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 2 of 6 — ← Prev page 1 [2] 3 4 5 6 Next page →
| From | dlzc <dlzc1@cox.net> |
|---|---|
| Date | 2018-12-12 09:12 -0800 |
| Message-ID | <df11b232-228b-4250-9a6a-c33f3ed2271b@googlegroups.com> |
| In reply to | #479690 |
Dear tjrob137: On Wednesday, December 12, 2018 at 9:28:42 AM UTC-7, tjrob137 wrote: > On 12/7/18 1:13 PM, dlzc wrote: > > In the quantum realm, the mass of a particle / system is a quantum > > number times the magnetic moment. > > This is just plain wrong. There are massive particles > with zero magnetic moment, such as pions (and all > other scalar particles). https://arxiv.org/abs/1309.2225 > It's not even true for leptons: the muon and electron > magnetic moments are equal to within a few parts per > million, but their masses differ by a factor of 200. "mass is proportional to" does not imply exactly equal to, Tom. > Your apparent obsession with magnetic moment is > completely unwarranted. Maybe. Thanks for pitching in. Can you find the "first grader" formula that I so vaguely recall, or are you just going to snipe here? David A. Smith
[toc] | [prev] | [next] | [standalone]
| From | Tom Roberts <tjroberts137@sbcglobal.net> |
|---|---|
| Date | 2018-12-14 12:21 -0600 |
| Message-ID | <NZadnTblubU4aY7BnZ2dnUU7_8zNnZ2d@giganews.com> |
| In reply to | #479693 |
On 12/12/18 11:12 AM, dlzc wrote: > On Wednesday, December 12, 2018 at 9:28:42 AM UTC-7, tjrob137 wrote: >> On 12/7/18 1:13 PM, dlzc wrote: >>> In the quantum realm, the mass of a particle / system is a >>> quantum number times the magnetic moment. >> >> This is just plain wrong. There are massive particles with zero >> magnetic moment, such as pions (and all other scalar particles). > > https://arxiv.org/abs/1309.2225 That's irrelevant here; it discusses the difference between the theoretical and experimental values of the muon g-2 -- currently they differ by about 3 sigma. This reference discusses attempts to reduce the errorbar in the theoretical value; there is an experiment at Fermilab working to reduce the experimental errorbar. Three sigma is on the borderline between being believable as a purely statistical quirk, and believable as a significant difference. Hence the current interest. >> It's not even true for leptons: the muon and electron magnetic >> moments are equal to within a few parts per million, but their >> masses differ by a factor of 200. > > "mass is proportional to" does not imply exactly equal to, Tom. They are not anywhere close to proportional. Just look up the masses and magnetic moments for e- and mu- and compute the ratios. In the PDG tables they only list the anomalies, so you must convert those to the magnetic moments themselves. Hint: the anomaly is (g-2)/2; solve for g and then multiply by the spin (1/2 h-bar for both). > Can you find the "first grader" formula that I so vaguely recall, There is no formula relating mass to magnetic moment, because as far as anybody knows they are independent. I have no idea what you "vaguely recall", but the notion that mass and magnetic moment are proportional is nonsense (counterexamples given previously). Tom Roberts
[toc] | [prev] | [next] | [standalone]
| From | "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> |
|---|---|
| Date | 2019-01-06 09:20 -0800 |
| Message-ID | <ff3ed9e5-8a28-4851-88e3-2a6f7d06ebe3@googlegroups.com> |
| In reply to | #479204 |
dlzc escrito The mass of one free electron + the mass of one free proton - mass of a single H1 atom = (13.6 eV + some recoil in the atom to conserve momentum)/c² ((((13.6878866 eV) / (electron mass + proton mass - (1 atomic mass unit)))^0.5) * (4e-7pi)) / (376.730313462 * 137.035999) = 1e-5 m / s ((electron mass + proton mass - (1 atomic mass unit)) * (c^2)) / ((Boltzmann constant * kelvin * 299792458) / ((4e-6 * pi)^0.5)) = 1.00018087
[toc] | [prev] | [next] | [standalone]
| From | "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> |
|---|---|
| Date | 2019-01-06 09:37 -0800 |
| Message-ID | <58c39798-8fbe-4956-8308-3b6396f75d7b@googlegroups.com> |
| In reply to | #481196 |
