Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]
Groups > sci.physics.relativity > #479202 > unrolled thread
| Started by | RichD <r_delaney2001@yahoo.com> |
|---|---|
| First post | 2018-12-06 10:49 -0800 |
| Last post | 2018-12-12 13:44 -0500 |
| Articles | 20 on this page of 108 — 17 participants |
Back to article view | Back to sci.physics.relativity
the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-06 10:49 -0800
Re: the famous formula pnalsing@gmail.com - 2018-12-06 11:30 -0800
Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-06 11:43 -0800
Re: the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-07 10:15 -0800
Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-07 11:13 -0800
Re: the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-09 00:02 -0800
Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-09 15:23 -0800
Re: the famous formula Libor Striz <poutnik@privacy.net> - 2018-12-10 09:54 +0100
Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-10 05:43 -0800
Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-12 11:08 -0600
Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-12 09:35 -0800
Re: the famous formula Rich Carpenter <lcug@kigcohc.pa> - 2018-12-12 19:20 +0000
Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-12 19:28 -0600
Re: the famous formula Scott Richman <sunn@idhnonow.iv> - 2018-12-13 16:13 +0000
Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2018-12-14 04:56 -0800
Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2018-12-14 05:23 -0800
Re: the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-19 13:06 -0800
Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-19 20:32 -0600
Re: the famous formula RichD <r_delaney2001@yahoo.com> - 2018-12-27 13:40 -0800
Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-12 10:28 -0600
Re: the famous formula dlzc <dlzc1@cox.net> - 2018-12-12 09:12 -0800
Re: the famous formula Tom Roberts <tjroberts137@sbcglobal.net> - 2018-12-14 12:21 -0600
Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2019-01-06 09:20 -0800
Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2019-01-06 09:37 -0800
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-06 20:50 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-06 22:47 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-07 12:06 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-07 16:15 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 14:18 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-08 13:43 +0000
Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2018-12-08 06:01 -0800
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 18:18 +0100
Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-08 19:26 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-09 19:35 +0100
Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-09 20:24 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-10 18:48 +0100
Re: the famous formula Rich Carpenter <lcug@kigcohc.pa> - 2018-12-10 19:37 +0000
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 22:27 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 22:35 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-12 16:45 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-13 17:21 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-09 19:03 +0100
Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-09 20:37 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-09 22:20 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-10 18:40 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 21:59 +0100
Re: the famous formula "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2018-12-18 05:33 -0800
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-22 14:51 +0100
Re: the famous formula "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2018-12-22 15:30 -0800
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-23 01:34 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-10 21:41 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-11 16:31 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-11 18:24 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-12 16:32 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-12 16:27 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-14 15:13 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-14 14:28 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-15 13:35 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-15 13:52 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-15 19:50 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-15 20:33 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-17 11:13 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-18 04:45 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-18 18:50 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-21 22:48 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-15 22:02 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-17 11:01 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-17 15:28 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-18 18:42 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2019-01-06 18:45 +0100
Re: the famous formula Steve BH <sbharris@gmail.com> - 2019-01-01 16:54 -0800
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2019-01-06 17:17 +0100
Re: the famous formula "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> - 2019-01-06 08:35 -0800
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2019-01-07 14:56 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2019-01-06 18:47 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-15 12:55 +0000
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-13 23:14 +0100
[OT] Work (was: the famous formula) Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-11 21:56 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-08 20:14 +0100
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-09 19:22 +0100
Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-07 17:57 +0100
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 14:03 +0100
Re: the famous formula Gary Harnagel <hitlong@yahoo.com> - 2018-12-08 05:29 -0800
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 18:43 +0100
[OT] Work (was: the famous formula) Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-08 21:53 +0100
Re: [OT] Work Thomas Heger <ttt_heg@web.de> - 2018-12-09 19:08 +0100
Re: [OT] Work Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-09 19:31 +0100
Re: [OT] Work Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-09 18:58 +0100
Re: [OT] Work Thomas Heger <ttt_heg@web.de> - 2018-12-10 18:52 +0100
Re: [OT] Work Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-10 22:28 +0100
Re: [OT] Work Thomas Heger <ttt_heg@web.de> - 2018-12-11 16:36 +0100
Re: [OT] Work Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-11 20:24 +0100
Re: the famous formula Gary Harnagel <hitlong@yahoo.com> - 2018-12-08 14:02 -0800
Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-08 16:56 +0100
Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 16:05 +0000
Re: the famous formula Thomas Heger <ttt_heg@web.de> - 2018-12-08 18:35 +0100
Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 17:58 +0000
Re: the famous formula Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> - 2018-12-08 19:34 +0100
Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 18:54 +0000
Re: the famous formula Ejí Utorka <edew@aewedoc.cl> - 2018-12-08 16:09 +0000
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-06 22:16 +0100
Re: the famous formula Steve BH <sbharris@gmail.com> - 2019-01-01 18:15 -0800
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2019-01-02 22:10 +0100
Re: the famous formula Odd Bodkin <bodkinodd@gmail.com> - 2018-12-07 16:07 +0000
Re: the famous formula Steve BH <sbharris@gmail.com> - 2019-01-01 17:22 -0800
Re: the famous formula Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-12-07 22:47 +0100
Re: the famous formula "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2018-12-09 15:06 -0800
Re: the famous formula benj <benj.fake@gmail.com> - 2018-12-12 13:44 -0500
Page 3 of 6 — ← Prev page 1 2 [3] 4 5 6 Next page →
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-13 17:21 +0100 |
| Message-ID | <693bbe47-497c-8153-824e-6d6edc03b9e2@PointedEars.de> |
| In reply to | #479685 |
Thomas Heger wrote: > Am 10.12.2018 um 22:27 schrieb Thomas 'PointedEars' Lahn: >>> Now you provided a mathematical proof, that collages produce brainwashed >>> zombies. >> Blithering fool, I’ll remind you of that the next time you are cluelessly >> babbling about the purely mathematical concept of Minkowski spacetime as >> formulated by Minkowski in 1908. > > My assumption is, that Minkowski's spacetime is real. Which is why your argument is inconsistent. Minkowski spacetime is originally a *purely* mathematical construct, yet you claim it is real. (Moreover, you claim that it is real while special relativity is not, ignoring that special relativity and Minkowski spacetime are *one and the same thing*.) Now, all observation indicates that Minkowski spacetime *is* real *locally*. But *so is the definition of work* as it has been given to you. *That* is the problem with your argument: You are not accepting *all* observable reality, only the *part* of observable reality that you *like* (because you are still able to understand it). That is NOT how science, NOT how a *scientist* *to be taken seriously*, are working. Scientists do NOT pick and choose, they formulate their conclusions based on *all* available information. It is only crackpots, *wannabe* scientists, who pick and choose. -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail.
