A Question About Einstein's Theory Of Relativity.
Roshard Davis
Greetings everyone. I have a question concerning Albert Einstein's
Theory of Relativity. I heard that the theory is split up into 2
sections call Special Relativity and General Relativity and I was
wondering, what is difference between these two and what compelled
Einstein to create to different sections on this Theory of
Relativity?Thanks.
Patrick Reany
The best, but long answer, is for you to read Einstein and Infeld's
book, The Evolution of Physics.
My own answer can be found at
http://ajnpx.com/html/Relativity-for-beginners.html
and
http://ajnpx.com/html/Relativity.html
both of which assume some college physics background for a complete
understanding.
In the future you should post your relativity questions to
sci.physics.relativity
The short answer is this: Einstein's first step in revamping the
foundations to physics (SR) was to generalize Newton's mechanics
similarly to how Lorentz had done it, yet not sacrifice the pure
principle of relativity in doing so, as Lorentz had done when he
assumed an ether which was the embodiment of an absolute space that
assigns absolute velocities to all particles in the universe. This
step left accelerations as absolute and made velocites relative as
they had been in Newton's theory, but were not in Lorentz's theory.
The second step (GR) removed the postulate that accelerations are
absolute.
cheers,
Patrick
Sam Wormley
Read These:
http://scienceworld.wolfram.com/biography/Einstein.html
http://scienceworld.wolfram.com/physics/SpecialRelativity.html
http://scienceworld.wolfram.com/physics/GeneralRelativity.html
Are There Any Good Books on Relativity Theory?
http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html
Randy Poe
numberedbea...@webtv.net (Roshard Davis) wrote in message
news:<f6a3a34f.0308...@posting.google.com>...
> Greetings everyone. I have a question concerning Albert Einstein's
Special Relativity came first (in 1905). Einstein worked out
what happens when two things are moving at some velocity
relative to each other, but it doesn't handle acceleration.
It is limited to what are called "inertial frames". Nor
did it handle gravity.
The General Theory was published in 1915 and required
considerable effort and some help from mathematicians with
working out the equations. It covers acceleration and
provides a theory of gravity as well. Most of the
really spectacular predictions (light bent around
heavy masses, black holes) that capture the imagination
come from GR.
Nothing "compelled" Einstein to split it up this way. It's
a fairly standard theoretical progression. First he got
his early theory, then he generalized it. He published
the work as he worked it out. GR built on top of SR.
- Randy
Uncle Al
http://fourmilab.to/etexts/einstein/specrel/specrel.pdf
Special Relativity, SR.
One tacitly assumes c=c, G=0, h=0 and goes on from there. Newton is
made to fit Maxwell's equations. Galilean transforms fall to
Lorentzian transforms. SR does not handle gravitation or accelerated
reference frames (one can tap dance a bit in the latter case).
General Relativity is SR plus the Equivalence Principle - all local
masses fall identically in vacuum. GR is gravitation. One tacitly
assumes c=c, G=G, h=0 and goes on from there. Everything else is
details (!).
Note that one can formulate equally predictive valid gravitation
theories by not assuming he Equivalence Principle (e.g.,
Weitzenboeck).
--
Uncle Al
http://www.mazepath.com/uncleal/eotvos.htm
(Do something naughty to physics)
C. E. S.
note that not everyone agrees (to say the least) with the notion that
special relativity can't handle accelerations. The FAQ entry
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
is relevant here. -TB]
In article <f6a3a34f.0308...@posting.google.com>,
numberedbea...@webtv.net (Roshard Davis) wrote:
Special relativity is about smooth, even, steady motion without any bumps,
jars or other acceleration forces. General relativity has to do with
gravity and other accerations (rate of change of speed).
--
Cogito ergo sum.
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
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Martin Hogbin
"Roshard Davis" <numberedbea...@webtv.net> wrote in message
news:f6a3a34f.0308...@posting.google.com...
Einstein's famous paper on what is now called special relativity
(SR) was first published in 1905. It was a theory of space and time
that explained some puzzling experimental observations of the
day.
Hermann Minkowski, one of Einstein's maths tutors, proposed
a geometric interpretation of Einstein's theory in which time
became a fourth dimension.
Einstein then developed Minkowski's geometric interpretation
over the next decade and in 1915 published a paper in
which he explained gravitation in terms of the curvature of
spacetime. This has become known as the general theory of
relativity (GR).
So GR includes SR, and the distinction between the two
theories is somewhat arbitrary. It is customary to
include in SR all of GR except gravitation. In other
words SR is the physics of flat (as opposed to curved
spacetime).
Mathematically, GR is much more complicated than SR.
Martin Hogbin
Gregory L. Hansen
The special theory of relativity was easy, he got that one out years
before the general theory.
Special relativity postulates the invariance of physical laws in
inertial reference frames and the constant speed of light in vacuum in
inertial reference frames, and does not include gravity. A lot can be
done with special relativity using high school algebra. The general
theory makes gravity a geometrical phenomenon, and requires some math
that is difficult today and was less widely used or understood back
then.
