Re: Does Cold Dark Matter Radiate 2.73K Heat?
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Jonathan Thornburg [remove -animal to reply]
[[in the context of arguing that it makes more sense to say that the
Earth goes around the Sun, rather than vice versa]]
| The mass of the planets compared to that of the Sun is very small,
| whether one subscribes to the Ptolemaic or Copernican or Tychonian
| system.
In article
<baefe111-ee51-410e...@e27g2000yqd.googlegroups.com>,
Chalky <chalk...@bleachboys.co.uk> wrote:
> However, this paradigm shift is largely negated by the general
> principle of relativity.
No, it's not negated by any aspect of general relativity (GR).
GR says that we can use any coordinate system (reference frame)
to do our calculations, and provides tools to do so. We should
(must!) always get the same physical results, regardless of the
coordinates we use in the calculations... but the physics of a
particular system may still single out a particular coordinate
system (reference frame).
The solar system has some "interesting" properties that are important
here:
(a) It's accurately modellable as an asymptotically flat spacetime.
That is, far from any solar-system body (but still much closer
than the nearest other star) spacetime is close to Minkowski
(where "close to" has a precise mathematical meaning which I'm
not going to describe here).
(b) It's a slow-motion weak-gravitational-field system: all the
massive bodies in it move much slower than the speed of light
(Mercury's orbital motion is around 1.5e-4 c), and the
dimensionless gravitational potential is on the order of 1e-6
or less anywhere in the solar system.
(c) The Sun's rest mass is about 300,000 times larger than the Earth's
rest mass.
(a) and (b) properties let us define a post-Newtonian approximation
to GR in a mathematically rigorous manner, so it's perfectly ok to
use Newtonian concepts like "inertial reference frame" in a GR as the
leading order of a series expansion. That is, we can prove that the
GR equations of motion can be expanded in power series in v/c,
[These power series are probably only asymptotically
convergent, but for our purposes that doesn't matter:
there's abundant experimental evidence that the leading
terms (up to (v/c)^2) provide a *very* good approximation
to solar-system dynamics.]
with Newtonian mechanics being the leading term in the series.
Thus it's perfectly ok-with-GR to say that
(1) The Sun's center-of-mass is much closer to an inertial coordinate
system than is the Earth's center-of-mass, and
(2) The solar system center-of-mass (defined up to (v/c)^2 accuracy)
is located much closer to the Sun's center-of-mass than it is to
the Earth center-of-mass.
(3) An observer out in the asymptotically-flat reference frame defined
earlier can measure the Sun's acceleration and the Earth's
acceleration, and find the latter to be a lot larger than the former.
(4) An observer sitting on each of the Earth and the Sun can measure
her respective acceleration with respect to the asymptotically-flat
reference frame defined earlier; the Earth observer will measure
a much larger number than the Sun observer.
If you turned the Sun and each planet into a black hole with the
same rest mass and center-of-mass motion [i.e., break property (b)
above, keeping (a) and (c)], then I think (2) and (3) would still hold,
but I'm not sure how you would rigorously define what you mean by
(1) and (4).
If you then pushed all those black holes a lot closer together, so
their typical orbital velocities were a significant fraction of the
speed of light, then I think (3) still survives. Numerical relativists
do this sort of thing routinely nowdays when they simulate asymmetric
black-hole coalescence. See, eg,
* Gonzalez, Sperhake, and Bruegmann,
"Black-hole binary simulations: The mass ratio 10:1",
Physical Review D volume 79, article 124006 (2009),
eprint arXiv:0811.3952.
* Berti et al
"Inspiral, merger and ringdown of unequal mass black hole binaries:
a multipolar analysis"
Physical Review D volume 76, article 064037 (2007),
eprint arXiv:gr-qc/0703053
But I emphasize again, the solar system is a slow-motion weak-field
system, so there's no problem in GR with using Newtonian notions like
"inertial coordinate system", and it's very clear that the Sun's
center-of-mass is a much better approximation to an inertial coordinate
system than is the Earth's center-of-mass.
--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam
Chalky
<jth...@astro.indiana-zebra.edu> wrote:
> Someone whose identity has overflowed my mental nested-quoting stack wrote:
> [[in the context of arguing that it makes more sense to say that the
> Earth goes around the Sun, rather than vice versa]]
> | The mass of the planets compared to that of the Sun is very small,
> | whether one subscribes to the Ptolemaic or Copernican or Tychonian
> | system.
That would be Phillip Helbig. However, I don't think we were arguing
about whether the Earth went round the Sun or vice versa, but whether
this solar dominance on the scale of the solar system justified an
assumption that the dynamism of the universe was similarly dominated
by the masses of stars. My argument that it didn't, was based on more
local considerations. However, this is largely a moot point, because
we all know that it doesn't based on still larger scale
considerations.
