On Sunday, 24 April 2022 at 10:41:21 UTC+2, J. J. Lodder wrote:
> RichD <r_dela...@yahoo.com> wrote:
>
> > On April 21, Thomas Heger wrote:
> > > Betrand Russel was head of the 'Fabians'.
> > > His nickname was 'Golem of Venice' and he was supposed to be a
> > > representaive of what is called 'Black Nobility'.
> > > One of Russels's 'achievements' was actually socialism, for what we
> > > common mortals are still 'thankful'.
> > > The Russel familly had held the title 'Earl of Tavistook', too.
> > > This name 'Tavistock' resurfaced together with Naziism and the so called
> > > 'New World Order' several times.
> > > The connection stems from the 'Tavistock Instutute for Human Relations'
> > > (aka: 'Scriptwriters of the New World Order').
> > > Befor that Institute got its current name, it was called 'Freud Hilton'.
> >
> > A hypothetical question: if a man is completely nuts, is it
> > biologically possible for him to recognize this fact?
> Good question indeed. While it cannot be excluded in general
> I guess this particular case is hopeless.

And your is hopeless for sure.

Am 24.04.2022 um 10:23 schrieb JanPB:

>>>> Before you crank out insults, you should at least ask google for the
>>>> meaning of 'Gallileo transformation'.
>>>>
>>>> The very first result would be this:
>>>>
>>>> https://en.wikipedia.org/wiki/Galilean_transformation
>>>>
>>>> Quote:
>>>>
>>>> " x' = x- v*t "
>>>>
>>> Just because the Galilean transformation is x' = x-vt doesn't mean every
>>> instance of "x'=x-vt" is the Galilean Transform.
>>> This is an example of the logical fallacy: If A then B implies If B then
>>> A. (If NOT B then NOT A is true, however)
>> You are absolute right, that this was not meant as Galilean
>> transformation, even if the equation of that transform was used.
>>
>> In fact Einstein made an error:
>>
>> he used 'x' in his equation "x' = x- v*t"
>
> No, this is correct.
>
>> Instead of that, actually:
>> x' = xsi - v*t
>> was meant with xsi=0.
>
> No.
>
>> The setting was: a light beam was sent from the zero spot of k to the
>> point x'. There was a mirror placed, which reflected the beam back to
>> the source.
>>
>> Now the zero spot of k has the xsi-coordinate xsi=0.
>>
>> This point moves along the X-axis of K with velocity v. Now we have:
>> x' = xsi - v*t (correcting Einstein's obvious error from above).
>
> No, this is wrong. x' is defined as x - vt. It's just a definition of a new
> variable called x', introduced for convenience. One could call this
> "Galilean transformation" only in the sense that the coordinates (x, y, z, t)
> are being changed to (x', y, z, t) the "Galilean" way but the physical
> context is wrong: the term "such-and-such transform" usually refers
> to two observers whose space and time coordinates are being related
> by the transformation. But here the context is that we have only one
> observer (K) who for convenience relabels one of the axes.

What did you mean by:

"It's just a definition of a new variable called x', introduced for
convenience."

To me this sounds like nonsense, because what kind of convenience is
required and for which purpose? Where was that purpose introduced?

This is the sentence of §3, page 6, which contains that equation.

"If we place x' = x −vt, it is clear that a point at rest in the system
k must have a system of values x', y, z, independent of time."

Now the question arises, what Einstein meant with his statement.

From the context I had assumed, he provided the mathematical
description to his 'thought experiment', where a beam of light is
emitted from the center of k towards a mirror placed somewhere on the
x-axis of K at a point with the name x'.

But if you disagree about this interpretation, I'm lost, because after
weeks of dealing with that particular part, I have not found any other
possible interpretation.

Therefore, I have to ask you, if you could be so kind and say, what
Einstein actually meant with the quoted text.

> Note that it's important that this relabelling be linear since the final
> transform (x, y, z, t) -> (ksi, eta, zeta, tau) is required to be linear.
>
>> Therefore the point x' must be the zero spot of K, which has the
>> x-coordinate x'=0.
>
> There is no "spot of K" corresponding to x'. What x' is is simply
> a mathematical formula involving K's x and t coordinates designed
> to remain a fixed number when x and t refer to a point moving at
> speed v wrt K.
>
>> This is all nice and possible, but would invalidate most of the
>> subsequent equations for a number of reasons.
>
> Nothing is invalidated, everything is corret there.
>
>> For instance del x' in the denominator would not make much sense, if x'=0.
>
> I'm compelled to ask at this point: do you know calculus?
>

To me it is totally idiotic to put zero into a denominator in a
differential equation.

But if x' denotes a fixed position, the value of del x'/del tau is zero.
Now the inverse would be infinite.

That can hardly be what Einstein had in mind.

So please be so kind and explain to me, what else he had in mind.

TH

On Sunday, April 24, 2022 at 2:08:12 AM UTC-7, Thomas Heger wrote:
> Am 24.04.2022 um 10:23 schrieb JanPB:
>
> >>>> Before you crank out insults, you should at least ask google for the
> >>>> meaning of 'Gallileo transformation'.
> >>>>
> >>>> The very first result would be this:
> >>>>
> >>>> https://en.wikipedia.org/wiki/Galilean_transformation
> >>>>
> >>>> Quote:
> >>>>
> >>>> " x' = x- v*t "
> >>>>
> >>> Just because the Galilean transformation is x' = x-vt doesn't mean every
> >>> instance of "x'=x-vt" is the Galilean Transform.
> >>> This is an example of the logical fallacy: If A then B implies If B then
> >>> A. (If NOT B then NOT A is true, however)
> >> You are absolute right, that this was not meant as Galilean
> >> transformation, even if the equation of that transform was used.
> >>
> >> In fact Einstein made an error:
> >>
> >> he used 'x' in his equation "x' = x- v*t"
> >
> > No, this is correct.
> >
> >> Instead of that, actually:
> >> x' = xsi - v*t
> >> was meant with xsi=0.
> >
> > No.
> >
> >> The setting was: a light beam was sent from the zero spot of k to the
> >> point x'. There was a mirror placed, which reflected the beam back to
> >> the source.
> >>
> >> Now the zero spot of k has the xsi-coordinate xsi=0.
> >>
> >> This point moves along the X-axis of K with velocity v. Now we have:
> >> x' = xsi - v*t (correcting Einstein's obvious error from above).
> >
> > No, this is wrong. x' is defined as x - vt. It's just a definition of a new
> > variable called x', introduced for convenience. One could call this
> > "Galilean transformation" only in the sense that the coordinates (x, y, z, t)
> > are being changed to (x', y, z, t) the "Galilean" way but the physical
> > context is wrong: the term "such-and-such transform" usually refers
> > to two observers whose space and time coordinates are being related
> > by the transformation. But here the context is that we have only one
> > observer (K) who for convenience relabels one of the axes.
> What did you mean by:
> "It's just a definition of a new variable called x', introduced for
> convenience."
> To me this sounds like nonsense, because what kind of convenience is
> required and for which purpose? Where was that purpose introduced?

The convenience is that if K refers to points moving along the x-axis
with speed v by means the variable x', its values will be constant.
So e.g. the emitter/receiver sits at x' equal to 0 and the mirror
sits at x' equal to some nonzero number which is likewise constant
(and denoted by x' by the standard abuse of notation).

If Einstein did not use x', the derivation would be bit more involved
because the emitter/receiver would be at the value of x equal to vt
(so t-dependent, not a constant like 0) and the mirror at the instant
of the ray emission would be at the value of x equal to d + v(t + delta)
where d is the separation between the emitter/receiver and the mirror,
as measured by K, and delta is the ray's flight time emitter->mirror,
again as measured by K.

The equation for (1/2)(tau_0 + tau_2) = tau_1 would thus have to be
a bit more complicated using (x, y, z, t) directly:

(1/2)*(tau(vt, 0, 0, t) + tau(v(t + d/(c-v) + d/(c+v)), 0, 0, t + d/(c-v) + d/(c+v))
= tau(d + v(t + d/(c-v)), 0, 0, t + d/(c-v))

Then differentiating with respect to d:

(1/2)*(v/(c-v) + v/(c+v))*dtau/dx + (1/(c-v) + 1/(c+v))*dtau/dt =

= (1 + v/(c-v))*dtau/dx + (1/(c-v))*dtau/dt

So a bit more complicated formula. Straightening it out yields:

dtau/dx = (-v/c^2)*dtau/dt

Since tau is presumed linear, we have (suppressing the y and
z dependence to save time):

tau = Ax + Bt

....for some v-dependent constants A and B, with:

A = dtau/dx and B = dtau/dt

So we have A = (-v/c^2)B, so:

tau = Bt - (vx/c^2)B = B(t - vx/c^2)

So it's a bit more work without the help of the x' trick.

Doodling in a similar way with the variables y and z
results in writing B as the product of 1/sqrt(1 - v^2/c^2)
and a still unknown function phi(v) which is a bit later
shown to be identically 1.

> This is the sentence of §3, page 6, which contains that equation.
>
> "If we place x' = x −vt, it is clear that a point at rest in the system
> k must have a system of values x', y, z, independent of time."
>
> Now the question arises, what Einstein meant with his statement.

He meant exactly what's written there: objects at rest in k have
their x'-numbers constant in time.

> From the context I had assumed, he provided the mathematical
> description to his 'thought experiment', where a beam of light is
> emitted from the center of k towards a mirror placed somewhere on the
> x-axis of K at a point with the name x'.

It's better to say "...placed somewhere on the x-axis of k" (not K) because
both the light source and the mirror are at rest in k.

> But if you disagree about this interpretation, I'm lost, because after
> weeks of dealing with that particular part, I have not found any other
> possible interpretation.
>
> Therefore, I have to ask you, if you could be so kind and say, what
> Einstein actually meant with the quoted text.

Pick any object at rest in k. It could be that light source or
the mirror, or anything else at rest in k. So at any time t that object
will be at position x (both t and x as measured by K).

If you write down a series of those corresponding pairs (t,x), you'll
notice that a particular combination of them always yields the same
number, that combination being x - vt.

That number is equal to the x-coordinate the object occupied
at time t = 0. Einstein sets things up so that the light source is
at x = 0 when t = 0, hence the light source has x - vt = 0 always.

Similarly, the mirror sits at some distance d from the origin of K at time
t = 0, so for the mirror x - vt = d always.

