"The Missing Symbol' (the second of three parts)
glird
Posting Two
The mathematical symbol that's missing from Einstein's 1905 paper is
shown here, and what it physically denotes is explained. It is then
proved that because it is missing, his fourth and fifth equations are
defective. The explanation of how he arrived at Equation 3 and why he
deleted the missing symbol is then begun. It will be completed in
Posting Three.
_________
The third equation (Eq 3) in Einstein's 1905 paper is,
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})
in which x' denotes the distance between clocks A and B of the moving
system k (xi, eta, zeta; tau) as measured by the stationary system K
(x, y, z; t). In order to examine the meaning of this and allied
equations we will let the clocks be at the ends of unit-rod r'AB of k,
so r'AB = xiB - xiA = 1 is the length of the path AB as measured by k.
If there is no physical change of lengths in a moving system, then
rAB = r'AB. The length of the path as measured by the two systems will
then be
rAB = xB - xA = r'AB = xiB - xiA = 1.
Accordingly, delta xi/delta x = r'AB/rAB = x' = x - vt = 1, in which
delta xi/delta x (the missing symbol) denotes the ratio of size of
units of length of the two systems in the X direction, and as of here,
x' = 1 is the length of the moving rod as measured by stationary
system K. Given "that the equations must be linear",
delta xi/delta x = dxi/dx = r'AB/rAB = 1.
_____
The fourth and fifth equations in Einstein's paper are
.5(1/{c - v} + 1/{c + v})delta tau/delta t
= delta tau/delta x' + (1/{c - v})delta tau/delta
t Eq 4
and
delta tau/delta x' + [v/(c2-v2)]delta tau/delta t =
0. Eq 5
It will now be shown that in terms of the values imposed by
Einstein's treatment to there, those two equations are defective.
(For simplicity, d will be substituted for delta.)
Let the velocity of system k be .6c in the X direction of system K.
Let end A of rod r'AB be at xi = x = 0 and end B at xi = 1 when a ray
emits from the coinciding origins at t = 0.
Letting x' = x - vt = 1 denote the length of the moving path AB as
measured by the stationary system K, it will take the ray
t1 = AB/(c - v) = x'/(c - v) = 1/.4 = 2.5 seconds
to reach end B; which will have moved vt = 1.5 to the right thus will
be at x = 2.5 when the ray reaches it. The return trip will take
t2 = AB/(c + v) = x'/(c + v) = 1/1.6 = .625 seconds.
The total time as measured by K will therefore be
tr = t1 + t2 = 1/(c - v) + 1/(c + v) = 3.125 seconds.
Putting these values into equation 4, the total time of the round-trip
as measured by the moving system k will be
taur = (1/{c - v} + 1/{c + v})dtau/dt = 3.125dtau/dt.
The return trip from B back to A will be
tau2 = dtau/dx' + (1/{c + v})dtau/dt = dtau/dx' + 2.5dtau/dt.
In order to obey the postulates that c = 1 as plotted by k, and that
the time each way must equal half the total time of the round trip, it
is required that
tau1 = .5(1/{c-v} + 1/{c+v})dtau/dt = (3.125dtau/dt)
= tau2 = dtau/dx' + [1/(c+v)]dtau/dt
= dtau/dx' +2.5dtau/dt
= 1;
thus that
.5(3.125dtau/dt) = dtau/dx' + 2.5dtau/dt = 1.
Since .5(3.125dtau/dt) = 1.5625 . dtau/dt = 1, it follows that
dtau/dt = 1/1.5625 = .64 = c2 - v2 = Q,
in which dtau/dt is the rate at which moving clocks beat compared to
the rate of stationary ones.
If we use dtau/dt = Q = .64 in dtau/dx' + 2.5dtau/dt = 1, we get
dtau/dx' + 2.5 x .64 = 1
wherefore
dtau/dx' = -.6,
in which dtau/dx' denotes the difference in the setting of one k clock
compared to another k clock that is x' away from it in the direction
of motion as measured by K.
Given that xi = x' = x - vt = 1, then dtau/dxi = dtau/dx' = -.6.
Accordingly, the total time plotted by the moving system will be
tauR = 2AB/c = 2 seconds
and the one way time will be
tau1 = tau2 = AB/c = 1 second,
which is exactly what postulate 2 requires. Similarly, since the ray
reached clock B at tau = 1 and returned to clock A at tau = 2,
equations 4 and 5 are also satisfied.
We thus see that if dtau/dt = c2 - v2 and successive clocks of a
moving system are offset by -vxi/c2 seconds compared to each other in
their direction of motion, everything Einstein postulated to here is
satisfied. Why, then, do I assert that equations 4 and 5 are
defective?
It's because they rule out the LTE, in which
dtau/dt = 1/(c2 - v2)1/2 = 1/q; not c2 - v2 = Q
and
dxi/dx = 1/q; not 1 (by omission)!
Demo: If we use the LTE's dtau/dt = 1/q = 1.25 in Eq 4, we get
.5(1/{c - v} + 1/{c + v})dtau/dt = dtau/dx' + (1/{c - v)dtau/
dt
= .5(2.5 + .625)1.25 = 1.953125 = -.6 + 2.5 . 1.25 = 2.525.
Using the LTE value in Eq 5, we get
dtau/dx' +(v/c2 - v2)dtau/dt = 0
= -.6 + (.6/.64)1.25 = -.6 + 1.17875 = .571875.
Therefore if we use the LTE value, equations 4 and 5 would both be
wrong! Furthermore, as we will now show, the symbol dtau/dt is
incorrectly used in Einstein's equations.
If a time interval is 5/3 seconds of t-time and rates of k run slow
by Q = .6, then tau = Qt = .6(5/3) = 1 shows that "one second" of k
time will be larger than "one second" in K time, so dtau/dt = 1/Q.
Therefore, if we let that interval be 1.5625 K seconds, and let Q = .
64, then in which
t(dtau/dt) = t/(1/Q) = (dt/dtau)t = Qt = 1,
in which
dt/dtau = Q = 1/(tau/dt).
(If anyone now objects that (dt/dtau)t disagrees with the rules of
calculus, the answer is: Einstein wasn't doing calculus there, he was
doing algebra. If they reply, "Then why, other than as a way to
introduce calculus into his next few equations, did he set x'
'infinitesimally small'?" the answer is: By doing that, he let x' be
smaller than the distance between any two k-clocks; thus delta tau
represented a time interval beat off by one k-clock, compared to the
same interval as plotted by a K clock. In effect, he thereby compared
the rate of the "general time" of k to that of K. Because calculus
blindfolds its practitioners from understanding the physical meaning
of their own equations -- which always equal zero "in the limit" --
most mathematicians will probably disagree. If any do, ask them.
"What is the relative size of one second of k, compared to one second
of K, in the equation, delta tau/delta t = Q?" If they give the right
answer, "dtau equals Q when dt equals 1", ask them to explain how one
second of k could be smaller than a K second, if clocks of k run
slower than those of K. (If k clocks ran faster then K clocks, one
second of k would be smaller then one of K, but neither the LTE nor
Einstein said that moving clocks speed up!)
Subtracting the right side of Eq 4 from its left side gives us
½(1/{c-v} + 1/{c+v})delta tau/delta t
- [delta tau/delta x' + (1/{c-v})delta tau/delta t] = 0.
Letting v = .6c and x' = .5, we get
.5(.5/.4 + .5/1.6)dtau/dt
- (dtau/dx' = tau - vxi/c2 = - .3) - (.5/.4)dtau/dt = 0,
i.e. .5(1.25 + .3125)dtau/dt + .3 - (1.25)delta tau/delta t = 0,
so (.78125 - 1.25)dtau/dt = -.3
i.e. (.46875)dtau/dt = .3,
wherefore
dtau/dt = .64 = c2 - v2.
If we let x' = .4, we get
.5(.4/{c-v} + .4/{c+v})dtau/dt
- [dtau/dx' + (.4/{c-v})dtau/dt] = 0
.5(1 + .25) - 1]dtau/dt - dtau/dx' = 0,
i.e. .375dtau/dt = .24,
wherefore
dtau/dt = .64 = c2 - v2.
We thus see that the value of dtau/dt (thus of delta tau/delta t) is
independent of the value of x'.
Indeed, the very fact that no one seems to have realized that the
only value of delta tau/delta t that allows equations 4 and 5 to be
correct is c2 - v2 proves that neither Einstein nor anyone else
understood the equations in his paper. (If they did, they would have
known _and publicized_ the fact that his derivation equations rule out
the deformations required and imposed by the LTE.)
I will now show how Einstein really obtained equations 4 and 5; and
then got equation 3 from them; and why he didn't understand what some
of their symbols mathematically represent or physically mean. To do
that, I will resurrect some of his initially submitted paper ("P1"
herein) including the relevant mental experiments he deleted from the
published version ("P2"). Along the way I will use d in place of
delta [and will add some new things in a different font, [as here.]
______________
Roused by the pulsating beat and now that they know someone is
listening at last, the P1 equations have begun to stir, to rearrange
themselves, to talk to me; an excited babble of eager voices anxious
to be heard. The living remnants of P1, splattered all over the bloody
pages of P2, are ready to come together again. In the following
reconstruction of Einstein's initial paper we will watch them be
reborn.
P1
ON THE ELECTRODYNAMICS
OF MOVING BODIES
In terms of Maxwell's equations the velocity of light is a constant
with respect to a luminiferous medium at rest in Newton's stationary
space. Therefore a ray of light should pass differently moving
systems at correspondingly different velocities. The Michelson-Morley
and allied experiments over the past few decades seem to disagree with
that conclusion.
Those unsuccessful attempts to discover any motion of Earth prompted
Poincare' to suggest that a new Physics should be developed in which
the velocity of light would be the same in all systems regardless of
their inertial motions. We will accept that as a postulate, namely,
that light is always propagated with a measured velocity c, which is
independent of the state of motion of the emitting body.
The theory to be developed is based on kinematics, since the
assertions of any such theory have to do with the relationships
between material bodies, clocks, and electromagnetic processes.
Insufficient consideration of these circumstances lies at the root of
the difficulties which the theories of the electrodynamics of moving
bodies encounter [2009 and counting].
1. ELECTRODYNAMICAL PART
§ 1. On the Nature of the Electromotive Forces
Occurring in a Moving System
Starting with the fundamental equations based on Lorentz's theory of
electrons, we will now examine the physical changes an atomic system
of reference undergoes as a consequence of its inertial motion. We
will assume that ponderable matter is made of separate atoms co-moving
through the luminiferous medium at rest in stationary space. We will
choose this space as our frame of reference. [Maxwell postulated that
the luminiferous medium, his "aether", consisted of tiny particles
suspended from each other by elastic springs. Other than for local
vibrations, they were taken as statistically at rest in Newton's
"empty space".]
Let there be a material system K (x, y, z) at rest in stationary
space, provided with a unit-long measuring-rod and a number of clocks.
Let the Maxwell-Hertz equations hold good so that we have
dX/cdt = dN/dy - dM/dz; dY/cdt = dL/dz - dN/dx; etc
dL/cdt = dY/dz - dZ/dy; dM/cdt = dZ/dx - dX/dz; etc Eqs 1
where (X,Y,Z) denote orthogonal components of the electric vector and
(L,M,N) those of the magnetic force.
Now let D be the dielectric displacement in empty space, H the
magnetic force, p the volume-density of the charge of an electron, v
the velocity of a point of such a particle, and F the repulsive
ponderomotive force on an electron. Then, as measured by stationary
Euclidean K at a time t,
div D = p, div H = 0,
curl H = (dD/dt + pv)/
c,
curl D = - dH/cdt,
Eqs 2
F = D + (v H)/c.
If these equations are referred to a physical system moving with
velocity v, to which co-ordinate system K (cs K) is now attached, and
if an electron has an additional local velocity u, then for that
electron,
vx = v + ux, vy = uy, vz = uz.= .6c,
and Eqs 2 become:
div D' = p', div H' = 0,
dH'z/dy - dH'y/dz = (1/c)(d/dt - v{d/dx})D'x
+ (1/c)p'(v + ux), etc,
F'x = Dx + (uyHz - uzHy)/c, Eqs 2'
F'y = Dy - [vHz + (uzHy - uxHz)]/c,
F'z = Dy - [vHy + (uxHy - uyHx)]/c
We will now consider the strength of these forces within the
moving system.
Y
d
. |
. |
. | ---> .6c
c b
. |
X p________a____ |
In the diagram the referent system and attached cs K are moving in
the X direction at a velocity of v = .6c. A spherically expanding
force, initiated at p when the origin of co-ordinates was there, rises
upon the moving Y axis; starting from p and reaching point d at t = 1.
Line a, which is vt = .6 units long, denotes the distance the Y axis
has moved in that one second. Line b denotes the distance the force
traveled on Y. Line c represents the length of the path of that
component of the force relative to a stationary system.
The Pythagorean Theorem gives us the length of line b, which is c2 =
b2 + (v/c)2. Therefore b = (c2 - v2/c2)1/2 = q. The relative
velocity of the force-ray upon Y is then given by the ratio of lengths
of lines b to c; i.e. by comparing (length b=q)/(t=1) to (length c=1)/
(t=1). Letting c'y,z denote the velocity of this force on the
perpendicular axes of the moving system, this yields c'y,z = (c2 - v2/
c2)1/2.
