Size of the sun?
Joris Dolderer
According to the barometric formula, the pressure in the earth's
atmosphere reduces (approximately) exponentially.
Does this also apply to the sun's gas?
If yes: -Why can I see it as a dish and not as a diffuse cloud?
-How is the suns' radius defined?
Thanks in advance
Joris
PD
> Hello,
>
> According to the barometric formula, the pressure in the earth's
> atmosphere reduces (approximately) exponentially.
>
> Does this also apply to the sun's gas?
> If yes: -Why can I see it as a dish and not as a diffuse cloud?
It IS a diffuse cloud.
http://images.google.com/images?gbv=2&q=sun+corona&btnG=Search+Images
> -How is the suns' radius defined?
Usually by the radius of the photosphere. Here the definition doesn't
have to do with where the outermost mass is, but where the temperature
and density of the solar mass falls enough that it doesn't shine
brightly. That's a bit sharper demarcation than where the mass stops.
>
> Thanks in advance
>
> Joris
Sam Wormley
The sun looks like it has an "edge" in the visible light
range because the photosphere changes from being almost
completely transparent to being almost opaque over a distance
of about 400 km.
Landy
"Joris Dolderer" <st...@123mail.cl> wrote in message
news:483d8b31$0$27450$9b4e...@newsspool4.arcor-online.net...
> Hello,
>
> According to the barometric formula, the pressure in the earth's
> atmosphere reduces (approximately) exponentially.
Reduces exponentially with what?
Androcles
"Landy" <no...@nowhere.net> wrote in message
news:g1kimo$qs3$1...@news-01.bur.connect.com.au...
it is exponential, logarithmic or linear is another matter but his followup
question isn't worth responding to, he needs to think that through for
himself instead of being told.
http://tinyurl.com/4rgqon
--
Androcles
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
http://www.androcles01.pwp.blueyonder.co.uk/
Landy
"Androcles" <Headm...@Hogwarts.physics> wrote in message
news:1Vk%j.20908$SA7....@newsfe09.ams2...
>
> "Landy" <no...@nowhere.net> wrote in message
> news:g1kimo$qs3$1...@news-01.bur.connect.com.au...
> |
> | "Joris Dolderer" <st...@123mail.cl> wrote in message
> | news:483d8b31$0$27450$9b4e...@newsspool4.arcor-online.net...
> | > Hello,
> | >
> | > According to the barometric formula, the pressure in the earth's
> | > atmosphere reduces (approximately) exponentially.
> |
> | Reduces exponentially with what?
> |
> I expect he means pressure reduces with increasing altitude.
Indeed. But his statement was abbreviated and assuming.
cheers
Bill
Androcles
--
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
1/2[tau(A)+tau(A')]= tau(B)
where
A = (0,0,0,t)
A' =(0,0,0,t+x'/(c-v) +x'/(c+v))
B = (x',0,0,t+x'/(c-v))
x' = x-vt
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
"Easy. He did not say that." - cretin van lintel.
Androcles
tadchem
> "Androcles" <Headmas...@Hogwarts.physics> wrote in message
The differential form of the barometric equation is exact:
dP = - rho*g*dh
where P is pressure, rho is density, and h is height.
The exponential form of the barometric equation, given here:
http://jcbmac.chem.brown.edu/baird/chem12/chem12-2004/Air/barometric.html
is developed with the assumptions that (1) gravity does not vary with
height, (2) that temperature does not vary with height, and (3) that
the gas is ideal, and of uniform composition. These assumptions are
necessary to perform the integration from the differential form.
The validity of the assumptions depends on the desired accuracy and
the limits of the integration (minimum and maximum h) to be
considered.
Tom Davidson
Richmond, VA
Androcles
"tadchem" <tad...@comcast.net> wrote in message
news:48441cf6-b895-407b...@m3g2000hsc.googlegroups.com...
You mean there is no jetstream? :-)
Actually the kid wanted to know why the Sun appears to have a well-defined
edge, not realising it doesn't.
--
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
1/2[tau(A)+tau(A')]= tau(B)
where
A = (0,0,0,t)
A' =(0,0,0,t+x'/(c-v) +x'/(c+v))
B = (x',0,0,t+x'/(c-v))
x' = x-vt
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
"Easy: he did NOT say that." - cretin harald.vanlin...@epfl.ch
Androcles
John Schutkeker
$9b4e...@newsspool4.arcor-online.net:
> Hello,
>
> According to the barometric formula, the pressure in the earth's
> atmosphere reduces (approximately) exponentially.
>
> Does this also apply to the sun's gas?
I doubt that anybody knows, because the physics of the outer layers has
buoyant convection, which may or may not dominate the simplified model.
I don't think that it does apply, because when density passes a certain
threshold, the plasma behaves more as an incompressible fluid (liquid)
than as a compressible fluid (gas). I can't tell you why I think this,
but just that people have always intuitively seen the sun as more of a
liquid than a gas.
This gets into phase transitions, which is a very complex area of
physics. Maybe there's even an undiscovered phase transition, similar
to condensation from a low density, gaseous plasma to a high density,
liquid plasma. That would be an interesting hypothesis to pursue, but I
have no idea how to get started proving or disproving it. Finding the
boundary between "low" and "high" densities would be a very important
discovery, if it turns out to be true.
If you want to do the calculation in the same way it's done for the
earth's oceans, that might be more interesting. In fact, it would be
interesting to see both calculations posted side by side, for neutral
fluids, to try to bracket the results for the more complex, solar fluid.
All these physics are very poorly understood, which means that you
probably have the beginnings of a good, amateur science problem. If you
can get it all finished, you might be able to get Scientific American to
print it.
You should definitely do the calculation and post it here, so that
others can try to figure out how to add crude models for convection,
radiation, electrostatic, magnetic and nuclear effects. One way to do
the problem, rather than trying to solve all the various dynamics at
once, is to create gedanken plasmas and add the effects one at a time,
independently of each other.
I don't know whether either of these have even been done. Intuitively,
it seems that they probably have, and I'd be surprised if they hadn't,
but I'm routinely surprised by the fairly simple looking things that
have never been done.
And if the calculations are buried in an obscure corner of the library,
then what they hey, you just reinvented the wheel, and there's no great
shame in that, although you can't publish the result as new work. The
other option, of course, is to go digging through the library for the
calculations. That can be an extremely difficult and unpleasant task,
and often self-defeating, if you don't have a world-class library at
your disposal. But everything good in life comes with a high cost
attached to it. ;(
> If yes: -Why can I see it as a dish and not as a diffuse cloud?
You mean "disk." That's also a good question. There is apparently a
value of radius at which the liquid plasma model gives way to a gaseous
plasma model. That would be the radius at which the phase transition
occurs, if my hypothesis is right. :]
> -How is the suns' radius defined?
The "edge" of the sun is the boundary between the chromosphere and the
corona. If there's a transition layer, it may be the chromosphere.
PD
wrote:> $9b4e6...@newsspool4.arcor-online.net:
Fabulous post.
John Schutkeker
news:342fd987-425e-492b...@r66g2000hsg.googlegroups.com:
Thank you. but after thinking about it, I'm becoming skeptical of the
phase transition conjecture. I'm more inclined to favor Chandrasekhar's
force balance model, where particles are either gravitationally confined
or expelled by some energetic process. It sort of raises the question
of what causes the solar wind, the corona and prominences.
John Schutkeker
news:Xns9AAE7129EBCB8lk...@207.115.33.102:
The question becomes, what is the origin of the supporting force in the
solar convection layer? I don't believe that it's radiative, because
that's the force in the radiation layer. If it isn't radiation, it must be
a plasmadynamic force. My gut reaction is that it can't be electrostatic,
because a plasma is quasi-neutral. It could be magnetic, at least in part,
but I suspect that's too simple, ;eaving the only alternative as simple
collisional forces, which are much more complex for plasmas than for
neutral gasses.
That brings us full circle, because the repulsive forces in neutral gasses
are electrostatic. So if it isn't possible to find a simple electrostatic
model as the first approximation to the repulsion, then my argument breaks
down completely. From my old job, I know that the convection layer is
dominated by the Kelvin-Helmholtz instability, somewhat like the Rayleigh-
Taylor instability in earth's atmosphere, but for long time scales of
electrostatic plasmas in the presence of the magnetic field. That was my
first job, back when I was a beginner, and didn't know what questions to
ask.
I have never seen the Rayleigh-Taylor equation, nor felt the need to look
it up, but a quick google search shows that there is a Kelvin-Helmholtz
equation. One site even refers to it as "well-known," but my search only
turned up 169 references to that phrase, and the equation is in none of
them. If anybody else can find something I've overlooked, it would be
greatly appreciated. :]
Landy
"tadchem" <tad...@comcast.net> wrote in message
news:48441cf6-b895-407b...@m3g2000hsc.googlegroups.com...
>> "Androcles" <Headmas...@Hogwarts.physics> wrote in message
>>
>> news:1Vk%j.20908$SA7....@newsfe09.ams2...
>>
>>
>>
>> > "Landy" <no...@nowhere.net> wrote in message
>> >news:g1kimo$qs3$1...@news-01.bur.connect.com.au...
>> > |
>> > | "Joris Dolderer" <st...@123mail.cl> wrote in message
>> > |news:483d8b31$0$27450$9b4e...@newsspool4.arcor-online.net...
>> > | > Hello,
>> > | >
>> > | > According to the barometric formula, the pressure in the earth's
>> > | > atmosphere reduces (approximately) exponentially.
>> > |
>> > | Reduces exponentially with what?
>> > |
>> > I expect he means pressure reduces with increasing altitude.
>>
>> Indeed. But his statement was abbreviated and assuming.
>> cheers
>> Bill
>
> The differential form of the barometric equation is exact:
> dP = - rho*g*dh
>
> where P is pressure, rho is density, and h is height.
I am well aware of that. My point was that there was absolutely no mention
of height in the original post.
cheers
Bill
Androcles
There was no mention of your awareness in your post.
--
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
1/2[tau(A)+tau(A')]= tau(B)
Landy
"Androcles" <Headm...@Hogwarts.physics> wrote in message
That was the intention. I was being facetious in my criticism of the
poster's lack of ability to communicate.
cheers
Bill
Androcles
http://aesopfables.com/cgi/aesop1.cgi?1&TheAssandHisShadow
(Please snip and do not respond to anything in excess of one sentence, you
pathetically stupid ignorant nitwit without the intelligence to figure out
what
the original poster intended.)
Landy
"Androcles" <Headm...@Hogwarts.physics> wrote in message
The original poster said "the pressure decreases". There are a multitude of
things that can do this you fuck-knuckle.
Thomas Smid
See my page http://www.plasmaphysics.org.uk/research/sun.htm for an
explanation.
In a nutshelll: the reason why you don't see the sun as a diffuse
cloud (which you would have in hydrostatic equilibrium) is the
existence of inelastic collisions of atoms above a certain height in
the sun. This leads to a local cooling of the gas which drastically
reduces the scale height for the radial density decrease.
Thomas
Androcles
You forgot to snip everything after
http://aesopfables.com/cgi/aesop1.cgi?1&TheAssandHisShadow ,
you ignorant nym-shifting cunt.
Landy
"Androcles" <Headm...@Hogwarts.physics> wrote in message
Unlike you, I don't take people's comments out of context by snipping
mid-sentence.
You snotty-faced heap of parrot droppings.
Is this the right room for an argument?
Androcles
Back again, shithead? I'd have thought you'd take the hint by now.
*plonk*
Landy
"Androcles" <Headm...@Hogwarts.physics> wrote in message
No, I don't take hints from arseholes. They're always worthless.
Plonk yourself.
Agent Smith
956a-c25...@w7g2000hsa.googlegroups.com:
I don't accept this, because you have golssed over *all* of the details
of this undefined, and almost certainly incorrect, inelastic scattering
process. ?:(
Steve Willner
tadchem <tad...@comcast.net> writes:
> The differential form of the barometric equation is exact:
> dP = - rho*g*dh
> where P is pressure, rho is density, and h is height.
In the stellar structure business, this is called the "equation of
hydrostatic equilibrium." It's exact only if there is no bulk
motion, but it's a fine approximation for the Sun.
> ... assumptions that (1) gravity does not vary with
> height, (2) that temperature does not vary with height, and (3) that
> the gas is ideal, and of uniform composition. These assumptions are
> necessary to perform the integration from the differential form.
No, not really. The equation can be integrated numerically even if
these assumptions don't hold. That's quite normal for stellar
atmosphere and interior models.
For the solar surface, 1 and 3 are excellent approximations, but
temperature does vary with height. However, the scale height of the
solar atmosphere (kT/g) is _much_ smaller than the solar radius, and
therefore the solar limb appears well defined.
--
Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)
Steve Willner
> > atmosphere reduces (approximately) exponentially.
> >
> > Does this also apply to the sun's gas?
In article <Xns9AADD309455D5lk...@207.115.17.102>,
John Schutkeker <jschu...@sbcglobal.net.nospam> writes:
> I doubt that anybody knows, because the physics of the outer layers has
> buoyant convection, which may or may not dominate the simplified model.
There are excellent numerical models of the solar atmosphere. An
exponential pressure model is a decent approximation, but it's far
from perfect because the temperature varies with height.
I did a brief web search for numerical models and didn't find
anything good. They are probably so standardized nowadays that
everyone in the field takes them for granted. One paper I did find:
http://arxiv.org/abs/astro-ph/0503496v1
uses standard models to work out solar chemistry. You could probably
trace through references to find the basic model they use.
> I don't think that it does apply, because when density passes a certain
> threshold, the plasma behaves more as an incompressible fluid (liquid)
> than as a compressible fluid (gas).
The ideal gas law is a good approximation for the equation of state
throughout the Sun.
> but just that people have always intuitively seen the sun as more of a
> liquid than a gas.
Where did you get this strange idea?
> The "edge" of the sun is the boundary between the chromosphere and the
> corona. If there's a transition layer, it may be the chromosphere.
More precisely, it's the height at which a tangential ray through the
Sun's atmosphere would have optical depth 1. That varies a bit
depending on exactly what wavelength you are looking at, but for
visible wavelengths the chromosphere is at the right height.
Greg Neill
news:g29ime$74u$1...@registered.motzarella.org
> I did a brief web search for numerical models and didn't find
> anything good. They are probably so standardized nowadays that
> everyone in the field takes them for granted. One paper I did find:
> http://arxiv.org/abs/astro-ph/0503496v1
> uses standard models to work out solar chemistry. You could probably
> trace through references to find the basic model they use.
You could do worse than to start by searching on the
term "polytrope".
Here's something to get started with:
Paul Schlyter
Steve Willner <wil...@cfa.harvard.edu> wrote:
>In article <48441cf6-b895-407b...@m3g2000hsc.googlegroups.com>,
> tadchem <tad...@comcast.net> writes:
>> The differential form of the barometric equation is exact:
>> dP = - rho*g*dh
>> where P is pressure, rho is density, and h is height.
>
>In the stellar structure business, this is called the "equation of
>hydrostatic equilibrium."
It's called so in other businesses as well.
>It's exact only if there is no bulk motion, but it's a fine approximation
>for the Sun.
Nah - to be exact, the gas must also be an ideal continuous gas which
never condenses. Real atmsopheres aren't continuous - they consist of
small discrete particles called atoms or molecules, which means there'll
always be very small deviations from the hydrostatic equilibrium at the
quantum level.
>> ... assumptions that (1) gravity does not vary with
>> height, (2) that temperature does not vary with height, and (3) that
>> the gas is ideal, and of uniform composition. These assumptions are
>> necessary to perform the integration from the differential form.
>
>No, not really. The equation can be integrated numerically even if
>these assumptions don't hold. That's quite normal for stellar
>atmosphere and interior models.
>
>For the solar surface, 1 and 3 are excellent approximations, but
>temperature does vary with height. However, the scale height of the
>solar atmosphere (kT/g) is _much_ smaller than the solar radius, and
>therefore the solar limb appears well defined.
The same holds for the Earth's atmosphere: the scale height at low
altitudes (approx. 8 km) is much smaller than the Earth's radius (6370 km).
Therefore the Earth's disk appears to have a sharp rather than a fuzzy
adge, when seen from space. True, the Earth is (mostly) solid, not
gaseous, however a daytime sky is about as bright as the ground on a
clear day, as seen from the Earth's surface.
>--
>Steve Willner Phone 617-495-7123 swil...@cfa.harvard.edu
>Cambridge, MA 02138 USA
>(Please email your reply if you want to be sure I see it; include a
>valid Reply-To address to receive an acknowledgement. Commercial
>email may be sent to your ISP.)
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stjarnhimlen dot se
WWW: http://stjarnhimlen.se/
Crown-Horned Snorkack
> In article <g29hte$3q...@registered.motzarella.org>,
>
> Steve Willner <will...@cfa.harvard.edu> wrote:
> >In article <48441cf6-b895-407b-8ceb-52f760230...@m3g2000hsc.googlegroups.com>,
A ray geometrically tangential to the sea surface is located below one
scale height for a length of over 600 km. How many tangential rays to
sea surface are not blocked by any clouds or haze?
Thomas
> > The "edge" of the sun is the boundary between the chromosphere and the
> > corona. If there's a transition layer, it may be the chromosphere.
>
> More precisely, it's the height at which a tangential ray through the
> Sun's atmosphere would have optical depth 1. That varies a bit
> depending on exactly what wavelength you are looking at, but for
> visible wavelengths the chromosphere is at the right height.
That is only a phenomenological criterion. Physically, the 'edge' of
the sun is defined by the height where the density becomes small
enough so that individual neutral atoms and thus inelastic collisions
can exist (if one takes 5*10^-9 cm for the size of an atom, this
density is about 10^23 cm^-3). If these inelastic collisions wouldn't
exist, the sun would just be a diffuse gas ball in hydrostatic
equilibrium and have a 1/r^2 density decrease (that is, if you could
see it all, as in the absence of inelastic collisions there wouldn't
be any emission of light either).
See my page http://www.plasmaphysics.org.uk/research/sun.htm for more.
Thomas
Agent Smith
news:g29ime$74u$1...@registered.motzarella.org:
>> > According to the barometric formula, the pressure in the earth's
>> > atmosphere reduces (approximately) exponentially.
>> >
>> > Does this also apply to the sun's gas?
>
> In article <Xns9AADD309455D5lk...@207.115.17.102>,
> John Schutkeker <jschu...@sbcglobal.net.nospam> writes:
>> I doubt that anybody knows, because the physics of the outer layers
>> has buoyant convection, which may or may not dominate the simplified
>> model.
>
> There are excellent numerical models of the solar atmosphere. An
> exponential pressure model is a decent approximation, but it's far
> from perfect because the temperature varies with height.
You're almost certainly right about this, but in the absebce of one of the
experts who writes these simulation, the information is unknown to the
denizens of this venue. Try looking for a good paper here
(http://www.iop.org/EJ/), because if you can't find it there then it might
as well be unknown. ;(
Actually these simulations are probably sufficiently well understood that
the material has been assembled into graduate or post-graduate level
textbooks. Unfortunately, they'll also cost $250, which is not in my
budget. :b
> I did a brief web search for numerical models and didn't find
> anything good. They are probably so standardized nowadays that
> everyone in the field takes them for granted.
People take a lot of stuff for granted that they shouldn't, especially
scientists. We always think that the other guy's specialty is better
understood than it actually is, and then when a beginner asks a simple
question, we all look glaringly stupid for not being able to answer it. If
I had a nickel for every time I've made that mistake, I'd be Bill gates,
and I've seen some of the top minds in the field make it, too. :b
> One paper I did find:
> http://arxiv.org/abs/astro-ph/0503496v1
> uses standard models to work out solar chemistry. You could probably
> trace through references to find the basic model they use.
I'm already behind on my reference tracing, and I probably won't have the
time this summer, but I'll save a copy of that paper, just in case.
>> I don't think that it does apply, because when density passes a
>> certain threshold, the plasma behaves more as an incompressible fluid
>> (liquid) than as a compressible fluid (gas).
>
> The ideal gas law is a good approximation for the equation of state
> throughout the Sun.
I can use that. :)
>> but just that people have always intuitively seen the sun as more of
>> a liquid than a gas.
>
> Where did you get this idea?