Einstein denotes the numbers x - vt by x'. He can do it because this
combination is a linear function, so after he solves the problem in terms
of (x', y, z, t), all he has left to do is substitute x' = x - vt back, without
screwing up the linearity.

He also observes, that the derivation works the same if the light source
were at a point other than the origin at t = 0. This is because the
change it would introduce to the constraining equation would
disappear after the differentiation, so the same formula for
dtau/dx and dtau/dt results regardless.

This is important because if that formula were valid only for only
one value of x, it would be useless as a transformation.

> > Note that it's important that this relabelling be linear since the final
> > transform (x, y, z, t) -> (ksi, eta, zeta, tau) is required to be linear.
> >
> >> Therefore the point x' must be the zero spot of K, which has the
> >> x-coordinate x'=0.
> >
> > There is no "spot of K" corresponding to x'. What x' is is simply
> > a mathematical formula involving K's x and t coordinates designed
> > to remain a fixed number when x and t refer to a point moving at
> > speed v wrt K.
> >
> >> This is all nice and possible, but would invalidate most of the
> >> subsequent equations for a number of reasons.
> >
> > Nothing is invalidated, everything is corret there.
> >
> >> For instance del x' in the denominator would not make much sense, if x'=0.
> >
> > I'm compelled to ask at this point: do you know calculus?
> >
> To me it is totally idiotic to put zero into a denominator in a
> differential equation.

But he does not put zero in that equation in the denominator (besides,
it's not a denominator, it's a half of the derivative symbol).

> But if x' denotes a fixed position,

Fixed in k.

> the value of del x'/del tau is zero.

We are calculating dtau/dx', not dx'/dtau. And at this point
we are *differentiating* with respect to x'. This means we
consider the entire experiment repeated for different
values of x' and then look at the limit of the relevant quotient
as x' goes to zero ("x' be chosen infinitesimally small" as they
used to say in the old-math-speak).

JanPB wrote:

>> > WHAT??
>> 'tau' was defined as a linear function, but was otherwise not
>> specified.
>
> Of course it was not specified, it's unknown at this point. Einstein is
> setting out to derive a formula for it.

are you telling me, the polakia is refusing immigrants from other
countries they bombed, with military support, was *mass_murdered* in
hundreds of thousands, see wikipedia etc, by the nazi "ukraine" along the
ww2 etc, knowingly the "ukraine" is nazi now, in politics, infrastructure,
military etc, and you still give asylum to the millions of nazi families
killing the polakians?? Ohh my, the polakia is nothing but a nazi
provincial territory stolen from Russia. The above serves as proofs.

On 4/24/2022 5:08 AM, Thomas Heger wrote:
> Am 24.04.2022 um 10:23 schrieb JanPB:
>
>>>>> Before you crank out insults, you should at least ask google for the
>>>>> meaning of 'Gallileo transformation'.
>>>>>
>>>>> The very first result would be this:
>>>>>
>>>>> https://en.wikipedia.org/wiki/Galilean_transformation
>>>>>
>>>>> Quote:
>>>>>
>>>>> " x' = x- v*t "
>>>>>
>>>> Just because the Galilean transformation is x' = x-vt doesn't mean
>>>> every
>>>> instance of "x'=x-vt" is the Galilean Transform.
>>>> This is an example of the logical fallacy: If A then B implies If B
>>>> then
>>>> A. (If NOT B then NOT A is true, however)
>>> You are absolute right, that this was not meant as Galilean
>>> transformation, even if the equation of that transform was used.
>>>
>>> In fact Einstein made an error:
>>>
>>> he used 'x' in his equation "x' = x- v*t"
>>
>> No, this is correct.
>>
>>> Instead of that, actually:
>>> x' = xsi - v*t
>>> was meant with xsi=0.
>>
>> No.
>>
>>> The setting was: a light beam was sent from the zero spot of k to the
>>> point x'. There was a mirror placed, which reflected the beam back to
>>> the source.
>>>
>>> Now the zero spot of k has the xsi-coordinate xsi=0.
>>>
>>> This point moves along the X-axis of K with velocity v. Now we have:
>>> x' = xsi - v*t (correcting Einstein's obvious error from above).
>>
>> No, this is wrong.  x'  is defined as  x - vt.  It's just a definition
>> of a new
>> variable called  x',  introduced for convenience.  One could call this
>> "Galilean transformation" only in the sense that the coordinates  (x,
>> y, z, t)
>> are being changed to  (x', y, z, t) the "Galilean" way but the physical
>> context is wrong: the term "such-and-such transform" usually refers
>> to two observers whose space and time coordinates are being related
>> by the transformation. But here the context is that we have only one
>> observer (K) who for convenience relabels one of the axes.
>
> What did you mean by:
>
> "It's just a definition of a new variable called  x',  introduced for
> convenience."
>
> To me this sounds like nonsense, because what kind of convenience is
> required and for which purpose? Where was that purpose introduced?

It's so a certain calculation or transformation performed repeatedly can
be done just once and assigned to a symbol, and the symbol used instead
of repeating the calculation/transformation again and again.

You are already familiar with this concept, it the same as writing "γ"
instead of "1/√(1-v²/c²) repeatedly in SR calculations. Similarly,
Einstein wrote "x'" = "x-vt" to make it clearer for his target audience
because it simplifies his math. (since you are not among his target
audience, it's understandable that you get confused)
>
>>> For instance del x' in the denominator would not make much sense, if
>>> x'=0.
>>
>> I'm compelled to ask at this point: do you know calculus?
>>
>
> To me it is totally idiotic to put zero into a denominator in a
> differential equation.

By writing that, it looks like you don't know calculus since "/dx'" or
"del x'" is just PART of a differentiation symbol.

On Sunday, April 24, 2022 at 5:14:32 AM UTC-7, I wrote:
>
> That number is equal to the x-coordinate the object occupied
> at time t = 0. Einstein sets things up so that the light source is
> at x = 0 when t = 0, hence the light source has x - vt = 0 always.

This is not correct, I worked too hard here. To reiterate correctly:

(a) the light source is at the origin of k (the "moving" system),
but it's NOT assumed that it's at the origin of K at the instant
of the emission of the ray,

(b) so it emits the light ray at some K-time t0 and K-position x0
(along the x-axis),

(c) because the transform we seek is presumed linear, it must
transform (x, y, z, t) = (0, 0, 0, 0) to (ksi, eta, zeta, tau) = (0, 0, 0, 0).
IOW, the origins of K and k coincide when both t = 0 and tau = 0,

(d) Hence, (a), (b), and (c) imply that x0 = v*t0,

(e) Hence, at all times t (not just at t = t0) the light source's
(x, t) coordinates satisfy: x = x0 + v*(t - t0),

(f) multiplying the above through, and using x0 = v*t0 yields:
x = x0 + v*t - v*t0 where x0 and v*t0 cancel out, so we end up
with: x = vt.

That's why Einstein only assumes (at first) that the light source
is at the origin of k but not necessarily at the origin of K. The
combination x - vt (which Einstein calls x') will be zero for
the light source.

All this then results in a formula for tau(t, x) which is valid only for
all (t, x) such that x = vt. But Einstein needs this formula for
ALL values of t and x. So he makes the comment that:

(g) "It is to be noted that instead of the origin of the
co-ordinates [of k] we might have chosen any other point for the
point of origin of the ray, and the equation just obtained
is therefore valid for all values of [t and x]".

He doesn't go into more details at this point because it's a very
easy exercise to rewrite the constraint equation for tau in such
a way that the light source is not at the origin of k. Such
rewritten equation nevertheless yields the same differential
equation for dtau/dx' and dtau/dt as before.

So the conclusion is that t and x' can be arbitrary in the formula
for tau.

--
Jan

JanPB wrote:

>> That number is equal to the x-coordinate the object occupied at time t
>> = 0. Einstein sets things up so that the light source is at x = 0 when
>> t = 0, hence the light source has x - vt = 0 always.
>
> This is not correct, I worked too hard here. To reiterate correctly:
> (a) the light source is at the origin of k (the "moving" system),
> but it's NOT assumed that it's at the origin of K at the instant of
> the emission of the ray,

Not true. Zelenske is a comedian actor, a nazi homo. The "president", if
such that do exists, is america. I suspect the nazi america will get
fucked up in a way or another. It can't play this gameshit, arming the
nazis.

JanPB wrote:

>> Some of these errors were extremely stupid, some very small, some
>> formal.
>
> There are no errors in Einstein's 1905 paper. It's all your imagination.
> (I'm assuming you mean by "error" what people cutomarily take it to
> mean.)

More proofs:

Die Wahrheit über den Schauspieler Selenski der Euch in den dritten
Weltkrieg führen will🤮 https://www.bitchute.com/video/qCCl5XnOCTP6/

Am 24.04.2022 um 10:52 schrieb JanPB:

>> But even if so, x' would still be zero.
>
> Yes, you get x' = 0 for the emitter/receiver because its location
> by construction of this experimental setup satisfies x = vt at
> all times t, hence the corresponding x' value is:
>
> x' = x - vt = 0
>
> throughout.

Here you actually agree, that x'=0 must be true.

> You get confused by the multiple uses of x' (the standard notational
> abuse). What Einstein does is he puts the emitter/receiver at x' = 0
> and the mirror at some nonzero value of x' which he also calls x'

'notational abuse' is in my world 'plain wrong'.

IOW: if a veriable is defined somehow, that definition remains valid,
unless the veriable is redefined with a new meaning explicitly.

So: we are not allowed to declare something is just 'notational abuse'.
Instead we have to use even the wrong definitions from the text, if
there are any, because we must not change a text to our own liking.

A text is what it is, whether we like it or not.

> (notational abuse). Then he varies the location of the mirror and
> compares the resulting differences, this is what the differentiation
> with respect to x' that follows does.

No, Einstein wrote, that the ray was emitted from the zero spot of k.
This is the moving system.

The zero spot of k has a coordinate in K, too, which is
xsi_0= x_0 + v*t

Since the emitter was located at a fixed point in k with cordinates in k
xsi=0, eta=0,m zeta=0, the mirror cannot be place at the same spot.

Contrary to your assumption, we are forced to assume, that x' was the
position of the MIRROR, even if Einstein had not written so.

There is actually no other possibility.

....

