Michael Moroney wrote:
> On 9/3/2021 8:09 PM, Robert Winn wrote:
>> On Friday, September 3, 2021 at 4:35:37 PM UTC-7, Michael Moroney wrote:
>>> On 9/3/2021 7:05 PM, Robert Winn wrote:
>>>> On Friday, September 3, 2021 at 1:56:59 PM UTC-7, Dono. wrote:
>>>>> On Friday, September 3, 2021 at 11:51:55 AM UTC-7, Robert Winn wrote:
>>>>>> x'=x-vt
>>>>>> y'=y
>>>>>> z'=z
>>>>>> t'=t-vx/c^2
>>>>>>
>>>>>> inverse equations
>>>>>>
>>>>>> x = x' - v't'
>>>>> Bzzt, wrong. You are unemployed again.
>>>> So go ahead and show the mistake, Dono. No?
>>> Dirk already did.
>>> But you blew him off.
>>>> Well, nice taking to you again.
>>>>
>> Dirk already did?  Dirk told me to derive x',y' z' and t' from the
>> other equations, which I did.  If Dirk could find an error in my
>> algebra, obviously, he would say what it was.
>
> And he did.  If you are unable to read his post, too bad for you, I
> guess you really are as dumb as those telephone poles you used to make.

And he's not going well, even how silly he uses to be I'm surprised he
cannot spot his mistake. sad.

On 9/3/2021 10:21 PM, Robert Winn wrote:
> On Friday, September 3, 2021 at 7:09:02 PM UTC-7, Michael Moroney wrote:
>> On 9/3/2021 8:09 PM, Robert Winn wrote:
>>> On Friday, September 3, 2021 at 4:35:37 PM UTC-7, Michael Moroney wrote:
>>>> On 9/3/2021 7:05 PM, Robert Winn wrote:
>>>>> On Friday, September 3, 2021 at 1:56:59 PM UTC-7, Dono. wrote:
>>>>>> On Friday, September 3, 2021 at 11:51:55 AM UTC-7, Robert Winn wrote:
>>>>>>> x'=x-vt
>>>>>>> y'=y
>>>>>>> z'=z
>>>>>>> t'=t-vx/c^2
>>>>>>>
>>>>>>> inverse equations
>>>>>>>
>>>>>>> x = x' - v't'
>>>>>> Bzzt, wrong. You are unemployed again.
>>>>> So go ahead and show the mistake, Dono. No?
>>>> Dirk already did.
>>>> But you blew him off.
>>>>> Well, nice taking to you again.
>>>>>
>>> Dirk already did? Dirk told me to derive x',y' z' and t' from the other equations, which I did. If Dirk could find an error in my algebra, obviously, he would say what it was.
>> And he did. If you are unable to read his post, too bad for you, I
>> guess you really are as dumb as those telephone poles you used to make.
> Well, if you had something to say, you would say it. So you have nothing to say. Telephone poles hold up phone lines. At least they do something useful.
> Dirk told me to derive x',y',z', and t' from the last four equations I had shown, which I did.

And he pointed out where you were wrong.

> Then Dirk made his comment about telephone poles, and we have not heard from him since then. So you are all saying that there is some mysterious error in my algebra, but no one will say what it is. I do not believe there is an error.

Of course you don't. Because you can't read Dirk's post correcting you.
Telephone poles can't read it either, but at least they have an
excuse, they're telephone poles.

On Friday, September 3, 2021 at 4:40:07 PM UTC-3, 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

v't' ≠ -vt
x = x' + vt

As x = vt

t' = t-vx/c² = t.(1 - v²/c²) = t . 1/ɣ²

t = t'/(1 - v²/c²) = t' . ɣ²

and

x = x' + vt = x' + v . t' . ɣ² ≠ ɣ² . (x' + v . t')

ɣ² = 1/(1 - v²/c²)

So, what is wrong with your calculations, Dirk?

> > 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 )

<snip>

On Friday, September 3, 2021 at 9:49:30 PM UTC-7, Richard Hertz wrote:

> As x = vt
Err, no. x is a general variable.

Rest of your pathetic attempt below is now a collection of imbecilities:

> t' = t-vx/c² = t.(1 - v²/c²) = t . 1/ɣ²
>
> t = t'/(1 - v²/c²) = t' . ɣ²
>
> and
>
> x = x' + vt = x' + v . t' . ɣ² ≠ ɣ² . (x' + v . t')
>
> ɣ² = 1/(1 - v²/c²)
>
> So, what is wrong with your calculations, Dirk?

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.

On Saturday, September 4, 2021 at 2:35:48 AM UTC-3, Dono. wrote:

> On Friday, September 3, 2021 at 9:49:30 PM UTC-7, Richard Hertz wrote:
>
> > As x = vt

> Err, no. x is a general variable.

