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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 5 of 5 — ← Prev page 1 2 3 4 [5]
| From | Poutnik <Poutnik4NNTP@gmail.com> |
|---|---|
| Date | 2015-09-11 18:00 +0200 |
| Message-ID | <msutns$5pd$1@dont-email.me> |
| In reply to | #520312 |
On 09/11/2015 05:38 PM, Poutnik wrote: > There is no energy exchange between frames, > as frames are human constructs, helping us > to observe, describe and evaluate reality. P.S.> Within each frame, objects of nature are evaluated according to the same rules, but evaluation provides generally different values, with exception of invariant quantities. Invariant values are constant for all frames, as their value does not depend on used coordinate system. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes a great man humble, but a small man arrogant.
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2015-09-11 11:22 -0500 |
| Message-ID | <msuv3d$bom$1@speranza.aioe.org> |
| In reply to | #520312 |
On 9/11/2015 10:38 AM, Poutnik wrote: > On 09/11/2015 04:16 PM, jaymoseley@hotmail.com wrote: >> 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. > > The source in every frame gains the same power > as it spends by the emission, even if > this power is different in different frames. > >> >> 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. >> > Therefore energy is conserved in every frame. > > There is no energy exchange between frames, > as frames are human constructs, helping us > to observe, describe and evaluate reality. > I'm beginning to think there is a fundamental misconception about reference frames that I've seen from others. Some people imagine a source and a detector that are moving relative to each other. In their thinking, the source belongs in one frame and the detector belongs in another. There is then thinking that energy conservation implies that the energy emitted in the source frame should equal the energy received in the detector frame. This is of course wrong. In reality, the source and the detector are both in the source frame, with the source stationary in this frame and the detector moving in this frame. In this frame the energy emitted is the same as the energy received. Also in reality, the source and the detector are both in the detector frame, with the source moving in this frame and the detector stationary. In this frame, the energy emitted is the same as the energy received, though these two numbers are not the same as they are in the source frame. -- Odd Bodkin --- maker of fine toys, tools, tables
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-12 12:29 -0700 |
| Message-ID | <6eb355ae-0ce9-4a02-8cf3-bdbacba8a07b@googlegroups.com> |
| In reply to | #520318 |
Odd wrote... >In reality, the source and the detector are both in the source frame, with the source stationary in this frame and the detector moving in this frame. In this frame the energy emitted is the same as the energy received. Also in reality, the source and the detector are both in the detector frame, with the source moving in this frame and the detector stationary. In this frame, the energy emitted is the same as the energy received, though these two numbers are not the same as they are in the source frame. This doesnt quite add up to me.You say the numbers are different in each frame. If I eject a ball at speed x at a target that moves towards the ball ejector at speed y, the ball will hit the target at a combined speed of x+y. If I keep the target still and move the ball ejector towards the target at speed y and eject the ball at speed x. Then the ball hits the target at speed x+y. Thats the same number for both frames isnt it?