dlzc escrito The mass of one free electron + the mass of one free proton - mass of a single H1 atom = (13.6 eV + some recoil in the atom to conserve momentum)/c² ((((13.6878866 eV) / (electron mass + proton mass - (1 atomic mass unit)))^0.5) * (4e-7pi)) / (376.730313462 * 137.035999) = 1e-5 m / s ((electron mass + proton mass - (1 atomic mass unit)) * (c^2)) / ((Boltzmann constant * kelvin * 299792458) / ((4e-6pi)^0.5)) = 1.00018087 1 / ((((electron mass + proton mass - (1 atomic mass unit)) / (Boltzmann constant * kelvin)) / (2^(5 / 2))) * (c^2)) = 6.6877599e-11 Electron Temp = (299792458 * pi) kelvin ((Boltzmann constant / (9.1224509E+20 pascals)) * ((299792458 * pi) * kelvin)) / ((2.4263102367E-12 m)^3) = 0.997937129
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-06 20:50 +0100 |
| Message-ID | <g6tcrdFugh7U1@mid.individual.net> |
| In reply to | #479202 |
Am 06.12.2018 um 19:49 schrieb RichD: > Everybody knows E = mc², and figures it's well understood. > I disagree. It all depends on what '=' means. > > By analogy, consider the following cases: > Al and Bob are identical twins; i.e. Al = Bob > > i) They attend a party, each wearing a name tag. > Everybody can see they're identical, that the name tags are immaterial. > > ii) They attend a party, each wearing a different mask. > No one knows which is which, but all agree, they're > still identical, the masks are immaterial. > > iii) One of them dies, but no one knows which. Meanwhile, > the survivor takes the deceased's identity; that is, Al is > now Bob, or vice versa. Are they still identical? > > iv) On a daily basis, they display contrasting > personalities; on Monday, Al is outgoing, Bob withdrawn; > on Tuesday, they switch. Are they still identical? > > Now, which of the above - or some other analogy - applies > to the famous formula? What is the meaning of 'equal'? > The equation is meant as quantitative description of the nuclear decay: A radioactive sample looses mass as it sends out energy in the form of rays and particles. The numerical values of the measured quantities follow that equation. So: certain NUMBERS are equal. 'Equal' means, that the number on the left side of the equation sign is the same as on the right side, if all the algebraic calculations are carried out on bot sides (and differences in sign are leveled out). TH TH
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-06 22:47 +0100 |
| Message-ID | <38b03482-19b8-f0e3-1c69-5d00ece928fb@PointedEars.de> |
| In reply to | #479205 |
Thomas Heger wrote:
> Am 06.12.2018 um 19:49 schrieb RichD:
>> Everybody knows E = mc², and figures it's well understood.
>> I disagree. It all depends on what '=' means.
>
> The equation is meant as quantitative description of the nuclear decay:
Not only.
> A radioactive sample looses mass as it sends out energy in the form of
_loses_ (same pronunciation, different spelling, different meaning)
> rays and particles.
>
> The numerical values of the measured quantities follow that equation.
Yes, that was Einstein’s original idea in 1905 (“Does the Inertia of a Body
Depend Upon Its Energy Content?”). But that is not the end of it.
For example, the Sun produces energy (to sustain herself against her own
gravity, as a side effect spending heat to her surrounding planets) by
fusing nuclei because the resulting nucleus has less mass than the sum of
the masses of the progenitor nuclei (mass defect). This energy is so-called
binding energy; it would be required to tear these nuclei apart again.
So, IOW, the Sun converts mass to energy. Because they really two aspects
of the same thing, they are *equivalent* (hence the *equals* sign).
Likewise, when in nuclear *fission* reactors we *split nuclei (earlier than
they would do that naturally), we gain energy because (put simply, due to
the number of nucleons as compared to the fusion case) the mass of the
resulting nuclei is less than that of the progenitor nucleus.
> So: certain NUMBERS are equal.
_Quantities_, not numbers. Were it only numbers, physics would be much
easier (and much more boring).
The quantity “energy” (E) literally has *units* of mass times speed squared
(among other ways).
[You can understand this using Newtonian mechanics if you consider that
one way to say it is that the energy of a physical system is a measure
of the capability of that system to do work.
Work W is force F applied over a distance s:
W = F s.
Force F is mass m moved with acceleration “a”:
F = m a.