Tim S
> [Sci.physics.research moderator's note: Top-posting corrected. Also,
> note that not everyone agrees (to say the least) with the notion that
> special relativity can't handle accelerations. The FAQ entry
>
> http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
>
> is relevant here. -TB]
>
> In article <f6a3a34f.0308...@posting.google.com>,
> numberedbea...@webtv.net (Roshard Davis) wrote:
>
>> Greetings everyone. I have a question concerning Albert Einstein's
>> Theory of Relativity. I heard that the theory is split up into 2
>> sections call Special Relativity and General Relativity and I was
>> wondering, what is difference between these two and what compelled
>> Einstein to create to different sections on this Theory of
>> Relativity?Thanks.
>
> Special relativity is about smooth, even, steady motion without any bumps,
> jars or other acceleration forces. General relativity has to do with
> gravity and other accerations (rate of change of speed).
I'm puzzled.
Why do so many people say that special relativity can't handle non-inertial
reference frames? Or, even more amazingly, that it can't handle acceleration
at all (hence presumably involves only free particles and/or fields, with no
forces)?
The way I see it, the thing about SR is that it presumes that the metric is
Minkowski, irrespective of the distribution of matter. (Also usually that
the topology is R^4, though I'd count Minkowski metrics on funny topologies
as 'funny SR', not anything to do with GR).
By contrast, GR requires the metric to be constrained by the Einstein
equation, hence in particular to depend on the distribution of matter
(specifically the stress tensor).
Within these different constraints on the geometry, surely you're free to
pick whatever coordinate systems or frames that you like? It happens that
people didn't feel the need to express physical laws in coordinate-free form
until they were forced to by GR, but that's a historical accident, surely?
But it seems to have become embedded in pedagogical folklore in a very
unfortunate way, which misleads people into believing that acceleration has
something to do with curvature.
It would be much better, IMO, if people first encountered non-inertial
frames in SR. Or, better still, in Newtonian physics, which is the only
context where phrases like 'the reference frame of an accelerated observer'
or 'gravity is a ficitious force' can actually be given some sort of
intelligible meaning.
<grumble>
Tim
John Baez
Tim S <T...@timsilverman.demon.co.uk> wrote:
>I'm puzzled.
>
>Why do so many people say that special relativity can't handle non-inertial
>reference frames? Or, even more amazingly, that it can't handle acceleration
>at all (hence presumably involves only free particles and/or fields, with no
>forces)?
It's because some people mistakenly thought this once
upon a time, and the error found its way into a bunch
of popularizations and maybe even some textbooks. A
lot of us have spent a lot of time trying to stamp out
this mistake, but this appears to be impossible - it's
one of those memes that won't die. It would be interesting
(but depressing) to study the history of this mistake and
its spread.
One way people spread this mistake is by telling a superficially
plausible story like this:
[TOTAL BALONEY FOLLOWS:]
"Einstein came up with special relativity, which handles
only unaccelerated frames of reference [or worse: unaccelerated
motion]. Then he realized that an accelerated frame of reference
is indistinguishable from a gravitational field - his famous
`elevator' thought experiment. This led him to invent general
relativity, which handles general frames of reference and gravity."
This is superficially plausible, in part because even Einstein
was a bit mixed up about this stuff at first: he made up
the term "general relativity" in part to emphasize the
"general covariance" of the theory, i.e. the invariance
of the equation of motion under arbitrary coordinate
transformations. But in fact, special relativity is
already "generally covariant" in this sense! What matters
is that special relativity is not a "background-free" theory.
This distinction has taken a long time to sink in. See
http://math.ucr.edu/home/baez/background.html
for more.
It's mildly annoying when people say special relativity
can only handle unaccelerated (= inertial) frames of
reference. But it goes beyond annoying into hilarious
when people say it's unable to handle accelerated motion.
If that were true, it couldn't handle the concept of force!
It would be pretty much useless.
(I'll set followups to a more accelerated newsgroup.)
greywolf42
Tim S:
"Why do so many people say that special relativity can't handle non-inertial
reference frames? Or, even more amazingly, that it can't handle acceleration
at all (hence presumably involves only free particles and/or fields, with no
forces)?"
John Baez:
"It's because some people mistakenly thought this once upon a time, and the
error found its way into a bunch of popularizations and maybe even some
textbooks."
And hence, this is a good time for another quest for the origin of a claim.
Is it myth? Or is it science?
Where was this claim first made, and can we find the technical backup?
3 points will be awarded for the first poster who identifies ANY written
source for these (or other such) claims -- regardless of degree of backup.
5 points will be awarded for the first poster who identifies ANY written
source that includes an actual analysis (as opposed to pontification).
A bonus of 5 points will be awarded for the identification of the source
with the earliest publication date (after 2 weeks) that appears after
Einstein's 1905 paper.
