the reciprocity between two SR frame exchanging role:
K measures a clock at rest in K' as slower, and viceversa
K' measures a clock at rest in K as slower
SR in this cases, don't tell anything about which of the two clocks, the one at rest in K, or the one at rest in K', is running slower.

let's call k the clock at rest in K and k' the clock at rest in K'
comparing the rates of these two identical clocks, k and k', differently moving, they must enter in one of these 3 possibilities, rates must be:

k>k' or k<k' or k=k';

SR doesn't give any answer to this: you will never know which is slower, or which is faster or if they are running at same rate!

instead, an absolute comparison is possible that tells immediately which of the two clocks, k or k', is running faster:
if they emit light accordingly to their rates, any observer can visually looking at them, tell which one is running slower: the one that happears to him (or to a photocamera) redder.

I have already claimed that this is true in the case of an observer positioned transversely and far away from the two observed clocks,
now, I reaffirm that claim, extending it to any possible third observer, regardless his position or inertial speed

the claim in the new (hazardous) form is: given two identical light clocks, differently inertial moving, an observer, in any position, moving at any inertial speed, can tell by visually looking at them, which one (of the two k or k') is moving faster and therefore having its rate slower: the one that h appears (to him or to a camera) redder.

I have come to this deduction intuitively and using some graphs, so any
help from math expert is welcome: is there any of the possible position of the observer and/or its speed, that reverts the visual appearance, so that makes the slower(speedwise) clock appears redder??
my (azardous) answer/claim is: not!
aberration or doppler an any condition doesn't revert the absolute difference between the apparent colors of the light clocks, so the absolute visual comparison is always possible.
please, being the proposed comparison strictly visual, the use of frame is not needed
thanks in advance for any help
cheers
beda

On 6/16/2021 5:53 PM, beda pietanza wrote:
> the reciprocity between two SR frame exchanging role:
> K measures a clock at rest in K' as slower, and viceversa
> K' measures a clock at rest in K as slower
> SR in this cases, don't tell anything about which of the two clocks, the one at rest in K, or the one at rest in K', is running slower.
>
> let's call k the clock at rest in K and k' the clock at rest in K'
> comparing the rates of these two identical clocks, k and k', differently moving, they must enter in one of these 3 possibilities, rates must be:
>
> k>k' or k<k' or k=k';
>
> SR doesn't give any answer to this: you will never know which is slower, or which is faster or if they are running at same rate!

Nope. When K sees k' as slower than k, and K' sees k as slower than k',
they aren't comparing the same thing. So each seeing the other as
slower is not a contradiction.

In their own frames, k and k' tick at their usual rate, of course.

On 17-Jun-21 9:27 am, Michael Moroney wrote:
> On 6/16/2021 5:53 PM, beda pietanza wrote:
>> the reciprocity between two SR frame exchanging role:
>> K measures a clock at rest in K' as slower, and viceversa
>> K' measures a clock at rest in K as slower
>> SR in this cases, don't tell anything about which of the two clocks,
>> the one at rest in K, or the one at rest in K', is running slower.
>>
>> let's call k the clock at rest in K and k' the clock at rest in K'
>> comparing the rates of these two identical clocks,  k and k',
>> differently moving, they must enter in one of these 3 possibilities,
>> rates must be:
>>
>> k>k'  or   k<k'  or  k=k';
>>
>> SR doesn't give any answer to this: you will never know which is
>> slower, or which is faster or if they are running at same rate!
>
> Nope. When K sees k' as slower than k, and K' sees k as slower than k',
> they aren't comparing the same thing.  So each seeing the other as
> slower is not a contradiction.
>
> In their own frames, k and k' tick at their usual rate, of course.

And, of course, Beda imagines he's the first to spot a bleedingly
obvious contradiction, rather than supposing that he's simply
misunderstood something.

Sylvia.

Il giorno giovedì 17 giugno 2021 alle 01:27:54 UTC+2 Michael Moroney ha scritto:
> On 6/16/2021 5:53 PM, beda pietanza wrote:
> > the reciprocity between two SR frame exchanging role:
> > K measures a clock at rest in K' as slower, and viceversa
> > K' measures a clock at rest in K as slower
> > SR in this cases, don't tell anything about which of the two clocks, the one at rest in K, or the one at rest in K', is running slower.
> >
> > let's call k the clock at rest in K and k' the clock at rest in K'
> > comparing the rates of these two identical clocks, k and k', differently moving, they must enter in one of these 3 possibilities, rates must be:
> >
> > k>k' or k<k' or k=k';
> >
> > SR doesn't give any answer to this: you will never know which is slower, or which is faster or if they are running at same rate!
>
> Nope. When K sees k' as slower than k, and K' sees k as slower than k',
> they aren't comparing the same thing. So each seeing the other as
> slower is not a contradiction.
>
> In their own frames, k and k' tick at their usual rate, of course.
beda
you are completely wrong
SR doesn't compare a single clock in K to a single clock in K',
but is the SR frame K that measures the single clock in K'
a very different thing, you should give a better thought on this
by the way, I quickly deleted the post, how did you get to see it
in such a short time, I am not that sure of my claim in its hard form!
cheers
beda

Il giorno giovedì 17 giugno 2021 alle 03:20:12 UTC+2 Sylvia Else ha scritto:
> On 17-Jun-21 9:27 am, Michael Moroney wrote:
> > On 6/16/2021 5:53 PM, beda pietanza wrote:
> >> the reciprocity between two SR frame exchanging role:
> >> K measures a clock at rest in K' as slower, and viceversa
> >> K' measures a clock at rest in K as slower
> >> SR in this cases, don't tell anything about which of the two clocks,
> >> the one at rest in K, or the one at rest in K', is running slower.
> >>
> >> let's call k the clock at rest in K and k' the clock at rest in K'
> >> comparing the rates of these two identical clocks, k and k',
> >> differently moving, they must enter in one of these 3 possibilities,
> >> rates must be:
> >>
> >> k>k' or k<k' or k=k';
> >>
> >> SR doesn't give any answer to this: you will never know which is
> >> slower, or which is faster or if they are running at same rate!
> >
> > Nope. When K sees k' as slower than k, and K' sees k as slower than k',
> > they aren't comparing the same thing. So each seeing the other as
> > slower is not a contradiction.
> >
> > In their own frames, k and k' tick at their usual rate, of course.
> And, of course, Beda imagines he's the first to spot a bleedingly
> obvious contradiction, rather than supposing that he's simply
> misunderstood something.
>
> Sylvia.
well sylvia, I delete the post 5 minute later, because I am dubious about its content,
I am still working on it, in any case, what Michael M. claims is completely wrong:
SR doesn't compares a single clocl in K to a single clock in K', hope you are aware of this,
cgeers
beda