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-09 19:03 +0100 |
| Message-ID | <ce05465f-3b47-a93b-cb02-18232e25148f@PointedEars.de> |
| In reply to | #479472 |
Thomas Heger wrote: > Am 08.12.2018 um 19:26 schrieb Libor Striz: >> Thomas Heger <ttt_heg@web.de> Wrote in message: >>> Vectors have a direction. >>> A Joule for instance or a kilo-Watt-hour has no direction. >> >> Why do you then say work is a vector ? >> The unit of work is 1 Joule. > > We have a similar problem with 'speed' and 'velocity'. We do not have a problem. You have one. Your problem is that you have not the slightest clue what you are talking about, and that either you have an attention deficiency that lets you forget what you said before – or you are simply a pathological liar: ,-<g6v2h6Fb0lhU1@mid.individual.net> | | […] | | Work is a vectorial unit and has a direction and can eventually become | negative [sic!] -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail.
[toc] | [prev] | [next] | [standalone]
| From | Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> |
|---|---|
| Date | 2018-12-09 20:37 +0100 |
| Message-ID | <pujqui$prm$1@dont-email.me> |
| In reply to | #479472 |
Thomas Heger <ttt_heg@web.de> Wrote in message: > > > Energy flow has a direction and energy has not. Neither the work has. Read any textbook from a high school or better how the term work is defined on context of physics. You will realize it is a scalar. > So work and energy are not the same thing. There is no so, as the latter cannot be implied from the former, no matter if the latter is true or not. > > They are related, but not equal. > You confuse equal and identical. Non identical quantities can still be equal. And the are not identical, as work has 2 meanings: 1/ a phenomena of a process of transformation of energy to a mechanical energy (in narrow sense) or non-thermal energy ( wider sense, e.g electrochemical work ) 2/ the amount of this transformed energy. > > -- 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/
[toc] | [prev] | [next] | [standalone]
| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-12-09 22:20 +0000 |
| Message-ID | <puk4et$1kui$3@gioia.aioe.org> |
| In reply to | #479472 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 08.12.2018 um 19:26 schrieb Libor Striz: >> Thomas Heger <ttt_heg@web.de> Wrote in message: >> >>> >>> Vectors have a direction. >>> >>> A Joule for instance or a kilo-Watt-hour has no direction. >> >> Why do you then say work is a vector ? >> The unit of work is 1 Joule. > > > We have a similar problem with 'speed' and 'velocity'. > > Both have units like m/s , but one is a vector quantity (velocity) and > one is not (speed). > > Speed is the magnitude of velocity. > > https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html > >> The acting force has direction. >> The path along which the force is acting has direction as well >> But the work as integral of scalar product of these 2 vectors is a >> scalar, without direction. > > work is NOT 'forth times path' > > but ' force times distance'. No, it is not that either. It is the integral of the scalar product between two vectors, the force and the differential displacement. If the force varies over location, such as with a spring, then the force cannot be pulled out of the integral as a constant. If the displacement does not follow a straight line, then the cosine of the angle between the two vectors in the integrals is not constant, and again the answer may be not simple. It is only in the case where the force is independent of the position in the integral, and the displacement follows a straight line, then the work is a simple scalar product between the force vector and the displacement vector. And it is only in an even simpler case, when the force vector is in the same direction as the displacement, that the work is force times distance. You have incorrectly taken the extremely simple case to be the general definition of work, and that’s a bad mistake to make. > > Since force is a vector quantity and distance a scalar quantity, work > must be a vector quantity. >> >> What is a vector and does have direction is e.g an energy flow, >> e.g power per square meter. > > > Energy flow has a direction and energy has not. > > So work and energy are not the same thing. > > They are related, but not equal. > > TH > > -- Odd Bodkin -- maker of fine toys, tools, tables
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-10 18:40 +0100 |
| Message-ID | <g77moqF8191U1@mid.individual.net> |
| In reply to | #479502 |
Am 09.12.2018 um 23:20 schrieb Odd Bodkin: > Thomas Heger <ttt_heg@web.de> wrote: >> Am 08.12.2018 um 19:26 schrieb Libor Striz: >>> Thomas Heger <ttt_heg@web.de> Wrote in message: >>> >>>> >>>> Vectors have a direction. >>>> >>>> A Joule for instance or a kilo-Watt-hour has no direction. >>> >>> Why do you then say work is a vector ? >>> The unit of work is 1 Joule. >> >> >> We have a similar problem with 'speed' and 'velocity'. >> >> Both have units like m/s , but one is a vector quantity (velocity) and >> one is not (speed). >> >> Speed is the magnitude of velocity. >> >> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >> >>> The acting force has direction. >>> The path along which the force is acting has direction as well >>> But the work as integral of scalar product of these 2 vectors is a >>> scalar, without direction. >> >> work is NOT 'forth times path' >> >> but ' force times distance'. > > No, it is not that either. It is the integral of the scalar product between > two vectors, the force and the differential displacement. 'Work' is the name of a quantity in physics. It relates to the real world, hence is not a product of either one or two vectors. 'Displacement' is actually not the same thing as 'distance'. Displacement is a vector, since you have a position before and after that movement. But if you have - say - the distance between Berlin and New York, this does not include any movement. What the hack is so difficult about that? ... TH
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-10 21:59 +0100 |
| Message-ID | <2993b118-c9a8-0055-aa50-52b34850762b@PointedEars.de> |
| In reply to | #479540 |
Thomas Heger wrote:
> Am 09.12.2018 um 23:20 schrieb Odd Bodkin:
>> Thomas Heger <ttt_heg@web.de> wrote:
>>> work is NOT 'forth times path'
>>>
>>> but ' force times distance'.
>>
>> No, it is not that either. It is the integral of the scalar product between
>> two vectors, the force and the differential displacement.
