Re: The Fundamental Flaw of SR.

From: Benno Muilwyk (benno_at_muilwijk-met-wyk.nl)
Date: 08/02/04 

Date: Tue, 3 Aug 2004 00:32:59 +0200

"Androcles" <androc1es@nospamblueyonder.co.uk> schreef in bericht
news:7CpPc.12954$MJ.133627000@news-text.cableinet.net...
>
> "Benno Muilwyk" <benno@muilwijk-met-wyk.nl> wrote in message
> news:410d8fde_3@news3.prserv.net...
> |
> | "Androcles" <androc1es@nospamblueyonder.co.uk> schreef in bericht
> | news:b35Pc.12387$K46.126928064@news-text.cableinet.net...
> | >
> | > "Benno Muilwyk" <benno@muilwijk-met-wyk.nl> wrote in message
> | > news:410c201b_2@news3.prserv.net...
> | > |
> | > | "Androcles" <androc1es@nospamblueyonder.co.uk> schreef in bericht
> | > | news:bEIOc.11800$u11.120504414@news-text.cableinet.net...
> | > | >
> | > | > "Benno Muilwyk" <benno@muilwijk-met-wyk.nl> wrote in message
> | > | > news:410acfce_1@news3.prserv.net...
> | > | > |
> | > | > | "Androcles" <androc1es@nospamblueyonder.co.uk> schreef in
bericht
> | > | > | news:XBpOc.11285$kZ6.114233487@news-text.cableinet.net...
> | > | > | >
> | > | > | > "Benno Muilwyk" <benno@muilwijk-met-wyk.nl> wrote in message
> | > | > | > news:41096cd2_3@news3.prserv.net...
> | > | > | > | You never quit, do you?
> | > | > | >
> | > | > | > Nope.
> | > |
> | > | I know yoe never quit either.
> | > |
> | > | > | > |
> | > | > | > | You simply fail to understand the True Fundamentals of SR!
> | > | > | >
> | > | > | > I undertand them better than you.
> | > | > |
> | > | > | That remains to be proven.
> | > | >
> | > | > Handwaving is pointless.
> | > |
> | > | So why do you do it?
> | >
> | > Did I? You were the one that claimed I simply failed to understand the
> | "True
> | > Fundamentals of SR", without defining what they were, waving your
hands.
> | > What are these "true fundamentals" that I don't understand?
> | >
> | You are qouting from a response to Ken Seto.
> | Are you and Ken Seto one and the same person???
> | >
> | > |
> | > | > |
> | > | > | > | You always omit some or other fundamental part in your
> attempts
> | to
> | > | > | "prove"
> | > | > | > a
> | > | > | > | flaw in SR.
> | > | > | > | By not following SR consistently to the end it's only
logical
> | you
> | > | find
> | > | > | > | inconsistencies.
> | > | > | >
> | > | > | > a = b.... so multiplying by a,
> | > | > | > a^2 = ab... if we subtract b^2,
> | > | > | > a^2-b^2 = ab-b^2... we can factorize, so
> | > | > | > (a+b)(a-b) = b(a-b)... but a-b is now on both sides, se we'll
> | cancel
> | > | > | > a+b = b... and since a= b, given
> | > | > | > b+b = b, or
> | > | > | > 2b = b. Dividing by b,
> | > | > | > 2 = 1.
> | > | > |
> | > | > | That's an old one.
> | > | > Yes, but it's instructive if you want to learn.
> | > | >
> | > | >
> | > | > |
> | > | > | > By not following consistently to the end it's only logical you
> | find
> | > | > | > inconsistencies.
> | > | > | >
> | > | > | > Don't go back and look for the inconsistency, simply accept
it.
> | > | > | >
> | > | > | > Here's another.
> | > | > | > a = (a+b)/(1+b/a)
> | > | > | >
> | > | > | > Let a = 1 and b = 1 be the velocity of the gun and the
velocity
> of
> | > | > bullet.
> | > | > | > 1 = (1+1)/(1 + 1/1).
> | > | > | > Feel safe, I can't shoot you if you run away faster than me.