TH

Am 24.04.2022 um 18:56 schrieb Michael Moroney:
> On 4/24/2022 5:08 AM, Thomas Heger wrote:
>> Am 24.04.2022 um 10:23 schrieb JanPB:
>>
>>>>>> Before you crank out insults, you should at least ask google for the
>>>>>> meaning of 'Gallileo transformation'.
>>>>>>
>>>>>> The very first result would be this:
>>>>>>
>>>>>> https://en.wikipedia.org/wiki/Galilean_transformation
>>>>>>
>>>>>> Quote:
>>>>>>
>>>>>> " x' = x- v*t "
>>>>>>
>>>>> Just because the Galilean transformation is x' = x-vt doesn't mean
>>>>> every
>>>>> instance of "x'=x-vt" is the Galilean Transform.
>>>>> This is an example of the logical fallacy: If A then B implies If B
>>>>> then
>>>>> A. (If NOT B then NOT A is true, however)
>>>> You are absolute right, that this was not meant as Galilean
>>>> transformation, even if the equation of that transform was used.
>>>>
>>>> In fact Einstein made an error:
>>>>
>>>> he used 'x' in his equation "x' = x- v*t"
>>>
>>> No, this is correct.
>>>
>>>> Instead of that, actually:
>>>> x' = xsi - v*t
>>>> was meant with xsi=0.
>>>
>>> No.
>>>
>>>> The setting was: a light beam was sent from the zero spot of k to the
>>>> point x'. There was a mirror placed, which reflected the beam back to
>>>> the source.
>>>>
>>>> Now the zero spot of k has the xsi-coordinate xsi=0.
>>>>
>>>> This point moves along the X-axis of K with velocity v. Now we have:
>>>> x' = xsi - v*t (correcting Einstein's obvious error from above).
>>>
>>> No, this is wrong. x' is defined as x - vt. It's just a
>>> definition of a new
>>> variable called x', introduced for convenience. One could call this
>>> "Galilean transformation" only in the sense that the coordinates (x,
>>> y, z, t)
>>> are being changed to (x', y, z, t) the "Galilean" way but the physical
>>> context is wrong: the term "such-and-such transform" usually refers
>>> to two observers whose space and time coordinates are being related
>>> by the transformation. But here the context is that we have only one
>>> observer (K) who for convenience relabels one of the axes.
>>
>> What did you mean by:
>>
>> "It's just a definition of a new variable called x', introduced for
>> convenience."
>>
>> To me this sounds like nonsense, because what kind of convenience is
>> required and for which purpose? Where was that purpose introduced?
>
> It's so a certain calculation or transformation performed repeatedly can
> be done just once and assigned to a symbol, and the symbol used instead
> of repeating the calculation/transformation again and again.
>
> You are already familiar with this concept, it the same as writing "γ"
> instead of "1/√(1-v²/c²) repeatedly in SR calculations. Similarly,
> Einstein wrote "x'" = "x-vt" to make it clearer for his target audience
> because it simplifies his math. (since you are not among his target
> audience, it's understandable that you get confused)
>>
>>>> For instance del x' in the denominator would not make much sense, if
>>>> x'=0.
>>>
>>> I'm compelled to ask at this point: do you know calculus?
>>>
>>
>> To me it is totally idiotic to put zero into a denominator in a
>> differential equation.
>
> By writing that, it looks like you don't know calculus since "/dx'" or
> "del x'" is just PART of a differentiation symbol.

Ok, I made actually an error, because del tau referrs to the velocity v,
but seen from system k, not from K.

If the position of x' is at the zero spot of K, it would not move in K,
because it is always at the same (zero-)spot.

The position of the mirror moves only in k, in respect to the emitter at
the zero spot of k with velocity v*tau to the left (the direction of
smaller xsi-values).

Now the derivative in K dx'/dt=0, because x' remains zero.

Seen from k, the derivative dx'/d tau would be v and the inverse
d tau/dx'= 1/v

For partial differential equations I would see no need, because there is
actually only the very simple relation dx'/d tau =v.

TH

Am 24.04.2022 um 23:09 schrieb Dong Vassilikos:
> JanPB wrote:
>
>>> That number is equal to the x-coordinate the object occupied at time t
>>> = 0. Einstein sets things up so that the light source is at x = 0 when
>>> t = 0, hence the light source has x - vt = 0 always.
>>
>> This is not correct, I worked too hard here. To reiterate correctly:
>> (a) the light source is at the origin of k (the "moving" system),
>> but it's NOT assumed that it's at the origin of K at the instant of
>> the emission of the ray,
>
> Not true. Zelenske is a comedian actor,...

This is actually true. See here:

Volodymyr Zelensky: From actor and comedian to taking on Putin | 60
Minutes Australia

https://www.youtube.com/watch?v=iWntXrLFaSc

I had actually problems with his talks to the British premier Boris
Johnson in Kiev. To me the visit didn't make any sense at all, because
it is totally impossible for a premier to travell into an active war
zone, which is not controlled by the Brits.

So, why not thinking the other way round and let Zelinsky visit Boris
Johnson?

The relocation of the film to Kiev could be done with 'green screen' and
CGI technology.

So we should have a close look at the pictures from Kiev and search for
remainders of green backgrounds.

And: Next time the Brits should hire better actors. Zelensky was simply
a bad choice, because he resembles more a truck driver or a man from the
hardware store, rather than a president.

Even in c-class pictures, they had better presidents.

TH

On 4/25/2022 4:09 AM, Thomas Heger wrote:
> Am 24.04.2022 um 18:56 schrieb Michael Moroney:
>> On 4/24/2022 5:08 AM, Thomas Heger wrote:
>>> Am 24.04.2022 um 10:23 schrieb JanPB:
>>>
>>>>>>> Before you crank out insults, you should at least ask google for the
>>>>>>> meaning of 'Gallileo transformation'.
>>>>>>>
>>>>>>> The very first result would be this:
>>>>>>>
>>>>>>> https://en.wikipedia.org/wiki/Galilean_transformation
>>>>>>>
>>>>>>> Quote:
>>>>>>>
>>>>>>> " x' = x- v*t "
>>>>>>>
>>>>>> Just because the Galilean transformation is x' = x-vt doesn't mean
>>>>>> every
>>>>>> instance of "x'=x-vt" is the Galilean Transform.
>>>>>> This is an example of the logical fallacy: If A then B implies If B
>>>>>> then
>>>>>> A. (If NOT B then NOT A is true, however)
>>>>> You are absolute right, that this was not meant as Galilean
>>>>> transformation, even if the equation of that transform was used.
>>>>>
>>>>> In fact Einstein made an error:
>>>>>
>>>>> he used 'x' in his equation "x' = x- v*t"
>>>>
>>>> No, this is correct.
>>>>
>>>>> Instead of that, actually:
>>>>> x' = xsi - v*t
>>>>> was meant with xsi=0.
>>>>
>>>> No.
>>>>
>>>>> The setting was: a light beam was sent from the zero spot of k to the
>>>>> point x'. There was a mirror placed, which reflected the beam back to
>>>>> the source.
>>>>>
>>>>> Now the zero spot of k has the xsi-coordinate xsi=0.
>>>>>
>>>>> This point moves along the X-axis of K with velocity v. Now we have:
>>>>> x' = xsi - v*t (correcting Einstein's obvious error from above).
>>>>
>>>> No, this is wrong.  x'  is defined as  x - vt.  It's just a
>>>> definition of a new
>>>> variable called  x',  introduced for convenience.  One could call this
>>>> "Galilean transformation" only in the sense that the coordinates  (x,
>>>> y, z, t)
>>>> are being changed to  (x', y, z, t) the "Galilean" way but the physical
>>>> context is wrong: the term "such-and-such transform" usually refers
>>>> to two observers whose space and time coordinates are being related
>>>> by the transformation. But here the context is that we have only one
>>>> observer (K) who for convenience relabels one of the axes.
>>>
>>> What did you mean by:
>>>
>>> "It's just a definition of a new variable called  x',  introduced for
>>> convenience."
>>>
>>> To me this sounds like nonsense, because what kind of convenience is
>>> required and for which purpose? Where was that purpose introduced?
>>
>> It's so a certain calculation or transformation performed repeatedly can
>> be done just once and assigned to a symbol, and the symbol used instead
>> of repeating the calculation/transformation again and again.
>>
>> You are already familiar with this concept, it the same as writing "γ"
>> instead of "1/√(1-v²/c²) repeatedly in SR calculations. Similarly,
>> Einstein wrote "x'" = "x-vt" to make it clearer for his target audience
>> because it simplifies his math.  (since you are not among his target
>> audience, it's understandable that you get confused)
>>>
>>>>> For instance del x' in the denominator would not make much sense, if
>>>>> x'=0.
>>>>
>>>> I'm compelled to ask at this point: do you know calculus?
>>>>
>>>
>>> To me it is totally idiotic to put zero into a denominator in a
>>> differential equation.
>>
>> By writing that, it looks like you don't know calculus since "/dx'" or
>> "del x'" is just PART of a differentiation symbol.
>
>
> Ok, I made actually an error, because del tau referrs to the velocity v,
> but seen from system k, not from K.
>
> If the position of x' is at the zero spot of K, it would not move in K,
> because it is always at the same (zero-)spot.

Except that it's not, as has been explained to you.

On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
> Am 24.04.2022 um 10:52 schrieb JanPB:
>
> >> But even if so, x' would still be zero.
> >
> > Yes, you get x' = 0 for the emitter/receiver because its location
> > by construction of this experimental setup satisfies x = vt at
> > all times t, hence the corresponding x' value is:
> >
> > x' = x - vt = 0
> >
> > throughout.
> Here you actually agree, that x'=0 must be true.

For the *light source*.

The differentiation OTOH is taken with respect to the *mirror* position
(denoted by x' by the abuse of notation) which is nonzero (obviously).

> > You get confused by the multiple uses of x' (the standard notational
> > abuse). What Einstein does is he puts the emitter/receiver at x' = 0
> > and the mirror at some nonzero value of x' which he also calls x'
> 'notational abuse' is in my world 'plain wrong'.

No, it's the standard. The term "notational abuse" is a bit of
a joke, it does not indicate anything is wrong. This notation is so
frequently used because the alternative can make make formulas
harder to read.

> IOW: if a veriable is defined somehow, that definition remains valid,
> unless the veriable is redefined with a new meaning explicitly.

In some cases it's not mentioned explicitly by a hundreds-years old
convention. Everyone in the profession knows about it.

> So: we are not allowed to declare something is just 'notational abuse'.

Yes, we are and we do. This is a required part of everyone's education
in science, like the scientific notation, say, or the use of the Greek alphabet.

> Instead we have to use even the wrong definitions from the text, if
> there are any, because we must not change a text to our own liking.