Solar densities are **much** higher than terrestrial gasses and much closer
to those of terrestrial liquids.
>> The "edge" of the sun is the boundary between the chromosphere and
>> the corona. If there's a transition layer, it may be the
>> chromosphere.
>
> More precisely, it's the height at which a tangential ray through the
> Sun's atmosphere would have optical depth 1. That varies a bit
> depending on exactly what wavelength you are looking at, but for
> visible wavelengths the chromosphere is at the right height.
It's traditional to use the center of the detection band of the human eye,
ie. the center of the entire visible spectrum, which IIRC puts it smack-dab
in the middle of the yellow band.
Agent Smith
$9a6e...@news.newshosting.com:
For an idel gas, gamma=5/3, making the polytropic index n=3/2. This
leaves what I call "the ideal Lane-Emden equation,"
f''+ (2/r)f'+f^(3/2)=0,
which is nonlinear, and we can't solve it with a series solution. We
can define g=sqrt(f), to get
g''+ (g'/2)[g+(4/3)g]+g^3=0,
which is hardly any better, because we still can't solve it with a
series solution, or any other method I know of. ?;(
I am stumped for what to do next, because I'd like to reduce ILE into
something that solved in a series expansion of known functions, so if
anybody has any ideas that are better than just getting a brute force
solution from the computer, I'd love to hear them. TIA.
Thomas
wrote:
> "Greg Neill" <gneill...@OVEsympatico.ca> wrote in news:484858f7$0$22390
> $9a6e1...@news.newshosting.com:
>
>
>
> > "Steve Willner" <will...@cfa.harvard.edu> wrote in message
Any isothermal ideal gas ball in hydrostatic equilibrium should have a
radial density distribution ~ 1/r^2 (see my page
http://www.plasmaphysics.org.uk/research/starformation.htm ). So this
can't explain the 'edge' of the sun. The latter is related to the fact
that radiative losses by inelastic collisions in the photosphere
(where the particle density of the sun has decreased to a sufficiently
low value so that these processes can take place) locally cool the gas
from the initial 10^7 K (the 'gravitational temperature' of the sun)
to about 6000 K (as explained in more detail on my page
http://www.plasmaphysics.org.uk/research/sun.htm ).
You can vaguely compare this to the phenomenon of striations in glow
discharges (see http://www.plasmaphysics.org.uk/research/striatn.htm
). Here the sharp left edge of each glowing striation layer is caused
by the circumstance that the electrons have acquired the necessary
excitation energy for an atomic excitation at this point (having been
accelerated by the applied electric field). In case of the sun, the
point is defined instead by the circumstance that here the density
becomes small enough so that these collisional excitation processes
can occur in the first place (for smaller radii the density of the
plasma is so high that no individual atoms and thus no excitation can
exist at all).
Thomas
Thomas
> Any isothermal ideal gas ball in hydrostatic equilibrium should have a
> radial density distribution ~ 1/r^2
Just a small correction here: it should of course be
"Any *self-gravitating* isothermal ideal gas ball in hydrostatic
equilibrium should have a radial density distribution ~ 1/r^2
Thomas
Steve Willner
SW> is no bulk motion, but it's a fine approximation for the Sun.
In article <g2ajrn$db3$1...@merope.saaf.se>,
pau...@saaf.se (Paul Schlyter) writes:
> Nah - to be exact, the gas must also be an ideal continuous gas which
> never condenses.
Continuous, yes. The equation is derived for fluids, which means the
mean free path must be small compared to any scale on which physical
conditions such as temperature or pressure change.
There is no requirement that the fluid be a gas, let alone an ideal
one. Of course if you want to calculate using the EHE, you have to
know the equation of state for the relevant materials and conditions.
> Real atmsopheres aren't continuous - they consist of
> small discrete particles called atoms or molecules, which means there'll
> always be very small deviations from the hydrostatic equilibrium at the
> quantum level.
Indeed, but it's easy to estimate how important these deviations are.
As I'm sure Paul knows, they are not at all important for solar
models.
Steve Willner
Thomas <thoma...@gmail.com> writes:
> Physically, the 'edge' of
> the sun is defined by the height where the density becomes small
> enough so that individual neutral atoms and thus inelastic collisions
> can exist
I've been debating whether to respond to this, which represents
non-standard physics. As you see, I've succumbed to temptation.
1. In normal use, "edge" is defined by optical depth. That's related
to physical density, of course, but also to temperature and
composition.
2. Neutral atoms exist well below the Sun's surface. (See 4.)
3. Inelastic collisions don't require neutral atoms to be present.
(Look up "thermal bremsstrahlung.")
> (if one takes 5*10^-9 cm for the size of an atom, this density is
> about 10^23 cm^-3).
4. The solar model in _Astrophysical Quantities_ (Table 14.9) gives a
density of 1.2E17 cm^-3 at the solar surface. The table only goes
to optical depth 24, where the density is 1.4E17. At this depth,
the hydrogen is about 2% ionized, 98% neutral.
Solar models have to match a vast array of line and continuum data.
They aren't perfect, but they are not going to be grossly wrong about
temperature, density, or ionization fraction.
Steve Willner
Agent Smith <agent...@two-blocks-on-your-left.com> writes:
> Try looking for a good paper here (http://www.iop.org/EJ/)
The best source for astronomy papers (including preprints, conference
proceedings, and some books) is the Astrophysics Data System:
http://adsabs.harvard.edu/ .
Thomas
> In article <c890c613-12b3-431c-bd0a-281ad3d45...@j22g2000hsf.googlegroups.com>,
>
> Thomas <thomas.s...@gmail.com> writes:
> > Physically, the 'edge' of
> > the sun is defined by the height where the density becomes small
> > enough so that individual neutral atoms and thus inelastic collisions
> > can exist
>
> I've been debating whether to respond to this, which represents
> non-standard physics. As you see, I've succumbed to temptation.
>
> 1. In normal use, "edge" is defined by optical depth. That's related
> to physical density, of course, but also to temperature and
> composition.
>
> 2. Neutral atoms exist well below the Sun's surface. (See 4.)
>
> 3. Inelastic collisions don't require neutral atoms to be present.
> (Look up "thermal bremsstrahlung.")
Yes, but the question is what causes the gas temperature to drop so
rapidly (going outwards) at the point where the 'edge' is located.
That can only be due to collisional excitation of neutral atoms (above
the height where these can exist). Thermal bremsstrahlung would not be
able to create such a step-like drop of temperature (and thus
density).
And as I have shown on my page http://www.plasmaphysics.org.uk/research/sun.htm
, an 'edge'-density of 3*10^23 cm^-3 is consistent with the size and
mass of the sun (assuming an 1/^r^2 increase of the density towards
the center, i.e. hydrostatic equlibrium), and this in turn is
consistent with the mentioned threshold density for the existence of
individual atoms.
> > (if one takes 5*10^-9 cm for the size of an atom, this density is
> > about 10^23 cm^-3).
>
> 4. The solar model in _Astrophysical Quantities_ (Table 14.9) gives a
> density of 1.2E17 cm^-3 at the solar surface. The table only goes
> to optical depth 24, where the density is 1.4E17. At this depth,
> the hydrogen is about 2% ionized, 98% neutral.
>
> Solar models have to match a vast array of line and continuum data.
> They aren't perfect, but they are not going to be grossly wrong about
> temperature, density, or ionization fraction.
Well, the Ptolemaic (geocentric) model of the universe also matched a
vast array of observations, but still it was grossly wrong. If one is
just opportunistic enough to adjust certain elements of the theory
accordingly, one can apparently make most data fit. But the point is
that in this case they simply don't physically explain the sharp
temperature and density drop that leads to the sharply defined
appearance of the photosphere. As shown in an earlier newsgroup thread
(http://groups.google.co.uk/group/sci.astro/browse_thread/thread/
e30c2d8f9d4f6da2/2d9750c470b53e1a ), the 'standard' model also leads
to discrepancies for the ground and excited atom densities as derived
from Fraunhofer lines.
Thomas
Agent Smith
news:7132b48d-2a96-410f...@c65g2000hsa.googlegroups.com:
> On 19 Jun, 22:01, will...@cfa.harvard.edu (Steve Willner) wrote:
>> In article
>> <c890c613-12b3-431c-bd0a-281ad3d45...@j22g2000hsf.googlegroups.com>,
>>
>> Thomas <thomas.s...@gmail.com> writes:
>> > Physically, the 'edge' of
>> > the sun is defined by the height where the density becomes small
>> > enough so that individual neutral atoms and thus inelastic
>> > collisions can exist
>>
>> I've been debating whether to respond to this, which represents
>> non-standard physics. As you see, I've succumbed to temptation.
>>
>> 1. In normal use, "edge" is defined by optical depth. That's related
>> to physical density, of course, but also to temperature and
>> composition.
>>
>> 2. Neutral atoms exist well below the Sun's surface. (See 4.)
>>
>> 3. Inelastic collisions don't require neutral atoms to be present.
>> (Look up "thermal bremsstrahlung.")
>
> Yes, but the question is what causes the gas temperature to drop so
> rapidly (going outwards) at the point where the 'edge' is located.
As I understand it, the temperature rises substantially from the
chromosphere to the corona. The chromosphere temperature is about 5,600
K, while the corona temperature is about 100,000 K.
The question is - What causes the temperature to rise so significantly?
Paul Schlyter
An answer to that question can be found here:
http://en.wikipedia.org/wiki/Sun#Atmosphere
Thomas
> On 19 Jun, 22:01, will...@cfa.harvard.edu (Steve Willner) wrote:
>
>
>
> > In article <c890c613-12b3-431c-bd0a-281ad3d45...@j22g2000hsf.googlegroups.com>,
>
> > Thomas <thomas.s...@gmail.com> writes:
> > > Physically, the 'edge' of
> > > the sun is defined by the height where the density becomes small
> > > enough so that individual neutral atoms and thus inelastic collisions
> > > can exist
>
> > I've been debating whether to respond to this, which represents
> > non-standard physics. As you see, I've succumbed to temptation.
>
> > 1. In normal use, "edge" is defined by optical depth. That's related
> > to physical density, of course, but also to temperature and
> > composition.
>
> > 2. Neutral atoms exist well below the Sun's surface. (See 4.)
>
> > 3. Inelastic collisions don't require neutral atoms to be present.
> > (Look up "thermal bremsstrahlung.")
>
> Yes, but the question is what causes the gas temperature to drop so
> rapidly (going outwards) at the point where the 'edge' is located.
> That can only be due to collisional excitation of neutral atoms (above
> the height where these can exist). Thermal bremsstrahlung would not be
> able to create such a step-like drop of temperature (and thus
> density).
>
Just an addition to what I said in my previous post:
spectral lines will of course not come from the 'edge' region with a
density of about 10^23 cm^-3. At this density all lines would be
broadened to such an extent that they would merge into a continuum (in
fact, I believe the sun's continuum is actually formed this way). The
visible spectral lines are formed by the less dense regions above
that.
Thomas
Thomas
left.com> wrote:
Have a look at my already mentioned page http://www.plasmaphysics.org.uk/research/sun.htm
where I have explained the coronal temperature: basically, the corona
(and the solar wind) consists of particles that have penetrated the
photosphere without suffering cooling due to collisional excitation of
atoms. They thus reflect the internal 'gravitational' temperature of
the sun.
Thomas
Agent Smith
news:g3sojc$2bc2$1...@merope.saaf.se:
> In article <Xns9AC7B4E5BC851ag...@207.115.17.102>,
> Agent Smith <agent...@two-blocks-on-your-left.com> wrote:
>>Thomas <thoma...@gmail.com> wrote in
>>news:7132b48d-2a96-410f-bcc7-
619ca7...@c65g2000hsa.googlegroups.com:
I didn't say *I* had that question. I just said that's what the
question is. :]
Paul Schlyter
So why did you say so if you weren't interested in any answer ???
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
> On 19 Jun, 22:01, will...@cfa.harvard.edu (Steve Willner) wrote:
> > In article <c890c613-12b3-431c-bd0a-281ad3d45...@j22g2000hsf.googlegroups.com>,
> >
> > Thomas <thomas.s...@gmail.com> writes:
> > > Physically, the 'edge' of
> > > the sun is defined by the height where the density becomes small
> > > enough so that individual neutral atoms and thus inelastic collisions
> > > can exist
> >
> > I've been debating whether to respond to this, which represents
> > non-standard physics. As you see, I've succumbed to temptation.
> >
> > 1. In normal use, "edge" is defined by optical depth. That's related
> > to physical density, of course, but also to temperature and
> > composition.
> >
> > 2. Neutral atoms exist well below the Sun's surface. (See 4.)
> >
> > 3. Inelastic collisions don't require neutral atoms to be present.
> > (Look up "thermal bremsstrahlung.")
>
> Yes, but the question is what causes the gas temperature to drop so
> rapidly (going outwards) at the point where the 'edge' is located.
> That can only be due to collisional excitation of neutral atoms (above
> the height where these can exist). Thermal bremsstrahlung would not be
> able to create such a step-like drop of temperature (and thus
> density).
Huh? The photosphere of the sun (the "edge") has the temperature that
maintains thermal equilibrium. I.e. if the sun is generating a steady
internal luminosity L, the surface temperature T at radius R must
satisfy L = 4*Pi*R^2*sigma_b*T^4 so that heat does not build up. Put
another way, the surface of the sun radiates to cold space in one
direction and gains little from space in return [*]. On the other
hand, the interior layers can share heat with their neighboring layers
in both directions. The surface must be colder than the interior.
The Virial theorem argument on your web page is not relevant since the
Virial theorem describes the mean behavior of the entire star, not a
particular atom. Indeed, sun's interior is probably ~10^7 K, more in
line with expectations from the Virial theorem.
As to the *original* question about the apparent "edge" of the sun, I
agree with Steve W, it depends on the optical depth at the surface,
which is not necessarily trivial to compute.
CM
[*] the corona has a very low optical depth, and thus is nearly
negligible when considering heat transfer.
Steve Willner
Thomas <thoma...@gmail.com> writes:
> Yes, but the question is what causes the gas temperature to drop so
> rapidly (going outwards) at the point where the 'edge' is located.
Standard solar models have no rapid drop. The one in _Astrophysical
Quantities_ has T=6520 at optical depth 1 and T=4400 at optical depth
2E-4 (the temperature minimum). The reason for the drop is the
balance between heating and cooling: above optical depth 1, radiation
escapes more or less freely to space.
If people want to discuss non-standard models, that's fine with me.
Those who want to be taken seriously will be clear about what they
are doing.
> Well, the Ptolemaic (geocentric) model of the universe also matched a
> vast array of observations, but still it was grossly wrong.
The question is how many observations and how many adjustable
parameters the model allows. Modern solar observations are far more
extensive than early planet position measurements.
> that in this case they simply don't physically explain the sharp
> temperature and density drop that leads to the sharply defined
> appearance of the photosphere.
Standard models not only explain the sharp limb, they explain limb
darkening all across the disk. Further, they do so at a wide range
of wavelengths from radio to ultraviolet. And they explain such
things as the emission reversals inside the calcium absorption
doublet.
Thomas
<craigm...@REMOVEcow.physics.wisc.edu> wrote:
This equation obviously holds by default as the temperature is defined
by it. It doesn't provide a physical explanation of the latter
(namely, why it is so much lower than the 'virial temperature' of the
sun).
> Put another way, the surface of the sun radiates to cold space in one
> direction and gains little from space in return [*]. On the other
> hand, the interior layers can share heat with their neighboring layers
> in both directions. The surface must be colder than the interior.
Cold space? Only matter can have a temperature, not space (which is a
vacuum). And the matter above the photosphere has in fact a higher
temperature (although the density is of course so low that there might
as well just be space).
Anyway, it is incorrect to assume that the gas would be cooled by the
loss of radiation into space. The gas is indeed cooled by inelastic
collisions and the production of radiation is rather a consequence of
this. And the radiation has here subsequently virtually no influence
on the temperature of the gas, as light at optical frequencies is not
able to ionize anything but is merely being scattered.
In any case, the loss of radiation from the boundary of a volume would
merely reduce the radiation density by a factor of the order 1/2 in
this region (simply due to the fact that one half-space of radiating
gas is missing here), but not by a factor 1/1800 (the ratio of the
photospheric to the 'virial' temperature)
>
> The Virial theorem argument on your web page is not relevant since the
> Virial theorem describes the mean behavior of the entire star, not a
> particular atom. Indeed, sun's interior is probably ~10^7 K, more in
> line with expectations from the Virial theorem.
As we are dealing with a highly collisional system of particles
(collision time << dynamical time scale), the average properties are
practically identical with the local ones. In other words, in the
absence of any localized heat sources or sinks, the system should be
isothermal throughout. And even if you assume an additional production
of heat in the interior, it can obviously only further increase the
temperature, not reduce them. So the question regarding the physical
cause for the low temperature of the photosphere remains, and the only
cause I can see here is the existence of a collisional excitation
process (which obviously would be a threshold process as far the gas
density is concerned, as above a certain density no individual atoms
can exist).
>
> As to the *original* question about the apparent "edge" of the sun, I
> agree with Steve W, it depends on the optical depth at the surface,
> which is not necessarily trivial to compute.
It is trivial to compute if you follow the suggestion on my page
http://www.plasmaphysics.org.uk/research/sun.htm . Assuming an 'edge'
density of about 3*10^23 cm^-3 (the density above which no individual
atoms can exist) increasing inwards like 1/r^2, the radius at which
the 'edge' is located is simply given by R=5.5*10^-5*M^1/3 [km]
(with M in kg). This yields for the sun R=6.9*10^5 km, and e.g. for
the hydrogen planets: Jupiter R=6.8*10^4 km (actual value 71,492 km
(eq.)/66,854 km (pol) ), Uranus R=2.4*10^4 km (actual 25,559/24,973),
Neptune R=2.6*10^4 km (actual 24,764/24,341). Only Saturn for some
reason does not agree quite as well ( R=4.6*10^4 (actual
60,268/54,364)).
Thomas
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
You must be mistaken. That equation describes a the nature of a
spherical "black-body" radiator, and does not "define" temperature
(which is a thermodynamic property which can be measured). A
spherical body in quasi-equilibrium like the sun, with radius R, could
not be 1000 or 10000 times hotter at the surface without being
ridiculously luminous.
Of course, there are bodies with luminosities comparable to the sun
but are much hotter -- white dwarfs and neutron stars -- but they are
much smaller to compensate. So the real question is what determines
the radius "R", and that depends on the radiative and thermal
properties of body. A neutron star, with its highly conductive
interior, can release a lot of thermal energy without puffing up. On
the other hand, our sun is dominated by a different set of radiative
opacities. This effectively makes the sun a better insulator, and it
is puffier and cooler at the surface than a neutron star.
> > Put another way, the surface of the sun radiates to cold space in one
> > direction and gains little from space in return [*]. On the other
> > hand, the interior layers can share heat with their neighboring layers
> > in both directions. The surface must be colder than the interior.
>
> Cold space? Only matter can have a temperature, not space (which is a
> vacuum). ...
"Cold space" is jargon, but valid. The Sun's photosphere is ~6000 K
and radiates this thermal energy to space. The amount of energy
returned by the cosmic microwave background (2.73 K), starlight, and
the corona is negligible. As seen by the photosphere, space is thus
"cold". Meanwhile, the interior layers can share thermal energy in
both directions, so they are hotter.
> Anyway, it is incorrect to assume that the gas would be cooled by the
> loss of radiation into space. ...
There's really no other way to remove energy from an isolated body in
space.
> ... The gas is indeed cooled by inelastic
> collisions and the production of radiation is rather a consequence of
> this. And the radiation has here subsequently virtually no influence
> on the temperature of the gas, as light at optical frequencies is not
> able to ionize anything but is merely being scattered.
Note that ionization is not required for a gas to radiate energy.
> In any case, the loss of radiation from the boundary of a volume would
> merely reduce the radiation density by a factor of the order 1/2 in
> this region (simply due to the fact that one half-space of radiating
> gas is missing here), but not by a factor 1/1800 (the ratio of the
> photospheric to the 'virial' temperature)
Again, you are making an invalid assumption about how the Virial
Theorem is to be applied. It applies to an ensemble average, not any
particular element of the ensemble.
