ultimately, therefore, someone is
> going to point out that an "observer" can't be pushed up to lightspeed to
> move along with the photon
>
But in physics we often put as a though experiment an observer in places
that are physically impossible!
In this particular case, I remember a similar though experiment where
Einstein did postulate a thought experiment of him sitting on a photon.
Now what would he see?
In this case, I would say that an observer travelling at light speed
would experience no time and space. The question as such has no meaning.
josX
Thanks. The claim is for an observer, but also, and first and foremost,
for a /reference-frame/. I have one more contradiction of this level, it
is quite ... /bad/, coming in 2 days :).
What the relativists have done is make their theory bullet-proof by
outlawing everything that can lead to it's downfall, FTL travel and the
attaching of the above reference-frames, possibly with this batch of
contradictions more things will be outlawed, perhaps computing something
from another reference-frame then your own. We'll see, it is good to have
them on the run, like they should with such a theory.
--
jos
Spaceman
>A reference frame cannot be so attached. It is physically meaningless to
>do so.
<ROFLOL>
Sorry Bob,
you lose bigtime there
a camera on front of a train that is in motion
is a reference frame that IS attached.
you are lost!
James M Driscoll Jr
Spaceman
http://www.realspaceman.com
Spaceman
>Yes, and he made it. The concept of a "reference frame of the moving
>photon" is physically meaningless.
Only because the "photon" is physically meaningless
when you say it has 0 mass.
otherwise.
JosX has facts and you have religion.
>Relativity is mathematically self-consistent, given its initial
>postulates.
Parrots love saying such.
but.
It is only parrot speak.
and ignorance of "basic math"
and religion of a bad religion.
How many Earth revs old are the paradox twins WRT the sun.
refuse to do that math and compare it huh?
The math that kills relativity.
<LOL>
Spaceman
>In this case, I would say that an observer travelling at light speed
>would experience no time and space. The question as such has no meaning.
Your "observations are not worth anything but illusion"
to the traveling photon rider.
your observations would be limited to lightspeed.
so therefore the lights speed is what is causing
an effect of you seeing "stop motion" (time stopping)
when in reality.
there is no stoppage of time or space
from motion.
No speed will stop time.
Not one.
even
a billion miles every 1/4 second
will not stop time.
It will only "play with lights illusion of observation"
Spaceman
>How do you give a photon a reference frame when it has no mass?
How do you give it a momentum when it has no mass?
what was in momentum?
pure energy?
<LOL>
Dang Fool!
>"The laws of physics are the same in any inertial frame, regardless of
>position or velocity".
Yes
the LAWS,
not the theories.
sheesh!
Eric,
Do you know how clocks work?
I "observe" that you do not.
Is that a fact because I have "observed such"?
Spaceman
Hi Mitchell,
>***{Forget the fact that he used the word "attached." His point is that in
>the reference frame of the moving photon, its speed is obviously zero.
>Thus, unarguably, it is false to claim that the speed of light is the same
>in all reference frames. Hence Einstein's claim that the speed of light is
>the same in all reference frames is obviously wrong. Do you have a
>response to the actual point being made, or not? --MJ}***
Bob will never admit such a fact.
He has this "religion" where he refuses to accept any
outside books nor facts to change his faith.
:)
Robert Kolker
Mitchell Jones wrote:
> ***{Forget the fact that he used the word "attached." His point is that in
> the reference frame of the moving photon, its speed is obviously zero.
> Thus, unarguably, it is false to claim that the speed of light is the same
> in all reference frames. Hence Einstein's claim that the speed of light is
> the same in all reference frames is obviously wrong. Do you have a
> response to the actual point being made, or not? --MJ}***
How do you make measuements with massive devices which are moving at c
wrt to the origin of an intertial frame. This is physically impossible.
Why not ask what is North of the North Pole.
Bob Kolker
Robert Kolker
BernardZ wrote:
> But in physics we often put as a though experiment an observer in places
> that are physically impossible!
Physically infeasible. But an observer at light speed is -impossible-
because it means one has to divide by 0.
Bob Kolker
Spaceman
>***{Your argument is circular. It is a precept of the Einstein theory,
>which he is denying and you are defending, that the photon has no mass.
>Hence, in effect, you are saying that the Einstein theory can't be wrong
>about the speed of light, because that would mean it is also wrong about
>the mass of the photon. Well, since you are the proponent of the Einstein
>theory, that's your problem, not his.
YES!
and since the photon "can have mass"
They are actually "supporting easter bunny theories"
:)
and the "lack of cause eggs" are there proof.
magical "nothingness" in motion.
:)
It is sad.
but..
now that the CAT is out of the box
and "we know the cat is outside now"
It is funny.
Spaceman
>josX's argument is outside the bounds of SR.
Proof it does not include all "physics" then..
for if there is an outside SR tha you can explain at all.
it means SR is missing "facts" and too circular in logic.
dump it before it makes you smell even worse.
Time changing as a cause huh?
>ROFLOL>
Spaceman
>Relativity says you cannot attach a frame of reference to a photon,
Then relativity is full of photon massless crap.
and is also "frame bound"
The universe is not "frame bound"
Relativity is bologna circular logic at best..
Time changing as a cause huh?
Still funny!
Dirk Van de moortel
"Robert Kolker" <bobk...@attbi.com> wrote in message news:3DAEAC6B...@attbi.com...
Bob, the fact that one has to divide by 0 is not the reason *why*
it is impossible. For various reasons we postulate a theory that
assumes that it is impossible, *and* the theory uses equations
that are consistent with this impossibility.
Dirk Vdm
Spaceman
>Physically infeasible. But an observer at light speed is -impossible-
>because it means one has to divide by 0.
BULLSHIT!
and only with your bad math will it have anything to do with
deviding by 0.
Spaceman
>How do you make measuements with massive devices which are moving at c
>wrt to the origin of an intertial frame. This is physically impossible.
>Why not ask what is North of the North Pole.
If you have created an actual north.
the things that are north of such are "floating above the planet"
North,
does not end just because a planets size stops.
If you are 240 miles above the North Pole and 100 miles east..
Do you head "down" to head further north?
<LOL>
Dirk Van de moortel
"Spaceman" <agents...@aol.combination> wrote in message news:20021017085356...@mb-mv.aol.com...
> >From: Robert Kolker bobk...@attbi.com
>
> >How do you make measuements with massive devices which are moving at c
> >wrt to the origin of an intertial frame. This is physically impossible.
> >Why not ask what is North of the North Pole.
>
> If you have created an actual north.
> the things that are north of such are "floating above the planet"
Good dog!
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html#FloatingNorth
Title: "North of the North Pole is "floating above the planet""
Dirk Vdm
Robert Kolker
Dirk Van de moortel wrote:
> Bob, the fact that one has to divide by 0 is not the reason *why*
> it is impossible. For various reasons we postulate a theory that
> assumes that it is impossible, *and* the theory uses equations
> that are consistent with this impossibility.
Point taken.
Bob Kolker
Thomas Jones
Where in relativity does it say you can't attach a frame of reference to a
photon?
Dirk Van de moortel
"Robert Kolker" <bobk...@attbi.com> wrote in message news:3DAEBBA...@attbi.com...
Gladly offered :-)
Cheers,
Dirk Vdm
Spaceman
>Good dog!
So dirk.
The ISS .
to head north..
would have to fall huh?
<LOL>
you are a sad ass lying twisting puke.
and so far.
observation is proving it with each post you make.
AllYou!
"josX" <jo...@mraha.kitenet.net> wrote in message
news:aolaqn$83b$1...@news1.xs4all.nl...
> Reference-point contradiction in relativity:
>
> \ | /
> \|/
> --*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> /|\ |
> / | \ | -a- a--->
> -b- |
> |
>
> Relativity says that light's speed is the same in every reference-frame,
> however when we attach a reference-frame to a moving lightfront, it's
> speed can obviously not be lightspeed, but should instead be 0.
SR prdicts that the above is impossible. Therefore, it cannot be a
contradiction.
me...@cars3.uchicago.edu
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
AllYou!
North of the North Pole. I like that one. It's actually a great way to sum
it up. Can I use it?
Spaceman
>North of the North Pole. I like that one. It's actually a great way to sum
>it up. Can I use it?
And yet another that could not foy near the north pole
for if you tell him to head north,
He will crash the plane instead of actually "fly more north"
:)
Do you think the north magnetic feild ends at the Earth surface?
<LOL>
Jon Hurwitz
Santa's Grotto, of course. Perhaps he knows what it's like to go faster
than light then. I wonder if that's how he delivers all the presents in one
evening.
Jon
Robert Kolker
AllYou! wrote:
>
> North of the North Pole. I like that one. It's actually a great way to sum
> it up. Can I use it?
>
By all means, and in good health too.
Bob Kolker
Franz Heymann
"Mitchell Jones" <mjo...@jump.net> wrote in message
news:mjones-1710...@66-105-229-26-aus-02.cvx.algx.net...
> In article <sggsqusm6aehsea54...@4ax.com>, Eric Gisse
> <kseggR...@uas.alaska.edu> wrote:
>
> > On 17 Oct 2002 03:29:27 GMT, jo...@mraha.kitenet.net (josX) wrote:
> >
> > >Reference-point contradiction in relativity:
> > >
> > >\ | /
> > > \|/
> > >--*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> > > /|\ |
> > >/ | \ | -a- a--->
> > > -b- |
> > > |
> > >
> > >Relativity says that light's speed is the same in every
reference-frame,
> > >however when we attach a reference-frame to a moving lightfront,
it's
> > >speed can obviously not be lightspeed, but should instead be 0.
> >
>
Mitchell Jones is incapable of distinguishing between the concepts of
internal consistency and circularity.
If his powers of ratiocination are as poor as that, one may safely
ignore anything else he has to say, so here goes the snippage.
[...]