The definition is not wrong. Only a small change of context occurs
silently. Everyone understands that sort of thing.

> A text is what it is, whether we like it or not.

I do like it, and the text is correct.

> > (notational abuse). Then he varies the location of the mirror and
> > compares the resulting differences, this is what the differentiation
> > with respect to x' that follows does.
> No,

Yes.

> Einstein wrote, that the ray was emitted from the zero spot of k.
> This is the moving system.

OK.

> The zero spot of k has a coordinate in K, too, which is
> xsi_0= x_0 + v*t

ksi is not K's coordinate.

> Since the emitter was located at a fixed point in k with cordinates in k
> xsi=0, eta=0,m zeta=0, the mirror cannot be place at the same spot.

The mirror is not at the same spot.

> Contrary to your assumption, we are forced to assume, that x' was the
> position of the MIRROR, even if Einstein had not written so.

No, that's wrong.

> There is actually no other possibility.

No, incorrect.

I'm beginning to have some doubts you'll ever begin to understand this.
It would be probably possible to clear this up face to face in front of
a whiteboard. A forum like this is just not interactive enough to
catch your mistakes in real time, they pile up and up to infinity at each
iteration. Every time something is cleared up, you come up with 10x as
many even more basic mistakes.

And now we are at the level of debating the fine points of a *notational*
tradition that goes back, uninterrupted and in constant use, since the
18th century, or earlier.

And now you attack it because of its funny *name*.

--
Jan

Am 25.04.2022 um 21:22 schrieb JanPB:
> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
>> Am 24.04.2022 um 10:52 schrieb JanPB:
>>
>>>> But even if so, x' would still be zero.
>>>
>>> Yes, you get x' = 0 for the emitter/receiver because its location
>>> by construction of this experimental setup satisfies x = vt at
>>> all times t, hence the corresponding x' value is:
>>>
>>> x' = x - vt = 0
>>>
>>> throughout.
>> Here you actually agree, that x'=0 must be true.
>
> For the *light source*.

No, this is wrong because Einstein had a different setting:

[quote]

"From the origin of system k let a ray be emitted at the time τ'along
the X-axis to x' and at the time τ_1 be reflected thence to the origin of
the coordinates, ..."

[/quote]

The system k was the moving system, hence Einstein wanted a moving
emitter and a stationary mirror.

The mirror should be located at x', because the text says so.

Because the zero spot of k has the x-coordinate in K of
xsi_0 = x_0 * v*t
we end up at the zero spot of K, if we subtract v*t from xsi_0.

> The differentiation OTOH is taken with respect to the *mirror* position
> (denoted by x' by the abuse of notation) which is nonzero (obviously).

No, x' is OBVIOUSLY the zero spot of K.

It makes no sense to write partial differential equations for a problem,
where only the x-axis is involved. But still we cannot differentiate a
fixed position, at least not in K.

Seen from the emitter, the mirrors moves and we have the usual relation
v= x'/d tau.

Now Einstein used the inverse 1/v in his equation, what is blatant nonsense.

This is nonsense, because the possible case of v=0 would produce an
infinite term.

I have complained about this fact a number of times, but instead of a
positive reply I was called a nutcase.

>>> You get confused by the multiple uses of x' (the standard notational
>>> abuse). What Einstein does is he puts the emitter/receiver at x' = 0
>>> and the mirror at some nonzero value of x' which he also calls x'
>> 'notational abuse' is in my world 'plain wrong'.
>
> No, it's the standard. The term "notational abuse" is a bit of
> a joke, it does not indicate anything is wrong. This notation is so
> frequently used because the alternative can make make formulas
> harder to read.

OK, I cannot decode insider jokes, because I'm actually an outsider.

So: to me a "notational abuse" is an error, period.

>> IOW: if a veriable is defined somehow, that definition remains valid,
>> unless the veriable is redefined with a new meaning explicitly.
>
> In some cases it's not mentioned explicitly by a hundreds-years old
> convention. Everyone in the profession knows about it.

Is this also an insider joke?

>> So: we are not allowed to declare something is just 'notational abuse'.
>
> Yes, we are and we do. This is a required part of everyone's education
> in science, like the scientific notation, say, or the use of the Greek alphabet.

I have actually occasionally the impression, that education of
physicists includes brainwashing.

>> Instead we have to use even the wrong definitions from the text, if
>> there are any, because we must not change a text to our own liking.
>
> The definition is not wrong. Only a small change of context occurs
> silently. Everyone understands that sort of thing.

???????????

>> A text is what it is, whether we like it or not.
>
> I do like it, and the text is correct.
>
>>> (notational abuse). Then he varies the location of the mirror and
>>> compares the resulting differences, this is what the differentiation
>>> with respect to x' that follows does.
>> No,
>
> Yes.
>
>> Einstein wrote, that the ray was emitted from the zero spot of k.
>> This is the moving system.
>
> OK.
>
>> The zero spot of k has a coordinate in K, too, which is
>> xsi_0= x_0 + v*t
>
> ksi is not K's coordinate.

Sure, the zero spot of k has also a coordinate in K.

The relation is totally simple: xsi_0= x_0 * v*t

>
>> Since the emitter was located at a fixed point in k with cordinates in k
>> xsi=0, eta=0,m zeta=0, the mirror cannot be place at the same spot.
>
> The mirror is not at the same spot.

Of course the mirror cannot coincide with the emitter.

>> Contrary to your assumption, we are forced to assume, that x' was the
>> position of the MIRROR, even if Einstein had not written so.
>
> No, that's wrong.
>
>> There is actually no other possibility.
>
> No, incorrect.
>

TH

Am 25.04.2022 um 17:51 schrieb Michael Moroney:

>>>>>> For instance del x' in the denominator would not make much sense, if
>>>>>> x'=0.
>>>>>
>>>>> I'm compelled to ask at this point: do you know calculus?
>>>>>
>>>>
>>>> To me it is totally idiotic to put zero into a denominator in a
>>>> differential equation.
>>>
>>> By writing that, it looks like you don't know calculus since "/dx'" or
>>> "del x'" is just PART of a differentiation symbol.
>>
>>
>> Ok, I made actually an error, because del tau referrs to the velocity
>> v, but seen from system k, not from K.
>>
>> If the position of x' is at the zero spot of K, it would not move in
>> K, because it is always at the same (zero-)spot.
>
> Except that it's not, as has been explained to you.

Now you disagree to an obvious consequence of Einstein's text.

Your assumption is wrong, because Einstein had a different setting:

[quote from §3, page 6]

"From the origin of system k let a ray be emitted at the time τ'along
the X-axis to x' and at the time τ_1 be reflected thence to the origin of
the coordinates, ..."

[/quote]

The system k was the moving system, hence Einstein wanted a moving
emitter and a stationary mirror at point x'.

Einstein used this x' in the subsequent equation inside a four-vector.

The four vector was the argument of a function named 'tau', which was
assumed to be linear and which produced four-vectors as output.

Then he used an astonishing trick.

To see, that his method was rather obscure, I change the equation a bit,
because we only need one spatial axis, not three.

The independent variable of the function is actually t, while the
dependent variable is x.

That would be a description of a moving point along the x-axis, where x
depends on t.

Now we take a function - say- f(t)=x

This is a linear function, because x was defined as x=v*t, with v as a
constant.

The first derivative of this function is obviously v, while Einstein
wrote something like del x/ del tau = del t/del tau.

TH

On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
> Am 25.04.2022 um 21:22 schrieb JanPB:
> > On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
> >> Am 24.04.2022 um 10:52 schrieb JanPB:
> >>
> >>>> But even if so, x' would still be zero.
> >>>
> >>> Yes, you get x' = 0 for the emitter/receiver because its location
> >>> by construction of this experimental setup satisfies x = vt at
> >>> all times t, hence the corresponding x' value is:
> >>>
> >>> x' = x - vt = 0
> >>>
> >>> throughout.
> >> Here you actually agree, that x'=0 must be true.
> >
> > For the *light source*.
> No, this is wrong because Einstein had a different setting:
>
> [quote]
>
> "From the origin of system k let a ray be emitted at the time τ'along
> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
> the coordinates, ..."
>
> [/quote]
>
> The system k was the moving system, hence Einstein wanted a moving
> emitter and a stationary mirror.

Both the light source and the mirror are stationary in k because that's how
the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).

> The mirror should be located at x', because the text says so.

Yes.

> Because the zero spot of k has the x-coordinate in K of
> xsi_0 = x_0 * v*t

If you say "x-coordinate", then don't say "ksi". So again: if we assume,
as Einstein did, that:

(1) the ray is emitted from the origin of k,

(2) the origins of K and k coincide when t = 0 and tau = 0,

(3) the ray is emitted at some K-time value of t_0,

THEN:

(a) the K x-coordinate x_0 of the emission satisfies: x_0 = v*t_0,

(b) the K x-coordinate x of the light ray pulse at subsequent times
t satisfies:L x = x_0 + v*(t - t_0).

(Note that for t = t_0 we get x = x_0, as expected.)

But x = x_0 + v*(t - t_0) means that x - v*t = x_0 - v*t_0.

By item (a) the RHS is equal to zero. Hence the combination x - v*t
is equal to zero for all times t following the emission (until the
reflection which alters this pattern). Einstein labels this combination
by x'.

Similar consideration yields x' = constant (NOT equal to zero) for the mirror.

> we end up at the zero spot of K, if we subtract v*t from xsi_0.
> > The differentiation OTOH is taken with respect to the *mirror* position
> > (denoted by x' by the abuse of notation) which is nonzero (obviously).
> No, x' is OBVIOUSLY the zero spot of K.
>
> It makes no sense to write partial differential equations for a problem,
> where only the x-axis is involved.

Both sides are functions of x' so they can be differentiated
with respect to x'.

> But still we cannot differentiate a
> fixed position, at least not in K.

When you perform differentiation, you consider multiple instances of the
function values and evaluate the resulting difference quotient limit.

In this case one considers instances of the experiment with different
values of x'.

> Seen from the emitter, the mirrors moves and we have the usual relation
> v= x'/d tau.

Again, a typo?

> Now Einstein used the inverse 1/v in his equation, what is blatant nonsense.

There is no 1/v there.

> This is nonsense, because the possible case of v=0 would produce an
> infinite term.

Sure but there is 1/(c-v) and 1/(c+v) there.

> I have complained about this fact a number of times, but instead of a
> positive reply I was called a nutcase.

You claim something that's not there.