> > The Virial theorem argument on your web page is not relevant since the
> > Virial theorem describes the mean behavior of the entire star, not a
> > particular atom. Indeed, sun's interior is probably ~10^7 K, more in
> > line with expectations from the Virial theorem.
>
> As we are dealing with a highly collisional system of particles
> (collision time << dynamical time scale), the average properties are
> practically identical with the local ones. In other words, in the
> absence of any localized heat sources or sinks, the system should be
> isothermal throughout. ...
This is an unsubstantiated conclusion. While your statement about
timescales may be true, that would only argue that the sun is
*time-stationary*, and not necessarily isothermal.
In fact, the sun is supported by hydrostatic equilibrium, and
accordingly the pressure *must* be greater at the center (and the
temperature will be greater as well).
...
> > As to the *original* question about the apparent "edge" of the sun, I
> > agree with Steve W, it depends on the optical depth at the surface,
> > which is not necessarily trivial to compute.
>
> It is trivial to compute if you follow the suggestion on my page
> http://www.plasmaphysics.org.uk/research/sun.htm . Assuming an 'edge'
> density of about 3*10^23 cm^-3 (the density above which no individual
> atoms can exist)
(assumption number 1)
> increasing inwards like 1/r^2,
(assumption number 2)
...
If you make unsubstantiated assumptions, then your conclusions will be
irrelevant. There is no reason for "no individual atoms [to] exist"
in the Sun's photosphere, nor is there no reason for the density
follow a 1/r^2 behavior. On the other hand, stellar interior
calculations show that the radiative processes within the sun are
quite complex.
CM
John Schutkeker
3a64f1...@m44g2000hsc.googlegroups.com:
Your page has an extremely basic mistake on it.
You have completely ignored the hydrostatic pressure, P, and you balanced
forces according to m_1*omega^2*r=G*m_1*m_2/r^2, where m_1 is the mass
inside the sphere where a differential element resides, and m_2 is the mass
of the differential element itself. This is just plain wrong.
You want to balance forces according to F_p=m_1*omega^2*r-G*m_1*m_2/r^2,
where F_p is the hydrostatic pressure force. I can't believe that a PhD
could make a first grade mistake like this one, because you've completely
ignored the fluid pressure, as though there were no such thing.
Your equations are only accurate for systems where pressure is negligible,
but the sun is certainly not like that, and I'm not sure any such system
exists anywhere in the universe. You need a reality check.
A gas cloud would have to be extremely diffuse to allow us to neglect
hydrostatic pressure, like the various interstellar clouds we have in the
galaxy. But those clouds hardly rotate, meaning that there's no
application for the equations you've derived, in real life.
You flubbed it, and that's putting it politely, because I can't believe
that a PhD could make such a basic mistake. You apparently never took a
fluid dynamics class when you were in school, and if you value your
reputation, you need to break open a Sophomore level text and teach
yourself this material. You have to stop making yourself look bad in front
of people who know basic physics.
I learned basic fluid dynamics in a class called 2.20 at MIT, but they've
renumbered the classes since then. The closest match I can find is 2.016 -
Hydrodynamics, at http://student.mit.edu/catalog/m2a.html . This class
contains a lot of geophysical stuff that we didn't cover, but the material
you care about is covered in the very first part, described as "Principles
of conservation of mass, momentum and energy in fluid mechanics." I expect
that it's also covered under the part called "Hydrostatics," but we didn't
have that back in '81, and I've just lately been teaching it to myself. :]
Caltech also has a class called ME 19a at this site
http://www.me.caltech.edu/academics/courses.html,
but unfortunately, everybody seems to have pulled their class links,
because it's summer. Otherwise I would have found you links to the
syllabi, telling you which specific texts they're teaching from. What
you'll have to do is to wait until September and then find out which texts
MIT and Caltech are using for those two classes, to read the chapters on
momentum conservation, ie. the Navier-Stokes Equation. The study of
pressures and forces is the study of conservation of momentum, so that's
what will give you a feel for how pressure works.
You also really need to work through the math at the site for the
University of Turku in Finland. That's the link that Greg Neill posted, but
I post it again, because it's quite instructive.
http://www.astro.utu.fi/~cflynn/Stars/l4.html
That link is just two posts ahead of yours, in this thread, and you really
should have looked at it before posting the link to your incorrect site.
AFAIK, the UTU derivation is state of the art. Nobody seems to have yet
solved the polytropes for a polytropic index of n=3/2, ie. gamma=5/3, and
that's the next logical step. That's what I'm trying to solve right now,
and it's messy, but I think that I can get through it. I'll keep working
on that idea as long as somebody can't name a specific reference to show
that it's already been done.
Thomas
<craigm...@REMOVEcow.physics.wisc.edu> wrote:
> > As we are dealing with a highly collisional system of particles
> > (collision time << dynamical time scale), the average properties are
> > practically identical with the local ones. In other words, in the
> > absence of any localized heat sources or sinks, the system should be
> > isothermal throughout. ...
>
> This is an unsubstantiated conclusion. While your statement about
> timescales may be true, that would only argue that the sun is
> *time-stationary*, and not necessarily isothermal.
What do you need to have substantiated? The fact that the solar matter
is collisionally dominated (which actually is automatically implied if
you use the notion of a particle 'temperature' rather than energy)? Or
that a collisional medium in a state of equlibrium will necessarily be
isothermal in the absence of any heat sources or sinks?
> In fact, the sun is supported by hydrostatic equilibrium, and
> accordingly the pressure *must* be greater at the center (and the
> temperature will be greater as well).
The pressure is greater because the density is greater, not the
temperature.
> > > > Yes, but the question is what causes the gas temperature to drop so
> > > > rapidly (going outwards) at the point where the 'edge' is located.
> > > > That can only be due to collisional excitation of neutral atoms (above
> > > > the height where these can exist). Thermal bremsstrahlung would not be
> > > > able to create such a step-like drop of temperature (and thus
> > > > density).
>
> > > Huh? The photosphere of the sun (the "edge") has the temperature that
> > > maintains thermal equilibrium. I.e. if the sun is generating a steady
> > > internal luminosity L, the surface temperature T at radius R must
> > > satisfy L = 4*Pi*R^2*sigma_b*T^4 so that heat does not build up.
>
> > This equation obviously holds by default as the temperature is defined
> > by it. It doesn't provide a physical explanation of the latter
> > (namely, why it is so much lower than the 'virial temperature' of the
> > sun).
>
> You must be mistaken. That equation describes a the nature of a
> spherical "black-body" radiator, and does not "define" temperature
> (which is a thermodynamic property which can be measured). A
> spherical body in quasi-equilibrium like the sun, with radius R, could
> not be 1000 or 10000 times hotter at the surface without being
> ridiculously luminous.
The point I am making is that if the photospheric temperature is just
1/1800 of the 'gravitational' temperature (that it should have on the
basis of the virial theorem) then there must be some cooling processes
going on there. The Stefan-Boltzmann relationship can not address this
question.
> "Cold space" is jargon, but valid. The Sun's photosphere is ~6000 K
> and radiates this thermal energy to space. The amount of energy
> returned by the cosmic microwave background (2.73 K), starlight, and
> the corona is negligible. As seen by the photosphere, space is thus
> "cold". Meanwhile, the interior layers can share thermal energy in
> both directions, so they are hotter.
As I said already, the lack of a layer in one direction could only
reduce the temperature by about a factor 1/2 (not 1/1800), and even
this would in case of radiation require that the radiation field is
thermally coupled to the gas, which is not the case here as visible
radiation is merely being scattered, i.e. it can not change the
kinetic energy of gas atoms (and thus it doesn't matter for the gas
whether the radiation escapes or not).
So the low temperature of the solar photosphere can only be explained
by inelastic collisions (which is then accompanied by the emission of
radiation). This cooling causes then the apparent 'edge' of the sun
(and not the other way around).
> > > As to the *original* question about the apparent "edge" of the sun, I
> > > agree with Steve W, it depends on the optical depth at the surface,
> > > which is not necessarily trivial to compute.
>
> > It is trivial to compute if you follow the suggestion on my page
> >http://www.plasmaphysics.org.uk/research/sun.htm. Assuming an 'edge'
> > density of about 3*10^23 cm^-3 (the density above which no individual
> > atoms can exist)
> (assumption number 1
It isn't actually an assumption as such. Given the accepted value for
the size of neutral atoms, it follows from elementary arithmetics that
no individual atoms can exist for atom volume densities higher than
about 3*10^23 cm^-3 .
> increasing inwards like 1/r^2,
>
> (assumption number 2)
This isn't an assumption either. An isothermal gas volume (see above)
in hydrostatic equilibrium must have a radial density dependence like
1/r^2 (see my page http://www.plasmaphysics.org.uk/research/starformation.htm
).
And as figured out above, the masses and sizes of the sun as well as
the hydrogen planets in our solar system very much confirm this
scenario.
Thomas
Thomas
John Schutkeker wrote:
It isn't a mistake. I merely derived the 1/r^2 density distribution
from a dynamical consideration of force equilibrium. As mentioned on
my web page, this is also consistent with the equation for hydrostatic
equilibrium (see Eq.(9)). Both methods are equivalent (after all,
Newton's laws still apply for a collisional system as well).
Thomas
Thomas
sun?'
Thomas
John Schutkeker
news:94cfc87e-f456-44a7...@26g2000hsk.googlegroups.com:
How you got a PhD in astronomy, I don't know, because you're making a
very basic mistake. Your boss shouldn't have passed you through if
you're this bad with sophomore physics.
You simply don't understand the concept of hydrodynamic pressure, so you
just left it out of the equation. If you'd put it in, like Greg Neill's
UTU site, you'd have the right physics, but you've discarded the entire
left hand side of the controlling ODE, which is the most important part
of the equation. That's the part yuou want to solve for, in terms of
everything else.
Have you got any more web sites with wrong physics that we should be
reading? You're not the guy obsessed with neutral particles controlling
the pressure/density relationship, are you. His stuff is wrong, too.
Lane Emden is canonical, and the references to it go back directly and
immediately to Chandrasekhar in the fifties. You can't piss on
Chandrasekhar, because he's worth a million like you. He's a legend,
just like Einstein or Kolmogorov.
Pissing on Chandrasekhar ranks right up there with the people in here
who piss on Einstein, and say relativity is wrong. You're eyes don't
work right, and you should pluck them out.
Thomas
wrote:
> Thomas <thomas.s...@gmail.com> wrote innews:94cfc87e-f456-44a7...@26g2000hsk.googlegroups.com:
>
>
>
>
>
> > John Schutkeker wrote:
>
> >> 3a64f18ad...@m44g2000hsc.googlegroups.com:
You should have read more carefully what I said:
I am assuming an ideal gas sphere in a state of hydrostatic
equilibrium in the absence of any heat sources or sinks. This means
that the sphere must be isothermal, and hence the hydrostatic equation
becomes
1/n(r)*dP(r)/dr = kT/n(r)*dn(r)/dr = -GmM(r)/r^2 .
Now since for a spherical mass distribution M(r) is given by the
integral (from 0 to r) over n(r')*r'^2 , this means that n(r) must be
proportional to 1/r^2 (both sides of the equation are then
proportional to 1/r).
Thomas
Thomas
wrote:
>
>
>
>
>
> > John Schutkeker wrote:
>
> >> 3a64f18ad...@m44g2000hsc.googlegroups.com:
>
John Schutkeker
news:d3f7d010-65f5-49e3...@w7g2000hsa.googlegroups.com:
I see dP/dr now, which means that I owe you an apology for shouting at you.
Now I feel bad for all those things I said. ;(
Your calculation points up the limitations of the isothermal model, since
Neill's Finnish UTU page clearly shows the profiles to be much more complex
than 1/r^2, although all the curves do all look something like r raised to
some negative exponent. You might want to overlay your profile onto their
data, to see where it fits, compared to all the rest of their curves. If I
can finish my n=3/2 model, I can also put it in with those curves, and make
one big happy family. :]
What you wrote down suggests the question that, having gotten the density
profile from Lane Emden, I can then work backwards to the pressure profile,
using P=K*rho^gamma, and from there to the temperature profile, using
P=nkT.
That temperature profile won't even marginally resemble the real one,
because the heat coming out from the core will strongly affect the
temperature profile. Since it's not dominated by the gravitational
pressure balance, I need to find a way to put new terms into my analysis to
account for heat generation in the core. It looks like I need to put into
a lot more work into my project before it's a real model. :o
Thomas
wrote:
The fact that the mass-radius relation for the sun and all the of
hydrogen planets is quite accurately represented by a R~M^1/3 law (I
gave the actual values already earlier in this thread) indicates that
all of them are pretty much isothermal throughout (this excludes
obviously the temperature of the 'surface', which can not be treated
as an ideal gas anyway due to the presence of inelastic collisions).
Now as mentioned above, the isothermal case results in a density
decrease proportional to 1/r^2. Other temperature profiles will result
in different density profiles and different mass-radius
relationships.
You can actually get the density profile from the general pressure-
density relation P(r)~n(r)^g (note that I use n for the number
density here, and g is the usual 'gamma' exponent).
Assume that n(r) can be represented by some function n(r)~r^-k (with k
an as yet unknown exponent).
If you insert this into the equation for hydrostatic equilibrium of a
self-gravitating sphere
1/n(r)*dP(r)/dr = -GmM(r)/r^2
(with M(r) given by the integral
(from 0 to r) over n(r')*r'^2 ),
you get
r^(k*(1-g)-1) ~ r^(1-k) .
Since this should hold for all r, the exponents must be identical,
i.e.
k*(1-g)-1 = 1-k
or
k=2/(2-g).
Now for g=1 (isothermal case) you have k=2 (i.e. n(r)~r^-2) , but for
g=4/3 you get k=3 (n(r)~r^-3) and for g=5/3 you get k=6 (n(r)~r^-6).
The point is that you can't even integrate the mass distributions for
the latter two cases as the integral over n(r')*r'^2 diverges at zero
(logarithmically for k=3 and even stronger for k=6). So these cases
are physically not possible anyway unless you make further assumptions
about the density at the center.
Thomas
Thomas
> The fact that the mass-radius relation for the sun and all the of
> hydrogen planets is quite accurately represented by a R~M^1/3 law (I
> gave the actual values already earlier in this thread) indicates that
> all of them are pretty much isothermal throughout (this excludes
> obviously the temperature of the 'surface', which can not be treated
> as an ideal gas anyway due to the presence of inelastic collisions).
> Now as mentioned above, the isothermal case results in a density
> decrease proportional to 1/r^2. Other temperature profiles will result
> in different density profiles and different mass-radius relationships
Just a correction here: a different mass-radius relationship for
different polytropic coefficients is claimed to arise from the usual
Lane-Emden theory (namely R~M^(1-n)/(3-n); see for instance
http://www.astro.princeton.edu/~gk/A403/polytrop.pdf ), but it should
be evident from my approach above that it should always be R~M^1/3
whatever the density profile (as indicated by the value of k).
However, values of k different from k=2 (i.e. n(r)~1/r^2)) would not
result in the required surface density n(R)=3*10^23 cm^-3 for a given
mass (see the arguments earlier in this thread).
Thomas
John Schutkeker
news:cee1566a-ef70-4f3e...@a1g2000hsb.googlegroups.com:
Have you overlooked the possibility that R~M^1/3 might be true for a
variety of thermal profiles, as well as a variety of mass profiles?
Thomas
wrote:
> Thomas <thomas.s...@gmail.com> wrote innews:cee1566a-ef70-4f3e...@a1g2000hsb.googlegroups.com:
>
>
>
> > On 10 Jul., 18:49, Thomas <thomas.s...@gmail.com> wrote:
> >> The fact that the mass-radius relation for the sun and all the of
> >> hydrogen planets is quite accurately represented by a R~M^1/3 law (I
> >> gave the actual values already earlier in this thread) indicates that
> >> all of them are pretty much isothermal throughout (this excludes
> >> obviously the temperature of the 'surface', which can not be treated
> >> as an ideal gas anyway due to the presence of inelastic collisions).
> >> Now as mentioned above, the isothermal case results in a density
> >> decrease proportional to 1/r^2. Other temperature profiles will
> >> result in different density profiles and different mass-radius
> >> relationships
>
> > Just a correction here: a different mass-radius relationship for
> > different polytropic coefficients is claimed to arise from the usual
> > Lane-Emden theory (namely R~M^(1-n)/(3-n); see for instance
> > be evident from my approach above that it should always be R~M^1/3
> > whatever the density profile (as indicated by the value of k).
> > However, values of k different from k=2 (i.e. n(r)~1/r^2)) would not
> > result in the required surface density n(R)=3*10^23 cm^-3 for a given
> > mass (see the arguments earlier in this thread).
>
> Have you overlooked the possibility that R~M^1/3 might be true for a
> variety of thermal profiles, as well as a variety of mass profiles?
The mass contained within the radius R of a sphere with a power law
density function n(r)=n(R)*(R/r)^k is
M(R)= 4pi*n(R)* Int[dr*r^2*(R/r)^k](0..R) =
= 4pi*n(R)*R^k* Int[dr*r^(2-k)](0..R) =
= 4pi*n(R)*R^k *1/(3-k)* [r^(3-k](0..R)
This immediately rules out values k>=3 as the total mass would become
infinite due to the singularity at 0. For smaller values of k we have
then
M(R)= 4pi*n(R)*R^k*1/(3-k)*R^(3-k)
=4pi*n(R)*1/(3-k)*R^3
i.e.
R=[M(R)*(3-k)/4pi/n(R)]^1/3
so R~M^1/3 , but it depends also on k.
As mentioned earlier, k=2 corresponds to an isothermal atmosphere, so
values k<2 would actually mean that the temperature has to *increase*
with increasing r (e.g. ~r if k=1), which seems to be physically a
rather unlikely scenario in this context.
So realistically only values 2<=k<3 should be possible for the radial
density behaviour.
Thomas
Thomas
> The mass contained within the radius R of a sphere with a power law
> density function n(r)=n(R)*(R/r)^k is
>
> M(R)= 4pi*n(R)* Int[dr*r^2*(R/r)^k](0..R) =
> = 4pi*n(R)*R^k* Int[dr*r^(2-k)](0..R) =
> = 4pi*n(R)*R^k *1/(3-k)* [r^(3-k](0..R)
>
> This immediately rules out values k>=3 as the total mass would become
> infinite due to the singularity at 0. For smaller values of k we have
> then
>
> M(R)= 4pi*n(R)*R^k*1/(3-k)*R^(3-k)
> =4pi*n(R)*1/(3-k)*R^3
>
> i.e.
>
> R=[M(R)*(3-k)/4pi/n(R)]^1/3
I forgot actually the atomic mass m here (n(R) is the number density),
so it should have been
M(R)= m*4pi*n(R)* Int[dr*r^2*(R/r)^k](0..R)
=m*4pi*n(R)*1/(3-k)*R^3
and thus
R=[M(R)*(3-k)/4pi/n(R)/m]^1/3 .
> so R~M^1/3 , but it depends also on k.
>
> As mentioned earlier, k=2 corresponds to an isothermal atmosphere, so
> values k<2 would actually mean that the temperature has to *increase*
> with increasing r (e.g. ~r if k=1),
the temperature profile for k=1 should actually be T~r*ln(r)
(T~r^2 for k=0 (constant density)). This follows directly from the
hydrostatic equilibrium equation
1/n(r)*dP(r)/dr = GmM(r)/r^2 = Gm^2*4pi/(3-k)*r*n(r) , i.e.
dP(r)/dr = Gm^2*4pi/(3-k)*r*n^2(r)
and the ideal gas equation
P(r)=K*n(r)*T(r) (where K is here the Boltzmann constant here)
assuming n(r)~r^-k .
Thomas
John Schutkeker
3ca6a8...@x41g2000hsb.googlegroups.com:
My project isn't looking so good right now. I just learned that terms
can't be grouped when the multinomial series is used to solve nonlinear
ODE's with integer exponents. Unless I can find the right series in
Gradshteyn and Rhyshik, the project is finished, and really I hope that
it doesn't rain tomorrow, so I can test ride my bike for a visit to the
library. With my arthritis I haven't visited the library in years, and
have consequently dropped the ball on several juicy projects.