> | > | > | > Dont go back and look for the inconsistency, simply accept it.
> | > | > | >
> | > | > | Introduction to following c = (c+v)/(1 + v/c), I presume?
> | > | >
> | > | > Don't go back and look for the inconsistency, simply accept it.
> | > | > You simply fail to understand the True Fundamentals of SR!
> | > | >
> | > | >
> | > | >
> | > | > | >
> | > | > | > | Follow SR right to the end and you will be surprised to find
> | that
> | > SR
> | > | > is
> | > | > | > not
> | > | > | > | incosistent at all
> | > | > | > | and that all inconsistencies you found earlier were simply
> | caused
> | > by
> | > | > | flaws
> | > | > | > | in your own understanding of SR!
> | > | > | >
> | > | > | > Of course, follow a=b right to the end and 2 = 1 isn't at all
> | > | > | inconsistent,
> | > | > | > as I proved. Don't go back and look for an error. Simply
accept
> | it.
> | > | > | >
> | > | > | > |
> | > | > | > | I wish you would put more effort in understanding SR
> completely
> | > and
> | > | > | > learning
> | > | > | > | how to apply it correctly instead of posting your flaws to
> this
> | > news
> | > | > | group
> | > | > | > | each time you overlook something.
> | > | > | > |
> | > | > | > Wishes won't change Einstein's divide-by-zero that leads to 2
=
> 1,
> | > | > called
> | > | > | > c = (c+v)/(1 + v/c).
> | > | > | > If you think I've made a flaw, show where.
> | > | > |
> | > | > | You show me where the divide-by-zero occurs.
> | > | > |
> | > | > | > | Good luck,
> | > | > | > |
> | > | > | > | Benno
> | > | > | >
> | > | > | > I don't think luck can be relied upon in physics or
mathematics.
> | > | > |
> | > | > | Ken could use some luck in finding a good text book on SR which
he
> | can
> | > | > | understand.
> | > | > | And that's probably an understatement.
> | > | > |
> | > | > | > v = x/t. If you deny it, show why.
> | > | > | > x' = x-vt. Einstein says so.
> | > | > | In particular case, yes. So did Lorentz.
> | > | > | > x' = 0, simple algebra.
> | > | > | > When t = 0, x' = x, simple algebra.
> | > | > | > Now, what is it you want me to follow that leads to 2=1?
> | > | > | > What is the fundamental part that I always omit to prove a
flaw
> in
> | > SR?
> | > | > | > Show me.
> | > | > | > The truth is, you and many others always omit, i.e fail to
spot,
> | the
> | > | > | > divide-by-zero. You want to pretend it isn't there. It is.
> | > | > |
> | > | > | Enlighten me! Where is it?
> | > | >
> | > | > Right here:
> | > | >
> | > | > Einstein writes: "If we place x' = x-vt..."
> |
> | So you are quoting from an English translation of Einsteins original
1905
> | paper. That was not clear to me earlier.
>
> Nothing is clear to you, is it?
>
> |
> | > | > So, x' = x-x and that must equal 0.
> | > | > Of course this is simply the coordinate of the origin of the
moving
> | > frame,
> | > | > which is always at 0-vt in the stationary frame.
> |
> | Not in the context of the paragraph from which you are quoting!
> | Not far above your quote it says: "To any system of values x, y, z, t,
> which
> | completely defines the place and time of an event in the stationary
> | system..."
> | So x refers to the coordinate of an (arbitrary) event, which does not
> | necessarily take place at the origin of the moving frame!
>
> Wrong. x' = x-vt and x' is where the reflection takes place.