> In article
> <baefe111-ee51-410e-8cdc-280ad0e41...@e27g2000yqd.googlegroups.com>,
You raise some interesting points which, afaict can be applied equally
well to the universe at large, which leads me to the following
question:
Would it be fair to additionally say that standard theory also
describes the dynamism of the universe relative to such inertial
coordinate systems (at least for very nearly flat cosmological
solutions)?
Chalky
<jth...@astro.indiana-zebra.edu> wrote:
> Thus it's perfectly ok-with-GR to say that
> (1) The Sun's center-of-mass is much closer to an inertial coordinate
> system than is the Earth's center-of-mass, and
> (2) The solar system center-of-mass (defined up to (v/c)^2 accuracy)
> is located much closer to the Sun's center-of-mass than it is to
> the Earth center-of-mass.
> (3) An observer out in the asymptotically-flat reference frame defined
> earlier can measure the Sun's acceleration and the Earth's
> acceleration, and find the latter to be a lot larger than the former.
> (4) An observer sitting on each of the Earth and the Sun can measure
> her respective acceleration with respect to the asymptotically-flat
> reference frame defined earlier; the Earth observer will measure
> a much larger number than the Sun observer.
I would not want to argue with that statement (in its appropriate
context), but, afaict, that is only one of several possible ways of
looking at things. It also seems to me that a large part of the
attraction of that approach is that it achieves maximum backward
compatibility with the preceding (Newtonian) paradigm.
However, iiuc, the 'preferred' reference frame in GR is the null
geodesic, ie the ref frame of observers in free fall under the
influence of gravity.
In this context, unless my mental arithmetic is wildly out of kilter,
the Earthward directed acceleration of such an observer, just above
the Earth's surface, is far in excess of the corresponding Sunward
directed acceleration of observers at 1au, when rotation of such
observers is excluded (to keep the physics simple).
Expressed the other way round, the power of a jetpack required to keep
us floating just above the Earth, is far in excess of the jetpack
power required to keep us (without circular motion) at 1 au from the
Sun.
This leads me to ask an additional question (which I really am asking
because I genuinely do not know the answer):
Within the context of standard GR, are such null geodesic observers
treated as 'inertial', or not?
eric gisse
[...]
> Within the context of standard GR, are such null geodesic observers
> treated as 'inertial', or not?
Never. Only time-like paths can be inertial.
Jonathan Thornburg
| Thus it's perfectly ok-with-GR to say that
| (1) The Sun's center-of-mass is much closer to an inertial coordinate
| system than is the Earth's center-of-mass, and
| (2) The solar system center-of-mass (defined up to (v/c)^2 accuracy)
| is located much closer to the Sun's center-of-mass than it is to
| the Earth center-of-mass.
| (3) An observer out in the asymptotically-flat reference frame defined
| earlier can measure the Sun's acceleration and the Earth's
| acceleration, and find the latter to be a lot larger than the former.
| (4) An observer sitting on each of the Earth and the Sun can measure
| her respective acceleration with respect to the asymptotically-flat
| reference frame defined earlier; the Earth observer will measure
| a much larger number than the Sun observer.
Chalky <chalk...@bleachboys.co.uk> wrote:
> I would not want to argue with that statement (in its appropriate
> context), but, afaict, that is only one of several possible ways of
> looking at things. It also seems to me that a large part of the
> attraction of that approach is that it achieves maximum backward
> compatibility with the preceding (Newtonian) paradigm.
In general relativity you can use any nonsingular coordinate system
you want, and you'll always get physically equivalent answers. Thus
people usually try to choose coordinates so as to simplify the
calculation. (Such choices are of course widespread in many areas
of physics.) In practice, the easy-to-use coordinates do somewhat
resemble quasi-Newtonian ones... but their key property is being
easy-to-use, not being quasi-Newtonian.
I should also point out that different researchers also may have
different ideas of what coordinates are easiest to use (corresponding
to different calculation techniques). But the mathematical structure
of GR guarantees that they should all get physically equivalent
answers. (Explicitly checking this can be a lot of work, but it's
also a great way to check that lengthly calculutions are in fact
correct.)
> However, iiuc, the 'preferred' reference frame in GR is the null
> geodesic, ie the ref frame of observers in free fall under the
> influence of gravity.
It's *sometimes* convenient to use such a reference frame.
It's *sometimes* convenient to use a reference frame defined by
freely falling photons (a current research project of mine does this).
