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Groups > sci.physics.relativity > #478177 > unrolled thread
| Started by | Thomas Heger <ttt_heg@web.de> |
|---|---|
| First post | 2018-11-26 16:01 +0100 |
| Last post | 2018-12-01 07:34 -0800 |
| Articles | 20 on this page of 79 — 18 participants |
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One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-26 16:01 +0100
Re: One way speed of light Lee Primer <itoo@wodinwnc.cg> - 2018-11-26 15:50 +0000
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-11-26 21:48 +0100
Re: One way speed of light "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2018-11-27 07:07 -0800
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-11-29 02:09 +0100
Re: One way speed of light JanPB <filmart@gmail.com> - 2018-11-26 13:04 -0800
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-26 21:29 +0000
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-27 15:31 +0100
Re: One way speed of light Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-11-27 15:59 +0100
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-27 16:26 +0100
Re: One way speed of light moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-11-27 16:47 +0000
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-27 16:54 +0000
Re: One way speed of light Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-11-27 18:01 +0100
Re: One way speed of light Lee Primer <itoo@wodinwnc.cg> - 2018-11-28 00:55 +0000
Re: One way speed of light Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-11-28 05:14 +0100
Re: One way speed of light Lee Primer <itoo@wodinwnc.cg> - 2018-11-28 15:13 +0000
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-27 16:44 +0000
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-28 16:34 +0100
Re: One way speed of light Lee Primer <itoo@wodinwnc.cg> - 2018-11-28 15:40 +0000
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-28 17:20 +0000
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-29 20:43 +0100
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-29 20:53 +0000
Re: One way speed of light Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-11-29 22:21 +0100
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-30 21:15 +0100
Re: One way speed of light moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-11-30 22:46 +0000
Re: One way speed of light Tom Roberts <tjroberts137@sbcglobal.net> - 2018-11-30 17:23 -0600
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-01 00:34 +0100
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-30 23:40 +0000
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-12-01 16:17 +0100
Re: One way speed of light "Dono," <sa_ge@comcast.net> - 2018-12-01 07:32 -0800
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-12-02 13:50 +0100
Re: One way speed of light Gary Harnagel <hitlong@yahoo.com> - 2018-12-02 05:11 -0800
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-12-03 12:23 +0100
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-12-03 18:53 +0000
Re: One way speed of light mitchrae3323@gmail.com - 2018-12-03 11:25 -0800
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-12-04 13:04 +0100
Re: One way speed of light Gary Harnagel <hitlong@yahoo.com> - 2018-12-03 11:57 -0800
Re: One way speed of light alsor@interia.pl - 2018-12-03 13:28 -0800
Re: One way speed of light mitchrae3323@gmail.com - 2018-12-03 13:56 -0800
Re: One way speed of light alsor@interia.pl - 2018-12-03 14:13 -0800
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-02 14:44 +0100
Re: One way speed of light "Dono," <sa_ge@comcast.net> - 2018-12-02 07:13 -0800
Re: One way speed of light Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-02 10:28 -0600
Re: One way speed of light Mole Santiago <msb@awwnuvs.an> - 2018-12-02 18:42 +0000
Re: One way speed of light alsor@interia.pl - 2018-12-02 11:08 -0800
Re: One way speed of light moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-12-02 18:49 +0000
Re: One way speed of light Mole Santiago <msb@awwnuvs.an> - 2018-12-02 18:57 +0000
Re: One way speed of light Mole Santiago <msb@awwnuvs.an> - 2018-12-02 18:58 +0000
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-12-02 21:23 +0000
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-01 18:20 +0100
Re: One way speed of light Tom Roberts <tjroberts137@sbcglobal.net> - 2018-11-29 16:03 -0600
Re: One way speed of light mlwozniak@wp.pl - 2018-11-29 15:03 -0800
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-30 21:18 +0100
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-12-01 15:43 +0000
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-02 14:58 +0100