I would be glad if you would confirm which of the following are the
reason for the inclusion of Stephen Speicher and Mati Meron in your
killfike list:
(1) You are not mentally equipped to understand their arguments.
or
(2) They beat you hollow in one or more discussions, and you do not
have the intellectual integrity to admit it.
Franz Heymann
-----------------------------------------------------------------
> Killfile inmates: Charles Cagle, Stephen Speicher, Mati Meron.
Franz Heymann
> In article <3DAE2FCC...@attbi.com>, bobk...@attbi.com wrote:
>
> > josX wrote:
> > > Reference-point contradiction in relativity:
> > >
> > > \ | /
> > > \|/
> > > --*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> > > /|\ |
> > > / | \ | -a- a--->
> > > -b- |
> > > |
> > >
> > > Relativity says that light's speed is the same in every
reference-frame,
> > > however when we attach a reference-frame to a moving lightfront,
it's
> > > speed can obviously not be lightspeed, but should instead be 0.
> >
meaningless to
> > do so.
> >
>
> ***{Forget the fact that he used the word "attached." His point is
that in
> the reference frame of the moving photon, its speed is obviously
zero.
It is not possible to transform to some putativr rest frame of a photon,
unless you are prepared to give a unique value to the quantity 0/0.
> Thus, unarguably, it is false to claim that the speed of light is the
same
> in all reference frames. Hence Einstein's claim that the speed of
light is
> the same in all reference frames is obviously wrong. Do you have a
> response to the actual point being made, or not? --MJ}***
The contents of your killfile defines you as being bereft of powers of
discrimination..
Franz Heymann
>
> ===============================================
Eric Gisse
wrote:
>In article <sggsqusm6aehsea54...@4ax.com>, Eric Gisse
><kseggR...@uas.alaska.edu> wrote:
>
>> On 17 Oct 2002 03:29:27 GMT, jo...@mraha.kitenet.net (josX) wrote:
>>
>> >Reference-point contradiction in relativity:
>> >
>> >\ | /
>> > \|/
>> >--*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
>> > /|\ |
>> >/ | \ | -a- a--->
>> > -b- |
>> > |
>> >
>> >Relativity says that light's speed is the same in every reference-frame,
>> >however when we attach a reference-frame to a moving lightfront, it's
>> >speed can obviously not be lightspeed, but should instead be 0.
>>
>> How do you give a photon a reference frame when it has no mass?
>
>***{Your argument is circular. It is a precept of the Einstein theory,
>which he is denying and you are defending, that the photon has no mass.
>Hence, in effect, you are saying that the Einstein theory can't be wrong
>about the speed of light, because that would mean it is also wrong about
>the mass of the photon. Well, since you are the proponent of the Einstein
>theory, that's your problem, not his.
Jos doesn't like the theory of SR.
In order to prove SR wrong, you have to find either an internal
contradiction or empericial evidence countering it.
Jos is trying to find an internal contradiction, but in reality he is
retarded. In SR a photon has zero mass and travels at c, and only
*inertial* bodies can have reference frames. I am saying he is full of
shit, in other words.
Im not really a proponent because Im not aware of any alternatives
that are viable. Sortof like how the people of Iraq "re-elect" Saddam
when he is the only one on the ballot.
>In addition to your argument being circular, it is based on a premise that
>is not merely false, but ludicrous: there is no basis whatsoever for your
>assumption that the origin of a coordinate system must have a mass-bearing
>particle at its center.
<snipo>
Hey, we are talking about SR.
Is a photon a valid interial reference frame?
Why do people get confused about inertia...its not that hard.
>
>--Mitchell Jones}***
>
>[snarling and gnashing of teeth deleted]
Wraaa!
Eric Gisse
wrote:
>>From: Eric Gisse kseggR...@uas.alaska.edu
>
>>How do you give a photon a reference frame when it has no mass?
>
>How do you give it a momentum when it has no mass?
>
>what was in momentum?
>pure energy?
><LOL>
>Dang Fool!
>
>
E=hf
Solar sails, comet tails, Orion. Physics laughs at you, and so do I.
>>"The laws of physics are the same in any inertial frame, regardless of
>>position or velocity".
>
>Yes
>the LAWS,
>not the theories.
>sheesh!
>
>Eric,
>Do you know how clocks work?
>I "observe" that you do not.
>Is that a fact because I have "observed such"?
When have I talked about clocks, james?
You once asked me if I would reasearch 5 different types of clocks and
present the results. I asked if you would listen to what I had to say,
you said no.
What is with you and your clock fetish, james? Is it some weird
masicist fantasy you have had since your daddy hit you with your alarm
clock?
Spaceman
>Jos is trying to find an internal contradiction,
No moron,
He has found many along with others.
and you are just too brainwashed to get any of them.
Don't tell me,
you say the paradoxes are not paradoxes huh?
<LOL>
Hey fool.
the clock god is dead.
too bad you have no unbrainwashed grey-matter to get it.
>Hey, we are talking about SR.
No,
you are worshiping SR as a religion with a infallable god.
(atomic clock)
Eric,
Wake up dingbat.
Eric Gisse
wrote:
Not expicitly, but in the postulates.
Inertia requires mass.
Does a photon have mass?
Not within several factors of "no"
Spaceman
>E=hf
h is a magic hat.
and a massless joke.
too bad you don't get the "trick"
>Solar sails, comet tails, Orion. Physics laughs at you, and so do I.
actually
only brainwashed morons laugh at me,
non brainwashed can see the wrongs.
you are a poor sad ass brainwashed dingbat.
and .
I still laugh with every single clock maker and engineer
that has brains at you.
>When have I talked about clocks, james?
>
You refuse to research then so you also refuse to talk about them huh?
>You once asked me if I would reasearch 5 different types of clocks and
>present the results. I asked if you would listen to what I had to say,
>you said no.
BULLSHIT!
I said I would listen if you actually researched the different clock
and not 5 atomic clocks.
>What is with you and your clock fetish, james? Is it some weird
>masicist fantasy you have had since your daddy hit you with your alarm
>clock?
What is with you and your "time changing is the cause for time changing"
moronic circular logic?
Still won't attempt to prove anything wrong huh?
Why won't you?
Brainwashed religions huh?
Sad..
just sad..
Still no glue at highspeed in 1 atomsphere cargo bay
calculations huh?
<LOL>
Eric.
you are a dupe.
and you might also be a scam artist too.
You should
seek help.
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
> >From: "Dirk Van de moortel"
dirkvand...@ThankS-NO-SperM.hotmail.com
>
> >Good dog!
>
> So dirk.
> The ISS .
> to head north..
> would have to fall huh?
> <LOL>
>
> you are a sad ass lying twisting puke.
> and so far.
> observation is proving it with each post you make.
>
You are the epitome of what is wrong with the concept of free speech.
Congratulations for achieving this unique honour.
Franz Heymann
Franz Heymann
"BernardZ" <Bern...@LargeMailBox.com to reply delete Large and Box>
wrote in message news:MPG.181943acb5baf98698a33a@news...
> In article
<mjones-1710...@66-105-229-12-aus-02.cvx.algx.net>,
> mjo...@jump.net says...
> > Ultimately, therefore, someone is
> > going to point out that an "observer" can't be pushed up to
lightspeed to
> > move along with the photon
> >
>
> But in physics we often put as a though experiment an observer in
places
> that are physically impossible!
I am afraid the proposed frame is not only physically, but also
mathematically impossible.
Franz Heymann
Spaceman
>Jimmy, or may I call you Spacemannikin?
What you call me shows your state of mind.
both you chose prove you are not even trying to
think about what is stated.
>You are the epitome of what is wrong with the concept of free speech.
>Congratulations for achieving this unique honour.
Thanks,
That is actually a compliment when it comes from such a brainwashed fool.
Too bad you still "just don't get anything I state"
Still on the "time changing wagon" huh?
<LOL>
Still on the wagon but drunk as a skunk!
<LOL>
Stage one denial all the way folks!
:)
Spaceman
>I am afraid the proposed frame is not only physically, but also
>mathematically impossible.
<ROFLOL>
refuse to try it huh?
<LOL>
What does a camera on the front of a train represent?
Gues what Franz,
It represents that "frame" view you say can not exist physically.
Never did any 3D work that uses camara views huh?
sheesh!
That is "real" sad.
No wonder you are lost.
you never actually have used anything 3D in space before huh?
Try buying animation master
http://www.hash.com
and weep when you see the facts hit you real hard.
about "time and path lines"
and even some fun particle motion stuff.
(water waves too, if you get the hang of it and all)
rain ..
fog,
fire..
etc..
all the "reals"
.
and..
nevermind the "lighting capabilites"
I really feel sorry for any physicist who has never
used a 3D animation program yet.
There is no space time.
There is space and time.
It is that simple.
too bad you "just refuse to even find out"
Dang flat worlders..
<LOL>
Eric Gisse
wrote:
>>From: Eric Gisse kseggR...@uas.alaska.edu
>
>>E=hf
>
>h is a magic hat.
>and a massless joke.
>
>too bad you don't get the "trick"
OK, fair enough. You say my method is crap. How do i calculate the
energy of a 450 nm photon using your method, whatever it may be?
>>Solar sails, comet tails, Orion. Physics laughs at you, and so do I.
>
>actually
>only brainwashed morons laugh at me,
>non brainwashed can see the wrongs.
>you are a poor sad ass brainwashed dingbat.
>and .
>I still laugh with every single clock maker and engineer
>that has brains at you.
Now you speak for clockmakers.
Can you produce a contact for a high-precision atomic clock
manufacturer?
>
>>When have I talked about clocks, james?
>>
>
>You refuse to research then so you also refuse to talk about them huh?
No james, you are the one who brings up "clock god" day in day out.
Are you schitzophrenic? Do you hear voices when you read my posts?
I refuse to research because I know what you will do. When I do not
produce results that agree with your bullshit metaphysics, you will
scream, cry, and whine about how im worshiping something that does not
exist.
