Groups > sci.physics.relativity > #389108 > unrolled thread

Slow clock transport and the measurement of one way light speed.

Started by	Larry Harson <johnmcandrew66@gmail.com>
First post	2016-08-03 11:07 -0700
Last post	2016-08-14 13:05 -0700
Articles	17 on this page of 97 — 18 participants

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#389889

From	"HGWilson, DSc." <hgw@....>
Date	2016-08-13 08:58 +1000
Message-ID	<nolkav$13rm$2@gioia.aioe.org>
In reply to	#389880

On 13/08/16 07:11, rotchm wrote:
> On Friday, August 12, 2016 at 3:52:37 PM UTC-4, The Starmaker wrote:
>
>> Okay, I understand it does not show the same values...
>>
>> but what are the values, what are the numbers...
>
> Found by correctly applying t'=(t-xv/c^2)g

...so which one is ahead?
It looks as though they are both ahead of each other!!!
HAHAHHHAHAHHAHHHAHAA!

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#389892

From	rotchm <rotchm@gmail.com>
Date	2016-08-12 16:35 -0700
Message-ID	<57be8e5c-30ec-4945-b552-8099178ae2b0@googlegroups.com>
In reply to	#389889

On Friday, August 12, 2016 at 6:58:43 PM UTC-4, HGWilson, DSc. wrote:
> On 13/08/16 07:11, rotchm wrote:

> 
> ...so which one is ahead?
> It looks as though they are both ahead of each other!!!
> HAHAHHHAHAHHAHHHAHAA!

If course idiot wilson thinks that t & t' are exactly the same symbol; he cant notice the difference. This shows that idiot wilson does not have ahead.

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#389894

From	"HGWilson, DSc." <hgw@....>
Date	2016-08-13 09:58 +1000
Message-ID	<nolnqg$18ee$1@gioia.aioe.org>
In reply to	#389892

On 13/08/16 09:35, rotchm wrote:
> On Friday, August 12, 2016 at 6:58:43 PM UTC-4, HGWilson, DSc.
> wrote:
>> On 13/08/16 07:11, rotchm wrote:
>
>>
>> ...so which one is ahead? It looks as though they are both ahead of
>> each other!!! HAHAHHHAHAHHAHHHAHAA!
>
> If course idiot wilson thinks that t & t' are exactly the same
> symbol; he cant notice the difference. This shows that idiot wilson
> does not have ahead.
>

HAHAHHHAHHA! THE IDIOT Rotchm cannot see that his equation applies 
equally to both! HAHAHHHAHHAHHA!

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#389896

From	rotchm <rotchm@gmail.com>
Date	2016-08-12 17:15 -0700
Message-ID	<5851b4d7-bab7-4c01-9f1b-d31deda1680b@googlegroups.com>
In reply to	#389894

On Friday, August 12, 2016 at 7:58:12 PM UTC-4, HGWilson, DSc. wrote:

> HAHAHHHAHHA! THE IDIOT Rotchm cannot see that his equation applies 
> equally to both! HAHAHHHAHHAHHA!


Didnt you understand when I said 
"Found by correctly applying t'=(t-xv/c^2)g "

So you are admitting that you dont know how to CORRECTLY apply the LT's ?

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#389939

From	"HGWilson, DSc." <hgw@....>
Date	2016-08-14 06:21 +1000
Message-ID	<nonvg6$75e$1@gioia.aioe.org>
In reply to	#389896

On 13/08/16 10:15, rotchm wrote:
> On Friday, August 12, 2016 at 7:58:12 PM UTC-4, HGWilson, DSc. wrote:
>
>> HAHAHHHAHHA! THE IDIOT Rotchm cannot see that his equation applies
>> equally to both! HAHAHHHAHHAHHA!
>
>
> Didnt you understand when I said
> "Found by correctly applying t'=(t-xv/c^2)g"
>
> So you are admitting that you dont know how to CORRECTLY apply the LT's ?

The equation is the same for both clocks...
You seem to think that direction is absolute. Are you still stupid 
enough to believe in the one absolute aether?