Re: One way speed of light moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-11-29 23:38 +0000
Re: One way speed of light mlwozniak@wp.pl - 2018-11-29 16:40 -0800
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-30 03:22 +0000
Re: One way speed of light benj <benj.fake@gmail.com> - 2018-11-29 22:41 -0500
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-30 12:13 +0000
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-11-30 21:21 +0100
Re: One way speed of light Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-11-30 22:10 +0100
Re: One way speed of light Thomas Heger <ttt_heg@web.de> - 2018-12-01 16:26 +0100
Re: One way speed of light moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-11-30 23:01 +0000
Re: One way speed of light Odd Bodkin <bodkinodd@gmail.com> - 2018-11-30 23:03 +0000
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-01 00:43 +0100
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-11-29 02:01 +0100
Re: One way speed of light Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-11-29 02:04 +0100
Re: One way speed of light alsor@interia.pl - 2018-11-26 15:51 -0800
Re: One way speed of light Gary Harnagel <hitlong@yahoo.com> - 2018-11-26 18:53 -0800
Re: One way speed of light alsor@interia.pl - 2018-11-27 09:49 -0800
Re: One way speed of light numbernumber1964@gmail.com - 2018-11-28 15:11 -0800
Re: One way speed of light mitchrae3323@gmail.com - 2018-11-29 14:19 -0800
Re: One way speed of light carleto4157990662@gmail.com - 2018-11-29 14:36 -0800
Re: One way speed of light benj <benj.fake@gmail.com> - 2018-11-29 22:28 -0500
Re: One way speed of light mitchrae3323@gmail.com - 2018-11-29 20:08 -0800
Re: One way speed of light carleto4157990662@gmail.com - 2018-11-29 15:10 -0800
Re: One way speed of light alsor@interia.pl - 2018-11-30 12:48 -0800
Alex Sorokin, persistent imbecile "Dono," <sa_ge@comcast.net> - 2018-12-01 07:34 -0800
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| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-11-26 16:01 +0100 |
| Subject | One way speed of light |
| Message-ID | <g62g6cFm3jU1@mid.individual.net> |
Hi NG in the famous Michelson-Morley experiment, they have tried to find out, whether or not the speed of light is influenced by an assumed 'ether wind'. Problematic was the used methods to measure such variations of the speed of light, since those included mirrors and therefore a two-way speed of light measurement. This two way measurement could easily cancel out any effect (with or without an ether) because acceleration of one trip would be balanced by deceleration on the way back. To measure one-way speed of light I would suggest this experiment: a axis of 2 meter length has two rotors - one at each end. Each wheel has a diameter of 1 m. In the middle the axis has a turbine, which could spin the rotor to any desired velocity (required is more than 1250 per second). The wheels have a fine grid etched into the polished rim. It should be 60.000 stripes along the rim. (actually other combinations of dimensions of the device are possible, like finer grid, longer axle and lower velocity.) A laser beam is focused at the rotating grid and reflected to the other wheel. Now the laser gets modulated with a certain frequency, that is determined by the number of lines on the grid and the speed of the rotation. The length is known between the wheels. If the rotor is spun at very high speed and then left unaccelerated, the rotor gets slower through friction. The modulation of the laser and the spin velocity is continuously measured (at the second wheel). The wheel would produce a modulation of the beam and that would generate a wave with 4m wavelength at a certain rotation rate. Then the reflected beam would hit a pit at the other wheel and the twice reflected beam has its lowest intensity. Since all the measurements could be made at one side (modulation and frequency of the rotating grid), the speed of light could be figured out for a one way trip. TH
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| From | Lee Primer <itoo@wodinwnc.cg> |
|---|---|
| Date | 2018-11-26 15:50 +0000 |
| Message-ID | <pth4nq$oig$2@gioia.aioe.org> |
| In reply to | #478177 |
Thomas Heger wrote: > Hi NG > in the famous Michelson-Morley experiment, they have tried to find out, > whether or not the speed of light is influenced by an assumed 'ether > wind'. Not at all, but a stationary aether as a medium. A wind cannot exists, but this is the least of your worries, since as stationary also will introduce an unfrogivable paradox. Actually a bigger paradox compared to those already existent in relativity.
