Robert Winn <rbwinn3@gmail.com> wrote:
> On Saturday, September 4, 2021 at 7:49:17 AM UTC-7, bodk...@gmail.com wrote:
>> Robert Winn <rbw...@gmail.com> wrote:
>>> On Friday, September 3, 2021 at 11:38:59 PM UTC-7, Sylvia Else wrote:
>>>> On 04-Sep-21 4:51 am, Robert Winn wrote:
>>>>> x'=x-vt
>>>>> y'=y
>>>>> z'=z
>>>>> t'=t-vx/c^2
>> Well let’s just do some algebra. Start from last equation.
>>
>> t=t’+vx/c^2
>>
>> Stick in first equation.
>>
>> x=x’+vt=x’+v(t’+vx^2)=x’+vt’+xv^2/c^2
>> x(1-v^2/c^2)=x’+vt’
>>
>> Well look at that. There’s the gamma thing.
>
> Not quite. Lorentz got x = (x-vt)/sqrt(1-v^2/c^2).

Yes I know. But I started with the equations you wrote. So one doesn’t
expect to arrive at the Lorentz equations.

That (1-v^2/c^2) factor in the equation I obtained using algebra from your
equations is the reciprocal of gamma squared. As I said, there’s that gamma
thing.

On the other hand, someone else (neither Lorentz nor Einstein) tried the
equations you posted. There were problems with them, as I mentioned.

>>>>>
>> Odd Bodkin — Maker of fine toys, tools, tables
>

--
Odd Bodkin — Maker of fine toys, tools, tables

Robert Winn <rbwinn3@gmail.com> wrote:
> On Saturday, September 4, 2021 at 10:07:45 AM UTC-7, bodk...@gmail.com wrote:
>> Robert Winn <rbw...@gmail.com> wrote:
>>> On Saturday, September 4, 2021 at 7:53:55 AM UTC-7, bodk...@gmail.com wrote:
>>>> Robert Winn <rbw...@gmail.com> wrote:
>>>>> On Saturday, September 4, 2021 at 6:09:50 AM UTC-7, bodk...@gmail.com wrote:
>>>>>> Robert Winn <rbw...@gmail.com> wrote:
>>>>>>> On Friday, September 3, 2021 at 7:27:59 PM UTC-7, Python wrote:
>>>>>>>> Robert Winn wrote:
>>>>>>>>> On Friday, September 3, 2021 at 6:20:26 PM UTC-7, Python wrote:
>>>>>>>>>> Robert Winn wrote:
>>>>>>>>>>> On Friday, September 3, 2021 at 4:33:21 PM UTC-7, Python wrote:
>>>>>>>>>>>> Robert Winn wrote:
>>>>>>>>>>>>> On Friday, September 3, 2021 at 4:04:17 PM UTC-7, Python wrote:
>>>>>>>>>>>>>> Robert Winn wrote:
>>>>>>>>>>>>>>> On Friday, September 3, 2021 at 12:40:07 PM UTC-7, Dirk Van
>>>>>>>>>>>>>>> de moortel wrote:
>>>>>>>>>>>>>>>> Op 03-sep.-2021 om 20:51 schreef Robert
>>>>>>>>>>>>>>> Winn:
>>>>>>>>>>>>>>>>> x'=x-vt
>>>>>>>>>>>>>>>>> y'=y
>>>>>>>>>>>>>>>>> z'=z
>>>>>>>>>>>>>>>>> t'=t-vx/c^2
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> inverse equations
>>>>>>>>>>>>>>>> When I solve these equations for x, y, z, t, I get
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> x = g^2 ( x' + v t' )
>>>>>>>>>>>>>>>> y = y'
>>>>>>>>>>>>>>>> z = z'
>>>>>>>>>>>>>>>> t = g^2 ( t' + v x / c^2 )
>>>>>>>>>>>>>>>> where
>>>>>>>>>>>>>>>> g = 1 / sqrt( 1 - v^2/c^2 )
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> But you have this:
>>>>>>>>>>>>>>>>> x = x' - v't'
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> v't' = -vt
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> x = x' + vt
>>>>>>>>>>>>>>>>> y = y'
>>>>>>>>>>>>>>>>> z = z'
>>>>>>>>>>>>>>>>> t = t' + vx/c^2
>>>>>>>>>>>>>>>> Please solve these last four equations for x', y', z', t'.
>>>>>>>>>>>>>>>> Whet do you get?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Dirk Vdm
>>>>>>>>>>>>>>> I know how difficult this must seem.
>>>>>>>>>>>>>>> x = x' + vt
>>>>>>>>>>>>>>> x' = x - vt
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> y=y'
>>>>>>>>>>>>>>> y'=y
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> z = z'
>>>>>>>>>>>>>>> z' = z
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> t = t' + vx/c^2
>>>>>>>>>>>>>>> t' = t - vx/c^2
>>>>>>>>>>>>>>> So what is the problem you see?
>>>>>>>>>>>>>> Look carefully, at some point you made a basic algebraic mistake.
>>>>>>>>>>>>> I made a basic algebraic mistake? Well, go ahead and show the
>>>>>>>>>>>>> mistake, Dono. Don't be so bashful.
>>>>>>>>>>>> Now you made two mistakes, one is that I'm not Dono. Look harder,
>>>>>>>>>>>> you may find the other one.
>>>>>>>>>>> No, can't find a mistake. Go ahead and point it out.
>>>>>>>>>> You didn't look hard enough. The mistake is so elementary that
>>>>>>>>>> it is really a bad sign for you, Robert. Try harder.
>>>>>>>>> Baloney. The equations are continuous, unlike the Lorentz equations,
>>>>>>>>> which are discontinuous.
>>>>>>>> I wonder what you mean by continuous in the context. Anyway, your
>>>>>>>> mistake is an obvious one, and you've got an hint. How come you cannot
>>>>>>>> spot it?
>>>>>>> Continuous means that the equations describe reality. In other words, if
>>>>>>> an observer in a moving frame of reference has a slower or faster clock
>>>>>>> than an observer in a frame of reference that is not moving, both
>>>>>>> observers see the same thing. Einstein's interpretation of the Lorentz
>>>>>>> equations has the opposite observer seeing a clock that does the opposite
>>>>>>> of what the other observer sees. That is what I mean by discontinuous.
>>>>>> Ah somebody else here who likes to take physics terms and just assign some
>>>>>> private meaning to them.
>>>>>>> There is no mistake in the algebra. You are just doing what you do by
>>>>>>> habit, pretending to be superior in some way.
>>>>>>>
>>>>>> Think that’s a bit of a complex on your part maybe, always thinking others
>>>>>> are trying to prove you stupid.
>>>>>> --
>>>>>> Odd Bodkin — Maker of fine toys, tools, tables
>>>>> Well, according to you I am supposed to be at the library looking for a
>>>>> book that does not exist.
>>>> Well I found a book that does exist and I’m a woodworker. So what you’re
>>>> telling me is that welders are not as industrious or self-reliant as
>>>> woodworkers and need someone else to find things for them.
>>>>> I do not assign some private meaning.
>>>> You’re the only person I’ve ever met that used “continuous” and
>>>> “discontinuous” the way you did, and you did say this is what *you* mean by
>>>> those words, not what anybody else means by them. That sure sounds like a
>>>> private meaning to me.
>>>>> Einstein said, and all scientists believe, that if there is a moving
>>>>> frame of reference, an observer in a frame of reference will see a clock
>>>>> in the moving frame of reference as being slower than a clock in his
>>>>> frame of reference, but then he says that an observer in the moving frame
>>>>> of reference would see a clock in the frame of reference at rest as being
>>>>> slower. The Galilean transformation equations show that both observers
>>>>> would see the same clock as being slower.
>>>>>
>>>> --
>>>> Odd Bodkin — Maker of fine toys, tools, tables
>>> So if the book exists, what is the name of the book?
>> I found it myself. Surely you’re not going to say a woodworker can figure
>> things out that a welder can’t.
>>> I know how difficult this all must seem.
>>> I know all about how scientists define continuous. They claim the
>>> Lorentz equations are continuous.
>> And what do they mean when they say that?
>
> So now you are saying that this book is some sort of great mystery that I
> am supposed to find.