The average relative velocity of a radiating electromagnetic force
in an apparatus attached to Earth would therefore be
c'x = (c2 - v2) = Q
in the axis of Earth's motion, and the relative velocity in the
perpendicular axes would be
c'y,z = (c2 - v2)1/2 = q.
Accordingly, such forces will have traveled a different distance per
second to x or y or z than they would in a stationary system. The
repulsive forces between the particles will therefore be
correspondingly weaker at all such points.
Let the electrons be locally stationary so u = 0, and the strength
of the repulsive ponderomotive force in the moving atomic system to
which cs K is attached becomes
F'x = (c2 - v2)Fx = QFx,
F'y = [(c2 - v2).5]Fy = qFy,
F'z = qFz.
Because a ponderable material point can be made into an electron (in
our sense of the word) by the addition of an electric charge, no
matter how small; and the repulsive force becomes Q weaker in the X
directions and q weaker in the Y and Z directions, the particles so
affected will end up closer together than they were when their system
was stationary.
A point at the right end of a unit rod lying on X, whose left end is
at x = 0, would be at x = Q when plotted by attached Euclidean cs K.
If another such rod was lying on Y or Z with its bottom end at the
origin of the cs, its top would be at y or z = q. A rigid body of no
matter what shape will therefore be shortened in the ratio of [(c2 -
v2) : 1] = Q in the direction of its motion, and by [(c2 - v2).5 : 1]
= q in the perpendicular axes.
These Q, q, q length-contracted moving systems are the ONLY group
for which no rate change is needed in order to hold the round-trip
speed of light equal to c. [As Lorentz said in 1904, "the only effect"
of the translation must have been a contraction of the whole moving
system of particles.]
We will now demonstrate that in M&M type experiments the amount of
time required for a ray of light to roundtrip the legs of the
apparatus would be identical in all directions. Let the length per
leg be l = rAB = 1 unit when the apparatus is stationary in the ether
and let "one second" be the time it takes a ray of light to travel
that far in a vacuum. Now let the apparatus be given a velocity of .6c
in the X direction. Because of the force changes, l will shrink to c2
- v2 = Q units long in that direction and to (c2 - v2)1/2 = q units
long in the Y or Z direction.
Let a ray of light now emit from A toward B per rod, and reflect
back to A. In the X directed leg it will take
l/(c-v) + l/(c+v) = 2Q/Q = 2 seconds
for a round-trip. In a Y or Z directed leg it will take
l/q = q/q = 1 second
each way, thus 2 seconds for a round-trip.
Therefore the rays would take identical amounts of time to perform a
round-trip in any direction in the apparatus. That's why there were
no interference patterns even though the speed of light is variable
relative to the apparatus and despite the fact that the Earth was
moving relative to the luminiferous medium at rest in stationary
space.
§ 2. On the Relativity of Co-ordinates and Times of a
System in Uniform Motion in Stationary Space
Let there be given a stationary rigid rod rAB and let it be 1 unit
long as measured by a measuring rod which is also stationary. We now
imagine the axis of the rod lying along the axis of x of the
stationary system of co-ordinates, and that a uniform motion of
translation with velocity v = .6c in the direction of increasing x is
then imparted to the rod. We now inquire as to the length of the
moving rod, and imagine its length to be ascertained by the following
two operations:-
(a) The observer moves together with the given measuring rod and the
rod to be measured, and measures "the length of the rod" by directly
superposing the measuring rod, in just the same way as if all three
were at rest.
(b) By means of synchronized clocks set up in the stationary system
the observer ascertains at what points of the stationary system the
two ends of the moving rod are located at a definite time, and
measures the distance between these two points, which is a length
which may be designated as "the length of the (moving) rod", by
directly superposing the measuring rod which in this case is
stationary.
Current kinematics tacitly assumes that the lengths determined by
these two operations are precisely equal, or in other words, that a
moving rigid body may in geometrical respects be perfectly represented
by the same body at rest in a definite position. This we shall now
determine and we shall find that the length of the moving rod found by
method (a) differs from the length of the same rod found by method
(b).
Let rod rAB be moving at v = .6c. We will now examine the length of
this rod as measured via methods (a) and (b).
(a) Since this method lets the measurement be done by a co-moving
thus identically deformed measuring rod, it will be one unit long when
measured this way.
(b) The moving rod initially extends from
x = 0 to x = c2 - v2 = .64
as plotted by observers on Euclidean system K at t = 0. At any later
time the two ends will have moved identical distances and the rod
would remain .64 units long no matter what the value of t might be.
Since the length of a moving rod is different when measured via
method (a) then it is via method (b), we see that the measured length
of a given rod depend on how it is measured.
As will now be demonstrated, the time it will take a ray to traverse
a moving rod depends on the direction and speed of the rod's motion
relative to the measuring system even though the round-trip time is
identical in all directions.
Let a unit-rod rAB be given a velocity of .6c and it will shrink to
Q = .64 units long in the direction of v. It will take a ray
t1 = AB/(c - v) = .64/.4 = 1.6 seconds
for the outbound leg in that direction and
t2 = .64/1.6 = .4 seconds
for the return leg.
Although the length of the moving rod measured by a stationary
system is .64 units, it will remain 1 unit long as measured by the co-
moving system, via method (a). Measured by the moving system, the
outward trip will therefore take
t1 = 1/c' = 1.6 seconds
and the return trip will take
t2 = 1/c' = .4 seconds.
Since c' = distance/time, the moving system would find that
c' = 1/1.6 = .625c
in the outbound leg and
c' = 1/.4 = 2.5c
in the return leg.
In the perpendicular legs the relative velocity of light is q, so it
will take t = q/q = 1 second each way. The co-moving system would
measure this as t = 1/c' = 1, wherefore c' = c each way in those
directions.
Hence we see that as measured by differently moving systems, the
time is relative to the speed and direction of motion of the viewed
system; and so is the one-way speed of light.
____
We will now examine what would happen if someone invents a way to
measure the one way speed of light, and uses the relative speeds of
light, c - v and c + v in the axis of motion, to discover the value of
Earth's velocity, v.
Suppose unit-rod rAB is at rest on X of system K which is moving in
the +x direction at .6c. Imagine that at ends A and B of the rod
clocks are placed and that with each clock there is an observer.
Imagine further that these observers synchronize the clocks by setting
their indications to be identical with those of a stationary system
k' (x', y', z').
Let a ray of light depart from A at the time tA and let it be
reflected at B at the time tB and reach A again at the time tA'.
Taking into consideration the constancy of the velocity of light in
the stationary system and the dx/dx' = Q contraction of the moving rod
we find that
tB - tA = dx/dx'/(c - v) = .64/.4 = 1.6
and tA' - tB = dx/dx'/(c + v) = .64/1.6 = .4
in which dx/dx' denotes the ratio of size of units of length of the
two systems as measured by stationary system k'. Therefore
tB - tA =/= tA' - tB.
[As shown now, we'd reach the same conclusion even if there were no
deformations of a moving rod, wherefore dx/dx' = 1. We'd find that
tB - tA = 1/(c - v) = 2.5 and tA' - tB = 1/(c + v) = .625
wherefore tB - tA =/= tA' - tB.
From tB = .64/(c - v) = 1.6 we get .64 = 1.6(c - v) and from that we
get
c - v = .64/1.6 --> v = c - .4c = .6c.]
Therefore the value of Earth's inertial velocity could easily be
found and so could the relative velocity of light:
c' = c +/- v =/= c.
Our first postulate would thus be proven false!
In order to overcome this difficulty we will now postulate that the
following definition of "synchronous" is universally applicable.
Clocks of a given system are synchronous if the "time" required by
light to travel from A to B is equal to the "time" it requires to
travel from B to A.
Using Poincare's method of adjusting clocks, the observers on the
moving system now reset their clocks in accord with this definition.
To do that, they turn their clock settings back by -vx/c2 seconds, in
which x is the distance between two clocks as measured by the moving
system itself. Accordingly, when the origin clock registers t, a
clock at each successive point on X will indicate that t_x = t - vx/
c2, in which v is the velocity of the given system.
When setting their clocks by this method, the observers don’t need
to know the velocity of their system in the luminiferous ether, nor
the amount of length or rate deformations. They merely set their
clocks to whatever time is needed in order to obey this operational
definition.
Demonstration: Let t denote the time of one clock and t' denote the
time of another clock of the same system. The observers on any given
system, moving or not, set their clocks so that t' = t - vx/c2, in
which x is the distance between the clocks as measured by the given
system itself.
Because the time per clock of a moving system may differ from that
of a stationary one at the same point, we will hereafter let t' denote
"the 'time' of the moving system".
Let a ray start at t’A from A towards B, arrive there and be
reflected at t’B, and arrive back at A at t’A'. In accordance with
the above definition, the two clocks are synchronous if
t’B - t’A = t’A' - t’B.
We assume that this definition of synchronism is free from
contradictions and possible for any number of points, and that the
following relations are universally valid:-
1. If the clock at B synchronizes with the clock at A, then the clock
at A synchronizes with the clock at B.
2. If the clock at A synchronizes with the clock at B and also with
the clock at C, the clocks at B and C also synchronize with each
other. Clocks that have been set this way are called "esynched".
Let an event occur at A at t'A as marked by the observer at A and
another event occur at B at t'B as marked by the observer at B. We
now establish by definition that these events are "simultaneous" if
t'A equals t'B.
Let A occur at the origins of the two systems when they coincide at
tA = t'A = 0. Let B simultaneously occur at x' = 1. The stationary
system will plot them as occurring at t = 0 in both places. As
determined by the esynched moving system K, however, t'B = t'A - vx/c2
= 0 - .6. = -.6. Hence, as measured in the two systems,
tA = tB = t'A =/= t'B .
Therefore system k' would find the events simultaneous, while the
moving system K would declare them non-simultaneous.
So we see that we cannot attach any absolute significance to the
concept of simultaneity, but that two events which are simultaneous,
as viewed from a stationary system, can not be looked upon as
simultaneous events when plotted by an esynched moving system.
Thus with the help of these imaginary experiments we have settled
what is to be understood by “esynched” clocks located at different
places in a given system, and have evidently obtained a definition of
"time" and of "synchronous" and of "simultaneous".
In physics, the "time" of an event is that which is given
simultaneously with the event by a clock located at the place of the
event.
zzz Whether or not the events occurred at the same instant, if they
happen at the same time, as marked by esynched clocks, they are called
"simultaneous" if
tB - tA = tA' – tB.
wo events are called “simultaneous” if the time at those events is
identical as marked by esynched clocks of a given system.
We will prove that our first postulate is now valid in all
directions. Let the moving system send a ray from A to B of its unit-
rod rAB, whose clock A is at the origin and clock B at x = 1; and let
it be reflected back to A.
It will take the ray rAB/(c - v) to reach B and rAB/c + v) for the
return trip. Since rAB is Q light-units long [whether or not it is
measured by any system], the total time will be
(1/{c - v} + 1/{c + v})dx/dx' = 2rAB/(c2 - v2)
= 2Q/Q = 2 seconds,
in which the amount of time it took to reach clock B is
tB - tA = rAB/(c - v) = .64/.4 = 1.6 seconds.
Cs K would now measure this outbound time as taking
t'B = t'1 = [1/(c -v)]dx/dx' + dtau/dx
= t + -vx = 1.6 - .6 = 1 second.
The return trip will take
tA' - tB = t2 = Q/(c + v) = .64/1.6 = .4 seconds.
Cs K would now measure this as taking
t'2 = t + vx = 1 second.
In the axis of motion, therefore,
t'A' = t'1 + t'2 = 1 + 1 = 2 seconds.
On a q-contracted Y or Z rod a ray will take
tB =.8/.8 = 1 second
each way, so tB = 1 and tA' = 2.
Since cs K finds its own rods 1 unit long in all directions, it
would find that
c' = 2rAB/(t'A' - t'A) = 2units/2sec = 1 unit/second = c.
Hence, in agreement with our first postulate, both the one-way and two-
way speed of light now remain c as determined by any esynched system;
whether or not the system is moving.
We thus see that the introduction of a “luminiferous ether" has now
become superfluous, since the view herein developed does not require
an “absolutely stationary space” provided with special properties, nor
assign a velocity vector to a point of the space in which
electromagnetic processes take place. ANY point may now be chosen as
referent and any point B as target; and the velocity of light, c, will
remain 1 unit/second in all directions as determined by an esynched co-
ordinate system attached to any frame of reference, whether or not it
is stationary.
_____
Although P1 would have continued, and might have contained the other
two articles that were published in the same journal, I will now
interrupt it in order to show what happened when Poincare's Sur le
Dynamique arrived and Einstein ran home to put its LTE into P1.
First, he deleted everything about a Q, q, q change in lengths.
That allowed the typesetters to change his proof copy with a minimum
of trouble. It also gave him some room to add an entirely new section,
which became P2's §3.
Then, when he realized that the + vt in Poincare's equation, "x' =
beta(x + vt)", means that the viewed system is moving to the left at -
v, rather than to the right at +v, as in P1's treatment; he switched
from k' as the stationary system on which K moved at v, to K (x,y,z)
as "the 'stationary' system" on which k (xi, eta,zeta) moved at v.