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#389944

From	rotchm <rotchm@gmail.com>
Date	2016-08-13 13:48 -0700
Message-ID	<f1948f1c-1cc6-417a-9668-fe1b60dea1ec@googlegroups.com>
In reply to	#389939

On Saturday, August 13, 2016 at 4:21:30 PM UTC-4, HGWilson, DSc. wrote:

> The equation is the same for both clocks...

Yes. But in the enunciation, t & t' are DIFFERENT PARAMETERS; different clocks. 

> You seem to think that direction is absolute.

Nope. If you think that, then you are very confused. Then again, what can be expected from an idiot such as you who cant even solve *basic* math & physics problems.

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#389947

From	"HGWilson, DSc." <hgw@....>
Date	2016-08-14 07:16 +1000
Message-ID	<noo2o7$c2l$1@gioia.aioe.org>
In reply to	#389944

On 14/08/16 06:48, rotchm wrote:
> On Saturday, August 13, 2016 at 4:21:30 PM UTC-4, HGWilson, DSc.
> wrote:
>
>> The equation is the same for both clocks...
>
> Yes. But in the enunciation, t & t' are DIFFERENT PARAMETERS;
> different clocks.
>
>> You seem to think that direction is absolute.
>
> Nope. If you think that, then you are very confused. Then again, what
> can be expected from an idiot such as you who cant even solve *basic*
> math & physics problems.

You poor fool, you don't even understand what relativity means.

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#389118

From	larry harson <larryharson66@gmail.com>
Date	2016-08-03 13:12 -0700
Message-ID	<1e0a904e-c670-4c9e-93fc-13fdf96cec64@googlegroups.com>
In reply to	#389111

On Wednesday, August 3, 2016 at 7:22:54 PM UTC+1, rotchm wrote:
> On Wednesday, August 3, 2016 at 2:07:18 PM UTC-4, Larry Harson wrote:
> 
> > I'm using slow clock transport to keep time dilation negligible so the
> >  clocks remains synchronized with one another within experimental error.
> 
> 
> Ok. Notice however, this will maintain e-synch. The distant clocks will not be "absolutely" synched.  
> 
> > > Note that your exp is essentially a TWLS exp. 
> > 
> > That's not my intention. 
> 
> I said that in the sense that since you had the moving clocks coming back to the origin (as I understood it) then ...it came back. That made your setup a TWLS; you went both ways. But irrelevant now, since in your explanation below, it doesnt come back. Ok... 
> 
> 
> > The clock at the origin transmits a pulse at its local time t1 
> > to the clock at +L. The clock at +L records a local time t2 when 
> > it receives it. Remembering the clocks are synchronized via slow 
> > clock transport, the one-way light speed here is therefore L/(t2-t1).
> 
> Yes, you can define OWLS that way (there are two common & different meanings for OWLS). 

> But how is L measured (in SR)? 

Using a tape measure or "measuring rod" as Einstein called it.

>By sending a lightpulse from 0 to L and back to 0. That makes it a TWLS. All OWLS, as interpreted by SR, used somewhere a returning signal. In that sense, its not an OWLS but a TWLS.

I don't recall Einstein emphasizing the measurement of distance this way; his emphasis was on clocks and measuring rods measuring time and distance AFAIR.

LH

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#389119

From	rotchm <rotchm@gmail.com>
Date	2016-08-03 14:00 -0700
Message-ID	<e17ed1c7-3fc2-48ae-97e2-4d5abade8cd8@googlegroups.com>
In reply to	#389118

On Wednesday, August 3, 2016 at 4:12:39 PM UTC-4, larry harson wrote:
> On Wednesday, August 3, 2016 at 7:22:54 PM UTC+1, rotchm wrote:

> > But how is L measured (in SR)? 
> 
> Using a tape measure or "measuring rod" as Einstein called it.

Well, E did say it that way for simplicity, but that was not what he meant, nor what was detailed. The length of a rod is done via clocks (as E said & as Poincare specified a few years earlier); take the two ends of the rod *simultaneously* (or send light to & fro, take total time of flight T and the length is thus defined as c*T/2.). So, the length measurement of a rod uses a TWLS process. 