So far so good.

> That is NOT arbitrary.

Yes it is.

> It fixes x in K very precisely at the origin (0') of k.

No, it doesn't.
x' = x - vt
x = x' + vt
The values of x' and vt are independent of each other: x' could be 1, for
example, while vt = 2.
Then x = 3.
In other words, at the moment that the origin of k is at 2 (in K) the light
ray could be reflected at x=3 (in K), still following the same example, so
the coordinate of reflection in k is x' = x - vt = 3 - 2 = 1.

> You are obviously incapable of understanding Einstein's paper or basic
> algebra.
>
>
> |
> | > | I'd say it's at 0+vt in the stationary frame.
> | >
> | > That would depend on the direction of v and which frame you are
looking
> | > from.
> |
> | In the paragraph from which you are quoting it says: "Now to the origin
of
> | one of the two systems (k) let a constant velocity v be imparted in the
> | direction of the increasing x of the other stationary system (K) ..."
> | This means that the origin of the moving frame (system k) has an
> | x-coordinate in the other frame of x_0+vt.
>
> We don't call it x_0. We call it 0' to avoid obfuscation, and 0' = 0-vt
> because x' = x-vt is given. Also at t = 0, x' = x for all x, including 0.

x is a coordinate in K and x' is the corresponding coordinate in k.
Since system k is moving at a constant velocity v "in the direction of the
increasing x of the other stationary system (K)", the origin of k has moved
+vT at time t=T since time t=0 in system K. If the origins of the two
systems coincided at time t=0, an event that occurs at time t=T and
coordinate x in K will have a corresponding coordinate x' = x - vT in k
because the origin of k is at +vT in K: 0 + vT + x' = 0 + x

> You are obviously incapable of understanding Einstein's paper or basic
> algebra.
>
>
>
> | > If we use the convention that positive numbers monotonically increase
in
> | > value toward the right, then k moves with positive v in the K frame
and
> K
> | > moves with -ve in the k-frame, which is the PoR. A lot of people are
> | > confused by Einstein's definition about the laws of physics being the
> same
> | > in all frames, but really the PoR is simply reduced to the fact that
if
> | I'm
> | > flying in a plane, from my point of view the ground is passing beneath
> me
> | > and going backwards. v is a vector and therefore has direction.
> | >
> | >
> | > | But x=vt is only true in the special case where x refers to the
> | > stationary
> | > | coordinate of the moving frame's origin. In general x can refer to
> the
> | > | coordinate of any event in the stationary frame, in which case x' =
> x-vt
> | > | would refer to the coordinate of the same event in the moving frame.
> | > | Of course, this is all classical mechanics, as gamma is not
mentioned.
> | >
> | > Of course gamma is not mentioned. It hasn't been derived yet.
> | > Once you introduce gamma you are in violation of the PoR. Actually
> Eugene
> | > Shubert's derivation is based on the assumption that u and v (he uses
> | > u --actually mu -- and v to represent the ground beneath the plane
> moving
> | > backwards as seen for the air and and the plane flying overhead as
seen
> | from
> | > the ground) do not have equal magnitude. He claims
> | > (ref: http://www.everythingimportant.org )
> | > v = u/[1+u.a(u)] where a() is some as yet undetermined function.
> | > He is trying to prove SR without recourse to light's speed by assuming
> the
> | > PoR is invalid to being with. His argument is of course circular.
> | >
> | > | > So far so good.
> | > | > But then Einstein goes on...
> | > | >
> | > | > "From the origin of system k let a ray be emitted at the time tau0
> | along
> | > | the
> | > | > X-axis to x',"
> | > | >
> | > | > But as we've just seen, the origin of system k and x' are one and
> the
> | > | same.
> |
> | Nope. Your claim that x=vt is not valid in the quoted paragraph.
> | In this quotation x' simply refers to the coordinate where the emitted
ray
> | is reflected at time tau1.
> | Nowhere is stated, nor can it be deduced, that this occurs at the origin
> of
> | system k.
> Sure it can. x' = x-vt, so 0' = 0-vt.
> The origin of system k is therefore at -vt. However, when Einstein writes
> his equation:
> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
> he fails to use 0', so either he needs glasses or he has his frames mixed
> up.

You idiot! Einstein does not use 0 for the origin of K, nor 0' for the
origin of k.
He only uses 0 for the numerical value zero. He defined tau as a function
of x', y, z, and t.
So tau(0,0,0,t) simply refers to the value of tau at time t and coordinates
(x'=0, y=0, z=0) which happens to be the origin of k where the ray was
emitted!
Likewise, tau(x',0,0,t+x'/(c-v)) refers to the value of tau at time
t+x'/(c-v) and coordinates (x'=x', y=0, z=0) which is whereever the ray is
reflected (anywhere along the X-axis).

The rest of your arguments still fall like dominoes.

You don't even know how to read common function notation, or you need
(better?) glasses to read the function definition.
And you claim you understand it??? LOL!!!