NOTE: i forgot to add that's is valid for an observer not moving in the frame E'(x',t').

Being E(x,t) considered at rest and E'(x',t') moving at constant v from (0,0), it applies that x =vt.

An observer standing at x=0 and time t, emitting a pulse of light towards the observer standing at x'=0, knows
that E' has moved a distance x = v.t, being t the time measured at his clock since t=0, when both frames
are synchronized at the instant that E(0,0) and E'(0,0) occupy the same space-time (pure Einstein, isn't it?).

The light pulse travels a distance c.t = x/v to reach the position x'=0 at the time t'.

The elapsed time t-t' for the light pulse to reach x'=0, since t'=t= 0, is t-t' =x.v/c², as measured from the frame at rest.

t-t' =x.v/c² = x/c . v/c = (time for light to reach x') x (ratio of velocities of x' and of c). If v=c, then (according to SR),
the difference would be only t-t' = x/c, because c is the absolute maximum velocity.

So, the observer standing at x'=0 (is not moving in the frame E'), measures a time t'

t' = t - x.v/c²

Happy now, Dono? I forgot to express x = v.t for x'=0.

The rest follows as it was written.

t' = t-vx/c² = t.(1 - v²/c²) = t . 1/ɣ²

t = t'/(1 - v²/c²) = t' . ɣ²

and

x = x' + vt = x' + v . t' . ɣ² ≠ ɣ² . (x' + v . t')

ɣ² = 1/(1 - v²/c²)

Op 04-sep.-2021 om 06:49 schreef Richard Hertz:
> On Friday, September 3, 2021 at 4:40:07 PM UTC-3, 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
>
> v't' ≠ -vt
> x = x' + vt
>
> As x = vt

Idiot.
Pilchard likes physics and mathematics, and then miserably
fails to understand the physical meanings of the variables,
and has no idea about the meaning and usage of linear
transformations.
He "likes physics and mathematics" :-)

Dirk Vdm

>
> t' = t-vx/c² = t.(1 - v²/c²) = t . 1/ɣ²
>
> t = t'/(1 - v²/c²) = t' . ɣ²
>
> and
>
> x = x' + vt = x' + v . t' . ɣ² ≠ ɣ² . (x' + v . t')
>
> ɣ² = 1/(1 - v²/c²)
>
> So, what is wrong with your calculations, Dirk?
>
>
>>> 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 )
>
> <snip>
>

Robert Winn <rbwinn3@gmail.com> wrote:
> On Friday, September 3, 2021 at 7:04:20 PM UTC-7, bodk...@gmail.com wrote:
>> Robert Winn <rbw...@gmail.com> wrote:
>>> On Friday, September 3, 2021 at 5:09:55 PM UTC-7, bodk...@gmail.com wrote:
>>>> Robert Winn <rbw...@gmail.com> wrote:
>>>>> On Friday, September 3, 2021 at 1:40:58 PM UTC-7, bodk...@gmail.com wrote:
>>>>>> Robert Winn <rbw...@gmail.com> 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
>>>>>> These equations were proposed about 120 years ago by another scientist (not
>>>>>> Einstein). It’s probably worth your time finding out who did them before
>>>>>> you, and how it was determined that they are not right.
>>>>>>>
>>>>>>> So the equations are in the form scientists say they insist on. t is the
>>>>>>> time of a clock in S, t' is the time of a clock in S'. The equations
>>>>>>> show reality as it actually exists. If a pilot in an airplane has a
>>>>>>> clock that is slower than a clock on the ground, as Einstein proposed,
>>>>>>> then t' is the time of the clock in the airplane, and v' is the speed of
>>>>>>> the airplane according to the time of the clock in the airplane that
>>>>>>> shows t'. Since there is no length contraction in the Galilean
>>>>>>> transformation equations, the distance the airplane travels is the same
>>>>>>> as observed from either frame of reference. This enables us to say
>>>>>>>
>>>>>>> x = x' - v't'
>>>>>>> ct = ct' - v't'
>>>>>>> ct = ct' +vt
>>>>>>> t = t' + vt/c
>>>>>>> t = t' + vct/c^2
>>>>>>> t = t' + vx/c^2
>>>>>>>
>>>>>>> which is the same equation we got for the times using
>>>>>>>
>>>>>>> x'=x-vt
>>>>>>>
>>>>>>> ct' = ct -vt
>>>>>>> t' = t - vt/c
>>>>>>> t' = t - vx/c^2
>>>>>>>
>>>>>> --
>>>>> Well, since you know all about these equations, go ahead and give the details. No?
>>>>> So how is it that I am supposed to find this information?
>>>> A library. A book.
>>>>>> Odd Bodkin -- maker of fine toys, tools, tables
>>>>>
>>>> --
>>>> Odd Bodkin — Maker of fine toys, tools, tables
>>> Not enough information.
>> It is for people who do physics.
>>
>> You’re trying to do physics.
>>> So since you know all about it, how about saying something? No? Well,
>>> that is nothing new.
>>>
>> --
>> Odd Bodkin — Maker of fine toys, tools, tables
> Well, so relativity is physics. Don't want to discuss relativity?

