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Groups > sci.physics > #519359 > unrolled thread
| Started by | Fabian Russell <root@localhost.localdomain> |
|---|---|
| First post | 2015-09-06 20:21 +0000 |
| Last post | 2015-09-14 09:31 -0500 |
| Articles | 20 on this page of 100 — 11 participants |
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Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-06 20:21 +0000
Re: Relativistic Radio Double-A <double-a3@hush.com> - 2015-09-06 14:34 -0700
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-06 21:47 +0000
Re: Relativistic Radio pcardinale@volcanomail.com - 2015-09-06 17:36 -0700
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-07 01:56 +0000
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-07 07:15 +0200
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-07 05:44 +0000
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-07 08:50 +0200
Re: Relativistic Radio pcardinale@volcanomail.com - 2015-09-07 15:16 -0700
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-08 10:04 -0500
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-08 16:59 +0000
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-08 12:41 -0500
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-08 18:24 +0000
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-08 21:22 +0200
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-08 17:43 +0000
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-08 12:47 -0500
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-08 19:26 +0000
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-10 01:58 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-10 19:16 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-10 19:42 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-11 00:53 -0700
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-11 06:07 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-11 15:08 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-11 15:28 +0200
Re: Relativistic Radio Mahipal <mahipal7638@gmail.com> - 2015-09-11 06:31 -0700
Re: Relativistic Radio "Phony McNymster" <invalid@example.com> - 2015-09-11 07:42 -0700
Re: Relativistic Radio Mahipal <mahipal7638@gmail.com> - 2015-09-11 07:59 -0700
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-11 07:16 -0700
Re: Relativistic Radio Sam Wormley <swormley1@gmail.com> - 2015-09-11 09:41 -0500
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-11 09:49 -0500
Re: Relativistic Radio Fabian Russell <root@localhost.localdomain> - 2015-09-11 18:18 +0000
Re: Relativistic Radio noTthaTguY <abu.kuanysh05@gmail.com> - 2015-09-11 11:24 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-12 11:20 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-12 06:21 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 02:04 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-13 00:07 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 09:22 +0200
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 09:59 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-13 03:44 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-14 07:33 +0200
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-14 07:49 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 01:07 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-14 10:40 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 03:40 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-14 13:19 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 04:55 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-14 15:01 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-14 16:36 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 12:18 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-15 08:29 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 02:41 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 12:11 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 06:06 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 15:40 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 06:06 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 15:59 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 09:39 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 18:47 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 10:10 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-15 19:57 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 11:14 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-15 20:23 +0200
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-15 20:31 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 12:21 -0700
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-15 13:30 -0500
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 00:57 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-14 14:50 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 12:57 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-15 08:37 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 03:06 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 13:10 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 13:15 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-15 06:43 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 16:11 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 18:26 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 18:35 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 16:44 +0200
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-14 08:15 +0200
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-14 08:44 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-11 17:38 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-11 18:00 +0200
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-11 11:22 -0500
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-12 12:29 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 01:16 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-12 23:09 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 08:35 +0200
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 08:44 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-12 23:50 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 09:14 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-13 05:24 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-14 07:35 +0200
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 01:01 -0700
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-14 10:39 +0200
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-14 09:34 -0500
Re: Relativistic Radio jaymoseley@hotmail.com - 2015-09-14 12:31 -0700
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-15 08:53 +0200
Re: Relativistic Radio Poutnik <Poutnik4NNTP@gmail.com> - 2015-09-15 10:22 +0200
Re: Relativistic Radio Poutnik <poutnik4nntp@gmail.com> - 2015-09-13 09:08 +0200
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-14 09:32 -0500
Re: Relativistic Radio Odd Bodkin <bodkinodd@gmail.com> - 2015-09-14 09:31 -0500
Page 2 of 5 — ← Prev page 1 [2] 3 4 5 Next page →
| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-11 00:53 -0700 |
| Message-ID | <36ba79e3-27e5-4072-a70b-6216bc783def@googlegroups.com> |
| In reply to | #520116 |
Poutnik wrote... > As if a source is moving toward you, emission of a photon toward you decelerates it and the energy >is added to the photon energy. >OTOH, if the source is moving away of you, emission of a photon toward you accelerates >it and the energy is subtracted from the photon energy. >In both cases, the extra energy increases with the source speed. I can understand the above argument. But in that case how do you explain the same example, if you calculate the energy as *waves* not photons? As it is blueshifted for the observor moving towards the source. Because as I showed in my previous post you dont need to assume energy is lost or gained to the source, as the emitted radiation, if blueshifted to half its original wavelength, is observed for only half the time, and ends up being observed to be the same amount of energy as the source frames longer wavelength non blueshifted wave radiation. In other words you show how energy is being conserved *between frames*, without having to say its lost or gained to the source. So your argument contradicts itself depending on whether you use photons or waves. Energy is lost/ gained for the source only if light is a photon?