Correct. (AISB)
> 'Work' is the name of a quantity in physics.
>
> It relates to the real world, hence is not a product of either one or
> two vectors.
I am getting tired of telling you this: You have not the slightest clue what
you are talking about.
Work is *precisely* one of those physical quantities that are *calculated*,
i. e. that can NOT be observed, and can NOT be measured directly.
> 'Displacement' is actually not the same thing as 'distance'.
>
> Displacement is a vector, since you have a position before and after
> that movement.
>
> But if you have - say - the distance between Berlin and New York, this
> does not include any movement.
“I find your argument strewn with gaping defects in logic.”
–Spock
> What the hack is so difficult about that?
This is what you should be asking yourself, about yourself.
Why is it so hard for you to understand these basic concepts of physics?
Why is it so hard for you to take up a textbook and actually *read* about
them *before* you are talking about them?
In general: Why do you keep talking about things as if you knew about them
*although* you do NOT (and probably, deep down, you *know* that you do NOT
know)?
--
PointedEars
Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
[toc] | [prev] | [next] | [standalone]
| From | "Ross A. Finlayson" <ross.finlayson@gmail.com> |
|---|---|
| Date | 2018-12-18 05:33 -0800 |
| Message-ID | <fa644bda-82f1-43ec-8f63-03947d54f22d@googlegroups.com> |
| In reply to | #479561 |
On Monday, December 10, 2018 at 12:59:27 PM UTC-8, Thomas 'PointedEars' Lahn wrote: > Thomas Heger wrote: > > Am 09.12.2018 um 23:20 schrieb Odd Bodkin: > >> Thomas Heger <ttt_heg@web.de> wrote: > >>> work is NOT 'forth times path' > >>> > >>> but ' force times distance'. > >> > >> No, it is not that either. It is the integral of the scalar product between > >> two vectors, the force and the differential displacement. > > Correct. (AISB) > > > 'Work' is the name of a quantity in physics. > > > > It relates to the real world, hence is not a product of either one or > > two vectors. > > I am getting tired of telling you this: You have not the slightest clue what > you are talking about. > > Work is *precisely* one of those physical quantities that are *calculated*, > i. e. that can NOT be observed, and can NOT be measured directly. > > > 'Displacement' is actually not the same thing as 'distance'. > > > > Displacement is a vector, since you have a position before and after > > that movement. > > > > But if you have - say - the distance between Berlin and New York, this > > does not include any movement. > > “I find your argument strewn with gaping defects in logic.” > > –Spock > > > What the hack is so difficult about that? > > This is what you should be asking yourself, about yourself. > > Why is it so hard for you to understand these basic concepts of physics? > > Why is it so hard for you to take up a textbook and actually *read* about > them *before* you are talking about them? > > In general: Why do you keep talking about things as if you knew about them > *although* you do NOT (and probably, deep down, you *know* that you do NOT > know)? > > -- > PointedEars > > Twitter: @PointedEars2 > Please do not cc me. / Bitte keine Kopien per E-Mail. Just because he's informal, doesn't mean he's not formal.
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-22 14:51 +0100 |
| Message-ID | <3cc28e41-8d46-fa2e-f787-123609fb3c68@PointedEars.de> |
| In reply to | #480126 |
Ross A. Finlayson wrote: > Just because he's informal, > doesn't mean he's not formal. I have no idea what you are referring to, but JFYI: “informal” *means* “not formal”. -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail.
[toc] | [prev] | [next] | [standalone]
| From | "Ross A. Finlayson" <ross.finlayson@gmail.com> |
|---|---|
| Date | 2018-12-22 15:30 -0800 |
| Message-ID | <585aea32-e3c6-45e2-b4a8-8964b18bdddc@googlegroups.com> |
| In reply to | #480422 |
On Saturday, December 22, 2018 at 5:51:24 AM UTC-8, Thomas 'PointedEars' Lahn wrote: > Ross A. Finlayson wrote: > > Just because he's informal, > > doesn't mean he's not formal. > > I have no idea what you are referring to, but JFYI: > > “informal” *means* “not formal”. > -- > PointedEars > > Twitter: @PointedEars2 > Please do not cc me. / Bitte keine Kopien per E-Mail. So, my idea is basically about ubiquitous ordinals for the unbounded in both the horizontal and vertical following Zermelo, one universe with everything in it (including itself). "Physics is always on the search for a mathematical "theory of everything" (ToE), for mathematics it's the same, a foundation. (A formalist's general mathematical logic.) " These I copied and pasted, these paragraphs, I lifted them just now from a post, where, relevant to opinion, I copy it here, and it continues and maintains what point I was making here, that now pointing to here.
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-23 01:34 +0100 |
| Message-ID | <e39458e5-66ab-f7b3-09b6-821242d55077@PointedEars.de> |
| In reply to | #480441 |
Ross A. Finlayson wrote: > On Saturday, December 22, 2018 at 5:51:24 AM UTC-8, Thomas 'PointedEars' Lahn wrote: >> Ross A. Finlayson wrote: >>> Just because he's informal, >>> doesn't mean he's not formal. >> >> I have no idea what you are referring to, but JFYI: >> >> “informal” *means* “not formal”. > > So, my idea is […] Nobody asked for that. Learn to quote. -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail.