> It should be (unarguably, since x is primed on the RHS)
> ½[tau(0',0,0,t)+tau(0',0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
>
> So:
> ½[tau(-vt,0,0,t)+tau(-vt,0,0,t+0/(c-v)+0/(c+v))] = tau(0,0,0,t+x'/(c-v))
> since the tip of the ray travels a distance vt.
> Hence
> ½[tau(-vt,0,0,t)+tau(-vt,0,0,t)] = tau(0,0,0,t)
>
>
>
>
>
> |
> | > | What do you mean "as we've just seen"? You omitted to define system
> k.
> | > | Or, since you are quoting Einstein, you omitted to qoute how he
> defined
> | > | system k.
> | >
> | > I do not expect to reproduce ALL of Einstein's text in a discussion
> here.
> |
> | No, but a link or at least the title and year of the publication would
> help.
> | Thanks for giving the link later on, but it wasn't clear to me you were
> | quoting from his original 1905 paper.
> |
> | > | You also omitted to specify from which publication you are
> | > | quoting, so I can't even check it myself.
> | >
> | > You say to me "You simply fail to understand the True Fundamentals of
> SR!"
> | > and you don't even know the paper it comes from?
> |
> | Do you expect me to know all his his publications by heart?
>
> Not all, but it was you that claimed "You simply fail to understand the
True
> Fundamentals of SR!" so I DO expect you to know "On the Electrodynamics of
> Moving Bodies" by heart, or at the least recognize it instantly.
> I do.
>
>
>
> | You could have been quoting from any of his papers, as you didn't tell
me
> at
> | first.
> |
> | > That is arrogance in the
> | > extreme, exceeded only by Dinky van de Torquemada the Deranged, who
said
> | > after Gerald Kellerher quoted Einstein (same paper),
> | > "What is this?
> | > Some kind of quote of some post?
> | > An introduction to the *** you produce later on?
> | > *** that you expect someone will bother reading?"
> | >
> | >
> | > Read it for yourself at
> | > http://www.fourmilab.ch/etexts/einstein/specrel/www/, and quit being
so
> | > lazy.
> | >
> | Thanks for the link this time. I'm not too lazy to read a paper myself
> when
> | you point me to it.
> | But don't expect me to find out which paper you are quoting from while
you
> | don't mention it.
> | >
> | > | Without knowing the full text of
> | > | the article from which you are quoting I can't even verify whether
the
> | > | last quotation relates to the special case where x' actually refers
to
> | > | the origin of the moving frame.
> | >
> | > So you admit you know nothing about the True Fundamentals of SR or
where
> | > they come from? But you are arrogant enough to claim I do not?
> |
> | There are many text books and internet sites from many authors from
which
> | one can learn SR and many of them are easier to read than Einstein's.
>
> Who cares about the tail wagging the dog? Certainly not me. How else is
> anyone going to get to "the true fundamentals of SR", which, it is now
> apparent, I understand and you do not.
>
> | On the other hand, knowing what Einstein wrote does not guarantee proper
> | understanding of SR.
> | Claiming that you spotted a fundamental error like zero-divide, which no
> | serious scientist would have spotted in 99 years is what I call
arrogance.
>
> You may call it what you wish, I call it proof. Arrogance is when you make
> claims such as "You simply fail to understand the True Fundamentals of
SR!"
> without understanding a word I've said.
>
>
> |
> | > | So, you have _not_ shown that the origin of system k and x' are one
> and
> | > the
> | > | same.
> | > |
> | > | Therefore, the rest of your arguments fall like an array of
dominoes,
> | > | unless you can demonstrate the truth of your last statement above.
> |
> | You still have _not_ shown that the origin of system k and x' are one
and
> | the same, so the rest of your arguments still fall like domonioes!
>
> Sorry if you cannot think or read, but that's your problem, not mine.
>
> |
> | > x' = x-vt... What happens when t = 0, Mr Lazy Incompetent Arrogance?
> | > does x' = x?
> | > Can I put two number lines side - by - side and let them sit there as
> time
> | > passes?
> | > Can I start a stop-watch as I slide one past the other?
> | > Can I sent a ray of light from the coincident origins to x'?
> | > Does x' move from x as the light travels?
> | > How far?
> | > Is it
> | > a) ct?
> | > b) vt?
> | > c) both?
> | > d) neither?
> | > Let us have your answer. Seems to me you don't know the True
> Fundamentals
> | > of anything.
>
> No answer?
> Cannot think or read?