In Posting Three I will show what happened next, and will then prove
the following things:
1. His statement, "the equations must be linear on account of the
properties of homogeneity we attribute to space and time" is
nonsense. The real reason for the linearity of the equations, thus
why delta xi/delta x = dxi/dx, is that the ratio of lengths for a
given velocity is constant. (The value of dxi/dx may be a function of
v, in which case it will change if v does, but it remains constant for
a given value of v.)
2. In light of his treatment in P2, where he had substituted a for
delta tau/delta t, his statement that "a is a function phi(v) at
present unknown" proves that he didn't understand that in Poincare's
treatment as well as his own, phi(v) denotes the ratio of lengths in
the perpendicular axes Y and Z; rather than the ratio of rates of the
two systems.
3. Although he may have borrowed the phrase from Lorentz's 1904
paper, his appeal to "reasons of symmetry" as a way to prove that
phi(v) = 1 was useless. That proves that he didn't understand either
Lorentz's use or his own.
4. Since P2 has no symbol for the ratio of lengths in the axis of
motion, and since the LTE demand that dxi/dx = q (or 1/q) it is clear
that his derivational equations along the way are inadequate. Indeed,
by omission the value of dxi/dx is unity, rather than the Lorentzian
value he ended up with.
5. Prior to writing Poincare's LTE, he wrote "c2 - v2"; but after
inserting Poincare's equivalent expression, "1 - v2/c2", he used that
form everywhere but once in the rest of his paper. This proves two
things:
a: He copied the LTE and the value of beta directly from Poincare's
Sur la Dynamique.
b: The fact that his prior expression appears one more time, many
pages later, proves that he pasted it there while revising the entire
proof copy of his paper.
Taken together, these things prove that in 1905 Einstein, who was a
3rd class clerk yet to be accepted into college, didn't understand his
own mathematics, let alone that of Lorentz and Poincare'; nor,
therefore, the physical meanings of his own equations. The fact that
no one heretofore ever recognized these things proves that no one else
did either.
Fortunately, as time passed and he did go to college and learn how
to do calculus and other forms of mathematics, he never stopped trying
to understand the physical meanings imposed by his and everyone else's
equations.
He came closer than most people to doing exactly that!
Dono.
> The Missing Symbol
> idiocies snipped <
Still slurring Einstein, shithead?
Uncle Al
>
> The Missing Symbol
> Posting Two
>
> The mathematical symbol that's missing from Einstein's 1905 paper is
> shown here, and what it physically denotes is explained. It is then
> proved that because it is missing, his fourth and fifth equations are
> defective.
<http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html>
Experimental constraints on Special Relativity
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
spudnik
a precis would be appreciated.
and, please, don't just *say* that Michelson, Morley et al go "no"
results,
because they actually got some (small) seasonal anomalies --
that goes for you, two, Uncle Al!
> Experimental constraints on Special Relativity
--Light: A History!
http://wlym.com
Uncle Al
>
> I respect his right to wait til "P3" is out, but
> a precis would be appreciated.
>
> and, please, don't just *say* that Michelson, Morley et al go "no"
> results,
> because they actually got some (small) seasonal anomalies --
> that goes for you, two, Uncle Al!
Uncle Al
>
> I respect his right to wait til "P3" is out, but
> a precis would be appreciated.
>
> and, please, don't just *say* that Michelson, Morley et al go "no"
> results,
> because they actually got some (small) seasonal anomalies --
> that goes for you, two, Uncle Al!
Hey fucking stooopid,
A 2007 study sensitive to 10^(-16) relative employed two simultaneous
interferometers over a year's observation: Optical in Berlin, Germany
at 52�31'N 13�20'E and microwave in Perth, Australia at 31�53'S
115�53E. An aether background could never be at rest relative to both
of them.
http://arXiv.org/abs/0706.2031
idiot
spudnik
get out of the frying pan:
http://21stcenturysciencetech.com/sample.html
thus:
re "spheres of nanothermite,"
do either of you know the primary use of thermite, and
what it is made of?
thus:
I respect his right to wait til "P3" is out, but
a precis would be appreciated.
and, please, don't just *say* that Michelson, Morley et al go "no"
results,
because they actually got some (small) seasonal anomalies --
that goes for you, two, Uncle Al!
> Experimental constraints on Special Relativity
glird
> On Mar 8, 7:44 am, glird <gl...@aol.com> wrote:
>
> > The Missing Symbol
> idiocies snipped <
Most of the first 1/3 of the "idiocies" you snipped was Einstein's
words.
> Still slurring Einstein, shithead?
Einstein wasn't a shithead,
Roused by the pulsating beat and now that they know someone is
listening at last, the P1 equations have begun to stir, to rearrange
themselves, to talk to me; an excited babble of eager voices anxious
to be heard. The living remnants of P1, splattered all over the bloody
pages of P2, are ready to come together again. In the following
reconstruction of Einstein's initial paper we will watch them be
reborn.
glird
glird
> glird wrote:
>
> > The Missing Symbol
> > Posting Two
>
> > The mathematical symbol that's missing from Einstein's 1905 paper is
> > shown here, and what it physically denotes is explained. It is then
> > proved that because it is missing, his fourth and fifth equations are
> > defective.
>
> [snip 600 lines of crap]
About 150 of those lines were written by Einstein.
Here is some of what the idiot Uncle Al snipped out:
"Roused by the pulsating beat and now that they know someone is
listening at last, the P1 equations have begun to stir, to rearrange
themselves, to talk to me; an excited babble of eager voices anxious
to be heard. The living remnants of P1, splattered all over the bloody
pages of P2, are ready to come together again. In the following
reconstruction of Einstein's initial paper we will watch them be
reborn."
glird
BURT
What is the symbol for speed? We have one for acceleration.
Mitch Raemsch
glird
> I respect his right to wait til "P3" is out, but
> a precis would be appreciated.
The first 40% of Posting 2 examined Einstein's 1905 paper and
worked out the meanings of every symbol in his equations by
putting in numerical values for most of them in each equation,
and then solving for any that remained symbols.
After mathematically proving that some of his key symbols,
such as (delta tau)/(delta t), have values that disagree with the
well known values in the LTE, the second posting starts to
reconstruct his initially submitted paper,"P1", Here is the beginning
of tht reconstruction:
"Roused by the pulsating beat and now that they know someone is
listening at last, the P1 equations have begun to stir, to rearrange
themselves, to talk to me; an excited babble of eager voices anxious
to be heard. The living remnants of P1, splattered all over the bloody
pages of P2, are ready to come together again. In the following
reconstruction of Einstein's initial paper we will watch them be
reborn."
If you want to become a midwife, let me know and I will show you the
baby, in Posting number 3.
glird
Dono.
>
> > Still slurring Einstein, shithead?
>
> Einstein wasn't a shithead,
>
> Roused by the pulsating beat and now that they know someone is
No, imbecile, you, Lebau are the shithead. Don't you get it?
BURT
What is the history of light in the distance?
spudnik
are you avoiding the issue,
that M&M et al did not get null results,
as proclaimed by non-et al, and Al?
now, perhaps this study refutes them, or
perhaps it does not -- silly & stupid,
for a guy who does actual experiments.
> A 2007 study sensitive to 10^(-16) relative employed two simultaneous
> interferometers over a year's observation: Optical in Berlin, Germany
> at 52°31'N 13°20'E and microwave in Perth, Australia at 31°53'S
> 115°53E. An aether background could never be at rest relative to both
> of them.
>
> http://arXiv.org/abs/0706.2031
--Light: A History!
http://wlym.com
--Weber's electron, Moon's nucleus!
http://21stcenturysciencetech.com/sample.html
--Stop Cheeny, Rice, Waxman and the ICC's 3rd British invasion of
Sudan!
http://laroucehpub.com
spudnik
The Perth setup compares the frequencies of two orthogonally
orientated high Q-factor (2 × 108) cryogenically
cooled ( 6K) microwave resonators. Each consists
of a sapphire crystal mounted inside a metallic shield.
The crystal is 3 cm diameter and height, and is machined
with its crystal axis in line with the cylindrical
axis. Each resonator is excited in the whispering gallery
WGH8,0,0 mode at approximately 10GHz by two separate
Pound stabilized oscillator circuits, with a difference
frequency of about 226 kHz. The WGH8,0,0 mode has
dominant electric and magnetic fields in the axial and
radial directions respectively, corresponding to a Poynting
vector around the circumference. 98% of the electromagnetic
energy is confined within the sapphire crystal
so it is subject to the index of refraction of sapphire
(n⊥ = 3.04, nk = 3.37 at 6K). The two resonators are
oriented with their cylindrical axis perpendicular to each
other in the horizontal plane and placed inside a vacuum
chamber and cryogenic dewar, which is mounted in
a turntable and rotated at 1/(18 s) about its vertical axis.
thus:
why does a wave in the "vacuum" need anything else,
to wave transversally?... there is no absolute vacuum, and
their are no rocks o'light!
thus:
you refer to the great-circle path on the God-am map!...
as I already stated, aircraft do not do "level" flight,
in order to minimize use of feul; it's a trajectory, but
it can't be symmetrical, up & down, due to drag & so forth.
thus:
now, perhaps this study refutes M&M et al, or
perhaps it does not -- silly & stupid,
for a guy who does actual experiments.
spudnik
> After mathematically proving that some of his key symbols,
> such as (delta tau)/(delta t), have values that disagree with the
> well known values in the LTE, the second posting starts to
> reconstruct his initially submitted paper,"P1", Here is the beginning
> of tht reconstruction:
thus:
isn't searching on some thing that you typed,
rather recently, an example of "positive feedback?" anyway,
perhaps, the real deal is that the googolplex tracks
the usage from particular servers, thereby stuffing one
into a sort of virtual ghetto of would-be correspondents. well,
elsewise it'd be completely unmanageable as an enterprize;
nay?... anyway, sorry, for wasting my time & yours!
thus:
why does a wave in the "vacuum" need anything else,
to wave transversally?... there is no absolute vacuum, and
their are no rocks o'light!
thus:
you refer to the great-circle path on the God-am map!...
as I already stated, aircraft do not do "level" flight,
in order to minimize use of fuel; it's a trajectory, but
it can't be symmetrical, up & down, due to drag & so forth.
thus:
now, perhaps this study refutes M&M et al, or
perhaps it does not -- silly & stupid,
for a guy who does actual experiments.
> http://arXiv.org/abs/0706.2031
--Light: A History!
http://wlym.com
--Weber's electron, Moon's nucleus!
glird
> P1, talk to your new daddy!... what's the LTE?
Heddo, peepa. D LTE r "The Lorentz Transformations".
Hear,peep,is d big ining of Posting 3:
________
The Missing Symbol
Posting Three
This posting will show what happened after Einstein read Poincare's
"Sur la Dynamique des l' Electron" and unsuccessfully tried to derive
its LTE. It will also show that the following things are
misunderstood:
a) In equations containing dxi/dx or dtau/dt it is implied by
calculus that those expressions denote the ratio of size of units of
measure of the given dimension, x for length and t for time. However,
in transformation compared to deformation equations the numerical
value per expression is its own reciprocal; and in inverse cases
compared to direct cases, the numerical values of the ratio of size of
units of length and time switch places with each other; and so do the
symbols in expressions such as dj/dk -- i.e. dj/dk in the direct
equation becomes dk/dj in the inverse equation.
b) Einstein was doing deformation mathematics as per Lorentz, not
transformations as per Poincare'.
In deformation equations, dxi/dx denotes the ratio of size of units of
length of the two systems; so if dxi=1 is Q = (c2-v2) shorter than
dx=1, the deformation relation is dxi/dx = (c2-v2)/1 = Q. The
transformation equation for this case is xi = [1/(dxi/dx)]x = x/Q
because a Q-shorter unit length of Xi will fit 1/Q more times into a
given distance than would a unit length of X.
Similarly, if clocks of moving systems run slow by Q then dtau = 1
second is 1/Q larger than dt = 1 second, so the deformation is dtau/dt
= [1/(c2 - v2)]/1 = 1/Q; in which dtau/dt denotes the ratio of size of
units of time of the two systems, as measured by the system whose time
is t. The inverse deformation for this case is dt/dtau = 1/(dtau/dt)
= Q; because dt/dtau is numerically, thus symbolically, inverse to
dtau/dt.
The transformation equations wrt time will be discussed in the body
of this posting.
c) He thought a moving system deforms as plotted by an attached
Euclidean cs, and tried to use the Galilean transformations to get the
values plotted by a differently moving Euclidean cs! So he never
understood the LTE even though he eventually DID understand the
physics they require and impose.
d) His statement, "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" proves that he didn't understand the LTE,
which rule out the homogeneity of space and time. (If they were
homogeneous, as in Euclidean systems, then units of length and rate
would be constant in all directions regardless of velocities and the
following relations would apply,
dtau/dt = a = dxi/dx = phi(v) = deta/dy = dzeta/dz = 1;
dtau/dx' = 0.
Indeed, the fact that nobody heretofore realized and publicized these
things proves that no one else understood the LTE or the relativity
theory they quantify.)
________
> why does a wave in the "vacuum" need anything else,
> to wave transversally?... there is no absolute vacuum, and
> their are no rocks o'light!
In the compressible matter that fills a vacuum, "light" (which is
the visible portion of the spectrum of em waves) is a pressure that
radiates at c. Whether that pressure is above (+) or below (-) the
average level of the matter it is in, it will automatically cause a
flow of matter in order to neutralize it. If it is a + pressure, the
flow will be away from it and if it is a - p the inflow will be toward
it.