I also dont dive into open-ended research. What do you want me to
research specifically? Im sure I could find what I want to know about
many types of clocks with google.
Oh yes, im also atheist. No religion, no god. Comprende?
>
>>You once asked me if I would reasearch 5 different types of clocks and
>>present the results. I asked if you would listen to what I had to say,
>>you said no.
>
>BULLSHIT!
>I said I would listen if you actually researched the different clock
>and not 5 atomic clocks.
Your right, my memory was mistaken. Just my natural bias creeping up
against you. What would you do if the results were not in your favor?
>
>>What is with you and your clock fetish, james? Is it some weird
>>masicist fantasy you have had since your daddy hit you with your alarm
>>clock?
>
>What is with you and your "time changing is the cause for time changing"
>moronic circular logic?
No james, that is philosophy. Science does not research "why".
>Still won't attempt to prove anything wrong huh?
>Why won't you?
>Brainwashed religions huh?
>Sad..
>just sad..
>
>Still no glue at highspeed in 1 atomsphere cargo bay
>calculations huh?
><LOL>
Chemical processes would work normally, relative to the ship. They
would run slower/faster depending on how fast the observer is going.
If your in the cargo bay watching the glue dry, /_\v would be 0. No
time dilation, no problem.
>Eric.
>you are a dupe.
>and you might also be a scam artist too.
>You should
>seek help.
Care to recommend someone? I bet you know a few people...
Eric Gisse
wrote:
>>From: Eric Gisse kseggR...@uas.alaska.edu
>
>>Jos is trying to find an internal contradiction,
>
>No moron,
>He has found many along with others.
>and you are just too brainwashed to get any of them.
Repeat after me.
"Doman of applicibility"
"Doman of applicibility"
"Doman of applicibility"
Jos uses examples that go outside SR, he also uses examples that have
been solved.
Its not my fault he doesnt have the intellectual horsepower to climb
up that hill.
>
>Don't tell me,
>you say the paradoxes are not paradoxes huh?
><LOL>
>
>Hey fool.
>the clock god is dead.
>too bad you have no unbrainwashed grey-matter to get it.
There you go with your clock god speil again. Must be a fetish or some
kind of recurring mental defect. Intermittent failures are the worst
to diagnose.
>
>
>>Hey, we are talking about SR.
>
>No,
>you are worshiping SR as a religion with a infallable god.
>(atomic clock)
But we are *still* talking about it, regardless of the context of the
discussion.
>
>Eric,
>Wake up dingbat.
James,
Take your meds, retard.
Eric Gisse
wrote:
>>From: "Franz Heymann" Franz....@btopenworld.com
>
>>I am afraid the proposed frame is not only physically, but also
>>mathematically impossible.
>
><ROFLOL>
>refuse to try it huh?
><LOL>
>
Whats sqrt(0), james?
Thats why its impossible, for one.
m4r...@xs4a11.nl
> OK, fair enough. You say my method is crap. How do i calculate the
> energy of a 450 nm photon using your method, whatever it may be?
First, you convert the photon to a number of tires, so that Driscoll
can work with them. Then you multiply by a negative, yielding a positive.
After that, you start talking about puke and crap, and finally you
end with "LOL" or "sheesh!". This is called the Driscoll-method of
deductive reasoning.
Aardvark
Now that's just unfair.
Spacegoof is still working on last week's problem, 3 + (-2) = ?, he
can't be expected to complete two such complex calculations
simultaneously.
Jem
Dirk Van de moortel <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in
message news:Vwwr9.165940$8o4....@afrodite.telenet-ops.be...
>
> "Mitchell Jones" <mjo...@jump.net> wrote in message
news:mjones-1710...@66-105-229-19-aus-02.cvx.algx.net...
> > ***{Forget the fact that he used the word "attached." His point is that
in
> > the reference frame of the moving photon, its speed is obviously zero.
>
> Bob's point is that according to the theory, the photon cannot
> have a reference frame to begin with, so there is no way to
> define, let alone measure "speed wrt a photon". Saying that
> its own speed is obviously zero, does not make sense.
Dirk, your comment makes me wonder whether there's been any attempt to
create a
model in which motion at speed "c" is treated as stationary (i.e. a geometry
in which "photons" serve as points)? Seems like a resonable thing to
consider in view of the fact that under the standard model such objects
"move" along geodesics whose points are separated by zero "distance".
Spaceman
>OK, fair enough. You say my method is crap. How do i calculate the
>energy of a 450 nm photon using your method, whatever it may be?
find the mass of the photon first
be "finding the mass" instead of accepting Plancks "lucky guess"
Planck already "thought light had no mass"
so he did not look "for mass at all"
If you use his work,
You CAN find the mass.
and then
Newton actually takes over from there
along with Maxwell etc..
REALS.
not joke nothingnesses in motion.
>Now you speak for clockmakers.
Why not,
I know how to make them.
and have "researched all types"
and have even made one out of Legos "just for fun"
Have you researched clocks at all yet?
I know you have not.
Or you would not give me any crap about them at all.
>Can you produce a contact for a high-precision atomic clock
>manufacturer?
If I had the machine and instructions just like the guy that is making them
now.
Yes!
But,
that is irrelevant.
and you still are ignoring "doing the research"
>No james, you are the one who brings up "clock god" day in day out.
>Are you schitzophrenic? Do you hear voices when you read my posts?
Not even close Eric.
you must have a god complex?
It is the joke of
clocks regulating time that you still don't get.
<LOL>
>I refuse to research because I know what you will do. When I do not
>produce results that agree with your bullshit metaphysics, you will
>scream, cry, and whine about how im worshiping something that does not
>exist.
<ROFLOL>
sad ass copout..
<LOL>
>No james, that is philosophy. Science does not research "why".
You are very wrong
Science must have "a why" for a cause.
and you are lost if you have no "why" for a cause.
and,
you are lost..
you have no "physical causes" being found for
lots of things still happening.
I have the causes.
and they are simply.
too simple for complex brainwashed brains (like yours I guess)
to even get it.
>Chemical processes would work normally, relative to the ship. They
>would run slower/faster depending on how fast the observer is going.
Wrong,
observations from different points of view,
would be slower/faster (not the "running itsefl)
the ACTUAL reaction does not change rate at all.
you really do need to research more.
It is sad.
The reaction would not "run" slower at all.
and constant moving clocks do not slow down
at any speed known yet.
>If your in the cargo bay watching the glue dry, /_\v would be 0. No
>time dilation, no problem.
No shit.
clocks work the same way in such a bay.
Time dilation is a clock malfunction.,
NOT TIME CHANGING RATE.
sheesh!
Why are you so ignorant?
>Care to recommend someone? I bet you know a few people...
Fuch Off
research or insult like a fool.
I don't care.
tis your own ship you sink.
Spaceman
>Repeat after me.
>
>"Doman of applicibility"
>"Doman of applicibility"
>"Doman of applicibility"
Parrots sure are funny!
<LOL>
>Jos uses examples that go outside SR, he also uses examples that have
>been solved.
Domain of the Universe.
To explain the Universe.
You CAN go out of SR and use
anything in the UNIVERSE.
DUH!
Domain of morons.
<LOL>
>James,
>Take your meds, retard.
>>
Eric,
take yours.
you "REALLY" need them now..
Spaceman
>Whats sqrt(0), james?
>
What does the sqrt represent in reality Eric.
What does the (0) represent in reality Eric
If you have no reals.
you have already started with a "nothing and a fanstasy at best"
If I am doing 185,000 mps and I am able to
add another 20,000 mps within 1 second.
how fast would I be going and "would my ship contract"?
and..
how slow would the clock run?
Would time slowdown a bit?
Eric,
Time, does not exist.
You have been fooled.
Dirk Van de moortel
"Jem" <x...@xxx.com> wrote in message news:rvTr9.233698$TX5.9...@news1.east.cox.net...
I guess that many have been thinking about it, perhaps even
having tried advanced and exotic alternative mathematical
models. No idea...
If you take the results of SR and take the limits for v --> c, you
can think about the few results that still make at least some sense.
The FAQ has a bit of fun (but not very useful) reading:
http://www.physics.iastate.edu/sci.physics/faq/Relativity/SpeedOfLight/headlights.html
Dirk Vdm
Dirk Van de moortel
"Spaceman" <agents...@aol.combination> wrote in message news:20021018092418...@mb-mq.aol.com...
[snip]
> Eric,
> Time, does not exist.
> You have been fooled.
Spaceman, the people over here say that time is what we
read on our clocks. Do you think that it is impossible that
we read numbers on our clocks?
Just a yes or no, could you manage that?
Dirk Vdm
Spaceman
>Spaceman, the people over here say that time is what we
>read on our clocks. Do you think that it is impossible that
>we read numbers on our clocks?
>Just a yes or no, could you manage that?
Dirk,
The numbers you read on your clocks are
looping ratios.
and nothing more than such.
It is too bad you think a ratio is a "thing" all by itself
I really think it is funny
that your site has the words,
immortal...
too bad you don't get it!
<LOL>
The Queen must have had such a scroll back in the flat worlder days too!
<LOL>
Dirk Van de moortel
"Spaceman" <agents...@aol.combination> wrote in message news:20021018100036...@mb-dh.aol.com...
Dirk Vdm
Robert Kolker
Spaceman wrote:
> If I am doing 185,000 mps and I am able to
> add another 20,000 mps within 1 second.
But you can't add another 20,000 mps. The amount of energy needed to get
you just to light speed becomes infinte. But there is only a finite
amount of energy in the universe, and certainly a finite amount of
energy that is available to the likes of us.
Bob Kolker
Robert Kolker
Spaceman wrote:
> The numbers you read on your clocks are
> looping ratios.
> and nothing more than such.
ratios are numbers, so your answer should be yes.
For example the ratio of 7 to 5 is a rational number (by definition).