And there are cases where neither of these is convenient, or even
*possible*. (What happens if the worldlines of those freely-falling
observers or photons cross? Then we get a coordinate singularity.
Ick. Not nice. That's usually a reason to pick a different
coordinate system.) It's *sometimes* convenient to use completely
different reference frames.
For doing post-Newtonian solar system dynamics, it's generally most
convenient to use coordinates based on the (approximate) asymptotic
flatness and slow-motion weak-gravitational-field nature of the
physics. See, for example, Nordtvedt's gr-qc/0301024, pages 5 and 7.
> In this context, unless my mental arithmetic is wildly out of kilter,
> the Earthward directed acceleration of such an observer, just above
> the Earth's surface, is far in excess of the corresponding Sunward
> directed acceleration of observers at 1au, when rotation of such
> observers is excluded (to keep the physics simple).
>
> Expressed the other way round, the power of a jetpack required to keep
> us floating just above the Earth, is far in excess of the jetpack
> power required to keep us (without circular motion) at 1 au from the
> Sun.
True, but that's not necessarily the "right" measure of "strength of
gravitational field". For example, the parameter in post-Newtonian
expansions is usually interpreted as a dimensionless gravitational
potential.
[It's roughly GM/(rc^2) summed over all masses in the
approximately-asymptotically-flat system; this can
also be written as the square of the
escape-velocity-to-the-asymptotically-flat-region
as a fraction of the speed of light.]
At the Earth's surface, the Sun contributes around 2e-8, while the
Earth contributes around 1.5e-9, i.e., the Sun dominates.
> Within the context of standard GR, are such null geodesic observers
> treated as 'inertial', or not?
No. Read gr-qc/0301024 for an overview of how post-Newtonian
calculations are done for a slow-motion weak-gravitational-field
asymptotically flat system in standard GR. If you're not in such
a system, then lots of different kinds of coordinates are used,
and the word "inertial" is usually not very meaningful.
Jonathan Thornburg [remove -animal to reply]
| GR says that we can use any coordinate system (reference frame)
| (must!) always get the same physical results, regardless of the
| coordinates we use in the calculations... but the physics of a
| particular system may still single out a particular coordinate
| system (reference frame).
|
| The solar system has some "interesting" properties that are important
| here:
| (a) It's accurately modellable as an asymptotically flat spacetime.
| ? ? than the nearest other star) spacetime is close to Minkowski
| ? ? (where "close to" has a precise mathematical meaning which I'm
| ? ? not going to describe here).
| (b) It's a slow-motion weak-gravitational-field system: all the
| ? ? (Mercury's orbital motion is around 1.5e-4 c), and the
| ? ? dimensionless gravitational potential is on the order of 1e-6
| ? ? or less anywhere in the solar system.
| (c) The Sun's rest mass is about 300,000 times larger than the Earth's
[[...]]
| But I emphasize again, the solar system is a slow-motion weak-field
| system, so there's no problem in GR with using Newtonian notions like
| "inertial coordinate system", and it's very clear that the Sun's
| center-of-mass is a much better approximation to an inertial coordinate
| system than is the Earth's center-of-mass.
Chalky <chalk...@bleachboys.co.uk> wrote:
> You raise some interesting points which, afaict can be applied equally
> well to the universe at large, which leads me to the following
> question:
>
> Would it be fair to additionally say that standard theory also
> describes the dynamism of the universe relative to such inertial
> coordinate systems (at least for very nearly flat cosmological
> solutions)?
Not really, for a couple of different reasons:
First, something that I should perhaps have clarified in my previous
posting: General relativity says we can use any nonsingular coordinate
system we want to do our calculations, and we'll always get physically
equivalent answers. This doesn't change the fact that for any given
problem, some coordinate systems may be (much) easier or harder than
others to use. It also doesn't change the fact that some physical
systems have special properties, which might single out certain types
of coordinates as "physically preferred". (A simple example is a
system with (say) spherical symmetry, where it's natural to use
coordinates compatible with that symmetry, i.e., to write our
spacetime as M x S^2 for some 2-dimensional Lorentzian manifold M.)
Next, Newtonian dynamics is embedded in general relativity as the
slow-motion weak-gravitational-field limit. More precisely, for a
slow-motion weak-gravitational-field limit one can expand general
relativity in any of several different types of power series in
(v/c). The details are horribly messy, but one of these types of
power series (a "post-Newtonian" expansion) has Newtonian dynamics
as its leading terms.
[Another type of such series (a "post-Minkowskian"
expansion) has Minkowski-spacetime special relativity
as its leading terms.]
It's in this sense that we can recover the Newtonian notion of an
inertial coordinate system (a.k.a. inertial reference frame).