> I don't recall Einstein emphasizing the measurement of distance this way;
>  his emphasis was on clocks and measuring rods measuring time and 
> distance AFAIR.

True. But you reading further and not only him, you get to find out what is the operational definition of length.

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#389120

From	larry harson <larryharson66@gmail.com>
Date	2016-08-03 15:01 -0700
Message-ID	<127e6acf-3056-41db-8a82-e66d151c1d43@googlegroups.com>
In reply to	#389119

On Wednesday, August 3, 2016 at 10:01:00 PM UTC+1, rotchm wrote:
> On Wednesday, August 3, 2016 at 4:12:39 PM UTC-4, larry harson wrote:
> > On Wednesday, August 3, 2016 at 7:22:54 PM UTC+1, rotchm wrote:
> 
> > > But how is L measured (in SR)? 
> > 
> > Using a tape measure or "measuring rod" as Einstein called it.
> 
> Well, E did say it that way for simplicity, but that was not what he meant, nor what was detailed.

I've found a copy of his original paper:
http://einsteinpapers.press.princeton.edu/vol2-trans/154

A search for "measuring rod" gives 6 hits in the paper, such as on page 144:

"Let there be given a rigid rod at rest, its length measured by a *measuring rod" that is also at rest, shall be L".

I can't see the bit where he measures distance using the propagation of light.

>The length of a rod is done via clocks (as E said & as Poincare specified a few years earlier); take the two ends of the rod *simultaneously* (or send light to & fro, take total time of flight T and the length is thus defined as c*T/2.). So, the length measurement of a rod uses a TWLS process.

Yes, but I'm not doing it that way. I'm using a tape measure.

> > I don't recall Einstein emphasizing the measurement of distance this way;
> >  his emphasis was on clocks and measuring rods measuring time and 
> > distance AFAIR.

> True. But you reading further and not only him, you get to find out what is the operational definition of length.

I don't see how this invalidates measuring distances using a measuring tape.

LH

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#389127

From	dlzc <dlzc1@cox.net>
Date	2016-08-03 16:04 -0700
Message-ID	<189026bb-394e-4df6-a6fb-59a23863c378@googlegroups.com>
In reply to	#389120

Dear larry harson:

On Wednesday, August 3, 2016 at 3:01:54 PM UTC-7, larry harson wrote:
...
> "Let there be given a rigid rod at rest, its
> length measured by a *measuring rod" that is
> also at rest, shall be L".
> 
> I can't see the bit where he measures distance
> using the propagation of light.

It's just metrology.  Time-of-flight from end to end,and produce a calibration certificate.  Or move a mirror from end to end, and count maxima and minima from a monochromatic source.

> Yes, but I'm not doing it that way. I'm using
> a tape measure.

The tape measure is calibrated (finally) by using TWLS measurement.

David A. Smith

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#389135

From	rotchm <rotchm@gmail.com>
Date	2016-08-03 18:35 -0700
Message-ID	<93ded214-db26-4bde-9427-8ec9f4fbbb16@googlegroups.com>
In reply to	#389120

On Wednesday, August 3, 2016 at 6:01:54 PM UTC-4, larry harson wrote:
> On Wednesday, August 3, 2016 at 10:01:00 PM UTC+1, rotchm wrote:

> I can't see the bit where he measures distance using the propagation of light.

It takes a little more reading; not just that source of his. 

> Yes, but I'm not doing it that way. I'm using a tape measure.

If you define lengths via a tape measure, then OWLS *in that sense*, is an "owls". But then you need to specify how that behaves when moved, whenever you move it; you will need a model. The only viable one will be one where you postulate a length contraction, which will lead you back to square one...

And note, that in modern physics "tape measures" or even "gratings" are calibrated/constructed via light signals, via the TWLS procedure. 

> I don't see how this invalidates measuring distances using a measuring tape.

In SR, it doesnt, since 1) the tape measure is constructed via a TWLS procedure and 2) SR implies that such a rod will be "length contracted" by the same value given by the simultaneity definition. Thats why we can use "tape measures" in principle. And thats why your setup, is considered a TWLS measurement rather than a OWLS measurement.