[toc] | [prev] | [next] | [standalone]
| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-12-10 21:41 +0000 |
| Message-ID | <pummhj$vf2$2@gioia.aioe.org> |
| In reply to | #479540 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 09.12.2018 um 23:20 schrieb Odd Bodkin: >> Thomas Heger <ttt_heg@web.de> wrote: >>> Am 08.12.2018 um 19:26 schrieb Libor Striz: >>>> Thomas Heger <ttt_heg@web.de> Wrote in message: >>>> >>>>> >>>>> Vectors have a direction. >>>>> >>>>> A Joule for instance or a kilo-Watt-hour has no direction. >>>> >>>> Why do you then say work is a vector ? >>>> The unit of work is 1 Joule. >>> >>> >>> We have a similar problem with 'speed' and 'velocity'. >>> >>> Both have units like m/s , but one is a vector quantity (velocity) and >>> one is not (speed). >>> >>> Speed is the magnitude of velocity. >>> >>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >>> >>>> The acting force has direction. >>>> The path along which the force is acting has direction as well >>>> But the work as integral of scalar product of these 2 vectors is a >>>> scalar, without direction. >>> >>> work is NOT 'forth times path' >>> >>> but ' force times distance'. >> >> No, it is not that either. It is the integral of the scalar product between >> two vectors, the force and the differential displacement. > > > 'Work' is the name of a quantity in physics. As is velocity. > > It relates to the real world, hence is not a product of either one or > two vectors. I disagree. There are LOTS of real world quantities that are related in some way to other real world quantities. Velocity (a vector) is related to differential displacement and differential time. Momentum is related to mass and velocity. > > 'Displacement' is actually not the same thing as 'distance'. That is correct. And work is characterized by its relationship to DISPLACEMENT in general, not distance. The relation to distance only applies in a simple sub case. > > Displacement is a vector, since you have a position before and after > that movement. > > But if you have - say - the distance between Berlin and New York, this > does not include any movement. > > What the hack is so difficult about that? > > ... > > > TH > -- Odd Bodkin — Maker of fine toys, tools, tables
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-11 16:31 +0100 |
| Message-ID | <g7a3hoFoc5dU1@mid.individual.net> |
| In reply to | #479567 |
Am 10.12.2018 um 22:41 schrieb Odd Bodkin: > Thomas Heger <ttt_heg@web.de> wrote: >> Am 09.12.2018 um 23:20 schrieb Odd Bodkin: >>> Thomas Heger <ttt_heg@web.de> wrote: >>>> Am 08.12.2018 um 19:26 schrieb Libor Striz: >>>>> Thomas Heger <ttt_heg@web.de> Wrote in message: >>>>> >>>>>> >>>>>> Vectors have a direction. >>>>>> >>>>>> A Joule for instance or a kilo-Watt-hour has no direction. >>>>> >>>>> Why do you then say work is a vector ? >>>>> The unit of work is 1 Joule. >>>> >>>> >>>> We have a similar problem with 'speed' and 'velocity'. >>>> >>>> Both have units like m/s , but one is a vector quantity (velocity) and >>>> one is not (speed). >>>> >>>> Speed is the magnitude of velocity. >>>> >>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >>>> >>>>> The acting force has direction. >>>>> The path along which the force is acting has direction as well >>>>> But the work as integral of scalar product of these 2 vectors is a >>>>> scalar, without direction. >>>> >>>> work is NOT 'forth times path' >>>> >>>> but ' force times distance'. >>> >>> No, it is not that either. It is the integral of the scalar product between >>> two vectors, the force and the differential displacement. >> >> >> 'Work' is the name of a quantity in physics. > > As is velocity. > >> >> It relates to the real world, hence is not a product of either one or >> two vectors. > > I disagree. There are LOTS of real world quantities that are related in > some way to other real world quantities. Velocity (a vector) is related to > differential displacement and differential time. Momentum is related to > mass and velocity. This is correct. But I had something against this statement: ".. It is the integral of the scalar product between two vectors..." I said: no, it is not, since a vector is a mathematical construct. A number is also a mathematical construct. And you cannot multiply two brainchilds and hope to get something real. First statement: a quantity in physics (work) is defined as being the product of one quantity (force)and some sort of length. We could take the integral of tiny displacements and regard this as vector (for instance). This would be ok. But mainly the vector is a representation of something real, not real itself. I wanted to address the question, wether the second component of this product is a vector or a scalar. A vector (vectorial quantity) has a direction and a value, while a scalar has no direction and only a value. In my opinion work is a quantity, which essential has a direction. This is so because it is a product of force (which has a direction) and something with length (distance or path). I suggested to take distance and not path, because then work become a vectorial quantity and energy the associated scalar. This would have benefits, since it is more like our understanding of the terms. As example I take this: you lift a heavy box by 1m. This is work (done by you) and the work applied had a direction (up) and an amount (in Nm - depending on the weight). If work is a scalar, then the information of direction is discarded for no obvious real. But more serious is the interpretation of work as form of energy. This would then create work as 'kind of invisible stuff'. And this would raise some illogic questions, like: where did the work go? Sure: it went into the box. But that is not true, since the work is not done by the box, but by the person that lifted it. To make the box contain that work, work had to become kind of 'stuff'. TH
[toc] | [prev] | [next] | [standalone]
| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-12-11 18:24 +0000 |
| Message-ID | <puovcb$1mbf$1@gioia.aioe.org> |
| In reply to | #479610 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 10.12.2018 um 22:41 schrieb Odd Bodkin: >> Thomas Heger <ttt_heg@web.de> wrote: >>> Am 09.12.2018 um 23:20 schrieb Odd Bodkin: >>>> Thomas Heger <ttt_heg@web.de> wrote: >>>>> Am 08.12.2018 um 19:26 schrieb Libor Striz: >>>>>> Thomas Heger <ttt_heg@web.de> Wrote in message: >>>>>> >>>>>>> >>>>>>> Vectors have a direction. >>>>>>> >>>>>>> A Joule for instance or a kilo-Watt-hour has no direction. >>>>>> >>>>>> Why do you then say work is a vector ? >>>>>> The unit of work is 1 Joule. >>>>> >>>>> >>>>> We have a similar problem with 'speed' and 'velocity'. >>>>> >>>>> Both have units like m/s , but one is a vector quantity (velocity) and >>>>> one is not (speed). >>>>> >>>>> Speed is the magnitude of velocity. >>>>> >>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >>>>> >>>>>> The acting force has direction. >>>>>> The path along which the force is acting has direction as well >>>>>> But the work as integral of scalar product of these 2 vectors is a >>>>>> scalar, without direction. >>>>> >>>>> work is NOT 'forth times path' >>>>> >>>>> but ' force times distance'. >>>> >>>> No, it is not that either. It is the integral of the scalar product between >>>> two vectors, the force and the differential displacement. >>> >>> >>> 'Work' is the name of a quantity in physics. >> >> As is velocity. >> >>> >>> It relates to the real world, hence