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Link between dark matter and baryonic matter
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wor...@yandex.ru
Sep 28, 2016, 5:54:40 AM9/28/16
to
Hello!
I just read the work: https://arxiv.org/pdf/1609.05917v1.pdf
The authors conclusion: "The dark and baryonic mass are strongly coupled."
Only now they have noticed this?
8 years ago I came to the same conclusion. One can even read this ratio directly from rotation curves: they oscillate. Now, the authors speculate what may be the cause. And I have the explanation already ready ;)
https://sites.google.com/site/testsofphysicaltheories/English/dark-matter
"The dark matter is to be concentrated in the spiral arms. This contradicts the assumption that it consists from particles that surround the galaxy like a halo. On the other hand, this is strongly reminiscent of the asteroid belt in our solar system or rings around Saturn and Jupiter. Therefore, it may act on dark matter around small cosmic bodies, such as rocks, asteroids, planets, comets, and snowballs. They are small to be seen from a distance, but in principle they can make up the bulk of our galaxy and other galaxies."
Or see here: https://groups.google.com/d/msg/sci.astro/VG5gJbLN59k/be691fU5sxIJ
Best regards
Walter Orlov
I just read the work: https://arxiv.org/pdf/1609.05917v1.pdf
The authors conclusion: "The dark and baryonic mass are strongly coupled."
Only now they have noticed this?
8 years ago I came to the same conclusion. One can even read this ratio directly from rotation curves: they oscillate. Now, the authors speculate what may be the cause. And I have the explanation already ready ;)
https://sites.google.com/site/testsofphysicaltheories/English/dark-matter
"The dark matter is to be concentrated in the spiral arms. This contradicts the assumption that it consists from particles that surround the galaxy like a halo. On the other hand, this is strongly reminiscent of the asteroid belt in our solar system or rings around Saturn and Jupiter. Therefore, it may act on dark matter around small cosmic bodies, such as rocks, asteroids, planets, comets, and snowballs. They are small to be seen from a distance, but in principle they can make up the bulk of our galaxy and other galaxies."
Or see here: https://groups.google.com/d/msg/sci.astro/VG5gJbLN59k/be691fU5sxIJ
Best regards
Walter Orlov
Yousuf Khan
Sep 28, 2016, 9:04:56 AM9/28/16
to
On 9/28/2016 5:54 AM, wor...@yandex.ru wrote:
> Hello!
>
> I just read the work: https://arxiv.org/pdf/1609.05917v1.pdf
> The authors conclusion: "The dark and baryonic mass are strongly coupled."
>
> Only now they have noticed this?
It sounds like they are rediscovering MOND.
> Hello!
>
> I just read the work: https://arxiv.org/pdf/1609.05917v1.pdf
> The authors conclusion: "The dark and baryonic mass are strongly coupled."
>
> Only now they have noticed this?
Yousuf Khan
wor...@yandex.ru
Sep 28, 2016, 9:38:37 AM9/28/16
to
Am Mittwoch, 28. September 2016 15:04:56 UTC+2 schrieb Yousuf Khan:
>
> It sounds like they are rediscovering MOND.
>
If one definitely want a new physics, then comes into question only the old MOND :)
>
> It sounds like they are rediscovering MOND.
>
dlzc
Sep 28, 2016, 10:55:10 AM9/28/16
to
Dear wor...:
On Wednesday, September 28, 2016 at 2:54:40 AM UTC-7, wor...@yandex.ru wrote:
> Hello!
>
> I just read the work: https://arxiv.org/pdf/1609.05917v1.pdf
> The authors conclusion: "The dark and baryonic mass
> are strongly coupled."
>
> Only now they have noticed this?
Actually there are papers at the same source, that agree that all Dark Matter could be normal, baryonic matter, but it creates the problem of evaluating "by some other means" the amount of matter near the rim, since luminosity drops off so quickly (due to dust, etc.).
Recall that "Dark" originated from establishing a mass / luminosity ratio at the center of a spiral galaxy. We have since found out what a special place that location is, as compared to elsewhere in a spiral's disc... no significant dust, stars without much in the way of a photosphere (due to tidal effects), and likely very few planets of any size.
Since we no longer rely on mass / luminosity, but can in most cases verify rotation curves with microlensing (at least locally), we no longer need to use failed methods.
Neither MOND nor exotic matter are *required*, nor do they show up in our own solar system. So we are NOT in a special place after all.
If we are into opinions here...
David A. Smith
On Wednesday, September 28, 2016 at 2:54:40 AM UTC-7, wor...@yandex.ru wrote:
> Hello!
>
> I just read the work: https://arxiv.org/pdf/1609.05917v1.pdf
> The authors conclusion: "The dark and baryonic mass
> are strongly coupled."
>
> Only now they have noticed this?
Recall that "Dark" originated from establishing a mass / luminosity ratio at the center of a spiral galaxy. We have since found out what a special place that location is, as compared to elsewhere in a spiral's disc... no significant dust, stars without much in the way of a photosphere (due to tidal effects), and likely very few planets of any size.
Since we no longer rely on mass / luminosity, but can in most cases verify rotation curves with microlensing (at least locally), we no longer need to use failed methods.
Neither MOND nor exotic matter are *required*, nor do they show up in our own solar system. So we are NOT in a special place after all.
If we are into opinions here...
David A. Smith
Yousuf Khan
Oct 16, 2016, 1:45:21 AM10/16/16
to
On 9/28/2016 10:54 AM, dlzc wrote:
> Actually there are papers at the same source, that agree that all Dark Matter could be normal, baryonic matter, but it creates the problem of evaluating "by some other means" the amount of matter near the rim, since luminosity drops off so quickly (due to dust, etc.).
>
> Recall that "Dark" originated from establishing a mass / luminosity ratio at the center of a spiral galaxy. We have since found out what a special place that location is, as compared to elsewhere in a spiral's disc... no significant dust, stars without much in the way of a photosphere (due to tidal effects), and likely very few planets of any size.
>
> Since we no longer rely on mass / luminosity, but can in most cases verify rotation curves with microlensing (at least locally), we no longer need to use failed methods.
>
> Neither MOND nor exotic matter are *required*, nor do they show up in our own solar system. So we are NOT in a special place after all.
>
> If we are into opinions here...
Maybe the issue here is not to find a new modification of Newtonian
> Actually there are papers at the same source, that agree that all Dark Matter could be normal, baryonic matter, but it creates the problem of evaluating "by some other means" the amount of matter near the rim, since luminosity drops off so quickly (due to dust, etc.).
>
> Recall that "Dark" originated from establishing a mass / luminosity ratio at the center of a spiral galaxy. We have since found out what a special place that location is, as compared to elsewhere in a spiral's disc... no significant dust, stars without much in the way of a photosphere (due to tidal effects), and likely very few planets of any size.
>
> Since we no longer rely on mass / luminosity, but can in most cases verify rotation curves with microlensing (at least locally), we no longer need to use failed methods.
>
> Neither MOND nor exotic matter are *required*, nor do they show up in our own solar system. So we are NOT in a special place after all.
>
> If we are into opinions here...
gravity, but perhaps our reliance on still using Newtonian gravity even
100 years after we found a better theory of gravity might be the problem
here? We're still using Newtonian gravity after all of these years,
because it's frankly much easier to calculate with than General
Relativity. And we're still confidant in its validity, because we
imagine that it is "still good enough". GR isn't calculating
easy-to-understand force-distance relationships, instead it's
calculating curvatures in spacetime. But in a many-body system such as
stars in a galaxy or galaxies in a universe, those simple
inverse-distance squared relationships simply don't work out anymore?
We're still using Newtonian gravity in this day and age because we still
don't have computers strong enough to do a GR calculation for an entire
galaxy. Using even our strongest supercomputers we can do perhaps a
simulation of only a few million stars in a galaxy using GR, but our
galaxy contains perhaps as much as 400 billion stars, so we keep
approximating with Newton. If one day we can do a full simulation of the
Milky Way with all of its entire 400 billion stars, then likely we'll
see surprising results coming out of GR that are inconsistent with
Newton, and then we'll be finally shaken of our illusion that Newton is
"still good enough".
Yousuf Khan
dlzc
Oct 16, 2016, 1:05:44 PM10/16/16
to
Dear Yousuf Khan:
On Saturday, October 15, 2016 at 10:45:21 PM UTC-7, Yousuf Khan wrote:
...
> Maybe the issue here is not to find a new
> modification of Newtonian gravity, but perhaps
> our reliance on still using Newtonian gravity
> even 100 years after we found a better theory
> of gravity might be the problem here?
I find it more likely that a nearly 100 year old assumption that luminosity is directly proportional to the amount of mass present, when it has long been known that luminosity drops off rapidly with surface temperature. If you have cooler objects, they simply don't put out as much light... especially in the visible light bands.
> We're still using Newtonian gravity after all
> of these years, because it's frankly much easier
> to calculate with than General Relativity.
Paper on this subject for a "simple" galaxy, and evaluating the possible error between Newtonian gravity-as-a-force and GR, and in that galaxy, it is a 1% (or so) error, not the necessary 300% error.
> But in a many-body system such as stars in a
> galaxy or galaxies in a universe, those simple
> inverse-distance squared relationships simply
> don't work out anymore?
They do work out "well enough", for simple gravitation.
But we are "blind as bats" at these scales, and have a full complement of "flatlander fallacies" that we have to divest ourselves of.
> We're still using Newtonian gravity in this day
> and age because we still don't have computers
> strong enough to do a GR calculation for an
> entire galaxy.
False. The amount of computer time might still be abysmally long for an interesting galaxy, but it would still be doable. After all, Nature does this math in real time...
> Using even our strongest supercomputers we can
> do perhaps a simulation of only a few million
> stars in a galaxy using GR, but our galaxy contains
> perhaps as much as 400 billion stars, so we keep
> approximating with Newton.
Yet, even small spirals show a need for Dark Matter. Globular clusters, essentially don't.
> If one day we can do a full simulation of the
> Milky Way with all of its entire 400 billion stars,
> then likely we'll see surprising results coming out
> of GR that are inconsistent with Newton, and then
> we'll be finally shaken of our illusion that Newton
> is "still good enough".
Maybe. But the speeds and curvature on something the size of a galaxy, even the Milky Way, should present minimal error in using Newton.
Now what I wonder is, if the "perfectly mirrored, massless box, containing photons", which has rest mass, exists between a star and the gases/dust/planets that give that star a background temperature higher than the CMBR. So some Dark Matter (probably less than 1%) might still be photons in transit between intersystem objects...?
David A. Smith
On Saturday, October 15, 2016 at 10:45:21 PM UTC-7, Yousuf Khan wrote:
...