I'm really getting sick of running in place and am just about ready to
hang it up and get a day job as a programmer again. I can't tell you
how frustrating it is to have one project after another die in your
hands, and the only good side-effect is that I did complete one
important project, which should hopefully be enough to clear the PhD off
my endless list of responsibilities. At 48, I'm only 23 years behind
schedule on that annoying little task.
Well, at least I finally have the damn bike and don't have to be laid up
for three days, every time I do my literature search. Jeez, it just
never friggin' ends. ;(
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
This is an assumption which implies an unphysical boundary condition
at the center of the star. Thus, it is likely that the conclusions
based on this assumption are irrelevant.
...
> R=[M(R)*(3-k)/4pi/n(R)]^1/3
>
> so R~M^1/3 , but it depends also on k.
The *actual* mass radius relationship for main sequence stars is
something more like (ref. Zombeck)
R ~ M^(1.3)
Of course for special stars like white dwarfs or neutron stars, the
physics within the star is dominated by different processes than main
sequence stars, so the mass-radius relationship is different.
CM
References
Zombeck, Handbook of Space Astronomy and Astrophysics, p. 72.
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
> On 4 Jul, 23:35, Craig Markwardt
> <craigm...@REMOVEcow.physics.wisc.edu> wrote:
> > > As we are dealing with a highly collisional system of particles
> > > (collision time << dynamical time scale), the average properties are
> > > practically identical with the local ones. In other words, in the
> > > absence of any localized heat sources or sinks, the system should be
> > > isothermal throughout. ...
> >
> > This is an unsubstantiated conclusion. While your statement about
> > timescales may be true, that would only argue that the sun is
> > *time-stationary*, and not necessarily isothermal.
>
> What do you need to have substantiated? The fact that the solar matter
> is collisionally dominated (which actually is automatically implied if
> you use the notion of a particle 'temperature' rather than energy)? Or
> that a collisional medium in a state of equlibrium will necessarily be
> isothermal in the absence of any heat sources or sinks?
What I "need" is irrelevant. I am simply disputing your arguments on
the basis that they are totally unsubstantiated. I note that you
could have substantiated them but did not. By your arguments and
assumptions, the earth's troposphere should be isothermal, and yet it
is not. Clearly there is something more complicated going on in the
troposphere. And one can make a similar argument for the sun, namely
that the radiation transport and thermal transport processes are
non-trivial, and in fact differ in different parts of the interior.
(example, Hansen, Kawaler & Trimble, 1994, *Stellar Interiors*,
Springer).
> > In fact, the sun is supported by hydrostatic equilibrium, and
> > accordingly the pressure *must* be greater at the center (and the
> > temperature will be greater as well).
>
> The pressure is greater because the density is greater, not the
> temperature.
It is both.
Since the Virial theorem addresses only the *mean* kinetic energy of a
system in equilibrium, and not a particular atom or subset of the
system, you cannot use it support your conclusions about the
surface(-only) layer of the sun. Of course there is a cooling process
at the surface of the sun -- radiation of thermal energy to space --
for which the Stefan Boltzmann equation is perfectly applicable. *My*
point was that for a given solar radius, the surface temperature is
fixed by this equation... there is no freedom. How the heat is
distributed throughout the interior of the star is then a matter of
radiation transport processes.
> > "Cold space" is jargon, but valid. The Sun's photosphere is ~6000 K
> > and radiates this thermal energy to space. The amount of energy
> > returned by the cosmic microwave background (2.73 K), starlight, and
> > the corona is negligible. As seen by the photosphere, space is thus
> > "cold". Meanwhile, the interior layers can share thermal energy in
> > both directions, so they are hotter.
>
> As I said already, the lack of a layer in one direction could only
> reduce the temperature by about a factor 1/2 (not 1/1800), and even
> this would in case of radiation require that the radiation field is
> thermally coupled to the gas, which is not the case here as visible
> radiation is merely being scattered, i.e. it can not change the
> kinetic energy of gas atoms (and thus it doesn't matter for the gas
> whether the radiation escapes or not).
There are several unsubtantiated claims here. First, there is no
basis for a body surface radiating to space to only be "half" the
temperature of the body interior. This *totally* depends on the
radiation and thermal transport within the body. If the body is a
highly effective insulator, for example, then the outer surface can be
much colder than the interior. Whether something is an "effective
insulator" or not will depend on the specific heat and thermal
processes which dominate for that particular state of the body.
(i.e. it depends on the *physics* of the problem, and not your
unsubstantiated suppositions.)
Second, *of course* the radiation field is coupled to the gas. You've
just finished arguing that the system is in equilibrium, and now you
are arguing that it is not? The color of the sun, yellow, corresponds
to a color temperature ~5000 K. Which, *SURPRISE*, is the approximate
temperature of the gas in the solar photosphere!
> So the low temperature of the solar photosphere can only be explained
> by inelastic collisions (which is then accompanied by the emission of
> radiation). This cooling causes then the apparent 'edge' of the sun
> (and not the other way around).
Since your conclusion is based on unsubstantiated premises, the
conclusions are likely irrelevant.
> > > > As to the *original* question about the apparent "edge" of the sun, I
> > > > agree with Steve W, it depends on the optical depth at the surface,
> > > > which is not necessarily trivial to compute.
> >
> > > It is trivial to compute if you follow the suggestion on my page
> > >http://www.plasmaphysics.org.uk/research/sun.htm. Assuming an 'edge'
> > > density of about 3*10^23 cm^-3 (the density above which no individual
> > > atoms can exist)
> > (assumption number 1
>
> It isn't actually an assumption as such. Given the accepted value for
> the size of neutral atoms, it follows from elementary arithmetics that
> no individual atoms can exist for atom volume densities higher than
> about 3*10^23 cm^-3 .
I was challenging your assumption that the "edge" (solar photosphere)
density was 3e23 cm^{-3}, which was not substantiated. Furthermore,
your "explanation" reveals another unsubstantiated assumption, that
somehow something "magical" happens at densities of 3e23 cm^{-3}.
Whether or not atoms can exist, a thermal plasma is happy to exist at
those densities -- and higher.
> > increasing inwards like 1/r^2,
> >
> > (assumption number 2)
>
> This isn't an assumption either. An isothermal gas volume (see above)
> in hydrostatic equilibrium must have a radial density dependence like
> 1/r^2 (see my page http://www.plasmaphysics.org.uk/research/starformation.htm
> ).
Unsubstantiated assumption: isothermal gas volume.
Unphysical conclusion: density follows 1/r^2, which reaches a
singularity at the center. If you were talking about a black hole,
that might be relevant, but you are not, and it is not relevant.
CM
Thomas
<craigm...@REMOVEcow.physics.wisc.edu> wrote:
The troposphere *is* virtually isothermal. Its temperature decreases
only by about 10% from the surface value of 300K. And this even though
the heating is essentially only from the earth's surface. So the
cooling is actually even much less here than factor 1/2 I mentioned
above with regard to the effect of 'radiating away' energy at the
atmospheric boundary.
As I said already, for a collisional system in equilibrium, the
temperature will be equal to the average temperature everywhere (in
the absence of any heat sources or sinks), so the Virial theorem
becomes locally valid.
>Of course there is a cooling process
> at the surface of the sun -- radiation of thermal energy to space --
> for which the Stefan Boltzmann equation is perfectly applicable. *My*
> point was that for a given solar radius, the surface temperature is
> fixed by this equation... there is no freedom. How the heat is
> distributed throughout the interior of the star is then a matter of
> radiation transport processes.
You still seem to be missing the point: it is not about the
consistency (or not) between the luminosity of the sun and the surface
temperature (according to the Stefan-Boltzmann equation), but about
the question why the 'surface' temperature of the sun is only 1/1800
of the temperature that a gas ball with the mass of the sun should
have in hydrostatic equilibrium.
> > > "Cold space" is jargon, but valid. The Sun's photosphere is ~6000 K
> > > and radiates this thermal energy to space. The amount of energy
> > > returned by the cosmic microwave background (2.73 K), starlight, and
> > > the corona is negligible. As seen by the photosphere, space is thus
> > > "cold". Meanwhile, the interior layers can share thermal energy in
> > > both directions, so they are hotter.
>
> > As I said already, the lack of a layer in one direction could only
> > reduce the temperature by about a factor 1/2 (not 1/1800), and even
> > this would in case of radiation require that the radiation field is
> > thermally coupled to the gas, which is not the case here as visible
> > radiation is merely being scattered, i.e. it can not change the
> > kinetic energy of gas atoms (and thus it doesn't matter for the gas
> > whether the radiation escapes or not).
>
> There are several unsubtantiated claims here. First, there is no
> basis for a body surface radiating to space to only be "half" the
> temperature of the body interior. This *totally* depends on the
> radiation and thermal transport within the body. If the body is a
> highly effective insulator, for example, then the outer surface can be
> much colder than the interior. Whether something is an "effective
> insulator" or not will depend on the specific heat and thermal
> processes which dominate for that particular state of the body.
> (i.e. it depends on the *physics* of the problem, and not your
> unsubstantiated suppositions.)
What sort of insulating material should there be in the sun?
Essentially, the sun consists only of individual protons and electrons
throughout, i.e. there is maximum transmission of thermal energy in
the contact of neighbouring gas volumes. Also, remember that we would
need an insulator which reduces the temperature from 10^7K to 6000K.
> Second, *of course* the radiation field is coupled to the gas. You've
> just finished arguing that the system is in equilibrium, and now you
> are arguing that it is not? The color of the sun, yellow, corresponds
> to a color temperature ~5000 K. Which, *SURPRISE*, is the approximate
> temperature of the gas in the solar photosphere!
By means of what processes should visible radiation thermally couple
to a gas? It doesn't even in the earth's atmosphere. The latter is
only heated by ultraviolet light (in the upper regions, where it
causes ionization and dissociation) or infrared light (in the lower
regions, where molecules are present so that vibrational and
rotational modes can be excited (which then can be transferred to
thermal energies in the course of collisions).
> > So the low temperature of the solar photosphere can only be explained
> > by inelastic collisions (which is then accompanied by the emission of
> > radiation). This cooling causes then the apparent 'edge' of the sun
> > (and not the other way around).
>
> Since your conclusion is based on unsubstantiated premises, the
> conclusions are likely irrelevant.
>
> > > > > As to the *original* question about the apparent "edge" of the sun, I
> > > > > agree with Steve W, it depends on the optical depth at the surface,
> > > > > which is not necessarily trivial to compute.
>
> > > > It is trivial to compute if you follow the suggestion on my page
> > > >http://www.plasmaphysics.org.uk/research/sun.htm. Assuming an 'edge'
> > > > density of about 3*10^23 cm^-3 (the density above which no individual
> > > > atoms can exist)
> > > (assumption number 1
>
> > It isn't actually an assumption as such. Given the accepted value for
> > the size of neutral atoms, it follows from elementary arithmetics that
> > no individual atoms can exist for atom volume densities higher than
> > about 3*10^23 cm^-3 .
>
> I was challenging your assumption that the "edge" (solar photosphere)
> density was 3e23 cm^{-3}, which was not substantiated. Furthermore,
> your "explanation" reveals another unsubstantiated assumption, that
> somehow something "magical" happens at densities of 3e23 cm^{-3}.
> Whether or not atoms can exist, a thermal plasma is happy to exist at
> those densities -- and higher.
A thermal plasma is happy to exist at those densities, but not any
atomic excitation processes (which require isolated neutral (or
partially ionized) atoms).
>
> > > increasing inwards like 1/r^2,
>
> > > (assumption number 2)
>
> > This isn't an assumption either. An isothermal gas volume (see above)
> > in hydrostatic equilibrium must have a radial density dependence like
> > 1/r^2 (see my pagehttp://www.plasmaphysics.org.uk/research/starformation.htm
> > ).
>
> Unsubstantiated assumption: isothermal gas volume.
> Unphysical conclusion: density follows 1/r^2, which reaches a
> singularity at the center. If you were talking about a black hole,
> that might be relevant, but you are not, and it is not relevant.
There is no problem with a n(r)~1/r^2 singularity, as the total mass
(which is given by the integral over r^2*n(r) ) is still finite. Only
density profiles n(r)~1/r^3 and stronger would be impossible in this
respect. And as I showed a couple of posts above, anything departing
substantially from a 1/r^2 density profile (i.e. from an isothermal
gas) would either lead indeed to an infinite mass, or alternatively
result in an increasing temperature with increasing r.
Thomas
Steve Willner
Craig Markwardt <crai...@REMOVEcow.physics.wisc.edu> writes:
> The *actual* mass radius relationship for main sequence stars is
> something more like (ref. Zombeck)
> R ~ M^(1.3)
_Astrophysical Quantities_ gives the exponent as 0.917 from 0.1 to
1.3 solar masses and 0.64 from 1.3 to 20. Neither is at all close to
the value 0.33 suggested by Smid.
I'm not sure what causes the change in slope at 1.3 solar masses. A
lot of stellar properties change around that mass, and it isn't clear
to me which is most important for determining the radius.
Thomas
> In article <m2prph5slb....@phloem.local>,
> Craig Markwardt <craigm...@REMOVEcow.physics.wisc.edu> writes:
>
> > The *actual* mass radius relationship for main sequence stars is
> > something more like (ref. Zombeck)
> > R ~ M^(1.3)
>
> _Astrophysical Quantities_ gives the exponent as 0.917 from 0.1 to
> 1.3 solar masses and 0.64 from 1.3 to 20. Neither is at all close to
> the value 0.33 suggested by Smid.
These values are presumably not based on actually measured radii, but
on inferred ones on the basis of the Stefan-Boltzmann equation and an
empirical mass-luminosity relationship.
If one takes the masses of our sun and the hydrogen planets in our
solar system (where the masses and radii are known accurately) then
this fits indeed indeed very well a R~M^1/3 law: the theoretical
radii for Jupiter, Saturn, Uranus, Neptune are then (Sun=1) : 0.099,
0.066, 0.035, 0.037 . The actual values are 0.1, 0.085, 0.037, 0.035.
An exponent 0.9 on the other hand would result in : 0.002, 0.00064,
0.00012, 0.00014.
Thomas
Craig Markwardt
wil...@cfa.harvard.edu (Steve Willner) writes:
> In article <m2prph5...@phloem.local>,
> Craig Markwardt <crai...@REMOVEcow.physics.wisc.edu> writes:
> > The *actual* mass radius relationship for main sequence stars is
> > something more like (ref. Zombeck)
> > R ~ M^(1.3)
>
> _Astrophysical Quantities_ gives the exponent as 0.917 from 0.1 to
> 1.3 solar masses and 0.64 from 1.3 to 20. Neither is at all close to
> the value 0.33 suggested by Smid.
OK, fair enough, I was just plotting the Zombeck numbers and fitting a
line through it in log-log space.
> I'm not sure what causes the change in slope at 1.3 solar masses. A
> lot of stellar properties change around that mass, and it isn't clear
> to me which is most important for determining the radius.
Craig
--
--------------------------------------------------------------------------
Craig B. Markwardt, Ph.D. EMAIL: crai...@REMOVEcow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
--------------------------------------------------------------------------
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
> On 16 Jul, 22:43, will...@cfa.harvard.edu (Steve Willner) wrote:
> > In article <m2prph5slb....@phloem.local>,
> > Craig Markwardt <craigm...@REMOVEcow.physics.wisc.edu> writes:
> >
> > > The *actual* mass radius relationship for main sequence stars is
> > > something more like (ref. Zombeck)
> > > R ~ M^(1.3)
> >
> > _Astrophysical Quantities_ gives the exponent as 0.917 from 0.1 to
> > 1.3 solar masses and 0.64 from 1.3 to 20. Neither is at all close to
> > the value 0.33 suggested by Smid.
>
> These values are presumably not based on actually measured radii, but
> on inferred ones on the basis of the Stefan-Boltzmann equation and an
> empirical mass-luminosity relationship.
On which scientific study are you basing that claim?
CM
Thomas
<craigm...@REMOVEcow.physics.wisc.edu> wrote:
On the figures given in the second half of my post (that you chose to
delete here):
the radii for the hydrogen masses in the solar system (Sun, Jupiter,
Saturn, Uranus, Neptune) follow the sequence 1, 0.1, 0.085, 0.037,
0.035, which quite accurately corresponds to an exponent 1/3 in the
mass-radius relationship, but not to any of those claimed above.
Thomas
John Park
Something to do with the fact that they're not doing nuclear fusion?
--John Park
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
> the heating is essentially only from the earth's surface. ...
Ah, *virtually* isothermal. A key retreat word. *Virtually*
isothermal means not "necessarily isothermal" as you described above.
But let's look at how virtually isothermal the earth's troposphere is.
Consider the International Standard Atmosphere; the difference in
temperatures between the base of the troposphere and the tropopause is
about 71 C [*], which is more like 25%, not 10%. I suppose that's
"virtually" the same as 10% for you?
But let's consider this a little more. Might it be possible that if
the troposphere were twice as thick, the temperature would decrease to
(.75)x(.75) = 56% of the base? [*] Or (.75)^3 = 42% for a triply
thick troposphere? The earth's troposphere is ~11 km thick, or about
0.17% of the earth's radius. Compare this to the convective layer of
the sun, which is about 5% thick. That's (5% / 0.17%) = 30 scale
heights, so in principle the photosphere temperature could decrease to
(0.75)^(30) = 1/5600th of the base temp. Huh! Of *COURSE* I don't
think this is how the solar atmosphere actually works, but I'm
pointing out how your "virtually isothermal" double standard.
According to your claim, a gas which is "collisionally dominated" must
*necessarily* be isothermal, but you also claim that it need only be
"virtually" isothermal. I'm sure it's quite convenient for you to
define any standard you wish. The fact is, if you stack enough thin
layers of atmosphere, this can have a significant effect on the
radiation transport.
Ah, but then you also allow that the troposphere is heated from below
by the earth's surface, which creates a temperature gradient. Did you
ever stop to think that the *sun* might be heated from below???
[*] Height 0 km: Temp = +15 C; Height 11 km: Temp = -56.5 C.
> ... So the
> cooling is actually even much less here than factor 1/2 I mentioned
> above with regard to the effect of 'radiating away' energy at the
> atmospheric boundary.
Of course, you are ignoring the differences in optical depth between
the earth's atmosphere and the solar atmosphere.
Don't you mean "virtually equal to the average temperature?" But as
you allowed that when there is a heat source the earth's troposphere
has a temperature gradient, mightn't you allow the same for the sun,
which has a much more powerful heat source? I.e. the stellar
atmosphere is not necessarily isothermal.
> >Of course there is a cooling process
> > at the surface of the sun -- radiation of thermal energy to space --
> > for which the Stefan Boltzmann equation is perfectly applicable. *My*
> > point was that for a given solar radius, the surface temperature is
> > fixed by this equation... there is no freedom. How the heat is
> > distributed throughout the interior of the star is then a matter of
> > radiation transport processes.
>
> You still seem to be missing the point: it is not about the
> consistency (or not) between the luminosity of the sun and the surface
> temperature (according to the Stefan-Boltzmann equation), but about
> the question why the 'surface' temperature of the sun is only 1/1800
> of the temperature that a gas ball with the mass of the sun should
> have in hydrostatic equilibrium.
But my point still stands. For sun's radius and energy output rate
today, how could the surface temperature be anything other than
~6000K? It must be that temperature, because it's basic conservation
of energy (i.e. the balance between energy generation rate with
Thermal radiation). You are right, it's interesting to know the
surface temperature is so much lower than the core temperature... it
tells us how heat flows through the stellar interior.
Note, I did not say an "insulating material," I said an "effective
insulator," which I then qualified by the radiation/thermal properties
of the medium. Such effects may include thermal conductivity, which
will be driven by density and electron temperature; radiative
transport, which is driven by opacity, density and ionization state;
convection, which is driven by local energy density, and gas density.
The gas density itself will be governed by the gas equation of state,
which may change depending on the ionization state. In other words,
there are a complicated set of coupled systems which constrain how
much energy is able to escape the various layers of the sun. If these
processes are not very efficient, then the star will puff up to larger
but cooler radius to compensate.