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-13 01:16 +0200 |
| Message-ID | <mt2bkn$nmt$1@dont-email.me> |
| In reply to | #520526 |
Dne 12/09/2015 v 21:29 jaymoseley@hotmail.com napsal(a): > Odd wrote... >> In reality, the source and the detector are both in the source frame, with the source stationary in this frame and the detector moving in this frame. In this frame the energy emitted is the same as the energy received. Also in reality, the source and the detector are both in the detector frame, with the source moving in this frame and the detector stationary. In this frame, the energy emitted is the same as the energy received, though these two numbers are not the same as they are in the source frame. > > This doesnt quite add up to me.You say the numbers are different in each frame. > If I eject a ball at speed x at a target that moves towards the ball ejector at speed y, > the ball will hit the target at a combined speed of x+y. > If I keep the target still and move the ball ejector towards the target at speed y > and eject the ball at speed x. Then the ball hits the target at speed x+y. > Thats the same number for both frames isnt it? > Yes, as it relates not to speed, but to closing speed. Closing speed is in Galileo transformation context invariant, in contrary to energy or speed. In the frame where the ball ejector is still, the ball gains kinetic energy 1/2 . m . x^2. In the frame where the ball ejector moves by speed y, the ball gains kinetic energy 1/2 . m . [(x+y)^2 - y^2] = 1/2 . m . [x^2 + 2.x.y], having initially already energy 1/2 . m . y^2 and finally energy 1/2 . m . (x+y)^2. -- 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 23:09 -0700 |
| Message-ID | <0733dab1-fdab-43ae-8d06-113152095cd8@googlegroups.com> |
| In reply to | #520551 |
Poutnik wrote... > In the frame where the ball ejector is still, the ball gains kinetic energy 1/2 . m . x^2. >In the frame where the ball ejector moves by speed y, the ball gains kinetic energy 1/2 . m . [(x+y)^2 - y^2] = 1/2 . m . [x^2 + 2.x.y], having initially already >energy 1/2 . m . y^2 and finally energy 1/2 . m . (x+y)^2. In your first example above where the ball ejector is still you have the ball gaining kinetic energy 1/2.m.x^2. But you forgot to take into account the fact that the target is moving towards the ball ejector at speed y in this example. Therefore ball hits the moving target at speed x+y . Not at speed x So shouldnt the calculation for the first example be the same as the second example? Ie... 1/2 . m . (x+y)^2. The important speed is not how fast the ball moves relative to the ejector, but rather how fast it hits the target.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-13 08:35 +0200 |
| Message-ID | <mt35ci$lbc$1@dont-email.me> |
| In reply to | #520623 |
Dne 13/09/2015 v 08:09 jaymoseley@hotmail.com napsal(a): > Poutnik wrote... >> In the frame where the ball ejector is still, the ball gains kinetic energy 1/2 . m . x^2. >> In the frame where the ball ejector moves by speed y, the ball gains kinetic energy 1/2 . m . [(x+y)^2 - y^2] = 1/2 . m . [x^2 + 2.x.y], having initially already >energy 1/2 . m . y^2 and finally energy 1/2 . m . (x+y)^2. > > In your first example above where the ball ejector is still you have the ball gaining > kinetic energy 1/2.m.x^2. > But you forgot to take into account the fact that the target is moving towards > the ball ejector at speed y in this example. Therefore ball hits the moving target > at speed x+y . Not at speed x No, I did not. You confuse speed and closing speed. Speed of the ball is x, while closing speed bass vs target is x+y. Kinetic energy of an object is defined by its speed, not by its closing speed wrt other object. > So shouldnt the calculation for the first example be the same as the second example? > Ie... 1/2 . m . (x+y)^2. No, it should not. The ball energy in the still ejector case is 1/2 . m . x^2. > The important speed is not how fast the ball moves relative to the ejector, but > rather how fast it hits the target. It is the same, unless there is acceleration, what is not the case. Both are the same quantity, closing speed ball vs target. Speed is how fast an object move wrt a frame. Closing speed is how fast an object moves wrt other object. If the other object is coincidentally still in the frame, both quantities are numerically equal. The difference is both frames is, that in the still ejector case, energy released at collision comes from both the ball and target, while in the still target case all the collision energy comes from the ball. Also, in both frames the objects have different energy after collision. -- 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 08:44 +0200 |
| Message-ID | <mt35tg$mi5$1@dont-email.me> |
| In reply to | #520624 |