> Maybe the issue here is not to find a new
> modification of Newtonian gravity, but perhaps
> our reliance on still using Newtonian gravity
> even 100 years after we found a better theory
> of gravity might be the problem here?
> We're still using Newtonian gravity after all
> of these years, because it's frankly much easier
> to calculate with than General Relativity.
> But in a many-body system such as stars in a
> galaxy or galaxies in a universe, those simple
> inverse-distance squared relationships simply
> don't work out anymore?
But we are "blind as bats" at these scales, and have a full complement of "flatlander fallacies" that we have to divest ourselves of.
> We're still using Newtonian gravity in this day
> and age because we still don't have computers
> strong enough to do a GR calculation for an
> entire galaxy.
> Using even our strongest supercomputers we can
> do perhaps a simulation of only a few million
> stars in a galaxy using GR, but our galaxy contains
> perhaps as much as 400 billion stars, so we keep
> approximating with Newton.
> If one day we can do a full simulation of the
> Milky Way with all of its entire 400 billion stars,
> then likely we'll see surprising results coming out
> of GR that are inconsistent with Newton, and then
> we'll be finally shaken of our illusion that Newton
> is "still good enough".
Now what I wonder is, if the "perfectly mirrored, massless box, containing photons", which has rest mass, exists between a star and the gases/dust/planets that give that star a background temperature higher than the CMBR. So some Dark Matter (probably less than 1%) might still be photons in transit between intersystem objects...?
David A. Smith
Steve Willner
Oct 19, 2016, 3:06:01 PM10/19/16
to
In article <jKCdna6p3aF3jp7F...@giganews.com>,
gravity provided gravity is not "strong" and speeds are "low"
compared to the speed of light. The magnitude of the errors can be
quantified for the actual gravitational potentials and speeds in any
particular calculation.
The real problems are 1) even Newtonian gravity is too hard to
calculate when the system has too many bodies, and 2) treating stars
and gas together in a single calculation is difficult. These
problems require approximations to be made, and in some cases these
approximations may be inaccurate enough to affect results. That's
not to say every calculation ever done is hopelessly wrong, but any
result that depends critically on what happens when two stars come
very close together is uncertain.
People who do these calculations are well aware of the uncertainties,
but those seldom get transmitted in popular articles.
--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
Cambridge, MA 02138 USA
Yousuf Khan <bbb...@spammenot.yahoo.com> writes:
> We're still using Newtonian gravity in this day and age because we still
> don't have computers strong enough to do a GR calculation for an entire
> galaxy.
That's not the problem. GR is mathematically identical to Newtonian
> We're still using Newtonian gravity in this day and age because we still
> don't have computers strong enough to do a GR calculation for an entire
> galaxy.
gravity provided gravity is not "strong" and speeds are "low"
compared to the speed of light. The magnitude of the errors can be
quantified for the actual gravitational potentials and speeds in any
particular calculation.
The real problems are 1) even Newtonian gravity is too hard to
calculate when the system has too many bodies, and 2) treating stars
and gas together in a single calculation is difficult. These
problems require approximations to be made, and in some cases these
approximations may be inaccurate enough to affect results. That's
not to say every calculation ever done is hopelessly wrong, but any
result that depends critically on what happens when two stars come
very close together is uncertain.
People who do these calculations are well aware of the uncertainties,
but those seldom get transmitted in popular articles.
--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
Cambridge, MA 02138 USA
Steve Willner
Oct 19, 2016, 3:15:35 PM10/19/16
to
In article <6c2ab24e-fdd0-442e...@googlegroups.com>,
> luminosity= is directly proportional to the amount of mass
> present, when it has long b= een known that luminosity drops off
> rapidly with surface temperature.
As you say, the dependence of luminosity on temperature -- more
generally on stellar population -- is well known. It is taken into
account as well as possible given the data available, and the
resulting uncertainties are understood.
> If y= ou have cooler objects, they simply don't put out as much
doesn't eliminate the uncertainties altogether, though.
> Now what I wonder is, if the "perfectly mirrored, massless box,
> containing = photons", which has rest mass, exists between a star
Photons have energy, which contributes to gravitation, but they don't
have rest mass.
> and the gases/dust/pla= nets that give that star a background
> temperature higher than the CMBR. So= some Dark Matter (probably
> less than 1%) might still be photons in transit= between
> intersystem objects...?
Light has to be considered separately from matter in the cosmological
equations because its energy density decreases as the fourth power of
scale factor. The energy density of light has been less than that of
matter since the first several minutes of cosmic time, and its energy
density is negligible at later epochs.
> luminosity= is directly proportional to the amount of mass
> present, when it has long b= een known that luminosity drops off
> rapidly with surface temperature.
As you say, the dependence of luminosity on temperature -- more
generally on stellar population -- is well known. It is taken into
account as well as possible given the data available, and the
resulting uncertainties are understood.
> If y= ou have cooler objects, they simply don't put out as much
> light... especially in the visible light bands.
As you indicate, working in the infrared helps quite a bit. It
doesn't eliminate the uncertainties altogether, though.
> Now what I wonder is, if the "perfectly mirrored, massless box,
Photons have energy, which contributes to gravitation, but they don't
have rest mass.
> and the gases/dust/pla= nets that give that star a background
> temperature higher than the CMBR. So= some Dark Matter (probably
> less than 1%) might still be photons in transit= between
> intersystem objects...?
Light has to be considered separately from matter in the cosmological
equations because its energy density decreases as the fourth power of
scale factor. The energy density of light has been less than that of
matter since the first several minutes of cosmic time, and its energy
density is negligible at later epochs.
dlzc
Oct 19, 2016, 7:30:12 PM10/19/16
to
Dear Steve Willner:
On Wednesday, October 19, 2016 at 12:15:35 PM UTC-7, Steve Willner wrote:
> In article <6c2ab24e-fdd0-442e-97bb-c1a1321b***c...@googlegroups.com>,
> dlzc <dl***@cox.net> writes:
...
> > and the gases/dust/pla= nets that give that star a background
> > temperature higher than the CMBR. So= some Dark Matter (probably
> > less than 1%) might still be photons in transit= between
> > intersystem objects...?
>
> Light has to be considered separately from matter in the
> cosmological equations because its energy density decreases
> as the fourth power of scale factor. The energy density of
> light has been less than that of matter since the first several
> minutes of cosmic time, and its energy density is negligible at
> later epochs.
As I said, I did not expect it to be even 1%, much less what is required to be Dark Matter.
Thank you.
David A. Smith
On Wednesday, October 19, 2016 at 12:15:35 PM UTC-7, Steve Willner wrote:
> In article <6c2ab24e-fdd0-442e-97bb-c1a1321b***c...@googlegroups.com>,
> dlzc <dl***@cox.net> writes:
...
> > Now what I wonder is, if the "perfectly mirrored, massless box,
> > containing = photons", which has rest mass, exists between a star
>
> Photons have energy, which contributes to gravitation, but
> they don't have rest mass.
Individually, no. But in groups, with a center of momentum frame, they do have rest mass. And hence gravitational mass.
> > containing = photons", which has rest mass, exists between a star
>
> Photons have energy, which contributes to gravitation, but
> they don't have rest mass.
> > and the gases/dust/pla= nets that give that star a background
> > temperature higher than the CMBR. So= some Dark Matter (probably
> > less than 1%) might still be photons in transit= between
> > intersystem objects...?
>
> Light has to be considered separately from matter in the
> cosmological equations because its energy density decreases
> as the fourth power of scale factor. The energy density of
> light has been less than that of matter since the first several
> minutes of cosmic time, and its energy density is negligible at
> later epochs.
Thank you.
David A. Smith
Yousuf Khan
Oct 20, 2016, 2:46:24 AM10/20/16
to
On 10/16/2016 1:05 PM, dlzc wrote:
> Dear Yousuf Khan:
>
> On Saturday, October 15, 2016 at 10:45:21 PM UTC-7, Yousuf Khan
> wrote: ...
>> Maybe the issue here is not to find a new modification of
>> Newtonian gravity, but perhaps our reliance on still using
>> Newtonian gravity even 100 years after we found a better theory of
>> gravity might be the problem here?
>
> I find it more likely that a nearly 100 year old assumption that
> luminosity is directly proportional to the amount of mass present,
> when it has long been known that luminosity drops off rapidly with
> surface temperature. If you have cooler objects, they simply don't
> put out as much light... especially in the visible light bands.
But they do still glow in the cooler invisible light bands like IR and
> Dear Yousuf Khan:
>
> On Saturday, October 15, 2016 at 10:45:21 PM UTC-7, Yousuf Khan
> wrote: ...
>> Maybe the issue here is not to find a new modification of
>> Newtonian gravity, but perhaps our reliance on still using
>> Newtonian gravity even 100 years after we found a better theory of
>> gravity might be the problem here?
>
> I find it more likely that a nearly 100 year old assumption that
> luminosity is directly proportional to the amount of mass present,
> when it has long been known that luminosity drops off rapidly with
> surface temperature. If you have cooler objects, they simply don't
> put out as much light... especially in the visible light bands.
microwave and radio.
>> We're still using Newtonian gravity after all of these years,
>> because it's frankly much easier to calculate with than General
>> Relativity.
>
> Paper on this subject for a "simple" galaxy, and evaluating the
> possible error between Newtonian gravity-as-a-force and GR, and in
> that galaxy, it is a 1% (or so) error, not the necessary 300% error.
models, rather than full galaxy models.
>> But in a many-body system such as stars in a galaxy or galaxies in
>> a universe, those simple inverse-distance squared relationships
>> simply don't work out anymore?
>
> They do work out "well enough", for simple gravitation.
>
> But we are "blind as bats" at these scales, and have a full
> complement of "flatlander fallacies" that we have to divest
> ourselves of.
of those flatlander fallacies.
>> We're still using Newtonian gravity in this day and age because we
>> still don't have computers strong enough to do a GR calculation
>> for an entire galaxy.
>
> False. The amount of computer time might still be abysmally long
> for an interesting galaxy, but it would still be doable. After all,
> Nature does this math in real time...
We can barely put two qubits together yet.
>> Using even our strongest supercomputers we can do perhaps a
>> simulation of only a few million stars in a galaxy using GR, but
>> our galaxy contains perhaps as much as 400 billion stars, so we
>> keep approximating with Newton.
>
> Yet, even small spirals show a need for Dark Matter. Globular
> clusters, essentially don't.
dwarf galaxies.
>> If one day we can do a full simulation of the Milky Way with all
>> of its entire 400 billion stars, then likely we'll see surprising
>> results coming out of GR that are inconsistent with Newton, and
>> then we'll be finally shaken of our illusion that Newton is "still
>> good enough".
>
> Maybe. But the speeds and curvature on something the size of a
> galaxy, even the Milky Way, should present minimal error in using
> Newton.
assumption is wrong?