As this +/- pressure node progresses in a given direction, say X of
a local coordinate system, the resulting in or out flow of matter will
be PERPENDICULAR to X.
Therefore, even though a +/- pressure might travel linearly, the co-
traveling density gradient will be perpendicular to that line, as
plotted on a given coordinate system.
> now, perhaps this study refutes M&M et al, or
> perhaps it does not -- silly & stupid,
> for a guy who does actual experiments.
It is independent of the M&M results; although it totally explains
them. So did Lorentz, in 1904.
glird
waldofj
> The Missing Symbol
> Posting Two
>
> The mathematical symbol that's missing from Einstein's 1905 paper is
> shown here, and what it physically denotes is explained. It is then
> proved that because it is missing, his fourth and fifth equations are
> defective. The explanation of how he arrived at Equation 3 and why he
> deleted the missing symbol is then begun. It will be completed in
> Posting Three.
> _________
>
> The third equation (Eq 3) in Einstein's 1905 paper is,
> 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})
> in which x' denotes the distance between clocks A and B of the moving
> system k (xi, eta, zeta; tau) as measured by the stationary system K
> (x, y, z; t). In order to examine the meaning of this and allied
> equations we will let the clocks be at the ends of unit-rod r'AB of k,
> so r'AB = xiB - xiA = 1 is the length of the path AB as measured by k.
> If there is no physical change of lengths in a moving system,
You're already messing up here. One of the things you're trying to
find out is if there is a change of length or not in a moving system.
It's not sensible to start out by assuming one way or the other. Here
you say "if" but later on you use it as if it has already been
established that there is no change of length.
Equation 4 is partial differential equation and equation 5 is just a
rewrite of 4.
Equation 5 has to solved for tau and you don't do that by plugging in
numbers like you're doing here.
It turns out this equation has a non-linear solution as well as a
linear one. When Einstein says
"since tau is a linear function" he just means he's only interested in
the linear solution.
Sorry, but the only thing missing here is your understanding of
calculus.
Dono.
>
>
> Sorry, but the only thing missing here is your understanding of
> calculus.
You are absolutely right but the Lebau imbecile has been at it for
years, you don't expect him to get it , now that he's approaching 80,
do you?
Androcles
"waldofj" <wal...@verizon.net> wrote in message
news:c8fb6c1a-ca4e-4775...@19g2000yqu.googlegroups.com...
==============================================
It is NOT ok to ignore what you don't like.
Sorry, but the only thing missing here is your brain.
Sorry, but you are a fuckin' lunatic.
Sorry... hmm... no, I'm not sorry at all, I lied. Drop dead, you
are fuckin' useless, a mental wanker. Why are posting to sci.logic?
spudnik
about the deployment of reciprocals by Einstein, but I'd need more
of a qualitative wording of "what went wrong,
penintimately or penultimately."
no; M&M got no null result. the experiment that Al cited, although
it shows no results to its degree of precision, appaerently,
is completetly different (as in, "dewars).
things are simpler, using waves:
spherical ones, not "linear" ones!
> Therefore, even though a +/- pressure might travel linearly, the co-
> traveling density gradient will be perpendicular to that line, as
> plotted on a given coordinate system.
> It is independent of the M&M results; although it totally explains
> them. So did Lorentz, in 1904.
thus:
why don't you just look it up,
the practice of commercial pilots?
> GET lost! — NE —-
thus:
science is about refining a hypothesis,
which doesn't have to be one's own. most of "global" warming is,
strictly, computerized simulacra & very selective reporting:
the "hole" in the ozone is really, "the sky is glowing!"
> Science is not about showing that somebody was wrong; it's about showing
> what is right. Do your own work, using your own data, and derive an
> analysis of your own. Then publish your result. THAT is science.
--Light: A History!
http://wlym.com
--Weber's electron, Moon's nucleus!
http://21stcenturysciencetech.com/sample.html
--Stop Cheeny, Rice, Waxman, ICC's 3rd British invasion of Sudan!
http://laroucehpub.com
glird
> On Mar 8, 10:44 am, glird <gl...@aol.com> wrote:
>
><< The third equation (Eq 3) in Einstein's 1905 paper is
= tau(x',0,0,t + x'/{c-v})
in which x' denotes the distance between clocks A and B of the moving
system k (xi, eta, zeta; tau) as measured by the stationary system K
(x, y, z; t). In order to examine the meaning of this and allied
equations we will let the clocks be at the ends of unit-rod r'AB of k,
so
r'AB = xiB - xiA = 1
is the length of the path AB as measured by k.
If there is no physical change of lengths in a moving system, >>
>
>< You're already messing up here. One of the things you're trying to find out is if there is a change of length or not in a moving system. >
I am NOT trying to find out what happens in moving systems; I am
trying to explain what Einstein's EQUATIONS say.
>< It's not sensible to start out by assuming one way or the other. Here you say "if" but later on you use it as if it has already been established that there is no change of length. >
"If" there is no change in length, then "this is what follows".
Unless and until Einstein changes the "If", what's wrong with that?
>< Equation 4 is a partial differential equation and equation 5 is just a rewrite of 4.
Equation 5 has to solved for tau and you don't do that by plugging in
numbers like you're doing here. >
I did solve it. If you don't like my solution, show how you would
solve for tau via calculus. (!)
>< It turns out this equation has a non-linear solution as well as a linear one. When Einstein says "since tau is a linear function" he just means he's only interested in
the linear solution.>
I found the value(s) of tau by algebra, using numbers.
> Sorry, but the only thing missing here is your
> understanding of calculus.
Challenge: Using calculus, show how he got from eq 3 to
"tau = a(t - vx'/{c^2 - v^2})".
If you can't do it by calculus, try to show it by any method at all.
(Having done it years ago, I am fairly sure that you can't!)
glird
Dono.
>
> Challenge: Using calculus, show how he got from eq 3 to
> "tau = a(t - vx'/{c^2 - v^2})".
> If you can't do it by calculus, try to show it by any method at all.
> (Having done it years ago, I am fairly sure that you can't!)
>
Are you still strugglng to understand Taylor expansion? After all
these years?
spudnik
Minkowski, like anyone else, put his pants on ...
one lightcone at a a time!
> Are you still strugglng to understand Taylor expansion? After all
> these years?
thus:
that should be, "Fossilized Fuel (TM)" and *sic*;
it is simply a trade-name of oilolgical mythology
(sediments piling-up in the ocean create enough pressure
to create oil ... even though this happens, continuously,
since time immemorial).
thus:
read _The Big Bang Never Happened" by Eric Lerner,
student of the late Johannes Alfven (I have not).
thus:
what a bunch of silliness. the only real question is,
how much energy was in "de planes,"
compared to the rather small amount that is required
for a "controlled demo?" that is to say,
were the planes not adequate bombs?
thus:
the "official" report (NIST) does have interesting stuff in it
-- I linked to it via the link to Wiki --
for example, "Figure 9.3. Minimum heating of reinforced heavy columns
to initiate global collapse," which shows "temperature range
for a 50% redicution in steel strength," as opposed
to the typical desideratum of "melting" that is promoted.
thus:
please, don't bother
with the pro-hominemania of your supposed status
as a practicing physicist and/or trained netdoggy!
proabably most of the interpretation of the EPR "paradox"
results,
a la Alain Aspect et al, is due to the ideal of a photon,
in assinging all of the energy of the wave-front
as a "mass" (electron-voltage, say) of a particle, whence
the wave-energy was somehow collected
by the photoeletrical device. here are two ways to get over this: a)
just consider the practice of audio quantization, the phonon; b)
show how the photoelectrical device is actually tuned
to absorb a particular frequency of light.
so, is the "phonon" just one cycle of the period of the sound,
and
like-wise, is the photon just one cycle of the frequency?
--Light: A History!
http://wlym.com
--Weber's electron, Moon's nucleus!
http://www.21stcenturysciencetech.com/
--The Ides of March Are Coming:
Pro-Impeachment Democrat
Wins Nomination in Texas!
http://larouchepub.com/pr_lar/2010/lar_pac/100303kesha_victory.html
glird
Nobody cares what Dono wrote. As Androcles said,
Drop dead, you are fuckin' useless, a mental wanker.
Why are posting to sci.logic?
glird
waldofj
> Challenge: Using calculus, show how he got from eq 3 to
> "tau = a(t - vx'/{c^2 - v^2})".
> If you can't do it by calculus, try to show it by any method at all.
> (Having done it years ago, I am fairly sure that you can't!)
>
> glird
ok, I accept the challenge and of course you do it with calculus!
I did this 17 years ago while I was taking multi-variable calculus in
college. And no, it's not like riding a bike! I had to crack the old
text books and refresh my memory. I did run into a problem which I'll
elaborate on below.
First of all I hope you understand when E says "if X' be chosen
infinitesimally small" he means take the partial derivative of tau
with respect to X' (I'll use @ for the partial derivative symbol)
You need the chain rule for this so:
@tau/@X' = (@tau/@X')(@/@X')X' + (@tau/@Y)(@/@X')Y + (@tau/@Z)(@/@X')Z
+ (@tau/@T)(@/@X')T
(I hate using ascii text like this, you might want to write this stuff
out on paper to see it more clearly)
The first expression needs some explanation because the X' symbol has
a double meaning (I remember in another thread someone was complaining
about this and I agree, it makes it a little confusing)
The X' in the denominator of the expression is the variable of
differentiation. The X' on the end is the first parameter of tau, it
takes on the first value in the various expression of tau.
so, for the first term:
(@/@X')tau[0,0,0,T]
= (@tau/@X')(@/@X')0 + (@tau/@Y)(@/@X')0 + (@tau/@Z)(@/@X')0 + (@tau/
@T)(@/@X')T
(keep in mind the derivative of a constant is 0 and in this partial
derivative T is considered a constant)
= (@tau/@X')0 + (@tau/@Y)0 + (@tau/@Z)0 + (@tau/@T)0
= 0
the next term:
(@/@X')tau[0,0,0,T + X'/(c - v) + X'/(c + v)]
= (@tau/@X')(@/@X')0 + (@tau/@Y)(@/@X')0 + (@tau/@Z)(@/@X')0 + (@tau/
@T)(@/@X')(T + X'/(c - v) + X'/(c + v))
= (@tau/@X')0 + (@tau/@Y)0 + (@tau/@Z)0 + (@tau/@T)(0 + 1/(c - v) + 1/
(c + v))
= (@tau/@T)(1/(c - v) + 1/(c + v))
for the third term:
(@/@X')tau[X',0,0,T + X'/(c - v)]
= (@tau/@X')(@/@X')X' + (@tau/@Y)(@/@X')0 + (@tau/@Z)(@/@X')0 + (@tau/
@T)(@/@X')(T + X'/(c - v))
= @tau/@X' + (@tau/@Y)0 + (@tau/@Z)0 + (@tau/@T)(0 + 1/(c - v))
= @tau/@X' + (@tau/@T)(1/(c - v))
put these results together and you get Eq 4 and a little algebra gives
Eq 5
Here is where I ran into a problem. 17 years ago I taught myself how
to solve such an Eq but I don't remember what resource I used. A
handout in one of my classes, a book from the library, whatever, I
don't remember how to derive a general solution.
btw I need to take a moment to comment, this is a partial differential
equation, you absolutely CANNOT solve it using algebra. I don't know
where in the world you got the idea that you could.
I do remember the general solution has many non-linear solutions, in
fact whole families of non-linear solutions, but, we're only
interested in a linear solution.
After staring at it for a while I realized I could come up with a
linear solution just by examination as follows:
rearrange Eq 5 like this:
(@tau/@X') / (@tau/@T) = -v/(c^2 - v^2)
What this means is the ratio of these two partial derivatives is a
constant. So as X' and or T varies this ratio has to remain constant.
The simplest solution for this is a function who's partial derivatives
are constants. so start with:
@tau/@X' = c1
integrate both sides (wrt X') to get
tau(X') = X'c1 + c2
and:
@tau/@T = c3
integrate both sides (wrt T) to get
tau(T) = Tc3 + c4
put this together:
tau(X', T) = (X'c1 + c2 + Tc3 + c4)
Of course the next step is to determine what these constants are but I
want to pause here and introduce the function a. As E says, a is a
function of v and 17 years ago I thought I had a reason to go along
with that. But now I look at it and, at least in the case of a general
solution to Eq 5, a can be a function of ANY combination of variables
as long as it doesn't include X' and T. In any case it is moot point
as later E shows it is just a constant function equal to 1. But in the
mean time we have to include it here.
so:
tau = a(X'c1 + c2 + Tc3 + c4)
First there is an initial condition, T = tau = X = X' = 0. Plug this
in to get
0 = a(c2 + c4) these two constants cancel each other out and can be
set to zero so
tau = a(X'c1 + Tc3)
To get these two constants take the partial derivatives wrt X' and T
and plug them back into Eq 5
this gives:
ac1 + avc3/(c^2 - v^2) = 0
c1/c3 = -v/(c^2 - v^2)
these constants can't constrained any further. I can do something
like:
set c3 = 1 then c1 = -v/(c^2 - v^2) which gives
tau = a(T - vX'/(c^2 - v^2))
or
set c1 = -v then c3 = (c^2 - v^2) which gives:
tau = a((c^2 - v^2)T - vX')
or
set c1 = 1 then c3 = (c^2 - v^2)/-v which gives:
tau = a(X' - T(c^2 - v^2)/v)
All three of these are solutions to Eq 5 so you have to look at other
aspects of the problem you're solving to determine which one is
appropriate to use.