On Saturday, September 4, 2021 at 6:01:58 AM UTC-3, Dirk Van de moortel wrote:

1) Cites Robert Winn:

x'=x-vt
y'=y
z'=z
t'=t-vx/c^2

They are 1887 Voigt's transforms in modern notation:
x'=x-vt
y'=y/ɣ
z'=z/ɣ
t'=t-vx/c²

For v/c << 1 , like v/c < 0.1 (30,000 Km/sec), then ɣ ≈ 1 (like Einstein used to do). Then,

x'=x-vt
y'=y
z'=z
t'=t-vx/c²
and that is exactly what Robert Winn wrote on his OP. He used Voigt/Lorentz transforms for low velocities.

2) From them, it can be derived that:

t=t'+vx/c² = t'+v/c².x'+v/c².vt

t.(1- v²/c²) = (t'+v.x'/c²)
t = ɣ² (t'+v.x'/c²)

x=x'+vt = x' + ɣ² .v.(t'+v.x'/c²) = x' + ɣ² ..v.t'+ x'. ɣ² .v²/c² = x' (1+ɣ² .v²/c²) + ɣ² .v.t'

(1+ɣ² .v²/c²) = (1+v²/c²/(1- v²/c²)) = 1/(1- v²/c²) = ɣ²

x=x'+vt = ɣ².(x'+v.t')

You are right, Dirk, and I was wrong. They are Voigt/Lorentz transforms for v << c.
This is what you wrote, using my symbols.

x = ɣ².(x' + v.t')
y = y'
z = z'
t = ɣ² (t'+v.x'/c²)

Good job.

On Saturday, September 4, 2021 at 2:52:34 PM UTC-3, Richard Hertz wrote:

I FORGOT TO REDUCE THEM WHEN v << c, which gives ɣ ≈ 1

> x = ɣ².(x' + v.t')
> y = y'
> z = z'
> t = ɣ² (t'+v.x'/c²)

is reduced to:

x = x' + v.t'
y = y'
z = z'
t = t'+v.x'/c²

which are symmetric to (depending on the observer at E or E') the original set in the OP:

x' = x - v.t
y' = y
z' = z
t' = t - v.x/c²

On Saturday, September 4, 2021 at 9:13:41 AM UTC-3, Sylvia Else wrote:

<snip>

> There's no time dilation either, so t = t'.
>
> Then you end up with the simple Galilean transform which is the low
> velocity limit of the Lorentz transform, and has no more significance
> than that.
>
> Sylvia.

Sylvia, actually they are Voigt/Lorentz transforms but for low speeds (v << c), so ɣ ≈ 1.

Then, time is only adjusted due the concept of simultaneity. In my understanding, it's about
having different local times for two observers, who are separated by a distance x/c, which
increases due to the velocity of the E' frame, respect to the E frame at rest.

I think that the concept behind this is that different local times only can be verified by communicating
time at the speed of light plus the effect of v.

So, even when clocks at E and E' are still synchronized since t = t' = 0 and x = x' = 0, the "simultaneity" is
only perceived with a time gap |t - t'| = v.x'/c².

What do you think about this? It's relativistic, after all.

Richard Hertz wrote:

> I think that the concept behind this is that different local times only
> can be verified by communicating time at the speed of light plus the
> effect of v.

relativity effects is not including the communication of the states.

On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
> Richard Hertz wrote:
>
> > I think that the concept behind this is that different local times only
> > can be verified by communicating time at the speed of light plus the
> > effect of v.

> relativity effects is not including the communication of the states.

Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
Which does it means, besides my interpretation?

Remember that Voigt's early relativity was based on the Doppler effect.

Imagine that you are on a train, moving at v and without acceleration. A siren sounds at a given instant t,
for the observer at rest. If the speed of sound is W, when an observer at a train start to hear the pitch?

The observer at rest, next to the sound source hears it instantaneously.