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| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-11-26 21:48 +0100 |
| Message-ID | <178f3809-bc7d-0acd-0600-3e400d20805b@PointedEars.de> |
| In reply to | #478177 |
Thomas Heger wrote: > in the famous Michelson-Morley experiment, they have tried to find out, > whether or not the speed of light is influenced by an assumed 'ether wind'. Correct. The assumption was that there is a stationary ether, and as Earth would move through that ether, there should be an “ether wind” blowing against the direction of motion. > Problematic was the used methods to measure such variations of the speed > of light, since those included mirrors and therefore a two-way speed of > light measurement. That was not problematic, but the only way that it can be done reliably. > This two way measurement could easily cancel out any effect (with or > without an ether) because acceleration of one trip would be balanced by > deceleration on the way back. No. Shut up and calculate. -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail.
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| From | "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> |
|---|---|
| Date | 2018-11-27 07:07 -0800 |
| Message-ID | <32eae9c4-2818-41c4-959a-454bb670d1b2@googlegroups.com> |
| In reply to | #478204 |
Thomas 'PointedEars' Lahn Thomas Heger wrote: > in the famous Michelson-Morley experiment, they have tried to find out, > whether or not the speed of light is influenced by an assumed 'ether wind'. Correct. The assumption was that there is a stationary ether, and as Earth would move through that ether, there should be an “ether wind” blowing against the direction of motion. There IS an “Aether Wind” It is Relativistic Mass & the Acceleration Of Gravity
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| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-11-29 02:09 +0100 |
| Message-ID | <08bd8966-7283-a26d-7906-18e8dd87f257@PointedEars.de> |
| In reply to | #478257 |
Learn to quote. David (Kronos Prime) Fuller wrote: > Thomas 'PointedEars' Lahn >> Thomas Heger wrote: >>> in the famous Michelson-Morley experiment, they have tried to find out, >>> whether or not the speed of light is influenced by an assumed 'ether wind'. >> >> Correct. The assumption was that there is a stationary ether, and as Earth >> would move through that ether, there should be an “ether wind” blowing >> against the direction of motion. > > There IS an “Aether Wind” No, there is not. Michelson had not found any in 1881, Michelson and Morley have not found any in 1887, and nobody after them who has tried (and there were many sophisticated attempts) has found any. You can do the experiment yourself. I dare you to find any “aether wind”. > It is Relativistic Mass & the Acceleration Of Gravity Not even wrong. -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail.
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| From | JanPB <filmart@gmail.com> |
|---|---|
| Date | 2018-11-26 13:04 -0800 |
| Message-ID | <51d502aa-5dd2-4028-9712-80c759933c5c@googlegroups.com> |
| In reply to | #478177 |
On Monday, November 26, 2018 at 7:01:34 AM UTC-8, Thomas Heger wrote: > Hi NG > > in the famous Michelson-Morley experiment, they have tried to find out, > whether or not the speed of light is influenced by an assumed 'ether wind'. > > Problematic was the used methods to measure such variations of the speed > of light, since those included mirrors and therefore a two-way speed of > light measurement. > > This two way measurement could easily cancel out any effect (with or > without an ether) because acceleration of one trip would be balanced by > deceleration on the way back. No, actually. It's a high-school-level exercise to see that this is not so. -- Jan
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-11-26 21:29 +0000 |
| Message-ID | <pthokn$1v5o$2@gioia.aioe.org> |
| In reply to | #478206 |