Acceleration “a” is speed (change) v per time (change) t:
a = v/t
Speed v is distance s traveled per time t:
v = s/t
Then energy/work is
E = W = F s = m a s = m v/t s = m s/t² s = m s²/t² = m (s/t)² = m v².
The far right-hand side should ring a bell if you think of kinetic
energy, the energy of motion: T = E_kin = 1/2 m v² (Newtonian).
c is the *speed* of light, so E₍₀₎ = m c² makes sense dimensionally.]
> 'Equal' means, that the number on the left side of the equation sign is
> the same as on the right side, if all the algebraic calculations are
> carried out on bot sides (and differences in sign are leveled out).
No, differences in sign are NOT "leveled out".
--
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 | 2018-12-07 12:06 +0100 |
| Message-ID | <g6v2h6Fb0lhU1@mid.individual.net> |
| In reply to | #479211 |
Am 06.12.2018 um 22:47 schrieb Thomas 'PointedEars' Lahn: .. > > Then energy/work is > > E = W = F s = m a s = m v/t s = m s/t² s = m s²/t² = m (s/t)² = m v². > > The far right-hand side should ring a bell if you think of kinetic > energy, the energy of motion: T = E_kin = 1/2 m v² (Newtonian). > > c is the *speed* of light, so E₍₀₎ = m c² makes sense dimensionally.] Energy and work are not the same thing. Energy is a scalar unit, hence cannot become negative and has no direction. Work is a vectorial unit and has a direction and can eventually become negative > >> 'Equal' means, that the number on the left side of the equation sign is >> the same as on the right side, if all the algebraic calculations are >> carried out on bot sides (and differences in sign are leveled out). > > No, differences in sign are NOT "leveled out". > Sure they are. Since energy is positive and the LOSS of mass is negative, we need to add a negative sign somewhere. But I personally think, the equation (E=m*c²) needs also 'deltas', since 'mass lost' addresses a difference, for what you should use a delta. TH
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-07 16:15 +0100 |
| Message-ID | <22bed404-b9e0-9995-d1bf-aa333a5dd7df@PointedEars.de> |
| In reply to | #479228 |
Thomas Heger wrote:
> Am 06.12.2018 um 22:47 schrieb Thomas 'PointedEars' Lahn:
>> Then energy/work is
>>
>> E = W = F s = m a s = m v/t s = m s/t² s = m s²/t² = m (s/t)² = m v².
>>
>> The far right-hand side should ring a bell if you think of kinetic
>> energy, the energy of motion: T = E_kin = 1/2 m v² (Newtonian).
>>
>> c is the *speed* of light, so E₍₀₎ = m c² makes sense dimensionally.]
>
> Energy and work are not the same thing.
Dimensionally they are exactly the same because of how energy is defined
(AISB in the part that you did not quote).
> Energy is a scalar unit,
Energy is a _quantity_, not a unit.
> hence cannot become negative
But it can.
> and has no direction.
A matter of interpretation. There are processes that transport energy in a
particular direction.
> Work is a vectorial unit
Nonsense.
> and has a direction and can eventually become negative
No, that the value of work can become negative is indicative instead of work
performed *on* a system as compared to work performed *by* a system.
This is only a *convention*.
For example, if you push a piston to reduce the volume of the gas in a
cylinder, then you perform work *on* that system, and the convention is
that then this work has a positive sign.
If instead the piston is pushed outwards, increasing the volume of the gas
in a cylinder, then work is performed *by* the system on the environment,
and it follows that this work must have a negative sign.
Thus, the first law of thermodynamics is
dU = δQ + δW = δQ − p dV
with
dU – infinitesimal change in inner energy
δQ − infinitesimal change in heat
δW – infinitesimal change in work/infinitesimal work performed
p – pressure
dV – infinitesimal change in volume
[One writes δX instead of dX to indicate that X is not a state variable
but depends on the process performed; to get from δX to X you have to
integrate over a path.]
>>> 'Equal' means, that the number on the left side of the equation sign is
>>> the same as on the right side, if all the algebraic calculations are
>>> carried out on bot sides (and differences in sign are leveled out).
>> No, differences in sign are NOT "leveled out".
>
> Sure they are.
No, they are not.
> Since energy is positive and the LOSS of mass is negative, we need to
> add a negative sign somewhere.
Milchmädchenrechnung.