E went with the first one, it makes the rest of his derivations work
out right.
This is all just freshman calculus, no big deal.
Androcles
> Challenge: Using calculus, show how he got from eq 3 to
> "tau = a(t - vx'/{c^2 - v^2})".
> If you can't do it by calculus, try to show it by any method at all.
> (Having done it years ago, I am fairly sure that you can't!)
>
> glird
ok, I accept the challenge and of course you do it with calculus!
I did this 17 years ago while I was taking multi-variable calculus in
college. And no, it's not like riding a bike! I had to crack the old
text books and refresh my memory. I did run into a problem which I'll
elaborate on below.
First of all I hope you understand when E says "if X' be chosen
infinitesimally small" he means take the partial derivative of tau
with respect to X'
=====================================================
Second of all I hope you understand when E says "if X' be chosen
infinitesimally small" he means x-vt is infinitessimal small, but of course
you don't.
Third of all I hope you understand c = dx'/dt but of course you don't.
Fourth of all I hope you understand f'(x) = [f(x+h)-f(x)]/h but of course
you've forgotten that.
======================================================
==================================================
1/sqrt(1-v^2/c^2) = 1/sqrt ( c^2/c^2 - v^2/c^2)
= 1/sqrt ( (c^2- v^2)/c^2)
= 1/sqrt ( (c-v)*(c+v) /c^2)
cannot be solved with algebra...
==================================================
I don't know where in the world you got the idea that you could.
===================================================
Hardly surprising... you don't anything.
glird
>
> > Challenge: Using calculus, show how he got from eq 3 to "tau = a(t - vx'/{c^2 - v^2})".
If you can't do it by calculus, try to show it by any method at all.
(Having done it years ago, I am fairly sure that you can't!)
glird >>
>
> ok, I accept the challenge and of course you do it with >calculus!
Thank you for trying. (You are the first one in about 2 years who
even tried.)
>< I did this 17 years ago while I was taking multi-variable calculus in college. And no, it's not like riding a bike! I had to crack the old text books and refresh my memory. I did run into a problem which I'll
elaborate on below.
First of all I hope you understand when E says "if x' be chosen
infinitesimally small" he means take the partial derivative of tau
with respect to x' (I'll use @ for the partial derivative symbol.) >
It is evident from what you say further down that you don't
understand WHY he set x' "infinitesimally small" rather than
"infinitely small".
>< You need the chain rule for this so:
@tau/@X' = (@tau/@X')(@/@X')X' + (@tau/@Y)(@/@X')Y
+ (@tau/@Z)(@/@X')Z + (@tau/@T)(@/@X')T.
(I hate using ascii text like this, you might want to write this stuff
out on paper to see it more clearly) >
I did. Took me about half an hour to change some of the symbols to
symbol font. Along the way, after putting in several comments such as
{= 0] and {BS??}, I realized that you did it wrong.
>< The first expression needs some explanation because the X' symbol has a double meaning (I remember in another thread someone was complaining about this and I agree, it makes it a little confusing.)>
Though it isn't confusing, x' has THREE different meanings in these
equations. As I posted on these groups years ago, they are:
1. It denotes the position of clock A on X. (It is at
x = 0 at t = tau = 0; thus at x' = x - vt = 0 thereafter.)
2. It denotes the length of the moving path AB as plotted by the
stationary system.
3. It denotes the "infinitesimally small" distance between two moving
clocks that are set to measure the speed of light as a constant.
Note that had he set x' infinitely small, as it would be if it was a
point, then the value of @tau/@x' in eqs 4 and 5 would have been zero.
Note further, that although that might have been invisible to a
calculus expert, it is almost self evident to anyone doing this by
algebra.
>< The X' in the denominator of the expression is the variable of differentiation. The X' on the end is the first parameter of tau, it takes on the first value in the various expression of tau.
So, for the first term:
(@/@X')tau[0,0,0,T] =
(@tau/@X')(@/@X')0 + (@tau/@Y)(@/@X')0
+ (@tau/@Z)(@/@X')0 + (@tau/@T)(@/@X')T >
There's where I put my three [= 0] expressions, to show that the first
three of yours equal zero and algebraically disappear.
> (keep in mind the derivative of a constant is 0 and in
> this partial derivative T is considered a constant)
That's where I put in my first {Bullshit!](in a red font).
If @tau/@t equaled zero, it would disappear from ALL of E's
equations, including the Lorentz Transformations, in which @tau/@t = 1/
sqrt(c^2-v^2).
> = (@tau/@X')0 + (@tau/@Y)0 + (@tau/@Z)0 + (@tau/@T)0
> = 0
Similarly, if x' was an infinitely small point, then -- as you said
prior to that part of your equation --
@tau/@x' = @tau/@t = 0
in all of his equations, whereupon eqs 4, 5 and 7 become meaningless
at best; and false in fact.
>< the next term:
(@/@X')tau[0,0,0,T + X'/(c - v) + X'/(c + v)]
= (@tau/@X')(@/@X')0 + (@tau/@Y)(@/@X')0 + (@tau/@Z)(@/@X')0 + (@tau/
@T)(@/@X')(T + X'/(c - v) + X'/(c + v))
= (@tau/@X')0 + (@tau/@Y)0 + (@tau/@Z)0 + (@tau/@T)(0
+ 1/(c - v) + 1/(c + v))
= (@tau/@T)(1/(c - v) + 1/(c + v)) [= 0!!!] >
Since it was obvious by now that your "calculus" was giving defective
answers, i stopped looking at my copy and returned to this internet
page to continue my review.
> for the third term:
> (@/@X')tau[X',0,0,T + X'/(c - v)]
[= @tau/@X' = 0] times whatever...
>< = (@tau/@X')(@/@X')X' + (@tau/@Y)(@/@X')0 + (@tau/@Z)(@/@X')0 + (@tau/@T)(@/@X')(T + X'/(c - v))
= @tau/@X' + (@tau/@Y)0 + (@tau/@Z)0 + (@tau/@T)(0 + 1/(c - v)) =
@tau/@X' + (@tau/@T)(1/(c - v))
put these results together and you get Eq 4 and a little algebra
gives Eq 5.>
A little algebra shows that your calculus ends up with
0 + 0 + 0 = 0 + 0 + 0 + 0 + 0 = 0 + 0 = ZILCH.
Even so, the challenge was "Using calculus, show how he got from eq
3 to tau = a(t - vx'/{c^2 - v^2})." That you may have done it wrong to
there, let's see how you replied to the actual challenge.
>< Here is where I ran into a problem. 17 years ago I taught myself how to solve such an Eq but I don't remember what resource I used. A handout in one of my classes, a book from the library, whatever, I don't remember how to derive a general solution.
btw I need to take a moment to comment, this is a partial
differential equation, you absolutely CANNOT solve it using algebra. I
don't know where in the world you got the idea that you could.>
By DOING it!
>< I do remember the general solution has many non-linear solutions, in fact whole families of non-linear solutions, but, we're only interested in a linear solution.
After staring at it for a while I realized I could come up with a
linear solution just by examination as follows:
rearrange Eq 5 like this:
(@tau/@X') / (@tau/@T) = -v/(c^2 - v^2)>
BULLshit!! Even if @tau/@X' is a real number, since (according to
what you previously said) @tau/@X' = 0,
then (via simple algebra)
"(@tau/@X') / zero = -v/(c^2 - v^2)"
is bullshit.
>< What this means is the ratio of these two partial derivatives is a constant. So as X' and or T varies this ratio has to remain constant. >
BS^2. (Although the numerical value of tau in @tau/@x' does depend
on the value of x' = x - vt, and in @tau/@t it does depend on the
value of t; it is easy to show that "the ratio of these two" algebraic
expressions is variable.
> The simplest solution for this is a function who's
> partial derivatives are constants.
"(keep in mind the derivative of a constant is 0 and in
this partial derivative T is considered a constant)"
> so start with:
@tau/@X' = c1
integrate both sides (wrt X') to get
tau(X') = X'c1 + c2
and:
@tau/@T = c3 >
Did you forget what you said? If you were right when you said it,
then c3 = 0.
>< integrate both sides (wrt T) to get
tau(T) = Tc3 + c4
put this together:
tau(X', T) = (X'c1 + c2 + Tc3 + c4)
Of course the next step is to determine what these constants are but
I want to pause here and introduce the function a. >
Ahhh... at last you are going to answer the challenge's question,
Where did the symbol _a_ in E's eq 7 come from.
><As E says, a is a function of v and 17 years ago I thought I had a reason to go along with that. But now I look at it and, at least in the case of a general solution to Eq 5, a can be a function of ANY combination of variables as long as it doesn't include X' and T. In any case it is moot point as later E shows it is just a constant function equal to 1. But in the mean time we have to include it here.>
Ah shit. You aren't going to answer the question after all. You are
just going to "include" _a_ in eq 7 even though it was NOT present in
eqs 3, 4, or 5!
(For anyone who wants to know, Einstein replaced
(delta tau/(delta t)
in eqs 4 and 5 with the symbol _a_ in eq 7.
> so:
So my wife is calling me to come help her in the kitchen.
Rather than having her get increasingly angry, and since it is obvious
that you can't answer the challenge by calculus or any other method
you might be familiar with, I am going to end my reply in 1 more
minute.
snip
> This is all just freshman calculus, no big deal.
Not even a little deal, since it was unable to do what algebra can.
Bye now.
glird
glird
>
> (For anyone who wants to know, Einstein replaced
> (delta tau/(delta t)
>
> > so:
>
> So my wife is calling me to come help her in the kitchen.
> Rather than having her get increasingly angry, and since it is obvious
> that you can't answer the challenge by calculus or any other method
> you might be familiar with, I am going to end my reply in 1 more
> minute.
She wanted me to slice the corned beef for her to put in the pot of
boiling cb & cabbage soup. Did it and she said i have to be back
there at 6:30. That gives me 15 minutes to correct the misleading
comment put into my prior posting:
"(For anyone who wants to know, Einstein replaced
(delta tau/(delta t)
in eqs 4 and 5 with the symbol _a_ in eq 7.)"
For those who REALLY want to know, E replaced the _a_ in eq 7 with
the symbol (delta tau/(delta t) in eqs 5 and 4 which he got FROM eq
7.
(This is shown in the third posting of The Missing Symbol; which
has yet to neput on the internet.)
glird
waldofj
well, since you don't know calculus I didn't really expect you to
understand the answer. Just thought I would show it to you anyway.
Nice exercise for me, dusting off the old books and all.
glird
"(For anyone who wants to know, Einstein replaced
(delta tau/(delta t)
in eqs 4 and 5 with the symbol _a_ in eq 7.)"
For those who REALLY want to know, E replaced the _a_ in eq 7 with
the symbol (delta tau/(delta t) in eqs 5 and 4 which he got FROM eq 7.
(This is shown in the third posting of The Missing Symbol; which has
Yummm!!! It was DElicious.
However, i realized that my "corrected" comment was equally
misleading. Without looking at the exact wording in The Missing
Symbol, i will try again to say it right.
Einstein didn't get the symbol _a_ from the symbol (delta tau/(delta
t) in eqs 5 and 4 NOR did he get (delta tau/(delta t) in 4 and 5 from
the _a_ in eq 7.
He got both of them from P1, where he was trying to encorporate
Poincare's LTE into his paper. As will be shown in TMS, the seventh
equation in his paper wasn't "derived" from anywhere at all!
It was the bimp, the link between the steps to there and the steps
that followed after. (A 'bimp' is an almost invisible step that seems
to bridge an unbridgeable gap between elements of a logical
progression from any here to any there.)
To see it in detail, ask and it will be shown.
glird
Dono.
To remind you that you are the same imbecile. Since you have
Alzheimer, you tend to forget....
glird
> "waldofj" <wald...@verizon.net> wrote
> On Mar 8, 10:44 am, glird <gl...@aol.com> wrote:
>
> > > If there is no physical change of lengths in a moving system,
>
> << You're already messing up here. One of the things you're trying to find out is if there is a change of length or not in a moving system.
It's not sensible to start out by assuming one way or the other. Here
you say "if" but later on you use it as if it has already been
established that there is no change of length.
Equation 4 is partial differential equation and equation 5 is just a
rewrite of 4. Equation 5 has to solved for tau and you don't do that
by plugging in numbers like you're doing here. It turns out this
equation has a non-linear solution as well as a linear one. When
Einstein says
"since tau is a linear function" he just means he's only interested
in
the linear solution.
Sorry, but the only thing missing here is your understanding of
calculus. >>
> ==============================================
> It is NOT ok to ignore what you don't like.
Hello Johnnie, it's so good to finally hear your voice.....
How about taking part in this discussion for real,
rather than as a distant supporter!
As I said and still believe despite contrary opinions by
respected sources; YOU are one of the few people on
these newsgroups able to fully understand and appreciate
my "Postings". Therefore, I invite you to openly discuss
these things here, especially if you disagree with any of
them.
glird
glird
= 1/sqrt ( (c^2- v^2)/c^2)
= 1/sqrt ( (c-v)*(c+v) /c^2)
cannot be solved with algebra... >
I'll give it a try;
Let c = 1 unit = the distance light travels per second. Let v = .6c.