JanPB <filmart@gmail.com> wrote: > On Monday, November 26, 2018 at 7:01:34 AM UTC-8, Thomas Heger wrote: >> Hi NG >> >> in the famous Michelson-Morley experiment, they have tried to find out, >> whether or not the speed of light is influenced by an assumed 'ether wind'. >> >> Problematic was the used methods to measure such variations of the speed >> of light, since those included mirrors and therefore a two-way speed of >> light measurement. >> >> This two way measurement could easily cancel out any effect (with or >> without an ether) because acceleration of one trip would be balanced by >> deceleration on the way back. > > No, actually. It's a high-school-level exercise to see that this is not so. > > -- > Jan > To wit: Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides to swim in a river that has a current of 1 mile per hour. He swims upstream a quarter mile relative to the shore, and then he turns around and swims downstream a quarter mile to his starting point at the shore, a total of a half mile swum. What is his average speed relative to the shore for the trip? -- Odd Bodkin -- maker of fine toys, tools, tables
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| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-11-27 15:31 +0100 |
| Message-ID | <g652qdFi448U1@mid.individual.net> |
| In reply to | #478208 |
Am 26.11.2018 um 22:29 schrieb Odd Bodkin: >>> This two way measurement could easily cancel out any effect (with or >>> without an ether) because acceleration of one trip would be balanced by >>> deceleration on the way back. >> >> No, actually. It's a high-school-level exercise to see that this is not so. >> >> -- >> Jan >> > > To wit: > Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides > to swim in a river that has a current of 1 mile per hour. He swims upstream > a quarter mile relative to the shore, and then he turns around and swims > downstream a quarter mile to his starting point at the shore, a total of a > half mile swum. What is his average speed relative to the shore for the > trip? > The swimmer swims faster downstream and slower upstream, because we had to add the velocity of the water in one case and subtract it in the other. The time is an eighth of an hour up and a sixteenth downstream and that is 675 seconds for a half mile. The result is roughly 1.19 m/s TH
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| From | Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> |
|---|---|
| Date | 2018-11-27 15:59 +0100 |
| Message-ID | <ptjm4c$cgb$1@dont-email.me> |
| In reply to | #478253 |
Thomas Heger <ttt_heg@web.de> Wrote in message: > Am 26.11.2018 um 22:29 schrieb Odd Bodkin: > >>>> This two way measurement could easily cancel out any effect (with or >>>> without an ether) because acceleration of one trip would be balanced by >>>> deceleration on the way back. >>> >>> No, actually. It's a high-school-level exercise to see that this is not so. >>> >>> -- >>> Jan >>> >> >> To wit: >> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >> to swim in a river that has a current of 1 mile per hour. He swims upstream >> a quarter mile relative to the shore, and then he turns around and swims >> downstream a quarter mile to his starting point at the shore, a total of a >> half mile swum. What is his average speed relative to the shore for the >> trip? >> > > > The swimmer swims faster downstream and slower upstream, because we had > to add the velocity of the water in one case and subtract it in the other. > > The time is an eighth of an hour up and a sixteenth downstream and that > is 675 seconds for a half mile. The result is roughly 1.19 m/s > The point is, the total time up and downstream, and therefore the average speed as well, depends on the stream speed. The same as for the supposed aether stream speed in the MMX. As the average speed along the same distance is not an arithmetic mean of one way speeds, but a harmonic mean v=2/(1/v1+1/v2). This means, with the same arithmetic mean of both speeds, their harmonic mean gets lower with growing difference of the one way speeds. -- Libor Striz aka Poutnik ( a pilgrim/wanderer/wayfarer) "Humour is the only effective weapon against stupidity." Miloš Forman ----Android NewsGroup Reader---- http://usenet.sinaapp.com/
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| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-11-27 16:26 +0100 |
| Message-ID | <g6560nFiqfdU1@mid.individual.net> |
| In reply to | #478256 |