> But I personally think, the equation (E=m*c²) needs also 'deltas', since
> 'mass lost' addresses a difference, for what you should use a delta.
If you knew what you are talking about, you would know that Einstein already
basically suggested
∆m = ∆E∕c²
in the 1905 paper that you cited (he used words instead of writing an equation).
--
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 | 2018-12-08 14:18 +0100 |
| Message-ID | <g71uloFubu2U1@mid.individual.net> |
| In reply to | #479239 |
Am 07.12.2018 um 16:15 schrieb Thomas 'PointedEars' Lahn: > Thomas Heger wrote: >> Am 06.12.2018 um 22:47 schrieb Thomas 'PointedEars' Lahn: >>> Then energy/work is >>> >>> E = W = F s = m a s = m v/t s = m s/t² s = m s²/t² = m (s/t)² = m v². >>> >>> The far right-hand side should ring a bell if you think of kinetic >>> energy, the energy of motion: T = E_kin = 1/2 m v² (Newtonian). >>> >>> c is the *speed* of light, so E₍₀₎ = m c² makes sense dimensionally.] >> >> Energy and work are not the same thing. > > Dimensionally they are exactly the same because of how energy is defined > (AISB in the part that you did not quote). > >> Energy is a scalar unit, > > Energy is a _quantity_, not a unit. Well, yes. English is a foreign language for me and occasionally I make mistakes. > >> hence cannot become negative > > But it can. There is a difference between mathematics and the real world. If we want to regard physics as science, we must always prefer the 'real world perspective' (over the mathematical). So a quantity in nature is called 'energy' and this cannot become negative (by definition of the term 'energy'). > >> and has no direction. > > A matter of interpretation. There are processes that transport energy in a > particular direction. The '*direction of transport' seems to correspond to 'arrow of time'. So a positive direction of time should always be regarded as positive, hence the path determines the direction, where signs are positive. Since you cannot create (or transport) something, that absorbs energy without a trace, you must regard the path of something as positive and its energetic features also as positive. > >> Work is a vectorial unit > > Nonsense. > >> and has a direction and can eventually become negative > > No, that the value of work can become negative is indicative instead of work > performed *on* a system as compared to work performed *by* a system. > > This is only a *convention*. > > For example, if you push a piston to reduce the volume of the gas in a > cylinder, then you perform work *on* that system, and the convention is > that then this work has a positive sign. > > If instead the piston is pushed outwards, increasing the volume of the gas > in a cylinder, then work is performed *by* the system on the environment, > and it follows that this work must have a negative sign. You address the problem of pulling a piston out and creating a lower pressure. But this is - of course - also positive work. If the piston pushes outwards against some kind of resistance, then we have also positive work, done by that piston. Work is therefor positive, if seen from one side and negative, if seen from the other. This is why work should be a vector, while energy is not. .. TH
[toc] | [prev] | [next] | [standalone]
| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2018-12-08 13:43 +0000 |
| Message-ID | <pughq5$118f$1@gioia.aioe.org> |
| In reply to | #479322 |
Thomas Heger <ttt_heg@web.de> wrote: > Am 07.12.2018 um 16:15 schrieb Thomas 'PointedEars' Lahn: >> Thomas Heger wrote: >>> Am 06.12.2018 um 22:47 schrieb Thomas 'PointedEars' Lahn: >>>> Then energy/work is >>>> >>>> E = W = F s = m a s = m v/t s = m s/t² s = m s²/t² = m (s/t)² = m v². >>>> >>>> The far right-hand side should ring a bell if you think of kinetic >>>> energy, the energy of motion: T = E_kin = 1/2 m v² (Newtonian). >>>> >>>> c is the *speed* of light, so E₍₀₎ = m c² makes sense dimensionally.] >>> >>> Energy and work are not the same thing. >> >> Dimensionally they are exactly the same because of how energy is defined >> (AISB in the part that you did not quote). >> >>> Energy is a scalar unit, >> >> Energy is a _quantity_, not a unit. > > Well, yes. English is a foreign language for me and occasionally I make > mistakes. > >> >>> hence cannot become negative >> >> But it can. > > There is a difference between mathematics and the real world. > > If we want to regard physics as science, we must always prefer the 'real > world perspective' (over the mathematical). > > So a quantity in nature is called 'energy' and this cannot become > negative (by definition of the term 'energy'). > >> >>> and has no direction. >> >> A matter of interpretation. There are processes that transport energy in a >> particular