> > Second, *of course* the radiation field is coupled to the gas. You've
> > just finished arguing that the system is in equilibrium, and now you
> > are arguing that it is not? The color of the sun, yellow, corresponds
> > to a color temperature ~5000 K. Which, *SURPRISE*, is the approximate
> > temperature of the gas in the solar photosphere!
>
> By means of what processes should visible radiation thermally couple
> to a gas?
You mean like free-free, bound-free, free-bound, bound-bound emission,
among others? The sun's photosphere is a sea of plasma and radiation
in near-equilibrium. By the equipartition theorem the energy density
of the gas and radiation should be equal.
> ... It doesn't even in the earth's atmosphere. The latter is
> only heated by ultraviolet light (in the upper regions, where it
> causes ionization and dissociation) or infrared light (in the lower
> regions, where molecules are present so that vibrational and
> rotational modes can be excited (which then can be transferred to
> thermal energies in the course of collisions).
Huh? The equilibrium temperature of the earth is ~300 K, so the
radiation is infrared, not optical.
I note you again did not substantiate your claim that the photosphere
density was 3e23 cm^{-3}. I accept your claim that "not any" atomic
excitation processes function at high densities. However, *some* do.
This is still irrelevant since the solar photosphere density is not
3e23 cm^{-3}.
> > > > increasing inwards like 1/r^2,
> >
> > > > (assumption number 2)
> >
> > > This isn't an assumption either. An isothermal gas volume (see above)
> > > in hydrostatic equilibrium must have a radial density dependence like
> > > 1/r^2 (see my pagehttp://www.plasmaphysics.org.uk/research/starformation.htm
> > > ).
> >
> > Unsubstantiated assumption: isothermal gas volume.
> > Unphysical conclusion: density follows 1/r^2, which reaches a
> > singularity at the center. If you were talking about a black hole,
> > that might be relevant, but you are not, and it is not relevant.
>
> There is no problem with a n(r)~1/r^2 singularity, as the total mass
> (which is given by the integral over r^2*n(r) ) is still finite. ...
I note this claim is unsubstantiated. In reality, even for finite
total mass, an infinite local density would be unphysical.
> ... Only
> density profiles n(r)~1/r^3 and stronger would be impossible in this
> respect. And as I showed a couple of posts above, anything departing
> substantially from a 1/r^2 density profile (i.e. from an isothermal
> gas) would either lead indeed to an infinite mass, or alternatively
> result in an increasing temperature with increasing r.
*OR*, your assumption about a pure power law is invalid. There are an
infinitude of functions which are not power laws. I guess you meant
"virtually" a power law?
CM
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
> On 19 Jul, 15:51, Craig Markwardt
> <craigm...@REMOVEcow.physics.wisc.edu> wrote:
> > Thomas <thomas.s...@gmail.com> writes:
> > > On 16 Jul, 22:43, will...@cfa.harvard.edu (Steve Willner) wrote:
> > > > In article <m2prph5slb....@phloem.local>,
> > > > Craig Markwardt <craigm...@REMOVEcow.physics.wisc.edu> writes:
> >
> > > > > The *actual* mass radius relationship for main sequence stars is
> > > > > something more like (ref. Zombeck)
> > > > > R ~ M^(1.3)
> >
> > > > _Astrophysical Quantities_ gives the exponent as 0.917 from 0.1 to
> > > > 1.3 solar masses and 0.64 from 1.3 to 20. Neither is at all close to
> > > > the value 0.33 suggested by Smid.
> >
> > > These values are presumably not based on actually measured radii, but
> > > on inferred ones on the basis of the Stefan-Boltzmann equation and an
> > > empirical mass-luminosity relationship.
> >
> > On which scientific study are you basing that claim?
>
> On the figures given in the second half of my post (that you chose to
> delete here):
Your figures are irrelevant to your claim that "these [stellar mass
and radius values listed in Zombeck or Allen] are presumably not based
on actually measured radii, but on inferred ones on the basis of the
Stefan-Boltzmann equation and an empirical mass-luminosity
relationship."
On the basis of what scientific study would you "presume" that? You
made the claim: did you just pull it out of nowhere?
CM
John Park
analogous despite the facts that: one has a surface temperature a good
5500 K greater than the others; in four the hydrogen is present as
molecules or a metallic phase, while in the other it is monatomic or in a
plasma; and in two cases hydrogen probably makes up only about 15% of
the total mass anyway.
--John Park
Thomas Smid
> John Park (af...@FreeNet.Carleton.CA) writes:
On 20 Jul, 02:20, af...@FreeNet.Carleton.CA (John Park) wrote:
> John Park (af...@FreeNet.Carleton.CA) writes:
It is in the first place a hard observational fact that the radius for
these bodies follows a R~M^1/3 law. Whatever you make out of this
with regard to their physics has to be consistent with it. It
definitely suggests that the radial density function is pretty much
identical in all cases (if you assume an ideal gas in hydrostatic
equilibrium, then (as shown above) the radius is given by R=[M(R)*(3-
k)/4pi/n(R)/m]^1/3 , where k is the power index for the radial
density function n(r)~r^-k ; the figures show that k differs not by
more than about 10% between the sun and the hydrogen planets (Saturn
excluded for some reason)); so even if there is any fusion occurring
in the sun, it has an insignificant effect on its structure).
Thomas
John Park
[...]
>>
>> >> the radii for the hydrogen masses in the solar system (Sun, Jupiter,
>> >> Saturn, Uranus, Neptune) follow the sequence 1, 0.1, 0.085, 0.037,
>> >> 0.035, which quite accurately corresponds to an exponent 1/3 in the
>> >> mass-radius relationship, but not to any of those claimed above.
I replied:
>>
>> > Something to do with the fact that they're not doing nuclear fusion?
>>
>> More to the point, you treat these five bodies as though they were
>> analogous despite the facts that: one has a surface temperature a good
>> 5500 K greater than the others; in four the hydrogen is present as
>> molecules or a metallic phase, while in the other it is monatomic or in a
>> plasma; and in two cases hydrogen probably makes up only about 15% of
>> the total mass anyway.
>
> It is in the first place a hard observational fact that the radius for
> these bodies follows a R~M^1/3 law. Whatever you make out of this
> with regard to their physics has to be consistent with it. It
> definitely suggests that the radial density function is pretty much
> identical in all cases (if you assume an ideal gas in hydrostatic
^^^^^^^^^^^^^^^^^^^^^^^^^^
> equilibrium, then (as shown above) the radius is given by R=[M(R)*(3-
> k)/4pi/n(R)/m]^1/3 , where k is the power index for the radial
> density function n(r)~r^-k ; the figures show that k differs not by
> more than about 10% between the sun and the hydrogen planets (Saturn
> excluded for some reason)); so even if there is any fusion occurring
> in the sun, it has an insignificant effect on its structure).
But it's pretty clearly not the _same_ ideal gas (plasma/H vs H2/CH4 &c); and
for the outer planets it's questionable whether it makes sense to regard
them as gases at all. You've enough data to find a coincidence;
not enough to show a connection. (Note that Uranus and Neptune are so
similar they're effectively the same point.)
Put it another way: if Uranus and Neptune had the same masses but were 75%
hydrogen instead of 15% and had orbits inside Venus', would you still expect
them to fit your curve?
--John Park
Thomas Smid
I think it makes sense to describe all these planets as consisting
largely of an ideal gas, because even for Neptune, the internal
temperature (as given by the gravitational potential over the virial
theorem) is still about 15,000 K, which should make it impossible for
organized structures of the matter in the form of a fluid or a solid
to develop. So essentially, we are still dealing here (like in the
case of the sun) with a 'plasma soup' of nuclei and electrons. Because
no atoms (and thus no excitation of atomic transitions) can exist at
this temperature and density, one can consider it as an ideal gas (in
fact more ideal than gases in the usual sense).
> You've enough data to find a coincidence;
> not enough to show a connection. (Note that Uranus and Neptune are so
> similar they're effectively the same point.)
Well, the point is that these 'coincidences' confirm a mass-radius
relationship R~M^1/3, but invalidate other exponents.
>
> Put it another way: if Uranus and Neptune had the same masses but were 75%
> hydrogen instead of 15% and had orbits inside Venus', would you still expect
> them to fit your curve?
The fact that they fit the curve suggests that they *do* consist
largely of hydrogen.
Thomas
Thomas Smid
> I think it makes sense to describe all these planets as consisting
> largely of an ideal gas, because even for Neptune, the internal
> temperature (as given by the gravitational potential over the virial
> theorem) is still about 15,000 K, which should make it impossible for
> organized structures of the matter in the form of a fluid or a solid
> to develop. So essentially, we are still dealing here (like in the
> case of the sun) with a 'plasma soup' of nuclei and electrons. Because
> no atoms (and thus no excitation of atomic transitions) can exist at
> this temperature and density, one can consider it as an ideal gas (in
> fact more ideal than gases in the usual sense).
>
> > You've enough data to find a coincidence;
> > not enough to show a connection. (Note that Uranus and Neptune are so
> > similar they're effectively the same point.)
>
> Well, the point is that these 'coincidences' confirm a mass-radius
> relationship R~M^1/3, but invalidate other exponents.
>
>
>
> > Put it another way: if Uranus and Neptune had the same masses but were 75%
> > hydrogen instead of 15% and had orbits inside Venus', would you still expect
> > them to fit your curve?
>
> The fact that they fit the curve suggests that they *do* consist
> largely of hydrogen
By the way, the smaller planets also follow a R~M^1/3 law for
themselves, but have a consistently smaller radius by about a factor
0.6 (Mars 0.7) (this agrees with their higher average density compared
to the sun and giant planets).
Thomas
John Park
Could you elaborate on that? What I've read suggest that the outer planets
are mostly slushes (with rocky cores). Which "virial theorem" are you
referring to, the one relating kinetic and potential energies?
> organized structures of the matter in the form of a fluid or a solid
> to develop. So essentially, we are still dealing here (like in the
> case of the sun) with a 'plasma soup' of nuclei and electrons. Because
> no atoms (and thus no excitation of atomic transitions) can exist at
> this temperature and density, one can consider it as an ideal gas (in
> fact more ideal than gases in the usual sense).
>
>> You've enough data to find a coincidence;
>> not enough to show a connection. (Note that Uranus and Neptune are so
>> similar they're effectively the same point.)
>
> Well, the point is that these 'coincidences' confirm a mass-radius
> relationship R~M^1/3, but invalidate other exponents.
They don't confirm anything. They suggest a mass-radius relationship, which,
if true, would imply that all other variables, such as composition,
temperature and thermonuclear activity, have no effect. That is
counter-intuitive, to say the least.
There must be lots of stellar mass/radius data from eclipsing binaries. Why
don't you use those results?
>>
>> Put it another way: if Uranus and Neptune had the same masses but were 75%
>> hydrogen instead of 15% and had orbits inside Venus', would you still expect
>> them to fit your curve?
>
> The fact that they fit the curve suggests that they *do* consist
> largely of hydrogen.
>
You haven't got nearly enough data to claim that.
>By the way, the smaller planets also follow a R~M^1/3 law for
>themselves, but have a consistently smaller radius by about a factor
>0.6 (Mars 0.7) (this agrees with their higher average density compared
>to the sun and giant planets).
>
So the 1/3 power law applies to rocky bodies as well as ideal gases? How
do you rationalise that?
--John Park
Thomas Smid
> > I think it makes sense to describe all these planets as consisting
> > largely of an ideal gas, because even for Neptune, the internal
> > temperature (as given by the gravitational potential over the virial
> > theorem) is still about 15,000 K, which should make it impossible for
>
> Could you elaborate on that? What I've read suggest that the outer planets
> are mostly slushes (with rocky cores).
Yes, I have read this as well, but it is evidently wrong (an average
density of 1.2g/cm^3 clearly shows that hydrogen is the main
constituent).
> Which "virial theorem" are you
> referring to, the one relating kinetic and potential energies?
Yes.
> > Well, the point is that these 'coincidences' confirm a mass-radius
> > relationship R~M^1/3, but invalidate other exponents.
>
> They don't confirm anything. They suggest a mass-radius relationship, which,
> if true, would imply that all other variables, such as composition,
> temperature and thermonuclear activity, have no effect. That is
> counter-intuitive, to say the least.
Again, the mass-radius relationship R~M^1/3 for solar system objects
(including the sun) stands as an observational fact, independently of
any physical theory. If a theory implies a different relationship,
then it is inconsistent with these observations. Unless you have an
explanation for this inconsistency, it invalidates thus the theory.
And given the formula I mentioned above R=[M(R)*(3-k)/4pi/n(R)/
m]^1/3 , it is apparent that the radius *does* indeed also depend on
the composition (as it depends on the atomic mass m) ,and potentially
the structure (through the index k for the radial density
distribution), but the data clearly indicate that the density
structure is identical for all the objects (including the sun).
> There must be lots of stellar mass/radius data from eclipsing binaries. Why
> don't you use those results?
Because they are not reliable and accurate enough. The determination
of masses and radii for binary stars depends on a host of assumptions
with regard to distance, orbit, limb darkening etc. These assumptions
are likely to be biased such that they conform with standard theories
regarding stellar structure, so they are by no means direct and
independent measurements (like they are possible for solar system
objects).
> >>>> Put it another way: if Uranus and Neptune had the same masses but were 75%
> >> hydrogen instead of 15% and had orbits inside Venus', would you still expect
> >> them to fit your curve?
>
> > The fact that they fit the curve suggests that they *do* consist
> > largely of hydrogen.
>
> You haven't got nearly enough data to claim that.
You can't really blame me that there not more planets in the solar
system (which is to date the only place where we can measure masses
and radii of objects accurately), but the point is that all objects
present obey a R~M^1/3 relationship.
> >By the way, the smaller planets also follow a R~M^1/3 law for
> >themselves, but have a consistently smaller radius by about a factor
> >0.6 (Mars 0.7) (this agrees with their higher average density compared
> >to the sun and giant planets).
>
> So the 1/3 power law applies to rocky bodies as well as ideal gases? How
> do you rationalise that?
Even for the smaller planets, the temperature in the interior is still
a couple of thousand degrees. Apparently, this is still sufficient to
make the material compressible like an ideal gas (which will thus also
result in a 1/r^2 density increase towards the center).
Thomas
John Park
> On 21 Jul, 11:12, af...@FreeNet.Carleton.CA (John Park) wrote:
>
>> > I think it makes sense to describe all these planets as consisting
>> > largely of an ideal gas, because even for Neptune, the internal
>> > temperature (as given by the gravitational potential over the virial
>> > theorem) is still about 15,000 K, which should make it impossible for
>>
>> Could you elaborate on that? What I've read suggest that the outer planets
>> are mostly slushes (with rocky cores).
>
> Yes, I have read this as well, but it is evidently wrong (an average
> density of 1.2g/cm^3 clearly shows that hydrogen is the main
> constituent).
It's not clear to me. That's quite a bit denser than Saturn, which has
several times the mass.
>
>> Which "virial theorem" are you
>> referring to, the one relating kinetic and potential energies?
>
> Yes.
>
Under the assumption that Uranus was made of hydrogen I used the VT to
estimate a mean temperature of 5000 K, which may or may not be consistent
with your core temperature of 15,000 K. But clearly the surface temperature
is much lower than that, and molecular gases--especially hddrogen--are quite
transparent in the optical range. So why aren't these planets emitting
viisble light?
Incidentally I also estimated a VT temperature for the Earth, assuming it
was made of iron: 40,000 K. I wouldn't trust either estimate.
>> > Well, the point is that these 'coincidences' confirm a mass-radius
>> > relationship R~M^1/3, but invalidate other exponents.
>>
>> They don't confirm anything. They suggest a mass-radius relationship, which,
>> if true, would imply that all other variables, such as composition,
>> temperature and thermonuclear activity, have no effect. That is
>> counter-intuitive, to say the least.
>
> Again, the mass-radius relationship R~M^1/3 for solar system objects
> (including the sun) stands as an observational fact, independently of
> any physical theory. If a theory implies a different relationship,
> then it is inconsistent with these observations. Unless you have an
> explanation for this inconsistency, it invalidates thus the theory.
> And given the formula I mentioned above R=[M(R)*(3-k)/4pi/n(R)/
> m]^1/3 , it is apparent that the radius *does* indeed also depend on
> the composition (as it depends on the atomic mass m) ,and potentially
> the structure (through the index k for the radial density
> distribution), but the data clearly indicate that the density
> structure is identical for all the objects (including the sun).
You haven't enough points to exclude other possibilities.
>> There must be lots of stellar mass/radius data from eclipsing binaries. Why
>> don't you use those results?
>
> Because they are not reliable and accurate enough. The determination
> of masses and radii for binary stars depends on a host of assumptions
> with regard to distance, orbit, limb darkening etc. These assumptions
> are likely to be biased such that they conform with standard theories
> regarding stellar structure, so they are by no means direct and
> independent measurements (like they are possible for solar system
> objects).
I'm not an expert, but frankly I don't believe you. And what about
interferometric or other direct determinations of stellar radii?
>
>> >>>> Put it another way: if Uranus and Neptune had the same masses but were 75%
>> >> hydrogen instead of 15% and had orbits inside Venus', would you still expect
>> >> them to fit your curve?
>>
>> > The fact that they fit the curve suggests that they *do* consist
>> > largely of hydrogen.
>>
>> You haven't got nearly enough data to claim that.
>
> You can't really blame me that there not more planets in the solar
> system (which is to date the only place where we can measure masses
> and radii of objects accurately), but the point is that all objects
> present obey a R~M^1/3 relationship.
>
I can blame you for leaping to an extravagant conclusion based on inadequate
data.
>> >By the way, the smaller planets also follow a R~M^1/3 law for
>> >themselves, but have a consistently smaller radius by about a factor
>> >0.6 (Mars 0.7) (this agrees with their higher average density compared
>> >to the sun and giant planets).
>>
>> So the 1/3 power law applies to rocky bodies as well as ideal gases? How
>> do you rationalise that?
>
> Even for the smaller planets, the temperature in the interior is still
> a couple of thousand degrees. Apparently, this is still sufficient to
> make the material compressible like an ideal gas (which will thus also
> result in a 1/r^2 density increase towards the center).
>
So how compressible is iron at 2000 K? And what's your evidence for the
temperatures?
--John Park
Thomas
I presume it simply because the given exponents are about what
standard theoretical models would predict, and because of the fact
that these are not consistent with solar system data (where masses and
radii are precisely and unambiguously known).
Thomas
Thomas
<craigm...@REMOVEcow.physics.wisc.edu> wrote:
> Thomas <thomas.s...@gmail.com> writes:
> > The troposphere *is* virtually isothermal. Its temperature decreases
> > only by about 10% from the surface value of 300K. And this even though
> > the heating is essentially only from the earth's surface. ...
> Ah, *virtually* isothermal. A key retreat word. *Virtually*
> isothermal means not "necessarily isothermal" as you described above.
> But let's look at how virtually isothermal the earth's troposphere is.
> Consider the International Standard Atmosphere; the difference in
> temperatures between the base of the troposphere and the tropopause is
> about 71 C [*], which is more like 25%, not 10%. I suppose that's
> "virtually" the same as 10% for you?#
Yes, it is virtually the same because I said *about* 10% (meaning of
the order of 10%). But even a 25% percent temperature drop still means
that cooling processes and the related loss of energy at the top of
the atmosphere are not overly important here, and that equipartition
of energy by means of elastic collisions dominates.
> But let's consider this a little more. Might it be possible that if
> the troposphere were twice as thick, the temperature would decrease to
> (.75)x(.75) = 56% of the base? [*] Or (.75)^3 = 42% for a triply
> thick troposphere? The earth's troposphere is ~11 km thick, or about
> 0.17% of the earth's radius. Compare this to the convective layer of
> the sun, which is about 5% thick. That's (5% / 0.17%) = 30 scale
> heights, so in principle the photosphere temperature could decrease to
> (0.75)^(30) = 1/5600th of the base temp. Huh! Of *COURSE* I don't
> think this is how the solar atmosphere actually works, but I'm
> pointing out how your "virtually isothermal" double standard.
Such a scenario is debatable because it all depends on what
assumptions you make regarding the details of the cooling processes.
Without any inelastic collisions that can destroy the kinetic energy
of the atoms or molecules there simply won't be any cooling at all in
a bound system of particles.