Am 27.11.2018 um 15:59 schrieb Libor Striz: > Thomas Heger <ttt_heg@web.de> Wrote in message: >> Am 26.11.2018 um 22:29 schrieb Odd Bodkin: >> >>>>> This two way measurement could easily cancel out any effect (with or >>>>> without an ether) because acceleration of one trip would be balanced by >>>>> deceleration on the way back. >>>> >>>> No, actually. It's a high-school-level exercise to see that this is not so. >>>> >>>> -- >>>> Jan >>>> >>> >>> To wit: >>> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >>> to swim in a river that has a current of 1 mile per hour. He swims upstream >>> a quarter mile relative to the shore, and then he turns around and swims >>> downstream a quarter mile to his starting point at the shore, a total of a >>> half mile swum. What is his average speed relative to the shore for the >>> trip? >>> >> >> >> The swimmer swims faster downstream and slower upstream, because we had >> to add the velocity of the water in one case and subtract it in the other. >> >> The time is an eighth of an hour up and a sixteenth downstream and that >> is 675 seconds for a half mile. The result is roughly 1.19 m/s >> > The point is, the total time up and downstream, > and therefore the average speed as well, > depends on the stream speed. Yes, but the mean speed is not really interesting. Instead you could compare the 'momentary mean velocity'. That is v_mean_ momentary= (speed upstream + speed downstream) /2 and this is independent from the speed of the water. TH
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| From | moroney@world.std.spaamtrap.com (Michael Moroney) |
|---|---|
| Date | 2018-11-27 16:47 +0000 |
| Message-ID | <ptjsf6$5m3$2@pcls7.std.com> |
| In reply to | #478259 |
Thomas Heger <ttt_heg@web.de> writes: >>>> To wit: >>>> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >>>> to swim in a river that has a current of 1 mile per hour. He swims upstream >>>> a quarter mile relative to the shore, and then he turns around and swims >>>> downstream a quarter mile to his starting point at the shore, a total of a >>>> half mile swum. What is his average speed relative to the shore for the >>>> trip? >>>> >>> >>> >>> The swimmer swims faster downstream and slower upstream, because we had >>> to add the velocity of the water in one case and subtract it in the other. >>> >>> The time is an eighth of an hour up and a sixteenth downstream and that >>> is 675 seconds for a half mile. The result is roughly 1.19 m/s >>> >> The point is, the total time up and downstream, >> and therefore the average speed as well, >> depends on the stream speed. >Yes, but the mean speed is not really interesting. >Instead you could compare the 'momentary mean velocity'. Odd's point is that the total speed or travel time depends on the speed of the river. >That is >v_mean_ momentary= (speed upstream + speed downstream) /2 >and this is independent from the speed of the water. And is _wrong_. This is what trips up those who believe that light moves at c+v in one direction and c-v on the return trip and therefore they say light moves at c overall for the round trip because the v's cancel. If this (light moving at c+v/c-v) were true, the overall time is always _less than_ c, as long as v is nonzero. Michelson/Morley knew this. They expected that for certain alignments of their apparatus, light along one path would travel at c+v one way and c-v on the return trip and they'd get a fringe shift from light on the right angle path (c both ways). From the amount of shift they could determine v (the aether wind speed). Because light actually traveled at c all the time, they always got a null result (v=0 no matter how their apparatus happened to be aligned in space).
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-11-27 16:54 +0000 |
| Message-ID | <ptjss2$1f8e$1@gioia.aioe.org> |