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-11 06:07 -0700 |
| Message-ID | <47ab64c9-120b-48bf-bcc0-c969a70b93d4@googlegroups.com> |
| In reply to | #520248 |
On Friday, 11 September 2015 08:53:40 UTC+1, jaymo...@hotmail.com wrote: > Poutnik wrote... > > As if a source is moving toward you, emission of a photon toward you decelerates it and the energy > >is added to the photon energy. > > >OTOH, if the source is moving away of you, emission of a photon toward you accelerates > >it and the energy is subtracted from the photon energy. > > >In both cases, the extra energy increases with the source speed. (Im reposting my last post again below...) I can understand the above argument. But in that case how do you explain the same example, if you calculate the energy in *waves* not photons. Although the emitted radiation is blueshifted to half its original wavelength (and at a higher energy per second), because the observer is moving at speed towards the source the waves emitted are also only visible for 1/2 the time to the observer. Hence the total blueshifted emr energy is still the same amount in both frames. (The two frames being: 1)the observer and source are not moving relative to each other and 2)the observer moves towards the source) The energy is being conserved *between frames*, in the wave version without having to show any lost or gained at the source. So your argument contradicts itself depending on whether you use photons or waves. With light modelled as photons you need to explain energy gained or lost to the source. With light as waves the same total energy is seen in any frame. No energy is lost to the source. You cant have both.
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| From | Poutnik <Poutnik4NNTP@gmail.com> |
|---|---|
| Date | 2015-09-11 15:08 +0200 |
| Message-ID | <msujlo$qpg$1@dont-email.me> |
| In reply to | #520248 |
On 09/11/2015 09:53 AM, jaymoseley@hotmail.com wrote: > In other words you show how energy is being conserved *between frames*, > without having to say its lost or gained to the source. So your argument > contradicts itself depending on whether you use photons or waves. > Energy is lost/ gained for the source only if light is a photon? No, you have misunderstood what I wrote. Energy is generally never conserved between frames, but within a frame. In frame A the energy EA is provided to produce a photon of energy EA. In frame B the energy EB is provided to produce a photon of energy EB. Energy conservation law is valid within a frame, not between frames. Linear momentum conservation law as well. Angular momentum conservation law as well. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes a great man humble, but a small man arrogant.
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| From | Poutnik <Poutnik4NNTP@gmail.com> |
|---|---|
| Date | 2015-09-11 15:28 +0200 |
| Message-ID | <msukr2$vs7$1@dont-email.me> |
| In reply to | #520277 |
On 09/11/2015 03:08 PM, Poutnik wrote: > On 09/11/2015 09:53 AM, jaymoseley@hotmail.com wrote: >> In other words you show how energy is being conserved *between frames*, >> without having to say its lost or gained to the source. So your argument >> contradicts itself depending on whether you use photons or waves. >> Energy is lost/ gained for the source only if light is a photon? > > No, you have misunderstood what I wrote. > Energy is generally never conserved between frames, but within a frame. > > In frame A the energy EA is provided to produce a photon of energy EA. > In frame B the energy EB is provided to produce a photon of energy EB. > Similarly, at wave macroscale level: In frame A the power PA is provided to produce wave of total power PA, equal to integral over the all solid angles of energy flow Eflow_A [J/s/steradian]. In frame B the power PB is provided to produce wave of total power PB, equal to integral over the all solid angles of energy flow Eflow_B [J/s/steradian]. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes a great man humble, but a small man arrogant.