That is why we are able to do fractional measures with our meter sticks.
Bob Kolker
Spaceman
>But you can't add another 20,000 mps. The amount of energy needed to get
>you just to light speed becomes infinte.
Won't do the math huh?
thanks for proof.
:)
>But there is only a finite
>amount of energy in the universe, and certainly a finite amount of
>energy that is available to the likes of us.
I am only askingg for a "finite change"
your "needed" infinites are a joke.
nothing infinite is needed to add "any finite speed"
me...@cars3.uchicago.edu
ain't true. You can't get there because there ain't "there" there.
It is not that the speed exists, only is inaccessible. It *does not*
exist. This is reflected in the Lorentz transformations and the
divergence of energy when you approach c is the *consequence* not
*cause* of this fact.
Furthermore, somehow, people (even those who should know better, keep
forgetting that speed (and velocity) is relative. There is no such
thing as "a body moving at 0.9c" or so, there is only "moving at v
relative to ...". So, lets look back at the question above and the
proper representation.
Lets take a spaceship moving away from Earth at 0.9c (relative to
Earth). Now, lets say that the ship drops overboard a marker
(which'll serve as a new reference) moving parallel to it at same
speed, then start accelerating. Can it get to a speed of 0.9c
relative to the marker. Sure, of course. The frame of the marker is
a nice inertial frame just like the Earth, it doesn't come with a
sticker saying "you've only 0.1c left". So, yes, you can get to 0.9c
relative to the marker which, itself, is doing 0.9c relative to the
Earth. Only, your speed relative to the Earth, now, is not 1.8c, it
is just (0.9c + 0.9c)/(1 + 0.9*0.9) = 1.8c/1.81 = 0.95c
(approximately). The law for combination of velocities *is not*
addition.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Dirk Van de moortel
<me...@cars3.uchicago.edu> wrote in message news:%AWr9.50$P4....@news.uchicago.edu...
Very nicely formulated quibble :-)
One more to reinstate a bit of confusion: according to an
observer on the marker in between Earth and ship, the distance
between Earth and the ship will increase like 1.8c.
Sorry, couldn't resist ;-)
Dirk Vdm
Eric Gisse
wrote:
>>From: Eric Gisse kseggR...@uas.alaska.edu
>
>>OK, fair enough. You say my method is crap. How do i calculate the
>>energy of a 450 nm photon using your method, whatever it may be?
>
Show, don't tell.
<snip retch>
josX
No, there is no such thing as >c relative velocity anywhere in relativity.
You arrived at your number using the obvious and intuitive (true)
V' = V + v0, but it doesn't work that way in relativity, not even
for a man-in-the-middle.
A man in the middle has the Earth at -.9c, and the ship at +.9c,
this will result Galileanly in
-.9c = V + +.9c
-.9c -.9c = V = -1.8c
You arrived at your answer with Galilean Math Dirk. Something for your
fumbles ?
The "correct" speed addition formula is
V' = (V + v0)/(1 + V * v0/c^2)
Remember?
NOT V' = V + v0
Which will never ever result in >c speeds.
V' = (V + v0)/(1 + V * v0/c^2)
-.9 = (V + .9)/(1 + V * .9)
-.9 - .81V = V + .9
-.9 -.9 = V + .81V
-1.8c = 1.81V
V = 1.8c/1.81 = .9944751c
Getting a feeling for the absurdity of SR already ?
No?
You gotta love it: anti-SR folk reading relativists the lesson on how
their own theory they are defending works...
--
jos
Dirk Van de moortel
"josX" <jo...@mraha.kitenet.net> wrote in message news:aopjk6$h18$1...@news1.xs4all.nl...
[snip]
> No
Kst.
The dog must stay under the table and keep quiet.
Dirk Vdm
Randy Poe
> No, there is no such thing as >c relative velocity anywhere in relativity.
> You arrived at your number using the obvious and intuitive (true)
> V' = V + v0, but it doesn't work that way in relativity, not even
> for a man-in-the-middle.
Not surprisingly, you misunderstand something very basic.
The man-in-the-middle can say the distance between two particles
is growing at a rate of 1.8c in his rest frame. That does not
violate relativity because it is not the velocity of anything.
This is NOT a Galilean transform. It isn't asking how particle
1 looks to particle 2.
> A man in the middle has the Earth at -.9c, and the ship at +.9c,
> this will result Galileanly in
> -.9c = V + +.9c
> -.9c -.9c = V = -1.8c
> You arrived at your answer with Galilean Math Dirk. Something for your
> fumbles ?
>
> The "correct" speed addition formula is
> V' = (V + v0)/(1 + V * v0/c^2)
Correct formula, incorrectly applied. This tells you how
to calculate the speed of the ship as measured from the
Earth, or the speed of the Earth as measured from the ship.
If you apply the correct formula, you will correctly
predict that the ship and the earth each measure the
other moving at 0.9945c. IN THEIR OWN FRAMES.
The "separation rate" as measured by the man-in-the-middle
is something else entirely.
The velocity addition formula tells you how the velocity
of some particle will be measured in some frame. 1.8c does
not represent such a quantity. The man in the middle does
not predict anything is moving at 1.8c in his frame.
What do you think relativity says about the separation
rate of two light beams, one pointing left and one
pointing right, according to the guy in the middle?
- Randy
josX
>josX wrote:
>> No, there is no such thing as >c relative velocity anywhere in relativity.
>> You arrived at your number using the obvious and intuitive (true)
>> V' = V + v0, but it doesn't work that way in relativity, not even
>> for a man-in-the-middle.
>
>Not surprisingly, you misunderstand something very basic.
>
>The man-in-the-middle can say the distance between two particles
>is growing at a rate of 1.8c in his rest frame. That does not
>violate relativity because it is not the velocity of anything.
>
>This is NOT a Galilean transform. It isn't asking how particle
>1 looks to particle 2.
No?
How else then:
A = B + C
-.9c = B + .9c
B = -1.8c
How did you compute it, doc ?
How did you reach your "1.8c" result.
Where in relativity is there a V1 = V2 + V3 formula that makes it possible
for you to get your 1.8c result.
WHERE!
>> A man in the middle has the Earth at -.9c, and the ship at +.9c,
>> this will result Galileanly in
>> -.9c = V + +.9c
>> -.9c -.9c = V = -1.8c
>> You arrived at your answer with Galilean Math Dirk. Something for your
>> fumbles ?
>>
>> The "correct" speed addition formula is
>> V' = (V + v0)/(1 + V * v0/c^2)
>
>Correct formula, incorrectly applied. This tells you how
>to calculate the speed of the ship as measured from the
>Earth, or the speed of the Earth as measured from the ship.
>
>If you apply the correct formula, you will correctly
>predict that the ship and the earth each measure the
>other moving at 0.9945c. IN THEIR OWN FRAMES.
>
>The "separation rate" as measured by the man-in-the-middle
>is something else entirely.
Because you can't get your head around this in SR ?
You are using V' = V + v0 here.
>The velocity addition formula tells you how the velocity
>of some particle will be measured in some frame. 1.8c does
>not represent such a quantity. The man in the middle does
>not predict anything is moving at 1.8c in his frame.
I'm sorry dude, but you can only wed one wife, and when there's two,
you gotta chose one. You chose relativity, that means Galilean math
is off limits for you.
>What do you think relativity says about the separation
>rate of two light beams, one pointing left and one
>pointing right, according to the guy in the middle?
In reality: 2times the speed of light from ANY reference point.
In relativity it is 1c.
Do the math: V1 = (V2 + V3)/(1 + V2 * V3/c^2)
Look how you are getting your 2c result:
V' = V + v0.
That is Galileo.
My goodness, relativists don't even know their own theory, that really
does it, doesn't it?
What math did you use!
Oh my gosh, you guys are fucking up.
--
jos
josX
>Randy Poe <rp...@nospam.com> wrote:
>>josX wrote:
>>> No, there is no such thing as >c relative velocity anywhere in relativity.
>>> You arrived at your number using the obvious and intuitive (true)
>>> V' = V + v0, but it doesn't work that way in relativity, not even
>>> for a man-in-the-middle.
>>
>>Not surprisingly, you misunderstand something very basic.
>>
>>The man-in-the-middle can say the distance between two particles
>>is growing at a rate of 1.8c in his rest frame. That does not
>>violate relativity because it is not the velocity of anything.
>>
>>This is NOT a Galilean transform. It isn't asking how particle
>>1 looks to particle 2.
>
>How else then:
>
>A = B + C
>-.9c = B + .9c
>B = -1.8c
>
>How did you compute it, doc ?
>How did you reach your "1.8c" result.
>
>Where in relativity is there a V1 = V2 + V3 formula that makes it possible
>for you to get your 1.8c result.
>
>WHERE!
>
>>> this will result Galileanly in
>>> -.9c = V + +.9c
>>> -.9c -.9c = V = -1.8c
>>> You arrived at your answer with Galilean Math Dirk. Something for your
>>> fumbles ?
>>>
>>> The "correct" speed addition formula is
>>> V' = (V + v0)/(1 + V * v0/c^2)
>>
>>Correct formula, incorrectly applied. This tells you how
>>to calculate the speed of the ship as measured from the
>>Earth, or the speed of the Earth as measured from the ship.
>>
>>If you apply the correct formula, you will correctly
>>predict that the ship and the earth each measure the
>>other moving at 0.9945c. IN THEIR OWN FRAMES.
>>
>>The "separation rate" as measured by the man-in-the-middle
>>is something else entirely.
>
>
>>The velocity addition formula tells you how the velocity
>>of some particle will be measured in some frame. 1.8c does
>>not represent such a quantity. The man in the middle does
>>not predict anything is moving at 1.8c in his frame.
>
>you gotta chose one. You chose relativity, that means Galilean math
>is off limits for you.