No points will be awarded for attempts to provide new definitions of the
problem.
No points will be awarded for attempts to prove there is no problem.
Please do not clutter this thread with your own views on the "problem". This
thread is simply a hunt for the origins of this particular "standard
argument." Not a forum for discussing the merits of the argument.
greywolf42
ubi dubium ibi libertas
Gregory L. Hansen
greywolf42 <min...@sim-ss.com> wrote:
>The following exchange occured on sci.physics.relativity:
>
>Tim S:
>"Why do so many people say that special relativity can't handle non-inertial
>reference frames? Or, even more amazingly, that it can't handle acceleration
>at all (hence presumably involves only free particles and/or fields, with no
>forces)?"
>
>John Baez:
>"It's because some people mistakenly thought this once upon a time, and the
>error found its way into a bunch of popularizations and maybe even some
>textbooks."
>
>
>And hence, this is a good time for another quest for the origin of a claim.
>Is it myth? Or is it science?
>
>
>
>Where was this claim first made, and can we find the technical backup?
>
>3 points will be awarded for the first poster who identifies ANY written
>source for these (or other such) claims -- regardless of degree of backup.
This might be good for 1.5 points.
In Matolesi's "Spacetime without reference frames: An applicaton to
synchronizations on a rotating disk", _Foundations of Physics_ 28, 1685
(1998),
"As an introductory remark to the absolute (without reference frames)
formulation of spacetime, we note that the frequently states assertion
that special relativity is the theory of inertial reference frames and
general relativity is the theory of arbitrary reference frames [9] is to
be substituted with the one that general relativity describes gravitation
and special relativity concerns the lack of gravitation [10, 11, 7]"
I haven't looked up the cited references, but they are
[7] Matolesi, _Spacetime without Reference Frames_ (Akademiai Kiado,
Budapest, 1993)
[9] Moller, _The Theory of Relativity_, 2nd edn. (Clarendon Press, Oxford,
1972.)
[10] Synge, _Relativity: The Special Theory_ (North-Holland, Amsterdam,
1955), Secs. I.9, I.8, II.7.
[11] Synge, _Relativity: The General Theory_ (North-Holland, Amsterdam,
1964), Indroduction, Sec. III.5.
Maybe the references have better references.
>
>5 points will be awarded for the first poster who identifies ANY written
>source that includes an actual analysis (as opposed to pontification).
>
>A bonus of 5 points will be awarded for the identification of the source
>with the earliest publication date (after 2 weeks) that appears after
>Einstein's 1905 paper.
That won't be me.
--
"A good plan executed right now is far better than a perfect plan
executed next week."
-Gen. George S. Patton
shuba
> In article <vkf8ifq...@corp.supernews.com>,
> greywolf42 <min...@sim-ss.com> wrote:
> >The following exchange occured on sci.physics.relativity:
Typical of the bogus scholarship of the Bastard greywolf, this is
incorrect. The exchange did *not* take place in s.p.relativity,
though follow-ups were set there by Baez. I have again set the
follow-ups there. By the way, the "Tim S" referred to below was
not me.
> >Tim S:
> >"Why do so many people say that special relativity can't handle non-inertial
> >reference frames? Or, even more amazingly, that it can't handle acceleration
> >at all (hence presumably involves only free particles and/or fields, with no
> >forces)?"
> >
> >John Baez:
> >"It's because some people mistakenly thought this once upon a time, and the
> >error found its way into a bunch of popularizations and maybe even some
> >textbooks."
> >
> >
> >And hence, this is a good time for another quest for the origin of a claim.
> >Is it myth? Or is it science?
What babble! Baez simply made a factual statement. See the nice
post of Dirk Van de moortel (actually in s.p.relativity)
<a3M1b.82977$F92....@afrodite.telenet-ops.be>
to see how SR handles acceleration.
> In Matolesi's "Spacetime without reference frames: An applicaton to
> synchronizations on a rotating disk", _Foundations of Physics_ 28, 1685
> (1998),
>
> "As an introductory remark to the absolute (without reference frames)
> formulation of spacetime, we note that the frequently states assertion
> that special relativity is the theory of inertial reference frames and
> general relativity is the theory of arbitrary reference frames [9] is to
> be substituted with the one that general relativity describes gravitation
> and special relativity concerns the lack of gravitation [10, 11, 7]"
>
> I haven't looked up the cited references, but they are
>
> [7] Matolesi, _Spacetime without Reference Frames_ (Akademiai Kiado,
> Budapest, 1993)
>
> [9] Moller, _The Theory of Relativity_, 2nd edn. (Clarendon Press, Oxford,
> 1972.)
>
> [10] Synge, _Relativity: The Special Theory_ (North-Holland, Amsterdam,
> 1955), Secs. I.9, I.8, II.7.
>
> [11] Synge, _Relativity: The General Theory_ (North-Holland, Amsterdam,
> 1964), Indroduction, Sec. III.5.
>
> Maybe the references have better references.