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| From | Mahipal <mahipal7638@gmail.com> |
|---|---|
| Date | 2015-09-11 06:31 -0700 |
| Message-ID | <e157106a-d2c1-4558-80b1-eed8fcec142f@googlegroups.com> |
| In reply to | #520277 |
On Friday, September 11, 2015 at 9:08:59 AM UTC-4, Poutnik wrote: > On 09/11/2015 09:53 AM, jaymoseley@hotmail.com wrote: > > In other words you show how energy is being conserved *between frames*, > > without having to say its lost or gained to the source. So your argument > > contradicts itself depending on whether you use photons or waves. > > Energy is lost/ gained for the source only if light is a photon? > > No, you have misunderstood what I wrote. Big surprise that, given what you Pouting Poutnik write. > Energy is generally never conserved between frames, but within a frame. BS. > In frame A the energy EA is provided to produce a photon of energy EA. > In frame B the energy EB is provided to produce a photon of energy EB. It's written by a wondering Poutnik, so it must be true, as stated. > Energy conservation law is valid within a frame, not between frames. > Linear momentum conservation law as well. > Angular momentum conservation law as well. Thanks? > -- > Poutnik ( the Czech word for a wanderer ) > > Knowledge makes a great man humble, but a small man arrogant. Hey Potluck, which Phycology department issued your Chemistry degree? -- Mahipal
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| From | "Phony McNymster" <invalid@example.com> |
|---|---|
| Date | 2015-09-11 07:42 -0700 |
| Message-ID | <msup7p$sqs$1@speranza.aioe.org> |
| In reply to | #520282 |
> Hey Potluck, which Phycology department issued your Chemistry degree? > > -- Mapihal > Learn how to spell 'psychology', you blithering fuckwit. http://dictionary.reference.com/browse/psychology?s=t
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| From | Mahipal <mahipal7638@gmail.com> |
|---|---|
| Date | 2015-09-11 07:59 -0700 |
| Message-ID | <7d341317-3598-447b-81bc-d889b1e5a783@googlegroups.com> |
| In reply to | #520299 |
On Friday, September 11, 2015 at 10:42:08 AM UTC-4, Phony McNymster wrote: > > Hey Potluck, which Phycology department issued your Chemistry degree? > > > > -- Mapihal > > > Learn how to spell 'psychology', you blithering fuckwit. > > http://dictionary.reference.com/browse/psychology?s=t Why? I meant Phycology, a noun, the branch of botany concerned with seaweeds and other algae. -- Mahipal
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-11 07:16 -0700 |
| Message-ID | <ab664cc6-1513-4186-90f6-10be3af0e660@googlegroups.com> |
| In reply to | #520277 |
Poutnik wrote... > Energy is generally never conserved between frames, but within a frame. >in frame A the energy EA is provided to produce a photon of energy EA. In frame B the energy EB is provided to produce a photon of energy EB. >energy conservation law is valid within a frame, not between frames. Nonetheless as I have described in my last post, the source that emits the radiation does not need to lose or gain energy, regardless of the observors motion relative to the source. Calculating this where light is treated as a wave proves my point. In fact, contrary to your claim of conservation within frames, I would say that your arguments mean no conservation of energy in the source frame. Because other frames, like the observor frame, take it away.