is not a product of either one or >>> two vectors. >> >> I disagree. There are LOTS of real world quantities that are related in >> some way to other real world quantities. Velocity (a vector) is related to >> differential displacement and differential time. Momentum is related to >> mass and velocity. > > This is correct. But I had something against this statement: > > ".. It is the integral of the scalar product between two vectors..." > > I said: no, it is not, since a vector is a mathematical construct. I disagree. A vector is a way to characterize a physical quantity. Velocity is a vector, momentum is a vector, force is a vector. All of those are real. > > A number is also a mathematical construct. And you cannot multiply two > brainchilds and hope to get something real. > > > First statement: a quantity in physics (work) is defined as being the > product of one quantity (force)and some sort of length. “the product” is too vague. Given that the two things you are multiplying are vectors, you have to be precise about what you mean by “product”. In this case “product” means scalar- or dot-product. > > We could take the integral of tiny displacements and regard this as > vector (for instance). > > This would be ok. > > But mainly the vector is a representation of something real, not real > itself. > > I wanted to address the question, wether the second component of this > product is a vector or a scalar. > > A vector (vectorial quantity) has a direction and a value, while a > scalar has no direction and only a value. > > In my opinion work is a quantity, which essential has a direction. Your opinion is wrong here. > > This is so because it is a product of force (which has a direction) and > something with length (distance or path). > > I suggested to take distance and not path, because then work become a > vectorial quantity and energy the associated scalar. And you are trying to rig things to defend your misconception. That is not the way work is defined or calculated. How it is properly defined and calculated is well documented in beginner textbooks. That is not as you described. > > This would have benefits, since it is more like our understanding of the > terms. > > As example I take this: > > you lift a heavy box by 1m. > > This is work (done by you) and the work applied had a direction (up) and > an amount (in Nm - depending on the weight). No. The work does not have a direction. The force has a direction (up). The displacement has a direction (up). In this case, the force is constant and always collinear with the displacement. This is the simplest of cases but is by no means the general one. Here, though the force and the displacement have directions, the work itself does not by DEFINITION. You have simply incorrectly ascribed that direction (up) to the work. This is a beginner error. > > If work is a scalar, then the information of direction is discarded for > no obvious real. > > But more serious is the interpretation of work as form of energy. This > would then create work as 'kind of invisible stuff'. > > And this would raise some illogic questions, like: where did the work go? > > Sure: it went into the box. > > But that is not true, since the work is not done by the box, but by the > person that lifted it. > > To make the box contain that work, work had to become kind of 'stuff'. > > > TH > > > > -- Odd Bodkin — Maker of fine toys, tools, tables
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-12 16:32 +0100 |
| Message-ID | <g7co0oFbh39U1@mid.individual.net> |
| In reply to | #479618 |
Am 11.12.2018 um 19:24 schrieb Odd Bodkin: >>>>>> >>>>>> We have a similar problem with 'speed' and 'velocity'. >>>>>> >>>>>> Both have units like m/s , but one is a vector quantity (velocity) and >>>>>> one is not (speed). >>>>>> >>>>>> Speed is the magnitude of velocity. >>>>>> >>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >>>>>> >>>>>>> The acting force has direction. >>>>>>> The path along which the force is acting has direction as well >>>>>>> But the work as integral of scalar product of these 2 vectors is a >>>>>>> scalar, without direction. >>>>>> >>>>>> work is NOT 'forth times path' >>>>>> >>>>>> but ' force times distance'. >>>>> >>>>> No, it is not that either. It is the integral of the scalar product between >>>>> two vectors, the force and the differential displacement. >>>> >>>> >>>> 'Work' is the name of a quantity in physics. >>> >>> As is velocity. >>> >>>> >>>> It relates to the real world, hence is not a product of either one or >>>> two vectors. >>> >>> I disagree. There are LOTS of real world quantities that are related in >>> some way to other real world quantities. Velocity (a vector) is related to >>> differential displacement and differential time. Momentum is related to >>> mass and velocity. >> >> This is correct. But I had something against this statement: >> >> ".. It is the integral of the scalar product between two vectors..." >> >> I said: no, it is not, since a vector is a mathematical construct. > > I disagree. A vector is a way to characterize a physical quantity. Velocity > is a vector, momentum is a vector, force is a vector. All of those are > real. There is still a difference between 'vector' and 'physical quantity characterized by a vector'. A vector is a mathematical construct, while quantities have units and a physical meaning. You may characterize that quantity by an appropriate vector, but this does not make a vector a quantity. >> >> A number is also a mathematical construct. And you cannot multiply two >> brainchilds and hope to get something real. >> >> >> First statement: a quantity in physics (work) is defined as being the >> product of one quantity (force)and some sort of length. > > “the product” is too vague. Given that the two things you are multiplying > are vectors, you have to be precise about what you mean by “product”. In > this case “product” means scalar- or dot-product. I was trying to show, that the product of a force and a path is not the right idea. You are correct, since it is always done that way. But it is wrong, in my opinion. So let me tell you, why work should be a vectorial quantity: Force is a quantity (represented by a vector) with a direction (in space). This vector contains enough information about the direction. A path is actually a construct composed out of points in Euclidean space, which the object in question reaches at certain points in time. First problem: the points are unimportant and do not belong into the calculation. IOW: if you lift a heavy stone by one meter, the work should be equal, if you lift the same stone a little bit later at a different position (same gravity assumed). So the location does not belong into the equation and also time (absolute time) does not belong there neither. If we kick that out of the equation, we have got left 'distance' (in the direction of the movement), which is something we need. This distance is a scalar. If we multiply the scalar value of distance with a vector for force, then we get a good and satisfying result. The benefit is, that the quantity would match to the usual understanding of the term 'work'. It is also matched by a counterpart, similar to Newton's actio equals reactio. The work is then done by the active part (in case of the stone: by you) and balanced by the passive part (the stone). If work is a scalar quantity, you cannot easily do this, Instead it would require to invent something like 'working stuff'. But work is not a substance, but an activity (done by e.g. material objects). ... TH