The main application of these "post-Newtonian" is high-precision
calculations of asymptotically flat systems,
[recall from my previous posting that the solar system
is (very) well approximated as being asymptotically flat]
so people normally only do these calculations under the assumption
of asymptotic flatness. (More precisely, I have never seen a
post-Newtonian expansion done for a non-asymptotically-flat system.
But, I am not an expert in post-Newtonian expansions, and it may be
that experts do such things and I've just not heard about them.)
Finally, the universe as a whole is NOT well-approximated as an
asymptotically flat system,
[The modern "flat" cosmologies have approximately
flat *spatial* *slices*, but that's not the same as
*spacetime* being flat.]
so when doing cosmology we can't use standard post-Newtonian expansions.
Chalky
Since you snipped "ie the ref frame of observers in free fall under
the influence of gravity" I guess your point is that when the free
falling body of reference is a photon (or pack of them) their
trajectories are no longer time-like, and thus no longer inertial.
This leads me to ask a related question. When describing the observed
universe in spherical polar coordinates, I would like to say, in the
context of my own analysis, that the radial distance (not to be
confused with comoving radial distance) is time-like, in the sense
that a distance of 100 light years means, in context, that objects at
that distance are also located 100 years in our past.
Does the above identified distinction between "null" for photon
trajectories and "time-like" for everything else, also mean I am not
even allowed to describe that radial distance as time-like?
Jenny
> However, iiuc, the 'preferred' reference frame in GR is the null
> geodesic, ie the ref frame of observers in free fall under the
> influence of gravity.
I think you have that wrong.
A "null geodesic" is the path of light in "free fall".
For observers with mass, AFAIK, the geodesic is "extremal". That is
to say that observers following geodesics will measure an extremal
time between events.
> This leads me to ask an additional question (which I really am asking
> because I genuinely do not know the answer):
> Within the context of standard GR, are such null geodesic observers
> treated as 'inertial', or not?
By *your* definition (" the null geodesic, ie the ref frame of
observers in free fall under the influence of gravity"), the answer
would be "yes" - for the simple reason that you've *defined* it that
way.
But, in the conventional meaning of the term, only massless observers
follow "null geodesic" paths. In the absence of massless observers,
the answer would be "no".
Love,
Jenny
Roger L Hale
eric gisse <jowr.pi...@gmail.com> wrote:
>Chalky wrote:
>
>[...]
>
>> Within the context of standard GR, are such null geodesic observers
>> treated as 'inertial', or not?
Yes, observers in free fall count as 'inertial'.
>
>Never. Only time-like paths can be inertial.
I believe there's some confusion here. By 'null geodesic' the original
poster meant 'in free fall' (and thus, yes, 'inertial'), and not
'light-like'.
(Further, in what sense is a light-like geodesic not inertial? Although
without mass, an amount of light bears momentum and energy, which
qualifies in my book as a 'quantity of motion', the conservation of
which is 'inertia'.)
With best regards,
Roger Hale
'spinclad'
r...@theworld.com
eric gisse
Inertial paths are specifically and unambiguously defined as massive, by
virtue of the word "inertial".
Yes light carries energy and thus mass-equivalent in its' little journey
through time and space, but that is not the same thing as being imbued
with mass.
eric gisse
>> Chalky wrote:
>>
>> [...]
>>
>> > Within the context of standard GR, are such null geodesic observers
>> > treated as 'inertial', or not?
>>
>> Never. Only time-like paths can be inertial.
>
> Since you snipped "ie the ref frame of observers in free fall under
> the influence of gravity" I guess your point is that when the free
> falling body of reference is a photon (or pack of them) their
> trajectories are no longer time-like, and thus no longer inertial.
Pretty much. I specifically distinguish between 'inertial' trajectories, ie:
time-like geodesics, and geodesic trajectories in general as they are not
the same.
>
> This leads me to ask a related question. When describing the observed
> universe in spherical polar coordinates, I would like to say, in the
> context of my own analysis, that the radial distance (not to be
> confused with comoving radial distance) is time-like, in the sense
> that a distance of 100 light years means, in context, that objects at
> that distance are also located 100 years in our past.
Caaareful.
First off, coordinates are irrelevant. Dispense of that notion as soon as
possible. It will only serve to confuse you.
Spatial distance can not be quantified with "time-like" or anything such as
that because when one refers to "distance" in relativity the usual reference
is to the distance between two events as measured using the metric ds^2 =
(whatever). A spatial distance neglects the distance in time, which makes
the exercise futile.
>
> Does the above identified distinction between "null" for photon
> trajectories and "time-like" for everything else, also mean I am not
> even allowed to describe that radial distance as time-like?