Then 1/sqrt(1-v^2/c^2) = 1/(sqrt(1 - .36/1) = 1/sqrt(.64) = 1/.8 =
1.25;
and 1/sqrt (c^2/c^2 - v^2/c^2) = 1/sqrt(1/1 - .36) = 1/.8 = 1.25;
and 1/sqrt ((c^2- v^2)/c^2) = 1/sqrt(1/1 - .64/1) = 1/.8 = 1.25;
and 1/sqrt ( (c-v)*(c+v)/c^2) = 1/[sqrt(1 -.6)*(1 + .6)]/1
=1/ sqrt(.4 * 1.6)/1 = 1/sqrt .64 =
1.25.
Therefore <1/sqrt(1-v^2/c^2) = 1/sqrt ( c^2/c^2 - v^2/c^2)
= 1/sqrt ( (c^2- v^2)/c^2)
= 1/sqrt ( (c-v)*(c+v) /c^2).
Put in any other number for v and the algebra (with a little help
via a hand calculator) would still prove the conclusion is correct.
Once upon a time I graduated from Junior High School and entered the
tenth grade of High School. There were 5 junior Highs and one High
School in the city. The top 10 math pupils in each Jr High were then
picked out and separated into two classes, which were held intact
throughout the three years of High School.
one day, when we were seniors, I was playing pool with the top three
guys in the other class. One of them said to me, "We were in maath
class the othre day and the teacher gave us a geometry problem to
solve. Although the entire class tried to solve it via making
suggestions, we couldn't do it. If you, the leader of the other class,
are so smart, try to solve it now." Whikle the other guys snirked he
then told me the problem. I'd been shooting pool balls while he was
talking, and when I finally missed a shot -- which ended my turn as
shooter -- i stood up and told him the answer to the problem.
Although i don't remember the problem, i do remembre that in order
to solve it i extended one of the lines and then treated the angles
that now existed for the first time. Being smart boys, the three guys
instantly agreed that my solution was right.
Perhaps the same thing happened now. Perhaps nobody else thought of
the simple method of algebraically solving the above equations by
putting in numerical values for c and v. If so, they may be surprised
at how easy it is to solve ALL of Einstein's 1905 equations that same
way; as i did.
glird
glird
> > "tau = a(t - vx'/{c^2 - v^2})".
> > If you can't do it by calculus, try to show it by any method at all.
> > (Having done it years ago, I am fairly sure that you can't!)
> > glird
In today's previous reply I showed how to solve a few equations
Waldorf had written prior to saying, "btw I need to take a moment to
comment, this is a partial differential equation, you absolutely
CANNOT solve it using algebra", Then I said, "Perhaps nobody else
thought of the simple method of algebraically solving the above
equations by putting in numerical values for c and v. If so, they may
be surprised at how easy it is to solve ALL of Einstein's 1905
equations that same way; as i did."
Although that is correct, it might mislead you into thinking that you
can therefore algebraically show how E got from eq 3 to eq 7, which is
tau = a(t - vx'/{c^2-v^2}).
If you could, that would answer my challenge!
Why, then, did i say that i'm fairly sure that you can't?
> I don't know where in the world you got the idea that you could.
Try it like this and perhaps you'll see far more than you expected:
tau = a(t - vx'/{c^2 - v^2}) = a(t - .6x'/.64
in which t = 1 is the time when the ray reaches B at x' = x - vt = x
- .6 and tau is either the time of clock A or clock B, which was at x
when the ray left A at t = 0.
If, as E said later, a = phi(v) = 1, we can solve the equation by
letting the distance between the two clocks be dx = 1 as plotted by
the stationary system. Thus clock A is at x = 0 + vt = .6 when the ray
reaches B at x = 1 + vt = 1.6. Letting Q = {c^2 - v^2} = 1 - .36 = .
64, we get:
tau_A = 1(1 - .6(x + vt - vt)/Q = (1 - .6(.6 - # + # = .6))/.64
= .64/.64 = 1
when the ray reachs clock B; and
tau_B = 1(1 - .6(1 + # - # = 1))/.64 = (1 - .6)/.64 = 5/8;
which is not the value given by STR's equations.
If we let clock B be at x' = x - vt = Q when the ray emits from A at
t = 0, thus at x' = .64 forever, this gives us
tau_B = a(t - vx'/{c^2-v^2})= 1(1 - .6 * .64)/.64 = .9625.
If you still don't like the answer try letting dx = x' = x - vt = .8.
(You'd get tau_B = (1 - .48)/.64 = .8125.) If you still want to get
the "right answer" you have to first know what that answer is! Here's
a way to find it via Voight's local time equation. We'd have
tauB = tauA - vxi/c^2 = 1 - .6xi = .625.
Therefore clock B had to be at xi = 1 when the ray reached it.
Anyone who is still with me will have begun to see the reason why i
looked for the symbol "delta xi/delta x" in Einstein's paper. It isn't
there! It is totally missing!!
If you want to know why it is needed in order for _a_ to equal 1;
look it up in A flower for Einstein. And if you still can't figure out
WHY it's missing; wait for posting #3.
glird
Dono.
>snip imbecilities<
Lebau, old fart
Are you still struggling with understanding Taylor expansion and
simple partial differential equations?
waldofj
> On Mar 13, 4:26 pm, waldofj <wald...@verizon.net> wrote:
>
> <1/sqrt(1-v^2/c^2) = 1/sqrt ( c^2/c^2 - v^2/c^2)
> = 1/sqrt ( (c^2- v^2)/c^2)
> = 1/sqrt ( (c-v)*(c+v) /c^2)
> cannot be solved with algebra... >
>
those words were snipped from a post by androcles and yet you present
them here as if I said them.
How DARE you present someone else's words as my own, especially that
garbled nonsense coming from androcles. That is a deep insult and I
severely resent it.
Androcles
===================================================
Start resenting, you fucking stupid cunt! That is a DELIBERATE deep insult
and you can resent it as much as you like.
glird
I apologize for the unintentional insult!
Because you did say "btw I need to take a moment to comment,
this is a partial differential equation,
you absolutely CANNOT solve it using algebra. I don't know
where in the world you got the idea that you could"
I thought the entire passage was yours.
On checking through all of your messages in this thread,
though, I see that only the words in the last
sentence -- which is the one to which I actually replied -- are close
to what you actually said.
Accordingly, although it was not deliberate nor did I know
it, I apologize for insulting you.
Meanwhile, I notice that you ignored all of my comments
concerning the defect in your mathematics. In case you are
willing to address them -- rather than dismiss them with
comments about my mathematical inadequacies -- here are
some of them:
gl: It is evident from what you say further down that you don't
understand WHY he set x' "infinitesimally small" rather than
"infinitely small".
wa: (keep in mind the derivative of a constant is 0 and in this
partial derivative T is considered a constant)
gl: If @tau/@t equaled zero, it would disappear from ALL of E's
equations, including the Lorentz Transformations, in which @tau/@t = 1/
sqrt(c^2-v^2).
wa: @/@X')tau[0,0,0,T]
= (@tau/@X')0 + (@tau/@Y)0 + (@tau/@Z)0 + (@tau/@T)0
= 0
gl: Similarly, if x' was an infinitely small point, then -- as you
said prior to that part of your equation --
@tau/@x' = @tau/@t = 0
in all of his equations, whereupon eqs 4, 5 and 7 become meaningless
at best; and false in fact.
gl: A little algebra shows that your calculus ends up with
0 + 0 + 0 = 0 + 0 + 0 + 0 + 0 = 0 + 0 = ZILCH.
Even so, the challenge was "Using calculus, show how he got from eq 3
to tau = a(t - vx'/{c^2 - v^2})." Though you may have done it wrong to
there, let's see how you replied to the actual challenge.
wa: you absolutely CANNOT solve it using algebra. I don't know where
in the world you got the idea that you could.
gl: By DOING it!
wa: I could come up with a linear solution just by examination as
follows:
rearrange Eq 5 like this:
(@tau/@X') / (@tau/@T) = -v/(c^2 - v^2)>
gl: Even if @tau/@X' is a real number, since
(according to what you previously said)
@tau/@X' = 0,
then (via simple algebra)
"(@tau/@X')/zero = -v/(c^2 - v^2)"
is bullshit.
wa: What this means is the ratio of these two partial derivatives is a
constant. So as X' and or T varies this ratio has to remain constant.
>
gl: BS^2. (Although the numerical value of tau in @tau/@x' does
depend on the value of x' = x - vt, and in @tau/@t it does depend on
the value of t; it is easy to show that "the ratio of these two"
algebraic expressions is variable.
wa: The simplest solution for this is a function who's partial
derivatives are constants.
gl: "(keep in mind the derivative of a constant is 0 and in
this partial derivative T is considered a constant)"
wa: so start with:
@tau/@X' = c1
integrate both sides (wrt X') to get
tau(X') = X'c1 + c2
and:
@tau/@T = c3 >
gl: Did you forget what you said? If you were right when you said
it, then c3 = 0.
wa: I want to pause here and introduce the function a. >
gl: Ahhh... at last you are going to answer the challenge's
question, Where did the symbol _a_ in E's eq 7 come from.
wa: As E says, a is a function of v and 17 years ago I thought I had
a reason to go along with that. But now I look at it and, at least in
the case of a general solution to Eq 5, a can be a function of ANY
combination of variables as long as it doesn't include X' and T. In
any case it is moot point as later E shows it is just a constant
function equal to 1. But in the mean time we have to include it
here.>
gl: Ah shit. You aren't going to answer the question after all. You
are just going to "include" _a_ in eq 7 even though it was NOT present
in eqs 3, 4, or 5!
(For anyone who wants to know, Einstein replaced
(delta tau/(delta t)
in eqs 4 and 5 with the symbol _a_ in eq 7.
glird
glird
Why don't you argue with me, instead of
with people who ignore legitimate arguments
that prove their views are defective?
Casting insulting remarks in response to
those they make only puts you with them, in
the same alt.shithead category as they are.
glird
waldofj
ok, I accept your apology. Thank you
But try to be more careful in the future ok?
>
> Meanwhile, I notice that you ignored all of my comments
> concerning the defect in your mathematics. In case you are
> willing to address them -- rather than dismiss them with
> comments about my mathematical inadequacies -- here are
> some of them:
The problem is there are no defects in my mathematics. What I posted
is exactly correct. You didn't understand a single word of it because
you don't know calculus. A perfect example is the function _a_ that
shows up in Eq 7. You say this is not in Eq's 3, 4, or 5 and yet it
just suddenly pops up in Eq 7? Then you go on to say it comes from
@tau/@t somehow. According to you it's just a bunch of hand waving
bullshit.
Well, it does kinda pop up from nowhere, but, if you knew calculus you
would understand what it's about and why it needs to be there. I tried
to explain this to you and you didn't understand a single word of it,
BECAUSE YOU DON'T KNOW CALCULUS!
And why is that? Newton invented calculus as a mathematical tool for
solving problems in physics and, as it turns out, it is one of the
most powerful mathematical tools ever devised. And yet you just turn
your nose up at it.
Why is that?
There's another problem, you tried plugging some numbers into Eq 7 and
got the wrong answers. That is as it should be, Eq 7 is not correct!
Referring to page 45 of my Dover paperback edition "The Principle of
Relativity" Eq 7 is at the top of the page and this is at the bottom
"Substituting for x' its value, we obtain" and you flip the page and
see four equations at the top of page 46. So lets do that. Start with
tau = a(t - vx'/(c^2 - v^2))
= a(t - v(x - vt)/(c^2 - v^2))
= a(t -(vx - tv^2)/(c^2 - v^2))
= a((tc^2 - tv^2 - vx + tv^2)/(c^2 - v^2))
= a((tc^2 - vx)/(c^2 - v^2))
= a((t - vx/c^2)/(1 - v^2/c^2))
= a(t - vx/c^2)gamma^2 (here I'm using the new term gamma for 1/sqrt(1
- v^2/c^2))
compare this to the equation at the top of page 46 and you see that a
factor of gamma has been divided out. All four equations have a factor
of gamma divided out. That step isn't shown nor is any explanation
given for it. A whole page is missing!
From my own work with the LTE I know why it's done. The equations on
page 45 are compatible with the second postulate (constancy of the
speed of light) but they aren't compatible with the first postulate
(principle of relativity). Once a factor of gamma is divided out the
equations become compatible with both postulates.
It would have been nice if they had left that in.
Androcles
§ 4. Physical Meaning of the Equations Obtained in Respect to Moving Rigid
Bodies and Moving Clocks
"Between the quantities x, t, and tau, which refer to the position of the
clock, we have, evidently, x=vt"
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
= a(t - vvt/c^2)/(1 - v^2/c^2))
= at(1 - v^2/c^2)/(1 - v^2/c^2))
= at
compare this to the equation at the top of page 46 and you see that a
factor of gamma has been divided out. All four equations have a factor
of gamma divided out. That step isn't shown nor is any explanation
given for it. A whole page is missing!
From my own work with the LTE
==============================================
Bwhahahahahahahahahaha!
==============================================
I know why it's done. The equations on
page 45 are compatible with the second postulate (constancy of the
speed of light) but they aren't compatible with the first postulate
(principle of relativity). Once a factor of gamma is divided out the
equations become compatible with both postulates.
It would have been nice if they had left that in.
================================================
Nice? "They"?
What the fuck does "nice" have to do with the price of rice ?