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| From | Sam Wormley <swormley1@gmail.com> |
|---|---|
| Date | 2015-09-11 09:41 -0500 |
| Message-ID | <89mdnVb2ioq8dW_InZ2dnUU7-XGdnZ2d@giganews.com> |
| In reply to | #520294 |
On 9/11/15 9:16 AM, jaymoseley@hotmail.com wrote: > Nonetheless as I have described in my last post, the source that emits the radiation > does not need to lose or gain energy, regardless of the observors motion relative to > the source. Calculating this where light is treated as a wave proves my point. Impossible -- Law of Conservation of Energy applies. -- sci.physics is an unmoderated newsgroup dedicated to the discussion of physics, news from the physics community, and physics-related social issues.
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2015-09-11 09:49 -0500 |
| Message-ID | <msupm2$toc$1@speranza.aioe.org> |
| In reply to | #520294 |
On 9/11/2015 9:16 AM, jaymoseley@hotmail.com wrote: > In fact, contrary to your claim of conservation within frames, I would say that your > arguments mean no conservation of energy in the source frame. Because other frames, > like the observor frame, take it away. Jay, you can't go to a different frame to assess whether energy has been taken away from the source frame. This is what Poutnik is trying to explain -- energy is not expected to be the same in different frames. It never has been. Take even a basic example. A child throws a 1 kg apple at his brother inside a moving car at 2 m/s. Let's look at this in the frame of the source: the child's frame. The apple has a kinetic energy of (1/2)(1 kg)(2 m/s)^2 = 2 J. But now let's look at this very same apple in the frame of someone standing on the street, with the car going by at 20 m/s. Now the kinetic energy of the apple is (1/2)(1 kg)(22 m/s)^2 = 242 J. Now, did the observer on the street add 240 J to the apple? No. So how is it that 242 J vs. 2 J doesn't violate energy conservation? What's your answer to that question? -- Odd Bodkin --- maker of fine toys, tools, tables
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| From | Fabian Russell <root@localhost.localdomain> |
|---|---|
| Date | 2015-09-11 18:18 +0000 |
| Message-ID | <pan.2015.09.11.18.17.30@localhost.localdomain> |
| In reply to | #520301 |
On Fri, 11 Sep 2015 09:49:40 -0500, Odd Bodkin wrote: > > energy is not expected to be the same in different frames. It > never has been. > As long as no net momentum is carried away by the radiation, which is true for a radiating dipole, then the POWER is Lorentz invariant. Read carefully the reference (or any other discussion of relativistic electrodynamics): http://farside.ph.utexas.edu/teaching/em/lectures/node106.html Power is not energy. Power = d_energy/d_time, but for EM waves the power is more practically significant.
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| From | noTthaTguY <abu.kuanysh05@gmail.com> |
|---|---|
| Date | 2015-09-11 11:24 -0700 |
| Message-ID | <d9f416b4-d800-42c8-aa42-aec4916c174b@googlegroups.com> |
| In reply to | #520347 |
power is rate times elapsed time, watt-hours e.g > Read carefully the reference (or any other discussion of > relativistic electrodynamics): > > http://farside.ph.utexas.edu/teaching/em/lectures/node106.html > > Power is not energy. Power = d_energy/d_time, but for EM waves > the power is more practically significant.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-12 11:20 +0200 |
| Message-ID | <mt0qm3$31k$1@dont-email.me> |
| In reply to | #520347 |
Dne 11/09/2015 v 20:18 Fabian Russell napsal(a): > On Fri, 11 Sep 2015 09:49:40 -0500, Odd Bodkin wrote: > >> >> energy is not expected to be the same in different frames. It >> never has been. >> > > As long as no net momentum is carried away by the radiation, > which is true for a radiating dipole, then the POWER is Lorentz > invariant. > As long as... -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-12 06:21 -0700 |
| Message-ID | <21f2e5f7-8c5e-425f-95a5-caf4ca52c49c@googlegroups.com> |
| In reply to | #520301 |