| In reply to | #478259 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 27.11.2018 um 15:59 schrieb Libor Striz: >> Thomas Heger <ttt_heg@web.de> Wrote in message: >>> Am 26.11.2018 um 22:29 schrieb Odd Bodkin: >>> >>>>>> This two way measurement could easily cancel out any effect (with or >>>>>> without an ether) because acceleration of one trip would be balanced by >>>>>> deceleration on the way back. >>>>> >>>>> No, actually. It's a high-school-level exercise to see that this is not so. >>>>> >>>>> -- >>>>> Jan >>>>> >>>> >>>> To wit: >>>> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >>>> to swim in a river that has a current of 1 mile per hour. He swims upstream >>>> a quarter mile relative to the shore, and then he turns around and swims >>>> downstream a quarter mile to his starting point at the shore, a total of a >>>> half mile swum. What is his average speed relative to the shore for the >>>> trip? >>>> >>> >>> >>> The swimmer swims faster downstream and slower upstream, because we had >>> to add the velocity of the water in one case and subtract it in the other. >>> >>> The time is an eighth of an hour up and a sixteenth downstream and that >>> is 675 seconds for a half mile. The result is roughly 1.19 m/s >>> >> The point is, the total time up and downstream, >> and therefore the average speed as well, >> depends on the stream speed. > > > Yes, but the mean speed is not really interesting. > > Instead you could compare the 'momentary mean velocity'. > > That is > v_mean_ momentary= (speed upstream + speed downstream) /2 This quantity pertains to nothing measurable. > > and this is independent from the speed of the water. > > > TH > > -- Odd Bodkin — Maker of fine toys, tools, tables
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| From | Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> |
|---|---|
| Date | 2018-11-27 18:01 +0100 |
| Message-ID | <ptjt8n$pgq$1@dont-email.me> |
| In reply to | #478259 |
Thomas Heger <ttt_heg@web.de> Wrote in message: > Am 27.11.2018 um 15:59 schrieb Libor Striz: >> Thomas Heger <ttt_heg@web.de> Wrote in message: >>> Am 26.11.2018 um 22:29 schrieb Odd Bodkin: >>> >>>>>> This two way measurement could easily cancel out any effect (with or >>>>>> without an ether) because acceleration of one trip would be balanced by >>>>>> deceleration on the way back. >>>>> >>>>> No, actually. It's a high-school-level exercise to see that this is not so. >>>>> >>>>> -- >>>>> Jan >>>>> >>>> >>>> To wit: >>>> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >>>> to swim in a river that has a current of 1 mile per hour. He swims upstream >>>> a quarter mile relative to the shore, and then he turns around and swims >>>> downstream a quarter mile to his starting point at the shore, a total of a >>>> half mile swum. What is his average speed relative to the shore for the >>>> trip? >>>> >>> >>> >>> The swimmer swims faster downstream and slower upstream, because we had >>> to add the velocity of the water in one case and subtract it in the other. >>> >>> The time is an eighth of an hour up and a sixteenth downstream and that >>> is 675 seconds for a half mile. The result is roughly 1.19 m/s >>> >> The point is, the total time up and downstream, >> and therefore the average speed as well, >> depends on the stream speed. > > > Yes, but the mean speed is not really interesting. But it is, as it is the mean speed differences the MMX indirectly registered via the fringe shift > > Instead you could compare the 'momentary mean velocity'. Not in the MMX > > That is > v_mean_ momentary= (speed upstream + speed downstream) /2 Momentary and mean are mutually exclusive words. At any infinitesimal distance interval, the mean speed is still harmonic and not arithmetic mean. Btw, it you insist on velocity mean, it goes toward zero for speed difference going toward zero . > > and this is independent from the speed of the water. Yes, but it is useless in MMX context. -- Libor Striz aka Poutnik ( a pilgrim/wanderer/wayfarer) "Humour is the only effective weapon against stupidity." Miloš Forman ----Android NewsGroup Reader---- http://usenet.sinaapp.com/
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| From | Lee Primer <itoo@wodinwnc.cg> |
|---|---|
| Date | 2018-11-28 00:55 +0000 |
| Message-ID | <ptkp26$vee$3@gioia.aioe.org> |
| In reply to | #478256 |
Libor Striz wrote: > As the average speed along the same distance > is not an arithmetic mean of one way speeds, > but a harmonic mean v=2/(1/v1+1/v2). This means, > with the same arithmetic mean of both speeds, > their harmonic mean gets lower with growing difference of the one way > speeds. Go watch what a "harmonic" stands for in physics, before bringing your country in shame.