direction. > > The '*direction of transport' seems to correspond to 'arrow of time'. > > So a positive direction of time should always be regarded as positive, > hence the path determines the direction, where signs are positive. > > > Since you cannot create (or transport) something, that absorbs energy > without a trace, you must regard the path of > something as positive and its energetic features also as positive. >> >>> Work is a vectorial unit >> >> Nonsense. >> >>> and has a direction and can eventually become negative >> >> No, that the value of work can become negative is indicative instead of work >> performed *on* a system as compared to work performed *by* a system. >> >> This is only a *convention*. >> >> For example, if you push a piston to reduce the volume of the gas in a >> cylinder, then you perform work *on* that system, and the convention is >> that then this work has a positive sign. >> >> If instead the piston is pushed outwards, increasing the volume of the gas >> in a cylinder, then work is performed *by* the system on the environment, >> and it follows that this work must have a negative sign. > > You address the problem of pulling a piston out and creating a lower > pressure. > > But this is - of course - also positive work. > > If the piston pushes outwards against some kind of resistance, then we > have also positive work, done by that piston. > > Work is therefor positive, if seen from one side and negative, if seen > from the other. > > This is why work should be a vector, while energy is not. > > .. > > TH > You make a couple of mistakes. First of all, energy is not positive-definite. I don’t know why you think it is. Secondly, a scalar can have either a positive or negative sign. Being signed doesn’t make it a vector. -- Odd Bodkin -- maker of fine toys, tools, tables
[toc] | [prev] | [next] | [standalone]
| From | "David (Kronos Prime) Fuller" <fuller.david@hotmail.com> |
|---|---|
| Date | 2018-12-08 06:01 -0800 |
| Message-ID | <2fc5dd4e-1d1b-41bc-a65b-b6389a593107@googlegroups.com> |
| In reply to | #479325 |
Odd Bodkin escrito You make a couple of mistakes. First of all, energy is not positive-definite. I don’t know why you think it is. Secondly, a scalar can have either a positive or negative sign. Being signed doesn’t make it a vector. Using “current mainstream Dimensionally accurate laws of physics”, The Entire universe is a giant violation of the law of conservation of Energy & all the laws of physics. That’s why the mainstream is furiously groping in the dark for (dark energy & matter) to balance their error. To imply “Negative Energy as valid” is mysticism & witch doctoring
[toc] | [prev] | [next] | [standalone]
| From | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-08 18:18 +0100 |
| Message-ID | <g72cn4F2stjU1@mid.individual.net> |
| In reply to | #479326 |
Am 08.12.2018 um 15:01 schrieb David (Kronos Prime) Fuller: > Odd Bodkin escrito > > You make a couple of mistakes. > > First of all, energy is not positive-definite. I don’t know why you think it is. > > Secondly, a scalar can have either a positive or negative sign. > Being signed doesn’t make it a vector. Vectors have a direction. A Joule for instance or a kilo-Watt-hour has no direction. Numbers often have signs, what does not turn them into vectors. > Using “current mainstream Dimensionally accurate laws of physics”, The Entire universe is a giant violation of the law of conservation of Energy & all the laws of physics. > > That’s why the mainstream is furiously groping in the dark for (dark energy & matter) to balance their error. > > To imply “Negative Energy as valid” is mysticism & witch doctoring > Energy must be a scalar quantity and always positive. You can - nevertheless - subtract energy, which is somehow similar to negative energy. But in the real word there should be a thing or something similar, which has an attribute, that you could eventually call 'negative energy'. Such an 'energy absorber' should absorb energy and add that to its own 'negative energy' to nothing (zero). But I have never heard of such a thing and actually cannot imagine one. Sure, things can absorb heat, for instance. But then (positive) thermal energy is transferred from a donator to an absorber (or: from source to sink, if you prefer these terms). There is still no negative energy involved. TH
[toc] | [prev] | [next] | [standalone]
| From | Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> |
|---|---|
| Date | 2018-12-08 19:26 +0100 |
| Message-ID | <puh2dj$ei8$1@dont-email.me> |
| In reply to | #479345 |
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. 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. What is a vector and does have direction is e.g an energy flow, e.g power per square meter. -- 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 | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-09 19:35 +0100 |