> Now what I wonder is, if the "perfectly mirrored, massless box,
> containing photons", which has rest mass, exists between a star and
> the gases/dust/planets that give that star a background temperature
> higher than the CMBR. So some Dark Matter (probably less than 1%)
> might still be photons in transit between intersystem objects...?
Yousuf Khan
dlzc
Oct 20, 2016, 11:40:26 AM10/20/16
to
Dear Yousuf Khan:
On Wednesday, October 19, 2016 at 11:46:24 PM UTC-7, Yousuf Khan wrote:
> On 10/16/2016 1:05 PM, dlzc wrote:
> > Dear Yousuf Khan:
> >
> > On Saturday, October 15, 2016 at 10:45:21 PM UTC-7, Yousuf Khan
> > wrote: ...
> >> Maybe the issue here is not to find a new modification of
> >> Newtonian gravity, but perhaps our reliance on still using
> >> Newtonian gravity even 100 years after we found a better
> >> theory of gravity might be the problem here?
> >
> > I find it more likely that a nearly 100 year old
> > assumption that luminosity is directly proportional
> > to the amount of mass present, when it has long been
> > known that luminosity drops off rapidly with surface
> > temperature. If you have cooler objects, they simply
> > don't put out as much light... especially in the
> > visible light bands.
>
> But they do still glow in the cooler invisible light
> bands like IR and microwave and radio.
At a *much* lower luminosity. Remember, they use luminosity, essentially watts, and calibrate to normal-mass-present.
> >> We're still using Newtonian gravity after all of these
> >> years, because it's frankly much easier to calculate
> >> with than General Relativity.
> >
> > Paper on this subject for a "simple" galaxy, and
> > evaluating the possible error between Newtonian
> > gravity-as-a-force and GR, and in that galaxy, it
> > is a 1% (or so) error, not the necessary 300%
> > error.
>
> That's the point I'm trying to make, they are using
> "simple" galaxy models, rather than full galaxy models.
Even dwarf spiral galaxies need Dark Matter, however. And they have a few billion stars. This should be doable soon.
GR really kicks in:
- to handle light,
- to handle advancement of perihelion (for close objects),
- to handle gravitational radiation.
> >> But in a many-body system such as stars in a galaxy or
> >> galaxies in a universe, those simple inverse-distance
> >> squared relationships simply don't work out anymore?
> >
> > They do work out "well enough", for simple gravitation.
> >
> > But we are "blind as bats" at these scales, and have
> > a full complement of "flatlander fallacies" that we
> > have to divest ourselves of.
>
> So then we're basically agreeing on this. Newtonian
> gravity might be one of those flatlander fallacies.
Remove the *serious* errors of (normal-mass / luminosity) calibration, and then see if you think a further 5 or 10% (max) correction is necessary.
> >> We're still using Newtonian gravity in this day and
> >> age because we still don't have computers strong
> >> enough to do a GR calculation for an entire galaxy.
> >
> > False. The amount of computer time might still be
> > abysmally long for an interesting galaxy, but it
> > would still be doable. After all, Nature does
> > this math in real time...
>
> Nature has its own entire universe-sized quantum
> computer to work with. We can barely put two qubits
> together yet.
But GR (like Newton), is a classical solution. GR simplifies to Newton, under the right circumstances, circumstances suitable to galaxies "in the large".
I don't think GR explains "Dark Matter", better than Newton does. They both have to accept that there is more matter that our myopic vision cannot detect (except via gravity).
> >> Using even our strongest supercomputers we can do
> >> perhaps a simulation of only a few million stars in
> >> a galaxy using GR, but our galaxy contains perhaps
> >> as much as 400 billion stars, so we keep
> >> approximating with Newton.
> >
> > Yet, even small spirals show a need for Dark Matter.
> > Globular clusters, essentially don't.
>
> Then we need to investigate where the globular clusters
> differ from dwarf galaxies.
There is no significant rotation in a globular cluster, so the normal mass present, is explained by microlensing, and other methods that apply equally well to a spiral's nucleus, or a globular cluster (expected to be ancient cores of spiral galaxies).
> >> If one day we can do a full simulation of the Milky Way
> >> with all of its entire 400 billion stars, then likely
> >> we'll see surprising results coming out of GR that are
> >> inconsistent with Newton, and then we'll be finally
> >> shaken of our illusion that Newton is "still good enough".
> >
> > Maybe. But the speeds and curvature on something the
> > size of a galaxy, even the Milky Way, should present
> > minimal error in using Newton.
>
> Well, that's been our assumption all along hasn't it?
> Maybe our assumption is wrong?
We *know* it is still a classical theory, however.
> > Now what I wonder is, if the "perfectly mirrored,
> > massless box, containing photons", which has rest
> > mass, exists between a star and the gases / dust /
> > planets that give that star a background temperature
> > higher than the CMBR. So some Dark Matter (probably
> > less than 1%) might still be photons in transit
> > between intersystem objects...?
>
> Or even neutrinos.
Amen. Absolutely "dark" too, just not very massive, and in order to stay in the halo (as we observe), would have to be moving damned slowly... so would have to be too numerous to be all of Dark Matter.
David A. Smith
On Wednesday, October 19, 2016 at 11:46:24 PM UTC-7, Yousuf Khan wrote:
> On 10/16/2016 1:05 PM, dlzc wrote:
> > Dear Yousuf Khan:
> >
> > On Saturday, October 15, 2016 at 10:45:21 PM UTC-7, Yousuf Khan
> > wrote: ...
> >> Maybe the issue here is not to find a new modification of
> >> Newtonian gravity, but perhaps our reliance on still using
> >> Newtonian gravity even 100 years after we found a better
> >> theory of gravity might be the problem here?
> >
> > I find it more likely that a nearly 100 year old
> > assumption that luminosity is directly proportional
> > to the amount of mass present, when it has long been
> > known that luminosity drops off rapidly with surface
> > temperature. If you have cooler objects, they simply
> > don't put out as much light... especially in the
> > visible light bands.
>
> But they do still glow in the cooler invisible light
> bands like IR and microwave and radio.
> >> We're still using Newtonian gravity after all of these
> >> years, because it's frankly much easier to calculate
> >> with than General Relativity.
> >
> > Paper on this subject for a "simple" galaxy, and
> > evaluating the possible error between Newtonian
> > gravity-as-a-force and GR, and in that galaxy, it
> > is a 1% (or so) error, not the necessary 300%
> > error.
>
> That's the point I'm trying to make, they are using
> "simple" galaxy models, rather than full galaxy models.
GR really kicks in:
- to handle light,
- to handle advancement of perihelion (for close objects),
- to handle gravitational radiation.
> >> But in a many-body system such as stars in a galaxy or
> >> galaxies in a universe, those simple inverse-distance
> >> squared relationships simply don't work out anymore?
> >
> > They do work out "well enough", for simple gravitation.
> >
> > But we are "blind as bats" at these scales, and have
> > a full complement of "flatlander fallacies" that we
> > have to divest ourselves of.
>
> So then we're basically agreeing on this. Newtonian
> gravity might be one of those flatlander fallacies.
> >> We're still using Newtonian gravity in this day and
> >> age because we still don't have computers strong
> >> enough to do a GR calculation for an entire galaxy.
> >
> > False. The amount of computer time might still be
> > abysmally long for an interesting galaxy, but it
> > would still be doable. After all, Nature does
> > this math in real time...
>
> Nature has its own entire universe-sized quantum
> computer to work with. We can barely put two qubits
> together yet.
I don't think GR explains "Dark Matter", better than Newton does. They both have to accept that there is more matter that our myopic vision cannot detect (except via gravity).
> >> Using even our strongest supercomputers we can do
> >> perhaps a simulation of only a few million stars in
> >> a galaxy using GR, but our galaxy contains perhaps
> >> as much as 400 billion stars, so we keep
> >> approximating with Newton.
> >
> > Yet, even small spirals show a need for Dark Matter.
> > Globular clusters, essentially don't.
>
> Then we need to investigate where the globular clusters
> differ from dwarf galaxies.
> >> If one day we can do a full simulation of the Milky Way
> >> with all of its entire 400 billion stars, then likely
> >> we'll see surprising results coming out of GR that are
> >> inconsistent with Newton, and then we'll be finally
> >> shaken of our illusion that Newton is "still good enough".
> >
> > Maybe. But the speeds and curvature on something the
> > size of a galaxy, even the Milky Way, should present
> > minimal error in using Newton.
>
> Well, that's been our assumption all along hasn't it?
> Maybe our assumption is wrong?
> > Now what I wonder is, if the "perfectly mirrored,
> > massless box, containing photons", which has rest
> > mass, exists between a star and the gases / dust /
> > planets that give that star a background temperature
> > higher than the CMBR. So some Dark Matter (probably
> > less than 1%) might still be photons in transit
> > between intersystem objects...?
>
> Or even neutrinos.
David A. Smith
Yousuf Khan
Oct 25, 2016, 2:18:24 AM10/25/16
to
On 10/20/2016 11:40 AM, dlzc wrote:
> Dear Yousuf Khan:
>
> On Wednesday, October 19, 2016 at 11:46:24 PM UTC-7, Yousuf Khan
> wrote:
>> On 10/16/2016 1:05 PM, dlzc wrote:
>>> I find it more likely that a nearly 100 year old assumption that
>>> luminosity is directly proportional to the amount of mass
>>> present, when it has long been known that luminosity drops off
>>> rapidly with surface temperature. If you have cooler objects,
>>> they simply don't put out as much light... especially in the
>>> visible light bands.
>>
>> But they do still glow in the cooler invisible light bands like IR
>> and microwave and radio.
>
> At a *much* lower luminosity. Remember, they use luminosity,
> essentially watts, and calibrate to normal-mass-present.
Well obviously they did that because humans naturally favor those
> Dear Yousuf Khan:
>
> On Wednesday, October 19, 2016 at 11:46:24 PM UTC-7, Yousuf Khan
> wrote:
>> On 10/16/2016 1:05 PM, dlzc wrote:
>>> I find it more likely that a nearly 100 year old assumption that
>>> luminosity is directly proportional to the amount of mass
>>> present, when it has long been known that luminosity drops off
>>> rapidly with surface temperature. If you have cooler objects,
>>> they simply don't put out as much light... especially in the
>>> visible light bands.
>>
>> But they do still glow in the cooler invisible light bands like IR
>> and microwave and radio.
>
> At a *much* lower luminosity. Remember, they use luminosity,
> essentially watts, and calibrate to normal-mass-present.
wavelengths that we can see. We also didn't take into account the higher
frequency UV, X-ray, and gamma scales. These higher frequency radiation
would be higher luminosity than visible light.