On 6/17/2021 7:01 AM, beda pietanza wrote:
> Il giorno giovedì 17 giugno 2021 alle 01:27:54 UTC+2 Michael Moroney ha scritto:
>> On 6/16/2021 5:53 PM, beda pietanza wrote:
>>> the reciprocity between two SR frame exchanging role:
>>> K measures a clock at rest in K' as slower, and viceversa
>>> K' measures a clock at rest in K as slower
>>> SR in this cases, don't tell anything about which of the two clocks, the one at rest in K, or the one at rest in K', is running slower.
>>>
>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>> comparing the rates of these two identical clocks, k and k', differently moving, they must enter in one of these 3 possibilities, rates must be:
>>>
>>> k>k' or k<k' or k=k';
>>>
>>> SR doesn't give any answer to this: you will never know which is slower, or which is faster or if they are running at same rate!
>>
>> Nope. When K sees k' as slower than k, and K' sees k as slower than k',
>> they aren't comparing the same thing. So each seeing the other as
>> slower is not a contradiction.
>>
>> In their own frames, k and k' tick at their usual rate, of course.
> beda
> you are completely wrong
> SR doesn't compare a single clock in K to a single clock in K',
> but is the SR frame K that measures the single clock in K'
> a very different thing, you should give a better thought on this

You didn't complete your idea. The observer in K compares a single
clock k to a series of clocks k1', k2', ... as they pass k, while the
observer in K' compares a single clock k' to a series of clocks in K,
k1, k2, ... which is why I said they aren't measuring the same thing.

> by the way, I quickly deleted the post, how did you get to see it
> in such a short time, I am not that sure of my claim in its hard form!

I just told someone else the following just yesterday:

You use Giggle Groups. Giggle Groups does not play well with Usenet
proper. Specifically, when you delete a post using Giggle Groups, you
delete it ONLY from Giggle Groups' Usenet interface. It is not deleted
from the rest of Usenet. Keep that in mind before posting something,
there really is no way to delete a post. In my case I saw your post
about 90 minutes after you posted it, and for me, your "deleted" post is
still there because I don't use Giggle Groups.
> cheers
> beda
>

Op 16-jun.-2021 om 23:53 schreef beda pietanza:
> the reciprocity between two SR frame exchanging role: K measures a
> clock at rest in K' as slower, and viceversa K' measures a clock at
> rest in K as slower SR in this cases, don't tell anything about which
> of the two clocks, the one at rest in K, or the one at rest in K', is
> running slower.
>
> let's call k the clock at rest in K and k' the clock at rest in K'
> comparing the rates of these two identical clocks, k and k',
> differently moving, they must enter in one of these 3 possibilities,
> rates must be:
>
> k>k' or k<k' or k=k';
>

Here we have a schoolbook example of lack of precision in
expression and in thought.
You did not properly or meaningfully describe the physical
meanings of k and k'.
You call:
" k the clock at rest in K "
and:
" k' the clock at rest in K' ",
and then you compare k and k' as if they are numbers:
" k > k' etc... "

Think about that. This is completely wrong.
A *clock* "greater than" another *clock*?

Here's how --in SR-- the physical meanings of things are
properly described.
We call:
" k the time as measured on a clock at rest in K,
between two events that happen at the same place
in K, for instance, AT THAT K-CLOCK "
and:
" k' the time as measured on a clock at rest in K',
between two events that happen at the same place
in K', for instance, AT THAT K'-CLOCK ",
and then we clearly see that it is completely useless
to even think about whether
" k > k' or k < k' or k = k' ".

Of course, if we also specify that the event pairs are
consecutive ticks on the respective clocks, and we
assume that the clocks are sound, then, for really good
and reliable clocks, the only possibility should trivially
be that
" k = k' ".
Otherwise the clocks are completely useless -- or at least
one of them.

But more importantly, we immediately see that the quantity k
pertains to a pair of specific events (at the same place
on a K-clock), where k' pertains to a an entirely
*different* pair of events (at the same place on a K'-clock).

Now, if we want to compare things and have equations
describe what we mean, we need more variables.

For instance, if we want to talk about SR-time dilation,
there are two ways:

(1) Let's use the the pair of events that happen at the same
place at the K-clock. We called k the time between the
events as measured on that clock. If the K'-clock measures
the time between THESE two events, they will find that,
as measured with the K'-clock, it takes a time
T' = gamma * k.
DO NOT CALL THAT TIME k', because that is a time pertaining
to a different pair of events -- see (2)!

(2) Let's use the the pair of events that happen at the same
place at the K'-clock. We Called k' the time between the
events as measured on that clock. If the K-clock measures
the time between THESE two events, they will find that,
as measured with the K-clock, it takes a time
T = gamma * k'.
DO NOT CALL THAT TIME k, because that is a time pertaining
to a different pair of events -- see (1)!

In the first case we have
T' / k = gamma.

In the second case we have
T / k' = gamma.
And *that* is what reciprocity of time dilation means in SR.

Hundreds, if not thousands, of people have made exactly the
same mistake. Even Dingle:
http://users.telenet.be/vdmoortel/dirk/Physics/Dingle/DinglesTrivialFumble.html

Dirk Vdm

Il giorno giovedì 17 giugno 2021 alle 17:37:58 UTC+2 Michael Moroney ha scritto:
> On 6/17/2021 7:01 AM, beda pietanza wrote:
> > Il giorno giovedì 17 giugno 2021 alle 01:27:54 UTC+2 Michael Moroney ha scritto:
> >> On 6/16/2021 5:53 PM, beda pietanza wrote:
> >>> the reciprocity between two SR frame exchanging role:
> >>> K measures a clock at rest in K' as slower, and viceversa
> >>> K' measures a clock at rest in K as slower
> >>> SR in this cases, don't tell anything about which of the two clocks, the one at rest in K, or the one at rest in K', is running slower.
> >>>
> >>> let's call k the clock at rest in K and k' the clock at rest in K'
> >>> comparing the rates of these two identical clocks, k and k', differently moving, they must enter in one of these 3 possibilities, rates must be:
> >>>
> >>> k>k' or k<k' or k=k';
> >>>
> >>> SR doesn't give any answer to this: you will never know which is slower, or which is faster or if they are running at same rate!
> >>
> >> Nope. When K sees k' as slower than k, and K' sees k as slower than k',
> >> they aren't comparing the same thing. So each seeing the other as
> >> slower is not a contradiction.
> >>
> >> In their own frames, k and k' tick at their usual rate, of course.
> > beda
> > you are completely wrong
> > SR doesn't compare a single clock in K to a single clock in K',
> > but is the SR frame K that measures the single clock in K'
> > a very different thing, you should give a better thought on this
> You didn't complete your idea. The observer in K compares a single
> clock k to a series of clocks k1', k2', ... as they pass k, while the
> observer in K' compares a single clock k' to a series of clocks in K,
> k1, k2, ... which is why I said they aren't measuring the same thing.
beda
ok, michael, but you could be more precise, the measured clock (k') in K'
is compared to the elapsed time of a row of clocks positioned along the x axis in K
whose elapsed time is the combine of the rate of the clocks(identical)plus the offset
of the clock made by the Esynchro.
or reverse the passing by row of clocks along the axis in k' is compared as you say to
the single clock(k) in K