> >>> You get confused by the multiple uses of x' (the standard notational
> >>> abuse). What Einstein does is he puts the emitter/receiver at x' = 0
> >>> and the mirror at some nonzero value of x' which he also calls x'
> >> 'notational abuse' is in my world 'plain wrong'.
> >
> > No, it's the standard. The term "notational abuse" is a bit of
> > a joke, it does not indicate anything is wrong. This notation is so
> > frequently used because the alternative can make make formulas
> > harder to read.
> OK, I cannot decode insider jokes, because I'm actually an outsider.
>
> So: to me a "notational abuse" is an error, period.

No. Error is something different. If you want to call everything you
don't know about "error", then you have no arguments left.

> >> IOW: if a veriable is defined somehow, that definition remains valid,
> >> unless the veriable is redefined with a new meaning explicitly.
> >
> > In some cases it's not mentioned explicitly by a hundreds-years old
> > convention. Everyone in the profession knows about it.
> Is this also an insider joke?

No. It's a simple boring fact. It's used in high school math all the time.
How could you not know about it?

> >> So: we are not allowed to declare something is just 'notational abuse'..
> >
> > Yes, we are and we do. This is a required part of everyone's education
> > in science, like the scientific notation, say, or the use of the Greek alphabet.
> I have actually occasionally the impression, that education of
> physicists includes brainwashing.

It's a standard crank line, yes. Another favourite is "regurgitation".
You are not saying anything we haven't seen on this NG before, I'm afraid.

> >> Instead we have to use even the wrong definitions from the text, if
> >> there are any, because we must not change a text to our own liking.
> >
> > The definition is not wrong. Only a small change of context occurs
> > silently. Everyone understands that sort of thing.
> ???????????

Well, this is simply a fact. It's triviality. How could you not know of it?

--
Jan

Am 24.04.2022 um 10:41 schrieb J. J. Lodder:
> RichD <r_delaney2001@yahoo.com> wrote:
>
>> On April 21, Thomas Heger wrote:
>>> Betrand Russel was head of the 'Fabians'.
>>> His nickname was 'Golem of Venice' and he was supposed to be a
>>> representaive of what is called 'Black Nobility'.
>>> One of Russels's 'achievements' was actually socialism, for what we
>>> common mortals are still 'thankful'.
>>> The Russel familly had held the title 'Earl of Tavistook', too.
>>> This name 'Tavistock' resurfaced together with Naziism and the so called
>>> 'New World Order' several times.
>>> The connection stems from the 'Tavistock Instutute for Human Relations'
>>> (aka: 'Scriptwriters of the New World Order').
>>> Befor that Institute got its current name, it was called 'Freud Hilton'.
>>
>> A hypothetical question: if a man is completely nuts, is it
>> biologically possible for him to recognize this fact?
>
> Good question indeed. While it cannot be excluded in general
> I guess this particular case is hopeless.
>
> TH lives in a fancy world of his own making,
> and he cannot escape it,

Well, actually quite a few people have guessed, that the so called
'Fabians' have promoted (national-) socialism.

Naziism (aka national-socialism) is actually closely related to
'Eugenics', occultism and racism.

So we can search for racists in connections to 'Eugenics'.

That is actually easy:

Theosophie and their leaders (like e.g. Rudolph Steiner) were very
racist. (also related to occultism, of course)

Churchil, Darwin and Thomas Henry Huxley were pronounced racists.

So we need a hypothetical point, where the life-lines of related people
have met.

That could be the elite universities like Cambridge and Oxford and
especially certain groups there, like e.g. the Cambridge 'Apostles'.

That group had a conncetion to the real Hitler, because Ludwig
Wittgenstein was both: a member of the 'Apostles' and related to the
real Hitler (appeared on a class-photo together with Hitler).

TH

Am 26.04.2022 um 08:52 schrieb JanPB:
> On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
>> Am 25.04.2022 um 21:22 schrieb JanPB:
>>> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
>>>> Am 24.04.2022 um 10:52 schrieb JanPB:
>>>>
>>>>>> But even if so, x' would still be zero.
>>>>>
>>>>> Yes, you get x' = 0 for the emitter/receiver because its location
>>>>> by construction of this experimental setup satisfies x = vt at
>>>>> all times t, hence the corresponding x' value is:
>>>>>
>>>>> x' = x - vt = 0
>>>>>
>>>>> throughout.
>>>> Here you actually agree, that x'=0 must be true.
>>>
>>> For the *light source*.
>> No, this is wrong because Einstein had a different setting:
>>
>> [quote]
>>
>> "From the origin of system k let a ray be emitted at the time τ'along
>> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
>> the coordinates, ..."
>>
>> [/quote]
>>
>> The system k was the moving system, hence Einstein wanted a moving
>> emitter and a stationary mirror.
>
> Both the light source and the mirror are stationary in k because that's how
> the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).

The coordinates in k had small Greek letters as names.

Therefore x' must be a coordinate from system K.

On the other hand, the origin of the ray needs a coordinate in k, too,
which had to be xsi=0.

If now the mirror is stationary in respect to xsi=0, it cannot have a
fixed position in K, but should move to 'the right' with velocity v.

So, your statement cannot be true, even if your argument with the
equation you have quoted is correct.

But a mirror stationary in respect to the emitter cannot possibly be
meant, because in this case we would not have a use for velocity v.

Actually the system K could be left away entirely in this case.

So, something is wrong, because every possible interpretation would
require to change something a little.

In effect we are lost and should think about an interpretation, where
the number of errors is minimized.

Your proposal would fit to the equation (1/2)*(tau_0 +tau_2) = tau_1,
but would render the subsequent equations useless.

Also missing is in this case a defintion of x'.

Emitter and mirror should not coincide, but should have a little
distance. But what distance is appropriate?

The next question would be, how we make use of system K in a scenario,
which does not include system K.

>> The mirror should be located at x', because the text says so.
>
> Yes.
>
>> Because the zero spot of k has the x-coordinate in K of
>> xsi_0 = x_0 * v*t
>
> If you say "x-coordinate", then don't say "ksi". So again: if we assume,

The naming convention of Einstein was, that the moving system k had
small Greek letters as symbols for the coordinate values and the system
K had small Latin letters.

So xsi would be the equivalent to x, but in system k, while x belongs to
K, with xsi=x+v*t.

xsi=x+v*t could be transformed to x= xsi - v*t, what looks quite
similar to Einstein's equation x'=x-v*t, especially if we exchange the x
for xsi and the x' for x.

But you are right, that in this case the equation

(1/2)*(tau_0 +tau_2) = tau_1

cannot be kept, if v is not equal to 0.

So, there is actually no possibility, which would leave the text intact,
hence we need to decide for a ssolution, which minimzes errors.

> as Einstein did, that:
>
> (1) the ray is emitted from the origin of k,
>
> (2) the origins of K and k coincide when t = 0 and tau = 0,
>
> (3) the ray is emitted at some K-time value of t_0,

No, that is not a possibility.

If the ray is emitted at t =tau=0, the ray would point away from the
zero spot of K, where the mirror should be placed. That would exclude
the possibility of a reflection.

The reason: if emitter and mirror coincide, the ray would point away
from the mirror, wherever the ray is pointed to.

Also: rays are not momentary events, but have a duration, what does not
fit to a time t=0.

Also: time should remaind the independent variable, hence all values for
time should be possible.

>
> THEN:
>
> (a) the K x-coordinate x_0 of the emission satisfies: x_0 = v*t_0,
>
> (b) the K x-coordinate x of the light ray pulse at subsequent times
> t satisfies:L x = x_0 + v*(t - t_0).

To apply a duration of the light beam would be a very bad idea!

The equation was not about a pulse. Meant was the duration of the travel
of the ray between emitter and a mirror and on the way back. The length
of the pulse was irrelevant.

We are still discussing the question, where we want to place the mirror
and which state of motion this mirror had.

My proposal was, that Einstein meant a scenario, where the emitter is
placed at point xsi=0 in k and the mirror is placed at x'=0 in K.

Your proposal was, that the emitter was placed at xsi=0 in k and the
mirror at a point x'=xsi_mirror= -something (a negative xsi-value,
unknown, but fixed).

Unfortunately both proposals would cause trouble in subsequent parts of
the text, though different trouble in different parts.

> (Note that for t = t_0 we get x = x_0, as expected.)
>
> But x = x_0 + v*(t - t_0) means that x - v*t = x_0 - v*t_0.
>
> By item (a) the RHS is equal to zero. Hence the combination x - v*t
> is equal to zero for all times t following the emission (until the
> reflection which alters this pattern). Einstein labels this combination
> by x'.
>
> Similar consideration yields x' = constant (NOT equal to zero) for the mirror.

You should say, which coordinate system you mean.

Above you wrote, that mirror and emitter would not move inrespect to
each other. That would say, that x' is actually a xsi coordinate in k.
That would need to have a negative value, because the mirror should be
placed in the direction of the beam, which was assumed to point towards
smaller values (to the left).

In this scenario, the system K could be left away entirely, because it
would not be involved in the situation anymore.

That in turn would exclude the use of coordinates from K (x, y, z and t)
and from velocity v.

And that would leave a lot of empty space in the following text.

>> we end up at the zero spot of K, if we subtract v*t from xsi_0.
>>> The differentiation OTOH is taken with respect to the *mirror* position
>>> (denoted by x' by the abuse of notation) which is nonzero (obviously).
>> No, x' is OBVIOUSLY the zero spot of K.
>>
>> It makes no sense to write partial differential equations for a problem,
>> where only the x-axis is involved.
>
> Both sides are functions of x' so they can be differentiated
> with respect to x'.

No. The function tau is a function of time, which produces a value for x
at time t, with:
tau(t)=x=v*t

You could also mean a function tau(x)=t, but that would not make much
sense, because you cannot possibly say, that time would be depending on
the coordinates of a point.

>> But still we cannot differentiate a
>> fixed position, at least not in K.
>
> When you perform differentiation, you consider multiple instances of the
> function values and evaluate the resulting difference quotient limit.

Sure, but how do you do this with zero?

If you have only a fixed value of zero, there would be nothing to change.

We can change time, however, and let the clocks run for some time, say a
day.

After that day the position x had still the value zero.

That is, btw, the reason, why time must be the independent variable and
not the x-coordinate of the mirror.

> In this case one considers instances of the experiment with different
> values of x'.

Sure. It is possible to chose other places for the mirror, as long as
the mirror remains at that position.