>
>>rate of two light beams, one pointing left and one
>>pointing right, according to the guy in the middle?
>
>In relativity it is 1c.
>Do the math: V1 = (V2 + V3)/(1 + V2 * V3/c^2)
>
>Look how you are getting your 2c result:
>V' = V + v0.
>That is Galileo.
>
>My goodness, relativists don't even know their own theory, that really
>does it, doesn't it?
Look:
U = (V + W)/(1 + V * W/c^2) versus U = V + W
U = 30m/sec
V = -30m/sec
What is the speed difference?
U = V + W
30 = -30 + W
W = 60 m/sec (in rational Galilean math)
What is it in relativity?
U = (V + W)/(1 + V * W/c^2 )
30 = (-30 + W)/(1 + -30 * W/3e8^2)
30 - 30 * 30 * W/ 3e8^2 = -30 + W
30 + 30 - 900* W/ 3e8^2 = W
30 + 30 = W + 900 / 3e8^2 W
60 = 1 900/3e8^2 W
60 / (1 900/3e8^2) = W
W = 59.999999999999999333333333333333340740740740740740658436213991769548...
That is the famous "almost intuitive" result at low speeds. The old
"indistinguishable", remember? At low speeds?
See that your formula works near perfectly at low speeds?
What's up Randy, can't accept that near lightspeed things "act differently",
as relativists are always so proud of proclaiming ?
Why are you suddenly reverting to Galileo when we are getting at higher
speeds, where is your sudden U = V + W coming from???
Don't tell me you came to your senses.
You are now effectively on our side? euh... welcome i guess ;).
--
jos
Randy Poe
> Randy Poe <rp...@nospam.com> wrote:
>
>>josX wrote:
>>
>>>No, there is no such thing as >c relative velocity anywhere in relativity.
>>>You arrived at your number using the obvious and intuitive (true)
>>>V' = V + v0, but it doesn't work that way in relativity, not even
>>>for a man-in-the-middle.
>>
>>Not surprisingly, you misunderstand something very basic.
>>
>>The man-in-the-middle can say the distance between two particles
>>is growing at a rate of 1.8c in his rest frame. That does not
>>violate relativity because it is not the velocity of anything.
>>
>>This is NOT a Galilean transform. It isn't asking how particle
>>1 looks to particle 2.
>
>
> No?
It isn't asking how particle X is moving relative to any rest frame.
Period.
It is not a transformation of velocities from one frame to
another.
> How else then:
>
> A = B + C
> -.9c = B + .9c
> B = -1.8c
>
> How did you compute it, doc ?
> How did you reach your "1.8c" result.
>
> Where in relativity is there a V1 = V2 + V3 formula that makes it possible
> for you to get your 1.8c result.
>
> WHERE!
d1 = 0.9ct
d2 = -0.9ct
These are particle positions. SR specifies how they transform
between frames.
Define d = d1-d2 = the separation between particles in my
frame.
SR tells me that since d1 = 0.9ct and d2 = 0.9ct, then the
quantity d = d1-d2 is given by 0.9ct - (-0.9ct) = 1.8ct.
This is not the position of any particle in any rest frame.
It is merely a quantity I have defined in terms of two
positions.
It's derivative is NOT a velocity of anything. It's derivative
is d/dt[1.8ct] = 1.8c. But so what?
> What math did you use!
I used simple addition in the rest frame, to calculate this
new and somewhat meaningless quantity d, which does not
correspond to anything in particular.
I could also define a quantity d1*sin(d2) and tell you what
its derivative is in time. I could tell you how it transforms
in different frames, because I have rules for how d1 and d2
transform in different frames.
> Oh my gosh, you guys are fucking up.
No, you're still screwed up on the basics.
The derivative of d is not a velocity of anything. d is a function
of d1 and d2. The derivatives of d1 and d2 are subject to the
Lorentz, not the Galilean transform. There is nothing that
says "d" is a quantity that has to be calculated by the
relativistic addition formula because the relativistic
addition formula tells you how to TRANSFORM PARTICLE
VELOCITIES WHEN YOU CHANGE FRAMES.
It doesn't apply here because:
(a) The derivative of d is not a particle velocity and
(b) I'm not changing frames.
You have no idea what I'm saying, I know. But that is the
answer to this particular "paradox".
I can also define other quantities in my rest frame, like
"how fast does a particle recede from its image in a mirror?"
Guess what? That's not a Lorentz transform either.
- Randy
Dirk Van de moortel
"josX" <jo...@mraha.kitenet.net> wrote in message news:aoprgs$7c2$1...@news1.xs4all.nl...
> Randy Poe <rp...@nospam.com> wrote:
> >josX wrote:
> >> No, there is no such thing as >c relative velocity anywhere in relativity.
> >> You arrived at your number using the obvious and intuitive (true)
> >> V' = V + v0, but it doesn't work that way in relativity, not even
> >> for a man-in-the-middle.
> >
> >Not surprisingly, you misunderstand something very basic.
> >
> >The man-in-the-middle can say the distance between two particles
> >is growing at a rate of 1.8c in his rest frame. That does not
> >violate relativity because it is not the velocity of anything.
> >
> >This is NOT a Galilean transform. It isn't asking how particle
> >1 looks to particle 2.
>
> No?
> How else then:
>
> A = B + C
> -.9c = B + .9c
> B = -1.8c
>
> How did you compute it, doc ?
> How did you reach your "1.8c" result.
>
> Where in relativity is there a V1 = V2 + V3 formula that makes it possible
> for you to get your 1.8c result.
>
> WHERE!
sounds angry.
religious problem somewhere?
does it hurt?
excellent.
Dirk Vdm
Franz Heymann
"Spaceman" <agents...@aol.combination> wrote in message
> >From: "Franz Heymann" Franz....@btopenworld.com
>
> >Jimmy, or may I call you Spacemannikin?
>
> What you call me shows your state of mind.
> both you chose prove you are not even trying to
> think about what is stated.
>
> >You are the epitome of what is wrong with the concept of free speech.
> >Congratulations for achieving this unique honour.
>
> Thanks,
> That is actually a compliment when it comes from such a brainwashed
fool.
> Too bad you still "just don't get anything I state"
But I do, Spacemannikin, I do. I do better than you do. You see, I can
see that you post crap and you cannot see that.
>
> Still on the "time changing wagon" huh?
> <LOL>
> Still on the wagon but drunk as a skunk!
> <LOL>
> Stage one denial all the way folks!
Franz Heymann
me...@cars3.uchicago.edu
that "relative velocity" has stricter meaning in relativity than in
Newtonian mechanics.
josX
>josX wrote:
>> Randy Poe <rp...@nospam.com> wrote:
>>
>>>josX wrote:
>>>
>>>>No, there is no such thing as >c relative velocity anywhere in relativity.
>>>>You arrived at your number using the obvious and intuitive (true)
>>>>V' = V + v0, but it doesn't work that way in relativity, not even
>>>>for a man-in-the-middle.
>>>
>>>Not surprisingly, you misunderstand something very basic.
>>>
>>>The man-in-the-middle can say the distance between two particles
>>>is growing at a rate of 1.8c in his rest frame. That does not
>>>violate relativity because it is not the velocity of anything.
>>>
>>>This is NOT a Galilean transform. It isn't asking how particle
>>>1 looks to particle 2.
>>
>>
>> No?
>
>It isn't asking how particle X is moving relative to any rest frame.
>
>Period.
>
>It is not a transformation of velocities from one frame to
>another.
You are adding velocities.
>> How else then:
>>
>> A = B + C
>> -.9c = B + .9c
>> B = -1.8c
>>
>> How did you compute it, doc ?
>> How did you reach your "1.8c" result.
>>
>> Where in relativity is there a V1 = V2 + V3 formula that makes it possible
>> for you to get your 1.8c result.
>>
>> WHERE!
>
>d1 = 0.9ct
>d2 = -0.9ct
>
>These are particle positions. SR specifies how they transform
>between frames.
>
>Define d = d1-d2 = the separation between particles in my
>frame.
>
>SR tells me that since d1 = 0.9ct and d2 = 0.9ct, then the
>quantity d = d1-d2 is given by 0.9ct - (-0.9ct) = 1.8ct.
Yes, that is the Galilean velocity addition.
>This is not the position of any particle in any rest frame.
>It is merely a quantity I have defined in terms of two
>positions.
>
>It's derivative is NOT a velocity of anything. It's derivative
>is d/dt[1.8ct] = 1.8c. But so what?
1.8c, but not a velocity of anything?
>> What math did you use!
>
>I used simple addition in the rest frame, to calculate this
>new and somewhat meaningless quantity d, which does not
>correspond to anything in particular.
doesn't correspond to anything in particular?
Boy, that is a cheap defense isn't it.
"oh sorry, my number didn't apply to anything in particular when i
used the Galilean velocity /addition/, please forget about this please??"
>I could also define a quantity d1*sin(d2) and tell you what
>its derivative is in time. I could tell you how it transforms
>in different frames, because I have rules for how d1 and d2
>transform in different frames.
>
>> Oh my gosh, you guys are fucking up.
>
>No, you're still screwed up on the basics.
No i'm not, but this isn't the first time relativity is getting away
with total contradictions, it's a state of mind.
>The derivative of d is not a velocity of anything. d is a function
>of d1 and d2. The derivatives of d1 and d2 are subject to the
>Lorentz, not the Galilean transform. There is nothing that
>says "d" is a quantity that has to be calculated by the
>relativistic addition formula because the relativistic
>addition formula tells you how to TRANSFORM PARTICLE
>VELOCITIES WHEN YOU CHANGE FRAMES.
>
>It doesn't apply here because:
> (a) The derivative of d is not a particle velocity and
> (b) I'm not changing frames.
>
>You have no idea what I'm saying, I know. But that is the
>answer to this particular "paradox".
>
>I can also define other quantities in my rest frame, like
>"how fast does a particle recede from its image in a mirror?"