But the comparison of the sun with the troposphere is misleading here
anyway: as indicated already earlier, the troposphere consists of
individual molecules that can absorb and re-emit infrared radiation.
No such coupling between matter and radiation could exist in the solar
interior, because a) no individual molecules and atoms can exist there
due to the high density (and temperature), and b) visible radiation
(which constitutes the bulk of the solar energy output) can not be
absorbed by atoms anyway.
Furthermore, the sun is, in contrast to the troposphere, a self-
gravitating system that can gain energy from gravitational
contraction: if you assume a self-gravitating gas ball in hydrostatic
equilibrium, and somehow suddenly remove the kinetic energy of the
particles in some sub-volume, then this sub-volume will begin to
collapse at a free-fall rate until it has gained enough energy so that
a new equilibrium is established. So if you have a region where
cooling takes place (like in the solar photosphere), the information
about this cooling can never propagate to the sun's interior as
hydrostatic equilibrium is locally established much faster than the
heat can travel. So the effects of the cooling will only be restricted
to the photosphere itself, but not extend to regions below (where we
have the original 'virial' temperature).
> According to your claim, a gas which is "collisionally dominated" must
> *necessarily* be isothermal, but you also claim that it need only be
> "virtually" isothermal. I'm sure it's quite convenient for you to
> define any standard you wish. The fact is, if you stack enough thin
> layers of atmosphere, this can have a significant effect on the
> radiation transport.
What I said was that a collisionally dominated gas (in a state of
equilibrium) must be necessarily isothermal *in the absence of any
heat sources or sinks*. And in that case it will be *strictly*
isothermal. Of course this is an ideal condition which in practice
does not strictly hold, but according to my argument above, it should
hold quite closely for the interior of the sun, as the effect of
gravity will level out any temperature gradients potentially arising
from the cooling in the photosphere.
> Ah, but then you also allow that the troposphere is heated from below
> by the earth's surface, which creates a temperature gradient. Did you
> ever stop to think that the *sun* might be heated from below???
No, not really. In contrast to the troposphere, the sun doesn't need a
separate heating to be stable as it is a self-gravitating system and
thus produces the required energy itself.
Also, the fact that the sun obeys the same R~M^1/3 mass-radius
relationship as the giant planets in the solar system, indicates that
there is no difference in their density (and thus temperature-)
structure.
> > You still seem to be missing the point: it is not about the
> > consistency (or not) between the luminosity of the sun and the surface
> > temperature (according to the Stefan-Boltzmann equation), but about
> > the question why the 'surface' temperature of the sun is only 1/1800
> > of the temperature that a gas ball with the mass of the sun should
> > have in hydrostatic equilibrium.
> But my point still stands. For sun's radius and energy output rate
> today, how could the surface temperature be anything other than
> ~6000K? It must be that temperature, because it's basic conservation
> of energy (i.e. the balance between energy generation rate with
> Thermal radiation). You are right, it's interesting to know the
> surface temperature is so much lower than the core temperature... it
> tells us how heat flows through the stellar interior.
I quote from http://www.ap.smu.ca/~guenther/Level01/solar/what_is_ssm.html
:
"Nearly all stellar evolutionary calculations are calibrated with
respect to the standard solar model. The standard solar model is
derived from the conservation laws and energy transport equations of
physics, applied to a spherically symmetric gas (plasma) sphere and
constrained by the luminosity, radius, age and composition of the
Sun".
It should be obvious that if you apply physical parameters as
constraints to a model (like the luminosity (i.e. the surface
temperature)), then the model can not give an explanation for the
parameters, because they serve on the contrary as input values.
As I said already repeatedly, the temperature of the particles in the
sun can only be lowered below the 'virial' temperature if inelastic
collisions are present. The standard solar model can't address this
point in any way, and therefore it is not a valid physical model. On
the other hand, I only need the mass here to *derive* both the radius
and the surface temperature (the latter being 1/1800 of the 'virial
temperature').
> > By means of what processes should visible radiation thermally couple
> > to a gas?
> You mean like free-free, bound-free, free-bound, bound-bound emission,
> among others? The sun's photosphere is a sea of plasma and radiation
> in near-equilibrium. By the equipartition theorem the energy density
> of the gas and radiation should be equal.
You can only affect the kinetic energy of atoms/ions if you can
produce ionization (or dissociation or rotational/vibrational
excitation in case of molecules). Visible radiation can't produce
either of those.
> > ... It doesn't even in the earth's atmosphere. The latter is
> > only heated by ultraviolet light (in the upper regions, where it
> > causes ionization and dissociation) or infrared light (in the lower
> > regions, where molecules are present so that vibrational and
> > rotational modes can be excited (which then can be transferred to
> > thermal energies in the course of collisions).
> Huh? The equilibrium temperature of the earth is ~300 K, so the
> radiation is infrared, not optical.
Yes, that's what I said.
> > > I was challenging your assumption that the "edge" (solar photosphere)
> > > density was 3e23 cm^{-3}, which was not substantiated. Furthermore,
> > > your "explanation" reveals another unsubstantiated assumption, that
> > > somehow something "magical" happens at densities of 3e23 cm^{-3}.
> > > Whether or not atoms can exist, a thermal plasma is happy to exist at
> > > those densities -- and higher.
> > A thermal plasma is happy to exist at those densities, but not any
> > atomic excitation processes (which require isolated neutral (or
> > partially ionized) atoms).
> I note you again did not substantiate your claim that the photosphere
> density was 3e23 cm^{-3}. I accept your claim that "not any" atomic
> excitation processes function at high densities. However, *some* do.
> This is still irrelevant since the solar photosphere density is not
> 3e23 cm^{-3}.
I said the *threshold density* is 3*10^23 cm^-3 , so the photospheric
density must be below that in order for atomic excitation processes to
be able to take place. This threshold density defines the 'edge' of
the sun. We may strictly speaking not be seeing it because there are a
few kilometers of less dense material sitting on top of it, but
nevertheless it defines the edge (and the mass-radius data for the sun
as well as the planets confirm this).
> > There is no problem with a n(r)~1/r^2 singularity, as the total mass
> > (which is given by the integral over r^2*n(r) ) is still finite. ...
> I note this claim is unsubstantiated. In reality, even for finite
> total mass, an infinite local density would be unphysical.
The electrostatic force for a point charge also diverges like 1/r^2 at
the origin, but nevertheless it is a valid and successful theoretical
model.
Of course I am not saying that the density is infinite at the center
of the sun. The assumption of an ideal gas in hydrostatic equilibrium
will surely break down if the density reaches the nuclear density for
instance, but this should only apply within 20m or so of the sun's
centre, so it will barely make any difference for the calculation.
> > ... Only
> > density profiles n(r)~1/r^3 and stronger would be impossible in this
> > respect. And as I showed a couple of posts above, anything departing
> > substantially from a 1/r^2 density profile (i.e. from an isothermal
> > gas) would either lead indeed to an infinite mass, or alternatively
> > result in an increasing temperature with increasing r.
> *OR*, your assumption about a pure power law is invalid. There are an
> infinitude of functions which are not power laws. I guess you meant
> "virtually" a power law?
If you want, you can do a series expansion in terms of power laws. The
argument would still then apply to each term. But there is hardly a
point to do this as the mass-radius data for the sun and the planets
are consistent with a 1/r^2 density decrease (at any rate an identical
structure)).
Thomas
Thomas Smid
> Thomas Smid (thomas.s...@gmail.com) writes:
> > On 21 Jul, 11:12, af...@FreeNet.Carleton.CA (John Park) wrote:
>
> >> > I think it makes sense to describe all these planets as consisting
> >> > largely of an ideal gas, because even for Neptune, the internal
> >> > temperature (as given by the gravitational potential over the virial
> >> > theorem) is still about 15,000 K, which should make it impossible for
>
> >> Could you elaborate on that? What I've read suggest that the outer planets
> >> are mostly slushes (with rocky cores).
>
> > Yes, I have read this as well, but it is evidently wrong (an average
> > density of 1.2g/cm^3 clearly shows that hydrogen is the main
> > constituent).
>
> It's not clear to me. That's quite a bit denser than Saturn, which has
> several times the mass.
> >>> Which "virial theorem" are you
> >> referring to, the one relating kinetic and potential energies?
>
> > Yes.
>
> Under the assumption that Uranus was made of hydrogen I used the VT to
> estimate a mean temperature of 5000 K,
My calculation according to E=G*M*m/R/2 yields about 13000 K for
Uranus.
> which may or may not be consistent
> with your core temperature of 15,000 K. But clearly the surface temperature
> is much lower than that, and molecular gases--especially hddrogen--are quite
> transparent in the optical range. So why aren't these planets emitting
> viisble light?
The surface is only there in the fist place because collisional
processes cool the gas if the density is below a certain threshold. It
is cooled down by about a factor 1/1800 (which according to the
suggestion on my page http://www.plasmaphysics.org.uk/research/sun.htm
is the electron/proton mass ratio)
>
> Incidentally I also estimated a VT temperature for the Earth, assuming it
> was made of iron: 40,000 K. I wouldn't trust either estimate.
I wouldn't insert the atomic weight of iron here. The mass density of
the earth is only 5 times higher than for the hydrogen planets. So as
an effective value I would insert 5 times the hydrogen mass here. This
gives about 17,000 K.
Do you at least agree that the sun together with the hydrogen planets
in the solar system fits more or less exactly a R~M^1/3 mass-radius
relationship, and that this is based on accurate and unambiguous
figures?
> And what about
> interferometric or other direct determinations of stellar radii?
There are no 'direct' determinations of stellar radii in the sense of
the word. They are all quite complicated and convoluted procedures
that make a large number of assumptions both with regard to the stars
as well as the measuring instruments and the intervening medium.
>
> >> >>>> Put it another way: if Uranus and Neptune had the same masses but were 75%
> >> >> hydrogen instead of 15% and had orbits inside Venus', would you still expect
> >> >> them to fit your curve?
>
> >> > The fact that they fit the curve suggests that they *do* consist
> >> > largely of hydrogen.
>
> >> You haven't got nearly enough data to claim that.
>
> > You can't really blame me that there not more planets in the solar
> > system (which is to date the only place where we can measure masses
> > and radii of objects accurately), but the point is that all objects
> > present obey a R~M^1/3 relationship.
>
> I can blame you for leaping to an extravagant conclusion based on inadequate
> data.
I am not making any conclusions. I am just stating that the sun fits
with the hydrogen planets into a R~M^1/3 relationship.
>
> >> >By the way, the smaller planets also follow a R~M^1/3 law for
> >> >themselves, but have a consistently smaller radius by about a factor
> >> >0.6 (Mars 0.7) (this agrees with their higher average density compared
> >> >to the sun and giant planets).
>
> >> So the 1/3 power law applies to rocky bodies as well as ideal gases? How
> >> do you rationalise that?
>
> > Even for the smaller planets, the temperature in the interior is still
> > a couple of thousand degrees. Apparently, this is still sufficient to
> > make the material compressible like an ideal gas (which will thus also
> > result in a 1/r^2 density increase towards the center).
>
> So how compressible is iron at 2000 K? And what's your evidence for the
> temperatures?
The melting point of iron is about 1800K and the boiling point 3100K,
so at least for the latter value one should expect that all lattice
and other molecular structures in the material become destroyed. At
this point the matter should simply become a plasma of ions and
electrons and should behave like an ideal gas.
Direct measurements of the internal temperature are obviously not
possible. The figures one can find are merely estimates from model
calculations.
Thomas
John Park
> On 21 Jul, 21:37, af...@FreeNet.Carleton.CA (John Park) wrote:
>> Thomas Smid (thomas.s...@gmail.com) writes:
>> > On 21 Jul, 11:12, af...@FreeNet.Carleton.CA (John Park) wrote:
>>
>> >> > I think it makes sense to describe all these planets as consisting
>> >> > largely of an ideal gas, because even for Neptune, the internal
>> >> > temperature (as given by the gravitational potential over the virial
>> >> > theorem) is still about 15,000 K, which should make it impossible for
>>
>> >> Could you elaborate on that? What I've read suggest that the outer planets
>> >> are mostly slushes (with rocky cores).
>>
>> > Yes, I have read this as well, but it is evidently wrong (an average
>> > density of 1.2g/cm^3 clearly shows that hydrogen is the main
>> > constituent).
The actual values appear to be 1.3 and 1.6 g/cc--Heptune is more than
twice as dense as Saturn, and more than three times as dense as
liquid methane. You haven't addressed this point.
[...]
[re Virial Theorem:]
>>
>> Under the assumption that Uranus was made of hydrogen I used the VT to
>> estimate a mean temperature of 5000 K,
>
> My calculation according to E=G*M*m/R/2 yields about 13000 K for
> Uranus.
I may check that and get back to you later. (Incidentally I don't believe
that's quite the self-gravitational energy of a sphere. Integrating
spherical shells gives me a factor of 0.6 instead of 0.5.) In any case
applying the VT here is circular--it essentially assumes approximately
ideal gas behaviour.
>> which may or may not be consistent
>> with your core temperature of 15,000 K. But clearly the surface temperature
>> is much lower than that, and molecular gases--especially hddrogen--are quite
>> transparent in the optical range. So why aren't these planets emitting
>> viisble light?
>
> The surface is only there in the fist place because collisional
> processes cool the gas if the density is below a certain threshold. It
> is cooled down by about a factor 1/1800 (which according to the
> suggestion on my page http://www.plasmaphysics.org.uk/research/sun.htm
> is the electron/proton mass ratio)
How do collisional processes cool anything?
>>
>> Incidentally I also estimated a VT temperature for the Earth, assuming it
>> was made of iron: 40,000 K. I wouldn't trust either estimate.
>
> I wouldn't insert the atomic weight of iron here. The mass density of
> the earth is only 5 times higher than for the hydrogen planets. So as
> an effective value I would insert 5 times the hydrogen mass here. This
> gives about 17,000 K.
Are you saying you don't believe the core of the Earth is iron? (And see
comment at end.)
[...]
Incidentally I find it a bit easier to think about your model as implying
that density is constant.
>>
>> > Again, the mass-radius relationship R~M^1/3 for solar system objects
>> > (including the sun) stands as an observational fact, independently of
>> > any physical theory. If a theory implies a different relationship,
>> > then it is inconsistent with these observations. Unless you have an
>> > explanation for this inconsistency, it invalidates thus the theory.
>> > And given the formula I mentioned above R=[M(R)*(3-k)/4pi/n(R)/
>> > m]^1/3 , it is apparent that the radius *does* indeed also depend on
>> > the composition (as it depends on the atomic mass m) ,and potentially
>> > the structure (through the index k for the radial density
>> > distribution), but the data clearly indicate that the density
>> > structure is identical for all the objects (including the sun).
>>
>> You haven't enough points to exclude other possibilities.
>>
[...]
>>> The determination
>> > of masses and radii for binary stars depends on a host of assumptions
>> > with regard to distance, orbit, limb darkening etc. These assumptions
>> > are likely to be biased such that they conform with standard theories
>> > regarding stellar structure, so they are by no means direct and
>> > independent measurements (like they are possible for solar system
>> > objects).
>>
>> I'm not an expert, but frankly I don't believe you.
>
> Do you at least agree that the sun together with the hydrogen planets
> in the solar system fits more or less exactly a R~M^1/3 mass-radius
> relationship, and that this is based on accurate and unambiguous
> figures?
>
I believe there are two "hydrogen planets" in the solar system. One has
about the same density as the sun; one does not.
>> And what about
>> interferometric or other direct determinations of stellar radii?
>
> There are no 'direct' determinations of stellar radii in the sense of
> the word. They are all quite complicated and convoluted procedures
> that make a large number of assumptions both with regard to the stars
> as well as the measuring instruments and the intervening medium.
Again I'm sceptical.
>
[...]
>> I can blame you for leaping to an extravagant conclusion based on inadequate
>> data.
>
> I am not making any conclusions. I am just stating that the sun fits
> with the hydrogen planets into a R~M^1/3 relationship.
I thought you were inferring that that relationship applied to all bodies
made of hydrogen.
>>
>> >> >By the way, the smaller planets also follow a R~M^1/3 law for
>> >> >themselves, but have a consistently smaller radius by about a factor
>> >> >0.6 (Mars 0.7) (this agrees with their higher average density compared
>> >> >to the sun and giant planets).
The inner three do have about the same density; Mars is off the curve.
[...]>
>> > Even for the smaller planets, the temperature in the interior is still
>> > a couple of thousand degrees. Apparently, this is still sufficient to
>> > make the material compressible like an ideal gas (which will thus also
>> > result in a 1/r^2 density increase towards the center).
>>
>> So how compressible is iron at 2000 K? And what's your evidence for the
>> temperatures?
>
> The melting point of iron is about 1800K and the boiling point 3100K,
> so at least for the latter value one should expect that all lattice
> and other molecular structures in the material become destroyed. At
> this point the matter should simply become a plasma of ions and
> electrons and should behave like an ideal gas.
>
> Direct measurements of the internal temperature are obviously not
> possible. The figures one can find are merely estimates from model
> calculations.
OK. I'm having to re-educate myself a bit. But how do you deal with the fact
that seismic data indicate that only 15% of the Earth's volume (maybe
30% of the mass) is fluid--presumably liquid not gaseous--and the innermost
core is _solid_?
--John Park
Steve Willner
Thomas <thoma...@gmail.com> writes:
> [stellar mass-radius relation]
> on inferred ones on the basis of the Stefan-Boltzmann equation and an
> empirical mass-luminosity relationship.
If you are suggesting Stefan-Boltzmann is wrong, "new physics"
doesn't even to begin to describe what you are doing. Stellar radii
are based on models, interferometric measurements, and eclipsing
binaries.
Your later suggestion that all these data are somehow "biased" is
nothing more than ignoring evidence.
> If one takes the masses of our sun and the hydrogen planets in our
> solar system (where the masses and radii are known accurately) then
> this fits indeed indeed very well a R~M^1/3 law: the theoretical
> radii for Jupiter, Saturn, Uranus, Neptune are then (Sun=1) : 0.099,
> 0.066, 0.035, 0.037 . The actual values are 0.1, 0.085, 0.037, 0.035.
Your power law is equivalent to saying density is constant.
Reference to any standard table will show that is false. Densities
range from Saturn's 0.7 to Neptune's 1.76. The Sun's 1.41 is within
the range, but that's about all that can be said for your "law."
> An exponent 0.9 on the other hand would result in : 0.002, 0.00064,
> 0.00012, 0.00014.
The _stellar_ mass-radius relation cannot be used to predict
_planetary_ radii. The equation of state for planetary densities and
temperatures is not given by the ideal gas law.
Thomas
> > If one takes the masses of our sun and the hydrogen planets in our
> > solar system (where the masses and radii are known accurately) then
> > this fits indeed indeed very well a R~M^1/3 law: the theoretical
> > radii for Jupiter, Saturn, Uranus, Neptune are then (Sun=1) : 0.099,
> > 0.066, 0.035, 0.037 . The actual values are 0.1, 0.085, 0.037, 0.035.
>
> Your power law is equivalent to saying density is constant.
> Reference to any standard table will show that is false. Densities
> range from Saturn's 0.7 to Neptune's 1.76. The Sun's 1.41 is within
> the range, but that's about all that can be said for your "law."
That's a strange argument as the average density is exactly derived
from the mass and radius. Fact is that an exponent 1/3 in the mass-
radius relationship fits the major solar system bodies (including the
sun) but not an exponent 0.9 or 0.6.
> > An exponent 0.9 on the other hand would result in : 0.002, 0.00064,
> > 0.00012, 0.00014.
>
> The _stellar_ mass-radius relation cannot be used to predict
> _planetary_ radii.
Could it be that this is because it is incorrect and the correct
stellar mass-radius relation is R ~M^1/3?
Thomas
Thomas Smid
> Thomas Smid (thomas.s...@gmail.com) writes:
> >> > Yes, I have read this as well, but it is evidently wrong (an average
> >> > density of 1.2g/cm^3 clearly shows that hydrogen is the main
> >> > constituent).
>
> The actual values appear to be 1.3 and 1.6 g/cc--Heptune is more than
> twice as dense as Saturn, and more than three times as dense as
> liquid methane. You haven't addressed this point.