> You are an arrogant prick who claims to know the True Fundamentals of SR!
> and incapable of providing an answer to a simple question.
>
>
> | >
> | > |
> | > | > The ray has zero distance to travel.(Or we could suppose that the
> | origin
> | > | of
> | > | > system k is at -vt, from 0' = 0-vt, but that doesn't appear in
the
> | > | equation
> | > | > that follows.)
> | > | >
> | > | >
> | > | > "and at the time tau1 be reflected thence to the origin of the
> | > | co-ordinates,
> | > | > arriving there at the time tau2; we then must have ½(t0+ t2) =
t1,"
> | > | >
> | > | > Well, yes. We have ½(0+0) = 0. So what?
> | > | > x' is not some point remote from the origin of k where the
> reflection
> | > | takes
> | > | > place, it is AT the origin of k.
> | > | >
> | > | >
> | > | > Einstein proceeds:
> | > | >
> | > | > ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] =
> | tau(x',0,0,t+x'/(c-v))
> | > | >
> | > | > (Note that 0' = 0-vt is missing)
> | > | >
> | > | > and takes partial derivatives. To do this Einstein says
> | > | >
> | > | > "Hence, if x' be chosen infinitesimally small, "
> | > | >
> | > | > but it is already zero!
> | > | >
> | > | > 1/2 [1/(c-v) + 1/(c+v)]dtau/dt = dtau/dx' + 1/(c-v) dtau/dt,
> | > | >
> | > | > Which with a little manipulation is
> | > | >
> | > | > 1/2 [1/(c-v) + 1/(c+v)]dtau/dt - 1/(c-v) dtau/dt = dtau/dx'
> | > | >
> | > | > dtau/dt (1/2 [1/(c-v) + 1/(c+v)] - 1/(c-v)) = dtau/dx'
> | > | >
> | > | > = dtau/d0
> | > | >
> | > | > = dtau/0
> | > | >
> | > | > and we have a divide by zero.
> | > | >
> | > | > Here's the full text.
> | > | >
> | > | > "If we place x'=x-vt, it is clear that a point at rest in the
system
> k
> | > | must
> | > | > have a system of values x', y, z, independent of time. We first
> define
> | > tau
> | > | > as a function of x', y, z, and t. To do this we have to express in
> | > | equations
> | > | > that tau is nothing else than the summary of the data of clocks at
> | rest
> | > in
> | > | > system k, which have been synchronized according to the rule given
> in
> | §
> | > 1.
> | > | >
> | > | > From the origin of system k let a ray be emitted at the time tau0
> | along
> | > | the
> | > | > X-axis to x', and at the time tau1 be reflected thence to the
origin
> | of
> | > | the
> | > | > co-ordinates, arriving there at the time tau2; we then must have
> | > | > ½(tau0+tau2) =tau1, or, by inserting the arguments of the function
> tau
> | > and
> | > | > applying the principle of the constancy of the velocity of light
in
> | the
> | > | > stationary system:-
> | > | > ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] =
> | tau(x',0,0,t+x'/(c-v)).
> | > | >
> | > | > Hence, if x' be chosen infinitesimally small,
> | > | >
> | > | > 1/2 [1/(c-v) + 1/(c+v)]dtau/dt = dtau/dx' + 1/(c-v) dtau/dt,"
> | > | >
> | > | > Anyone that knows anything at all about mathematics would know
that
> | > | division
> | > | > by zero is undefined.
> | > | >
> | > | > You claimed "You always omit some or other fundamental part in
your
> | > | attempts
> | > | > to "prove" a flaw in SR."
> | > | > What did I omit?
> | > | > Einstein omitted 0' = 0-vt when he said x' = x-vt. His equation
> should
> | > be
> | > | > ½[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/(c-v)+x'/(c+v))] =
> | > | tau(x',0,0,t+x'/(c-v)).
> | > | > QED.
> | > | > Androcles.
> | > |
> | > | You have not demonstrated anything. Your flaws started right at the
> | > | beginning.
> | >
> | > You arrogant prick.
> | >
> | > | I stated your omissions in my comments above. At best you only
> | > demonstrated
> | > | you interpret quotations out of context; an error which usually
gives
> | > | incorrect
> | > | results. So your most fundamental omission was probably that you
> failed
> | > to
> | > | see Einsteins arguments in context.
> |
> | This has now been confirmed.
> All you have confirmed is your arrogance. You failed to respond to the
> question I put above.
>
> |
> | > | Benno
> | > You are obviously incapable of understanding reasoned argument.
> | > Androcles.
> |
> | You are obviously incapable of understanding Einsteins paper.
>
>
> | Benno
>
> You are obviously incapable of understanding Einstein's paper or basic
> algebra, and arrogant enough to accuse me of not being able to.
> You are an idiot, not worth bothering further with.
> *plonk*
> Androcles.

Does that mean you will shut up now? Good, I was just getting tired of your
BS.

Benno

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