Who the fuck are "they"?
glird
> On Mar 23, 3:05 pm, glird <gl...@aol.com> wrote:
>
> > Meanwhile, I notice that you ignored all of my comments
> > concerning the defect in your mathematics. In case you are
> > willing to address them -- rather than dismiss them with
> > comments about my mathematical inadequacies -- here are
> > some of them:
>
> The problem is there are no defects in my mathematics. What I posted
> is exactly correct. You didn't understand a single word of it because
> you don't know calculus. A perfect example is the function _a_ that
> shows up in Eq 7. You say this is not in Eq's 3, 4, or 5 and yet it
> just suddenly pops up in Eq 7? Then you go on to say it comes
> from @tau/@t somehow. According to you it's just a bunch of hand
> waving bullshit.
According to YOU i said that about eq 7.
> Well, it does kinda pop up from nowhere, but, if you knew calculus
> you would understand what it's about and why it needs to be there.
> I tried to explain this to you and you didn't understand a single word > of it, BECAUSE YOU DON'T KNOW CALCULUS!
BULLSHIT! It is YOUR hand-waving refusal to discuss the defects in
YOUR mathematics that I called bullshit -- and still do.
> There's another problem, you tried plugging some numbers into Eq
> 7 and got the wrong answers. That is as it should be, Eq 7 is not
> correct!
"Eq 7" is my term for EINSTEIN'S equation at the top of pg 45. If
you think his is incorrect, tell us what you think is wrong with it..
> Referring to page 45 of my Dover paperback edition "The Principle
> of Relativity" Eq 7 is at the top of the page and this is at the bottom
> "Substituting for x' its value, we obtain" and you flip the page and
> see four equations at the top of page 46. So lets do that. Start with
> tau = a(t - vx'/(c^2 - v^2)) [Eq 7]
> = a(t - v(x - vt)/(c^2 - v^2))
> = a(t -(vx - tv^2)/(c^2 - v^2))
> = a((tc^2 - tv^2 - vx + tv^2)/(c^2 - v^2))
> = a((tc^2 - vx)/(c^2 - v^2))
> = a((t - vx/c^2)/(1 - v^2/c^2))
> = a(t - vx/c^2)gamma^2 (here I'm using the new term gamma for
> 1/sqrt(1 - v^2/c^2)) [ = 1/q]
> compare this to the equation at the top of page 46 and you see that
> a factor of gamma has been divided out.
Your last step says that
tau = a((t - vx/c^2)/(1 - v^2/c^2)) = a(t - vx/c^2)gamma^2)
[ = betaphi(v)(t – vx/c^2) ]
in which beta = gamma = 1/q and a = dtau/dt = q = phi(v) = 1/gamma.
Accordingly,
tau = a((t - vx/c^2)/(1 - v^2/c^2)) = q(t - vx/c^2)gamma^2)
= q(t - vx/c^2)/q^2
= gamma (t - vx/c^2) = betaphi(v)(t - vx/c^2)
in which a = (delta tau)/(delta t) = dtau/dt = phi(v) = 1!
Note, then, that although the equations at the top of page 46 --
which I call the General Transformation Equations ("GTE") – ARE
compatible with the first postulate (principle of relativity) for ANY
value of phi(v), they do NOT fit the LTE!!
Ask yourself, therefore, Is THAT why E tried to prove that phi(v) =
1?
(The answer is YES!!)
Ask yourself, therefore, What was the value of _a_ B4 he tried that
failed proof? The answer, taken from the above ALGEBRA, was:
a = dtau/dt = q = 1/beta =/= the LTE!!
(If you don't understand this answer, look at my postings re The
Missing Symbol for details.)
> All four equations have a factor
> of gamma divided out. That step isn't shown nor is any explanation
> given for it. A whole page is missing!
Look again! (YOU missed page 47, on the way to the eqs on the top of
page 48!)
> From my own work with the LTE I know why it's done. The
> equations on page 45
[PAGE 46!!]
> are compatible with the second postulate (constancy of the
> speed of light) but they aren't compatible with the first postulate
> (principle of relativity).
Since that's the wrong reason, it is clear that although you may be
an expert at calculus, you are blind to the meanings of your own
equations; and of the simple algebra equations E was doing.
glird
spudnik
waldofj
> On Mar 27, 7:42 am, waldofj <wald...@verizon.net> wrote:
>
> > On Mar 23, 3:05 pm, glird <gl...@aol.com> wrote:
>
> > > Meanwhile, I notice that you ignored all of my comments
> > > concerning the defect in your mathematics. In case you are
> > > willing to address them -- rather than dismiss them with
> > > comments about my mathematical inadequacies -- here are
> > > some of them:
>
> > The problem is there are no defects in my mathematics. What I posted
> > is exactly correct. You didn't understand a single word of it because
> > you don't know calculus. A perfect example is the function _a_ that
> > shows up in Eq 7. You say this is not in Eq's 3, 4, or 5 and yet it
> > just suddenly pops up in Eq 7? Then you go on to say it comes
> > from @tau/@t somehow. According to you it's just a bunch of hand
> > waving bullshit.
>
> According to YOU i said that about eq 7.
true, my words. I was just paraphrasing.
>
> > Well, it does kinda pop up from nowhere, but, if you knew calculus
> > you would understand what it's about and why it needs to be there.
> > I tried to explain this to you and you didn't understand a single word > of it, BECAUSE YOU DON'T KNOW CALCULUS!
>
> BULLSHIT! It is YOUR hand-waving refusal to discuss the defects in
> YOUR mathematics that I called bullshit -- and still do.
No, what you're doing is calling calculus bullshit. This a calculus
problem which you obviously don't understand. So you apply some
algebra (not a valid technique here) and think you've discovered
something. All you've discovered is a lack of calculus.
You really should learn some calculus, it would do you a world of
good.
>
> > There's another problem, you tried plugging some numbers into Eq
> > 7 and got the wrong answers. That is as it should be, Eq 7 is not
> > correct!
>
> "Eq 7" is my term for EINSTEIN'S equation at the top of pg 45.
yes, that is what I was refering to.
> If
> you think his is incorrect, tell us what you think is wrong with it..
reread the following text. Like I said, it's off by a factor of gamma.
In fact all of the equations on page 45 are off by a factor of gamma.
>
> > Referring to page 45 of my Dover paperback edition "The Principle
> > of Relativity" Eq 7 is at the top of the page and this is at the bottom
> > "Substituting for x' its value, we obtain" and you flip the page and
> > see four equations at the top of page 46. So lets do that. Start with
> > tau = a(t - vx'/(c^2 - v^2)) [Eq 7]
> > = a(t - v(x - vt)/(c^2 - v^2))
> > = a(t -(vx - tv^2)/(c^2 - v^2))
> > = a((tc^2 - tv^2 - vx + tv^2)/(c^2 - v^2))
> > = a((tc^2 - vx)/(c^2 - v^2))
> > = a((t - vx/c^2)/(1 - v^2/c^2))
> > = a(t - vx/c^2)gamma^2 (here I'm using the new term gamma for
> > 1/sqrt(1 - v^2/c^2)) [ = 1/q]
> > compare this to the equation at the top of page 46 and you see that
> > a factor of gamma has been divided out.
>
> Your last step says that
> tau = a((t - vx/c^2)/(1 - v^2/c^2)) = a(t - vx/c^2)gamma^2)
> [ = betaphi(v)(t – vx/c^2) ]
> in which beta = gamma = 1/q and a = dtau/dt = q = phi(v) = 1/gamma.
you have added some things I didn't say but you're right, if you set a
= 1/gamma it fixes everything. Except that's not what E is doing here.
He leaves a (or phi(v), same thing) in place and yet a factor of gamma
has just disappeared from all four equations without any explanation.
> Accordingly,
> tau = a((t - vx/c^2)/(1 - v^2/c^2)) = q(t - vx/c^2)gamma^2)
> = q(t - vx/c^2)/q^2
> = gamma (t - vx/c^2) = betaphi(v)(t - vx/c^2)
> in which a = (delta tau)/(delta t) = dtau/dt = phi(v) = 1!
now wait a minute! you just got through setting a = phi(v) = 1/gamma.
Now you're saying 1 = phi(v) = 1?
Can't be both.
> Note, then, that although the equations at the top of page 46 --
> which I call the General Transformation Equations ("GTE") – ARE
> compatible with the first postulate (principle of relativity)
well, as it turns out they aren't. When I say they satisfy the Por
what I mean is the forward and reverse transforms should have the same
form and if you apply the reverse transform to the forward transform
you should get what you started with, i.e. x = x and t = t. When I did
that that's just what I got. Then I realized I did something I do so
automatically that I don't even think about it. I set phi(v) = 1. If
phi(v) is set to anything else these equations (top of page 46) don't
satisfy the Por. And with phi(v) = 1 they are then the LTE
> for ANY
> value of phi(v), they do NOT fit the LTE!!
other than 1 correct.
> Ask yourself, therefore, Is THAT why E tried to prove that phi(v) =
> 1?
> (The answer is YES!!)
of course yes!
> Ask yourself, therefore, What was the value of _a_ B4 he tried that
> failed proof?
what failed proof? He showed correctly that a must equal 1 for the
equations to work. And it never had a previous value. All prior
references say a is a function of v yet to be determined.
> The answer, taken from the above ALGEBRA, was:
> a = dtau/dt = q = 1/beta =/= the LTE!!
I think it has been pretty well established that if a doesn't equal 1
then no LTE.
> (If you don't understand this answer, look at my postings re The
> Missing Symbol for details.)
>
> > All four equations have a factor
> > of gamma divided out. That step isn't shown nor is any explanation
> > given for it. A whole page is missing!
>
> Look again! (YOU missed page 47, on the way to the eqs on the top of
> page 48!)
um, 47 comes after 46, not before
>
> > From my own work with the LTE I know why it's done. The
> > equations on page 45
>
> [PAGE 46!!]
no, at this point I was still trying to get from page 45 to the top of
page 46. Even you started to point out that to get there you have to
set a = 1/gamma. But E doesn't do that, he leaves it there and later
shows it has to be set to 1. So if I want to get from page 45 to the
top of page 46 only using E's methodology I have to do an extra step.
Divide out a factor of gamma
>
> > are compatible with the second postulate (constancy of the
> > speed of light) but they aren't compatible with the first postulate
> > (principle of relativity).
>
> Since that's the wrong reason
sorry, it's the right reason. It's what E is doing on pages 46 and 47.
It's not enough to simply follow the algebra, you have to understand
what he's trying to do in the first place. Clearly you are very
confused.
> it is clear that although you may be
> an expert at calculus, you are blind to the meanings of your own
> equations; and of the simple algebra equations E was doing.
>
one more thing. you use delta tau / delta t to refer to the symbols in
Eq 5.
When I see delta used like this I think of an interval like x2 - x1 =
delta x.
If that's how you intend to use it here you're making a big mistake.
The symbols used in Eq 5 (for which I substitute the @ symbol) are
differential operators, not ratio of variables or intervals. For
example, referring to the LTE
@tau/@t = gamma while
delta tau / delta t = 1/gamma
glird
>
>< The symbols used in Eq 5 (for which I substitute the @ symbol) are differential operators, not ratio of variables or intervals. For example, referring to the LTE
@tau/@t = gamma while
delta tau / delta t = 1/gamma >
About 5 years ago I had a similar discussion on s.p.relativity. One
of the calculus experts said almost the same thing you did here. When
i pointed out that Einstein had said these equations are _linear_ (!)
he withdrew his objection.
In any event, Waldoj, please excuse me for being impatient with you.
Since you are the only person in these groups who has been willing to
argue with me about the mathematics in E's paper, i will make a
deliberate effort to argue with you peacefully hereafter.
Tomorrow morning I will show you why Einstein's "proof" that phi(v)
- 1 was a failure. Meanwhile,
good night, and sleep well.
glird
waldofj
> On Mar 27, 8:21 pm, waldofj <wald...@verizon.net> wrote:
>
> >< The symbols used in Eq 5 (for which I substitute the @ symbol) are differential operators, not ratio of variables or intervals. For example, referring to the LTE
>
> @tau/@t = gamma while
> delta tau / delta t = 1/gamma >
>
> About 5 years ago I had a similar discussion on s.p.relativity. One
> of the calculus experts said almost the same thing you did here. When
> i pointed out that Einstein had said these equations are _linear_ (!)
> he withdrew his objection.
If he was an expert in calculus he should have said exactly what I
said. Unless he had some confusion over the LTE? And the fact that the
equations are linear (which indeed they are) makes no difference to
what I said. I curious what that discussion was, doesn't sound like
what we're discussing.
>
> In any event, Waldoj, please excuse me for being impatient with you.
Don't worry about that. I'm thick skinned, just hope you are as well.
I'm quite certain you won't like what I have to say.
However, if you see me getting nasty, just bitch slap me a few times.
> Since you are the only person in these groups who has been willing to
> argue with me about the mathematics in E's paper, i will make a
> deliberate effort to argue with you peacefully hereafter.
> Tomorrow morning I will show you why Einstein's "proof" that phi(v)
> - 1 was a failure.
already on shaky ground. He successfully derived the LTE (unless you
have a problem with the LTE?)
why do you keep calling that a failure?