One should be careful with using notions like 'liquid' or 'solids' for
those temperatures (10,000K+). I doubt that any of the molecular bonds
making up the liquid will survive under those conditions. What you
should have then is just a plasma of nuclei and electrons, and usual
density data will simply not be applicable here.
> >> Under the assumption that Uranus was made of hydrogen I used the VT to
> >> estimate a mean temperature of 5000 K,
>
> > My calculation according to E=G*M*m/R/2 yields about 13000 K for
> > Uranus.
>
> I may check that and get back to you later. (Incidentally I don't believe
> that's quite the self-gravitational energy of a sphere. Integrating
> spherical shells gives me a factor of 0.6 instead of 0.5.)
The factor 0.5 comes from the virial theorem which says that the
average kinetic energy is 1/2 of the average potential energy.
Also, you shouldn't forget the 1/r^2 density structure. The latter
could explain the discrepancy between your and my calculation.
> In any case
> applying the VT here is circular--it essentially assumes approximately
> ideal gas behaviour.
The Virial theorem holds for any bound system in a state of
equilibrium (or something that is close to an equilibrium).
>
> >> which may or may not be consistent
> >> with your core temperature of 15,000 K. But clearly the surface temperature
> >> is much lower than that, and molecular gases--especially hddrogen--are quite
> >> transparent in the optical range. So why aren't these planets emitting
> >> viisble light?
>
> > The surface is only there in the fist place because collisional
> > processes cool the gas if the density is below a certain threshold. It
> > is cooled down by about a factor 1/1800 (which according to the
> > suggestion on my pagehttp://www.plasmaphysics.org.uk/research/sun.htm
> > is the electron/proton mass ratio)
>
> How do collisional processes cool anything?
Yes, sorry, I wanted to say 'collisional excitation processes'. These
obviously cool the gas as particles lose energy by exciting atomic
transitions (which then results in radiation).
> >> Incidentally I also estimated a VT temperature for the Earth, assuming it
> >> was made of iron: 40,000 K. I wouldn't trust either estimate.
>
> > I wouldn't insert the atomic weight of iron here. The mass density of
> > the earth is only 5 times higher than for the hydrogen planets. So as
> > an effective value I would insert 5 times the hydrogen mass here. This
> > gives about 17,000 K.
>
> Are you saying you don't believe the core of the Earth is iron? (And see
> comment at end.)
> [...]
> Incidentally I find it a bit easier to think about your model as implying
> that density is constant.
As indicated above already, at these temperatures it isn't really
appropriate to think in terms of usual materials. At least as far as
the mass-radius relation is concerned, it appears to better (or at
least sufficient) to use some effective average atomic weight here (as
given by the average density).
But anyhow, I don't want to push the case too far here by including
the denser (and less massive) planets as well in my ideal gas model.
It would be difficult to justify here without getting into details of
the physics of the corresponding materials for these conditions. It is
still remarkable though that even here a R~M^1/3 relationship roughly
applies.
> >> > Even for the smaller planets, the temperature in the interior is still
> >> > a couple of thousand degrees. Apparently, this is still sufficient to
> >> > make the material compressible like an ideal gas (which will thus also
> >> > result in a 1/r^2 density increase towards the center).
>
> >> So how compressible is iron at 2000 K? And what's your evidence for the
> >> temperatures?
>
> > The melting point of iron is about 1800K and the boiling point 3100K,
> > so at least for the latter value one should expect that all lattice
> > and other molecular structures in the material become destroyed. At
> > this point the matter should simply become a plasma of ions and
> > electrons and should behave like an ideal gas.
>
> > Direct measurements of the internal temperature are obviously not
> > possible. The figures one can find are merely estimates from model
> > calculations.
>
> OK. I'm having to re-educate myself a bit. But how do you deal with the fact
> that seismic data indicate that only 15% of the Earth's volume (maybe
> 30% of the mass) is fluid--presumably liquid not gaseous--and the innermost
> core is _solid_?
Again, I can only say what I said above already: one should be careful
with using notions like 'liquid' or 'solids' for those temperatures
and densities. The behaviour of matter under these conditions isn't
really known, so those data are difficult to interpret.
The point is that apparently still a R~M^1/3 relationship roughly
applies even for heavier elements.
Thomas
John Park
> On 23 Jul, 15:08, af...@FreeNet.Carleton.CA (John Park) wrote:
>> Thomas Smid (thomas.s...@gmail.com) writes:
>
>> >> > Yes, I have read this as well, but it is evidently wrong (an average
>> >> > density of 1.2g/cm^3 clearly shows that hydrogen is the main
>> >> > constituent).
>>
>> The actual values appear to be 1.3 and 1.6 g/cc--Heptune is more than
>> twice as dense as Saturn, and more than three times as dense as
>> liquid methane. You haven't addressed this point.
>
> One should be careful with using notions like 'liquid' or 'solids' for
> those temperatures (10,000K+). I doubt that any of the molecular bonds
> making up the liquid will survive under those conditions. What you
> should have then is just a plasma of nuclei and electrons, and usual
> density data will simply not be applicable here.
The point is that if Saturn is a "hydrogen planet" it's easy to explain
Neptune's density as reflecting a different composition. If Neptune were a
"hydrogen planet" how could you explain Saturn's density?
>
>> >> Under the assumption that Uranus was made of hydrogen I used the VT to
>> >> estimate a mean temperature of 5000 K,
>>
>> > My calculation according to E=G*M*m/R/2 yields about 13000 K for
>> > Uranus.
>>
>> I may check that and get back to you later. (Incidentally I don't believe
>> that's quite the self-gravitational energy of a sphere. Integrating
>> spherical shells gives me a factor of 0.6 instead of 0.5.)
>
> The factor 0.5 comes from the virial theorem which says that the
> average kinetic energy is 1/2 of the average potential energy.
This is true for a 1/r potential which implies esentially an ideal gas. My
calculation gives V = k*G*m^2/R for the gravitational self-energy of a
sphere. If the density is constant k = 0.6; if the density goes as r^-2,
k = 0.3. The kinetic energy would be half these values (again for ideal
gases). These factors start to affect your temperatures significantly even
within your assumptions.
> Also, you shouldn't forget the 1/r^2 density structure. The latter
> could explain the discrepancy between your and my calculation.
>
>> In any case
>> applying the VT here is circular--it essentially assumes approximately
>> ideal gas behaviour.
>
> The Virial theorem holds for any bound system in a state of
> equilibrium (or something that is close to an equilibrium).
>
>>
>> >> which may or may not be consistent
>> >> with your core temperature of 15,000 K. But clearly the surface temperature
>> >> is much lower than that, and molecular gases--especially hddrogen--are quite
>> >> transparent in the optical range. So why aren't these planets emitting
>> >> viisble light?
>>
>> > The surface is only there in the fist place because collisional
>> > processes cool the gas if the density is below a certain threshold. It
>> > is cooled down by about a factor 1/1800 (which according to the
>> > suggestion on my pagehttp://www.plasmaphysics.org.uk/research/sun.htm
>> > is the electron/proton mass ratio)
>>
>> How do collisional processes cool anything?
>
> Yes, sorry, I wanted to say 'collisional excitation processes'. These
> obviously cool the gas as particles lose energy by exciting atomic
> transitions (which then results in radiation).
So where's the radiation?
Particularly if you're expecting ideal gas behaviour.
--John Park
John Park
I wrote:
>
> This is true for a 1/r potential which implies esentially an ideal gas. My
> calculation gives V = k*G*m^2/R for the gravitational self-energy of a
> sphere. If the density is constant k = 0.6; if the density goes as r^-2,
> k = 0.3.
>
correct.
But: In that case the mass is: integral 4*pi*rho*(r^2)*dr ~ integral 4*pi*dr
= 4*pi*r. So where is your m ~ r^3 relationship?
--John Park
Thomas Smid
> Thomas Smid (thomas.s...@gmail.com) writes:
> > On 23 Jul, 15:08, af...@FreeNet.Carleton.CA (John Park) wrote:
> >> Thomas Smid (thomas.s...@gmail.com) writes:
>
> >> >> > Yes, I have read this as well, but it is evidently wrong (an average
> >> >> > density of 1.2g/cm^3 clearly shows that hydrogen is the main
> >> >> > constituent).
>
> >> The actual values appear to be 1.3 and 1.6 g/cc--Heptune is more than
> >> twice as dense as Saturn, and more than three times as dense as
> >> liquid methane. You haven't addressed this point.
>
> > One should be careful with using notions like 'liquid' or 'solids' for
> > those temperatures (10,000K+). I doubt that any of the molecular bonds
> > making up the liquid will survive under those conditions. What you
> > should have then is just a plasma of nuclei and electrons, and usual
> > density data will simply not be applicable here.
>
> The point is that if Saturn is a "hydrogen planet" it's easy to explain
> Neptune's density as reflecting a different composition. If Neptune were a
> "hydrogen planet" how could you explain Saturn's density?
I don't know what causes Saturn to make an exception here. With
hydrogen being already the lightest element, it shouldn't really be
less dense. The only potential causes I can think of at the moment is
that either Saturn has an unusually thick atmosphere (which results in
the apparent radius being significantly larger than R (as used in the
equation above)), or that an internal magnetic field causes a
deviation from the simple gravity/pressure equilibrium.
> >> > The surface is only there in the fist place because collisional
> >> > processes cool the gas if the density is below a certain threshold. It
> >> > is cooled down by about a factor 1/1800 (which according to the
> >> > suggestion on my pagehttp://www.plasmaphysics.org.uk/research/sun.htm
> >> > is the electron/proton mass ratio)
>
> >> How do collisional processes cool anything?
>
> > Yes, sorry, I wanted to say 'collisional excitation processes'. These
> > obviously cool the gas as particles lose energy by exciting atomic
> > transitions (which then results in radiation).
>
> So where's the radiation?
Well, in case of the Sun it is obvious, Jupiter is also known to emit
radiation on its own, and for the other planets it may be just not
detectable due to its weakness.
> >> OK. I'm having to re-educate myself a bit. But how do you deal with the fact
> >> that seismic data indicate that only 15% of the Earth's volume (maybe
> >> 30% of the mass) is fluid--presumably liquid not gaseous--and the innermost
> >> core is _solid_?
>
> > Again, I can only say what I said above already: one should be careful
> > with using notions like 'liquid' or 'solids' for those temperatures
> > and densities. The behaviour of matter under these conditions isn't
> > really known, so those data are difficult to interpret.
>
> Particularly if you're expecting ideal gas behaviour.
At sufficiently high temperatures all substances should behave like an
ideal gas because all molecular and atomic bonds will be destroyed,
i.e. there will just be a high density plasma of individual nucleons
and electrons elastically colliding with each other. 10,000K or so may
probably not be enough to fully achieve this for iron, but it might
already partially behave like an ideal gas.
> I wrote:
>
> > This is true for a 1/r potential which implies esentially an ideal gas. My
> > calculation gives V = k*G*m^2/R for the gravitational self-energy of a
> > sphere. If the density is constant k = 0.6; if the density goes as r^-2,
> > k = 0.3.
>
> This was a mistake. If density goes as 1/r^2 your result for the energy seems
> correct.
>
> But: In that case the mass is: integral 4*pi*rho*(r^2)*dr ~ integral 4*pi*dr
> = 4*pi*r. So where is your m ~ r^3 relationship?
Your linear relationship describes the partial (relative) mass, not
the total mass. The latter is in general (for a density distribution
n(r)~r^-k) given by
M(R) =n(R)*m*4pi* Int[dr*r^2*(R/r)^k](0...R) =
=n(R)*m*4pi/(3-k)*R^3 ,
where n(R) is the number density at the edge R and m the atom mass.
Thomas
John Park
> On 25 Jul, 19:14, af...@FreeNet.Carleton.CA (John Park) wrote:
>>
>> The point is that if Saturn is a "hydrogen planet" it's easy to explain
>> Neptune's density as reflecting a different composition. If Neptune were a
>> "hydrogen planet" how could you explain Saturn's density?
>
> I don't know what causes Saturn to make an exception here. With
> hydrogen being already the lightest element, it shouldn't really be
> less dense. The only potential causes I can think of at the moment is
> that either Saturn has an unusually thick atmosphere (which results in
> the apparent radius being significantly larger than R (as used in the
> equation above)), or that an internal magnetic field causes a
> deviation from the simple gravity/pressure equilibrium.
I suggest it's much easier to believe that Saturn isn't an exception, but
the two outer plenets don't have the compositions you want them to have.
>> > Yes, sorry, I wanted to say 'collisional excitation processes'. These
>> > obviously cool the gas as particles lose energy by exciting atomic
>> > transitions (which then results in radiation).
>>
>> So where's the radiation?
>
> Well, in case of the Sun it is obvious, Jupiter is also known to emit
> radiation on its own, and for the other planets it may be just not
> detectable due to its weakness.
I find it hard to believe that a surface temperature of 60 K is compatible
with a hydrogen atmosphere and an average temperature of 13,000 K. But I'm
not ready to argue the point quantitatively.
>
> At sufficiently high temperatures all substances should behave like an
> ideal gas because all molecular and atomic bonds will be destroyed,
> i.e. there will just be a high density plasma of individual nucleons
> and electrons elastically colliding with each other. 10,000K or so may
> probably not be enough to fully achieve this for iron, but it might
> already partially behave like an ideal gas.
Might, partially ... ideal gases don't transmit transverse waves. And
solids and liquids don't obey the simple (V ~ 1/r) virial theorem on
which your temperatures are based.
>> I wrote:
>>
>> > This is true for a 1/r potential which implies esentially an ideal gas. My
>> > calculation gives V = k*G*m^2/R for the gravitational self-energy of a
>> > sphere. If the density is constant k = 0.6; if the density goes as r^-2,
>> > k = 0.3.
>>
>> This was a mistake. If density goes as 1/r^2 your result for the energy seems
>> correct.
>>
>> But: In that case the mass is: integral 4*pi*rho*(r^2)*dr ~ integral 4*pi*dr
>> = 4*pi*r. So where is your m ~ r^3 relationship?
>
> Your linear relationship describes the partial (relative) mass, not
> the total mass. The latter is in general (for a density distribution
> n(r)~r^-k) given by
>
> M(R) =n(R)*m*4pi* Int[dr*r^2*(R/r)^k](0...R) =
>
> =n(R)*m*4pi/(3-k)*R^3 ,
>
> where n(R) is the number density at the edge R and m the atom mass.
>
It then follows that for k = 2 the mean density is 3*n(R); in other
words the surface densities of the Sun and Jupiter would be over 0.4 g/cc.
I'm sure this must be in violent disagreement with observation--it
certainly contradicts solar models. (For a sun made of hydrogen it
predicts a surface pressure of the order of 100,000 atmospheres.)
--John Park
Thomas Smid
> Thomas Smid (thomas.s...@gmail.com) writes:
> > On 25 Jul, 19:14, af...@FreeNet.Carleton.CA (John Park) wrote:
>
> >> The point is that if Saturn is a "hydrogen planet" it's easy to explain
> >> Neptune's density as reflecting a different composition. If Neptune were a
> >> "hydrogen planet" how could you explain Saturn's density?
>
> > I don't know what causes Saturn to make an exception here. With
> > hydrogen being already the lightest element, it shouldn't really be
> > less dense. The only potential causes I can think of at the moment is
> > that either Saturn has an unusually thick atmosphere (which results in
> > the apparent radius being significantly larger than R (as used in the
> > equation above)), or that an internal magnetic field causes a
> > deviation from the simple gravity/pressure equilibrium.
>
> I suggest it's much easier to believe that Saturn isn't an exception, but
> the two outer plenets don't have the compositions you want them to have.
But Saturn is the only one of the big planets (and the sun) that does
not fit the R~M^1/3 law. So this suggests that there is something
special going on here.
>
> >> > Yes, sorry, I wanted to say 'collisional excitation processes'. These
> >> > obviously cool the gas as particles lose energy by exciting atomic
> >> > transitions (which then results in radiation).
>
> >> So where's the radiation?
>
> > Well, in case of the Sun it is obvious, Jupiter is also known to emit
> > radiation on its own, and for the other planets it may be just not
> > detectable due to its weakness.
>
> I find it hard to believe that a surface temperature of 60 K is compatible
> with a hydrogen atmosphere and an average temperature of 13,000 K. But I'm
> not ready to argue the point quantitatively.
First of all, I have to correct myself here: for Saturn, Uranus and
Neptune there is apparently a substantial excess heat observed as
well.
If you apply the method I mentioned before for getting the surface
temperature by dividing the internal temperature (as obtained over the
Virial theorem) by 1800 (the proton/electron mass ratio) in the case
of a hydrogen mass (which works well for stars), then one would obtain
from 13,000K a surface temperature of a mere 7K. But this obviously
neglects any solar heating, which is difficult to assess here as it
will depend on any cooling processes in the atmosphere which would
crucially depend on atmospheric composition
> >> But: In that case the mass is: integral 4*pi*rho*(r^2)*dr ~ integral 4*pi*dr
> >> = 4*pi*r. So where is your m ~ r^3 relationship?
>
> > Your linear relationship describes the partial (relative) mass, not
> > the total mass. The latter is in general (for a density distribution
> > n(r)~r^-k) given by
>
> > M(R) =n(R)*m*4pi* Int[dr*r^2*(R/r)^k](0...R) =
>
> > =n(R)*m*4pi/(3-k)*R^3 ,
>
> > where n(R) is the number density at the edge R and m the atom mass.
>
> It then follows that for k = 2 the mean density is 3*n(R); in other
> words the surface densities of the Sun and Jupiter would be over 0.4 g/cc.
> I'm sure this must be in violent disagreement with observation--it
> certainly contradicts solar models. (For a sun made of hydrogen it
> predicts a surface pressure of the order of 100,000 atmospheres.)
Yes, that would correspond for hydrogen to a number density of about
3*10^23 cm^-3, which is the density where the distance between two
atoms in a gas is equal to the size of an atom, i.e. it is the
threshold density above which no individual separated atoms can exist
anymore (and thus no collisional excitation of atoms and the related
cooling). This basically defines the base of the photosphere, which
however itself is not observable because the (less dense) photospheric
layer (where cooling and the production of radiation takes place) is
sitting on top of it.
Thomas
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
I see. So your presumption was not based on how the observations of
the stellar properties were actually performed, but rather your own
erroneous and usubstantiated speculation.
Since the solar system "data" refers to only one star -- the sun --
they are totally irrelevant to the question of how the masses and
radii were measured for large stellar populations.
CM
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
> On 19 Jul, 22:40, Craig Markwardt
> <craigm...@REMOVEcow.physics.wisc.edu> wrote:
> > Thomas <thomas.s...@gmail.com> writes:
>
> > > The troposphere *is* virtually isothermal. Its temperature decreases
> > > only by about 10% from the surface value of 300K. And this even though
> > > the heating is essentially only from the earth's surface. ...
> >
> > Ah, *virtually* isothermal. A key retreat word. *Virtually*
> > isothermal means not "necessarily isothermal" as you described above.
> > But let's look at how virtually isothermal the earth's troposphere is.
> > Consider the International Standard Atmosphere; the difference in
> > temperatures between the base of the troposphere and the tropopause is
> > about 71 C [*], which is more like 25%, not 10%. I suppose that's
> > "virtually" the same as 10% for you?#
>
> Yes, it is virtually the same because I said *about* 10% (meaning of
> the order of 10%). But even a 25% percent temperature drop still means
> that cooling processes and the related loss of energy at the top of
> the atmosphere are not overly important here, and that equipartition
> of energy by means of elastic collisions dominates.
Ahh, but the part of the discussion which you convenient deleted, was
the part where you said that the temperature of a gas will
"necessarily" be isothermal. Now you are allowing that there are ways
that it might not necessarily be isothermal, maybe just "virtually"
isothermal. I'm sure it's quite convenient for you to have it both
ways.
> > But let's consider this a little more. Might it be possible that if
> > the troposphere were twice as thick, the temperature would decrease to
> > (.75)x(.75) = 56% of the base? [*] Or (.75)^3 = 42% for a triply
> > thick troposphere? The earth's troposphere is ~11 km thick, or about
> > 0.17% of the earth's radius. Compare this to the convective layer of
> > the sun, which is about 5% thick. That's (5% / 0.17%) = 30 scale
> > heights, so in principle the photosphere temperature could decrease to
> > (0.75)^(30) = 1/5600th of the base temp. Huh! Of *COURSE* I don't
> > think this is how the solar atmosphere actually works, but I'm
> > pointing out how your "virtually isothermal" double standard.