Odd wrote.. > Take even a basic example. A child throws a 1 kg apple at his brother inside a moving car at 2 m/s. Let's look at this in the frame of the source: the child's frame. The apple has a kinetic energy of (1/2)(1 kg)(2 m/s)^2 = 2 J. But now let's look at this very same apple in the frame of someone standing on the street, with the car going by at 20 m/s. Now the kinetic energy of the apple is (1/2)(1 kg)(22 m/s)^2 = 242 J. Now, did the observer on the street add 240 J to the apple? No. Lets look.at your example in the context of the original question. If I shine a light on a detector for one time frame *only*( sec or whatever) At a specific frequency/ wavelength. What amount of energy does the detector recieve if, as fabian initially specified, all the emr emitted were detected? Now, ... move the detector towards the source as its emitting its one time frame of light.(At such a speed as to make the detectors observed wavelength halve). This should make the time frame at which the emitted light is observed to be observed by the detector for only * half* the amount of time. Now answer my second question: Does one second of observed radiation have the same amount of energy as one half of a second of observed radiation that is double the wavelength? (Of course Im assuming the conversion tables between wavelength and energy of emr correctly indicate that one half wavelength emr is double the energy of one wavelength of emr.) If you can answer that question for me then consider this question. Does the shortened wavelength give the same total energy to the detector as the double wavelength in half the time?, ( yes it should) and does your example of the child throwing the ball conflict with this conclusion?
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-13 02:04 +0200 |
| Message-ID | <mt2efc$vdf$1@dont-email.me> |
| In reply to | #520479 |
Dne 12/09/2015 v 15:21 jaymoseley@hotmail.com napsal(a): > Now, ... move the detector towards the source as its emitting its one time > frame of light.(At such a speed as to make the detectors observed wavelength > halve). This should make the time frame at which the emitted light > is observed to be observed by the detector for only * half* the amount > of time. Now answer my second question: > > Does one second of observed radiation have the same amount of energy > as one half of a second of observed radiation that is double the wavelength? > (Of course Im assuming the conversion tables between wavelength > and energy of emr correctly indicate that one half wavelength emr is double > the energy of one wavelength of emr.) You may have forgotten to include into consideration increased rate of absorption, moving toward the source, comparing to emission rate. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-13 00:07 -0700 |
| Message-ID | <17557737-c88e-4791-93f2-e8e8b5e2f108@googlegroups.com> |
| In reply to | #520560 |
Poutnik wrote... > You may have forgotten to include into consideration increased rate of absorption, moving toward the source, comparing to emission rate. I dont think so. I specified that in one frame the absorbtion rate was double because the emr had been blueshifted to half its wavelength. But,.. because it was only visible for half the amount of time the total energy absorbed by the moving detector is the same as when it wasnt moving relative to the source. In other words, the known conversion rates of nm - ev are not compatible with the photon model. No extra energy from momentum is gained by the detector regardless of how fast it moves relative to the source. Contrary to the photon model which erroneously claims that the source imparts extra energy to the detector if the source moves towards the detector. I know people will scream... "oh no, if you blueshift 200nm emr to 100 nm emr that will double the energy absorbed by the detector ..." But dont forget....if the source emits only 10 seconds of 200nm radiation that will be only observed for only 5 seconds if that emr is blueshifted to 100nm at the detector.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-13 09:22 +0200 |
| Message-ID | <mt383q$sd4$1@dont-email.me> |
| In reply to | #520627 |
Dne 13/09/2015 v 09:07 jaymoseley@hotmail.com napsal(a): > Poutnik wrote... >> You may have forgotten to include into consideration increased rate of absorption, moving toward the source, comparing to emission rate. > > I dont think so. I specified that in one frame the absorbtion rate was double > because the emr had been blueshifted to half its wavelength. But,.. because it > was only visible for half the amount of time the total energy absorbed by the > moving detector is the same as when it wasnt moving relative to the source. You omitted the fact the emission interval does not equal to absorption interval. So not only 1 photon contains more energy, but photon incoming rate is higher , with shorter interval, so total count of photons is the same, but total photon energy is double. The source provides more energy per photon, compared to still case, as you have already agreed previously. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-13 09:59 +0200 |