See above ^
Chalky
> Spatial distance can not be quantified with "time-like" or anything such as
> that because when one refers to "distance" in relativity the usual reference
> is to the distance between two events as measured using the metric ds^2 =
> (whatever).
True, but presuming you are talking about the 4 dimensional metric
element, this seems to result in the conclusion that the "distance"
between a light emission event and the corresponding light detection
event is always zero (in vacuum)
Not very helpful when describing the distance to an observed
supernova.
> A spatial distance neglects the distance in time, which makes
> the exercise futile.
What about a distance defined as c times light travel time?
eric gisse
> On Jan 14, 7:52 am, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>
>> Spatial distance can not be quantified with "time-like" or anything such
>> as that because when one refers to "distance" in relativity the usual
>> reference is to the distance between two events as measured using the
>> metric ds^2 = (whatever).
>
> True, but presuming you are talking about the 4 dimensional metric
> element, this seems to result in the conclusion that the "distance"
> between a light emission event and the corresponding light detection
> event is always zero (in vacuum)
Welcome to the land of null paths.
>
> Not very helpful when describing the distance to an observed
> supernova.
>
>> A spatial distance neglects the distance in time, which makes
>> the exercise futile.
>
> What about a distance defined as c times light travel time?
Look, you can't simultaneously define spatial distances independently of
time in relativity and have the moniker of light/time/space-like have any
meaning.
There is no answer I can give you that will allow you to do what you want.
Igor Khavkine
> On Jan 14, 7:52 am, eric gisse <jowr.pi.nos...@gmail.com> wrote:
>
> > Spatial distance can not be quantified with "time-like" or anything such as
> > that because when one refers to "distance" in relativity the usual reference
> > is to the distance between two events as measured using the metric ds^2
>
> True, but presuming you are talking about the 4 dimensional metric
> element, this seems to result in the conclusion that the "distance"
> between a light emission event and the corresponding light detection
> event is always zero (in vacuum)
The "distance" calculated in this way, in standard terminology, is
called the invariant interval. Please use it, so that others will not
be confused by what you mean. And yes, for points connected by light
rays it is always zero. Despite some seemingly odd features, the
invariant interval is the only kind of "distance" that does not depend
on the choice of coordinates.
> Not very helpful when describing the distance to an observed
> supernova.
Which is precisely the reason the invariant interval is not what
cosmologists mean when they cite distance measurements. They have a
specific coordinate choice in mind, where spatial distances are
computed within slices of constant cosmological time, the latter being
the relevant technical term.
> > A spatial distance neglects the distance in time, which makes
> > the exercise futile.
>
> What about a distance defined as c times light travel time?
This notion of defining distance is also coordinate dependent. Not
useless or futile, just meaningless without an accompanying specific
choice of coordinates.
Igor
Roger L Hale
eric gisse <jowr.pi...@gmail.com> wrote:
>Roger L Hale wrote:
...
>> (Further, in what sense is a light-like geodesic not inertial? Although
>> without mass, an amount of light bears momentum and energy, which
>> qualifies in my book as a 'quantity of motion', the conservation of
>> which is 'inertia'.)
>
>Inertial paths are specifically and unambiguously defined as massive, by
>virtue of the word "inertial".
This begs the questions, whose definition is this, and what virtue of the
word 'inertial' bears this specific implication of 'massive'. I believe
I gave a reasonable argument that a natural understanding of 'inertia'
does not bears this implication, and indeed I've found an indirect citation
of Einstein in MTW that appears to agree:
/All/ forms of energy possess inertia.
-- Albert Einstein, conclusion from his paper of September 26, 1905,
as summarized by von Laue in Schilpp (1949), p. 523.
-- Misner, Thorne, Wheeler, p. 460. While not Einstein's words in either
of his papers on Electrodynamics (published September 26) or equivalence
of inertia and energy content (received September 27) (conclusion: "If the
theory agrees with the facts, then radiation carries inertia between
emitting and absorbing bodies"), it is interesting that von Laue, at least,
put the matter in such unambiguous terms, which specifically _do not_
define inertia as massive, but as including light-like forms of energy.
>Yes light carries energy and thus mass-equivalent in its' little journey
>through time and space, but that is not the same thing as being imbued
>with mass.
Agreed, but I would continue to say that that is not the same thing as
being imbued with inertia.
With best regards,
Roger Hale
[[Mod. note -- Historically, the word "inertial" was used to mean
"a reference frame where Newton's 2nd law holds" before relativity
came along. If we now try to generalize this definition to work in
some general-relativity contexts, it's clear that null geodesics
aren't going to qualify in any reasonable sense.
-- jt]]