the result is a fictious measurement valid only within the SR peculiar procedure,
nothing to do with the real rates of k vs k', which remains unknowable with the
SR procedure,
unless you use a SR frame coincident with the absolute preferred frame at rest versus
the CMBR, hope you agree that that frame is a particular frame belonging to the SR set,
at same time is identical to an absolute preferred frame with absolute synchronized clocks
this frame assign to any objects their real absolute speed (and, in case of clocks, assign them
their real time rate)
> > by the way, I quickly deleted the post, how did you get to see it
> > in such a short time, I am not that sure of my claim in its hard form!
> I just told someone else the following just yesterday:
>
> You use Giggle Groups. Giggle Groups does not play well with Usenet
> proper. Specifically, when you delete a post using Giggle Groups, you
> delete it ONLY from Giggle Groups' Usenet interface. It is not deleted
> from the rest of Usenet. Keep that in mind before posting something,
> there really is no way to delete a post. In my case I saw your post
> about 90 minutes after you posted it, and for me, your "deleted" post is
> still there because I don't use Giggle Groups.
beda
thanks, a good reason to be more careful,
just a little question, if you know better, the standard procedure is
the row of k' measured against a clock in K
or the row in K measuring a clock in k'
it should be the same, what is the correct one?

about the content of the erased post, since you read it,
did you give a thought to my claim(??) that a direct visual observation by a third
observer can tell the difference between two different inertial moving light clocks,
by detecting their difference in the color of their lights, regardless the speed
and/or position of the observer?

thanks for you help,
cheers
beda

> > cheers
> > beda
> >

Il giorno giovedì 17 giugno 2021 alle 21:17:53 UTC+2 Dirk Van de moortel ha scritto:
> Op 16-jun.-2021 om 23:53 schreef beda pietanza:
> > the reciprocity between two SR frame exchanging role: K measures a
> > clock at rest in K' as slower, and viceversa K' measures a clock at
> > rest in K as slower SR in this cases, don't tell anything about which
> > of the two clocks, the one at rest in K, or the one at rest in K', is
> > running slower.
> >
> > let's call k the clock at rest in K and k' the clock at rest in K'
> > comparing the rates of these two identical clocks, k and k',
> > differently moving, they must enter in one of these 3 possibilities,
> > rates must be:
> >
> > k>k' or k<k' or k=k';
> >
> Here we have a schoolbook example of lack of precision in
> expression and in thought.
> You did not properly or meaningfully describe the physical
> meanings of k and k'.
> You call:
> " k the clock at rest in K "
> and:
> " k' the clock at rest in K' ",
> and then you compare k and k' as if they are numbers:
> " k > k' etc... "
>
> Think about that. This is completely wrong.
> A *clock* "greater than" another *clock*?
beda
you are right, but in the premises I wrote:
> > let's call k the clock at rest in K and k' the clock at rest in K'
> > comparing the rates of these two identical clocks, k and k',
> > differently moving, they must enter in one of these 3 possibilities,
> > rates must be:
> >
> > k>k' or k<k' or k=k';
in any case, to avoid misinterpretation, your suggestion is correct
>
>
> Here's how --in SR-- the physical meanings of things are
> properly described.
> We call:
> " k the time as measured on a clock at rest in K,
> between two events that happen at the same place
> in K, for instance, AT THAT K-CLOCK "
> and:
> " k' the time as measured on a clock at rest in K',
> between two events that happen at the same place
> in K', for instance, AT THAT K'-CLOCK ",
> and then we clearly see that it is completely useless
> to even think about whether
> " k > k' or k < k' or k = k' ".
>
> Of course, if we also specify that the event pairs are
> consecutive ticks on the respective clocks, and we
> assume that the clocks are sound, then, for really good
> and reliable clocks, the only possibility should trivially
> be that
> " k = k' ".
> Otherwise the clocks are completely useless -- or at least
> one of them.
>
>
> But more importantly, we immediately see that the quantity k
> pertains to a pair of specific events (at the same place
> on a K-clock), where k' pertains to a an entirely
> *different* pair of events (at the same place on a K'-clock).
>
> Now, if we want to compare things and have equations
> describe what we mean, we need more variables.
>
> For instance, if we want to talk about SR-time dilation,
> there are two ways:
>
> (1) Let's use the the pair of events that happen at the same
> place at the K-clock. We called k the time between the
> events as measured on that clock. If the K'-clock measures
> the time between THESE two events, they will find that,
> as measured with the K'-clock, it takes a time
> T' = gamma * k.
> DO NOT CALL THAT TIME k', because that is a time pertaining
> to a different pair of events -- see (2)!
>
> (2) Let's use the the pair of events that happen at the same
> place at the K'-clock. We Called k' the time between the
> events as measured on that clock. If the K-clock measures
> the time between THESE two events, they will find that,
> as measured with the K-clock, it takes a time
> T = gamma * k'.
> DO NOT CALL THAT TIME k, because that is a time pertaining
> to a different pair of events -- see (1)!
>
> In the first case we have
> T' / k = gamma.
>
> In the second case we have
> T / k' = gamma.
> And *that* is what reciprocity of time dilation means in SR.
>
> Hundreds, if not thousands, of people have made exactly the
> same mistake. Even Dingle:
beda
ok, your detailed explanation is correct, but to my opinion it is incomplete:
the two events that takes place in K and measured in K by clock(k) having a duration of a unit of local time in K
the same two events that takes in K' and measured in K' by clock(k') having a duration of a unit of local time in K'
these two local times, though identical in their local value, are absolutely different, this difference is partially detectable by any observer depending on his speed,
only a preferred frame at rest in the CMBR can assign to those two events, happening is the two frames K and K', their correct and absolute duration values.