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| From | Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> |
|---|---|
| Date | 2018-11-28 05:14 +0100 |
| Message-ID | <ptl4nj$56s$1@dont-email.me> |
| In reply to | #478295 |
Lee Primer <itoo@wodinwnc.cg> Wrote in message: > Libor Striz wrote: > >> As the average speed along the same distance >> is not an arithmetic mean of one way speeds, >> but a harmonic mean v=2/(1/v1+1/v2). This means, >> with the same arithmetic mean of both speeds, >> their harmonic mean gets lower with growing difference of the one way >> speeds. > > Go watch what a "harmonic" stands for in physics, before bringing your > country in shame. > Not before you learn what the harmonic mean means in mathematics and not before you stop bringing your country in shame by the annoying trolling and nymshifting. -- Libor Striz aka Poutnik ( a pilgrim/wanderer/wayfarer) "Humour is the only effective weapon against stupidity." Miloš Forman ----Android NewsGroup Reader---- http://usenet.sinaapp.com/
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| From | Lee Primer <itoo@wodinwnc.cg> |
|---|---|
| Date | 2018-11-28 15:13 +0000 |
| Message-ID | <ptmba4$1jre$1@gioia.aioe.org> |
| In reply to | #478302 |
Libor Striz wrote: > Lee Primer <itoo@wodinwnc.cg> Wrote in message: >> Libor Striz wrote: >> >>> As the average speed along the same distance >>> is not an arithmetic mean of one way speeds, >>> but a harmonic mean v=2/(1/v1+1/v2). This means, >>> with the same arithmetic mean of both speeds, >>> their harmonic mean gets lower with growing difference of the one way >>> speeds. >> >> Go watch what a "harmonic" stands for in physics, before bringing your >> country in shame. > > Not before you learn what the harmonic mean means in mathematics and not > before you stop bringing your country in shame by the annoying trolling > and nymshifting. Yes, especially in mathematics, as physics is tight to mathematics in a 1-1 way. Sad, Poland use to be such a wonderful country. Ie, when you go by train from Warshawa to Krakow etc. Now you brought you country in eternal shame.
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-11-27 16:44 +0000 |
| Message-ID | <ptjs95$1e1r$2@gioia.aioe.org> |
| In reply to | #478253 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 26.11.2018 um 22:29 schrieb Odd Bodkin: > >>>> This two way measurement could easily cancel out any effect (with or >>>> without an ether) because acceleration of one trip would be balanced by >>>> deceleration on the way back. >>> >>> No, actually. It's a high-school-level exercise to see that this is not so. >>> >>> -- >>> Jan >>> >> >> To wit: >> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >> to swim in a river that has a current of 1 mile per hour. He swims upstream >> a quarter mile relative to the shore, and then he turns around and swims >> downstream a quarter mile to his starting point at the shore, a total of a >> half mile swum. What is his average speed relative to the shore for the >> trip? >> > > > The swimmer swims faster downstream and slower upstream, because we had > to add the velocity of the water in one case and subtract it in the other. > > The time is an eighth of an hour up and a sixteenth downstream and that > is 675 seconds for a half mile. The result is roughly 1.19 m/s And is his average speed 3 miles per hour? Yes or no? > > TH > > -- Odd Bodkin — Maker of fine toys, tools, tables
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| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-11-28 16:34 +0100 |
| Message-ID | <g67qrbF68otU1@mid.individual.net> |
| In reply to | #478261 |
Am 27.11.2018 um 17:44 schrieb Odd Bodkin: > Thomas Heger <ttt_heg@web.de> wrote: >> Am 26.11.2018 um 22:29 schrieb Odd Bodkin: >> >>>>> This two way measurement could easily cancel out any effect (with or >>>>> without an ether) because acceleration of one trip would be balanced by >>>>> deceleration on the way back. >>>> >>>> No, actually. It's a high-school-level exercise to see that this is not so. >>>> >>>> -- >>>> Jan >>>> >>> >>> To wit: >>> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >>> to swim in a river that has a current of 1 mile per hour. He swims upstream >>> a quarter mile relative to the shore, and then he turns around and swims >>> downstream a quarter mile to his starting point at the shore, a total of a >>> half mile swum. What is his average speed relative to the shore for the >>> trip? >>> >> >> >> The swimmer swims faster downstream and slower upstream, because we had >> to add the velocity of the water in one case and subtract it in the other. >> >> The time is an eighth of an hour up and a sixteenth downstream and that >> is 675 seconds for a half mile. The result is roughly 1.19 m/s > > And is his average speed 3 miles per hour? Yes or no? How could you STILL use miles? There average is not 3 miles per hour, since you make an average as distance traveled by time needed. Since you gain less at higher speed (say: double velocity) than what you loose on lower speed, the average is influenced by the velocity. example: you need one minute with a for 1 km at 1km/min. But if you travel with 2km/min you save 30 seconds. If you travel with 0.5 km/min you need 2 min for one km at 0.5 km/min, hence need 60 seconds more. but 60 minus 30 is not zero. So THIS average is wrong, in connection to speed of light, since c is so much faster than the possible speed of movement through the 'ether'. TH so you need 30 seconds more, if you swim upstream an them downstream.
