Re: Clear writing about relativity


"blackhead" <larryharson@xxxxxxxxxxxx> wrote in message
news:f451a01d-1366-4912-876c-e1b89d0cc46a@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 27 June, 16:23, "Androcles" <Headmas...@xxxxxxxxxxxxxxxx> wrote:
"blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

news:a229cf5c-6a99-46c5-a597-339c260cb4dc@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 27 June, 14:31, "Androcles" <Headmas...@xxxxxxxxxxxxxxxx> wrote:





"blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

news:6f288024-100f-4e75-bcd9-6f8deee2ee06@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 26 June, 23:03, "Androcles" <Headmas...@xxxxxxxxxxxxxxxx> wrote:

"blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

news:bd3efb77-0773-4490-87e5-ef8010bf120f@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 26 June, 19:32, "Androcles" <Headmas...@xxxxxxxxxxxxxxxx> wrote:

"Uncle Ben" <b...@xxxxxxxxxxx> wrote in message

news:4cb4f187-1a2c-4fc1-8e3c-189b7762313c@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Through an unsatisfactory exchange with colleagues
here recently I was reminded of the casualness of
expression of most of us who are not
mathematicians in using mathematical variables in
algebraic discussions of physical things.

Example: When I say, "Let x = vt, then ---", I
will be understood to mean one thing, until I go
on to say, "where x represents the mass of an
object of density v and volume t." Most physicists
in this newsgroup would be astonished and
displeased. A mathematician or logician wouldn't
care, but a physicist might insist on "M = dV" or
"M=\rho\ v.

Our conventions let us abbreviate our discouse and
remember our definitions. They cause no trouble in
simple cases.

But when we start talking about
several frames of reference and need symbols for
the coordinates in each, we have to improvise
symbols that fit our habits and yet distinguish
different versions of similar things. Nowdays we
use primed and double-primed variables, whereas in
earlier times when classical learning was assumed
among the intelligensia, we would use greek
letters or even hebrew or arabic letters.

Einstein's 1905 paper on relativity was translated
into english more than once with more than one
degree of accuracy. Some translations even
improved on the orginal by correcting small errors
or oversights. The paper is not difficult to read,
although what is said is quite unconventional to
the ordinary mind.

If we focus just on length contraction in Section
4, we find the derivation quite unfamilar to
students using modern textbooks.

Then that would be "Lorentz dilation", thus rendering your use
of the term "Lorentz Transformation" highly unsatisfactory;
indeed, it is deliberately and maliciously designed to deceive
the unsuspecting student. It should be made quite clear to the
newbies that Einstein's change in length INCREASES with
increasing speed, AS SHOWN algebraically.

But if we edit
Einstein's words, using memorable terms and modern
rigor to resolve normally insignificant ambiguities --
in the minds of naive readers -- we may help these
readers comprehend the astonishing simplicity of
Einstein's demonstration.

Yes, I agree. The naive student would have to be astonishingly
stupid not to comprehend the English translation of Einstein's
ridiculous
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif

Astonishingly simple (and simple-minded) is:
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same, spewed out in that inequality
which purports to be an equation. The "=" sign is a LIE.

Most relativists hasten to say "Einstein did not say that", but he
did and it is there in black and white for those that can read
algebra, which a prerequisite for relativity.

For an example of better choice of terms, let us
describe a sphere

*** the sphere! You have no 'gamma' without
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
you useless old fart.

You LOST, Bonehead. Squirming around piss-poor terminology
can't save you. The exchange was highly satisfactory and so is
rubbing your nose in your worthless ***.

Wanna fence some more? Bring a battleaxe next time and derive gamma.

"In the first place it is clear that the equations must be linear on
account
of the properties of homogeneity which we attribute to space and
time." - -Einstein

In the second place the function tau() is not linear.

A theoretical physicist wouldn't care, but a mathematician or
logician
will insist on a proof that the function tau() is linear.

A competent electronic engineer should know enough maths to understand
why homogeneity of space and time imples linearity of tau().