aside the pretense of invariance of all laws of physics, symmetry and reciprocity for all possible speeds, excessively hazardous and very unlikely to hold physically for such large domain of applicability,
SR is conceptually incomplete, if not inserted correctly, into an vision that gives to the absolute its proper important role, since each SR frame (E)synchro setting, is the result of a precise sewed work of the absolute speed of light combined with the absolute speed of the SR frame, with the (E)synchro the SR frames carries with them the exact value of its absolute speed stamped in its (E)synchro ""asynchrony"",
let me know, I have with pleasure read from you after so many years,
>
> http://users.telenet.be/vdmoortel/dirk/Physics/Dingle/DinglesTrivialFumble.html
thanks for the link, I will ponder the contend

cheers
beda

>
> Dirk Vdm

Op 17-jun.-2021 om 22:10 schreef beda pietanza:
> Il giorno giovedì 17 giugno 2021 alle 21:17:53 UTC+2 Dirk Van de moortel ha scritto:
>> Op 16-jun.-2021 om 23:53 schreef beda pietanza:
>>> the reciprocity between two SR frame exchanging role: K measures a
>>> clock at rest in K' as slower, and viceversa K' measures a clock at
>>> rest in K as slower SR in this cases, don't tell anything about which
>>> of the two clocks, the one at rest in K, or the one at rest in K', is
>>> running slower.
>>>
>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>> comparing the rates of these two identical clocks, k and k',
>>> differently moving, they must enter in one of these 3 possibilities,
>>> rates must be:
>>>
>>> k>k' or k<k' or k=k';
>>>
>> Here we have a schoolbook example of lack of precision in
>> expression and in thought.
>> You did not properly or meaningfully describe the physical
>> meanings of k and k'.
>> You call:
>> " k the clock at rest in K "
>> and:
>> " k' the clock at rest in K' ",
>> and then you compare k and k' as if they are numbers:
>> " k > k' etc... "
>>
>> Think about that. This is completely wrong.
>> A *clock* "greater than" another *clock*?
> beda
> you are right, but in the premises I wrote:
>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>> comparing the rates of these two identical clocks, k and k',
>>> differently moving, they must enter in one of these 3 possibilities,
>>> rates must be:
>>>
>>> k>k' or k<k' or k=k';
> in any case, to avoid misinterpretation, your suggestion is correct
>>
>>
>> Here's how --in SR-- the physical meanings of things are
>> properly described.
>> We call:
>> " k the time as measured on a clock at rest in K,
>> between two events that happen at the same place
>> in K, for instance, AT THAT K-CLOCK "
>> and:
>> " k' the time as measured on a clock at rest in K',
>> between two events that happen at the same place
>> in K', for instance, AT THAT K'-CLOCK ",
>> and then we clearly see that it is completely useless
>> to even think about whether
>> " k > k' or k < k' or k = k' ".
>>
>> Of course, if we also specify that the event pairs are
>> consecutive ticks on the respective clocks, and we
>> assume that the clocks are sound, then, for really good
>> and reliable clocks, the only possibility should trivially
>> be that
>> " k = k' ".
>> Otherwise the clocks are completely useless -- or at least
>> one of them.
>>
>>
>> But more importantly, we immediately see that the quantity k
>> pertains to a pair of specific events (at the same place
>> on a K-clock), where k' pertains to a an entirely
>> *different* pair of events (at the same place on a K'-clock).
>>
>> Now, if we want to compare things and have equations
>> describe what we mean, we need more variables.
>>
>> For instance, if we want to talk about SR-time dilation,
>> there are two ways:
>>
>> (1) Let's use the the pair of events that happen at the same
>> place at the K-clock. We called k the time between the
>> events as measured on that clock. If the K'-clock measures
>> the time between THESE two events, they will find that,
>> as measured with the K'-clock, it takes a time
>> T' = gamma * k.
>> DO NOT CALL THAT TIME k', because that is a time pertaining
>> to a different pair of events -- see (2)!
>>
>> (2) Let's use the the pair of events that happen at the same
>> place at the K'-clock. We Called k' the time between the
>> events as measured on that clock. If the K-clock measures
>> the time between THESE two events, they will find that,
>> as measured with the K-clock, it takes a time
>> T = gamma * k'.
>> DO NOT CALL THAT TIME k, because that is a time pertaining
>> to a different pair of events -- see (1)!
>>
>> In the first case we have
>> T' / k = gamma.
>>
>> In the second case we have
>> T / k' = gamma.
>> And *that* is what reciprocity of time dilation means in SR.
>>
>> Hundreds, if not thousands, of people have made exactly the
>> same mistake. Even Dingle:

> beda

> ok, your detailed explanation is correct, but to my opinion it is
> incomplete: the two events that takes place in K and measured in K by
> clock(k) having a duration of a unit of local time in K
> the same two events that takes in K' and measured in K' by clock(k')
> having a duration of a unit of local time in K'

T H A T I S I M P O S S I B L E !
The former pair occur at the same place in K.
The latter pair occur at the same place in K'.
K and K' are in relative motion.
The former two events CANNOT be "the same two events" as the latter two
events. There is NO WAY for this to happen.
You are making exactly the same mistake -- again.

Dirk Vdm

On 17-Jun-21 9:08 pm, beda pietanza wrote:
> Il giorno giovedì 17 giugno 2021 alle 03:20:12 UTC+2 Sylvia Else ha scritto:
>> On 17-Jun-21 9:27 am, Michael Moroney wrote:
>>> On 6/16/2021 5:53 PM, beda pietanza wrote:
>>>> the reciprocity between two SR frame exchanging role:
>>>> K measures a clock at rest in K' as slower, and viceversa
>>>> K' measures a clock at rest in K as slower
>>>> SR in this cases, don't tell anything about which of the two clocks,
>>>> the one at rest in K, or the one at rest in K', is running slower.
>>>>
>>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>>> comparing the rates of these two identical clocks, k and k',
>>>> differently moving, they must enter in one of these 3 possibilities,
>>>> rates must be:
>>>>
>>>> k>k' or k<k' or k=k';
>>>>
>>>> SR doesn't give any answer to this: you will never know which is
>>>> slower, or which is faster or if they are running at same rate!
>>>
>>> Nope. When K sees k' as slower than k, and K' sees k as slower than k',
>>> they aren't comparing the same thing. So each seeing the other as
>>> slower is not a contradiction.
>>>
>>> In their own frames, k and k' tick at their usual rate, of course.
>> And, of course, Beda imagines he's the first to spot a bleedingly
>> obvious contradiction, rather than supposing that he's simply
>> misunderstood something.
>>
>> Sylvia.
> well sylvia, I delete the post 5 minute later, because I am dubious about its content,
> I am still working on it, in any case, what Michael M. claims is completely wrong:
> SR doesn't compares a single clocl in K to a single clock in K', hope you are aware of this,
> cgeers
> beda

This is not Facebook or YouTube. Usenet is a distributed system. Many
Usenet servers will not honour a delete. Once you post on Usenet, it's
out there.

Sylvia.