The observer at the train hears it with a time difference |t - t'| = v.x/W². As far as he goes, the difference increases.

What about the "simultaneity" of events heard by the two observers?

On Saturday, September 4, 2021 at 5:13:41 AM UTC-7, Sylvia Else wrote:
> On 04-Sep-21 9:43 pm, Robert Winn wrote:
> > On Friday, September 3, 2021 at 11:38:59 PM UTC-7, Sylvia Else wrote:
> >> On 04-Sep-21 4:51 am, Robert Winn wrote:
> >>> x'=x-vt
> >>> y'=y
> >>> z'=z
> >>> t'=t-vx/c^2
> >>>
> >>> inverse equations
> >>>
> >>> x = x' - v't'
> >>>
> >>> v't' = -vt
> >>>
> >>> x = x' + vt
> >>> y = y'
> >>> z = z'
> >>> t = t' + vx/c^2
> >>>
> >>> So the equations are in the form scientists say they insist on.
> >> In the first set, the primed variables are expressed in terms of the
> >> non-primed variables. So the inverse form has to express the non-primed
> >> variables in terms of the primed variables.
> >>
> >> Your "inverse" forms express the non-primed variables in terms of a
> >> mixture of primed and non-primed variables.
> >>
> >> Sylvia.
> > I explained what I was doing.
> > x' = x - vt
> > y'=y
> > z'=z
> > t'=t-vx/c^2
> >
> > inverse equations
> >
> > x = x' - v't'
> > Since there is no length contraction in the Galilean transformation equations,
> > v't' = -vt
> > Consequently,
> > x = x' + vt
> There's no time dilation either, so t = t'.
>
> Then you end up with the simple Galilean transform which is the low
> velocity limit of the Lorentz transform, and has no more significance
> than that.
>
> Sylvia.
Well, I just view them as two clocks with different rates, so it is really no different as far as the equations are concerned than when considering a clock that gives correct time and a clock that loses ten minutes per day, when looking at it from the perspective of the Galilean transformation equations. Here is the difference between a Galilean transformation equation perspective of relativity and the Lorentz one. With the Galilean transformation equations, we end up with

x'=x-vt
t'=t-vx/c^2
x = x' + vt
t = t' + vx/c^2

So the equations for t and t' are essentially the same. This coincides with reality because even though scientists and all their supporters believe that if an observer in one frame of reference sees a clock in another frame of reference as being slower than his, the observer in the other frame of reference will see the first observer's clock as being slower. Reality shows us that if one clock is slower than the other, both observers will see it as slower. Now, following Einstein's reasoning leads to some further problems, one of which is that x in the equation for time is not the same as x in x'=x-vt. x in the equation for distances can be any x on the x-axis. x in the equation for time was defined by Einstein to be x = ct or x = -ct. This being the case, if t is one second, then there are only two coordinates the x in the equation for time can be if t is one second. This x has to be either 186,000 miles or -186,000 miles. So it would appear we have to modify the equations from Lorentz's idea to t'=t-v(x2)/c^2 and
t = t' +(x2)/c^2, where (x2) = ct or -ct. That would give us

x'=x-vt
t'= t-v(x2)/c^2)
x = x' + vt
t = t' + v(x2)/c^2

Richard Hertz wrote:

> The observer at rest, next to the sound source hears it instantaneously.
>
> The observer at the train hears it with a time difference |t - t'| =
> v.x/W². As far as he goes, the difference increases.

relativity is not about air, but rather imply its absence. You guys, I
don't know what to say.

On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
> > Richard Hertz wrote:
> >
> > > I think that the concept behind this is that different local times only
> > > can be verified by communicating time at the speed of light plus the
> > > effect of v.
>
> > relativity effects is not including the communication of the states.
> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
> Which does it means, besides my interpretation?
>
> Remember that Voigt's early relativity was based on the Doppler effect.
>
> Imagine that you are on a train, moving at v and without acceleration. A siren sounds at a given instant t,
> for the observer at rest. If the speed of sound is W, when an observer at a train start to hear the pitch?
>
> The observer at rest, next to the sound source hears it instantaneously.
>
> The observer at the train hears it with a time difference |t - t'| = v.x/W². As far as he goes, the difference increases.
>
>
> What about the "simultaneity" of events heard by the two observers?
Simultaneity was the first disagreement I had with Einstein that I was able to express mathematically. We take Einstein's famous train and bolts of lightning problem. Lightning strikes ahead of and behind a moving train simultaneously as seen from the frame of reference of the railroad track equal distances from an observer on the train at the middle of the train. What will the observer on the train see? Einstein says he will see the lightning ahead of the train first.
This is easily shown to be wrong. Suppose we say that the two bolts of lightning strike the extreme front and rear of the train, simultaneous in the frame of reference of the train, also leaving marks on the railroad track. How far apart are the marks on the track?
Scientists say that they are less than the length of the train apart because the lightning at the rear of the train would have struck first in the frame of reference of the railroad track.
No, sorry, this is wrong. The marks on the tracks would be the length of the train apart, and both observers would see them as simultaneous. The reason for this is that even in the example Einstein used, the marks on the railroad track have nothing to do with when the observer on the train would see the lightning. If the distances were equal to the observer on the train when the lightning struck, then in his frame of reference, those were the distances the light had to travel, regardless of the movement of marks on the railroad track.