>>> Paper on this subject for a "simple" galaxy, and evaluating the
>>> possible error between Newtonian gravity-as-a-force and GR, and
>>> in that galaxy, it is a 1% (or so) error, not the necessary 300%
>>> error.
>>
>> That's the point I'm trying to make, they are using "simple" galaxy
>> models, rather than full galaxy models.
>
> Even dwarf spiral galaxies need Dark Matter, however. And they have
> a few billion stars. This should be doable soon.
likely going to find out the real differences between a GR model and a
Newtonian model.
> GR really kicks in: - to handle light, - to handle advancement of
> perihelion (for close objects), - to handle gravitational radiation.
inverse-squared distance relationship. The spacetime curvature may not
always equal the inverse-squared law, even at large distances. Maybe
especially at very large distances, such as galactic size ranges. There
were already attempts to model GR over larger scales, such as TeVeS.
>> So then we're basically agreeing on this. Newtonian gravity might
>> be one of those flatlander fallacies.
>
> Remove the *serious* errors of (normal-mass / luminosity)
> calibration, and then see if you think a further 5 or 10% (max)
> correction is necessary.
here. What I think is really going to make the difference is a new way
model the curvature of spacetime with the existing mass. I think when GR
is iterated over many self-dependent iterations, it will result in some
surprising relationships that we hadn't expected to see. Quantum
computing will make this task a lot easier.
One example is in the movie Interstellar. This is where they fed GR
equations into a movie graphics computer, and came up with an image of a
blackhole that nobody imagined in their own heads. The computer had no
preconceived notions of what it should see, it just ate the input data,
ran the equations, can came up with an output image.
>>>> We're still using Newtonian gravity in this day and age because
>>>> we still don't have computers strong enough to do a GR
>>>> calculation for an entire galaxy.
>>>
>>> False. The amount of computer time might still be abysmally long
>>> for an interesting galaxy, but it would still be doable. After
>>> all, Nature does this math in real time...
>>
>> Nature has its own entire universe-sized quantum computer to work
>> with. We can barely put two qubits together yet.
>
> But GR (like Newton), is a classical solution. GR simplifies to
> Newton, under the right circumstances, circumstances suitable to
> galaxies "in the large".
classical physics could be thought to emerge from macroscopic quantum
interactions. So far, we've only been considering the macroscopic
effects themselves, but no one has attempted to build up to a classical
solution from a series of quantum solutions.
>>>> Using even our strongest supercomputers we can do perhaps a
>>>> simulation of only a few million stars in a galaxy using GR,
>>>> but our galaxy contains perhaps as much as 400 billion stars,
>>>> so we keep approximating with Newton.
>>>
>>> Yet, even small spirals show a need for Dark Matter. Globular
>>> clusters, essentially don't.
>>
>> Then we need to investigate where the globular clusters differ from
>> dwarf galaxies.
>
> There is no significant rotation in a globular cluster, so the normal
> mass present, is explained by microlensing, and other methods that
> apply equally well to a spiral's nucleus, or a globular cluster
> (expected to be ancient cores of spiral galaxies).
in place with all of that gravity between them without there being
curved motion.
>>>> If one day we can do a full simulation of the Milky Way with
>>>> all of its entire 400 billion stars, then likely we'll see
>>>> surprising results coming out of GR that are inconsistent with
>>>> Newton, and then we'll be finally shaken of our illusion that
>>>> Newton is "still good enough".
>>>
>>> Maybe. But the speeds and curvature on something the size of a
>>> galaxy, even the Milky Way, should present minimal error in using
>>> Newton.
>>
>> Well, that's been our assumption all along hasn't it? Maybe our
>> assumption is wrong?
>
> We *know* it is still a classical theory, however.
it's quite a significant departure from Newton, just as Newton was a
departure from Aristotle's gravity.
>>> Now what I wonder is, if the "perfectly mirrored, massless box,
>>> containing photons", which has rest mass, exists between a star
>>> and the gases / dust / planets that give that star a background
>>> temperature higher than the CMBR. So some Dark Matter (probably
>>> less than 1%) might still be photons in transit between
>>> intersystem objects...?
>>
>> Or even neutrinos.
>
> Amen. Absolutely "dark" too, just not very massive, and in order to
> stay in the halo (as we observe), would have to be moving damned
> slowly... so would have to be too numerous to be all of Dark Matter.
LY across on its disk, probably 1 million LY across on its halo. So
neutrinos released now from the various stars of the galaxy will remain
within the borders of the galaxy for between 100,000 to 1 million years.
Enough time to remain a part of the galaxy's mass. Even photons will do
the same. So there's a large amount of time that energy will remain
within the borders of a galaxy, and even larger amount of time that
it'll remain within the borders of a galactic cluster or supercluster.
Now that we have detected gravitational waves, there's another even more
humongous source of energy (converting several solar masses into energy
at a time) that we know travels at only the speed of light. So lots of
energy stays locked into regions just due to the slow passage of time.
Yousuf Khan
dlzc
Oct 25, 2016, 12:43:13 PM10/25/16
to
Dear Yousuf Khan:
On Monday, October 24, 2016 at 11:18:24 PM UTC-7, Yousuf Khan wrote:
...
> Enough time to remain a part of the galaxy's mass.
> Even photons will do the same. So there's a large
> amount of time that energy will remain within the
> borders of a galaxy, and even larger amount of time
> that it'll remain within the borders of a galactic
> cluster or supercluster. Now that we have detected
> gravitational waves, there's another even more
> humongous source of energy (converting several solar
> masses into energy at a time) that we know travels
> at only the speed of light.
Do we *know* that, however? Have we always been able to correlate a wave detection with a visible "flash" (or loss of a periodic source) in the proper direction? Side issue...
> So lots of energy stays locked into regions just
> due to the slow passage of time.
Yes. So this is true of Newton, General Relativity, and its quantum replacement. But it isn't much, if the sum of each flavor of neutrinos is only 18 eV.
So is this contribution <1%, 10% or the necessary 600%? I'd say it was more along the 1% lines...
On Monday, October 24, 2016 at 11:18:24 PM UTC-7, Yousuf Khan wrote:
...
> Well, let's look at it practically. The Milky Way by
> itself is 100,000 LY across on its disk, probably 1
> million LY across on its halo. So neutrinos released
> now from the various stars of the galaxy will remain
> within the borders of the galaxy for between 100,000
> to 1 million years.
Well, they'd ("they" = neutrinos) be *average* about half that since they are generated across the disk, and if they were generated here, they'd be very energetic. If they were created much nearer the Big Bang, and are still hanging around, say in orbit, their total energy will be very close to their rest energy.
> itself is 100,000 LY across on its disk, probably 1
> million LY across on its halo. So neutrinos released
> now from the various stars of the galaxy will remain
> within the borders of the galaxy for between 100,000
> to 1 million years.
> Enough time to remain a part of the galaxy's mass.
> Even photons will do the same. So there's a large
> amount of time that energy will remain within the
> borders of a galaxy, and even larger amount of time
> that it'll remain within the borders of a galactic
> cluster or supercluster. Now that we have detected
> gravitational waves, there's another even more
> humongous source of energy (converting several solar
> masses into energy at a time) that we know travels
> at only the speed of light.
> So lots of energy stays locked into regions just
> due to the slow passage of time.
So is this contribution <1%, 10% or the necessary 600%? I'd say it was more along the 1% lines...
Steve Willner
Oct 25, 2016, 2:30:27 PM10/25/16
to
In article <d1490ae9-e7cc-4407...@googlegroups.com>,
> [photons] do have rest mass.
So the sum of a bunch of zeroes is non-zero? That's new physics.
I have no idea what "with a center of momentum frame" is supposed to
mean.
Standard physics says photons have momentum and energy but zero rest
mass. Photons react to gravity and (in principle, but I don't think
it has been measured) create gravity, but neither of those properties
requires rest mass.
> [photons] do have rest mass.
So the sum of a bunch of zeroes is non-zero? That's new physics.
I have no idea what "with a center of momentum frame" is supposed to
mean.
Standard physics says photons have momentum and energy but zero rest
mass. Photons react to gravity and (in principle, but I don't think
it has been measured) create gravity, but neither of those properties
requires rest mass.
dlzc
Oct 25, 2016, 5:22:20 PM10/25/16
to
Dear Steve Willner:
On Tuesday, October 25, 2016 at 11:30:27 AM UTC-7, Steve Willner wrote:
> In article <d1490ae9-e7cc-4407-bf7f-9df5d5f9c***@googlegroups.com>,
http://physics.stackexchange.com/questions/142672/center-of-mass-frame-for-massless-particles
... or just realize for two oppositely-directed, equal-energy photons:
E^2 = (sigma( p1, p2) = 0)^2 + (mc^2)^2
If the photons have non-zero energy (E=/=0), their momenta cancel, and they *must* have rest mass.
> I have no idea what "with a center of momentum frame"
> is supposed to mean.
>
> Standard physics says photons have momentum and energy
> but zero rest mass.
Individually, yes. As a system of particles (not all even have to be photons), no.
> Photons react to gravity and (in principle, but I
> don't think it has been measured) create gravity,
... and this is the discussion here, to what extent that photons that have not yet propagated past, say 10,000 light years from the center of a spiral galaxy, contribute to the "pull" outside that distance?
> but neither of those properties requires rest mass.
Gravitational mass, is actually the question, and so far gravitational mass = inertial mass = rest mass.
David A. Smith
On Tuesday, October 25, 2016 at 11:30:27 AM UTC-7, Steve Willner wrote:
> In article <d1490ae9-e7cc-4407-bf7f-9df5d5f9c***@googlegroups.com>,
> dlzc <dl***@cox.net> writes:
> > Individually, no. But in groups, with a center of
> > momentum frame, [photons] do have rest mass.
>
> So the sum of a bunch of zeroes is non-zero? That's
> new physics.
Quite old, actually. I first heard of it in "Spacetime Physics" by Taylor and Wheeler. You can read about it here:
> > Individually, no. But in groups, with a center of
> > momentum frame, [photons] do have rest mass.
>
> So the sum of a bunch of zeroes is non-zero? That's
> new physics.
http://physics.stackexchange.com/questions/142672/center-of-mass-frame-for-massless-particles
... or just realize for two oppositely-directed, equal-energy photons:
E^2 = (sigma( p1, p2) = 0)^2 + (mc^2)^2
If the photons have non-zero energy (E=/=0), their momenta cancel, and they *must* have rest mass.
> I have no idea what "with a center of momentum frame"
> is supposed to mean.
>
> Standard physics says photons have momentum and energy
> but zero rest mass.
> Photons react to gravity and (in principle, but I
> don't think it has been measured) create gravity,
> but neither of those properties requires rest mass.