Dne 13/09/2015 v 08:35 Poutnik napsal(a): > Also, in both frames the objects have different energy after collision. Particular values depend on if the collision is elastic ( the ball bounces ), or not elastic ( the ball is catched ) In both cases the final state is calculated form momentum conservation law and in case of elastic collision also from kinetic energy conservation law. ( saved in elastic, but not in not elastic collision, where part converts to different energy, usually thermal one. ) -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-12 23:50 -0700 |
| Message-ID | <90b771ad-a829-4473-8cc4-d5876b4d51e1@googlegroups.com> |
| In reply to | #520624 |
Poutnik wrote... > The difference is both frames is, that in the still ejector case, energy released at collision comes from both the ball and target, while in the still target case all the collision energy comes from the ball. Who cares how fast the ball moves in the still ejector case compared to how fast it moves when the ejector moves. Thats not being evaluated. Whats important is how fast the ball hits the target in both cases. And thats equal in both cases. Therefore the energy imparted from the ball to the target is equal in both cases.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-13 09:14 +0200 |
| Message-ID | <mt37km$r3d$1@dont-email.me> |
| In reply to | #520626 |
Dne 13/09/2015 v 08:50 jaymoseley@hotmail.com napsal(a): > Poutnik wrote... >> The difference is both frames is, that in the still ejector case, energy released at collision comes from both the ball and target, while in the still target case all the collision energy comes from the ball. > > Who cares how fast the ball moves in the still ejector case compared to > how fast it moves when the ejector moves. Thats not being evaluated. > Whats important is how fast the ball hits the target in both cases. > And thats equal in both cases. Therefore the energy imparted from the ball > to the target is equal in both cases. Even in case the ball is still and target moves x+y speed ? (smiling) Total collision energy is the same. Ratio of how 2 objects share this spent energy is not. In still target frame the ejector provides to the ball an extra energy, that is provided by the target in the frame with moving target. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-13 05:24 -0700 |
| Message-ID | <10e4dfee-ead9-4158-8f05-3ee372dd0fb0@googlegroups.com> |
| In reply to | #520629 |
Poutnik... > Even in case the ball is still and target moves x+y speed ? (smiling) Total collision energy is the same. Ratio of how 2 objects share this spent energy is not. In still target frame the ejector provides to the ball an extra energy, that is provided by the target in the frame with moving target. Im not sure what you mean here. But if a moving ball hits your hand at 30 kph there will be no diffeence in pain to you than if your moving hand hits a stationary ball at 30 mph
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-14 07:35 +0200 |
| Message-ID | <mt5m80$uj4$2@dont-email.me> |
| In reply to | #520655 |
Dne 13/09/2015 v 14:24 jaymoseley@hotmail.com napsal(a): > Poutnik... >> Even in case the ball is still and target moves x+y speed ? (smiling) > > Total collision energy is the same. Ratio of how 2 objects share this spent energy is not. > > In still target frame the ejector provides to the ball an extra energy, that is provided by the target in the frame with moving target. > > > Im not sure what you mean here. But if a moving ball hits your hand at 30 kph there will be no > diffeence in pain to you than if your moving hand hits a stationary ball at 30 mph > Sure, but in one frame the energy come from the ball, in the other frame the energy comes from your hand. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-14 01:01 -0700 |
| Message-ID | <c7293e8d-5749-4d29-81a0-3ab19f217743@googlegroups.com> |
| In reply to | #520954 |
Poutnik wrote.. > Sure, but in one frame the energy come from the ball, in the other frame the energy comes from your hand My original point was not to work out where one models it as coming from. But how much gets detected. And so in both frames the hand detects the same pain, regardless of where it comes from.
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| From | Poutnik <Poutnik4NNTP@gmail.com> |
|---|---|
| Date | 2015-09-14 10:39 +0200 |
| Message-ID | <mt610r$ut2$1@dont-email.me> |
| In reply to | #520964 |
On 09/14/2015 10:01 AM, jaymoseley@hotmail.com wrote: > Poutnik wrote.. >> Sure, but in one frame the energy come from the ball, in the other frame the energy comes from your hand > > My original point was not to work out where one models it as coming from. > But how much gets detected. And so in both frames the hand detects the same > pain, regardless of where it comes from. > We can agree with that. But it is not much about models. It is the classical mechanic from the 17th century. Note that if light and/or relativistic speeds are involved, only some analogies remains, as some things are different. -- Poutnik ( the Czech word for a wanderer ) Knowledge makes a great man humble, but a small man arrogant.