>Guess what? That's not a Lorentz transform either.
This is cutting close at the heart of the matter.
Relativists can do what they want with impunity. I ofcourse knew that
many relativists had a pet version that involved your above math, that
they claimed light going right and left had a 2c relative velocity in
that frame, even though it has a 1c relative velocity in the light's
own frame (do the math).
The reason is very simple ofcourse: nobody will buy into relativity if
they learn that relativity says what it says, so you gotta start using
the Galilean velocity addition here to boost relativity's credibility
to a high enough level that people will at least pay attention to their
brainwashing.
You may fool many, even the great majority, but you don't fool me.
You are using the Galilean velocity /addition/, clear cut V1 + V2,
this is inherently not relativity, not the unmixed version anyway that
will have big problems getting itself accepted. You don't have the
relative velocities discrepancy problem in relativity between frames,
the frames agree about the relative velocities between objects because
velocities always add like U = (V + W)/(1 + VW/cc), also in one frame
(as shown with the 30m/sec example, it works nicely for low velocities
(somewhat nicely anyway). Now that you are claiming Galilean velocity
addition must be used in cases where the speeds are both in the same
frame, you are opening up to a contradiction between what frames have
as the relative speed between objects, for instance a lightbeam will
have another in the oposite direction at 1c (do the math), and in
the middle frame between them it should in pure relativity also be
1c. Now however they suddenly disagree, it is 2c in the middle-frame.
So i ask you: what is the speed difference between a car going to the
right at 30m/sec and a car going to the left at 30m/sec.
Guess what!
You are actually going to use U = V + W.
30 = -30 + W
W = 60m/sec
Now a man walking in a train relative to us at 104m/sec, and the train
is going at 100m/sec, both to the right.
Question, what is the speed difference, how fast does the man walk in
the train!
U = V + W!!!
Same situation!
Both are in your frame!
104 = 100 + W
W = 4m/sec!
The man is walking at 4 m/sec in the train!
(Sorry for all the exclamation marks, but i already know what this means...)
So, if the man is walking at 4m/sec in the train... (we just computed that
didn't we?, check again: 104 = 100 + W -> W = 4), and the train is doing
100m/sec to us:
U = 100 + 4
V = 104
The man goes at 104m/sec relative to us!
No gamma bullshit anywhere Randy! Nowhere!
Check it again:
If the train is doing 100, and the man is doing 4 in the train
U = 4 + 100!!!!!
Because
104 = 4 + 100!!!
Because
104 = 100 + W
W = 4
Getting the picture? These things are /related/, you can't seperate them.
Stupid. :-)
And now, guess how i can spank you ass real bad with thiss, and show you
you don't know relativity but use your own pet version:
We increase the speeds a bit, first:
U = 2 + 2
U = 4
Right? That's what you said, because you said:
4 = 2 + W
Then W = 2 and 4 = 2 + 2 and U = 2 + 2 -> U = 4
Because... -2 = +2 + W -> W = -4, speeds in "the same frame".
But that means since we found -2 = +2 - 4, that U = +2 - 4, U = -2 !
No gamma Randy, no gamma.
So, here it comes:
U = 2 + 2...
U = 4 -> 4 = 2 + 2 because 4 = 2 + W, W = 2
U = 200,000,000 + 200,000,000
Want to fill in the blanks Randy ?
U = 400,000,000m/sec
Because...
400,000,000 = 200,000,000 + W
W = 200,000,000m/sec
Exactly as our train example with the two object relative to the tracks:
104 = 100 + W
No gamma Randy.
You know how this is called Randy?
"You are screwed."
Better yet, you screwed yourself. You are using your own pet version of
relativity because even you can't accept relativity for what it really
is, and that means accepting that the relative speed is determined by
U = (V + W)/(1 + VW/cc), and absolutely nothing else, like it or like it
not, for better or worse. You either accept it or you reject it, but if
you start using Galileo anywhere, then like i showed, you are wed to
it everywhere.
Relativity says that the relative speed should be computed from relativity's
math, and that will result in for two lightbeams, one in every direction,
+c and -c:
U = (V + W)/(1 + VW/cc)
(Let's first find out what everything means again...
U = V + W -> +c = -c + W, ok got it, U and V are +c and -c...(easy trick))
U = (V + W)/(1 + VW/cc)
+c= (-c + W)/(1 + -cW/cc)
+c= (-c + W)/(1 + -W/c)
+c(1 + -W/c)= -c + W
+c + -Wc/c= -c + W
+c + -W= -c + W
+c + c= + W + W
2c = 2W
W = c
I would be the first to tell you this is nonsense Randy, but it is what
relativity says, so either you are going to accept it, or reject it, but
you can't go on with your half-baked pet version of relativity that uses
Galileo where you actually still have some logic working in your head.
This is what relativity says, you can like it or hate it, but it is what
relativity says.
Did you understand what i said?
The man in the train and the train are both in your frame, the relative
speed is a measure of the speed of the man /relative to the train/.
Once you have defined that ala Galileo, you have no way back.
4 = 2 + W
W = 2
then
4 = 2 + 2
and
? = 2 + 2
? = 4
You're screwed.
I hope you understand any of this.
So you got a choice, you never used relativity in your life. Are you
going to accept relativity now you know what it is:
U = (V + W)/(1 + VW/cc)
"for better and for worse", or are you now saying: "no, this is too bizarre
for me, give me Galileo":
U = V + W
Good luck with your decision, never too late to turn.
--
jos
Mitchell Jones
> Mitchell Jones wrote:
> > ***{Forget the fact that he used the word "attached." His point is that in
> > the reference frame of the moving photon, its speed is obviously zero.
> > Thus, unarguably, it is false to claim that the speed of light is the same
> > in all reference frames. Hence Einstein's claim that the speed of light is
> > the same in all reference frames is obviously wrong. Do you have a
> > response to the actual point being made, or not? --MJ}***
>
> How do you make measuements with massive devices which are moving at c
> wrt to the origin of an intertial frame. This is physically impossible.
***{That's just another circular argument--which means: you deflect a
thrust made at the the Einstein theory by invoking a postulate of ... the
Einstein theory!
To elaborate, light moves at lightspeed, and, logically, the mass of a
photon is m = hf/c^2--which indicates rather unambiguously that a photon
has mass. And, of course, Einstein's principle of the equivalency of mass
and energy supports that conclusion as well. However, such an
interpretation would contradict Einstein's claim that no massive object
can reach lightspeed, and, thus, would permit an observer at lightspeed,
thereby demolishing his relativistic velocity addition formula. Result: he
arbitrarily postulated that a photon has no mass. But in that case, the
skeptic asks, what is the meaning of m = hf/c^2 for a photon? Why, that's
merely the "effective mass," and an effective mass is not real, Einstein
replies!
Bottom line: when Einstein found himself in a logical quandary, he simply
postulated his way out of it, without troubling himself very much about
whether the postulate made sense. That's why controversy has swirled
around his theory from its inception, and continues today, and it's also
why his supporters tend to favor insults over arguments: the theory is, in
fact, very difficult to defend, if one limits oneself to substantive
argumentation.
--Mitchell Jones}***
> Why not ask what is North of the North Pole?
***{Why not indeed, since the answer would be Polaris, of course! :-) --MJ}***
> Bob Kolker
===============================================
Killfile inmates: Charles Cagle, Stephen Speicher, Mati Meron.
Mitchell Jones
> BernardZ wrote:
> > But in physics we often put as a though experiment an observer in places
> > that are physically impossible!
>
> Physically infeasible. But an observer at light speed is -impossible-
> because it means one has to divide by 0.
***{Incorrect. Einstein's formula for the addition of velocities of
objects in uniform motion in the same direction is w' = (v + w)/(1 +
vw/c^2), where v is the velocity of one object relative to an inertial
frame, w is the velocity of a second object relative to the first, and w'
is the velocity of the second object relative to the inertial frame. If
the first object is moving at c and the second object is a photon emitted
in the forward direction by the first object, then we have w' = (v + w)/(1
+ vw/c^2) = (c + c)/(1 + c^2/c^2) = 2c/2 = c. No zero divide condition
arises, hence no infinities result. (You have apparently remembered the
denominator as 1 - vw/c^2, but that is wrong.) --MJ}***
Mitchell Jones
moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
> "Robert Kolker" <bobk...@attbi.com> wrote in message
news:3DAEAC6B...@attbi.com...
> >
> >
> > BernardZ wrote:
> > > But in physics we often put as a though experiment an observer in places
> > > that are physically impossible!
> >
> > Physically infeasible. But an observer at light speed is -impossible-
> > because it means one has to divide by 0.
>
> Bob, the fact that one has to divide by 0 is not the reason *why*
> it is impossible. For various reasons we postulate a theory that
> assumes that it is impossible, *and* the theory uses equations
> that are consistent with this impossibility.
***{Nope. Bob was mistaken in thinking that a zero-divide condition arises
in that equation, and, thus, you are mistaken when you agree with him.
--MJ}***
> Dirk Vdm
Mitchell Jones
<Franz....@btopenworld.com> wrote:
> "Mitchell Jones" <mjo...@jump.net> wrote in message
> news:mjones-1710...@66-105-229-26-aus-02.cvx.algx.net...
> > In article <sggsqusm6aehsea54...@4ax.com>, Eric Gisse
> > <kseggR...@uas.alaska.edu> wrote:
> >
> > > On 17 Oct 2002 03:29:27 GMT, jo...@mraha.kitenet.net (josX) wrote:
> > >
> > > >Reference-point contradiction in relativity:
> > > >
> > > >\ | /
> > > > \|/
> > > >--*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> > > > /|\ |
> > > >/ | \ | -a- a--->
> > > > -b- |
> > > > |
> > > >
> > > >Relativity says that light's speed is the same in every
> reference-frame,
> > > >however when we attach a reference-frame to a moving lightfront,
> it's
> > > >speed can obviously not be lightspeed, but should instead be 0.