What’s to discuss? You’ve written equations that were written by someone
else over a century ago and shown wrong. Why do you want to discuss those?

> Then you do not want to discuss physics.
>

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

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. There is no mistake in the algebra. You are just doing what you do by habit, pretending to be superior in some way.

On Friday, September 3, 2021 at 10:35:48 PM UTC-7, Dono. wrote:
> On Friday, September 3, 2021 at 9:49:30 PM UTC-7, Richard Hertz wrote:
>
> > As x = vt
> Err, no. x is a general variable.
>
> Rest of your pathetic attempt below is now a collection of imbecilities:
> > t' = t-vx/c² = t.(1 - v²/c²) = t . 1/ɣ²
> >
> > t = t'/(1 - v²/c²) = t' . ɣ²
> >
> > and
> >
> > x = x' + vt = x' + v . t' . ɣ² ≠ ɣ² . (x' + v . t')
> >
> > ɣ² = 1/(1 - v²/c²)
> >
> > So, what is wrong with your calculations, Dirk?
x = x' + vt
So where did you get x = vt?

t' = t - vx/c^2 = t(1-v^2/c^2)
Where did you get t(1-v^2/c^2)? There is no length contraction in the Galilean transformation equations. There is no gamma in anything I posted.

On Friday, September 3, 2021 at 10:35:48 PM UTC-7, Dono. wrote:
> On Friday, September 3, 2021 at 9:49:30 PM UTC-7, Richard Hertz wrote:
>
> > As x = vt
> Err, no. x is a general variable.

x= x' - v't'
Since there is no length contraction in the Galilean transformation equations, v't' = -vt.
That means
x = x' +vt, which is exactly the same as x' = x - vt.

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

To explain a little further for the world of science, I go back to the original problem I put together when Special Relativity was first explained to me. I imagined a clock in a flying airplane for the moving clock and a clock on the ground for the clock that was not moving. It was obvious to me that if the clock in the airplane was slower than the clock on the ground, then the pilot of the airplane would get a faster speed for the airplane than an observer on the ground would get using the clock on the ground. In the above equations that speed would be v'.

On Saturday, September 4, 2021 at 2:01:58 AM UTC-7, Dirk Van de moortel wrote:
> Op 04-sep.-2021 om 06:49 schreef Richard Hertz:
> > On Friday, September 3, 2021 at 4:40:07 PM UTC-3, 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
> >
> > v't' ≠ -vt
> > x = x' + vt
> >
> > As x = vt
> Idiot.
> Pilchard likes physics and mathematics, and then miserably
> fails to understand the physical meanings of the variables,
> and has no idea about the meaning and usage of linear
> transformations.
> He "likes physics and mathematics" :-)
>
> Dirk Vdm
> >
> > t' = t-vx/c² = t.(1 - v²/c²) = t . 1/ɣ²
> >
> > t = t'/(1 - v²/c²) = t' . ɣ²
> >
> > and
> >
> > x = x' + vt = x' + v . t' . ɣ² ≠ ɣ² . (x' + v . t')
> >
> > ɣ² = 1/(1 - v²/c²)
> >
> > So, what is wrong with your calculations, Dirk?
> >
> >
> >>> 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 )
> >
> > <snip>
> >
Yes, I know what you get. You are using the Lorentz equations. I am using the Galilean transformation equations.

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.

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 worked the problem that way, and these people did not like it. So I put it in terms of t and t', the way they said it had to be. The reason why the equations are not wrong is because I use the term v' for the velocity in S'. Isaac Newton had only v in his absolute time interpretation of the Galilean transformation equations.
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - (-v)t'
y=y'
z=z'
t = t'
Einstein was the one who theorized that a moving clock would be slower than a clock that was not moving. I just worked the math using the Galilean transformation equations. According to Hafele and Keating a moving clock can be slower or faster than a clock that is not moving, depending on the direction of motion relative to the rotation of the earth. The Galilean transformation equations can accommodate any clock rate. If the moving clock is faster, such as a clock in a GPS satellite, according to scientists, then an observer in the satellite would get a slower speed for the satellite than an observer on earth.