David A. Smith
dlzc
Oct 26, 2016, 10:01:44 AM10/26/16
to
Dear Yousuf Khan:
> So is this contribution <1%, 10% or the necessary
> 600%? I'd say it was more along the 1% lines...
Consider that our Sun loses about 1 part in 10^14 each year (I think), due to solar wind and "mass deficit" of fusion inside it. In even 100,000 years, that is 1 part in 10^9 exodus from the Milky Way, unless significant "greenhouse gas" effect reflects it back in.
David A. Smith
> So is this contribution <1%, 10% or the necessary
> 600%? I'd say it was more along the 1% lines...
David A. Smith
Nicolaas Vroom
Oct 28, 2016, 4:35:27 AM10/28/16
to
On Wednesday, 19 October 2016 21:06:01 UTC+2, Steve Willner wrote:
> In article <jKCdna6p3aF3jp7F...@giganews.com>,
> Yousuf Khan <bbb...@spammenot.yahoo.com> writes:
> > We're still using Newtonian gravity in this day and age because we still
> > don't have computers strong enough to do a GR calculation for an entire
> > galaxy.
>
> That's not the problem. GR is mathematically identical to Newtonian
> gravity provided gravity is not "strong" and speeds are "low"
> compared to the speed of light. The magnitude of the errors can be
> quantified for the actual gravitational potentials and speeds in any
> particular calculation.
In some sense I have a problem with your reply.
> In article <jKCdna6p3aF3jp7F...@giganews.com>,
> Yousuf Khan <bbb...@spammenot.yahoo.com> writes:
> > We're still using Newtonian gravity in this day and age because we still
> > don't have computers strong enough to do a GR calculation for an entire
> > galaxy.
>
> That's not the problem. GR is mathematically identical to Newtonian
> gravity provided gravity is not "strong" and speeds are "low"
> compared to the speed of light. The magnitude of the errors can be
> quantified for the actual gravitational potentials and speeds in any
> particular calculation.
GR is not mathematical identical to Newton's Law.
If you want to use GR the full Einstein equations are extremely difficult.
A typical document to study is: https://arxiv.org/abs/1203.5166
"Numerical simulations of compact object binaries" by Harald P. Pfeiffer
For a different document about my own investigations read this:
http://users.telenet.be/nicvroom/Numerical_relativity_documents.htm
In this document many articles about numerical relativity are discussed,
but mainly 2 or maximum 3 objects are considered.
I fully agree that in many cases with great succes Newton's Law
can be used.
> The real problems are 1) even Newtonian gravity is too hard to
> calculate when the system has too many bodies,
curves with 100 objects. I doubt if that is possible using the full
Einstein equations.
Nicolaas Vroom.
Steve Willner
Oct 28, 2016, 1:03:38 PM10/28/16
to
In article <dadadb14-bbfe-4afd...@googlegroups.com>,
the library. What page are you looking at (and in what edition)?
> You can read about it here:
>http://physics.stackexchange.com/questions/142672/center-of-mass-frame-for-massless-particles
Most of the "answers" there look wrong to me, though parts of some of
them are correct.
> ... or just realize for two oppositely-directed, equal-energy photons:
> E^2 = (sigma( p1, p2) = 0)^2 + (mc^2)^2
What are the symbols supposed to represent? And where does the
equation come from?
> If the photons have non-zero energy (E=/=0), their momenta cancel,
> and they *must* have rest mass.
For the system you describe, momentum is zero, energy is non-zero,
and rest mass is zero.
> Gravitational mass, is actually the question,
There's no such thing as "gravitational mass" in relativity unless
you want to define it redundantly as E/c^2. All energy contributes
to gravitation (or bends spacetime, if you prefer).
dlzc <dl...@cox.net> writes:
> Quite old, actually. I first heard of it in "Spacetime Physics" by
>Taylor and Wheeler.
I don't have that book on my shelf, but I can probably get it from
> Quite old, actually. I first heard of it in "Spacetime Physics" by
>Taylor and Wheeler.
the library. What page are you looking at (and in what edition)?
> You can read about it here:
>http://physics.stackexchange.com/questions/142672/center-of-mass-frame-for-massless-particles
them are correct.
> ... or just realize for two oppositely-directed, equal-energy photons:
> E^2 = (sigma( p1, p2) = 0)^2 + (mc^2)^2
equation come from?
> If the photons have non-zero energy (E=/=0), their momenta cancel,
> and they *must* have rest mass.
and rest mass is zero.
> Gravitational mass, is actually the question,
you want to define it redundantly as E/c^2. All energy contributes
to gravitation (or bends spacetime, if you prefer).
Steve Willner
Oct 28, 2016, 1:18:11 PM10/28/16
to
In article <nv00cg$s1a$2...@dont-email.me>,
filled with photons, the box would behave as if it had rest mass.
That is to say, the box would gravitationally attract other objects,
and it would require a force to accelerate the box. From the
outside, you couldn't tell whether the box contained photons or
fishing weights (really small ones to match any reasonable amount of
electromagnetic energy). That's not at all the same as saying
photons have rest mass, though.
wil...@cfa.harvard.edu (Steve Willner) writes:
> There's no such thing as "gravitational mass" in relativity unless
> you want to define it redundantly as E/c^2. All energy contributes
> to gravitation (or bends spacetime, if you prefer).
Maybe this is where you are going wrong... if you had a massless box
> There's no such thing as "gravitational mass" in relativity unless
> you want to define it redundantly as E/c^2. All energy contributes
> to gravitation (or bends spacetime, if you prefer).
filled with photons, the box would behave as if it had rest mass.
That is to say, the box would gravitationally attract other objects,
and it would require a force to accelerate the box. From the
outside, you couldn't tell whether the box contained photons or
fishing weights (really small ones to match any reasonable amount of
electromagnetic energy). That's not at all the same as saying
photons have rest mass, though.
dlzc
Oct 28, 2016, 9:23:44 PM10/28/16
to
Dear Steve Wilner:
On Friday, October 28, 2016 at 10:03:38 AM UTC-7, Steve Willner wrote:
> In article <dadadb14-bbfe-4afd-8085-04ccb919e***@googlegroups.com>,
David A. Smith
On Friday, October 28, 2016 at 10:03:38 AM UTC-7, Steve Willner wrote:
> In article <dadadb14-bbfe-4afd-8085-04ccb919e***@googlegroups.com>,
> dlzc <dl***@cox.net> writes:
> > Quite old, actually. I first heard of it in "Spacetime
> > Physics" by Taylor and Wheeler.
>
> I don't have that book on my shelf, but I can probably get
> it from the library. What page are you looking at (and in
> what edition)?
Second Edition, Section 8.4, Sample Problem 8-2. Quote: "A system consisting entirely of zero-mass photons can itself have nonzero mass!"
> > Quite old, actually. I first heard of it in "Spacetime
> > Physics" by Taylor and Wheeler.
>
> I don't have that book on my shelf, but I can probably get
> it from the library. What page are you looking at (and in
> what edition)?
David A. Smith
Dr J R Stockton
Oct 29, 2016, 6:44:46 PM10/29/16
to
In sci.astro message <nuo8bb$396$1...@dont-email.me>, Tue, 25 Oct 2016
18:30:03, Steve Willner <wil...@cfa.harvard.edu> posted:
>In article <d1490ae9-e7cc-4407...@googlegroups.com>,
> dlzc <dl...@cox.net> writes:
>> Individually, no. But in groups, with a center of momentum frame,
>> [photons] do have rest mass.
>
>So the sum of a bunch of zeroes is non-zero? That's new physics.
>
>I have no idea what "with a center of momentum frame" is supposed to
>mean.
>
>Standard physics says photons have momentum and energy but zero rest
>mass. Photons react to gravity and (in principle, but I don't think
>it has been measured) create gravity, but neither of those properties
>requires rest mass.
Wikipedia page "Kugelblitz (astrophysics)" strongly implies that John A
Wheeler considered photon energy to create gravity.
--
(c) John Stockton, Surrey, UK. 拯merlyn.demon.co.uk Turnpike v6.05 MIME.
Merlyn Web Site < > - FAQish topics, acronyms, & links.
18:30:03, Steve Willner <wil...@cfa.harvard.edu> posted:
>In article <d1490ae9-e7cc-4407...@googlegroups.com>,
> dlzc <dl...@cox.net> writes:
>> Individually, no. But in groups, with a center of momentum frame,
>> [photons] do have rest mass.
>
>So the sum of a bunch of zeroes is non-zero? That's new physics.
>
>I have no idea what "with a center of momentum frame" is supposed to
>mean.
>
>Standard physics says photons have momentum and energy but zero rest
>mass. Photons react to gravity and (in principle, but I don't think
>it has been measured) create gravity, but neither of those properties
>requires rest mass.
Wheeler considered photon energy to create gravity.
--
(c) John Stockton, Surrey, UK. 拯merlyn.demon.co.uk Turnpike v6.05 MIME.
Merlyn Web Site < > - FAQish topics, acronyms, & links.
wor...@yandex.ru
Nov 6, 2016, 5:45:11 AM11/6/16
to
Am Mittwoch, 28. September 2016 16:55:10 UTC+2 schrieb dlzc:
> Dear wor...:
Please refer Figure 12.
In addition to the halo, the scientists had added two rings of dark matter as part of the DM model...
Why not equal two rings of normal matter? The rings of rock and ice are nevertheless common and ordinary cosmic formation.
> Dear wor...:
>
> If we are into opinions here...
But the oscillations in the rotation curves can not be overlooked, fully established: https://arxiv.org/abs/1604.01216
> If we are into opinions here...
Please refer Figure 12.
In addition to the halo, the scientists had added two rings of dark matter as part of the DM model...
Why not equal two rings of normal matter? The rings of rock and ice are nevertheless common and ordinary cosmic formation.
Steve Willner
Nov 7, 2016, 5:41:41 PM11/7/16
to
In article <3902b6bc-a525-4de0...@googlegroups.com>,
http://mnras.oxfordjournals.org/content/463/3/2623.full
> In addition to the halo, the scientists had added two rings of dark
> matter as part of the DM model... Why not equal two rings of
> normal matter? The rings of rock and ice are nevertheless common
> and ordinary cosmic formation.
I'm not sure, but the dark-matter rings seem to date back to
a 2003 paper
http://www.sciencedirect.com/science/article/pii/S0370269303008633
The Huang et al. paper assumes cylindrical symmetry, and there's a
brief mention that non-axisymmetric structure could change the
conclusions. According to Table 4, the sum of the ring masses is
just over half the bulge+disk mass, so it's not obvious to me that
the rings have to be dark matter. However, this is based on just a
quick scan. I'm no expert on this subject, and a good answer would
take some work.
wor...@yandex.ru writes:
> But the oscillations in the rotation curves can not be overlooked,
> fully established: https://arxiv.org/abs/1604.01216
> Please refer Figure 12.