Matolesi's quote is quite correct, and exactly parallels Baez'
comment. I have seen the incorrect assertion about acceleration
many times. For instance, in Paul A. Tipler, _Physics_, 2nd ed.,
(Worth Publishers Inc., 1982), sec. 35-11, page 958:
"The generalization of relativity theory to noninertial
reference frames by Einstein in 1916 is known as the
general theory of relativity."
It's worth remembering that general relativity really occupies a
very small niche in our understanding of physics. It's no
accident that arxiv.org bundles it with quantum cosmology (gr-qc)
in its categorization scheme. It's rare that GR considerations
play a part in experiment and observation. Most physicists can
safely ignore GR in their laboratory experiments. It's therefore
not altogether surprising that a bogus assertion about GR can be
passed along in popular presentations and even textbooks that
attempt to give a broad overview of physics. Yes, our textbooks,
especially the elemetary-level ones, do contain errors of both
commision and ommision, and that's worth addressing. What's not
worth addressing is the crackpot agenda of the Bastard greywolf.
---Tim Shuba---
Ken S. Tucker
>The following exchange occured on sci.physics.relativity:
>
>Tim S:
>"Why do so many people say that special relativity can't handle non-inertial
>reference frames? Or, even more amazingly, that it can't handle acceleration
>at all (hence presumably involves only free particles and/or fields, with no
>forces)?"
>
>John Baez:
>"It's because some people mistakenly thought this once upon a time, and the
>error found its way into a bunch of popularizations and maybe even some
>textbooks."
>
>And hence, this is a good time for another quest for the origin of a claim.
>Is it myth? Or is it science?
From the first page, of THE FOUNDATION OF THE GENERAL
THEORY OF RELATIVITY, by A. EINSTEIN. (See Dover P of R),
I quote, "This postulate we call the 'special principle of relativity.'
The word "special" is meant to intimate that the principle is restricted
to the case when K' has a motion of uniform translation relatively to K,
but that the equivalence of K' and K does not extend to the case of non-
uniform motion of K' relatively to K." (1916).
In 1911, in the same Dover book, Einstein published,
"ON THE INFLUENCE OF GRAVITATION ON THE
PROPAGATION OF LIGHT", where he attempted to use
Special Relativity to predict the bending of light in accelerating
frames, and got the wrong answer.
This indicates one needs to know GR before one can
simplify acceleration problems to be *handled* by SR.
In effect an acceleration problem *handled* by SR is a
knowledgeable simplification and specialization of general
procedures prescribed by GR, as Einstein discovered.
Regards
Ken S. Tucker
Bill Hobba
> The following exchange occured on sci.physics.relativity:
>
> Tim S:
> "Why do so many people say that special relativity can't handle
non-inertial
> reference frames? Or, even more amazingly, that it can't handle
acceleration
> at all (hence presumably involves only free particles and/or fields, with
no
> forces)?"
>
> John Baez:
> "It's because some people mistakenly thought this once upon a time, and
the
> error found its way into a bunch of popularizations and maybe even some
> textbooks."
>
I have been a bit guilty of this myself. SR deals with constant velocity
only not acceleration. But acceleration can be looked on as a large number
of constant velocities and you can apply SR to each one of those. I have
lost my copy of MTW recently (no great loss I think as I never really liked
the book - but I may have to eventually replace it since it is widely used)
but I seem to recall it has a whole chapter devoted to this issue. Suffice
to say SR is enough to handle acceleration the same way it is done in
Newtonian mechanics - transform to an inertial frame work it out then
transform back. Come to think of it that basically is what GR does (for
local regions) so I suppose the question is a bit moot anyway.
Thanks
Bill
Bill Hobba
> From the first page, of THE FOUNDATION OF THE GENERAL
> THEORY OF RELATIVITY, by A. EINSTEIN. (See Dover P of R),
> I quote, "This postulate we call the 'special principle of relativity.'
> The word "special" is meant to intimate that the principle is restricted
> to the case when K' has a motion of uniform translation relatively to K,
> but that the equivalence of K' and K does not extend to the case of non-
> uniform motion of K' relatively to K." (1916).
>
> In 1911, in the same Dover book, Einstein published,
> "ON THE INFLUENCE OF GRAVITATION ON THE
> PROPAGATION OF LIGHT", where he attempted to use
> Special Relativity to predict the bending of light in accelerating
> frames, and got the wrong answer.
> This indicates one needs to know GR before one can
> simplify acceleration problems to be *handled* by SR.
> In effect an acceleration problem *handled* by SR is a
> knowledgeable simplification and specialization of general
> procedures prescribed by GR, as Einstein discovered.
>
Not quite Ken because an accelerated system can be broken down into a very
large number of inertial systems. Also one can transform from the
accelerated to the inertial system then work out the problem with SR then
transform back. When gravity is present the transformation can only be done
locally; but providing gravity is not present then we can use various tricks
to allows SR to handle it.