Androcles, victorious.- Hide quoted text -

- Show quoted text -

A schoolboy should know enough maths to understand a graph.
http://www.androcles01.pwp.blueyonder.co.uk/tAB=tBA.gif

Is this plot of tau against t linear? No.
Does x'/(c-v) = x'/(c+v) ? No.
Does tau[x'/(c-v)] = tau[x'/(c-v)]? No.
Does 1/2 tau[x'/(c-v) + x'/(c-v)] = tau[x'/(c-v)]? No.

In McCullough numbers and paraphrasing the idiot Einstein,
half of (16 second +4 seconds) = 16 seconds and the other half is 4
seconds.

1/2 tau(20) = tau (16), we'll just forget about the other tau(4)
seconds.

If tau (16) = 8, 1/2 tau(20) = 8 and tau(4) = 8
Hence tau(4) = tau(16)
Hence tau() is not linear.
Is x linear? Yes.
Is t linear? Yes.
Is v = x/t linear? Yes.
Is c = x/t linear? Yes.
That's good, because is clear that the equations must be linear on
account
of the properties of homogeneity which we attribute to space and time.

Is the function tau() linear? No.
Is t/tau linear? No.

Well, Larry thinks it should be.
Well, too bad, it fucking well isn't.
But Einstein said it must be. Why isn't it?
Because Einstein was not a competent electronic engineer,
he was a competent bullshitting ***!- Hide quoted text -

Write down what the Tau you're referring to, so I can have a look at
it.
===============================================

This tau(), boy:
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif

which is this tau, boy, when the length x' is reduced to nothing:

"Hence, if x' be chosen infinitesimally small,"
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img23.gif
"or"
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img24.gif

Just which of Einstein's many tau's were you planning on taking a look
at?
Read all about it here:
http://www.fourmilab.ch/etexts/einstein/specrel/www/

These expressions don't say anything about whether Tau is linear or
not.

============================================

A schoolboy should know enough maths to understand a graph.

THIS GRAPH!
http://www.androcles01.pwp.blueyonder.co.uk/tAB=tBA.gif

I'm not sure what you're doing. Your diagram shows B hitting the end
of the larger rod, it remains there with C continuing to move forward
so that the smaller rod and the clock on it becomes compressed. I
think this is why your plot of Tau against t s wrong because you seem
to have interpreted Einstein's derivation and hence Tau incorrectly.

=================================================
You mean you are not sure what Einstein was doing.
Let's try it this way and see if you have any sense at all.
http://tinyurl.com/c53cqo




Is this plot of tau against t linear? No.
Does x'/(c-v) = x'/(c+v) ? No.
Does tau[x'/(c-v)] = tau[x'/(c-v)]? No.
Does 1/2 tau[x'/(c-v) + x'/(c-v)] = tau[x'/(c-v)]? No.

It's your interpretation of Tau, which doesn't seem to be correct.
============================================
It's Einstein's interpretation of tau, and you are definitely incorrect.






In McCullough numbers and paraphrasing the idiot Einstein,
half of (16 second +4 seconds) = 16 seconds and the other half is 4
seconds.

1/2 tau(20) = tau (16), we'll just forget about the other tau(4) seconds.

If tau (16) = 8, 1/2 tau(20) = 8 and tau(4) = 8
Hence tau(4) = tau(16)
Hence tau() is not linear.
Is x linear? Yes.
Is t linear? Yes.
Is v = x/t linear? Yes.
Is c = x/t linear? Yes.
That's good, because is clear that the equations must be linear on account
of the properties of homogeneity which we attribute to space and time.

Is the function tau() linear? No.
Is t/tau linear? No.

Well, Larry thinks it should be.
Well, too bad, it fucking well isn't.
But Einstein said it must be. Why isn't it?
Because Einstein was not a competent electronic engineer,
he was a competent bullshitting ***!

Do let us know when you've learnt to read, boy.

============================================

They are additional properties of Tau. Tau being linear wrt t and
x means Tau(x,t) = Ax + Bt where A and B are constants and the
additional properties help narrow down what the constants A and B are.