Robert Winn <rbwinn3@gmail.com> wrote:
> On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
>> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
>>> Richard Hertz wrote:
>>>
>>>> I think that the concept behind this is that different local times only
>>>> can be verified by communicating time at the speed of light plus the
>>>> effect of v.
>>
>>> relativity effects is not including the communication of the states.
>> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
>> Which does it means, besides my interpretation?
>>
>> Remember that Voigt's early relativity was based on the Doppler effect.
>>
>> Imagine that you are on a train, moving at v and without acceleration. A
>> siren sounds at a given instant t,
>> for the observer at rest. If the speed of sound is W, when an observer
>> at a train start to hear the pitch?
>>
>> The observer at rest, next to the sound source hears it instantaneously.
>>
>> The observer at the train hears it with a time difference |t - t'| =
>> v.x/W². As far as he goes, the difference increases.
>>
>>
>> What about the "simultaneity" of events heard by the two observers?
> Simultaneity was the first disagreement I had with Einstein that I was
> able to express mathematically. We take Einstein's famous train and
> bolts of lightning problem. Lightning strikes ahead of and behind a
> moving train simultaneously as seen from the frame of reference of the
> railroad track equal distances from an observer on the train at the
> middle of the train. What will the observer on the train see? Einstein
> says he will see the lightning ahead of the train first.
> This is easily shown to be wrong. Suppose we say that the two bolts of
> lightning strike the extreme front and rear of the train, simultaneous in
> the frame of reference of the train, also leaving marks on the railroad
> track. How far apart are the marks on the track?
> Scientists say that they are less than the length of the train apart
> because the lightning at the rear of the train would have struck first in
> the frame of reference of the railroad track.
> No, sorry, this is wrong. The marks on the tracks would be the length of
> the train apart, and both observers would see them as simultaneous. The
> reason for this is that even in the example Einstein used, the marks on
> the railroad track have nothing to do with when the observer on the train
> would see the lightning. If the distances were equal to the observer on
> the train when the lightning struck, then in his frame of reference,
> those were the distances the light had to travel, regardless of the
> movement of marks on the railroad track.
>

I see you have missed the point entirely of what you read.

--
Odd Bodkin — Maker of fine toys, tools, tables

On Saturday, September 4, 2021 at 4:51:57 PM UTC-7, bodk...@gmail.com wrote:
> Robert Winn <rbw...@gmail.com> wrote:
> > On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
> >> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
> >>> Richard Hertz wrote:
> >>>
> >>>> I think that the concept behind this is that different local times only
> >>>> can be verified by communicating time at the speed of light plus the
> >>>> effect of v.
> >>
> >>> relativity effects is not including the communication of the states.
> >> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
> >> Which does it means, besides my interpretation?
> >>
> >> Remember that Voigt's early relativity was based on the Doppler effect..
> >>
> >> Imagine that you are on a train, moving at v and without acceleration. A
> >> siren sounds at a given instant t,
> >> for the observer at rest. If the speed of sound is W, when an observer
> >> at a train start to hear the pitch?
> >>
> >> The observer at rest, next to the sound source hears it instantaneously.
> >>
> >> The observer at the train hears it with a time difference |t - t'| =
> >> v.x/W². As far as he goes, the difference increases.
> >>
> >>
> >> What about the "simultaneity" of events heard by the two observers?
> > Simultaneity was the first disagreement I had with Einstein that I was
> > able to express mathematically. We take Einstein's famous train and
> > bolts of lightning problem. Lightning strikes ahead of and behind a
> > moving train simultaneously as seen from the frame of reference of the
> > railroad track equal distances from an observer on the train at the
> > middle of the train. What will the observer on the train see? Einstein
> > says he will see the lightning ahead of the train first.
> > This is easily shown to be wrong. Suppose we say that the two bolts of
> > lightning strike the extreme front and rear of the train, simultaneous in
> > the frame of reference of the train, also leaving marks on the railroad
> > track. How far apart are the marks on the track?
> > Scientists say that they are less than the length of the train apart
> > because the lightning at the rear of the train would have struck first in
> > the frame of reference of the railroad track.
> > No, sorry, this is wrong. The marks on the tracks would be the length of
> > the train apart, and both observers would see them as simultaneous. The
> > reason for this is that even in the example Einstein used, the marks on
> > the railroad track have nothing to do with when the observer on the train
> > would see the lightning. If the distances were equal to the observer on
> > the train when the lightning struck, then in his frame of reference,
> > those were the distances the light had to travel, regardless of the
> > movement of marks on the railroad track.
> >
> I see you have missed the point entirely of what you read.
> --
> Odd Bodkin — Maker of fine toys, tools, tables
I also did not look for the book you said I was supposed to find. How did I miss the point entirely?

Robert Winn <rbwinn3@gmail.com> wrote:
> On Saturday, September 4, 2021 at 4:51:57 PM UTC-7, bodk...@gmail.com wrote:
>> Robert Winn <rbw...@gmail.com> wrote:
>>> On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
>>>> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
>>>>> Richard Hertz wrote:
>>>>>
>>>>>> I think that the concept behind this is that different local times only
>>>>>> can be verified by communicating time at the speed of light plus the
>>>>>> effect of v.
>>>>
>>>>> relativity effects is not including the communication of the states.
>>>> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
>>>> Which does it means, besides my interpretation?
>>>>
>>>> Remember that Voigt's early relativity was based on the Doppler effect.
>>>>
>>>> Imagine that you are on a train, moving at v and without acceleration. A
>>>> siren sounds at a given instant t,
>>>> for the observer at rest. If the speed of sound is W, when an observer
>>>> at a train start to hear the pitch?
>>>>
>>>> The observer at rest, next to the sound source hears it instantaneously.
>>>>
>>>> The observer at the train hears it with a time difference |t - t'| =
>>>> v.x/W². As far as he goes, the difference increases.
>>>>
>>>>
>>>> What about the "simultaneity" of events heard by the two observers?
>>> Simultaneity was the first disagreement I had with Einstein that I was
>>> able to express mathematically. We take Einstein's famous train and
>>> bolts of lightning problem. Lightning strikes ahead of and behind a
>>> moving train simultaneously as seen from the frame of reference of the
>>> railroad track equal distances from an observer on the train at the
>>> middle of the train. What will the observer on the train see? Einstein
>>> says he will see the lightning ahead of the train first.
>>> This is easily shown to be wrong. Suppose we say that the two bolts of
>>> lightning strike the extreme front and rear of the train, simultaneous in
>>> the frame of reference of the train, also leaving marks on the railroad
>>> track. How far apart are the marks on the track?
>>> Scientists say that they are less than the length of the train apart
>>> because the lightning at the rear of the train would have struck first in
>>> the frame of reference of the railroad track.
>>> No, sorry, this is wrong. The marks on the tracks would be the length of
>>> the train apart, and both observers would see them as simultaneous. The
>>> reason for this is that even in the example Einstein used, the marks on
>>> the railroad track have nothing to do with when the observer on the train
>>> would see the lightning. If the distances were equal to the observer on
>>> the train when the lightning struck, then in his frame of reference,
>>> those were the distances the light had to travel, regardless of the
>>> movement of marks on the railroad track.
>>>
>> I see you have missed the point entirely of what you read.
>> --
>> Odd Bodkin — Maker of fine toys, tools, tables
> I also did not look for the book you said I was supposed to find. How
> did I miss the point entirely?
>

In this case it was something you read but now no longer remember
correctly, and you’ve captured that poor memory above. And now what are you
going to do?