Thanks
Bill
Ken S. Tucker
>Ken S. Tucker wrote:
>> From the first page, of THE FOUNDATION OF THE GENERAL
>> THEORY OF RELATIVITY, by A. EINSTEIN. (See Dover P of R),
>> I quote, "This postulate we call the 'special principle of relativity.'
>> The word "special" is meant to intimate that the principle is restricted
>> to the case when K' has a motion of uniform translation relatively to K,
>> but that the equivalence of K' and K does not extend to the case of non-
>> uniform motion of K' relatively to K." (1916).
>>
>> In 1911, in the same Dover book, Einstein published,
>> "ON THE INFLUENCE OF GRAVITATION ON THE
>> PROPAGATION OF LIGHT", where he attempted to use
>> Special Relativity to predict the bending of light in accelerating
>> frames, and got the wrong answer.
>> This indicates one needs to know GR before one can
>> simplify acceleration problems to be *handled* by SR.
>> In effect an acceleration problem *handled* by SR is a
>> knowledgeable simplification and specialization of general
>> procedures prescribed by GR, as Einstein discovered.
>>
>
>Not quite Ken because an accelerated system can be broken down into a very
>large number of inertial systems.
On the basis of this logic, can you find a means to predict
the deflection of light rays in a g-field?
> Also one can transform from the
>accelerated to the inertial system then work out the problem with SR then
>transform back. When gravity is present the transformation can only be done
>locally; but providing gravity is not present then we can use various tricks
>to allows SR to handle it.
The essential difference that GR recognizes that SR fails to
respect is the positional variation as a result of the spatial
metrics differnces within a field that are provided by
g11 ....g33. These metrics are physically proven by the
experimental results derived by the study of EM radiation
bent by the sun.
At this point, we should call into jury, or... respect those
reported results.
>Thanks
>Bill
Regards, Ken S. Tucker
me...@cars3.uchicago.edu
>greywolf42 wrote:
>> The following exchange occured on sci.physics.relativity:
>>
>> Tim S:
>> "Why do so many people say that special relativity can't handle
>non-inertial
>> reference frames? Or, even more amazingly, that it can't handle
>acceleration
>> at all (hence presumably involves only free particles and/or fields, with
>no
>> forces)?"
>>
>> John Baez:
>> "It's because some people mistakenly thought this once upon a time, and
>the
>> error found its way into a bunch of popularizations and maybe even some
>> textbooks."
>>
>
>I have been a bit guilty of this myself. SR deals with constant velocity
>only not acceleration.
Sigh. SR *does* deal with acceleration. Acceleration relative to
inertial frame is acceleration.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
ZZBunker
> greywolf42 wrote:
> > The following exchange occured on sci.physics.relativity:
> >
> > Tim S:
> > "Why do so many people say that special relativity can't handle
> non-inertial
> > reference frames? Or, even more amazingly, that it can't handle
> acceleration
> > at all (hence presumably involves only free particles and/or fields, with
> no
> > forces)?"
> >
> > John Baez:
> > "It's because some people mistakenly thought this once upon a time, and
> the
> > error found its way into a bunch of popularizations and maybe even some
> > textbooks."
> >
>
> I have been a bit guilty of this myself. SR deals with constant velocity
> only not acceleration.
SR deals with the constant velocity of *light*.
All other sci.relativity sci.fiction about
math concepts about gravity is bullshit.
Gregory L. Hansen
> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<bi8c8u$56h$1...@hood.uits.indiana.edu>...
>
> >The following exchange occured on sci.physics.relativity:
> >
> >Tim S:
> >"Why do so many people say that special relativity can't handle non-inertial
> >reference frames? Or, even more amazingly, that it can't handle acceleration
> >at all (hence presumably involves only free particles and/or fields, with no
> >forces)?"
> >
> >John Baez:
> >"It's because some people mistakenly thought this once upon a time, and the
> >error found its way into a bunch of popularizations and maybe even some
> >textbooks."
> >
> >And hence, this is a good time for another quest for the origin of a claim.
> >Is it myth? Or is it science?
>
> From the first page, of THE FOUNDATION OF THE GENERAL
> THEORY OF RELATIVITY, by A. EINSTEIN. (See Dover P of R),
> I quote, "This postulate we call the 'special principle of relativity.'
> The word "special" is meant to intimate that the principle is restricted
> to the case when K' has a motion of uniform translation relatively to K,
> but that the equivalence of K' and K does not extend to the case of non-
> uniform motion of K' relatively to K." (1916).
Back that far, huh?
>
> In 1911, in the same Dover book, Einstein published,
> "ON THE INFLUENCE OF GRAVITATION ON THE
> PROPAGATION OF LIGHT", where he attempted to use
> Special Relativity to predict the bending of light in accelerating
> frames, and got the wrong answer.
> This indicates one needs to know GR before one can
> simplify acceleration problems to be *handled* by SR.
No. It indicates one needs to know GR before one can handle problems
of gravitational interactions.