Now published in MNRAS (Huang et al. 2016)
> But the oscillations in the rotation curves can not be overlooked,
> fully established: https://arxiv.org/abs/1604.01216
> Please refer Figure 12.
http://mnras.oxfordjournals.org/content/463/3/2623.full
> In addition to the halo, the scientists had added two rings of dark
> matter as part of the DM model... Why not equal two rings of
> normal matter? The rings of rock and ice are nevertheless common
> and ordinary cosmic formation.
a 2003 paper
http://www.sciencedirect.com/science/article/pii/S0370269303008633
The Huang et al. paper assumes cylindrical symmetry, and there's a
brief mention that non-axisymmetric structure could change the
conclusions. According to Table 4, the sum of the ring masses is
just over half the bulge+disk mass, so it's not obvious to me that
the rings have to be dark matter. However, this is based on just a
quick scan. I'm no expert on this subject, and a good answer would
take some work.
wor...@yandex.ru
Nov 8, 2016, 3:39:21 AM11/8/16
to
Am Montag, 7. November 2016 23:41:41 UTC+1 schrieb Steve Willner:
> In article <3902b6bc-a525-4de0...@googlegroups.com>,
> wor...@yandex.ru writes:
>
> > In addition to the halo, the scientists had added two rings of dark
> > matter as part of the DM model... Why not equal two rings of
> > normal matter? The rings of rock and ice are nevertheless common
> > and ordinary cosmic formation.
>
> I'm not sure, but the dark-matter rings seem to date back to
> a 2003 paper
> http://www.sciencedirect.com/science/article/pii/S0370269303008633
>
Thanks for this link.
> In article <3902b6bc-a525-4de0...@googlegroups.com>,
> wor...@yandex.ru writes:
>
> > In addition to the halo, the scientists had added two rings of dark
> > matter as part of the DM model... Why not equal two rings of
> > normal matter? The rings of rock and ice are nevertheless common
> > and ordinary cosmic formation.
>
> I'm not sure, but the dark-matter rings seem to date back to
> a 2003 paper
> http://www.sciencedirect.com/science/article/pii/S0370269303008633
>
Quotes:"In the past, rises (or bumps) in galactic rotation curves have been interpreted as due to the presence of spiral arms [17]. Spiral arms may in fact cause some of the rises in rotation curves."
[17] C. Yuan, Astrophys. J. 158 (1969) 871; W.B. Burton, W.W. Shane, in: W. Becker, G.I. Kontopoulos (Eds.), Proceedings of the 38th IAU Symposium the Spiral Structure of our Galaxy, Reidel, Dordrecht, p. 397; W.W. Shane, Astron. Astrophys. 16 (1972) 118.
Yupi!!
Steve Willner
Nov 25, 2016, 5:37:50 PM11/25/16
to
In article <70651e27-fcfa-4ca6...@googlegroups.com>,
I'm looking at _Spacetime Physics_ by Taylor & Wheeler, Second
Edition, copyright 1992, problem 8-5. (Problem 8-2 is a simple
numerical calculation of E = mc^2.)
> Quote: "A system consisting entirely of zero-mass photons can
> itself have nonzero mass!"
I don't see that exact quote but perhaps am just missing it. There
are other words that amount to pretty much the same thought.
Some observations:
1. the only thing we are arguing about is terminology. Everyone
(well, everyone who actually understands the subject) agrees on what
the equations say and the results of calculations.
2. photons carry energy and momentum. Photon energy contributes to
gravitational attraction, i.e., bends spacetime, and photon momentum
must be included for conservation of momentum.
3. T&W use "mass" inconsistently. Sometimes they mean "relativistic
mass," E/c^2, and sometimes "rest mass." Modern terminology uses
"mass" to mean rest mass only. The term "energy" is used when older
texts would have used "relativistic mass."
4. at no place do I see T&W claiming a collection of photons somehow
acquires rest mass. Indeed, all their arguments (which are correct
so far as I can tell) are as valid for a single photon as for any
other amount of radiation.
I'm looking at _Spacetime Physics_ by Taylor & Wheeler, Second
Edition, copyright 1992, problem 8-5. (Problem 8-2 is a simple
numerical calculation of E = mc^2.)
> Quote: "A system consisting entirely of zero-mass photons can
> itself have nonzero mass!"
are other words that amount to pretty much the same thought.
Some observations:
1. the only thing we are arguing about is terminology. Everyone
(well, everyone who actually understands the subject) agrees on what
the equations say and the results of calculations.
2. photons carry energy and momentum. Photon energy contributes to
gravitational attraction, i.e., bends spacetime, and photon momentum
must be included for conservation of momentum.
3. T&W use "mass" inconsistently. Sometimes they mean "relativistic
mass," E/c^2, and sometimes "rest mass." Modern terminology uses
"mass" to mean rest mass only. The term "energy" is used when older
texts would have used "relativistic mass."
4. at no place do I see T&W claiming a collection of photons somehow
acquires rest mass. Indeed, all their arguments (which are correct
so far as I can tell) are as valid for a single photon as for any
other amount of radiation.
Nicolaas Vroom
Nov 26, 2016, 7:29:36 AM11/26/16
to
On Friday, 25 November 2016 23:37:50 UTC+1, Steve Willner wrote:
> 3. T&W use "mass" inconsistently. Sometimes they mean "relativistic
> mass," E/c^2, and sometimes "rest mass." Modern terminology uses
> "mass" to mean rest mass only. The term "energy" is used when older
> texts would have used "relativistic mass."
>
> 4. at no place do I see T&W claiming a collection of photons somehow
> acquires rest mass. Indeed, all their arguments (which are correct
> so far as I can tell) are as valid for a single photon as for any
> other amount of radiation.
>
> --
> Help keep our newsgroup healthy; please don't feed the trolls.
> Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
> Cambridge, MA 02138 USA
Just some thoughts.
> 3. T&W use "mass" inconsistently. Sometimes they mean "relativistic
> mass," E/c^2, and sometimes "rest mass." Modern terminology uses
> "mass" to mean rest mass only. The term "energy" is used when older
> texts would have used "relativistic mass."
>
> 4. at no place do I see T&W claiming a collection of photons somehow
> acquires rest mass. Indeed, all their arguments (which are correct
> so far as I can tell) are as valid for a single photon as for any
> other amount of radiation.
>
> --
> Help keep our newsgroup healthy; please don't feed the trolls.
> Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
> Cambridge, MA 02138 USA
IMO we should only use the word mass and not the word rest-mass.
My understanding is that objects exists and mass does not exist.
Mass is a purely a calculated value based on a model or theory.
The theory can be used Newton's law which starts with observations.
With the use of these observations (over a certain period) and with
the model the parameter mass can be calculated.
The use a different model and the same observations will result
in different mass values for the same objects.
Photons, using the same reasoning also have a mass.
The equation is m = E/c^2. The issue is what E ?
E is the amount of energy "released" when in some chemical reaction
a photon is released. This is in accordance with the law: Conservation
of energy. It is that "simple"
Gravitons using the same reasoning also have a mass.
The value is very low.
Nicolaas Vroom
dlzc
Nov 26, 2016, 11:00:54 AM11/26/16
to
Dear Nicolaas Vroom:
On Saturday, November 26, 2016 at 5:29:36 AM UTC-7, Nicolaas Vroom wrote:
> On Friday, 25 November 2016 23:37:50 UTC+1, Steve Willner wrote:
>
> > 3. T&W use "mass" inconsistently. Sometimes they
> > mean "relativistic mass," E/c^2, and sometimes "rest
> > mass." Modern terminology uses "mass" to mean rest
> > mass only. The term "energy" is used when older
> > texts would have used "relativistic mass."
> >
> > 4. at no place do I see T&W claiming a collection of
> > photons somehow acquires rest mass. Indeed, all
> > their arguments (which are correct so far as I can
> > tell) are as valid for a single photon as for any
> > other amount of radiation.
A collection of two "photons", exact same energy, one directed "left", and one directed "right". The center-of-momentum frame is at rest with respect to you. The system's frame has zero net momentum, and non-zero energy. Therefore, it has rest mass... until the two photons propagate out past your distance from their "origin center". At which time, their gravitational component becomes undifferentiable by you from background.
http://www.phy.duke.edu/~lee/P53/sys.pdf
> My understanding is that objects exists and mass
> does not exist.
"Object" is a macroscopic definition, in that it confines properties to a location. Mass is likewise a property assigned by the system Universe, a "coordinate" in some sense.
> Mass is a purely a calculated value based on a
> model or theory. The theory can be used Newton's
> law which starts with observations. With the use
> of these observations (over a certain period) and
> with the model the parameter mass can be
> calculated.
Which may or may not be why we cannot narrow G down to more than 6 sig figs (and unstable), but can get sqrt( G * M_sun ) to 11 sig figs and stable.
> The use a different model and the same observations
> will result in different mass values for the same
> objects.
Eotvos. We've tried that, and failed to find any variation.
> Photons, using the same reasoning also have a mass.
> The equation is m = E/c^2. The issue is what E ?
> E is the amount of energy "released" when in some
> chemical reaction a photon is released. This is in
> accordance with the law: Conservation of energy. It
> is that "simple"
(The mass of one proton) + (the mass of one electron) - (the mass of one H1 neutral atom) > the (13.6 eV/c^2 photon) known to be emitted... by a tiny bit, perhaps the recoil of the formed atom.
> Gravitons using the same reasoning also have a
> mass. The value is very low.
No, string theory makes a prediction for that, and it is "unbelievably" high. At no point have they found any hint of a graviton, at any energy level tried.
Not even sure they've found the same "Higg's boson" fingerprint when running at higher energy levels.
David A. Smith
On Saturday, November 26, 2016 at 5:29:36 AM UTC-7, Nicolaas Vroom wrote:
> On Friday, 25 November 2016 23:37:50 UTC+1, Steve Willner wrote:
>
> > 3. T&W use "mass" inconsistently. Sometimes they
> > mean "relativistic mass," E/c^2, and sometimes "rest
> > mass." Modern terminology uses "mass" to mean rest
> > mass only. The term "energy" is used when older
> > texts would have used "relativistic mass."
> >
> > 4. at no place do I see T&W claiming a collection of
> > photons somehow acquires rest mass. Indeed, all
> > their arguments (which are correct so far as I can
> > tell) are as valid for a single photon as for any
> > other amount of radiation.
> IMO we should only use the word mass and not the
> word rest-mass.
It was not Mr./Dr. Wilner's contention it was rest mass. But the mass at rest in the center of momentum frame, should be considered rest mass.
> word rest-mass.