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2015-09-14 09:34 -0500 |
| Message-ID | <mt6ltp$1pp$5@speranza.aioe.org> |
| In reply to | #520626 |
On 9/13/2015 1:50 AM, jaymoseley@hotmail.com wrote: > > Who cares how fast the ball moves in the still ejector case compared to > how fast it moves when the ejector moves. Thats not being evaluated. > Whats important is how fast the ball hits the target in both cases. No, that's not so, Jay. If you work through some elastic and inelastic collision problems in a freshman book, you'll see this. > And thats equal in both cases. Therefore the energy imparted from the ball > to the target is equal in both cases. > The momentum exchange is the same in both cases. The energy exchange is not. -- Odd Bodkin --- maker of fine toys, tools, tables
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| From | jaymoseley@hotmail.com |
|---|---|
| Date | 2015-09-14 12:31 -0700 |
| Message-ID | <ae88f055-35ee-49e0-b6ba-26ea9b064641@googlegroups.com> |
| In reply to | #521033 |
Odd wrote.. > The momentum exchange is the same in both cases. The energy exchange is not This is too vague, as it doesnt specify how much energy the target detects in each case. Does a detector measure the same amount of energy from a light source regardless of whether it moves or the source does? My understanding is yes it does. Your answer, couched in unwritten caveats, hyperbole, obfuscation and misinformation, suggests to me, that a target like a detector measures different energies from a source depending on whether it or the source moves.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-15 08:53 +0200 |
| Message-ID | <mt8f5t$1fc$1@dont-email.me> |
| In reply to | #521116 |
Dne 14/09/2015 v 21:31 jaymoseley@hotmail.com napsal(a): > Odd wrote.. >> The momentum exchange is the same in both cases. The energy exchange is not > > ...................Your answer, couched in unwritten > caveats, hyperbole, obfuscation and misinformation, suggests to me, > that a target like a detector measures different energies from a source > depending on whether it or the source moves. Yes, it does, as illustrated on laser Doppler cooling case in other post. https://en.wikipedia.org/wiki/Doppler_cooling -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | Poutnik <Poutnik4NNTP@gmail.com> |
|---|---|
| Date | 2015-09-15 10:22 +0200 |
| Message-ID | <mt8kd7$hgk$1@dont-email.me> |
| In reply to | #521233 |
On 09/15/2015 08:53 AM, Poutnik wrote: > Dne 14/09/2015 v 21:31 jaymoseley@hotmail.com napsal(a): >> Odd wrote.. >>> The momentum exchange is the same in both cases. The energy exchange is not >> >> ...................Your answer, couched in unwritten >> caveats, hyperbole, obfuscation and misinformation, suggests to me, >> that a target like a detector measures different energies from a source >> depending on whether it or the source moves. > > Yes, it does, as illustrated on laser Doppler cooling case > in other post. > > https://en.wikipedia.org/wiki/Doppler_cooling > There is even simpler, closely related example: thermal Doppler broadening of atomic absorption lines. Atoms have well defined transition energies, related to very narrow range of absorbed wavelength, with the width defined mostly by uncertainty principle. But their thermal motion causes Doppler effect. The consequence is, an atom approaching the light source absorbs a photon with energy lower than the nominal one for atomic transition. OTOH, an atom departing from the light source absorbs a photon with energy higher then nominal one for the atomic transition. This causes absorption line to be broadened to a shape very similar to Gaussian peak, widening with temperature. AND NOW : In the frame of the static light source, an approaching atom tops the energy deficit of lower-energy-than-needed photon by its kinetic energy. OTOH, a departing atom uses the extra energy of higher-energy-than-needed photon to increase its own kinetic energy. WHILE In the frame of the static absorbing atom, For the approaching atom from above, the light source is approaching. The lower-energy-than-needed photon observed in the first frame has in this frame the needed energy, as it is blue-shifted. OTOH, For the departing atom from above, the light source is departing. The higher-energy-than-needed photon observed in the first frame has in this frame the needed energy, as it is red-shifted. https://en.wikipedia.org/wiki/Doppler_broadening -- Poutnik ( the Czech word for a wanderer ) Knowledge makes a great man humble, but a small man arrogant. Eventual Wikipedia articles are provided with intention of a convenient reference, not as an evidence, argument, and usually not as a primary source of my knowledge.