Il giorno giovedì 17 giugno 2021 alle 22:59:35 UTC+2 Dirk Van de moortel ha scritto:
> Op 17-jun.-2021 om 22:10 schreef beda pietanza:
> > Il giorno giovedì 17 giugno 2021 alle 21:17:53 UTC+2 Dirk Van de moortel ha scritto:
> >> Op 16-jun.-2021 om 23:53 schreef beda pietanza:
> >>> the reciprocity between two SR frame exchanging role: K measures a
> >>> clock at rest in K' as slower, and viceversa K' measures a clock at
> >>> rest in K as slower SR in this cases, don't tell anything about which
> >>> of the two clocks, the one at rest in K, or the one at rest in K', is
> >>> running slower.
> >>>
> >>> let's call k the clock at rest in K and k' the clock at rest in K'
> >>> comparing the rates of these two identical clocks, k and k',
> >>> differently moving, they must enter in one of these 3 possibilities,
> >>> rates must be:
> >>>
> >>> k>k' or k<k' or k=k';
> >>>
> >> Here we have a schoolbook example of lack of precision in
> >> expression and in thought.
> >> You did not properly or meaningfully describe the physical
> >> meanings of k and k'.
> >> You call:
> >> " k the clock at rest in K "
> >> and:
> >> " k' the clock at rest in K' ",
> >> and then you compare k and k' as if they are numbers:
> >> " k > k' etc... "
> >>
> >> Think about that. This is completely wrong.
> >> A *clock* "greater than" another *clock*?
> > beda
> > you are right, but in the premises I wrote:
> >>> let's call k the clock at rest in K and k' the clock at rest in K'
> >>> comparing the rates of these two identical clocks, k and k',
> >>> differently moving, they must enter in one of these 3 possibilities,
> >>> rates must be:
> >>>
> >>> k>k' or k<k' or k=k';
> > in any case, to avoid misinterpretation, your suggestion is correct
> >>
> >>
> >> Here's how --in SR-- the physical meanings of things are
> >> properly described.
> >> We call:
> >> " k the time as measured on a clock at rest in K,
> >> between two events that happen at the same place
> >> in K, for instance, AT THAT K-CLOCK "
> >> and:
> >> " k' the time as measured on a clock at rest in K',
> >> between two events that happen at the same place
> >> in K', for instance, AT THAT K'-CLOCK ",
> >> and then we clearly see that it is completely useless
> >> to even think about whether
> >> " k > k' or k < k' or k = k' ".
> >>
> >> Of course, if we also specify that the event pairs are
> >> consecutive ticks on the respective clocks, and we
> >> assume that the clocks are sound, then, for really good
> >> and reliable clocks, the only possibility should trivially
> >> be that
> >> " k = k' ".
> >> Otherwise the clocks are completely useless -- or at least
> >> one of them.
> >>
> >>
> >> But more importantly, we immediately see that the quantity k
> >> pertains to a pair of specific events (at the same place
> >> on a K-clock), where k' pertains to a an entirely
> >> *different* pair of events (at the same place on a K'-clock).
> >>
> >> Now, if we want to compare things and have equations
> >> describe what we mean, we need more variables.
> >>
> >> For instance, if we want to talk about SR-time dilation,
> >> there are two ways:
> >>
> >> (1) Let's use the the pair of events that happen at the same
> >> place at the K-clock. We called k the time between the
> >> events as measured on that clock. If the K'-clock measures
> >> the time between THESE two events, they will find that,
> >> as measured with the K'-clock, it takes a time
> >> T' = gamma * k.
> >> DO NOT CALL THAT TIME k', because that is a time pertaining
> >> to a different pair of events -- see (2)!
> >>
> >> (2) Let's use the the pair of events that happen at the same
> >> place at the K'-clock. We Called k' the time between the
> >> events as measured on that clock. If the K-clock measures
> >> the time between THESE two events, they will find that,
> >> as measured with the K-clock, it takes a time
> >> T = gamma * k'.
> >> DO NOT CALL THAT TIME k, because that is a time pertaining
> >> to a different pair of events -- see (1)!
> >>
> >> In the first case we have
> >> T' / k = gamma.
> >>
> >> In the second case we have
> >> T / k' = gamma.
> >> And *that* is what reciprocity of time dilation means in SR.
> >>
> >> Hundreds, if not thousands, of people have made exactly the
> >> same mistake. Even Dingle:
>
> > beda
>
> > ok, your detailed explanation is correct, but to my opinion it is
> > incomplete: the two events that takes place in K and measured in K by
> > clock(k) having a duration of a unit of local time in K
> > the same two events that takes in K' and measured in K' by clock(k')
> > having a duration of a unit of local time in K'
> T H A T I S I M P O S S I B L E !
> The former pair occur at the same place in K.
> The latter pair occur at the same place in K'.
> K and K' are in relative motion.
> The former two events CANNOT be "the same two events" as the latter two
> events. There is NO WAY for this to happen.
> You are making exactly the same mistake -- again.
>
> Dirk Vdm
ok, I reformulate the concept,
the complete two way cycle of a local clock
in two different SR frames, takes the same duration of local time, but they correspond to
different absolute time seen from a preferred frame at rest versus the CMBR..
cheers
beda