> Meanwhile,
> good night, and sleep well.
likewise
Prophet Steering
profiles from calculations with the derived Pr distribution are shown
by continuous curves. Dashed curves correspond to the non-LTE profiles
from 9 theoretical line profiles for the derived mean non-LTE nitrogen
abundance are compared with the observations; excellent agreement is
found for almost ...
regarding the non+LTE line formation for Prii and Priii in A and Ap
starsby L profiles from calculations with the derived Pr distribution
are shown by continuous curves. Dashed curves correspond to the non-
LTE profiles from 9 theoretical line profiles for the derived mean non-
LTE nitrogen abundance are compared with the observations; excellent
agreement is found for almost
I hope you sleep well as well after all all is well ends well.
Musatov
(drying up the well)
glird
> On Mar 27, 9:19 pm, glird <gl...@aol.com> wrote:
>
> > On Mar 27, 8:21 pm, waldofj <wald...@verizon.net> wrote:
Tomorrow morning I will show you why Einstein's "proof"
>
> already on shaky ground. He successfully derived the LTE (unless you have a problem with the LTE?)
> why do you keep calling that a failure? >
Good Morning!!
Woke up last night and realized that "I WON!".
(I will tell you what i won -- and why -- in another message. This
one will answer your question.)
[[ Wait a second or so while I paste something here for that "another
message":
me: The General Transformation Equations ("GTE") – ARE
compatible with the first postulate (principle of relativity) >
<well, as it turns out they aren't. When I say they satisfy the Por
what I mean is the forward and reverse transforms should have the same
form and if you apply the reverse transform to the forward transform
you should get what you started with, i.e. x = x and t = t. When I did
that that's just what I got. Then I realized I did something I do so
automatically that I don't even think about it. I set phi(v) = 1. If
phi(v) is set to anything else these equations (top of page 46) don't
satisfy the Por. And with phi(v) = 1 they are then the LTE.
"for ANY value of phi(v)", they do NOT fit the LTE!! other than 1
correct. >
> Ask yourself, therefore, Is THAT why E tried to prove that phi(v) = 1? (The answer is YES!!) >
of course yes!
> Ask yourself, therefore, What was the value of _a_ B4 he tried that failed proof? >
what failed proof? He showed correctly that a must equal 1 for the
equations to work. And it never had a previous value. All prior
references say a is a function of v yet to be determined. ]]
Ok. ... Hold on another few minutes while i look up he proof in
my prior messages in this thread and if it's not there I will look it
up in A Flower for Einstein.
naaahhh .. even tho no "blindfolded" calculus experts ever saw it,
it's so simple that I'lll just do it here.
Look at pg 47 with me. See the equation in the ppg starting with
"Since". Notice that it gives the value of phi as viewed by TWO
systems, K and k viewing each other.
BUT!!! After continuing to treat a rod of cs k as viewed by k, he
then says "for reasons of symmetry" (which are inapplicable) blah de
blah. Now read his next sentence. (!) If v (OF SYSTEM k) is
interchanged with -v (OF an un-named THIRD system), BOTH of which are
viewed by system K IN THE MIDDLE (!!) ; then even if his "Hence ..."
conclusion is correct (see later comment), it has NOTHING = 0 = zip =
nadda = zilch in common with "the one previously found"!!
Therefore he "failed" to prove! "that phi(v) = 1". THAT is why I say
that he did NOT accurately derive the LTE; which, though valid, are
only one of an infinity of groups that fit the GTE, at the top of pg
46.
If you wish, W, I will provide the "later comment" later. meanwhile,
i will copy the stuff pasted within the [[stuff]] sign above, for use
in my next message.
regards,
glird
glird
> delta tau / delta t = 1/gamma
Given that the equations are linear (i.e each symbol has a given
value regardless of the unchanged value of v) WHY do you say that
delta tau / delta t = 1/(@tau/@t)?
glird
glird
glird
> On Mar 27, 9:57 pm, waldofj <wald...@verizon.net> wrote:
> ><He successfully derived the LTE (unless you
> have a problem with the LTE?)
> why do you keep calling that a failure? >
> Because he DID NOT "successfully" derive the LTE
> (with which i have no problem).
> glird
Here's why Einstein's "proof" that "phi(v)= 1" was a failure:
Look at the sentence on page 47 that starts "Since the r...". Note
that in the equation in his next sentence,
phi(v)phi(-v) = 1,
there is no third system! The v is the velocity of cs k as plotted by
K and -v is the velocity of cs K as plotted by k.
Keep reading. After appealing to irrelevant(*) "reasons of
symmetry" E went on the show that
phi(v)= phi(-v),
in which v is the velocity of k and -v is the velocity of an unnamed
THIRD SYSTEM; both of them as plotted by the stationary system K in
the middle.
Although the latter equation may (or may not -- see *) be valid, it
is totally unrelated to "the one previously found". Therefore, his
conclusion "that phi(v) = 1" is unproven.
* Let cs k move at v wrt "the stationary system" K (Einstein's words).
Let a third system, K', move to the right at v, as plotted by cs k.
That lets K' move at v and K move at -v, as plotted by _the moving_ cs
k in the middle. Therefore, if -- as E and the LTE required -- the
length of a unit rod of a given system is a function of its velocity,
say it shrinks by phi(v) = q, then rod rAB of k is shorter than that
of K and r'AB of K' is the shortest of the three. Hence it would
follow that
phi(v) =/= phi(-v)
as plotted by ANY of the three systems!
What really follows is that reasons of symmetry hold good if and only
if the system in the middle is ABSOLUTELY stationary (either in the
local luminiferous material, or in the hypothetical universally
stationary ether of classical physics, or in Einstein's equally
stationary and non-existing "empty space").
Since, as Newton pin-pointed out, there is no way to discover a
universally stationary referent for v, "reasons of symmetry" are
invalid in discovering a value for phi(v). Therefore neither Einstein
NOR ANYONE ELSE has ever "derived" the LTE. Indeed, as pointed out in
A Flower for Einstein and other prior books, it is impossible to
derive a value for (delta t'/delta t) or phi(+/- v) or The Missing
Symbol (delta xi/delta x) unless the value of at least one of them is
stipulated in advance.
glird
waldofj
that's not what linear means and that's not at all what I'm saying. In
this particular case, these are the results from using the equations
for tau from the LTE (i.e. the equation at the top of page 48)
if instead you refer to Eq 7 (top of page 45) you get
@tau/@t = a and
delta tau / delta t = a
now they're the same. There is no general relationship between them,
it depends on what equation you're working with.
Androcles
=================================================
Einstein decided a = 1.
Now
@tau/@t = delta tau /delta t =
@t/@tau =
delta t /delta tau =
dt/dtau =
dtau/dt = 1
Goodbye gamma.
Goodbye babbling cretin.
waldofj
> On Mar 28, 1:13 pm, glird <gl...@aol.com> wrote:
>
> > On Mar 27, 9:57 pm, waldofj <wald...@verizon.net> wrote:
> > ><He successfully derived the LTE (unless you
> > have a problem with the LTE?)
> > why do you keep calling that a failure? >
> > Because he DID NOT "successfully" derive the LTE
> > (with which i have no problem).
> > glird
>
> Here's why Einstein's "proof" that "phi(v)= 1" was a failure:
I was going to critique your statements one at a time but since every
thing you said is wrong it would just take too long. I will instead
try to explain what's on page 47. I have to admit page 47 is a mess.
There are two errors, an unnecessary complication, and some confusing
arguments.
The unnecessary complication is the introduction of a third system
(K'), the first error is the way it's done. I'm not even going to go
over that. The point E is trying to make here is that for these
transformation equations to be mathematically consistent, if you apply
(or insert) the forward LTEs into the reverse LTEs you should end up
with what you started with. IOW:
t = t
x = x
y = y
z = z
but if you do this with the equations at the top of page 46 you get
t = phi(v)phi(-v)t
x = phi(v)phi(-v)x
y = phi(v)phi(-v)y
z = phi(v)phi(-v)z
so these equations aren't valid unless phi(v)phi(-v) = 1
For thee second part E describes a rod in the moving system (k),
defines its end points (it is oriented perpendicular to the x axis and
parallel to the y axis) and then translates those coordinates to the
stationary system (K). Here is the second error and it's just a
nitpick really. The expression for y1 uses the forward transform. When
going from system k to system K you should use the reverse transform
so that should read y1 = Lphi(-v). It doesn't change his argument
however. In the first scenario he has the rod, and system k, moving to
the right (positive v wrt K). His point is, if you reverse the
direction of the rod, and system k,(make a note, he is not introducing
a third coordinate system here, just reversing the direction of k)
then the only thing that should change is the sign of v. He is arguing
"from reasons of symmetry" that the length of the rod, as seen from
the stationary system (K) should not depend on the direction it's
moving (i.e. left or right) but only on the magnitude of v. I'm not
really sure what he means by "from reasons of symmetry" but I think of
it like this: if it did make a difference what direction the rod is
moving (left or right) then that would mean the properties of space
would be different depending on whether you're going left or right.
But it is a basic assumption in all of physics that the properties of
space are the same in all directions, so it shouldn't matter if the
rod is going right or left. Just the magnitude of v matters. So Lphi(-
v) = Lphi(v) so phi(-v) = phi(v). So to satisfy both conditions phi(v)
= phi(-v) = 1 or phi(v) = phi(-v) = -1. However the negative solution
doesn't make sense physically so phi(v) = 1
glird
"waldofj" wrote in message
news:63b40fe3-940f-4081-979d-
a8edc1...@19g2000yqu.googlegroups.com...
<> Challenge: Using calculus, show how he got from eq 3 to
> "tau = a(t - vx'/{c^2 - v^2})".
> If you can't do it by calculus, try to show it by any method at all.
> (Having done it years ago, I am fairly sure that you can't!)
> glird
ok, I accept the challenge and of course you do it with calculus! >
On Apr 1, 8:00 am, "Androcles" <Headmas...@Hogwarts.physics_x> wrote:
> "waldofj" <wald...@verizon.net> wrote in message
Dear John Babble Walker,
The REASON "Einstein decided a = 1" is that in
Lorentz's 1904 paper, he (Lorentz!) said that the ONLY effect
of a system's motion would be a deformation of its lengths,
Given that, then there would be no change in rates of anything;
including the rate at which clocks beat,
Therefore, dt'/dt = 1 WOULD be the ratio of rates, dt' : dt, of
clocks
of two differently moving systems.
Given that his equations would "be linear on account of the
properties
of homogeneity" E attributed "to space and time", then you are right
insofar as "@tau/@t = delta tau /delta ... = 1" is concerned; and
that
(in P1!) @t/@tau = delta t /delta tau = dt/dtau = a = 1.
However, the little detail you slipped in and on is that rather
than
dt/dtau = dtau/dt - 1, as in your greasy equation (which is precisely
what tripped up Einstein); IF a = dt/dtau = 1 as viewed by system k',
then dtau/dt IS NOT equal to a = 1 as viewed by system K!!
You, John W, are the only one on these newsgroups who pinpointed the
bimp in Einstein's paper, "Hence, if x' be chosen infinitesimally
small,"
Although you still don't know it, eqs 4 and 5 were not derived from
eq 3;
nor was eq 7 derived from anything in P2. It was put there to bridge
the
unbridgeable gap (*) in his treatment.
Although it is the most brilliant equation in his entire paper, it
was and is
a bimp -- an invention to join what came before and what came after
it.
* Even though a = dt/dt' DID equal 1 in P1, and a = dtau/dt DOES
equal
sqrt(1-v^2/c^2) in P2, (both of which satisfy eq 7!); there is no
logical or
mathematical way to let it be both. .
Regards,
glird
glird
>
> Einstein decided a = 1.
> Now
> @tau/@t = delta tau /delta t =
> @t/@tau =
> delta t /delta tau =
> dt/dtau =
> dtau/dt = 1
> Goodbye gamma.
> Goodbye babbling cretin.
Dear John Babble Walker,
The REASON "Einstein decided a = 1" is that in
Lorentz's 1904 paper, he (Lorentz!) said that the ONLY effect
of a system's motion would be a deformation of its lengths,
Given that, then there would be no change in rates of anything;
including the rate at which clocks beat.
Therefore, dt'/dt = 1 WOULD be the ratio of rates, dt' : dt,
of clocks of two differently moving systems.
Given that his equations would "be linear on account of the
properties of homogeneity" E attributed "to space and time",
you are right insofar as "@tau/@t = delta tau /delta ... = 1"
is concerned.
In P1! @t/@tau = delta t /delta tau = dt/dtau = a = 1.
However, the little detail you slipped in and on is that rather
than dt/dtau = dtau/dt - 1, as in your greasy equation (which is
precisely
what tripped up Einstein), if a = dt/dtau = 1 as measured by system
k',
then dtau/dt IS NOT equal to a = 1 as measured by system K!!
You, John W, are the only one on these newsgroups who pinpointed the
small bimp in Einstein's paper, "Hence, if x' be chosen
infinitesimally
small,"
Although you still don't know it, eqs 4 and 5 were not derived from
eq 3; nor was eq 7 derived from anything in P2. It was put there to
bridge
the unbridgeable gap (*) in his treatment.
Although it is the most brilliant equation in his entire paper, it
was and is a HUGE {though heretofore invisible} bimp -- an invention
to join what came before and what came after it.
* Even though a = dt/dt' DID equal 1 in P1, and a = dtau/dt DOES
equal sqrt(1-v^2/c^2) in P2, (both of which satisfy eq 7!); there is
no
logical or mathematical way to let it be both.
Regards,
glird