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| From | Poutnik <poutnik4nntp@gmail.com> |
|---|---|
| Date | 2015-09-13 09:08 +0200 |
| Message-ID | <mt3791$q4q$1@dont-email.me> |
| In reply to | #520624 |
Dne 13/09/2015 v 08:35 Poutnik napsal(a): >> The important speed is not how fast the ball moves relative to the ejector, but >> rather how fast it hits the target. Both are not speeds, but closing speeds. Speed is defined as size of velocity vector. Closing speed is defined as size of difference of 2 velocity vectors. > It is the same, unless there is acceleration, what is not the case. > Both are the same quantity, closing speed ball vs target. I have misread your sentence, so I revoke this my response above. Both the ball *speed* ( not ball-vs ejector closing speed ) and ball-target closing speed are important. As the speed defines* kinetic energy of the ball and the target, while ball-target closing speed defines* for inelastic collision case, how much of *total* kinetic energy from *both* objects is converted. If the target was still, all collision energy comes from the moving ball. If the ball was still, all collision energy comes from the moving target. If both were moving, collision energy comes from both. In both frames, initial and final kinetic energy of objects are different, for both elastic and not elastic cases. * together with involved masses -- Poutnik ( the Czech word for a wanderer ) Knowledge makes great men humble, but small men arrogant.
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2015-09-14 09:32 -0500 |
| Message-ID | <mt6lqb$1pp$3@speranza.aioe.org> |
| In reply to | #520623 |
On 9/13/2015 1:09 AM, jaymoseley@hotmail.com wrote: > > In your first example above where the ball ejector is still you have the ball gaining > kinetic energy 1/2.m.x^2. > But you forgot to take into account the fact that the target is moving towards > the ball ejector at speed y in this example. Therefore ball hits the moving target > at speed x+y . Not at speed x > So shouldnt the calculation for the first example be the same as the second example? > Ie... 1/2 . m . (x+y)^2. > The important speed is not how fast the ball moves relative to the ejector, but > rather how fast it hits the target. > Jay, this is material that is covered in the collisions (elastic and inelastic) chapter of a freshman textbook. Have you tried working through some problems there? -- Odd Bodkin --- maker of fine toys, tools, tables
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2015-09-14 09:31 -0500 |
| Message-ID | <mt6lnb$1pp$1@speranza.aioe.org> |
| In reply to | #520526 |
On 9/12/2015 2:29 PM, jaymoseley@hotmail.com wrote: > Odd wrote... >> In reality, the source and the detector are both in the source frame, with the source >> stationary in this frame and the detector moving in this frame. In this frame the energy >> emitted is the same as the energy received. Also in reality, the source and the detector >> are both in the detector frame, with the source moving in this frame and the detector >> stationary. In this frame, the energy emitted is the same as the energy received, though >> these two numbers are not the same as they are in the source frame. > > This doesnt quite add up to me.You say the numbers are different in each frame. > If I eject a ball at speed x at a target that moves towards the ball ejector at speed y, > the ball will hit the target at a combined speed of x+y. > If I keep the target still and move the ball ejector towards the target at speed y > and eject the ball at speed x. Then the ball hits the target at speed x+y. > Thats the same number for both frames isnt it? > If you're in a frame where the ball-thrower is at rest and the ball-catcher is moving, then the ball has a kinetic energy of (1/2)m(x)^2 and the ball-cather has a kinetic energy of (1/2)M(y)^2. There is nothing in either frame that has kinetic energy that is proportional to (x+y)^2. To find out how much the energy of impact is (say in an inelastic collision), then you have to solve conservation of momentum equations to find out how much the post-collision system is moving and then you'll know how much kinetic energy was converted. -- Odd Bodkin --- maker of fine toys, tools, tables
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