[toc] | [prev] | [next] | [standalone]
| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-12-12 16:27 +0000 |
| Message-ID | <purcte$1tqu$2@gioia.aioe.org> |
| In reply to | #479684 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 11.12.2018 um 19:24 schrieb Odd Bodkin: > >>>>>>> >>>>>>> We have a similar problem with 'speed' and 'velocity'. >>>>>>> >>>>>>> Both have units like m/s , but one is a vector quantity (velocity) and >>>>>>> one is not (speed). >>>>>>> >>>>>>> Speed is the magnitude of velocity. >>>>>>> >>>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >>>>>>> >>>>>>>> The acting force has direction. >>>>>>>> The path along which the force is acting has direction as well >>>>>>>> But the work as integral of scalar product of these 2 vectors is a >>>>>>>> scalar, without direction. >>>>>>> >>>>>>> work is NOT 'forth times path' >>>>>>> >>>>>>> but ' force times distance'. >>>>>> >>>>>> No, it is not that either. It is the integral of the scalar product between >>>>>> two vectors, the force and the differential displacement. >>>>> >>>>> >>>>> 'Work' is the name of a quantity in physics. >>>> >>>> As is velocity. >>>> >>>>> >>>>> It relates to the real world, hence is not a product of either one or >>>>> two vectors. >>>> >>>> I disagree. There are LOTS of real world quantities that are related in >>>> some way to other real world quantities. Velocity (a vector) is related to >>>> differential displacement and differential time. Momentum is related to >>>> mass and velocity. >>> >>> This is correct. But I had something against this statement: >>> >>> ".. It is the integral of the scalar product between two vectors..." >>> >>> I said: no, it is not, since a vector is a mathematical construct. >> >> I disagree. A vector is a way to characterize a physical quantity. Velocity >> is a vector, momentum is a vector, force is a vector. All of those are >> real. > > There is still a difference between 'vector' and 'physical quantity > characterized by a vector'. Not in a physics discussion, normally. When I talk about velocity, I can refer to it as a vector and that means the physical quantity characterized as a vector. The two are synonymous and understood. Maybe not by you. > > A vector is a mathematical construct, while quantities have units and a > physical meaning. > > You may characterize that quantity by an appropriate vector, but this > does not make a vector a quantity. > > >>> >>> A number is also a mathematical construct. And you cannot multiply two >>> brainchilds and hope to get something real. >>> >>> >>> First statement: a quantity in physics (work) is defined as being the >>> product of one quantity (force)and some sort of length. >> >> “the product” is too vague. Given that the two things you are multiplying >> are vectors, you have to be precise about what you mean by “product”. In >> this case “product” means scalar- or dot-product. > > > I was trying to show, that the product of a force and a path is not the > right idea. If you want to redefine work, using the same term that has been used with a particular understanding for centuries, good luck with that. So many amateur hacks come into this group with a lack of background reading and understanding of basic concepts, and then they make mistakes because the concepts do not mean what they think those mean. And then they complain, “Well, it SHOULD mean what I think it means.” No, it should not. What SHOULD happen is that amateur hacks should educate themselves on the meaning of basic concepts before discussing them. > > You are correct, since it is always done that way. > > But it is wrong, in my opinion. So let me tell you, why work should be a > vectorial quantity: > > > Force is a quantity (represented by a vector) with a direction (in space). > > This vector contains enough information about the direction. > > A path is actually a construct composed out of points in Euclidean > space, which the object in question reaches at certain points in time. > > First problem: the points are unimportant and do not belong into the > calculation. > > IOW: if you lift a heavy stone by one meter, the work should be equal, > if you lift the same stone a little bit later at a different position > (same gravity assumed). > > So the location does not belong into the equation and also time > (absolute time) does not belong there neither. > > If we kick that out of the equation, we have got left 'distance' (in the > direction of the movement), which is something we need. > > This distance is a scalar. > > If we multiply the scalar value of distance with a vector for force, > then we get a good and satisfying result. > > > The benefit is, that the quantity would match to the usual understanding > of the term 'work'. > > It is also matched by a counterpart, similar to Newton's actio equals > reactio. > > The work is then done by the active part (in case of the stone: by you) > and balanced by the passive part (the stone). > > > If work is a scalar quantity, you cannot easily do this, > > Instead it would require to invent something like 'working stuff'. > > But work is not a substance, but an activity (done by e.g. material > objects). > > > ... > > > TH > -- Odd Bodkin — Maker of fine toys, tools, tables
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-14 15:13 +0100 |
| Message-ID | <g7hs52Ff4t7U1@mid.individual.net> |
| In reply to | #479689 |
Am 12.12.2018 um 17:27 schrieb Odd Bodkin: >>>>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >>>>>>>> >>>>>>>>> The acting force has direction. >>>>>>>>> The path along which the force is acting has direction as well >>>>>>>>> But the work as integral of scalar product of these 2 vectors is a >>>>>>>>> scalar, without direction. >>>>>>>> >>>>>>>> work is NOT 'forth times path' >>>>>>>> >>>>>>>> but ' force times distance'. >>>>>>> >>>>>>> No, it is not that either. It is the integral of the scalar product between >>>>>>> two vectors, the force and the differential displacement. >>>>>> >>>>>> >>>>>> 'Work' is the name of a quantity in physics. >>>>> >>>>> As is velocity. >>>>> >>>>>> >>>>>> It relates to the real world, hence is not a product of either one or >>>>>> two vectors. >>>>> >>>>> I disagree. There are LOTS of real world quantities that are related in >>>>> some way to other real world quantities. Velocity (a vector) is related to >>>>> differential displacement and differential time. Momentum is related to >>>>> mass and velocity. >>>> >>>> This is correct. But I had something against this statement: >>>> >>>> ".. It is the integral of the scalar product between two vectors..." >>>> >>>> I said: no, it is not, since a vector is a mathematical construct. >>> >>> I disagree. A vector is a way to characterize a physical quantity. Velocity >>> is a vector, momentum is a vector, force is a vector. All of those are >>> real. >> >> There is still a difference between 'vector' and 'physical quantity >> characterized by a vector'. > > Not in a physics discussion, normally. What you are apparently trying to say is, that if physicists chat at lunch brake or in a bar with a bear, they take the numbers for the thing and the vector for the quantity. Well, maybe.. In my view there is a difference between a model and the system it tries to model. There is also a difference between the mathematical representation of some type of behavior and the mechanisms acting in