> > >
> > > How do you give a photon a reference frame when it has no mass?
> >
> > ***{Your argument is circular.
>
> Mitchell Jones is incapable of distinguishing between the concepts of
> internal consistency and circularity.
> If his powers of ratiocination are as poor as that, one may safely
> ignore anything else he has to say, so here goes the snippage.
***{So Franz has spoken. Just insults so far, of course. Apparently he
has, at long last, jerked back a stump enough times to learn to avoid
substantive arguments! --MJ}***
> [...]
>
> I would be glad if you would confirm which of the following are the
> reason for the inclusion of Stephen Speicher and Mati Meron in your
> killfike list:
>
> (1) You are not mentally equipped to understand their arguments.
> or
> (2) They beat you hollow in one or more discussions, and you do not
> have the intellectual integrity to admit it.
***{Yup. They "beat me hollow" using the same tools--i.e., insults--that
you are using here. (Hey, we can call it the "jerked-back-stump" syndrome!
:-) --MJ}***
> Franz Heymann
> -----------------------------------------------------------------
Mitchell Jones
<Franz....@btopenworld.com> wrote:
> "Mitchell Jones" <mjo...@jump.net> wrote in message
> news:mjones-1710...@66-105-229-19-aus-02.cvx.algx.net...
> > In article <3DAE2FCC...@attbi.com>, bobk...@attbi.com wrote:
> >
> > > josX wrote:
> > > > Reference-point contradiction in relativity:
> > > >
> > > > \ | /
> > > > \|/
> > > > --*~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ --->
> > > > /|\ |
> > > > / | \ | -a- a--->
> > > > -b- |
> > > > |
> > > >
> > > > Relativity says that light's speed is the same in every
> reference-frame,
> > > > however when we attach a reference-frame to a moving lightfront,
> it's
> > > > speed can obviously not be lightspeed, but should instead be 0.
> > >
> > > A reference frame cannot be so attached. It is physically
> meaningless to
> > > do so.
> > >
> > > Bob Kolker
> >
> > ***{Forget the fact that he used the word "attached." His point is
> that in
> > the reference frame of the moving photon, its speed is obviously
> zero.
>
> It is not possible to transform to some putativr rest frame of a photon,
> unless you are prepared to give a unique value to the quantity 0/0.
***{Damn, Franz, when you did nothing but hurl insults in your last post,
I gave you credit for having finally recognized that substantive
discussions are out of your league. But now, in your very next post, you
have proven me wrong. Why do I say that? Because the above is a
substantive comment, and, as is usual for you, it is wrong! (For the
reason, see my response to Bob Kolker.) --MJ}***
> > Thus, unarguably, it is false to claim that the speed of light is the
> same
> > in all reference frames. Hence Einstein's claim that the speed of
> light is
> > the same in all reference frames is obviously wrong. Do you have a
> > response to the actual point being made, or not? --MJ}***
>
> The contents of your killfile defines you as being bereft of powers of
> discrimination..
>
> Franz Heymann
***{Nah. The people in my killfile are haters and dumbasses, like you.
____ _ _ _ _ _
| _ \ | | ___ _ __ | | __ | | | | | |
| |_)| | | / _ \ | '_ \ | |/ / | | | | | |
| __/ | | | (_) | | | | | | < |_| |_| |_|
|_| |_| \___/ |_| |_| |_|\_\ (_) (_) (_)
If you are unable to comprehend the above, ask one of your guards or
therapists to transfer it into a fixed-length font such as Courier.
Bye! :-)
--Mitchell Jones}***
Stephen Speicher
You are, to put it bluntly, an ignorant dolt, and a subjectivist
to boot. You do not understand what the theory is, so you make up
a version which suits your feelings. That is, for you, apparently
a much simpler approach than actually educating yourself.
Clearly you prefer ignorance which conforms to your feelings
rather than knowledge of the facts of reality.
You are hopeless, pathetically hopeless.
--
Stephen
s...@speicher.com
Ignorance is just a placeholder for knowledge.
Printed using 100% recycled electrons.
-----------------------------------------------------------
Stephen Speicher
> In article <3DAEAC6B...@attbi.com>, bobk...@attbi.com wrote:
>
> > BernardZ wrote:
> > > But in physics we often put as a though experiment an observer in places
> > > that are physically impossible!
> >
> > Physically infeasible. But an observer at light speed is -impossible-
> > because it means one has to divide by 0.
>
> ***{Incorrect. Einstein's formula for the addition of velocities of
> objects in uniform motion in the same direction is w' = (v + w)/(1 +
> vw/c^2), where v is the velocity of one object relative to an inertial
> frame, w is the velocity of a second object relative to the first, and w'
> is the velocity of the second object relative to the inertial frame. If
> the first object is moving at c and the second object is a photon emitted
> in the forward direction by the first object, then we have w' = (v + w)/(1
> + vw/c^2) = (c + c)/(1 + c^2/c^2) = 2c/2 = c. No zero divide condition
> arises, hence no infinities result. (You have apparently remembered the
> denominator as 1 - vw/c^2, but that is wrong.) --MJ}***
>
It is not your ignorance which is an embarassment to the human
race, but rather the fact that you do not care. You would rather
spout nonsense and pretend that you know, than exert the effort
to learn and actually gain knowledge.
Stephen Speicher
> In article <z4yr9.166051$8o4....@afrodite.telenet-ops.be>, "Dirk Van de
> moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>
> > "Robert Kolker" <bobk...@attbi.com> wrote in message
> news:3DAEAC6B...@attbi.com...
> > >
> > >
> > > BernardZ wrote:
> > > > But in physics we often put as a though experiment an observer in places
> > > > that are physically impossible!
> > >
> > > Physically infeasible. But an observer at light speed is -impossible-
> > > because it means one has to divide by 0.
> >
> > Bob, the fact that one has to divide by 0 is not the reason *why*
> > it is impossible. For various reasons we postulate a theory that
> > assumes that it is impossible, *and* the theory uses equations
> > that are consistent with this impossibility.
>
> ***{Nope. Bob was mistaken in thinking that a zero-divide condition arises
> in that equation, and, thus, you are mistaken when you agree with him.
> --MJ}***
>
Apparently, the only ones _not_ mistaken are MJ and josX.
What a pair of clowns these two characters are.
Stephen Speicher
[MJ plonks Franz Heymann.]
This MJ really deserves some recognition. As far as I can recall,
I do not think he has gotten even a _single_ thing right about
physics since he arrived, and his current best buddy is josX.
What a pair these two clowns make!
Dirk Van de moortel
<me...@cars3.uchicago.edu> wrote in message news:Oa0s9.65$P4....@news.uchicago.edu...
> In article <vWXr9.168114$8o4....@afrodite.telenet-ops.be>, "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
writes:
> >
[snip]
> >Very nicely formulated quibble :-)
> >
> >One more to reinstate a bit of confusion: according to an
> >observer on the marker in between Earth and ship, the distance
> >between Earth and the ship will increase like 1.8c.
> >Sorry, couldn't resist ;-)
> >
> It is a good thing you didn't. That's a nice example illustrating
> that "relative velocity" has stricter meaning in relativity than in
> Newtonian mechanics.
A nice example indeed, and it managed to confuse the Boersema
character even more. Hmmmm, I love it.
Dirk Vdm
Dirk Van de moortel
"Mitchell Jones" <mjo...@jump.net> wrote in message news:mjones-1910...@66-105-229-3-aus-02.cvx.algx.net...
> In article <z4yr9.166051$8o4....@afrodite.telenet-ops.be>, "Dirk Van de
> moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>
> > "Robert Kolker" <bobk...@attbi.com> wrote in message
> news:3DAEAC6B...@attbi.com...
> > >
> > >
> > > BernardZ wrote:
> > > > But in physics we often put as a though experiment an observer in places
> > > > that are physically impossible!
> > >
> > > Physically infeasible. But an observer at light speed is -impossible-
> > > because it means one has to divide by 0.
> >
> > Bob, the fact that one has to divide by 0 is not the reason *why*
> > it is impossible. For various reasons we postulate a theory that
> > assumes that it is impossible, *and* the theory uses equations
> > that are consistent with this impossibility.
>
> ***{Nope. Bob was mistaken in thinking that a zero-divide condition arises
> in that equation, and, thus, you are mistaken when you agree with him.
> --MJ}***
You are mistaken when you think I agreed with Bob.
Dirk Vdm
Dirk Van de moortel
"Mitchell Jones" <mjo...@jump.net> wrote in message news:mjones-1910...@66-105-229-3-aus-02.cvx.algx.net...
>
> > BernardZ wrote:
> > > But in physics we often put as a though experiment an observer in places
> > > that are physically impossible!
> >
> > Physically infeasible. But an observer at light speed is -impossible-
> > because it means one has to divide by 0.
>
> ***{Incorrect. Einstein's formula for the addition of velocities of
> objects in uniform motion in the same direction is w' = (v + w)/(1 +
> vw/c^2), where v is the velocity of one object relative to an inertial
> frame, w is the velocity of a second object relative to the first, and w'
> is the velocity of the second object relative to the inertial frame. If
> the first object is moving at c and the second object is a photon emitted
> in the forward direction by the first object, then we have w' = (v + w)/(1
> + vw/c^2) = (c + c)/(1 + c^2/c^2) = 2c/2 = c. No zero divide condition
> arises, hence no infinities result. (You have apparently remembered the
> denominator as 1 - vw/c^2, but that is wrong.) --MJ}***
(t',x') = ((t-vx)/sqrt(1-v^2),(x-vt)/sqrt(1-v^2)).
v = 1, zero division, no Lorentz transformation, no t' and x',
no observables, no observer. The theory uses the Lorentz
transformation, the theory does not allow v=1.
Dirk Vdm
Franz Heymann
You are quite right.