> In effect an acceleration problem *handled* by SR is a
> knowledgeable simplification and specialization of general
> procedures prescribed by GR, as Einstein discovered.
Boosts and rotations are part of the Lorentz group. Acceleration is a
succession of boosts. As long as you stay with the Lorentz group and
don't concern yourself with the stress-energy tensor as a source of
curvature, you're doing special relativity. That there's an
equivalence between gravitation and an accelerated reference frame
doesn't mean you need a theory of gravity to handle an accelerated
reference frame. But gravity doesn't let you use just one accelerated
reference frame because it has different strengths and directions in
different parts of space, and it needs to postulate a relation between
the acceleration and a source of gravity.
Working with accelerated frames in special relativity may bring up
some mathematical apparatus like Christoffel symbols, and that will
cause some people to immediately shout "Curvature! General
relativity!" But you get Christoffel symbols when you work in
accelerated frames, or non-Cartesian coordinates, in Newtonian
mechanics, too. Or you get derivatives of basis vectors, same thing.
You can get curvature tensors even in Newtonian mechanics if you do
mechanics on the surface of a sphere, for instance. Or "the constant
energy manifold" in a many-body problem. That doesn't mean anything,
there's nothing specifically general relativistic about that math
except that most students only ever encounter it in the context of
general relativity. The theory is defined by the postulates, not by
the mathematical apparatus that is ultimately used.
greywolf42
Gregory L. Hansen <glha...@indiana.edu> wrote in message
news:8ce5c97e.03082...@posting.google.com...
> dyna...@vianet.on.ca (Ken S. Tucker) wrote in message
news:<2202379a.03082...@posting.google.com>...
> > glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
news:<bi8c8u$56h$1...@hood.uits.indiana.edu>...
> >
Please do not clutter this thread with arguments about whether the
assertions are valid or not!
greywolf42
Please don't clutter this thread with discussions about the validity of the
argument. This is just to find out where the argument originated.
greywolf42
greywolf42
Please don't clutter this thread with claims about whether the argument is
valid or not. This thread is about the origin of the arguement.
greywolf42
greywolf42
news:tim.shuba-7A6AF...@corp.supernews.com...
>
> > >The following exchange occured on sci.physics.relativity:
>
> incorrect. The exchange did *not* take place in s.p.relativity,
> though follow-ups were set there by Baez. I have again set the
> follow-ups there. By the way, the "Tim S" referred to below was
> not me.
Timmy seems to have a personal problem. The post that I read *was* in SPR.
However, you have no standing to 'change' followups. Go play in your own
sandbox.
> > >Tim S:
> > >"Why do so many people say that special relativity can't handle
non-inertial
> > >reference frames? Or, even more amazingly, that it can't handle
acceleration
> > >at all (hence presumably involves only free particles and/or fields,
with no
> > >forces)?"
> > >
> > >John Baez:
> > >"It's because some people mistakenly thought this once upon a time, and
the
> > >error found its way into a bunch of popularizations and maybe even some
> > >textbooks."
> > >
> > >
> > >And hence, this is a good time for another quest for the origin of a
claim.
> > >Is it myth? Or is it science?
>
> What babble! Baez simply made a factual statement. See the nice
> post of Dirk Van de moortel (actually in s.p.relativity)
>
> <a3M1b.82977$F92....@afrodite.telenet-ops.be>
>
> to see how SR handles acceleration.
Please don't clutter this thread with attempts to prove or disprove the
argument. This is an attempt to find the origin of the statement.
greywolf42
greywolf42
Gregory L. Hansen <glha...@steel.ucs.indiana.edu> wrote in message
news:bi8c8u$56h$1...@hood.uits.indiana.edu...
Greg, this is actually good for 3 points. Even in 1998. It won't win the
'first' bonus.
> I haven't looked up the cited references, but they are
>
> [7] Matolesi, _Spacetime without Reference Frames_ (Akademiai Kiado,
> Budapest, 1993)
>
> [9] Moller, _The Theory of Relativity_, 2nd edn. (Clarendon Press, Oxford,
> 1972.)
>
> [10] Synge, _Relativity: The Special Theory_ (North-Holland, Amsterdam,
> 1955), Secs. I.9, I.8, II.7.
>
> [11] Synge, _Relativity: The General Theory_ (North-Holland, Amsterdam,
> 1964), Indroduction, Sec. III.5.
>
> Maybe the references have better references.
You've just given hints for possible points to everyone else.
greywolf42
Ken S. Tucker <dyna...@vianet.on.ca> wrote in message
news:2202379a.03082...@posting.google.com...
Well, that's worth 5 points to Ken.
> In 1911, in the same Dover book, Einstein published,
> "ON THE INFLUENCE OF GRAVITATION ON THE
> PROPAGATION OF LIGHT", where he attempted to use
> Special Relativity to predict the bending of light in accelerating
> frames, and got the wrong answer.
> This indicates one needs to know GR before one can
> simplify acceleration problems to be *handled* by SR.