that system. The statement, that I disagreed with was, that work is a scalar, because it is a product of two vectors. This is wrong, since it is not a product of two vectors, but a product of a physical quantity called 'force', which behaves like a vector, and something we are still debating. You said, this second term should be a vector, too, while I wanted to use a scalar. > When I talk about velocity, I can refer to it as a vector and that means > the physical quantity characterized as a vector. The two are synonymous and > understood. Maybe not by you. Sure. But where is the connection to the problem in question? >> >> A vector is a mathematical construct, while quantities have units and a >> physical meaning. >> >> You may characterize that quantity by an appropriate vector, but this >> does not make a vector a quantity. >> >> >>>> >>>> A number is also a mathematical construct. And you cannot multiply two >>>> brainchilds and hope to get something real. >>>> >>>> >>>> First statement: a quantity in physics (work) is defined as being the >>>> product of one quantity (force)and some sort of length. >>> >>> “the product” is too vague. Given that the two things you are multiplying >>> are vectors, you have to be precise about what you mean by “product”. In >>> this case “product” means scalar- or dot-product. >> >> >> I was trying to show, that the product of a force and a path is not the >> right idea. > > If you want to redefine work, using the same term that has been used with a > particular understanding for centuries, good luck with that. You must distinguish between the 'metaphysical' definition and the technical. If a certain term is used, we need to talk about its meaning. In case of 'work' we have something in mind. This would be somehow like "work is the quantity we have, if we lift a stone for a certain hight". That is a little rough, but gives an idea about what we are actually trying to define. A more technical definition would be like this: work for lifting a stone in Earth gravitational field is equal to weight times additional hight (by what the stone is lifted). This is still not very good, but defines, what we are trying to define: the physical quantity 'work'. Now we should look, whether or not a certain definition would actually fit to what we had in mind. 'additional height' is not a vector and it is also not a path. It is not a vector, because 'height' is meant as radius of the Earth surface at a certain point (where the thing to lift is right now). The larger radius with additional height is actually the former radius multiplied by a scalar. 'additional height' is also not a path, because 'path' is meant as set of points, which are reached one after the other at certain times. These points and the times, when they are passed are not interesting in this context and do not belong to the metaphysical construct 'work'. ... TH
[toc] | [prev] | [next] | [standalone]
| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-12-14 14:28 +0000 |
| Message-ID | <pv0en3$jmm$1@gioia.aioe.org> |
| In reply to | #479807 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 12.12.2018 um 17:27 schrieb Odd Bodkin: > >>>>>>>>> https://examples.yourdictionary.com/examples-vector-scalar-quantity-physics.html >>>>>>>>> >>>>>>>>>> The acting force has direction. >>>>>>>>>> The path along which the force is acting has direction as well >>>>>>>>>> But the work as integral of scalar product of these 2 vectors is a >>>>>>>>>> scalar, without direction. >>>>>>>>> >>>>>>>>> work is NOT 'forth times path' >>>>>>>>> >>>>>>>>> but ' force times distance'. >>>>>>>> >>>>>>>> No, it is not that either. It is the integral of the scalar product between >>>>>>>> two vectors, the force and the differential displacement. >>>>>>> >>>>>>> >>>>>>> 'Work' is the name of a quantity in physics. >>>>>> >>>>>> As is velocity. >>>>>> >>>>>>> >>>>>>> It relates to the real world, hence is not a product of either one or >>>>>>> two vectors. >>>>>> >>>>>> I disagree. There are LOTS of real world quantities that are related in >>>>>> some way to other real world quantities. Velocity (a vector) is related to >>>>>> differential displacement and differential time. Momentum is related to >>>>>> mass and velocity. >>>>> >>>>> This is correct. But I had something against this statement: >>>>> >>>>> ".. It is the integral of the scalar product between two vectors..." >>>>> >>>>> I said: no, it is not, since a vector is a mathematical construct. >>>> >>>> I disagree. A vector is a way to characterize a physical quantity. Velocity >>>> is a vector, momentum is a vector, force is a vector. All of those are >>>> real. >>> >>> There is still a difference between 'vector' and 'physical quantity >>> characterized by a vector'. >> >> Not in a physics discussion, normally. > > What you are apparently trying to say is, that if physicists chat at > lunch brake or in a bar with a bear, they take the numbers for the thing > and the vector for the quantity. > > Well, maybe.. > > In my view there is a difference between a model and the system it tries > to model. There is also a difference between the mathematical > representation of some type of behavior and the mechanisms acting in > that system. > > The statement, that I disagreed with was, that work is a scalar, because > it is a product of two vectors. > > This is wrong, since it is not a product of two vectors, but a product > of a physical quantity called 'force', which behaves like a vector, and > something we are still debating. The correct assessment is that you are debating it. This is in effect the same thing as you saying that it is open to debate what “penguin” means and that to you, a penguin is a large striped cat found in Asia. It isn’t really up to debate. The term “work” is defined clearly, has been defined clearly for a couple of centuries, and that definition contrasts with a meaning that you are saying makes sense to you. Surely you understand that what makes sense to you is largely irrelevant in the context of the definitions of basic concepts. Here, you are uneducated in the meanings of basics concepts, and you are attempting to make a case that, in the absence of education, what you come up just by “thinking about it” in a vacuum of ignorance should nevertheless be considered seriously. I disagree. Basic concepts are to be learned, not reinvented by “free thinkers” such as yourself. > > You said, this second term should be a vector, too, while I wanted to > use a scalar. > >> When I talk about velocity, I can refer to it as a vector and that means >> the physical quantity characterized as a vector. The two are synonymous and >> understood. Maybe not by you. > > Sure. > > But where is the connection to the problem in question? > >>> >>> A vector is a mathematical construct, while quantities have units and a >>> physical meaning. >>> >>> You may characterize that quantity by an appropriate vector, but this >>> does not make a vector a quantity. >>> >>> >>>>> >>>>> A number is also a mathematical construct. And you cannot multiply two >>>>> brainchilds and hope to get something real. >>>>> >>>>> >>>>> First statement: a quantity in physics (work) is defined as being the >>>>> product of one quantity (force)and some sort of length. >>>> >>>> “the product” is too vague. Given that the two things you are multiplying >>>> are vectors, you have to be precise about what you mean by “product”. In >>>> this case “product” means scalar- or