--
Odd Bodkin — Maker of fine toys, tools, tables

Robert Winn wrote:
> x in the equation for time was defined by Einstein to be x = ct or x = -ct.

No! x can take any value in the equation for time in Lorentz Equation.

x=ct or -ct can be considered as special cases in order to determine
the transformation equations but at the end x can take ANY value.

On Saturday, September 4, 2021 at 6:52:04 PM UTC-7, bodk...@gmail.com wrote:
> Robert Winn <rbw...@gmail.com> wrote:
> > On Saturday, September 4, 2021 at 4:51:57 PM UTC-7, bodk...@gmail.com wrote:
> >> Robert Winn <rbw...@gmail.com> wrote:
> >>> On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
> >>>> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
> >>>>> Richard Hertz wrote:
> >>>>>
> >>>>>> I think that the concept behind this is that different local times only
> >>>>>> can be verified by communicating time at the speed of light plus the
> >>>>>> effect of v.
> >>>>
> >>>>> relativity effects is not including the communication of the states..
> >>>> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
> >>>> Which does it means, besides my interpretation?
> >>>>
> >>>> Remember that Voigt's early relativity was based on the Doppler effect.
> >>>>
> >>>> Imagine that you are on a train, moving at v and without acceleration. A
> >>>> siren sounds at a given instant t,
> >>>> for the observer at rest. If the speed of sound is W, when an observer
> >>>> at a train start to hear the pitch?
> >>>>
> >>>> The observer at rest, next to the sound source hears it instantaneously.
> >>>>
> >>>> The observer at the train hears it with a time difference |t - t'| =
> >>>> v.x/W². As far as he goes, the difference increases.
> >>>>
> >>>>
> >>>> What about the "simultaneity" of events heard by the two observers?
> >>> Simultaneity was the first disagreement I had with Einstein that I was
> >>> able to express mathematically. We take Einstein's famous train and
> >>> bolts of lightning problem. Lightning strikes ahead of and behind a
> >>> moving train simultaneously as seen from the frame of reference of the
> >>> railroad track equal distances from an observer on the train at the
> >>> middle of the train. What will the observer on the train see? Einstein
> >>> says he will see the lightning ahead of the train first.
> >>> This is easily shown to be wrong. Suppose we say that the two bolts of
> >>> lightning strike the extreme front and rear of the train, simultaneous in
> >>> the frame of reference of the train, also leaving marks on the railroad
> >>> track. How far apart are the marks on the track?
> >>> Scientists say that they are less than the length of the train apart
> >>> because the lightning at the rear of the train would have struck first in
> >>> the frame of reference of the railroad track.
> >>> No, sorry, this is wrong. The marks on the tracks would be the length of
> >>> the train apart, and both observers would see them as simultaneous. The
> >>> reason for this is that even in the example Einstein used, the marks on
> >>> the railroad track have nothing to do with when the observer on the train
> >>> would see the lightning. If the distances were equal to the observer on
> >>> the train when the lightning struck, then in his frame of reference,
> >>> those were the distances the light had to travel, regardless of the
> >>> movement of marks on the railroad track.
> >>>
> >> I see you have missed the point entirely of what you read.
> >> --
> >> Odd Bodkin — Maker of fine toys, tools, tables
> > I also did not look for the book you said I was supposed to find. How
> > did I miss the point entirely?
> >
> In this case it was something you read but now no longer remember
> correctly, and you’ve captured that poor memory above. And now what are you
> going to do?
> --
> Odd Bodkin — Maker of fine toys, tools, tables
Well, I am going to say, if you think I remembered it incorrectly, go ahead and say what was correct. No?
Why not?

On Sunday, September 5, 2021 at 4:23:37 AM UTC-7, Python wrote:
> Robert Winn wrote:
> > x in the equation for time was defined by Einstein to be x = ct or x = -ct.
> No! x can take any value in the equation for time in Lorentz Equation.
>
> x=ct or -ct can be considered as special cases in order to determine
> the transformation equations but at the end x can take ANY value.
According to the Lorentz equations, right? So let's take an example. Suppose we have something traveling at the speed of the planet Mercury, 30 miles per second. So in one second, this moving frame of reference travels 30 miles. If the origins of S and S' coincided at t=0, then after one second, the origin of S' is at x=30 miles. The Galilean transformation equations would show this as
x' = x - vt
0 miles = 30 miles - 30 miles/sec( 1sec) So x = 30 miles, x' = 0 miles. Then we can say if x = 31 miles, x'=1 mile, if x =32 miles, x' = 2 miles, etc., until we finally get to x=ct, which would be x=186,000 miles and x'= 185,930.
But if we go with Lorentz's and Einstein's interpretation of this, then how do we reconcile all of these distances with time? They are using two different times if they say that x has the same value throughout their equations. They are saying that t= 1 sec. So suppose x=32 miles, for example, and they put 32 miles for x into their equation,
t'=(t-vx/c^2)/sqrt(1-v^2/c^2), they are introducing a new value for t because that part of Lorentz's equation was derived assuming that x=ct, and the time it takes light to go 32 miles is a lot less than the time it takes for it to go 186,000 miles. So maybe you would like to explain this from the scientific point of view.