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| From | Lee Primer <itoo@wodinwnc.cg> |
|---|---|
| Date | 2018-11-28 15:40 +0000 |
| Message-ID | <ptmcu0$1jre$4@gioia.aioe.org> |
| In reply to | #478312 |
Thomas Heger wrote: >> And is his average speed 3 miles per hour? Yes or no? > > How could you STILL use miles? > There average is not 3 miles per hour, since you make an average as > distance traveled by time needed. Excellent observation, including doctors, the don't understand this simple thing. It only works for constant speed. All data regarding the higher order is lost.
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-11-28 17:20 +0000 |
| Message-ID | <ptmipq$4pd$1@gioia.aioe.org> |
| In reply to | #478312 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 27.11.2018 um 17:44 schrieb Odd Bodkin: >> Thomas Heger <ttt_heg@web.de> wrote: >>> Am 26.11.2018 um 22:29 schrieb Odd Bodkin: >>> >>>>>> This two way measurement could easily cancel out any effect (with or >>>>>> without an ether) because acceleration of one trip would be balanced by >>>>>> deceleration on the way back. >>>>> >>>>> No, actually. It's a high-school-level exercise to see that this is not so. >>>>> >>>>> -- >>>>> Jan >>>>> >>>> >>>> To wit: >>>> Jimmy can swim at 3 miles per hour with respect to the water. Jimmy decides >>>> to swim in a river that has a current of 1 mile per hour. He swims upstream >>>> a quarter mile relative to the shore, and then he turns around and swims >>>> downstream a quarter mile to his starting point at the shore, a total of a >>>> half mile swum. What is his average speed relative to the shore for the >>>> trip? >>>> >>> >>> >>> The swimmer swims faster downstream and slower upstream, because we had >>> to add the velocity of the water in one case and subtract it in the other. >>> >>> The time is an eighth of an hour up and a sixteenth downstream and that >>> is 675 seconds for a half mile. The result is roughly 1.19 m/s >> >> And is his average speed 3 miles per hour? Yes or no? > > > How could you STILL use miles? > > > There average is not 3 miles per hour, since you make an average as > distance traveled by time needed. Exactly. And for the same reason, if you have a 2-way light speed measurement such as the MMX, then even if there were a luminiferous ether that the apparatus was moving through, so that upstream speed is c-v and downstream speed is c+v, the average speed for the round trip will not be c. Thus, though this was not technically available in 1887, it is now quite easily attainable to simply do a 2-way light traverse over, say, a couple hundred meters and TIME the flight of a light pulse and see if the time is different than c. Because if there were an ether that the apparatus was moving through with speed v, you’d expect to measure an average speed of c’ = c(1-(v/c)^2). For v/c = 1/20, then c’ would differ from c by 1/4%, which for a total distance of, say, 400 m, corresponds to a time resolution of 3ns, which is VERY doable. No one does it today of course, because the matter is already settled. But YOU could do it with a pulsed laser in a long darkened pipe with a couple mirrors, a photodiode, and an oscilloscope. > > Since you gain less at higher speed (say: double velocity) than what > you loose on lower speed, the average is influenced by the velocity. > > example: > > you need one minute with a for 1 km at 1km/min. > But if you travel with 2km/min you save 30 seconds. > > If you travel with 0.5 km/min you need 2 min for one km at 0.5 km/min, > hence need 60 seconds more. > > but 60 minus 30 is not zero. > > So THIS average is wrong, in connection to speed of light, since c is so > much faster than the possible speed of movement through the 'ether'. > > TH > > so you need 30 seconds more, if you swim upstream an them downstream. > > -- Odd Bodkin — Maker of fine toys, tools, tables
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