But the position was not yet specified, anyhow, hence any possition is
possible, as long it remains fixed.

>
>> Seen from the emitter, the mirrors moves and we have the usual relation
>> v= x'/d tau.
>
> Again, a typo?

No. I meant a velocity of the mirror, which is measured with measures
from system k.

You had a different setting, where v=0, while I used the defintion of v
as velocity of system k in respect to system K, where v<>0.

This velocity is v = x'/d tau

>
>> Now Einstein used the inverse 1/v in his equation, what is blatant nonsense.
>
> There is no 1/v there.

Sure, because he wrote del tau /del x'
in the equations in the middle of page 6

On Tuesday, April 26, 2022 at 1:24:06 PM UTC-7, Thomas Heger wrote:
> Am 26.04.2022 um 08:52 schrieb JanPB:
> > On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
> >> Am 25.04.2022 um 21:22 schrieb JanPB:
> >>> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
> >>>> Am 24.04.2022 um 10:52 schrieb JanPB:
> >>>>
> >>>>>> But even if so, x' would still be zero.
> >>>>>
> >>>>> Yes, you get x' = 0 for the emitter/receiver because its location
> >>>>> by construction of this experimental setup satisfies x = vt at
> >>>>> all times t, hence the corresponding x' value is:
> >>>>>
> >>>>> x' = x - vt = 0
> >>>>>
> >>>>> throughout.
> >>>> Here you actually agree, that x'=0 must be true.
> >>>
> >>> For the *light source*.
> >> No, this is wrong because Einstein had a different setting:
> >>
> >> [quote]
> >>
> >> "From the origin of system k let a ray be emitted at the time τ'along
> >> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
> >> the coordinates, ..."
> >>
> >> [/quote]
> >>
> >> The system k was the moving system, hence Einstein wanted a moving
> >> emitter and a stationary mirror.
> >
> > Both the light source and the mirror are stationary in k because that's how
> > the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).
> The coordinates in k had small Greek letters as names.
>
> Therefore x' must be a coordinate from system K.

Yes, it's an expression involving the K coordinates. But keep in mind
that the following three statements are simultaneously true:

(1) x' is an expression involving K coordinates,

(2) if x' denotes the values of this expression for the mirror
at different times t, then x' is constant and nonzero,

(3) the mirror is stationary in k (the lower-case k).
> On the other hand, the origin of the ray needs a coordinate in k, too,
> which had to be xsi=0.

Yes.

> If now the mirror is stationary in respect to xsi=0, it cannot have a
> fixed position in K, but should move to 'the right' with velocity v.

Yes, it does. You can think of the mirror being rigidly connected
to the light source.

> So, your statement cannot be true, even if your argument with the
> equation you have quoted is correct.

No, it's as I describe it.

> But a mirror stationary in respect to the emitter cannot possibly be
> meant, because in this case we would not have a use for velocity v.

It is being used for proving that the values of x' are constant
for both the light source (where x' = 0) and the mirror (where
x' is nonzero).

> Actually the system K could be left away entirely in this case.
No.

> So, something is wrong, because every possible interpretation would
> require to change something a little.

No, notuing is wrong and nothing needs any change.

> In effect we are lost and should think about an interpretation, where
> the number of errors is minimized.

There are no errors here, so nothing to "minimse".

[...]
> Emitter and mirror should not coincide, but should have a little
> distance. But what distance is appropriate?

We need to assume the setup is correctly described by the
constraining equation for all nonzero x' in some open interval containing 0.

> The next question would be, how we make use of system K in a scenario,
> which does not include system K.

This is a false conundrum, it doesn't exist.

> >> The mirror should be located at x', because the text says so.
> >
> > Yes.
> >
> >> Because the zero spot of k has the x-coordinate in K of
> >> xsi_0 = x_0 * v*t
> >
> > If you say "x-coordinate", then don't say "ksi". So again: if we assume,
> The naming convention of Einstein was, that the moving system k had
> small Greek letters as symbols for the coordinate values and the system
> K had small Latin letters.
>
> So xsi would be the equivalent to x, but in system k, while x belongs to
> K, with xsi=x+v*t.

This equation is incorrect. Think about it. It assumes we've calculated the
formula for ksi before calculating it. In fact ksi involves the
Lorentz gamma factor, as it turns out a bit later.

> So, there is actually no possibility, which would leave the text intact,
> hence we need to decide for a ssolution, which minimzes errors.

The text can be left intact, it's 100% correct.

> > as Einstein did, that:
> >
> > (1) the ray is emitted from the origin of k,
> >
> > (2) the origins of K and k coincide when t = 0 and tau = 0,
> >
> > (3) the ray is emitted at some K-time value of t_0,
> No, that is not a possibility.

Yes, it is exactly that.

> If the ray is emitted at t =tau=0, the ray would point away from the
> zero spot of K, where the mirror should be placed. That would exclude
> the possibility of a reflection.

Again, it's as I described it above in (1)-(3). t (emission time) may or may not
be zero. The mirror is at some distance from the origin of k (the light
source).

> The reason: if emitter and mirror coincide,

They don't (obviously).

> Also: rays are not momentary events, but have a duration, what does not
> fit to a time t=0.

If the ray is emitted at time t_0, then it hits the mirror at t_0 + x'/(c - v),
where x' is the (nonzero) constant describing the mirror.

> > THEN:
> >
> > (a) the K x-coordinate x_0 of the emission satisfies: x_0 = v*t_0,
> >
> > (b) the K x-coordinate x of the light ray pulse at subsequent times
> > t satisfies:L x = x_0 + v*(t - t_0).
> To apply a duration of the light beam would be a very bad idea!
>
> The equation was not about a pulse. Meant was the duration of the travel
> of the ray between emitter and a mirror and on the way back. The length
> of the pulse was irrelevant.

There is no length of the pulse. The equation x = x_0 + v*(t - t_0) describes
the motion of the light pulse (its position x in terms of the time t, both
are coordinates of K).

> We are still discussing the question, where we want to place the mirror
> and which state of motion this mirror had.

I've already described it. It's some distance away from the light source and
because ANY object that's stationary wrt k has its x' constant, so does
the mirror. Einstein denotes that number (the one "belonging" to the mirror)
by x'. It's some nonzero constant. The same combination (constant) for
the light source is equal to 0.

> My proposal was, that Einstein meant a scenario, where the emitter is
> placed at point xsi=0 in k

Yes.

> and the mirror is placed at x'=0 in K.

No, the mirror is some distance (nonzero) away from the light source, hence
its value of x' is a nonzero constant. It is the *light source* that sits at x' = 0 in K.

> Your proposal was, that the emitter was placed at xsi=0 in k and the
> mirror at a point x'=xsi_mirror

Don't write formulas involving the coordinates of K and k before
you calculate those formulas.

= -something (a negative xsi-value,
> unknown, but fixed).

Whatever, this is not even wrong.

> Unfortunately both proposals would cause trouble in subsequent parts of
> the text, though different trouble in different parts.

Yes, obviously. Thankfully, they are not needed.

> > (Note that for t = t_0 we get x = x_0, as expected.)
> >
> > But x = x_0 + v*(t - t_0) means that x - v*t = x_0 - v*t_0.
> >
> > By item (a) the RHS is equal to zero. Hence the combination x - v*t
> > is equal to zero for all times t following the emission (until the
> > reflection which alters this pattern). Einstein labels this combination
> > by x'.
> >
> > Similar consideration yields x' = constant (NOT equal to zero) for the mirror.
> You should say, which coordinate system you mean.
>
> Above you wrote, that mirror and emitter would not move inrespect to
> each other.

Yes.

> That would say, that x' is actually a xsi coordinate in k.

No. x' is just a formula involving x, v, and t for convenience.

[...]
> And that would leave a lot of empty space in the following text.
> >> we end up at the zero spot of K, if we subtract v*t from xsi_0.
> >>> The differentiation OTOH is taken with respect to the *mirror* position
> >>> (denoted by x' by the abuse of notation) which is nonzero (obviously)..
> >> No, x' is OBVIOUSLY the zero spot of K.
> >>
> >> It makes no sense to write partial differential equations for a problem,
> >> where only the x-axis is involved.
> >
> > Both sides are functions of x' so they can be differentiated
> > with respect to x'.
> No. The function tau is a function of time, which produces a value for x
> at time t, with:
> tau(t)=x=v*t

Am 27.04.2022 um 02:35 schrieb JanPB:
> On Tuesday, April 26, 2022 at 1:24:06 PM UTC-7, Thomas Heger wrote:
>> Am 26.04.2022 um 08:52 schrieb JanPB:
>>> On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
>>>> Am 25.04.2022 um 21:22 schrieb JanPB:
>>>>> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
>>>>>> Am 24.04.2022 um 10:52 schrieb JanPB:
>>>>>>
>>>>>>>> But even if so, x' would still be zero.
>>>>>>>
>>>>>>> Yes, you get x' = 0 for the emitter/receiver because its location
>>>>>>> by construction of this experimental setup satisfies x = vt at
>>>>>>> all times t, hence the corresponding x' value is:
>>>>>>>
>>>>>>> x' = x - vt = 0
>>>>>>>
>>>>>>> throughout.
>>>>>> Here you actually agree, that x'=0 must be true.
>>>>>
>>>>> For the *light source*.
>>>> No, this is wrong because Einstein had a different setting:
>>>>
>>>> [quote]
>>>>
>>>> "From the origin of system k let a ray be emitted at the time τ'along
>>>> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
>>>> the coordinates, ..."
>>>>
>>>> [/quote]
>>>>
>>>> The system k was the moving system, hence Einstein wanted a moving
>>>> emitter and a stationary mirror.
>>>
>>> Both the light source and the mirror are stationary in k because that's how
>>> the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).
>> The coordinates in k had small Greek letters as names.
>>
>> Therefore x' must be a coordinate from system K.
>
> Yes, it's an expression involving the K coordinates. But keep in mind
> that the following three statements are simultaneously true:
>
> (1) x' is an expression involving K coordinates,
>
> (2) if x' denotes the values of this expression for the mirror
> at different times t, then x' is constant and nonzero,
>
> (3) the mirror is stationary in k (the lower-case k).
>
>> On the other hand, the origin of the ray needs a coordinate in k, too,
>> which had to be xsi=0.
>
> Yes.
>
>> If now the mirror is stationary in respect to xsi=0, it cannot have a
>> fixed position in K, but should move to 'the right' with velocity v.
>
> Yes, it does. You can think of the mirror being rigidly connected
> to the light source.
>
>> So, your statement cannot be true, even if your argument with the
>> equation you have quoted is correct.
>
> No, it's as I describe it.
>
>> But a mirror stationary in respect to the emitter cannot possibly be
>> meant, because in this case we would not have a use for velocity v.
>
> It is being used for proving that the values of x' are constant
> for both the light source (where x' = 0) and the mirror (where
> x' is nonzero).

x' is moving in K to the direction of higher x-values with velocity v,
if the mirror rests at a certain spot in k.