As soon as I notice that a correspondent does not have the intellect to
folow an intellectually oriented argument, I start lampooning.
Particularly when it appears that said correspondent cannot appreciate
the niceties of algebraic signs, like you and Spacemannikin.
Anything else would be a waste of time.
Franz Heymann
Franz Heymann
"Mitchell Jones" <mjo...@jump.net> wrote in message
news:mjones-1910...@66-105-229-3-aus-02.cvx.algx.net...
>
> > BernardZ wrote:
> > > But in physics we often put as a though experiment an observer in
places
> > > that are physically impossible!
> >
> > Physically infeasible. But an observer at light speed
is -impossible-
> > because it means one has to divide by 0.
>
> ***{Incorrect. Einstein's formula for the addition of velocities of
> objects in uniform motion in the same direction is w' = (v + w)/(1 +
> vw/c^2), where v is the velocity of one object relative to an inertial
> frame, w is the velocity of a second object relative to the first, and
w'
> is the velocity of the second object relative to the inertial frame.
If
> the first object is moving at c and the second object is a photon
emitted
> in the forward direction by the first object, then we have w' = (v +
w)/(1
> + vw/c^2) = (c + c)/(1 + c^2/c^2) = 2c/2 = c. No zero divide condition
> arises, hence no infinities result. (You have apparently remembered
the
> denominator as 1 - vw/c^2, but that is wrong.) --MJ}***
Idiot.
Use the relative directions of the speeds involved corectly and come
again.
Franz Heymann
josX
><me...@cars3.uchicago.edu> wrote in message news:Oa0s9.65$P4....@news.uchicago.edu...
>
>[snip]
>
>>>Very nicely formulated quibble :-)
>>>
>>>One more to reinstate a bit of confusion: according to an
>>>observer on the marker in between Earth and ship, the distance
>>>between Earth and the ship will increase like 1.8c.
>>>Sorry, couldn't resist ;-)
>>
>>It is a good thing you didn't. That's a nice example illustrating
>>that "relative velocity" has stricter meaning in relativity than in
>>Newtonian mechanics.
>
>A nice example indeed, and it managed to confuse the Boersema
>character even more. Hmmmm, I love it.
Dirk, can you give me the name of this new form of relativity that has
C' = C + V ?
Let me show you imbecile:
Ship = +.9c
Earth = -.9c
.9c = -9c + V
V = 1.8c
Now, these are two object in /your/ frame, that is the rationale for
computing the relative speed like this aparently (see also Randy).
What does this mean if
Ship = +.9c
Earth = -.8c
.9c = -.8c + V
V = 1.7c
Ship = +.9c
Earth = +.8c
.9c = +.8c + V
V = .1c
oh oh Dirky Dirky... you shouldn't have made your humoristic mistake,
because now Randy and possibly the whole front of relativists is
labouring under it.
Let's not use the ship but a lightbeam...
Remember that you said a ligthbeam to the right and one to the left
means a relative velocity of 2c between them in that frame. Do you know
which formula that is?
U = V + W
So, now we have established you use this formula for objects and for
light, as long as both speeds are in the same frame "we are staying in
the same frame so thinks our Dirk".
Light = +1c
Earth = .8c
1c = .8c + V
V = .2c
It's the same formula dude.
You know what V stands for ?
/Lightspeed/
Lightspeed = .2c???
HUH? In relativity? I don't think so Dirk.
To make it a little bit more obvious for you:
V: the "speed of the object relative to another object", the "speed
difference between them", it is the SAME thing.
Why do you think Einstein (ripping off Lorentz) figured
U = (V + W)/(1 + VW/cc) ?
To keep lightspeed constant ofcourse, always.
Same situation in relativity now, so you can see how lightspeed constancy
is maintained, even though it are "objects in one frame, speeds given
relative to one frame, we are ``suposedly'' not leaving this frame".
Light = +1c
Earth = +.8c
C = (V + C')/(1 + VC'/cc)
1 = (.8 + C')/(1 + .8C'/1*1)
1(1 + .8C') = .8 + C'
1 + .8C' = .8 + C'
1 = .8 + C' - .8C'
1 - .8 = C' - .8C'
.2 = C' - .8C'
.2 = .2C'
.2/.2 = C'
1c = C'
See?
Lightspeed constancy is mathematically guaranteed because of the speed
addition formula relativity uses, as oposed to C = C' + V, even if speeds
are all given relative to one frame and we are "suposedly staying in
that frame".
This strange speed addition is the whole point of special relativity,
it gives the illusion of a lightspeed that can be constant (if we ignore
all contradictions ofcourse).
--
jos
Dirk Van de moortel
"josX" <jo...@mraha.kitenet.net> wrote in message news:aorkpg$mrp$1...@news1.xs4all.nl...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
> ><me...@cars3.uchicago.edu> wrote in message news:Oa0s9.65$P4....@news.uchicago.edu...
> >>In <vWXr9.168114$8o4....@afrodite.telenet-ops.be>, "Dirk Van de moortel":
> >
> >[snip]
> >
> >>>Very nicely formulated quibble :-)
> >>>
> >>>One more to reinstate a bit of confusion: according to an
> >>>observer on the marker in between Earth and ship, the distance
> >>>between Earth and the ship will increase like 1.8c.
> >>>Sorry, couldn't resist ;-)
> >>
> >>It is a good thing you didn't. That's a nice example illustrating
> >>that "relative velocity" has stricter meaning in relativity than in
> >>Newtonian mechanics.
> >
> >A nice example indeed, and it managed to confuse the Boersema
> >character even more. Hmmmm, I love it.
>
> Dirk,
Boersema, haven't you understood by now that I hardly ever
read what you write anymore?
I'm not interested in you anymore. Remember, you are just
an amusement factor.
Dirk Vdm
Gregory L. Hansen
Dirk Van de moortel <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>
>"josX" <jo...@mraha.kitenet.net> wrote in message
>news:aorkpg$mrp$1...@news1.xs4all.nl...
>Boersema, haven't you understood by now that I hardly ever
>read what you write anymore?
>I'm not interested in you anymore. Remember, you are just
>an amusement factor.
That's okay, I doubt he really reads what you write, either. Judging by
the quality of replies and the sheer volume, he can't be spending much
time reading and thinking about every one.
--
"A nice adaptation of conditions will make almost any hypothesis agree
with the phenomena. This will please the imagination but does not advance
our knowledge." -- J. Black, 1803.
Dirk Van de moortel
"Gregory L. Hansen" <glha...@steel.ucs.indiana.edu> wrote in message news:aormid$u9m$6...@rainier.uits.indiana.edu...
> In article <gZcs9.169287$8o4....@afrodite.telenet-ops.be>,
> Dirk Van de moortel <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
> >
> >"josX" <jo...@mraha.kitenet.net> wrote in message
> >news:aorkpg$mrp$1...@news1.xs4all.nl...
>
> >Boersema, haven't you understood by now that I hardly ever
> >read what you write anymore?
> >I'm not interested in you anymore. Remember, you are just
> >an amusement factor.
>
> That's okay, I doubt he really reads what you write, either. Judging by
> the quality of replies and the sheer volume, he can't be spending much
> time reading and thinking about every one.
I know for sure he sees everything I write ;-)
Dirk Vdm
Spaceman
>Lets take a spaceship moving away from Earth at 0.9c (relative to
>Earth). Now, lets say that the ship drops overboard a marker
>(which'll serve as a new reference) moving parallel to it at same
>speed, then start accelerating. Can it get to a speed of 0.9c
>relative to the marker. Sure, of course. The frame of the marker is
>a nice inertial frame just like the Earth, it doesn't come with a
>sticker saying "you've only 0.1c left". So, yes, you can get to 0.9c
>relative to the marker which, itself, is doing 0.9c relative to the
>Earth. Only, your speed relative to the Earth, now, is not 1.8c, it
>is just (0.9c + 0.9c)/(1 + 0.9*0.9) = 1.8c/1.81 = 0.95c
>(approximately). The law for combination of velocities *is not*
>addition.
@ spaceships heading towards each other
each having a relative to each other speed of
1.8c
and "Earth" as a passing point of the ships"
would read .9 for each ship realtvie to it.
C,mon
this limit shit is big big BULLSHIT now..
It is sadly piling up way too high and
smelling up physics REAL bad.
relative speeds above 1c exist.
"get over it!"
the limit is "killing what is actually good about relativity"
"relative positions and "actual relative speeds"
If you stop this silly invisible wall,
it would all start to fit "even better"
James M Driscoll Jr
Spaceman
http://www.realspaceman.com
Spaceman
>Getting a feeling for the absurdity of SR already ?
>No?
(Black Holes inside brains maybe?)
too dense to allow "different info to influence"
>You gotta love it: anti-SR folk reading relativists the lesson on how
>their own theory they are defending works...
It is funny..
that are lost with that stuff.
and are mad that you "do know it"
<LOL>
Even more scary.
They have no clue that I also know it.
and ..
know it's faults,
just as you do.
It is sad they just won't listen,
They are like mean parrots.
:)
Spaceman
dirkvand...@ThankS-NO-SperM.hotmail.com
>Kst.
>The dog must stay under the table and keep quiet.
There is no dog here you stupid parrot!
<LOL>
Franz Heymann
"Mitchell Jones" <mjo...@jump.net> wrote in message
news:mjones-1910...@66-105-229-3-aus-02.cvx.algx.net...
> ***{Incorrect. Einstein's formula for the addition of velocities of
> objects in uniform motion in the same direction is w' = (v + w)/(1 +
> vw/c^2), where v is the velocity of one object relative to an inertial
> frame, w is the velocity of a second object relative to the first, and
w'
> is the velocity of the second object relative to the inertial frame
You are not the first to have made the mistake of misunderstanding the
relative algebraic signs to be attached to v and w, and you will
probably unfortunately not be the last.
Franz Heymann