Op 18-jun.-2021 om 12:32 schreef beda pietanza:
> Il giorno giovedì 17 giugno 2021 alle 22:59:35 UTC+2 Dirk Van de moortel ha scritto:
>> Op 17-jun.-2021 om 22:10 schreef beda pietanza:
>>> Il giorno giovedì 17 giugno 2021 alle 21:17:53 UTC+2 Dirk Van de moortel ha scritto:
>>>> Op 16-jun.-2021 om 23:53 schreef beda pietanza:
>>>>> the reciprocity between two SR frame exchanging role: K measures a
>>>>> clock at rest in K' as slower, and viceversa K' measures a clock at
>>>>> rest in K as slower SR in this cases, don't tell anything about which
>>>>> of the two clocks, the one at rest in K, or the one at rest in K', is
>>>>> running slower.
>>>>>
>>>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>>>> comparing the rates of these two identical clocks, k and k',
>>>>> differently moving, they must enter in one of these 3 possibilities,
>>>>> rates must be:
>>>>>
>>>>> k>k' or k<k' or k=k';
>>>>>
>>>> Here we have a schoolbook example of lack of precision in
>>>> expression and in thought.
>>>> You did not properly or meaningfully describe the physical
>>>> meanings of k and k'.
>>>> You call:
>>>> " k the clock at rest in K "
>>>> and:
>>>> " k' the clock at rest in K' ",
>>>> and then you compare k and k' as if they are numbers:
>>>> " k > k' etc... "
>>>>
>>>> Think about that. This is completely wrong.
>>>> A *clock* "greater than" another *clock*?
>>> beda
>>> you are right, but in the premises I wrote:
>>>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>>>> comparing the rates of these two identical clocks, k and k',
>>>>> differently moving, they must enter in one of these 3 possibilities,
>>>>> rates must be:
>>>>>
>>>>> k>k' or k<k' or k=k';
>>> in any case, to avoid misinterpretation, your suggestion is correct
>>>>
>>>>
>>>> Here's how --in SR-- the physical meanings of things are
>>>> properly described.
>>>> We call:
>>>> " k the time as measured on a clock at rest in K,
>>>> between two events that happen at the same place
>>>> in K, for instance, AT THAT K-CLOCK "
>>>> and:
>>>> " k' the time as measured on a clock at rest in K',
>>>> between two events that happen at the same place
>>>> in K', for instance, AT THAT K'-CLOCK ",
>>>> and then we clearly see that it is completely useless
>>>> to even think about whether
>>>> " k > k' or k < k' or k = k' ".
>>>>
>>>> Of course, if we also specify that the event pairs are
>>>> consecutive ticks on the respective clocks, and we
>>>> assume that the clocks are sound, then, for really good
>>>> and reliable clocks, the only possibility should trivially
>>>> be that
>>>> " k = k' ".
>>>> Otherwise the clocks are completely useless -- or at least
>>>> one of them.
>>>>
>>>>
>>>> But more importantly, we immediately see that the quantity k
>>>> pertains to a pair of specific events (at the same place
>>>> on a K-clock), where k' pertains to a an entirely
>>>> *different* pair of events (at the same place on a K'-clock).
>>>>
>>>> Now, if we want to compare things and have equations
>>>> describe what we mean, we need more variables.
>>>>
>>>> For instance, if we want to talk about SR-time dilation,
>>>> there are two ways:
>>>>
>>>> (1) Let's use the the pair of events that happen at the same
>>>> place at the K-clock. We called k the time between the
>>>> events as measured on that clock. If the K'-clock measures
>>>> the time between THESE two events, they will find that,
>>>> as measured with the K'-clock, it takes a time
>>>> T' = gamma * k.
>>>> DO NOT CALL THAT TIME k', because that is a time pertaining
>>>> to a different pair of events -- see (2)!
>>>>
>>>> (2) Let's use the the pair of events that happen at the same
>>>> place at the K'-clock. We Called k' the time between the
>>>> events as measured on that clock. If the K-clock measures
>>>> the time between THESE two events, they will find that,
>>>> as measured with the K-clock, it takes a time
>>>> T = gamma * k'.
>>>> DO NOT CALL THAT TIME k, because that is a time pertaining
>>>> to a different pair of events -- see (1)!
>>>>
>>>> In the first case we have
>>>> T' / k = gamma.
>>>>
>>>> In the second case we have
>>>> T / k' = gamma.
>>>> And *that* is what reciprocity of time dilation means in SR.
>>>>
>>>> Hundreds, if not thousands, of people have made exactly the
>>>> same mistake. Even Dingle:
>>
>>> beda
>>
>>> ok, your detailed explanation is correct, but to my opinion it is
>>> incomplete: the two events that takes place in K and measured in K by
>>> clock(k) having a duration of a unit of local time in K
>>> the same two events that takes in K' and measured in K' by clock(k')
>>> having a duration of a unit of local time in K'
>> T H A T I S I M P O S S I B L E !
>> The former pair occur at the same place in K.
>> The latter pair occur at the same place in K'.
>> K and K' are in relative motion.
>> The former two events CANNOT be "the same two events" as the latter two
>> events. There is NO WAY for this to happen.
>> You are making exactly the same mistake -- again.
>>
>> Dirk Vdm

> ok, I reformulate the concept,
> the complete two way cycle of a local clock
> in two different SR frames,

T H A T I S N O N S E N S E !
(1) A "two way cycle of a clock" is gibberish.
(2) A clock cannot be local in two different frames.

Dirk Vdm

Dirk Van de moortel <dirkvandemoortel@notmail.com> wrote:
> Op 18-jun.-2021 om 12:32 schreef beda pietanza:
>> Il giorno giovedì 17 giugno 2021 alle 22:59:35 UTC+2 Dirk Van de moortel ha scritto:
>>> Op 17-jun.-2021 om 22:10 schreef beda pietanza:
>>>> Il giorno giovedì 17 giugno 2021 alle 21:17:53 UTC+2 Dirk Van de moortel ha scritto:
>>>>> Op 16-jun.-2021 om 23:53 schreef beda pietanza:
>>>>>> the reciprocity between two SR frame exchanging role: K measures a
>>>>>> clock at rest in K' as slower, and viceversa K' measures a clock at
>>>>>> rest in K as slower SR in this cases, don't tell anything about which
>>>>>> of the two clocks, the one at rest in K, or the one at rest in K', is
>>>>>> running slower.
>>>>>>
>>>>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>>>>> comparing the rates of these two identical clocks, k and k',
>>>>>> differently moving, they must enter in one of these 3 possibilities,
>>>>>> rates must be:
>>>>>>
>>>>>>> k' or k<k' or k=k';
>>>>>>
>>>>> Here we have a schoolbook example of lack of precision in
>>>>> expression and in thought.
>>>>> You did not properly or meaningfully describe the physical
>>>>> meanings of k and k'.
>>>>> You call:
>>>>> " k the clock at rest in K "
>>>>> and:
>>>>> " k' the clock at rest in K' ",
>>>>> and then you compare k and k' as if they are numbers:
>>>>> " k > k' etc... "
>>>>>
>>>>> Think about that. This is completely wrong.
>>>>> A *clock* "greater than" another *clock*?
>>>> beda
>>>> you are right, but in the premises I wrote:
>>>>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>>>>> comparing the rates of these two identical clocks, k and k',
>>>>>> differently moving, they must enter in one of these 3 possibilities,
>>>>>> rates must be:
>>>>>>
>>>>>>> k' or k<k' or k=k';
>>>> in any case, to avoid misinterpretation, your suggestion is correct
>>>>>
>>>>>
>>>>> Here's how --in SR-- the physical meanings of things are
>>>>> properly described.
>>>>> We call:
>>>>> " k the time as measured on a clock at rest in K,
>>>>> between two events that happen at the same place
>>>>> in K, for instance, AT THAT K-CLOCK "
>>>>> and:
>>>>> " k' the time as measured on a clock at rest in K',
>>>>> between two events that happen at the same place
>>>>> in K', for instance, AT THAT K'-CLOCK ",
>>>>> and then we clearly see that it is completely useless
>>>>> to even think about whether
>>>>> " k > k' or k < k' or k = k' ".
>>>>>
>>>>> Of course, if we also specify that the event pairs are
>>>>> consecutive ticks on the respective clocks, and we
>>>>> assume that the clocks are sound, then, for really good
>>>>> and reliable clocks, the only possibility should trivially
>>>>> be that
>>>>> " k = k' ".
>>>>> Otherwise the clocks are completely useless -- or at least
>>>>> one of them.
>>>>>
>>>>>
>>>>> But more importantly, we immediately see that the quantity k
>>>>> pertains to a pair of specific events (at the same place
>>>>> on a K-clock), where k' pertains to a an entirely
>>>>> *different* pair of events (at the same place on a K'-clock).
>>>>>
>>>>> Now, if we want to compare things and have equations
>>>>> describe what we mean, we need more variables.
>>>>>
>>>>> For instance, if we want to talk about SR-time dilation,
>>>>> there are two ways:
>>>>>
>>>>> (1) Let's use the the pair of events that happen at the same
>>>>> place at the K-clock. We called k the time between the
>>>>> events as measured on that clock. If the K'-clock measures
>>>>> the time between THESE two events, they will find that,
>>>>> as measured with the K'-clock, it takes a time
>>>>> T' = gamma * k.
>>>>> DO NOT CALL THAT TIME k', because that is a time pertaining
>>>>> to a different pair of events -- see (2)!
>>>>>
>>>>> (2) Let's use the the pair of events that happen at the same
>>>>> place at the K'-clock. We Called k' the time between the
>>>>> events as measured on that clock. If the K-clock measures
>>>>> the time between THESE two events, they will find that,
>>>>> as measured with the K-clock, it takes a time
>>>>> T = gamma * k'.
>>>>> DO NOT CALL THAT TIME k, because that is a time pertaining
>>>>> to a different pair of events -- see (1)!
>>>>>
>>>>> In the first case we have
>>>>> T' / k = gamma.
>>>>>
>>>>> In the second case we have
>>>>> T / k' = gamma.
>>>>> And *that* is what reciprocity of time dilation means in SR.
>>>>>
>>>>> Hundreds, if not thousands, of people have made exactly the
>>>>> same mistake. Even Dingle:
>>>
>>>> beda
>>>
>>>> ok, your detailed explanation is correct, but to my opinion it is
>>>> incomplete: the two events that takes place in K and measured in K by
>>>> clock(k) having a duration of a unit of local time in K
>>>> the same two events that takes in K' and measured in K' by clock(k')
>>>> having a duration of a unit of local time in K'
>>> T H A T I S I M P O S S I B L E !
>>> The former pair occur at the same place in K.
>>> The latter pair occur at the same place in K'.
>>> K and K' are in relative motion.
>>> The former two events CANNOT be "the same two events" as the latter two
>>> events. There is NO WAY for this to happen.
>>> You are making exactly the same mistake -- again.
>>>
>>> Dirk Vdm
>
>> ok, I reformulate the concept,
>> the complete two way cycle of a local clock
>> in two different SR frames,
>
> T H A T I S N O N S E N S E !
> (1) A "two way cycle of a clock" is gibberish.
> (2) A clock cannot be local in two different frames.
>
> Dirk Vdm
>