| Message-ID | <mt3a92$2ll$1@dont-email.me> |
| In reply to | #520630 |
Dne 13/09/2015 v 09:22 Poutnik napsal(a): > You omitted the fact the emission interval > does not equal to absorption interval. > > So not only 1 photon contains more energy, > but photon incoming rate is higher , with shorter interval, > so total count of photons is the same, > but total photon energy is double. P.S.:By other words, the speed how fast you fly against the source of finite count of photons or bullets aiming on you does not affect the count that hits you, but only time interval when it happens. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-13 03:44 -0700 |
| Message-ID | <5a208794-8599-4da6-b5f1-9bef74dbeb44@googlegroups.com> |
| In reply to | #520630 |
Poutnik wrote. >>> I dont think so. I specified that in one frame the absorbtion rate was double >>> because the emr had been blueshifted to half its wavelength. But,.. because it >>> was only visible for half the amount of time the total energy absorbed by the > moving detector is the same as when it wasnt moving relative to the >>>source. >You omitted the fact the emission interval does not equal to absorption interval. >So not only 1 photon contains more energy, but photon incoming rate is higher , with shorter interval, so total count of photons is the same, but total >photon energy is double. In the photon model those 10 emitted photons when observed by the detector (as the source moves towards the detector) have their energy increased. Therefore, in the photon model the total energy detected will increase from total energy emitted if the source moves towards the detector. I believe thats what you are saying. But in waves when you shorten a wave to half its length you preserve its energy. You DO NOT increase a waves energy when you blueshift it to half its length. (You only increase the energy *per second*) The observed conversion tables support this conclusion. If the photon model were correct then the well accepted observation that one second of 100 nm emr is double the energy of 1 second of 200 nm emr....is false!! Obviously the photon model is contradicted by observation and the nm to ev conversions which state NO energy is added when blueshifted. Dont forget, in the gedanken 10 seconds of emitted radiation, when blueshifted to half its length...only lasts for 5 seconds.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-14 07:33 +0200 |
| Message-ID | <mt5m49$uj4$1@dont-email.me> |
| In reply to | #520638 |
Dne 13/09/2015 v 12:44 jaymoseley@hotmail.com napsal(a): > Poutnik wrote. >>>> I dont think so. I specified that in one frame the absorbtion rate was double >>>> because the emr had been blueshifted to half its wavelength. But,.. because it >>>> was only visible for half the amount of time the total energy absorbed by the > moving detector is the same as when it wasnt moving relative to the >>>source. > >> You omitted the fact the emission interval does not equal to absorption interval. > >> So not only 1 photon contains more energy, but photon incoming rate is higher , with shorter interval, so total count of photons is the same, but total >photon energy is double. > > In the photon model those 10 emitted photons when observed by the detector (as > the source moves towards the detector) have their energy increased. > Therefore, in the photon model the total energy detected will increase from > total energy emitted if the source moves towards the detector. > I believe thats what you are saying. No. Both emitted and absorbed energies increase. > > But in waves when you shorten a wave to half its length you preserve its energy. > You DO NOT increase a waves energy when you blueshift it to half its length. > (You only increase the energy *per second*) > The observed conversion tables support this conclusion. If the photon > model were correct then the well accepted observation that one second > of 100 nm emr is double the energy of 1 second of 200 nm emr....is false!! Intensities are ignored here. > Obviously the photon model is contradicted by observation and the nm to ev > conversions which state NO energy is added when blueshifted. > > Dont forget, in the gedanken 10 seconds of emitted radiation, when blueshifted > to half its length...only lasts for 5 seconds. > There is missing evaluation of intensity, according to relativistic Doppler effect. The Lorentz invariant is intensity divided by frequency^3 -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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