A collection of two "photons", exact same energy, one directed "left", and one directed "right". The center-of-momentum frame is at rest with respect to you. The system's frame has zero net momentum, and non-zero energy. Therefore, it has rest mass... until the two photons propagate out past your distance from their "origin center". At which time, their gravitational component becomes undifferentiable by you from background.
http://www.phy.duke.edu/~lee/P53/sys.pdf
> My understanding is that objects exists and mass
> does not exist.
> Mass is a purely a calculated value based on a
> model or theory. The theory can be used Newton's
> law which starts with observations. With the use
> of these observations (over a certain period) and
> with the model the parameter mass can be
> calculated.
> The use a different model and the same observations
> will result in different mass values for the same
> objects.
> Photons, using the same reasoning also have a mass.
> The equation is m = E/c^2. The issue is what E ?
> E is the amount of energy "released" when in some
> chemical reaction a photon is released. This is in
> accordance with the law: Conservation of energy. It
> is that "simple"
> Gravitons using the same reasoning also have a
> mass. The value is very low.
Not even sure they've found the same "Higg's boson" fingerprint when running at higher energy levels.
David A. Smith
dlzc
Nov 26, 2016, 8:39:40 PM11/26/16
to
Dear Steve Willner:
On Friday, November 25, 2016 at 3:37:50 PM UTC-7, Steve Willner wrote:
> In article <70651e27-fcfa-4ca6-b284-aaf4f99a9***@googlegroups.com>,
David A. Smith
On Friday, November 25, 2016 at 3:37:50 PM UTC-7, Steve Willner wrote:
> In article <70651e27-fcfa-4ca6-b284-aaf4f99a9***@googlegroups.com>,
> dlzc <dl***@cox.net> writes:
> > Second Edition, Section 8.4, Sample Problem 8-2.
>
> I'm looking at _Spacetime Physics_ by Taylor &
> Wheeler, Second Edition, copyright 1992, problem
> 8-5. (Problem 8-2 is a simple numerical
> calculation of E = mc^2.)
No. Body of the chapter, worked example for section 8.4, labeled SAMPLE PROBLEM 8-2, which in my paperback copy is page 232.
> > Second Edition, Section 8.4, Sample Problem 8-2.
>
> I'm looking at _Spacetime Physics_ by Taylor &
> Wheeler, Second Edition, copyright 1992, problem
> 8-5. (Problem 8-2 is a simple numerical
> calculation of E = mc^2.)
David A. Smith
Steve Willner
Nov 28, 2016, 5:04:14 PM11/28/16
to
In article <ef9731ae-13a6-4f0a...@googlegroups.com>,
explanation. I have the hardbound edition (from the library), but
p 832 seems to be the same. Key text:
A photon has no rest energy--that is, no mass of its own. However,
a photon can contribute energy and momentum to a system of objects.
Hence the presence of one or more photons in a system can increase
the mass of that system. More: A system consisting entirely of
zero-mass photons can itself have non-zero mass!
I'll stand by my comments in my previous message. The issue is
terminology, not physics, and the text above is using "mass" in two
distinct ways. Nowadays, most physicists would not do that.
dlzc <dl...@cox.net> writes:
> Body of the chapter, worked example for section 8.4, labeled
> SAMPLE PROBLEM 8-2, which in my paperback copy is page 232.
OK, got it now. I was looking on pp 254ff, which has a more detailed
> Body of the chapter, worked example for section 8.4, labeled
> SAMPLE PROBLEM 8-2, which in my paperback copy is page 232.
explanation. I have the hardbound edition (from the library), but
p 832 seems to be the same. Key text:
A photon has no rest energy--that is, no mass of its own. However,
a photon can contribute energy and momentum to a system of objects.
Hence the presence of one or more photons in a system can increase
the mass of that system. More: A system consisting entirely of
zero-mass photons can itself have non-zero mass!
I'll stand by my comments in my previous message. The issue is
terminology, not physics, and the text above is using "mass" in two
distinct ways. Nowadays, most physicists would not do that.
dlzc
Nov 28, 2016, 7:50:41 PM11/28/16
to
Dear Steve Wilner:
On Monday, November 28, 2016 at 3:04:14 PM UTC-7, Steve Willner wrote:
...
Note that we were discussing "photon gas" trapped within a given radius from galactic center, and as a first blush it would be say 10,000 years (for a radius of 10,000 light years) of stellar output *assuming 1 part in 10^14, normal annual for our Sun), or about 1 part in 10^8 of the luminous mass. Far too low to be straining at anyway.
I think they are discussing only one mass, namely:
mass = rest mass = inertial mass = gravitational mass =/= relativistic mass, and I respect your belief to think they mean something else. I think you are wrong, but you have every right to it.
We can let this drop.
David A. Smith
On Monday, November 28, 2016 at 3:04:14 PM UTC-7, Steve Willner wrote:
...
> I'll stand by my comments in my previous message. The
> issue is terminology, not physics, and the text above
> is using "mass" in two distinct ways. Nowadays, most
> physicists would not do that.
OK.
> issue is terminology, not physics, and the text above
> is using "mass" in two distinct ways. Nowadays, most
> physicists would not do that.
Note that we were discussing "photon gas" trapped within a given radius from galactic center, and as a first blush it would be say 10,000 years (for a radius of 10,000 light years) of stellar output *assuming 1 part in 10^14, normal annual for our Sun), or about 1 part in 10^8 of the luminous mass. Far too low to be straining at anyway.
I think they are discussing only one mass, namely:
mass = rest mass = inertial mass = gravitational mass =/= relativistic mass, and I respect your belief to think they mean something else. I think you are wrong, but you have every right to it.
We can let this drop.
David A. Smith
Steve Willner
Nov 29, 2016, 3:26:53 PM11/29/16
to
[there were some strange characters in the quoted message. I've
replaced them with =, but if that wasn't what was intended, I may
have misunderstood the point.]
In article <ae50100c-c9da-49e8...@googlegroups.com>,
The first two would be the same thing in modern parlance. The last
three, if used at all, would now be considered synonyms for "energy."
In general they are not equal to rest mass. That's physics, not
terminology. Taylor & Wheeler used what is now obsolete terminology,
but they had the physics right.
replaced them with =, but if that wasn't what was intended, I may
have misunderstood the point.]
In article <ae50100c-c9da-49e8...@googlegroups.com>,
dlzc <dl...@cox.net> writes:
> I think they are discussing only one mass, namely:
> mass = rest mass = inertial mass = gravitational mass =
> relativistic mass,
> I think they are discussing only one mass, namely:
> mass = rest mass = inertial mass = gravitational mass =
The first two would be the same thing in modern parlance. The last
three, if used at all, would now be considered synonyms for "energy."
In general they are not equal to rest mass. That's physics, not
terminology. Taylor & Wheeler used what is now obsolete terminology,
but they had the physics right.
dlzc
Nov 29, 2016, 4:03:24 PM11/29/16
to
Dear Steve Willner:
On Tuesday, November 29, 2016 at 1:26:53 PM UTC-7, Steve Willner wrote:
> [there were some strange characters in the quoted
> message. I've replaced them with =, but if that
> wasn't what was intended, I may have misunderstood
> the point.]
That is the case. =/= was intended to mean "not equal to". Sometimes represented as #, or <>.
> > I think they are discussing only one mass, namely:
> > mass = rest mass = inertial mass = gravitational mass <>
> The last three, if used at all, would now be
> considered synonyms for "energy."
False.
> In general they are not equal to rest mass.
Still false, as Eotvos has shown.
> That's physics, not terminology. Taylor &
> Wheeler used what is now obsolete terminology,
> but they had the physics right.
On that last bit we agree.
David A. Smith
On Tuesday, November 29, 2016 at 1:26:53 PM UTC-7, Steve Willner wrote:
> [there were some strange characters in the quoted
> message. I've replaced them with =, but if that
> wasn't what was intended, I may have misunderstood
> the point.]
> > I think they are discussing only one mass, namely:
> > mass = rest mass = inertial mass = gravitational mass <>
> > relativistic mass,
>
> The first two would be the same thing in modern
> parlance.
... and the second two Eotvos has shown to be the same as the first two.
>
> The first two would be the same thing in modern
> parlance.
> The last three, if used at all, would now be
> considered synonyms for "energy."
> In general they are not equal to rest mass.
> That's physics, not terminology. Taylor &
> Wheeler used what is now obsolete terminology,
> but they had the physics right.
David A. Smith
Steve Willner
Dec 1, 2016, 2:01:32 PM12/1/16
to
In article <9372bcdb-6523-437f...@googlegroups.com>,
I still don't know where you used = and where =/=. That's a problem
with using non-ASCII character sets.
A quote from Taylor and Wheeler p. 256 may help:
The source of our difficulty is some confusion between two quite
different concepts: (1) energy, the time component of the
momentum-energy 4-vector, and (2) mass, the magnitude of this
4-vector.
Nowadays, physicists avoid confusion by using "energy" when they mean
(1). Terms used in the past for this concept include "relativistic
mass," "inertial mass," "gravitational mass," and most regrettably,
sometimes just "mass." (T&W did this, though it's usually clear in
context which concept they meant.) Nowadays the term "mass" is
reserved for (2), but "rest mass" or "proper mass" can be used if
there's any chance of confusion.
If you want to discuss physics, please be careful to use unambiguous
terminology. In particular, be careful to keep straight the two
concepts T&W described.
I still don't know where you used = and where =/=. That's a problem
with using non-ASCII character sets.
A quote from Taylor and Wheeler p. 256 may help:
The source of our difficulty is some confusion between two quite
different concepts: (1) energy, the time component of the
momentum-energy 4-vector, and (2) mass, the magnitude of this
4-vector.
Nowadays, physicists avoid confusion by using "energy" when they mean
(1). Terms used in the past for this concept include "relativistic
mass," "inertial mass," "gravitational mass," and most regrettably,
sometimes just "mass." (T&W did this, though it's usually clear in
context which concept they meant.) Nowadays the term "mass" is
reserved for (2), but "rest mass" or "proper mass" can be used if
there's any chance of confusion.
If you want to discuss physics, please be careful to use unambiguous
terminology. In particular, be careful to keep straight the two
concepts T&W described.
dlzc
Dec 1, 2016, 7:14:35 PM12/1/16
to
Dear Steve Willner:
On Thursday, December 1, 2016 at 12:01:32 PM UTC-7, Steve Willner wrote:
> In article <9372bcdb-6523-437f-915f-a550818d5***@googlegroups.com>,
I said these things are exactly the same:
mass, rest mass, inertial mass, gravitational mass.
(you correct this more clearly below.)
I said this was not the same as mass:
relativistic mass (which Einstein said not to teach).