dot-product. >>> >>> >>> I was trying to show, that the product of a force and a path is not the >>> right idea. >> >> If you want to redefine work, using the same term that has been used with a >> particular understanding for centuries, good luck with that. > > You must distinguish between the 'metaphysical' definition and the > technical. > > If a certain term is used, we need to talk about its meaning. > > In case of 'work' we have something in mind. This would be somehow like > "work is the quantity we have, if we lift a stone for a certain hight". > > That is a little rough, but gives an idea about what we are actually > trying to define. > > A more technical definition would be like this: > > work for lifting a stone in Earth gravitational field is equal to weight > times additional hight (by what the stone is lifted). > > > This is still not very good, but defines, what we are trying to define: > the physical quantity 'work'. > > > Now we should look, whether or not a certain definition would actually > fit to what we had in mind. > > 'additional height' is not a vector and it is also not a path. > > It is not a vector, because 'height' is meant as radius of the Earth > surface at a certain point (where the thing to lift is right now). > > The larger radius with additional height is actually the former radius > multiplied by a scalar. > > 'additional height' is also not a path, because 'path' is meant as set > of points, which are reached one after the other at certain times. > > These points and the times, when they are passed are not interesting in > this context and do not belong to the metaphysical construct 'work'. > > > > > ... > > > TH > -- Odd Bodkin -- maker of fine toys, tools, tables
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-15 13:35 +0100 |
| Message-ID | <g7kao5F12uiU1@mid.individual.net> |
| In reply to | #479808 |
Am 14.12.2018 um 15:28 schrieb Odd Bodkin: ... >> The statement, that I disagreed with was, that work is a scalar, because >> it is a product of two vectors. >> >> This is wrong, since it is not a product of two vectors, but a product >> of a physical quantity called 'force', which behaves like a vector, and >> something we are still debating. > > The correct assessment is that you are debating it. This is in effect the > same thing as you saying that it is open to debate what “penguin” means and > that to you, a penguin is a large striped cat found in Asia. > > It isn’t really up to debate. The term “work” is defined clearly, has been > defined clearly for a couple of centuries, and that definition contrasts > with a meaning that you are saying makes sense to you. The concept of vectors is much younger than the concept of the quantity 'work'. So you are incorrect in your assumption, that work is regarded as product of two vectors since many centuries. > .. Here, you are uneducated in > the meanings of basics concepts, and you are attempting to make a case > that, in the absence of education, what you come up just by “thinking about > it” in a vacuum of ignorance should nevertheless be considered seriously. I > disagree. Basic concepts are to be learned, not reinvented by “free Of course I am 'educated' in this field. But I wanted to express, that I do not agree to mainstream concepts about this term. I have written, why I think so, but you have not written any reply to my arguments and are spitting out statements of unjustified arrogance. TH
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-15 13:52 +0100 |
| Message-ID | <c1381a68-616c-fddc-ae08-64201ffbb300@PointedEars.de> |
| In reply to | #479859 |
Thomas Heger wrote: > Am 14.12.2018 um 15:28 schrieb Odd Bodkin: >> It isn’t really up to debate. The term “work” is defined clearly, has been >> defined clearly for a couple of centuries, and that definition contrasts >> with a meaning that you are saying makes sense to you. > > The concept of vectors is much younger than the concept of the quantity > 'work'. The *term* is younger, but that does not mean anything. > So you are incorrect in your assumption, that work is regarded as > product of two vectors since many centuries. The abstraction of a complex number as a pair of two numbers was introduced by William Rowan Hamilton (1805–1865). That was about two (“a couple of”) centuries ago now. The geometric concept of putting arrows atop each other is so basic that it was known earlier already; it is in the “Mechanics” of Heron of Alexandria (first century CE): <http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html> (first Google hit for “vector history”) >> .. Here, you are uneducated in the meanings of basics concepts, and you >> are attempting to make a case that, in the absence of education, what >> you come up just by “thinking about it” in a vacuum of ignorance should >> nevertheless be considered seriously. I disagree. Basic concepts are to >> be learned, not reinvented […] > > Of course I am 'educated' in this field. Once again you are using your silly straight-apostrophe quotes correctly. You are "educated" in that you do not have the slightest clue what you are talking about. -- PointedEars Twitter: @PointedEars2 Please do not cc me. / Bitte keine Kopien per E-Mail.
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-15 19:50 +0100 |
| Message-ID | <g7l0muF5qk3U1@mid.individual.net> |
| In reply to | #479860 |
Am 15.12.2018 um 13:52 schrieb Thomas 'PointedEars' Lahn:
>>> It isn’t really up to debate. The term “work” is defined clearly, has been
>>> defined clearly for a couple of centuries, and that definition contrasts
>>> with a meaning that you are saying makes sense to you.
>>
>> The concept of vectors is much younger than the concept of the quantity
>> 'work'.
>
> The *term* is younger, but that does not mean anything.
Newton wrote in the 17th century and vectors came into fashion in the
19th century.
https://www.researchgate.net/publication/244957729_A_History_of_Vector_Analysis
>
>> So you are incorrect in your assumption, that work is regarded as
>> product of two vectors since many centuries.
>
> The abstraction of a complex number as a pair of two numbers was introduced
> by William Rowan Hamilton (1805–1865). That was about two (“a couple of”)
> centuries ago now.
>
not quite. As far as I remember this was in the 1850th.
> The geometric concept of putting arrows atop each other is so basic that it
> was known earlier already; it is in the “Mechanics” of Heron of Alexandria
> (first century CE):
>
> <http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html>
>
> (first Google hit for “vector history”)
>
>>> .. Here, you are uneducated in the meanings of basics concepts, and you
>>> are attempting to make a case that, in the absence of education, what
>>> you come up just by “thinking about it” in a vacuum of ignorance should
>>> nevertheless be considered seriously. I disagree. Basic concepts are to
>>> be learned, not reinvented […]
>>
>> Of course I am 'educated' in this field.
>
> Once again you are using your silly straight-apostrophe quotes correctly.
No.
In German we use an active term for learning ('lernen'), while 'being
educated' is a passive phrase, which has no counterpart in German,
except maybe in prisons ('belehrt werden').
The entire concept of 'education' is actually British and adopted by
institutions in the USA.
TH
[toc] | [prev] | [next] | [standalone]
Page 3 of 6 — ← Prev page 1 2 [3] 4 5 6 Next page →
Back to top | Article view | sci.physics.relativity
csiph-web