============================================
tau(bananas, dog's breakfasts, railway time tables, x, y, z, t) is a time.
Oddly enough, the bananas, dog's breakfasts and railway time tables
vanish along with x, y and z when

"Hence, if x' be chosen infinitesimally small,"
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img23.gif

See, that inequality has no mention of bananas, dog's breakfasts,
railway time tables, x, y or z, and x' is infinitesimally small.
Why not?
Could it be they are irrelevant, put there by the idiot Einstein
to confuse poor Larry?
Oh wait... Einstein only put the x,y and z in it. Naughty Androcles
added the other irrelevancies.

Eisntein looks at the variation of Tau wrt each of the independent
variables one at a time to get a partial differential equation for
each. Using your example

Where @ is the partial symbol

@tau/@bananas = @tau/@dog's breakfasts = @tau/@railway time tables = 0

That's right.
And
@tau/@y = 0,
@tau/@z = 0.

x' is infinitessimally small, we can't compute [f(x'+h)+f(x')]/h because h
is
infinitessimally small also.

That means we have to compute the inverse velocity, 1/v = @t/@x

[f(t + h) - f(t)]/x' = @t/x' (which is NOT @t/@x', because x' is already
@x).

You are not sure what Einstein was doing. Einstein didn't have a fucking
clue
what he was doing.



Let us try this again, perhaps it will sink in:

A schoolboy should know enough maths to understand a graph.

THIS GRAPH!
http://www.androcles01.pwp.blueyonder.co.uk/tAB=tBA.gif

Your graph doesn't make sense to me.
====================================
Oh, it's simple enough.

"we establish by definition that the ``time'' required by light to travel
from A to B equals the ``time'' it requires to travel from B to A" --
Fuckwit Einstein.

1) The t-time is plotted along the horizontal axis from 0 to x'/(c-v).
That's the time it takes for light to get from A to B, a distance of x'.

2) The tau-time is plotted vertically.
That's the "time" it takes for light to get from A to B, a distance of x'.

NOTE that "time" is in quotation marks. That's very important, pure
Einstein.

3) the t-time along the horizontal axis continues from x'/(c-v) to
x'/(c-v)+x'/(c+v).
That's the time it takes for light to get from A to B and back again, a
distance of 2AB
or 2x'.

"In agreement with experience we further assume the quantity

2AB/(t-A-tA) = c
to be a universal constant--the velocity of light in empty space." --
Fuckwit Einstein


Do you see the word "assume" in there? Ass-u-me makes an ass out of -u- and
you,
but not me.


4) the tau-time (vertical height) doubles.

That's the "time" it takes for light to get from A to B and back again.

Why is it doubled?
"In agreement with experience we further assume... "
Nobody knows what this experience is, but it sounds good.

You must be an incredibly stupid ass not to understand that x'/(c-v) is
greater than x'/(c+v) and assume... oops...assuyou... that Einstein
was anything but an idiot making an ass of -u- but not me.

This competent electronic engineer knows enough maths to understand
why homogeneity of space and time does NOT imply linearity of tau(), BOY.


A schoolboy should know enough maths to understand a graph.
Just which of Einstein's many tau's were you planning on taking a look at?




Is this plot of tau against t linear? No.
Does x'/(c-v) = x'/(c+v) ? No.
Does tau[x'/(c-v)] = tau[x'/(c-v)]? No.
Does 1/2 tau[x'/(c-v) + x'/(c-v)] = tau[x'/(c-v)]? No.

One more time:

A schoolboy should know enough maths to understand a graph.

THIS GRAPH!
http://www.androcles01.pwp.blueyonder.co.uk/tAB=tBA.gif

Is this plot of tau against t linear? No.
Does x'/(c-v) = x'/(c+v) ? No.
Does tau[x'/(c-v)] = tau[x'/(c-v)]? No.
Does 1/2 tau[x'/(c-v) + x'/(c-v)] = tau[x'/(c-v)]? No.

"In the first place it is clear that the equations must be linear

read more »- Hide quoted text -

- Show quoted text -...


.