> A quote from Taylor and Wheeler p. 256 may help:
> The source of our difficulty is some confusion
> between two quite different concepts: (1)
> energy, the time component of the momentum-
> energy 4-vector, and (2) mass, the magnitude
> of this 4-vector.
>
> Nowadays, physicists avoid confusion by using
> "energy" when they mean (1). Terms used in the
> past for this concept include "relativistic
> mass," "inertial mass," "gravitational mass,"
> and most regrettably, sometimes just "mass."
> (T&W did this, though it's usually clear in
> context which concept they meant.) Nowadays the
> term "mass" is reserved for (2), but "rest mass"
> or "proper mass" can be used if there's any
> chance of confusion.
And is tantamount to a religious argument, with those that find no issue slinging relativistic mass around. Ignoring that it an infinite number for different scalar values, depending on in which axis a tiny bit of momentum might be applied, in relation to its line of motion.
> If you want to discuss physics, please be
> careful to use unambiguous terminology. In
> particular, be careful to keep straight the two
> concepts T&W described.
I find this harder and harder to do. Thanks.
David A. Smith
On Thursday, December 1, 2016 at 12:01:32 PM UTC-7, Steve Willner wrote:
> In article <9372bcdb-6523-437f-915f-a550818d5***@googlegroups.com>,
> dlzc <dl***@cox.net> writes:
> > That is the case. =/= was intended to mean "not equal to".
>
> I still don't know where you used = and where =/=.
> That's a problem with using non-ASCII character
> sets.
Sorry, but I *only* used ASCII character sets. I just presented it in a format you did not expect / decipher.
> > That is the case. =/= was intended to mean "not equal to".
>
> I still don't know where you used = and where =/=.
> That's a problem with using non-ASCII character
> sets.
I said these things are exactly the same:
mass, rest mass, inertial mass, gravitational mass.
(you correct this more clearly below.)
I said this was not the same as mass:
relativistic mass (which Einstein said not to teach).
> A quote from Taylor and Wheeler p. 256 may help:
> The source of our difficulty is some confusion
> between two quite different concepts: (1)
> energy, the time component of the momentum-
> energy 4-vector, and (2) mass, the magnitude
> of this 4-vector.
>
> Nowadays, physicists avoid confusion by using
> "energy" when they mean (1). Terms used in the
> past for this concept include "relativistic
> mass," "inertial mass," "gravitational mass,"
> and most regrettably, sometimes just "mass."
> (T&W did this, though it's usually clear in
> context which concept they meant.) Nowadays the
> term "mass" is reserved for (2), but "rest mass"
> or "proper mass" can be used if there's any
> chance of confusion.
> If you want to discuss physics, please be
> careful to use unambiguous terminology. In
> particular, be careful to keep straight the two
> concepts T&W described.
David A. Smith
Steve Willner
Dec 2, 2016, 4:58:23 PM12/2/16
to
In article <090fb92d-ffb1-4eda...@googlegroups.com>,
sure what actual quantity you mean by each of those terms. Let's
leave aside "mass," which is surely ambiguous. I think nearly
everyone (but possibly not you) takes "rest mass" to refer to the
invariant magnitude of the energy-momentum 4-vector. The point of
Taylor & Wheeler's discussion on pp 254ff is that gravitational
acceleration is related to the time component of the 4-vector.
That's what I'd mean by "gravitational mass," though I'd usually use
the term "energy."
The physical point to be made is that there are two different
quantities involved, and "rest mass" (meaning the invariant magnitude
of the 4-vector) is not the one directly related to gravity or
inertia.
dlzc <dl...@cox.net> writes:
> I said these things are exactly the same:
> mass, rest mass, inertial mass, gravitational mass.
This thread has shown that terminology can be a problem, so I'm not
> I said these things are exactly the same:
> mass, rest mass, inertial mass, gravitational mass.
sure what actual quantity you mean by each of those terms. Let's
leave aside "mass," which is surely ambiguous. I think nearly
everyone (but possibly not you) takes "rest mass" to refer to the
invariant magnitude of the energy-momentum 4-vector. The point of
Taylor & Wheeler's discussion on pp 254ff is that gravitational
acceleration is related to the time component of the 4-vector.
That's what I'd mean by "gravitational mass," though I'd usually use
the term "energy."
The physical point to be made is that there are two different
quantities involved, and "rest mass" (meaning the invariant magnitude
of the 4-vector) is not the one directly related to gravity or
inertia.
dlzc
Dec 2, 2016, 6:27:43 PM12/2/16
to
Dear Steve Willner:
On Friday, December 2, 2016 at 2:58:23 PM UTC-7, Steve Willner wrote:
> In article <090fb92d-ffb1-4eda-8969-325414355***@googlegroups.com>,
So you'd say that the photons represent "energy" or "gravitational mass" until they pass your position (Newton's shells).
How does that differ from some tiny amount of "rest mass" or "proper mass" disappearing (magic required of course), and "changes in the gravitational field" propagating outwards past your position at c (classical speed of gravity)?
And note one immediate difference is some light is absorbed and scattered by a medium (dust, gas) that is warmer than the CMBR, and located inboard of your position...
Again, this is far smaller than the amount of Dark Matter "orbiting a spiral galaxy", and probably could be entirely ignored.
David A. Smith
On Friday, December 2, 2016 at 2:58:23 PM UTC-7, Steve Willner wrote:
> In article <090fb92d-ffb1-4eda-8969-325414355***@googlegroups.com>,
> dlzc <dl***@cox.net> writes:
> > I said these things are exactly the same:
> > mass, rest mass, inertial mass, gravitational mass.
>
> This thread has shown that terminology can be
> a problem, so I'm not sure what actual quantity
> you mean by each of those terms. Let's leave
> aside "mass," which is surely ambiguous. I
> think nearly everyone (but possibly not you)
> takes "rest mass" to refer to the invariant
> magnitude of the energy-momentum 4-vector. The
> point of Taylor & Wheeler's discussion on pp
> 254ff is that gravitational acceleration is
> related to the time component of the 4-vector.
> That's what I'd mean by "gravitational mass,"
> though I'd usually use the term "energy."
>
> The physical point to be made is that there are
> two different quantities involved, and
> "rest mass" (meaning the invariant magnitude
> of the 4-vector) is not the one directly related
> to gravity or inertia.
OK, so in this thread, we are talking about all the photons currently wending their way outward, centered more-or-less on the center of a spiral galaxy, but inboard of your position in said galaxy. They originated as more lower-mass atoms, being fused into fewer higher-mass atoms plus light sprayed in all directions (on average).
> > I said these things are exactly the same:
> > mass, rest mass, inertial mass, gravitational mass.
>
> This thread has shown that terminology can be
> a problem, so I'm not sure what actual quantity
> you mean by each of those terms. Let's leave
> aside "mass," which is surely ambiguous. I
> think nearly everyone (but possibly not you)
> takes "rest mass" to refer to the invariant
> magnitude of the energy-momentum 4-vector. The
> point of Taylor & Wheeler's discussion on pp
> 254ff is that gravitational acceleration is
> related to the time component of the 4-vector.
> That's what I'd mean by "gravitational mass,"
> though I'd usually use the term "energy."
>
> The physical point to be made is that there are
> two different quantities involved, and
> "rest mass" (meaning the invariant magnitude
> of the 4-vector) is not the one directly related
> to gravity or inertia.
So you'd say that the photons represent "energy" or "gravitational mass" until they pass your position (Newton's shells).
How does that differ from some tiny amount of "rest mass" or "proper mass" disappearing (magic required of course), and "changes in the gravitational field" propagating outwards past your position at c (classical speed of gravity)?
And note one immediate difference is some light is absorbed and scattered by a medium (dust, gas) that is warmer than the CMBR, and located inboard of your position...
Again, this is far smaller than the amount of Dark Matter "orbiting a spiral galaxy", and probably could be entirely ignored.
David A. Smith
Steve Willner
Dec 5, 2016, 5:36:36 PM12/5/16
to
In article <08647f1f-5adc-452e...@googlegroups.com>,
that term) regardless of their position. The gravitational effect
they have depends on position; to the extent the photons form a
spherically symmetric distribution, the ones outside "your position"
won't have any net effect.
> How does that differ from some tiny amount of "rest mass" or
> "proper mass" disappearing (magic required of course),
As you pointed out (and I snipped), the photons originated from
proper mass "disappearing" by being converted to energy. The effect
(again assuming spherical symmetry) is a gradual decrease in
centripetal attraction as photons (on balance) fly past one's
position. That is, the fact that photons have energy gives them
exactly the same gravitational effect as the rest mass they replaced.
All that changes is the location; the photons move around faster than
does mass.
> And note one immediate difference is some light is absorbed and
> scattered by a medium (dust, gas) that is warmer than the CMBR,
> and located inboard of your position...
Energy is still conserved (in a suitable reference frame), and energy
translates to gravitational mass.
> Again, this is far smaller than the amount of Dark Matter "orbiting
> a spiral galaxy", and probably could be entirely ignored.
Oh indeed, the gravitational effect of radiation is utterly trivial
in every context I can think of in today's universe. It was
important, however, for some seconds after the Big Bang, when the
universe was radiation-dominated.
dlzc <dl...@cox.net> writes:
> OK, so in this thread, we are talking about all the photons
> currently wending their way outward, centered more-or-less on the
> center of a spiral galaxy, but inboard of your position in said
> galaxy.
> OK, so in this thread, we are talking about all the photons
> currently wending their way outward, centered more-or-less on the
> center of a spiral galaxy, but inboard of your position in said
> galaxy.
> So you'd say that the photons represent "energy" or "gravitational
> mass" until they pass your position (Newton's shells).
They represent energy (and "gravitational mass" if you want to use
> mass" until they pass your position (Newton's shells).
that term) regardless of their position. The gravitational effect
they have depends on position; to the extent the photons form a
spherically symmetric distribution, the ones outside "your position"
won't have any net effect.
> How does that differ from some tiny amount of "rest mass" or
> "proper mass" disappearing (magic required of course),
proper mass "disappearing" by being converted to energy. The effect
(again assuming spherical symmetry) is a gradual decrease in
centripetal attraction as photons (on balance) fly past one's
position. That is, the fact that photons have energy gives them
exactly the same gravitational effect as the rest mass they replaced.
All that changes is the location; the photons move around faster than
does mass.
> And note one immediate difference is some light is absorbed and
> scattered by a medium (dust, gas) that is warmer than the CMBR,
> and located inboard of your position...
translates to gravitational mass.
> Again, this is far smaller than the amount of Dark Matter "orbiting
> a spiral galaxy", and probably could be entirely ignored.
in every context I can think of in today's universe. It was
important, however, for some seconds after the Big Bang, when the
universe was radiation-dominated.
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