Therefore any value along the x-axis is possible, supposed the length of
the time of emission of the ray is long enough. which includes the zero
spot of K at x=0.

>> Actually the system K could be left away entirely in this case.
> No.

If system K is not involved in the scenario, it could be left away.

In that case k could be renamed to K and treated as at rest.

This would follow from the principle of relativity, that you cannot
distinguish inertial motion from being at rest.

https://en.wikipedia.org/wiki/Principle_of_relativity

quote:

"A consequence is that an observer in an inertial reference frame cannot
determine an absolute speed or direction of travel in space, and may
only speak of speed or direction relative to some other object. "

Since K contains no object of interest, it must be left away.

If K is left away, then k has nothing to base its state of motion upon,
hence can regard the own state of motion to be at rest.

This in turn would allow to set v=0, because in k only light moves with
c. All other entities are at rest.

>> So, something is wrong, because every possible interpretation would
>> require to change something a little.
>
> No, notuing is wrong and nothing needs any change.

Your interpretation is impossible, because if the mirror is at rest,
there is nothing moving anymore, besides of the system K in respect to
system k.

Since coordinate systems are 'brainchilds', hence can be arbitrarily
chosen, there is no need to chose an arbitrary system K, if nothing is
happening there.

Also time tau cannot possibly depend on an arbitrary velocity v.

>> In effect we are lost and should think about an interpretation, where
>> the number of errors is minimized.
>
> There are no errors here, so nothing to "minimse".
>
> [...]
>> Emitter and mirror should not coincide, but should have a little
>> distance. But what distance is appropriate?
>
> We need to assume the setup is correctly described by the
> constraining equation for all nonzero x' in some open interval containing 0.

The emitter is placed in k at xsi=0.

Now we let k and K coincide at time t=tau=0.

the mirror is placed initially at point x=-v*t

We then start the ray and let it shine for some time t.

After time t the mirror would coincide with the zero point of K, because
the emitter had moved to the right with velocity v to the point x=v*t
and the distance between mirror and emitter was assumed to be constant.

Therefore we cannot exclude x'=0 if we want allow a continous ray.

>> The next question would be, how we make use of system K in a scenario,
>> which does not include system K.
>
> This is a false conundrum, it doesn't exist.

If you say, that system K is necessary, than you should be able to say,
what it is good for, if nothing is happening there.

>>>> The mirror should be located at x', because the text says so.
>>>
>>> Yes.
>>>
>>>> Because the zero spot of k has the x-coordinate in K of
>>>> xsi_0 = x_0 * v*t
>>>
>>> If you say "x-coordinate", then don't say "ksi". So again: if we assume,
>> The naming convention of Einstein was, that the moving system k had
>> small Greek letters as symbols for the coordinate values and the system
>> K had small Latin letters.
>>
>> So xsi would be the equivalent to x, but in system k, while x belongs to
>> K, with xsi=x+v*t.
>
> This equation is incorrect. Think about it. It assumes we've calculated the
> formula for ksi before calculating it. In fact ksi involves the
> Lorentz gamma factor, as it turns out a bit later.

What????

>> So, there is actually no possibility, which would leave the text intact,
>> hence we need to decide for a ssolution, which minimzes errors.
>
> The text can be left intact, it's 100% correct.
>
....
>> and the mirror is placed at x'=0 in K.
>
> No, the mirror is some distance (nonzero) away from the light source, hence
> its value of x' is a nonzero constant. It is the *light source* that sits at x' = 0 in K.
>
>> Your proposal was, that the emitter was placed at xsi=0 in k and the
>> mirror at a point x'=xsi_mirror
>
> Don't write formulas involving the coordinates of K and k before
> you calculate those formulas.
>
> = -something (a negative xsi-value,
>> unknown, but fixed).
>
> Whatever, this is not even wrong.

In case of t=tau=0 the emitter starts emitting a the zero spot of K, too.

to hit the mirror, the ray should point to 'the left', which is the
direction of negative values for xsi.

Since the mirror should have some distance, the initial position had to
be in that negative value realm.

Since the amount of distance was not yet specified, the distance is
assumed to be arbitrary, but constant.

Placing the mirror in the other direction is actually also possible, but
would render the setting more questionable than it already is.

The reason is, that a one-dimensional movement along the x-axis into the
positive direction would look to an observer in K like a receeding point
with no diameter.

>> Unfortunately both proposals would cause trouble in subsequent parts of
>> the text, though different trouble in different parts.
>
> Yes, obviously. Thankfully, they are not needed.
>
>>> (Note that for t = t_0 we get x = x_0, as expected.)
>>>
>>> But x = x_0 + v*(t - t_0) means that x - v*t = x_0 - v*t_0.
>>>
>>> By item (a) the RHS is equal to zero. Hence the combination x - v*t
>>> is equal to zero for all times t following the emission (until the
>>> reflection which alters this pattern). Einstein labels this combination
>>> by x'.
>>>
>>> Similar consideration yields x' = constant (NOT equal to zero) for the mirror.
>> You should say, which coordinate system you mean.
>>
>> Above you wrote, that mirror and emitter would not move inrespect to
>> each other.
>
> Yes.
>
>> That would say, that x' is actually a xsi coordinate in k.
>
> No. x' is just a formula involving x, v, and t for convenience.
>
> [...]
>> And that would leave a lot of empty space in the following text.
>>>> we end up at the zero spot of K, if we subtract v*t from xsi_0.
>>>>> The differentiation OTOH is taken with respect to the *mirror* position
>>>>> (denoted by x' by the abuse of notation) which is nonzero (obviously).
>>>> No, x' is OBVIOUSLY the zero spot of K.
>>>>
>>>> It makes no sense to write partial differential equations for a problem,
>>>> where only the x-axis is involved.
>>>
>>> Both sides are functions of x' so they can be differentiated
>>> with respect to x'.
>> No. The function tau is a function of time, which produces a value for x
>> at time t, with:
>> tau(t)=x=v*t
>
> This is a made-up formula (happens to also incorrect).
> Again, you differentiate the constraining "tau" equations wrt x'.

Am 27.04.2022 um 02:35 schrieb JanPB:
> On Tuesday, April 26, 2022 at 1:24:06 PM UTC-7, Thomas Heger wrote:
>> Am 26.04.2022 um 08:52 schrieb JanPB:
>>> On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
>>>> Am 25.04.2022 um 21:22 schrieb JanPB:
>>>>> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
>>>>>> Am 24.04.2022 um 10:52 schrieb JanPB:
>>>>>>
>>>>>>>> But even if so, x' would still be zero.
>>>>>>>
>>>>>>> Yes, you get x' = 0 for the emitter/receiver because its location
>>>>>>> by construction of this experimental setup satisfies x = vt at
>>>>>>> all times t, hence the corresponding x' value is:
>>>>>>>
>>>>>>> x' = x - vt = 0
>>>>>>>
>>>>>>> throughout.
>>>>>> Here you actually agree, that x'=0 must be true.
>>>>>
>>>>> For the *light source*.
>>>> No, this is wrong because Einstein had a different setting:
>>>>
>>>> [quote]
>>>>
>>>> "From the origin of system k let a ray be emitted at the time τ'along
>>>> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
>>>> the coordinates, ..."
>>>>
>>>> [/quote]
>>>>
>>>> The system k was the moving system, hence Einstein wanted a moving
>>>> emitter and a stationary mirror.
>>>
>>> Both the light source and the mirror are stationary in k because that's how
>>> the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).
>> The coordinates in k had small Greek letters as names.
>>
>> Therefore x' must be a coordinate from system K.
>
> Yes, it's an expression involving the K coordinates. But keep in mind
> that the following three statements are simultaneously true:
>
> (1) x' is an expression involving K coordinates,
>
> (2) if x' denotes the values of this expression for the mirror
> at different times t, then x' is constant and nonzero,
>
> (3) the mirror is stationary in k (the lower-case k).

You should keep in mind, that it is absolutely impossible, that an
author requires from the readers, that they would marter their brain to
develop, what actually the author should have written.

Einstein himself had to write, which coordinate systems were used and
for which quantities. These quantities need to be defined properly.

But in contrast to common standards, we are now discussing the question,
whether the mirror moves (or not) in system K.

This is an information, an author needs to provide.

We are now discussing this question on the basis on other constraints
like other equations or descriptions in the text.

But I have tried hard and found no possible setting, which violated non
of these constraints somewhere.

Now we had to declare this an error of the text and not ours, because we
already did, what no author could rightfully demand, and tried to read
the authors mind from rare hints, but with no success.

....

TH

On Wednesday, April 27, 2022 at 10:14:03 PM UTC-7, Thomas Heger wrote:
> Am 27.04.2022 um 02:35 schrieb JanPB:
> > On Tuesday, April 26, 2022 at 1:24:06 PM UTC-7, Thomas Heger wrote:
> >> Am 26.04.2022 um 08:52 schrieb JanPB:
> >>> On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
> >>>> Am 25.04.2022 um 21:22 schrieb JanPB:
> >>>>> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
> >>>>>> Am 24.04.2022 um 10:52 schrieb JanPB:
> >>>>>>
> >>>>>>>> But even if so, x' would still be zero.
> >>>>>>>
> >>>>>>> Yes, you get x' = 0 for the emitter/receiver because its location
> >>>>>>> by construction of this experimental setup satisfies x = vt at
> >>>>>>> all times t, hence the corresponding x' value is:
> >>>>>>>
> >>>>>>> x' = x - vt = 0
> >>>>>>>
> >>>>>>> throughout.
> >>>>>> Here you actually agree, that x'=0 must be true.
> >>>>>
> >>>>> For the *light source*.
> >>>> No, this is wrong because Einstein had a different setting:
> >>>>
> >>>> [quote]
> >>>>
> >>>> "From the origin of system k let a ray be emitted at the time τ'along
> >>>> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
> >>>> the coordinates, ..."
> >>>>
> >>>> [/quote]
> >>>>
> >>>> The system k was the moving system, hence Einstein wanted a moving
> >>>> emitter and a stationary mirror.
> >>>
> >>> Both the light source and the mirror are stationary in k because that's how
> >>> the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).
> >> The coordinates in k had small Greek letters as names.
> >>
> >> Therefore x' must be a coordinate from system K.
> >
> > Yes, it's an expression involving the K coordinates. But keep in mind
> > that the following three statements are simultaneously true:
> >
> > (1) x' is an expression involving K coordinates,
> >
> > (2) if x' denotes the values of this expression for the mirror
> > at different times t, then x' is constant and nonzero,
> >
> > (3) the mirror is stationary in k (the lower-case k).
> You should keep in mind, that it is absolutely impossible, that an
> author requires from the readers, that they would marter their brain to
> develop, what actually the author should have written.