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#389948

From	astrofoton@interia.pl
Date	2016-08-13 14:25 -0700
Message-ID	<46110394-eab7-4fd3-bdd5-776b59d777fc@googlegroups.com>
In reply to	#389108

W dniu środa, 3 sierpnia 2016 20:07:18 UTC+2 użytkownik Larry Harson napisał:

> I'm using slow clock transport to keep time dilation negligible so the clocks remains synchronized with one another within experimental error. 

a clock can't be preserved during the slow transport, because:
t = L/v
this means:
the smaller v, the bigger time t is needed;
the L is fixed - const.

 
dt/t = 0.5 v^2/c^2 + ... 
...

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#389951

From	rotchm <rotchm@gmail.com>
Date	2016-08-13 15:01 -0700
Message-ID	<dd0c72a2-f8cf-4a87-8a2b-eb34c1076564@googlegroups.com>
In reply to	#389948

On Saturday, August 13, 2016 at 5:25:38 PM UTC-4, astro...@interia.pl wrote:

> > I'm using slow clock transport to keep time dilation negligible 
> > so the clocks remains synchronized with one another within
> > experimental error. 
> 
> a clock can't be preserved during the slow transport, because:

We know that. That is not in dispute. 

> t = L/v
> this means:
> the smaller v, the bigger time t is needed;
> the L is fixed - const.
> 
>  
> dt/t = 0.5 v^2/c^2 + ... 

Time to arrive is L/v. The moved clock will indicate (L/v)√(1-b^2).
The difference is 

(L/v)( 1 - √(1-b^2)) = (L/v)( v^2/(2c²) + o(4) ) = L( v/(2c²) + o(3) )
Thus, for v sufficiently small, the difference is Lv/(2c²). 
Suffice to chose a v such that Lv/(2c²) is smaller that the exp error(s).

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#389953

From	astrofoton@interia.pl
Date	2016-08-13 16:29 -0700
Message-ID	<a882d92d-fda0-42b4-b871-a7925cc863c2@googlegroups.com>
In reply to	#389951

W dniu niedziela, 14 sierpnia 2016 00:01:29 UTC+2 użytkownik rotchm napisał:
> On Saturday, August 13, 2016 at 5:25:38 PM UTC-4, astro...@interia.pl wrote:
> 
> > > I'm using slow clock transport to keep time dilation negligible 
> > > so the clocks remains synchronized with one another within
> > > experimental error. 
> > 
> > a clock can't be preserved during the slow transport, because:
> 
> We know that. That is not in dispute. 
> 
> > t = L/v
> > this means:
> > the smaller v, the bigger time t is needed;
> > the L is fixed - const.
> > 
> >  
> > dt/t = 0.5 v^2/c^2 + ... 
> 
> Time to arrive is L/v. The moved clock will indicate (L/v)√(1-b^2).
> The difference is 
> 
> (L/v)( 1 - √(1-b^2)) = (L/v)( v^2/(2c²) + o(4) ) = L( v/(2c²) + o(3) )
> Thus, for v sufficiently small, the difference is Lv/(2c²). 
> Suffice to chose a v such that Lv/(2c²) is smaller that the exp error(s).

no, there is possibility;
for small speed:
dt(L)/t(L) = 0.5 v^2/c^2 = const;

the correct result includes a number of odrer:
v^4/c^4, thus absolute irrelevant for the condition
of the slow transport: v/c -> 0.

You can measure something in the opposite case: v ~ c, only;
what is quite contradictory to the presumed conditions already.