> In effect an acceleration problem *handled* by SR is a
> knowledgeable simplification and specialization of general
> procedures prescribed by GR, as Einstein discovered.
A fascinating observation. But since this was not a claim that SR could
only be used in inertial frames, no points.
Ken S. Tucker
>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03082...@posting.google.com>...
>> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<bi8c8u$56h$1...@hood.uits.indiana.edu>...
Hi Gregory
Read your note several times, thank you, and posted
comments within context. But first I'd like to have some
cooperation on dilineating the boundary of GR and SR.
In my view, the covariant acceleration is described by
GR 's geodesic equation given by the familiar form,
dU^u/ds = GAMMA^u_ab U^a U^b
(U^u = dx^u/ds),
and this equation is GR, and not SR, and delineates
the difference, because it is generally covariant.
This equation corelates with the *absolute derivative*
DU^u = 0 ,
(DU^u = U^u;v U^v where ';' is covariant differentiation)
It is always possible to find a Coordinate System (CS)
where U^u =0 therefore DU^u =0 is always true.
What I just wrote is the General Theory of Relativity,
(gravitation is an application).
>> From the first page, of THE FOUNDATION OF THE GENERAL
>> THEORY OF RELATIVITY, by A. EINSTEIN. (See Dover P of R),
>> I quote, "This postulate we call the 'special principle of relativity.'
>> The word "special" is meant to intimate that the principle is restricted
>> to the case when K' has a motion of uniform translation relatively to K,
>> but that the equivalence of K' and K does not extend to the case of non-
>> uniform motion of K' relatively to K." (1916).
>
>Back that far, huh?
Greywolf, in the OP asked to go back as far as possible to suck
points.
>> In 1911, in the same Dover book, Einstein published,
>> "ON THE INFLUENCE OF GRAVITATION ON THE
>> PROPAGATION OF LIGHT", where he attempted to use
>> Special Relativity to predict the bending of light in accelerating
>> frames, and got the wrong answer.
>> This indicates one needs to know GR before one can
>> simplify acceleration problems to be *handled* by SR.
>
>No. It indicates one needs to know GR before one can handle problems
>of gravitational interactions.
Not just gravity, electrodynamics fails when accelerating
charges in Special Relativity. (See Dovers P of R,
ELECTRODYNAMICS, chp 10, entitled 'Dynamics of
the Slowly Accelerated Electron".)
In view of the Quantum Theory, the listed accelerations
derived by SR are wrong. However in GR, in view of
DU^u=0 these are correct, and are QT compatible.
(invariant mass m*DU^u=0 = DE^u=0, meaning NO
continuous change in energy is measureable).
>> In effect an acceleration problem *handled* by SR is a
>> knowledgeable simplification and specialization of general
>> procedures prescribed by GR, as Einstein discovered.
>
>Boosts and rotations are part of the Lorentz group. Acceleration is a
>succession of boosts.
>As long as you stay with the Lorentz group and
>don't concern yourself with the stress-energy tensor as a source of
>curvature, you're doing special relativity. That there's an
>equivalence between gravitation and an accelerated reference frame
>doesn't mean you need a theory of gravity to handle an accelerated
>reference frame. But gravity doesn't let you use just one accelerated
>reference frame because it has different strengths and directions in
>different parts of space, and it needs to postulate a relation between
>the acceleration and a source of gravity.
>
>Working with accelerated frames in special relativity may bring up
>some mathematical apparatus like Christoffel symbols, and that will
>cause some people to immediately shout "Curvature! General
>relativity!" But you get Christoffel symbols when you work in
>accelerated frames, or non-Cartesian coordinates, in Newtonian
>mechanics, too.
In view of your statement, can you provide the subsitutions into
the Christoffel symbol that account for centrifugal force without
using the geodesical equation DU^u =0?
In other words, derive the Christoffel metrics from SR
for centrifugal force.
>Or you get derivatives of basis vectors, same thing.
>You can get curvature tensors even in Newtonian mechanics if you do
>mechanics on the surface of a sphere, for instance. Or "the constant
>energy manifold" in a many-body problem. That doesn't mean anything,
>there's nothing specifically general relativistic about that math
>except that most students only ever encounter it in the context of
>general relativity.
>The theory is defined by the postulates, not by
>the mathematical apparatus that is ultimately used.
Agree 101%. But for brevity, I think, GR can be defined by,
DU^u =0 (GR)
and SR by
dx^i dx_i = 0 (SR)
(where i =1,2,3), and means absolute motion = zero.
A second glance at the SR equation says that a FoR
has no velocity relative to the same FoR, obviously.
Thanks again for your comments,
Sincerely
Ken S. Tucker
Gregory L. Hansen
Ken S. Tucker <dyna...@vianet.on.ca> wrote:
>glha...@indiana.edu (Gregory L. Hansen) wrote in message
>news:<8ce5c97e.03082...@posting.google.com>...