| Message-ID | <g755jaFlh6sU1@mid.individual.net> |
| In reply to | #479358 |
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'. 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
[toc] | [prev] | [next] | [standalone]
| From | Libor Striz <poutnik4REMOVEnntp@CAPITALSgmail.com.INVALID> |
|---|---|
| Date | 2018-12-09 20:24 +0100 |
| Message-ID | <pujq5r$kp4$1@dont-email.me> |
| In reply to | #479472 |
Thomas Heger <ttt_heg@web.de> Wrote in message: > > > We have a similar problem with 'speed' and 'velocity'. > There is no problem with them, but with people confusing both. > > > work is NOT 'forth times path' > > but ' force times distance'. Well, that is definition of work from physics lessons of the 8th grade of the basic school. Students of a high school are taught work is force times distance' times cosine of their angle. Students at a collage or a university already know that magnitude and direction of force can be variable and the path the force is acting along need not to be straight. Therefore the integral definition of work along a generally curved path - the endpoint distance can be even zero - is the path integral of force W = integral (path L)[ vec_F . vec_dL] Where . means scalar aka dot product of vector F and vector dL. Imagine you are walking across a forest along a curved path. You have always some direction of your motion ( vector dL) and the length of your path is bigger than distance of endpoints of your trip. > > 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 > > -- 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 | Thomas Heger <ttt_heg@web.de> |
|---|---|
| Date | 2018-12-10 18:48 +0100 |
| Message-ID | <g77n6uF84a3U1@mid.individual.net> |
| In reply to | #479482 |
Am 09.12.2018 um 20:24 schrieb Libor Striz: >> >> work is NOT 'forth times path' >> >> but ' force times distance'. > > Well, that is definition of work from physics lessons > of the 8th grade of the basic school. > > Students of a high school are taught > work is force times distance' times cosine of their angle. Sorry, no. The angle is meant as the angle between the applied force and the actual movement. Even if this is correct, the force in the equation work =force times distance is meant - of course - as the component of the force in the actual direction of movement. > > Students at a collage or a university already know that magnitude > and direction of force can be variable > and the path the force is acting along need not to be straight. > > Therefore the integral definition of work along a generally curved > path - the endpoint distance can be even zero > - is the path integral of force > W = integral (path L)[ vec_F . vec_dL] > Where . means scalar aka dot product of vector F and vector dL. > Now you provided a mathematical proof, that collages produce brainwashed zombies. > Imagine you are walking across a forest along a curved path. 'Scary'... So they scare the poor students, to believe in any nonsense they are supposed to swallow. TH
[toc] | [prev] | [next] | [standalone]
| From | Rich Carpenter <lcug@kigcohc.pa> |
|---|---|
| Date | 2018-12-10 19:37 +0000 |
| Message-ID | <pumfa9$brf$2@gioia.aioe.org> |
| In reply to | #479541 |
Thomas Heger wrote: >> Well, that is definition of work from physics lessons of the 8th grade >> of the basic school. >> Students of a high school are taught work is force times distance' >> times cosine of their angle. > > Sorry, no. > The angle is meant as the angle between the applied force and the actual > movement. Even if this is correct, the force in the equation Not the actual movement, but rather the next. As numerically it takes place in a loop piecewise. Thus it depends on the scheme used. Piece of cake, for the experienced. You guys are not even new beginners.
[toc] | [prev] | [next] | [standalone]
| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2018-12-10 22:27 +0100 |
| Message-ID | <d3673db0-83ef-8ffa-9ff6-90f55e300c96@PointedEars.de> |
| In reply to | #479541 |
Thomas Heger wrote:
> Am 09.12.2018 um 20:24 schrieb Libor Striz:
>>> work is NOT 'forth times path'
>>>
>>> but ' force times distance'.
>>
>> Well, that is definition of work from physics lessons
>> of the 8th grade of the basic school.
>>
>> Students of a high school are taught
>> work is force times distance' times cosine of their angle.
>
> Sorry, no.
Yes they are, for all schools comparable to the U.S. American “high school”.
Probably you never went to one.
> The angle is meant as the angle between the applied force and the actual
> movement.
“Movement” is not a physical quantity, you blithering fool.