Robert Winn <rbwinn3@gmail.com> wrote:
> On Saturday, September 4, 2021 at 6:52:04 PM UTC-7, bodk...@gmail.com wrote:
>> Robert Winn <rbw...@gmail.com> wrote:
>>> On Saturday, September 4, 2021 at 4:51:57 PM UTC-7, bodk...@gmail.com wrote:
>>>> Robert Winn <rbw...@gmail.com> wrote:
>>>>> On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
>>>>>> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
>>>>>>> Richard Hertz wrote:
>>>>>>>
>>>>>>>> I think that the concept behind this is that different local times only
>>>>>>>> can be verified by communicating time at the speed of light plus the
>>>>>>>> effect of v.
>>>>>>
>>>>>>> relativity effects is not including the communication of the states.
>>>>>> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
>>>>>> Which does it means, besides my interpretation?
>>>>>>
>>>>>> Remember that Voigt's early relativity was based on the Doppler effect.
>>>>>>
>>>>>> Imagine that you are on a train, moving at v and without acceleration. A
>>>>>> siren sounds at a given instant t,
>>>>>> for the observer at rest. If the speed of sound is W, when an observer
>>>>>> at a train start to hear the pitch?
>>>>>>
>>>>>> The observer at rest, next to the sound source hears it instantaneously.
>>>>>>
>>>>>> The observer at the train hears it with a time difference |t - t'| =
>>>>>> v.x/W². As far as he goes, the difference increases.
>>>>>>
>>>>>>
>>>>>> What about the "simultaneity" of events heard by the two observers?
>>>>> Simultaneity was the first disagreement I had with Einstein that I was
>>>>> able to express mathematically. We take Einstein's famous train and
>>>>> bolts of lightning problem. Lightning strikes ahead of and behind a
>>>>> moving train simultaneously as seen from the frame of reference of the
>>>>> railroad track equal distances from an observer on the train at the
>>>>> middle of the train. What will the observer on the train see? Einstein
>>>>> says he will see the lightning ahead of the train first.
>>>>> This is easily shown to be wrong. Suppose we say that the two bolts of
>>>>> lightning strike the extreme front and rear of the train, simultaneous in
>>>>> the frame of reference of the train, also leaving marks on the railroad
>>>>> track. How far apart are the marks on the track?
>>>>> Scientists say that they are less than the length of the train apart
>>>>> because the lightning at the rear of the train would have struck first in
>>>>> the frame of reference of the railroad track.
>>>>> No, sorry, this is wrong. The marks on the tracks would be the length of
>>>>> the train apart, and both observers would see them as simultaneous. The
>>>>> reason for this is that even in the example Einstein used, the marks on
>>>>> the railroad track have nothing to do with when the observer on the train
>>>>> would see the lightning. If the distances were equal to the observer on
>>>>> the train when the lightning struck, then in his frame of reference,
>>>>> those were the distances the light had to travel, regardless of the
>>>>> movement of marks on the railroad track.
>>>>>
>>>> I see you have missed the point entirely of what you read.
>>>> --
>>>> Odd Bodkin — Maker of fine toys, tools, tables
>>> I also did not look for the book you said I was supposed to find. How
>>> did I miss the point entirely?
>>>
>> In this case it was something you read but now no longer remember
>> correctly, and you’ve captured that poor memory above. And now what are you
>> going to do?
>> --
>> Odd Bodkin — Maker of fine toys, tools, tables
> Well, I am going to say, if you think I remembered it incorrectly, go
> ahead and say what was correct. No?
> Why not?
>

I wonder why you think it’s an amusement to say something wrong and hope
that people will fall over themselves to correct you? That would pose
endless weeks of fun for you, wouldn’t it? And so easy.

--
Odd Bodkin — Maker of fine toys, tools, tables

Op 04-sep.-2021 om 16:49 schreef Odd Bodkin:
> Robert Winn <rbwinn3@gmail.com> wrote:
>> On Friday, September 3, 2021 at 11:38:59 PM UTC-7, Sylvia Else wrote:
>>> On 04-Sep-21 4:51 am, Robert Winn wrote:
>>>> x'=x-vt
>>>> y'=y
>>>> z'=z
>>>> t'=t-vx/c^2
>
> Well let’s just do some algebra. Start from last equation.
>
> t=t’+vx/c^2
>
> Stick in first equation.
>
> x=x’+vt=x’+v(t’+vx^2)=x’+vt’+xv^2/c^2
> x(1-v^2/c^2)=x’+vt’
>
> Well look at that. There’s the gamma thing.

It's all red and sits in a corner.
A baby with a razor.

Dirk vdm

Robert Winn wrote:
> On Sunday, September 5, 2021 at 4:23:37 AM UTC-7, Python wrote:
>> Robert Winn wrote:
>>> x in the equation for time was defined by Einstein to be x = ct or x = -ct.
>> No! x can take any value in the equation for time in Lorentz Equation.
>>
>> x=ct or -ct can be considered as special cases in order to determine
>> the transformation equations but at the end x can take ANY value.
> According to the Lorentz equations, right?

Right.

[snip completely demented pseudo-use of actual equations]

On Sunday, September 5, 2021 at 3:43:40 PM UTC-7, Python wrote:
> Robert Winn wrote:
> > On Sunday, September 5, 2021 at 4:23:37 AM UTC-7, Python wrote:
> >> Robert Winn wrote:
> >>> x in the equation for time was defined by Einstein to be x = ct or x = -ct.
> >> No! x can take any value in the equation for time in Lorentz Equation.
> >>
> >> x=ct or -ct can be considered as special cases in order to determine
> >> the transformation equations but at the end x can take ANY value.
> > According to the Lorentz equations, right?
> Right.
>
> [snip completely demented pseudo-use of actual equations]
Well, if you do not want to discuss relativity, maybe some of these nice scientists would talk about the length contraction with you.

