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Re: Annotated version of SRT

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

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

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

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


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

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

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

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

quote:

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

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

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

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


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


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

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

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

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

The emitter is placed in k at xsi=0.

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

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

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

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

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

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


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


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

What????

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


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

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

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

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

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

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

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

tau is the independent variable, because in k the position of the mirror 
is fixed and you cannot use a single fixed value as independent variable.

The system K and all related variables play no role in that scenario.

The fixed position of the mirror would not allow to differentiate time 
after it, because that position has only one single value.

But we could place a point in K, call that x' and describe that point 
with measures from system k.

Since a fixed point in K would move with velocity v 'to the left', we 
could say d xsi_x'/ d tau=v.

Now in K this point xsi_x' has the name x', hence we could say dx'/d tau 
= v.

since Einstein used d tau/ d x', he also meant 1/v

But that is not allowed, because v=0 should be possible.

>
>>> When you perform differentiation, you consider multiple instances of the
>>> function values and evaluate the resulting difference quotient limit.
>> Sure, but how do you do this with zero?
>
> x'  is nonzero for the mirror.  Again, remember that you differentiate
> with respect to mirror positions.

No, the 'tandem' of emitter and mirror in k would move along the x-axis 
of K, hence could possibly pass the zero spot of that axis.

> [...]
>> That is, btw, the reason, why time must be the independent variable and
>> not the x-coordinate of the mirror.
>
> That's why Einstein makes that observation a bit later when he says that
> the light source need not be at the origin and the derivation would yield
> the same result. This is what decouples  x  from  t  and makes the result
> valid for any  x  and  t  combination (i.e., it makes them independent variables).


Actually that statement was wrong for other reasons.

quote

"It is to be noted that instead of the origin of the co-ordinates we 
might have chosen any other point for the point of origin of the ray, 
and the equation just obtained is therefore valid for all values of x, 
y, z."

This statement is wrong, because the requirement was, that the ray from 
the emitter hits the mirror and gets reflected back to its source.

This is only valid for certain relations between emitter and mirror. 
Especially useful would be, if the ray points towards the mirror, what 
cannot be guaranteed for arbitrary positions of the emitter.

Missing here was an appropriate statement about the position of the 
mirror relative to the emitted ray.


>>> In this case one considers instances of the experiment with different
>>> values of x'.
>> Sure. It is possible to chose other places for the mirror, as long as
>> the mirror remains at that position.
>>
>> But the position was not yet specified, anyhow, hence any possition is
>> possible, as long it remains fixed.
>
> Yes, that's all that's needed. The constraining equation is valid for any value
> of  x',  that's why we can differentiate wrt to  x'.
>
>>>> Seen from the emitter, the mirrors moves and we have the usual relation
>>>> v= x'/d tau.
>>>
>>> Again, a typo?
>> No. I meant a velocity of the mirror, which is measured with measures
>> from system k.
>
> But the  symbol  " x'/d tau "  is meaningless.


No.

dx/d t is usually understood as velocity.

since we are talking about relative motion with velocity v, such a term 
has actually a meaning.

It means:

infinitessimal change of the position of the mirror in respect to system 
K/ infinitessimal duration of that change

Now that duration can be measured with the time measure of K called t 
and with the time measure of k called tau.

In both case this would gain a quantity we can call velocity.

>> You had a different setting, where v=0,
>
> I never had  v = 0.

You wrote, that the mirror would be at rest in respect to the emitter, 
while I assumed, that v would be the velocity of the mirror in respect 
to system k.

But in fact your interpretation of the text was different then mine, 
hence you used v as velocity of system K in respect to system k.

But since nothing is happening in K, system K could be left away.

In this case v gets undefined, and you cannot use it anymore.

To use v as relative velocity between coordinates system is not 
possible, if one of the coordinate system is not used for anything.

The unused system can be left away entirely.


>> while I used the defintion of v
>> as velocity of system k in respect to system K, where v<>0.
>
> That's what I have.
>
>> This velocity is v = x'/d tau
>>>
>>>> Now Einstein used the inverse 1/v in his equation, what is blatant nonsense.
>>>
>>> There is no 1/v there.
>> Sure, because he wrote del tau /del x'
>> in the equations in the middle of page 6
>
> That's not  1/v.  1/v  would be  dt/dx.  You are mixing coordinates from
> different systems, too (this is not mathematically incorrect but
> this expression is not easily realted to  v,  not until we've calculated the
> Lorentz transformation coefficients).

What you want to do, is a coordinate transformation from k to system K.


If you want to transform observations from one system to another, the 
other system needs to be defined somehow.

That definition needs an 'anchor' and we only have the other system, 
which we could possibly use.

That would revert the situation and would require to regard the moving 
system k as stationary and base the position and velocity of K on system k.

Then we have a stationary system k and a moving system K.

But we don't need K, because in k emitter and mirror are stationary.


>> Since there is no 'del' in the position (because the position is fixed),
>> I took x'.
>
> You must put "del" there regardless. Of course then you'll get  v = 0
> which is incorrect (that alone should have alerted you to the fact you've
> made a mistake somewhere).


I have tried hard to find a solution, where all assumptions about 
Einstein's intentions fit and would not cause violations of known rules 
or Einstein's settings.

But unfortunately ALL interpretations I have found caused trouble somewhere.

E.g. this statement puzzled me and I have found no valid interpretation:

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


If I insert x'= x −vt into "...system of values x', y, z", I get

" ...system of values x−vt, y, z  "

which contains t, hence is NOT independent of time.


TH

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Re: Annotated version of SRT Thomas Heger <ttt_heg@web.de> - 2022-04-23 07:55 +0200
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        Re: Annotated version of SRT JanPB <filmart@gmail.com> - 2022-04-24 01:52 -0700
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                Re: Annotated version of SRT Thomas Heger <ttt_heg@web.de> - 2022-04-27 07:59 +0200
                Re: Annotated version of SRT Thomas Heger <ttt_heg@web.de> - 2022-04-28 07:14 +0200
                Re: Annotated version of SRT JanPB <filmart@gmail.com> - 2022-04-28 01:45 -0700
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                Re: Imbecile Thomas Heger perseveres Richard Hachel <r.hachel@tiscali.fr> - 2022-04-30 14:44 +0000
                Re: Imbecile Thomas Heger perseveres "Dono." <eggy20011951@gmail.com> - 2022-04-30 08:19 -0700
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                Re: Imbecile Thomas Heger perseveres Thomas Heger <ttt_heg@web.de> - 2022-04-30 20:15 +0200
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                Re: Annotated version of SRT JanPB <filmart@gmail.com> - 2022-05-02 00:15 -0700
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                Re: Annotated version of SRT Odd Bodkin <bodkinodd@gmail.com> - 2022-05-01 12:16 +0000
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                Re: Annotated version of SRT Thomas Heger <ttt_heg@web.de> - 2022-05-03 08:15 +0200
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      Re: Annotated version of SRT JanPB <filmart@gmail.com> - 2022-04-24 01:23 -0700
        Re: Annotated version of SRT Thomas Heger <ttt_heg@web.de> - 2022-04-24 11:08 +0200
          Re: Annotated version of SRT JanPB <filmart@gmail.com> - 2022-04-24 05:14 -0700
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