Robert Winn <rbwinn3@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

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. I do not assign some private meaning. 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.

On Friday, September 3, 2021 at 11:43:20 PM UTC-7, Richard Hertz wrote:
> On Saturday, September 4, 2021 at 2:35:48 AM UTC-3, Dono. wrote:
>
> > On Friday, September 3, 2021 at 9:49:30 PM UTC-7, Richard Hertz wrote:
> >
> > > As x = vt
>
> > Err, no. x is a general variable.
> NOTE: i forgot to add that's is valid for an observer not moving in the frame E'(x',t').
>
> Being E(x,t) considered at rest and E'(x',t') moving at constant v from (0,0), it applies that x =vt.
>
> An observer standing at x=0 and time t, emitting a pulse of light towards the observer standing at x'=0, knows
> that E' has moved a distance x = v.t, being t the time measured at his clock since t=0, when both frames
> are synchronized at the instant that E(0,0) and E'(0,0) occupy the same space-time (pure Einstein, isn't it?).
>
> The light pulse travels a distance c.t = x/v to reach the position x'=0 at the time t'.
>
> The elapsed time t-t' for the light pulse to reach x'=0, since t'=t= 0, is t-t' =x.v/c², as measured from the frame at rest.
>
> t-t' =x.v/c² = x/c . v/c = (time for light to reach x') x (ratio of velocities of x' and of c). If v=c, then (according to SR),
> the difference would be only t-t' = x/c, because c is the absolute maximum velocity.
>
> So, the observer standing at x'=0 (is not moving in the frame E'), measures a time t'
>
> t' = t - x.v/c²
>
> Happy now, Dono? I forgot to express x = v.t for x'=0.
>

Yes, I am very happy to see that you still entertain us with your utter cretinism.
Keep basking in your mistakes, imbecile.

Op 04-sep.-2021 om 13:45 schreef Robert Winn:
> On Saturday, September 4, 2021 at 2:01:58 AM UTC-7, Dirk Van de moortel wrote:
>> Op 04-sep.-2021 om 06:49 schreef Richard Hertz:
>>> On Friday, September 3, 2021 at 4:40:07 PM UTC-3, 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
>>>
>>> v't' ≠ -vt
>>> x = x' + vt
>>>
>>> As x = vt
>> Idiot.
>> Pilchard likes physics and mathematics, and then miserably
>> fails to understand the physical meanings of the variables,
>> and has no idea about the meaning and usage of linear
>> transformations.
>> He "likes physics and mathematics" :-)
>>
>> Dirk Vdm
>>>
>>> t' = t-vx/c² = t.(1 - v²/c²) = t . 1/ɣ²
>>>
>>> t = t'/(1 - v²/c²) = t' . ɣ²
>>>
>>> and
>>>
>>> x = x' + vt = x' + v . t' . ɣ² ≠ ɣ² . (x' + v . t')
>>>
>>> ɣ² = 1/(1 - v²/c²)
>>>
>>> So, what is wrong with your calculations, Dirk?
>>>
>>>
>>>>> 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 )
>>>
>>> <snip>
>>>
> Yes, I know what you get. You are using the Lorentz equations.

I'm not.
I'm using your (*YOUR*) first four equations and algebra.
Telephone pole.

> I am using the Galilean transformation equations.

You're not. Good grief.
Mega-telephone pole.

Go do some welding.

Dirk Vdm

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.

>>>
>>> inverse equations
>>>
>>> x = x' - v't'
>>>
>>> v't' = -vt
>>>
>>> x = x' + vt

That’s not what you’re looking for in an inverse relation. You want
unprimed x in terms of primed x’, t’.

>>> y = y'
>>> z = z'
>>> t = t' + vx/c^2

That’s not what you’re looking for in an inverse relation. You want
unprimed t in terms of primed x’, t’.

>>>
>>> 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
>
> To explain a little further for the world of science, I go back to the
> original problem I put together when Special Relativity was first
> explained to me. I imagined a clock in a flying airplane for the moving
> clock and a clock on the ground for the clock that was not moving. It
> was obvious to me that if the clock in the airplane was slower than the
> clock on the ground, then the pilot of the airplane would get a faster
> speed for the airplane than an observer on the ground would get using the
> clock on the ground. In the above equations that speed would be v'.
>

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

Robert Winn <rbwinn3@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

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 know how difficult this all must seem.
I know all about how scientists define continuous. They claim the Lorentz equations are continuous.

Robert Winn <rbwinn3@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?

>

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

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

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. Sorry, I just do not have time for that kind of errand.. I have no idea what scientists mean when they say that the Lorentz equations are continuous. They are discontinuous.