On Sunday, September 5, 2021 at 8:18:27 AM UTC-7, bodk...@gmail.com wrote:
> Robert Winn <rbw...@gmail.com> wrote:
> > On Saturday, September 4, 2021 at 6:52:04 PM UTC-7, bodk...@gmail.com wrote:
> >> Robert Winn <rbw...@gmail.com> wrote:
> >>> On Saturday, September 4, 2021 at 4:51:57 PM UTC-7, bodk...@gmail.com wrote:
> >>>> Robert Winn <rbw...@gmail.com> wrote:
> >>>>> On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
> >>>>>> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
> >>>>>>> Richard Hertz wrote:
> >>>>>>>
> >>>>>>>> I think that the concept behind this is that different local times only
> >>>>>>>> can be verified by communicating time at the speed of light plus the
> >>>>>>>> effect of v.
> >>>>>>
> >>>>>>> relativity effects is not including the communication of the states.
> >>>>>> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
> >>>>>> Which does it means, besides my interpretation?
> >>>>>>
> >>>>>> Remember that Voigt's early relativity was based on the Doppler effect.
> >>>>>>
> >>>>>> Imagine that you are on a train, moving at v and without acceleration. A
> >>>>>> siren sounds at a given instant t,
> >>>>>> for the observer at rest. If the speed of sound is W, when an observer
> >>>>>> at a train start to hear the pitch?
> >>>>>>
> >>>>>> The observer at rest, next to the sound source hears it instantaneously.
> >>>>>>
> >>>>>> The observer at the train hears it with a time difference |t - t'| =
> >>>>>> v.x/W². As far as he goes, the difference increases.
> >>>>>>
> >>>>>>
> >>>>>> What about the "simultaneity" of events heard by the two observers?
> >>>>> Simultaneity was the first disagreement I had with Einstein that I was
> >>>>> able to express mathematically. We take Einstein's famous train and
> >>>>> bolts of lightning problem. Lightning strikes ahead of and behind a
> >>>>> moving train simultaneously as seen from the frame of reference of the
> >>>>> railroad track equal distances from an observer on the train at the
> >>>>> middle of the train. What will the observer on the train see? Einstein
> >>>>> says he will see the lightning ahead of the train first.
> >>>>> This is easily shown to be wrong. Suppose we say that the two bolts of
> >>>>> lightning strike the extreme front and rear of the train, simultaneous in
> >>>>> the frame of reference of the train, also leaving marks on the railroad
> >>>>> track. How far apart are the marks on the track?
> >>>>> Scientists say that they are less than the length of the train apart
> >>>>> because the lightning at the rear of the train would have struck first in
> >>>>> the frame of reference of the railroad track.
> >>>>> No, sorry, this is wrong. The marks on the tracks would be the length of
> >>>>> the train apart, and both observers would see them as simultaneous. The
> >>>>> reason for this is that even in the example Einstein used, the marks on
> >>>>> the railroad track have nothing to do with when the observer on the train
> >>>>> would see the lightning. If the distances were equal to the observer on
> >>>>> the train when the lightning struck, then in his frame of reference,
> >>>>> those were the distances the light had to travel, regardless of the
> >>>>> movement of marks on the railroad track.
> >>>>>
> >>>> I see you have missed the point entirely of what you read.
> >>>> --
> >>>> Odd Bodkin — Maker of fine toys, tools, tables
> >>> I also did not look for the book you said I was supposed to find. How
> >>> did I miss the point entirely?
> >>>
> >> In this case it was something you read but now no longer remember
> >> correctly, and you’ve captured that poor memory above. And now what are you
> >> going to do?
> >> --
> >> Odd Bodkin — Maker of fine toys, tools, tables
> > Well, I am going to say, if you think I remembered it incorrectly, go
> > ahead and say what was correct. No?
> > Why not?
> >
> I wonder why you think it’s an amusement to say something wrong and hope
> that people will fall over themselves to correct you? That would pose
> endless weeks of fun for you, wouldn’t it? And so easy.
> --
> Odd Bodkin — Maker of fine toys, tools, tables
Oh, you are still upset that I did not go looking for that book.

On Sunday, September 5, 2021 at 8:39:33 AM UTC-7, Dirk Van de moortel wrote:
> Op 04-sep.-2021 om 16:49 schreef Odd Bodkin:
> > Robert Winn <rbw...@gmail.com> wrote:
> >> On Friday, September 3, 2021 at 11:38:59 PM UTC-7, Sylvia Else wrote:
> >>> On 04-Sep-21 4:51 am, Robert Winn wrote:
> >>>> x'=x-vt
> >>>> y'=y
> >>>> z'=z
> >>>> t'=t-vx/c^2
> >
> > Well let’s just do some algebra. Start from last equation.
> >
> > t=t’+vx/c^2
> >
> > Stick in first equation.
> >
> > x=x’+vt=x’+v(t’+vx^2)=x’+vt’+xv^2/c^2
> > x(1-v^2/c^2)=x’+vt’
> >
> > Well look at that. There’s the gamma thing.
> It's all red and sits in a corner.
> A baby with a razor.
>
> Dirk vdm
So what are you going to do with it? You can't get it to x'=(x-vt)/sqrt(1-v^2/c^2 like Lorentz did.

Crank Robert Winn wrote:
> On Sunday, September 5, 2021 at 3:43:40 PM UTC-7, Python wrote:
>> Robert Winn wrote:
>>> On Sunday, September 5, 2021 at 4:23:37 AM UTC-7, Python wrote:
>>>> Robert Winn wrote:
>>>>> x in the equation for time was defined by Einstein to be x = ct or x = -ct.
>>>> No! x can take any value in the equation for time in Lorentz Equation.
>>>>
>>>> x=ct or -ct can be considered as special cases in order to determine
>>>> the transformation equations but at the end x can take ANY value.
>>> According to the Lorentz equations, right?
>> Right.
>>
>> [snip completely demented pseudo-use of actual equations]
> Well, if you do not want to discuss relativity

How could one discusses relativity equations of any kind with
someone so confused about what transformation equations are about
that he could claim such an idiocy as "x in the equation for time was
defined by Einstein to be x = ct or x = -ct"? No way.

As Dirk wrote, you are a baby with a razor, you've always been Robert.