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#389955

From	rotchm <rotchm@gmail.com>
Date	2016-08-13 17:10 -0700
Message-ID	<7630b5e1-2bc2-4435-9da2-79edb882f41e@googlegroups.com>
In reply to	#389953

On Saturday, August 13, 2016 at 7:30:00 PM UTC-4, astro...@interia.pl wrote:
> W dniu niedziela, 14 sierpnia 2016 00:01:29 UTC+2 użytkownik rotchm napisał:

> > Time to arrive is L/v. The moved clock will indicate (L/v)√(1-b^2).
> > The difference is 
> > 
> > (L/v)( 1 - √(1-b^2)) = (L/v)( v^2/(2c²) + o(4) ) = L( v/(2c²) + o(3) )
> > Thus, for v sufficiently small, the difference is Lv/(2c²). 
> > Suffice to chose a v such that Lv/(2c²) is smaller that the exp error(s).
> 
> no, 

How so? Shw me where I erred in the math. Its all there in the 3 lines.


> there is possibility;

In english plz?

> for small speed:
> dt(L)/t(L) = 0.5 v^2/c^2 = const;

Wrong.

t(L) = L/v --> dt(L) = d(L)/v. 
Thus dt(L)/t(L) = (d(L)/v)/(L/v) = d(L)/L. 
So you are wrong. And why would you want to evaluate dt(L)/t(L) anyways ??



> the correct result includes a number of odrer:
> v^4/c^4, thus absolute irrelevant for the condition
> of the slow transport: v/c -> 0.

In english plz...

> You can measure something in the opposite case: v ~ c, only;

In english plz...

> what is quite contradictory to the presumed conditions already.

What you write makes no sense. So either you are an idiot/troll, or just very bad at explaining in english. Work on that before posting so that you may be understood.

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#389992

From	astrofoton@interia.pl
Date	2016-08-14 13:05 -0700
Message-ID	<8b3715fb-9dba-45ce-9d44-b9b671daae3d@googlegroups.com>
In reply to	#389955

W dniu niedziela, 14 sierpnia 2016 02:10:49 UTC+2 użytkownik rotchm napisał:
> On Saturday, August 13, 2016 at 7:30:00 PM UTC-4, astro...@interia.pl wrote:
> > W dniu niedziela, 14 sierpnia 2016 00:01:29 UTC+2 użytkownik rotchm napisał:
> 
> > > Time to arrive is L/v. The moved clock will indicate (L/v)√(1-b^2).
> > > The difference is 
> > > 
> > > (L/v)( 1 - √(1-b^2)) = (L/v)( v^2/(2c²) + o(4) ) = L( v/(2c²) + o(3) )
> > > Thus, for v sufficiently small, the difference is Lv/(2c²). 
> > > Suffice to chose a v such that Lv/(2c²) is smaller that the exp error(s).
> > 
> > no, 
> 
> How so? Shw me where I erred in the math. Its all there in the 3 lines.
> 
> 
> > there is possibility;
> 
> In english plz?
> 
> > for small speed:
> > dt(L)/t(L) = 0.5 v^2/c^2 = const;
> 
> Wrong.
> 
> t(L) = L/v --> dt(L) = d(L)/v. 
> Thus dt(L)/t(L) = (d(L)/v)/(L/v) = d(L)/L. 
> So you are wrong. And why would you want to evaluate dt(L)/t(L) anyways ??
> 
> 
> 
> > the correct result includes a number of odrer:
> > v^4/c^4, thus absolute irrelevant for the condition
> > of the slow transport: v/c -> 0.
> 
> In english plz...
> 
> > You can measure something in the opposite case: v ~ c, only;
> 
> In english plz...
> 
> > what is quite contradictory to the presumed conditions already.
> 
> What you write makes no sense. So either you are an idiot/troll, or just very bad at explaining in english. Work on that before posting so that you may be understood.

Indeed. I'm too smart for you. :)

dt/t = 0.5 v^2/c^2

but using a fact: L = vt, thus: v = L/t

dt/t = 0.5 L^2/t^2c^2 = 

dt = 0.5 L^2/c^2 / t

... and what is it?

tdt = 0.5 L^2/c^2 = A/2 = const


t^2 = At + C.
C = 0 ->
t = A = const (= inv). :)

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Slow clock transport and the measurement of one way light speed.

Contents

#389889

#389892

#389894

#389896

#389939

#389944

#389947

#389118

#389119

#389120

#389127

#389135

#389948

#389951

#389953

#389955

#389992