But the author has written all that. I only rewrote it.

> Einstein himself had to write, which coordinate systems were used and
> for which quantities. These quantities need to be defined properly.

They are.

> But in contrast to common standards, we are now discussing the question,
> whether the mirror moves (or not) in system K.

The mirror is stationary in k. This is obvious from the description as well as
from everything that went on before (e.g. the clock sync definition).

> This is an information, an author needs to provide.

It was provided.

> We are now discussing this question on the basis on other constraints
> like other equations or descriptions in the text.
>
> But I have tried hard and found no possible setting, which violated non
> of these constraints somewhere.

I've described the setting by paraphrasing he paper slightly to underline
the points we are discussing. There are no violations of anything there.

> Now we had to declare this an error of the text and not ours, because we
> already did, what no author could rightfully demand, and tried to read
> the authors mind from rare hints, but with no success.

I understand that it must be frustrating but Einstein made no errors in that paper.

--
Jan

JanPB wrote:

> On Wednesday, April 27, 2022 at 10:14:03 PM UTC-7, Thomas Heger wrote:
>> But in contrast to common standards, we are now discussing the
>> question, whether the mirror moves (or not) in system K.
>
> The mirror is stationary in k. This is obvious from the description as
> well as from everything that went on before (e.g. the clock sync
> definition).

nonsense. In capitalist america lying is called "intelligence" and killing
people in masses is called "healthcare". Now polakia, instead of free
natural gas from the russian federation, they have to dig deep underground
in kilometres to find coal. Amazing these idiots.

Thomas Heger <ttt_heg@web.de> wrote:
> Am 27.04.2022 um 02:35 schrieb JanPB:
>> On Tuesday, April 26, 2022 at 1:24:06 PM UTC-7, Thomas Heger wrote:
>>> Am 26.04.2022 um 08:52 schrieb JanPB:
>>>> On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
>>>>> Am 25.04.2022 um 21:22 schrieb JanPB:
>>>>>> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
>>>>>>> Am 24.04.2022 um 10:52 schrieb JanPB:
>>>>>>>
>>>>>>>>> But even if so, x' would still be zero.
>>>>>>>>
>>>>>>>> Yes, you get x' = 0 for the emitter/receiver because its location
>>>>>>>> by construction of this experimental setup satisfies x = vt at
>>>>>>>> all times t, hence the corresponding x' value is:
>>>>>>>>
>>>>>>>> x' = x - vt = 0
>>>>>>>>
>>>>>>>> throughout.
>>>>>>> Here you actually agree, that x'=0 must be true.
>>>>>>
>>>>>> For the *light source*.
>>>>> No, this is wrong because Einstein had a different setting:
>>>>>
>>>>> [quote]
>>>>>
>>>>> "From the origin of system k let a ray be emitted at the time τ'along
>>>>> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
>>>>> the coordinates, ..."
>>>>>
>>>>> [/quote]
>>>>>
>>>>> The system k was the moving system, hence Einstein wanted a moving
>>>>> emitter and a stationary mirror.
>>>>
>>>> Both the light source and the mirror are stationary in k because that's how
>>>> the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).
>>> The coordinates in k had small Greek letters as names.
>>>
>>> Therefore x' must be a coordinate from system K.
>>
>> Yes, it's an expression involving the K coordinates. But keep in mind
>> that the following three statements are simultaneously true:
>>
>> (1) x' is an expression involving K coordinates,
>>
>> (2) if x' denotes the values of this expression for the mirror
>> at different times t, then x' is constant and nonzero,
>>
>> (3) the mirror is stationary in k (the lower-case k).
>
>
> You should keep in mind, that it is absolutely impossible, that an
> author requires from the readers, that they would marter their brain to
> develop, what actually the author should have written.

Once again, Thomas, it is only up to the author to try to communicate to
his intended audience, which you are not a member of. This does not prevent
you from reading his communication to other people, but in so doing the
failure to communicate to you is your fault, not the author’s. In the same
way, you could purchase an airplane if you chose to, and you could mow your
back yard to have a landing strip. If you got into the airplane and tried
to fly it without being a trained pilot, your ensuing crash and death would
not be laid at the feet of the airplane maker.

>
> Einstein himself had to write, which coordinate systems were used and
> for which quantities. These quantities need to be defined properly.
>
> But in contrast to common standards, we are now discussing the question,
> whether the mirror moves (or not) in system K.
>
>
> This is an information, an author needs to provide.
>
>
> We are now discussing this question on the basis on other constraints
> like other equations or descriptions in the text.
>
> But I have tried hard and found no possible setting, which violated non
> of these constraints somewhere.
>
> Now we had to declare this an error of the text and not ours, because we
> already did, what no author could rightfully demand, and tried to read
> the authors mind from rare hints, but with no success.
>
>
> ...
>
>
> TH
>

--
Odd Bodkin -- maker of fine toys, tools, tables

Am 28.04.2022 um 10:45 schrieb JanPB:
> On Wednesday, April 27, 2022 at 10:14:03 PM UTC-7, Thomas Heger wrote:
>> Am 27.04.2022 um 02:35 schrieb JanPB:
>>> On Tuesday, April 26, 2022 at 1:24:06 PM UTC-7, Thomas Heger wrote:
>>>> Am 26.04.2022 um 08:52 schrieb JanPB:
>>>>> On Monday, April 25, 2022 at 10:02:19 PM UTC-7, Thomas Heger wrote:
>>>>>> Am 25.04.2022 um 21:22 schrieb JanPB:
>>>>>>> On Monday, April 25, 2022 at 12:52:14 AM UTC-7, Thomas Heger wrote:
>>>>>>>> Am 24.04.2022 um 10:52 schrieb JanPB:
>>>>>>>>
>>>>>>>>>> But even if so, x' would still be zero.
>>>>>>>>>
>>>>>>>>> Yes, you get x' = 0 for the emitter/receiver because its location
>>>>>>>>> by construction of this experimental setup satisfies x = vt at
>>>>>>>>> all times t, hence the corresponding x' value is:
>>>>>>>>>
>>>>>>>>> x' = x - vt = 0
>>>>>>>>>
>>>>>>>>> throughout.
>>>>>>>> Here you actually agree, that x'=0 must be true.
>>>>>>>
>>>>>>> For the *light source*.
>>>>>> No, this is wrong because Einstein had a different setting:
>>>>>>
>>>>>> [quote]
>>>>>>
>>>>>> "From the origin of system k let a ray be emitted at the time τ'along
>>>>>> the X-axis to x' and at the time τ_1 be reflected thence to the origin of
>>>>>> the coordinates, ..."
>>>>>>
>>>>>> [/quote]
>>>>>>
>>>>>> The system k was the moving system, hence Einstein wanted a moving
>>>>>> emitter and a stationary mirror.
>>>>>
>>>>> Both the light source and the mirror are stationary in k because that's how
>>>>> the (1/2)*(tau_0 +tau_2) = tau_1 criterion has been set up previously (in section 1).
>>>> The coordinates in k had small Greek letters as names.
>>>>
>>>> Therefore x' must be a coordinate from system K.
>>>
>>> Yes, it's an expression involving the K coordinates. But keep in mind
>>> that the following three statements are simultaneously true:
>>>
>>> (1) x' is an expression involving K coordinates,
>>>
>>> (2) if x' denotes the values of this expression for the mirror
>>> at different times t, then x' is constant and nonzero,
>>>
>>> (3) the mirror is stationary in k (the lower-case k).
>> You should keep in mind, that it is absolutely impossible, that an
>> author requires from the readers, that they would marter their brain to
>> develop, what actually the author should have written.
>
> But the author has written all that. I only rewrote it.
>
>> Einstein himself had to write, which coordinate systems were used and
>> for which quantities. These quantities need to be defined properly.
>
> They are.
>
>> But in contrast to common standards, we are now discussing the question,
>> whether the mirror moves (or not) in system K.
>
> The mirror is stationary in k. This is obvious from the description as well as
> from everything that went on before (e.g. the clock sync definition).
>
>> This is an information, an author needs to provide.
>
> It was provided.
>
>> We are now discussing this question on the basis on other constraints
>> like other equations or descriptions in the text.
>>
>> But I have tried hard and found no possible setting, which violated non
>> of these constraints somewhere.
>
> I've described the setting by paraphrasing he paper slightly to underline
> the points we are discussing. There are no violations of anything there.

In a scientific paper the author has to tell his story. It is not the
duty of the reader to 'fill the blancs'.

But if we actually try to do that (as I did), we find self-contradicting
requirements, which cannot be all equally fulfilled.

Einstein wrote about an emitter at the zero spot of the moving system k,
which sends out a ray, which is reflected by a mirror at the point x'.

The only introduction of x' that can be found in the text stems from the
equation x' =x - v*t.

Now we can assume, that Einstein wanted x' to be in constant distance to
the emitter, because otherwise the nect equation
(1/2)*(tau_0 +tau_2) = tau_1
cannot be fulfilled.

But -v*t is obviously negative, while x has (apparently) a fixed value.

Now adding a decreasing distance to a fixed position in K and an
increasing distance between x and emitter in k, would cause a doubled
velocity instead of zero velocity.

So the equation x'= x - v*t would reuire to replace x to xsi, what the
equation
(1/2)*(tau_0 +tau_2) = tau_1

forbids.

This is an oximoron, hence an uncurable error in the text.

>> Now we had to declare this an error of the text and not ours, because we
>> already did, what no author could rightfully demand, and tried to read
>> the authors mind from rare hints, but with no success.
>
> I understand that it must be frustrating but Einstein made no errors in that paper.
>
You can declare wrong to be right, if you like to do so.

But this would contradict a very important axiom is science:
that wrong statements are bad in science.

TH