The angle between the force vector (F⃗) and the distance vector (ds⃗, and
s⃗) is for motion on a straight line *equal* to the angle between the force
vector and the *velocity* (vector) v⃗, because both the distance vector and
the velocity vector are *tangent* vectors on the trajectory then.
Therefore (AISB), we *can* substitute
W = ∫ F⃗ · ds⃗ = ∫ F⃗ · ds⃗/dt dt = ∫ F⃗ · v⃗ dt.
And for a constant force F⃗, we can eliminate the constant from the integral
(because adding products with the same factor is the same as adding the
other factors and then multiplying the sum with the common factor –
basically *Grundschul*-arithmetics):
W = ∫ F · v⃗ dt = F ∫ ds⃗/dt dt = F r⃗.
For motion parallel to one axis:
W = ∫ F · v dt = F ∫ ds/dt dt = F ∫ ds = F s.
[See below for the proof.
I have used “s” now instead of ”r” to make it more obvious.]
> Even if this is correct,
It is.
> the force in the equation
>
> work =force times distance
>
> is meant - of course - as the component of the force in the actual
> direction of movement.
Which is why it is a vector, in general.
Only in the case where the motion is parallel to one of the coordinate axes
one can simplify and consider both it and the infinitesimal displacement a
scalar, because *then*
F⃗ · ds⃗ = (F_x, 0, 0) · (ds_x, 0, 0) = F_x · ds_x + 0 · 0 + 0 · 0
= F_x · ds_x = F ds
[(e.g. for motion parallel to the x axis)]
Because the scalar product (·) is defined in 3 spatial Euclidean dimensions as
a⃗ · b⃗ = (a_x, a_y, a_z) · (b_x, b_y, b_z) = a_x b_x + a_y b_y + a_z b_z
and
a⃗ · b⃗ = |a⃗| · |b⃗| · cos(∡(a⃗, b⃗)) = |a⃗| · |b⃗| · cos φ,
with φ being the angle between a⃗ and b⃗ (*highschool* geometry).
Which is *precisely* what Libor Striz wrote above.
>> Students at a collage or a university already know that magnitude
>> and direction of force can be variable
>> and the path the force is acting along need not to be straight.
>>
>> Therefore the integral definition of work along a generally curved
>> path - the endpoint distance can be even zero
>> - is the path integral of force
>> W = integral (path L)[ vec_F . vec_dL]
>> Where . means scalar aka dot product of vector F and vector dL.
>>
> 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.
--
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 | 2018-12-10 22:35 +0100 |
| Message-ID | <7d150345-9db1-b947-8ad4-7121c0c9477e@PointedEars.de> |
| In reply to | #479562 |
Thomas 'PointedEars' Lahn wrote: > Only in the case where the motion is parallel to one of the coordinate axes > one can simplify and consider both it and the infinitesimal displacement a > scalar, because *then* > > F⃗ · ds⃗ = (F_x, 0, 0) · (ds_x, 0, 0) = F_x · ds_x + 0 · 0 + 0 · 0 > = F_x · ds_x = F ds > > [(e.g. for motion parallel to the x axis)] I should say: “… where the motion is in the direction of the force, and we choose the orientation of the coordinate axes so that the direction of one coincides with the direction of either quantity.” [One can see that this is equally true if only either the y- or the z-components of the quantities are non-zero.] -- 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 | 2018-12-12 16:45 +0100 |
| Message-ID | <g7coovFbm2hU1@mid.individual.net> |
| In reply to | #479562 |
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. The world we observe is internal structure of this real physical entity. I call that concept 'structured spacetime'. I have written kind of book about this idea, which you find here: https://docs.google.com/present/view?id=dd8jz2tx_3gfzvqgd6 I had to change a few fundamental assumptions about our world, but the concept works quite well. E.g. I try to promote a certain mathematical construct as good representation. That is a certain algebra, based on bi-quaternions. (Those are also called 'complex four vectors'.) My assumption is, that nature uses at a certain fundamental level methods similar to mathematics and uses that specific algebra. This would build a field I call 'spacetime'. By being somewhere an observer sees the internal structure from that point. This is then like a 'cut', that generates the world we see. But this is only the world seen from that point, while infinitely many other points exist, from which other worlds could be seen. TH
[toc] | [prev] | [next] | [standalone]
Page 2 of 6 — ← Prev page 1 [2] 3 4 5 6 Next page →
Back to top | Article view | sci.physics.relativity
csiph-web