>
>
> But the comparison of the sun with the troposphere is misleading here
> anyway: as indicated already earlier, the troposphere consists of
> individual molecules that can absorb and re-emit infrared radiation. ...
While I wholeheartedly agree that a *direct* comparison between the
sun and troposphere is inappropriate, my point was that there are
mechanisms in gases which can preserve a temperature gradient. And
since the "atmosphere" of the sun is much deeper, non-trivial
temperature gradients might develop.
> ... No such coupling between matter and radiation could exist in the solar
> interior, because a) no individual molecules and atoms can exist there
> due to the high density (and temperature),
This point remains unsubstantiated, because (i) your argument about
"individual molecules and atoms" is itself unsubstantiated by any
theory or measurement, (ii) in the solar *interior* the radiation may
not be dominated by infrared radiation as you mentioned, (iii) there
are energy-matter couplings which do not require neutral atoms. I
note that you could have investigated these processes but did not.
> ... and b) visible radiation
> (which constitutes the bulk of the solar energy output) can not be
> absorbed by atoms anyway.
Again, this claim is unsubstantiated. In fact there are many
bound-bound hydrogen and helium transitions in the optical band. I
note that you could have reviewed such transitions but did not.
> Furthermore, the sun is, in contrast to the troposphere, a self-
> gravitating system that can gain energy from gravitational
> contraction: if you assume a self-gravitating gas ball in hydrostatic
> equilibrium, and somehow suddenly remove the kinetic energy of the
> particles in some sub-volume, then this sub-volume will begin to
> collapse at a free-fall rate until it has gained enough energy so that
> a new equilibrium is established. So if you have a region where
> cooling takes place (like in the solar photosphere), the information
> about this cooling can never propagate to the sun's interior as
> hydrostatic equilibrium is locally established much faster than the
> heat can travel. So the effects of the cooling will only be restricted
> to the photosphere itself, but not extend to regions below (where we
> have the original 'virial' temperature).
It's ironic that you've been claiming that a gas must "necessarily" be
isothermal, *except* for your own pet cooling process. That's silly.
In fact, radiation, convection and conduction processes *can*
communicate the outer layer temperature towards the center of a star,
although the transport may not be that efficient.
> > According to your claim, a gas which is "collisionally dominated" must
> > *necessarily* be isothermal, but you also claim that it need only be
> > "virtually" isothermal. I'm sure it's quite convenient for you to
> > define any standard you wish. The fact is, if you stack enough thin
> > layers of atmosphere, this can have a significant effect on the
> > radiation transport.
>
> What I said was that a collisionally dominated gas (in a state of
> equilibrium) must be necessarily isothermal *in the absence of any
> heat sources or sinks*. And in that case it will be *strictly*
> isothermal. Of course this is an ideal condition which in practice
> does not strictly hold, but according to my argument above, it should
> hold quite closely for the interior of the sun, as the effect of
> gravity will level out any temperature gradients potentially arising
> from the cooling in the photosphere.
So you are now shifting your claim to say that only the "interior" of
the sun must be isothermal? Incredible. I'm sure it continues to be
quite convenient for you to redefine your claims whenever expedient.
In any case your point is irrelevant since there is a *huge heat
source* near the center of the sun.
> > Ah, but then you also allow that the troposphere is heated from below
> > by the earth's surface, which creates a temperature gradient. Did you
> > ever stop to think that the *sun* might be heated from below???
>
> No, not really. In contrast to the troposphere, the sun doesn't need a
> separate heating to be stable as it is a self-gravitating system and
> thus produces the required energy itself.
So you never even stopped to consider the possibility that the sun
could be heated from the center? Again, incredible. This issue was
debated and settled in the late 1800s and early 1900s. You are in the
wrong century.
> Also, the fact that the sun obeys the same R~M^1/3 mass-radius
> relationship as the giant planets in the solar system, indicates that
> there is no difference in their density (and thus temperature-)
> structure.
The sun and the gas giant planets are obviously quite different, as
evidenced by their surface properties and compositions.
Also, correlation is not necessarily evidence of causation.
[ More in a reply to a separate post. ]
> > > You still seem to be missing the point: it is not about the
> > > consistency (or not) between the luminosity of the sun and the surface
> > > temperature (according to the Stefan-Boltzmann equation), but about
> > > the question why the 'surface' temperature of the sun is only 1/1800
> > > of the temperature that a gas ball with the mass of the sun should
> > > have in hydrostatic equilibrium.
> >
> > But my point still stands. For sun's radius and energy output rate
> > today, how could the surface temperature be anything other than
> > ~6000K? It must be that temperature, because it's basic conservation
> > of energy (i.e. the balance between energy generation rate with
> > Thermal radiation). You are right, it's interesting to know the
> > surface temperature is so much lower than the core temperature... it
> > tells us how heat flows through the stellar interior.
>
> I quote from http://www.ap.smu.ca/~guenther/Level01/solar/what_is_ssm.html
> :
>
> "Nearly all stellar evolutionary calculations are calibrated with
> respect to the standard solar model. The standard solar model is
> derived from the conservation laws and energy transport equations of
> physics, applied to a spherically symmetric gas (plasma) sphere and
> constrained by the luminosity, radius, age and composition of the
> Sun".
>
> It should be obvious that if you apply physical parameters as
> constraints to a model (like the luminosity (i.e. the surface
> temperature)), then the model can not give an explanation for the
> parameters, because they serve on the contrary as input values. ...
I note that this is not actually a relevant reply to the original
point, which was, "[H]ow could the surface temperature be anything
other than ~6000K? It must be that temperature, because it's basic
conservation of energy." There are no stellar "models" or
"explaining" required to apply the Stefan Boltzmann law and
conservation of energy, to the surface of an optically thick emitting
sphere. Doing that simple mathematical exercise, one finds that the
sun's surface *must* be ~6000 K. And the brightness temperature of
the sun is indeed ~6000 K.
Furthermore, your quotation refers to how *stellar* models
(i.e. models of other stars) are compared to the sun, and thus it is
inapplicable to your claims about *our sun* by itself.
> As I said already repeatedly, the temperature of the particles in the
> sun can only be lowered below the 'virial' temperature if inelastic
> collisions are present. The standard solar model can't address this
> point in any way, and therefore it is not a valid physical model. On
> the other hand, I only need the mass here to *derive* both the radius
> and the surface temperature (the latter being 1/1800 of the 'virial
> temperature').
Regardless of what you have said repeatedly, your claims are simply
irrelevant. There are many energy transport processes important in
the sun, not just the one you are obsessed with. I note you could
have investigated these but did not.
> > > By means of what processes should visible radiation thermally couple
> > > to a gas?
> >
> > You mean like free-free, bound-free, free-bound, bound-bound emission,
> > among others? The sun's photosphere is a sea of plasma and radiation
> > in near-equilibrium. By the equipartition theorem the energy density
> > of the gas and radiation should be equal.
>
> You can only affect the kinetic energy of atoms/ions if you can
> produce ionization (or dissociation or rotational/vibrational
> excitation in case of molecules). Visible radiation can't produce
> either of those.
Your claim is erroneous. See bound-bound emission. I note that you
could have educated yourself on emission processes but did not.
> > > ... It doesn't even in the earth's atmosphere. The latter is
> > > only heated by ultraviolet light (in the upper regions, where it
> > > causes ionization and dissociation) or infrared light (in the lower
> > > regions, where molecules are present so that vibrational and
> > > rotational modes can be excited (which then can be transferred to
> > > thermal energies in the course of collisions).
> >
> > Huh? The equilibrium temperature of the earth is ~300 K, so the
> > radiation is infrared, not optical.
>
> Yes, that's what I said.
My original claim: "the radiation field is coupled to the gas." Now
you understand that this means infrared radiation in the case of the
earth? [ and optical in the case of the sun's photosphere. ]
> > > > I was challenging your assumption that the "edge" (solar photosphere)
> > > > density was 3e23 cm^{-3}, which was not substantiated. Furthermore,
> > > > your "explanation" reveals another unsubstantiated assumption, that
> > > > somehow something "magical" happens at densities of 3e23 cm^{-3}.
> > > > Whether or not atoms can exist, a thermal plasma is happy to exist at
> > > > those densities -- and higher.
> >
> > > A thermal plasma is happy to exist at those densities, but not any
> > > atomic excitation processes (which require isolated neutral (or
> > > partially ionized) atoms).
> >
> > I note you again did not substantiate your claim that the photosphere
> > density was 3e23 cm^{-3}. I accept your claim that "not any" atomic
> > excitation processes function at high densities. However, *some* do.
> >
> > This is still irrelevant since the solar photosphere density is not
> > 3e23 cm^{-3}.
>
> I said the *threshold density* is 3*10^23 cm^-3 , so the photospheric
> density must be below that in order for atomic excitation processes to
> be able to take place. This threshold density defines the 'edge' of
> the sun. We may strictly speaking not be seeing it because there are a
> few kilometers of less dense material sitting on top of it, but
> nevertheless it defines the edge (and the mass-radius data for the sun
> as well as the planets confirm this).
You keep making these claims about "threshold density" and "individual
atoms and molecules." However, none of these claims is born out by
experiment or detailed theory. Firstly, these terms are vague,
undefined, and are not used in by researchers in the field. Second,
in the full continuum betweeen neutral gas and fully ionized plasma,
there are many possible emission and radiation transport mechanisms.
Finally, several measures of the solar interior (solor oscillations,
solar cycle, interior modeling), and *none* of them is consistent with
a density of 3e23 cm^{-3} anywhere near the surface.
The original question regarded why the sun we *see* had a well defined
edge... hence the *only* relevant topic is the nature of the
photosphere. This is by definition since the photosphere is the solar
surface we "see." If you are now claiming that the "edge" is not the
photosphere that we see, then your claims are simply irrelevant.
> > > There is no problem with a n(r)~1/r^2 singularity, as the total mass
> > > (which is given by the integral over r^2*n(r) ) is still finite. ...
> >
> > I note this claim is unsubstantiated. In reality, even for finite
> > total mass, an infinite local density would be unphysical.
>
> The electrostatic force for a point charge also diverges like 1/r^2 at
> the origin, but nevertheless it is a valid and successful theoretical
> model.
> Of course I am not saying that the density is infinite at the center
> of the sun. The assumption of an ideal gas in hydrostatic equilibrium
> will surely break down if the density reaches the nuclear density for
> instance, but this should only apply within 20m or so of the sun's
> centre, so it will barely make any difference for the calculation.
OK, but if you allow the 1/r^2 density profile and discount the
divergence at the very core... then you must also allow *all other*
1/r^{n} profiles as well since they only diverge at the central core
as well. I'm not arguing in favor of any 1/r^{n} density profile, but
merely pointing out that your claims are self-contradictory.
> > > ... Only
> > > density profiles n(r)~1/r^3 and stronger would be impossible in this
> > > respect. And as I showed a couple of posts above, anything departing
> > > substantially from a 1/r^2 density profile (i.e. from an isothermal
> > > gas) would either lead indeed to an infinite mass, or alternatively
> > > result in an increasing temperature with increasing r.
> >
> > *OR*, your assumption about a pure power law is invalid. There are an
> > infinitude of functions which are not power laws. I guess you meant
> > "virtually" a power law?
>
> If you want, you can do a series expansion in terms of power laws. The
> argument would still then apply to each term. ...
What I "want" is irrelevant, but your claims remain erroneous. In
fact, for an expansion in terms of r, r^2, r^3, etc, your "argument"
does *not* apply. *You* artificially chose to consider only power law
functions of the form 1/r^{n} (n>0), which lead to mathematical
problems except in special cases. There are still an infinitude of
other functions whose integral does not diverge, and most of them do
not have a power law form at all. I note you could have researched
these functions but did not.
I note that you conveniently deleted a portion of the conversation
directly related to the question of how energy is transported within
the sun:
: Smid:
: > What sort of insulating material should there be in the sun?
: > Essentially, the sun consists only of individual protons and electrons
: > throughout, i.e. there is maximum transmission of thermal energy in
: > the contact of neighbouring gas volumes. Also, remember that we would
: > need an insulator which reduces the temperature from 10^7K to 6000K.
:
: Markwardt:
: Note, I did not say an "insulating material," I said an "effective
: insulator," which I then qualified by the radiation/thermal properties
: of the medium. Such effects may include thermal conductivity, which
: will be driven by density and electron temperature; radiative
: transport, which is driven by opacity, density and ionization state;
: convection, which is driven by local energy density, and gas density.
: The gas density itself will be governed by the gas equation of state,
: which may change depending on the ionization state. In other words,
: there are a complicated set of coupled systems which constrain how
: much energy is able to escape the various layers of the sun. If these
: processes are not very efficient, then the star will puff up to larger
: but cooler radius to compensate.
Your "inelastic collisions" claim is vague and unsubstantiated. As
noted above, there are many processes that must be understood in order
to understand the structure of the solar interior.
CM
Craig Markwardt
Thomas <thoma...@gmail.com> writes:
> On 23 Jul, 19:07, will...@cfa.harvard.edu (Steve Willner) wrote:
>
> > > If one takes the masses of our sun and the hydrogen planets in our
> > > solar system (where the masses and radii are known accurately) then
> > > this fits indeed indeed very well a R~M^1/3 law: the theoretical
> > > radii for Jupiter, Saturn, Uranus, Neptune are then (Sun=1) : 0.099,
> > > 0.066, 0.035, 0.037 . The actual values are 0.1, 0.085, 0.037, 0.035.
> >
> > Your power law is equivalent to saying density is constant.
> > Reference to any standard table will show that is false. Densities
> > range from Saturn's 0.7 to Neptune's 1.76. The Sun's 1.41 is within
> > the range, but that's about all that can be said for your "law."
>
> That's a strange argument as the average density is exactly derived
> from the mass and radius. Fact is that an exponent 1/3 in the mass-
> radius relationship fits the major solar system bodies (including the
> sun) but not an exponent 0.9 or 0.6.
However, Steve's is a true argument. The densities of the gas giants
and the sun, range over a factor of ~2.5, contrary to your "law." In
your representation, M^{1/3} compresses the discrepancies so that they
are less visible (~35%), but they are still present. Meanwhile,
density depends on M/R^3, so the density discrepancies are much more
magnified.
Beyond that, your "correlation" is hardly a statement about causation.
Correlative studies are often skewed by sample selection biases unless
care is taken to control such biases (which you did not). In this
case, yes, the correlation between a few planets and the sun seems
suggestive, *BUT*, we know the compositions of these bodies are quite
different. Furthermore, the sun has an internal heat source while the
planets do not. Ice crystals and metals are known to form in the gas
planets, which does not occur in the sun. Thus, the physics is quite
different, and whatever tentative speculation we *might* have made
about them becomes yet more tenuous.
Finally, we have *actual* data about main sequence stellar masses and
radii (the previously quoted Zombeck and Allen references), and they
do *not* obey your putative R ~ M^{1/3} "law." You could have
researched how these measurements were made, but you did not, and
instead offered unsubstantiated speculation. Furthermore, your
tentative mass-radius correlation between the sun and planets does
*not* logically negate or discredit those measurements.
Finally, it's worth noting that quite a number of masses and radii
*have* been measured for low-mass stars and giant planets outside of
our solar system, via various independent techniques (e.g. Udalski et
al 2007, Pont et al 2005). These observations show quite conclusively
that the mass-radius relation for planets and low mass stars does
*not* follow the R~M^{1/3} "law" but does match theoretical
predictions based on physics (Baraffe et al 2003; Guillot 2005).
> > > An exponent 0.9 on the other hand would result in : 0.002, 0.00064,
> > > 0.00012, 0.00014.
> >
> > The _stellar_ mass-radius relation cannot be used to predict
> > _planetary_ radii.
>
> Could it be that this is because it is incorrect and the correct
> stellar mass-radius relation is R ~M^1/3?
No. See above. Your R~M^{1/3} relation is just a fanciful
coincidence, and falls apart when more measurements are taken into
account.
CM
References
Baraffe et al 2003 A&A 337 403
Guillot 2005 Ann Rev Earth Planetary Sci, 33, 493
Pont et al 2005 A&A L21-L24
Udalski et al 2007, A&A submitted, arXiv:0711.3978v1
John Park
> On 26 Jul, 14:51, af...@FreeNet.Carleton.CA (John Park) wrote:
>> Thomas Smid (thomas.s...@gmail.com) writes:
>> > On 25 Jul, 19:14, af...@FreeNet.Carleton.CA (John Park) wrote:
>>
>> >> The point is that if Saturn is a "hydrogen planet" it's easy to explain
>> >> Neptune's density as reflecting a different composition. If Neptune were a
>> >> "hydrogen planet" how could you explain Saturn's density?
>>
>> > I don't know what causes Saturn to make an exception here. With
>> > hydrogen being already the lightest element, it shouldn't really be
>> > less dense. The only potential causes I can think of at the moment is
>> > that either Saturn has an unusually thick atmosphere (which results in
>> > the apparent radius being significantly larger than R (as used in the
>> > equation above)), or that an internal magnetic field causes a
>> > deviation from the simple gravity/pressure equilibrium.
>>
>> I suggest it's much easier to believe that Saturn isn't an exception, but
>> the two outer plenets don't have the compositions you want them to have.
>
> But Saturn is the only one of the big planets (and the sun) that does
> not fit the R~M^1/3 law. So this suggests that there is something
> special going on here.
Or that the m^1/3 law is an illusion.
>
>>
>> >> > Yes, sorry, I wanted to say 'collisional excitation processes'. These
>> >> > obviously cool the gas as particles lose energy by exciting atomic
>> >> > transitions (which then results in radiation).
>>
>> >> So where's the radiation?
>>
>> > Well, in case of the Sun it is obvious, Jupiter is also known to emit
>> > radiation on its own, and for the other planets it may be just not
>> > detectable due to its weakness.
>>
>> I find it hard to believe that a surface temperature of 60 K is compatible
>> with a hydrogen atmosphere and an average temperature of 13,000 K. But I'm
>> not ready to argue the point quantitatively.
>
> First of all, I have to correct myself here: for Saturn, Uranus and
> Neptune there is apparently a substantial excess heat observed as
> well.
>
> If you apply the method I mentioned before for getting the surface
> temperature by dividing the internal temperature (as obtained over the
> Virial theorem) by 1800 (the proton/electron mass ratio) in the case
> of a hydrogen mass (which works well for stars), then one would obtain
> from 13,000K a surface temperature of a mere 7K. But this obviously
> neglects any solar heating, which is difficult to assess here as it
> will depend on any cooling processes in the atmosphere which would
> crucially depend on atmospheric composition
>
Reductio ad absurdum. Hydrogen at 7 K wouldn't give you a plasma; at
anything like normal pressures it wouldn't even be a gas. Ionisation will
absorb energy, but what you suggest looks out of the question. (And if
you''re forming ions, shouldn't the ionisation potential contribute to the
potential energy in the Virial Theorem?)
--John Park
Thomas
As I said already, there is only one star where the mass and radius is
accurately known and this is the sun (+ the planets in the solar
system). Everything else is based on assumptions and wishful thinking.
If you take for instance the measurements presented in
http://arxiv.org/PS_cache/astro-ph/pdf/0211/0211647v1.pdf , then a) it
is impossible to verify from the publication alone whether the radii
have been obtained appropriately and correctly from the data, and b)
the masses have actually not been measured at all, but are estimates
based on assumptions for a mass-luminosity relationship. If you
calculate the exponent in the mass-radius relations in Table 2 there
(by calculating ln(R)/ln(M) for each star) then you get the values
0.7, 1, 0.95, 0.91, 0.88, 1, 1.03, 1.27, 1.51. So there is no mass-
radius relationship visible here at all. On the other hand, in the
solar system (where there can be no doubts about the accuracy of the
actual masses and radii) you have the exponents (referred to
Jupiter) : Sun 0.33, Saturn 0.15, Uranus 0.36, Neptune 0.37, which
indicates quite clearly a R~M^1/3 law if you exclude Saturn.
Thomas