I have commented several times to Beda that it is nearly impossible to
maintain a conversation using terms, especially physics jargon terms, that
one person knows the meaning of and that another does not.

Beda’s stance is that these are all simple questions that should be
explainable to someone with no background in the subject, though he utters
this kind of sloppy jabberwocky all the time. He has no interest in
learning any physics, and he regards all complaints that he doesn’t know
enough to even discuss it intelligently as “insubstantial”.

I’ve given up trying to talk in his terms, because only HE knows what he
means by them, and they bear no resemblance to what the same or similar
words mean in physics.

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

Op 18-jun.-2021 om 20:27 schreef Odd Bodkin:
> Dirk Van de moortel <dirkvandemoortel@notmail.com> wrote:
>> Op 18-jun.-2021 om 12:32 schreef beda pietanza:
>>> Il giorno giovedì 17 giugno 2021 alle 22:59:35 UTC+2 Dirk Van de moortel ha scritto:
>>>> Op 17-jun.-2021 om 22:10 schreef beda pietanza:
>>>>> Il giorno giovedì 17 giugno 2021 alle 21:17:53 UTC+2 Dirk Van de moortel ha scritto:
>>>>>> Op 16-jun.-2021 om 23:53 schreef beda pietanza:
>>>>>>> the reciprocity between two SR frame exchanging role: K measures a
>>>>>>> clock at rest in K' as slower, and viceversa K' measures a clock at
>>>>>>> rest in K as slower SR in this cases, don't tell anything about which
>>>>>>> of the two clocks, the one at rest in K, or the one at rest in K', is
>>>>>>> running slower.
>>>>>>>
>>>>>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>>>>>> comparing the rates of these two identical clocks, k and k',
>>>>>>> differently moving, they must enter in one of these 3 possibilities,
>>>>>>> rates must be:
>>>>>>>
>>>>>>>> k' or k<k' or k=k';
>>>>>>>
>>>>>> Here we have a schoolbook example of lack of precision in
>>>>>> expression and in thought.
>>>>>> You did not properly or meaningfully describe the physical
>>>>>> meanings of k and k'.
>>>>>> You call:
>>>>>> " k the clock at rest in K"
>>>>>> and:
>>>>>> " k' the clock at rest in K' ",
>>>>>> and then you compare k and k' as if they are numbers:
>>>>>> " k > k' etc..."
>>>>>>
>>>>>> Think about that. This is completely wrong.
>>>>>> A *clock* "greater than" another *clock*?
>>>>> beda
>>>>> you are right, but in the premises I wrote:
>>>>>>> let's call k the clock at rest in K and k' the clock at rest in K'
>>>>>>> comparing the rates of these two identical clocks, k and k',
>>>>>>> differently moving, they must enter in one of these 3 possibilities,
>>>>>>> rates must be:
>>>>>>>
>>>>>>>> k' or k<k' or k=k';
>>>>> in any case, to avoid misinterpretation, your suggestion is correct
>>>>>>
>>>>>>
>>>>>> Here's how --in SR-- the physical meanings of things are
>>>>>> properly described.
>>>>>> We call:
>>>>>> " k the time as measured on a clock at rest in K,
>>>>>> between two events that happen at the same place
>>>>>> in K, for instance, AT THAT K-CLOCK "
>>>>>> and:
>>>>>> " k' the time as measured on a clock at rest in K',
>>>>>> between two events that happen at the same place
>>>>>> in K', for instance, AT THAT K'-CLOCK ",
>>>>>> and then we clearly see that it is completely useless
>>>>>> to even think about whether
>>>>>> " k > k' or k < k' or k = k' ".
>>>>>>
>>>>>> Of course, if we also specify that the event pairs are
>>>>>> consecutive ticks on the respective clocks, and we
>>>>>> assume that the clocks are sound, then, for really good
>>>>>> and reliable clocks, the only possibility should trivially
>>>>>> be that
>>>>>> " k = k' ".
>>>>>> Otherwise the clocks are completely useless -- or at least
>>>>>> one of them.
>>>>>>
>>>>>>
>>>>>> But more importantly, we immediately see that the quantity k
>>>>>> pertains to a pair of specific events (at the same place
>>>>>> on a K-clock), where k' pertains to a an entirely
>>>>>> *different* pair of events (at the same place on a K'-clock).
>>>>>>
>>>>>> Now, if we want to compare things and have equations
>>>>>> describe what we mean, we need more variables.
>>>>>>
>>>>>> For instance, if we want to talk about SR-time dilation,
>>>>>> there are two ways:
>>>>>>
>>>>>> (1) Let's use the the pair of events that happen at the same
>>>>>> place at the K-clock. We called k the time between the
>>>>>> events as measured on that clock. If the K'-clock measures
>>>>>> the time between THESE two events, they will find that,
>>>>>> as measured with the K'-clock, it takes a time
>>>>>> T' = gamma * k.
>>>>>> DO NOT CALL THAT TIME k', because that is a time pertaining
>>>>>> to a different pair of events -- see (2)!
>>>>>>
>>>>>> (2) Let's use the the pair of events that happen at the same
>>>>>> place at the K'-clock. We Called k' the time between the
>>>>>> events as measured on that clock. If the K-clock measures
>>>>>> the time between THESE two events, they will find that,
>>>>>> as measured with the K-clock, it takes a time
>>>>>> T = gamma * k'.
>>>>>> DO NOT CALL THAT TIME k, because that is a time pertaining
>>>>>> to a different pair of events -- see (1)!
>>>>>>
>>>>>> In the first case we have
>>>>>> T' / k = gamma.
>>>>>>
>>>>>> In the second case we have
>>>>>> T / k' = gamma.
>>>>>> And *that* is what reciprocity of time dilation means in SR.
>>>>>>
>>>>>> Hundreds, if not thousands, of people have made exactly the
>>>>>> same mistake. Even Dingle:
>>>>
>>>>> beda
>>>>
>>>>> ok, your detailed explanation is correct, but to my opinion it is
>>>>> incomplete: the two events that takes place in K and measured in K by
>>>>> clock(k) having a duration of a unit of local time in K
>>>>> the same two events that takes in K' and measured in K' by clock(k')
>>>>> having a duration of a unit of local time in K'
>>>> T H A T I S I M P O S S I B L E !
>>>> The former pair occur at the same place in K.
>>>> The latter pair occur at the same place in K'.
>>>> K and K' are in relative motion.
>>>> The former two events CANNOT be "the same two events" as the latter two
>>>> events. There is NO WAY for this to happen.
>>>> You are making exactly the same mistake -- again.
>>>>
>>>> Dirk Vdm
>>
>>> ok, I reformulate the concept,
>>> the complete two way cycle of a local clock
>>> in two different SR frames,
>>
>> T H A T I S N O N S E N S E !
>> (1) A "two way cycle of a clock" is gibberish.
>> (2) A clock cannot be local in two different frames.
>>
>> Dirk Vdm
>>
>
>
> I have commented several times to Beda that it is nearly impossible to
> maintain a conversation using terms, especially physics jargon terms, that
> one person knows the meaning of and that another does not.
>
> Beda’s stance is that these are all simple questions that should be
> explainable to someone with no background in the subject, though he utters
> this kind of sloppy jabberwocky all the time. He has no interest in
> learning any physics, and he regards all complaints that he doesn’t know
> enough to even discuss it intelligently as “insubstantial”.
>
> I’ve given up trying to talk in his terms, because only HE knows what he
> means by them, and they bear no resemblance to what the same or similar
> words mean in physics.