Robert Winn <rbwinn3@gmail.com> wrote:
> On Sunday, September 5, 2021 at 8:18:27 AM UTC-7, bodk...@gmail.com wrote:
>> Robert Winn <rbw...@gmail.com> wrote:
>>> On Saturday, September 4, 2021 at 6:52:04 PM UTC-7, bodk...@gmail.com wrote:
>>>> Robert Winn <rbw...@gmail.com> wrote:
>>>>> On Saturday, September 4, 2021 at 4:51:57 PM UTC-7, bodk...@gmail.com wrote:
>>>>>> Robert Winn <rbw...@gmail.com> wrote:
>>>>>>> On Saturday, September 4, 2021 at 2:00:38 PM UTC-7, Richard Hertz wrote:
>>>>>>>> On Saturday, September 4, 2021 at 5:32:27 PM UTC-3, H2O wrote:
>>>>>>>>> Richard Hertz wrote:
>>>>>>>>>
>>>>>>>>>> I think that the concept behind this is that different local times only
>>>>>>>>>> can be verified by communicating time at the speed of light plus the
>>>>>>>>>> effect of v.
>>>>>>>>
>>>>>>>>> relativity effects is not including the communication of the states.
>>>>>>>> Can you explain the difference |t - t'| = v.x'/c² when ɣ ≈ 1?
>>>>>>>> Which does it means, besides my interpretation?
>>>>>>>>
>>>>>>>> Remember that Voigt's early relativity was based on the Doppler effect.
>>>>>>>>
>>>>>>>> Imagine that you are on a train, moving at v and without acceleration. A
>>>>>>>> siren sounds at a given instant t,
>>>>>>>> for the observer at rest. If the speed of sound is W, when an observer
>>>>>>>> at a train start to hear the pitch?
>>>>>>>>
>>>>>>>> The observer at rest, next to the sound source hears it instantaneously.
>>>>>>>>
>>>>>>>> The observer at the train hears it with a time difference |t - t'| =
>>>>>>>> v.x/W². As far as he goes, the difference increases.
>>>>>>>>
>>>>>>>>
>>>>>>>> What about the "simultaneity" of events heard by the two observers?
>>>>>>> Simultaneity was the first disagreement I had with Einstein that I was
>>>>>>> able to express mathematically. We take Einstein's famous train and
>>>>>>> bolts of lightning problem. Lightning strikes ahead of and behind a
>>>>>>> moving train simultaneously as seen from the frame of reference of the
>>>>>>> railroad track equal distances from an observer on the train at the
>>>>>>> middle of the train. What will the observer on the train see? Einstein
>>>>>>> says he will see the lightning ahead of the train first.
>>>>>>> This is easily shown to be wrong. Suppose we say that the two bolts of
>>>>>>> lightning strike the extreme front and rear of the train, simultaneous in
>>>>>>> the frame of reference of the train, also leaving marks on the railroad
>>>>>>> track. How far apart are the marks on the track?
>>>>>>> Scientists say that they are less than the length of the train apart
>>>>>>> because the lightning at the rear of the train would have struck first in
>>>>>>> the frame of reference of the railroad track.
>>>>>>> No, sorry, this is wrong. The marks on the tracks would be the length of
>>>>>>> the train apart, and both observers would see them as simultaneous. The
>>>>>>> reason for this is that even in the example Einstein used, the marks on
>>>>>>> the railroad track have nothing to do with when the observer on the train
>>>>>>> would see the lightning. If the distances were equal to the observer on
>>>>>>> the train when the lightning struck, then in his frame of reference,
>>>>>>> those were the distances the light had to travel, regardless of the
>>>>>>> movement of marks on the railroad track.
>>>>>>>
>>>>>> I see you have missed the point entirely of what you read.
>>>>>> --
>>>>>> Odd Bodkin — Maker of fine toys, tools, tables
>>>>> I also did not look for the book you said I was supposed to find. How
>>>>> did I miss the point entirely?
>>>>>
>>>> In this case it was something you read but now no longer remember
>>>> correctly, and you’ve captured that poor memory above. And now what are you
>>>> going to do?
>>>> --
>>>> Odd Bodkin — Maker of fine toys, tools, tables
>>> Well, I am going to say, if you think I remembered it incorrectly, go
>>> ahead and say what was correct. No?
>>> Why not?
>>>
>> I wonder why you think it’s an amusement to say something wrong and hope
>> that people will fall over themselves to correct you? That would pose
>> endless weeks of fun for you, wouldn’t it? And so easy.
>> --
>> Odd Bodkin — Maker of fine toys, tools, tables
> Oh, you are still upset that I did not go looking for that book.
>

Heck no. I knew you weren’t going to look up anything. Anything that
involves getting up from your chair is too much effort.

--
Odd Bodkin — Maker of fine toys, tools, tables

Robert Winn <rbwinn3@gmail.com> wrote:
> On Sunday, September 5, 2021 at 8:39:33 AM UTC-7, Dirk Van de moortel wrote:
>> Op 04-sep.-2021 om 16:49 schreef Odd Bodkin:
>>> Robert Winn <rbw...@gmail.com> wrote:
>>>> On Friday, September 3, 2021 at 11:38:59 PM UTC-7, Sylvia Else wrote:
>>>>> On 04-Sep-21 4:51 am, Robert Winn wrote:
>>>>>> x'=x-vt
>>>>>> y'=y
>>>>>> z'=z
>>>>>> t'=t-vx/c^2
>>>
>>> Well let’s just do some algebra. Start from last equation.
>>>
>>> t=t’+vx/c^2
>>>
>>> Stick in first equation.
>>>
>>> x=x’+vt=x’+v(t’+vx^2)=x’+vt’+xv^2/c^2
>>> x(1-v^2/c^2)=x’+vt’
>>>
>>> Well look at that. There’s the gamma thing.
>> It's all red and sits in a corner.
>> A baby with a razor.
>>
>> Dirk vdm
> So what are you going to do with it? You can't get it to
> x'=(x-vt)/sqrt(1-v^2/c^2 like Lorentz did.
>

Not from your equations. That’s right.

--
Odd Bodkin — Maker of fine toys, tools, tables