Indeed.
Meaningful, effective, two-way communication seems impossible.
With pun and all.

Dirk Vdm

Am 17.06.2021 um 21:17 schrieb Dirk Van de moortel:
> Op 16-jun.-2021 om 23:53 schreef beda pietanza:
>> the reciprocity between two SR frame exchanging role: K measures a
>> clock at rest in K' as slower, and viceversa K' measures a clock at
>> rest in K as slower SR in this cases, don't tell anything about which
>> of the two clocks, the one at rest in K, or the one at rest in K', is
>> running slower.
>>
>> let's call k the clock at rest in K and k' the clock at rest in K'
>> comparing the rates of these two identical clocks, k and k',
>> differently moving, they must enter in one of these 3 possibilities,
>> rates must be:
>>
>> k>k' or k<k' or k=k';
>>
>
>
> Here we have a schoolbook example of lack of precision in
> expression and in thought.
> You did not properly or meaningfully describe the physical
> meanings of k and k'.
> You call:
> " k the clock at rest in K "
> and:
> " k' the clock at rest in K' ",
> and then you compare k and k' as if they are numbers:
> " k > k' etc... "
>
> Think about that. This is completely wrong.
> A *clock* "greater than" another *clock*?
>
>
> Here's how --in SR-- the physical meanings of things are
> properly described.
> We call:
> " k the time as measured on a clock at rest in K,
> between two events that happen at the same place
> in K, for instance, AT THAT K-CLOCK "
> and:
> " k' the time as measured on a clock at rest in K',
> between two events that happen at the same place
> in K', for instance, AT THAT K'-CLOCK ",
> and then we clearly see that it is completely useless
> to even think about whether
> " k > k' or k < k' or k = k' ".

Actually the letters 'K' and 'k' denote coordinate systems in Einstein's
'On the electrodynamics of moving bodies'

Quote:
"Now to the origin of one of the two systems (k) let a constant velocity
v be imparted in the direction of the increasing x of the other
stationary system(K), and ....'

The time measures in the coordinate systems had the names 'A-time' and
'B-time'.

The readings of a clock was called 'time' and symbolized by 't' in
coordinates system K and 'tau' in coordinate system k.

Therefore, a statement like k>k' did not make sense in Einstein's text.

The sign ' (prime) was used for inverted motion.

....

TH

Thomas Heger wrote:

> Therefore, a statement like k>k' did not make sense in Einstein's text.
> The sign ' (prime) was used for inverted motion.

not in this context. It's obvious that k' is the other k. The other k
being the general k', not necessarily a specific.

This is Microsoft telling you how they will use the 5G grid along with
the Covid19 jab to kill you
https://seed128.bitchute.com/gIudfF6VnKd0/x4xAYGJu5v56.mp4
biological computer (timers embedded) inside your living cells.

and

George Bush slips up on what he knows about 9/11
https://www.bitchute.com/video/QqlVbPLCb6hA/
controlled demolition explosives installed months in advance.

How would you possibly justify that, scientifically?

Am 19.06.2021 um 13:09 schrieb Hugh Jazz:
> Thomas Heger wrote:
>
>> Therefore, a statement like k>k' did not make sense in Einstein's text.
>> The sign ' (prime) was used for inverted motion.
>
> not in this context. It's obvious that k' is the other k. The other k
> being the general k', not necessarily a specific.
>
Einstein used four coordinate systems: K, k, K' and k'.

1) K was called stationary
2) k was moving with velocity v along the x-axis of K
3) K' was moving with velocity -v along the xsi-axis of k
4) k' shall express a coordinate system, which moves with velocity w
along the xsi-axis of k (parallel to k, but with other velocity w)

quote
"3. If in addition to the systems K and k figuring in § 3 we introduce
still another system of co-ordinates k' moving parallel to k, its
initial point moving on the axis of Ξ with the velocity w, ..."

TH