>
>>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message
>news:<2202379a.03082...@posting.google.com>...
>>> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
>news:<bi8c8u$56h$1...@hood.uits.indiana.edu>...
>
>Hi Gregory
>Read your note several times, thank you, and posted
>comments within context. But first I'd like to have some
>cooperation on dilineating the boundary of GR and SR.
> In my view, the covariant acceleration is described by
>GR 's geodesic equation given by the familiar form,
>
>dU^u/ds = GAMMA^u_ab U^a U^b
>
>(U^u = dx^u/ds),
>
>and this equation is GR, and not SR, and delineates
>the difference, because it is generally covariant.
>
>This equation corelates with the *absolute derivative*
>
>DU^u = 0 ,
Why would you make that the distinction between SR and GR? There's
nothing about it that conflicts with the postulates of special relativity,
and using it does not require additional postulates. It doesn't even
describe acceleration unless you're in an accelerated frame. The geodesic
equation is also used in *Newtonian* mechanics! The terms on the right
disappear in Cartesian coordinates, but try finding the geodesic in polar
coordinates. The geodesic in a rotating frame will give you the Coriolis
force terms. It's just geometry. What is the shortest path between two
points on a sphere? That's not a general relativity problem, it's just
the geometry of a surface in Euclidean space.
For that matter, I can plug a metric like
ds^2 = dt^2 + dx^2 - dy^2 - dz^2
into the geodesic equation and it won't be special relativity or general
relativity, it won't be anything.
The student typically doesn't see that math until he starts learning
general relativity. But general relativity is one of many applications of
differential geometry.
Some authors do seem to *define* general relativity as relativity with
acceleration. For instance, R. H. Good in "Uniformly accelerated
reference frame and the twin paradox" in Am. J. Phys. 50, 232 (1982). He
starts with the Lorentz transformations to find a composition of
velocities, takes a derivative of that to find acceleration, integrates
along paths... and at the end he says "When general relativity is
included in analysis of the traveling twin's trip..." But he never
brought any postulate of general relativity into the analysis. He didn't
use the equivalence principle, he didn't use G=8piT. All he did was
manipulate the Lorentz transformations in a space with a flat metric. So
calling that general relativity is just confused and spurious. The theory
is defined by the postulates, not by which problem you apply the
postulates to.
I have no idea what to think of the preceeding two paragraphs. But if
you're using GR in this sense as relativity with acceleration, then OF
COURSE electrodynamics would fail when accelerating charges in special
relativity because accelerating charges require acceleration.
>
>>> In effect an acceleration problem *handled* by SR is a
>>> knowledgeable simplification and specialization of general
>>> procedures prescribed by GR, as Einstein discovered.
>>
>>Boosts and rotations are part of the Lorentz group. Acceleration is a
>>succession of boosts.
>>As long as you stay with the Lorentz group and
>>don't concern yourself with the stress-energy tensor as a source of
>>curvature, you're doing special relativity. That there's an
>>equivalence between gravitation and an accelerated reference frame
>>doesn't mean you need a theory of gravity to handle an accelerated
>>reference frame. But gravity doesn't let you use just one accelerated
>>reference frame because it has different strengths and directions in
>>different parts of space, and it needs to postulate a relation between
>>the acceleration and a source of gravity.
>>
>>Working with accelerated frames in special relativity may bring up
>>some mathematical apparatus like Christoffel symbols, and that will
>>cause some people to immediately shout "Curvature! General
>>relativity!" But you get Christoffel symbols when you work in
>>accelerated frames, or non-Cartesian coordinates, in Newtonian
>>mechanics, too.
>
>In view of your statement, can you provide the subsitutions into
>the Christoffel symbol that account for centrifugal force without
>using the geodesical equation DU^u =0?
Can you find the shortest path between two points in plane polar
coordinates without using the geodesic equation? Does that make it GR?
Do you remember where the geodesic equation came from? The path length
between two points is given by
s = int sqrt(g_uv dx^u dx^v)
That's not general relativity or special relativity or anything, it's the
rule for finding a distance along a path in any metric space.
Parameterize the path.
s = int sqrt(g_uv dx^u/d(tau) dx^v/d(tau)) d(tau)
It's still not any kind of relativity since the only purpose tau serves so
far is to parameterize the path. Ultimately we'll chose a metric that
gives tau as an invariant length, and call it the proper time.
Now use the variational principle, integrate by parts, just think of s as
the Lagrangian L when you derive the Euler-Lagrange equations. Minimizing
the action is done the same way that minimizing the path length is done.
The stick s (or L) into the Euler-Lagrange equations and solve. Strictly
speaking, it's still not any kind of relativity until we choose a metric
and a physical interpretation.
>In other words, derive the Christoffel metrics from SR
>for centrifugal force.
Define rotating coordinate system
T = t t = T
X = x cos(wt) - y sin(wt) <=> x = X cos(wT) + Y sin(wT)
Y = x sin(wt) + y cos(wt) y = -X sin(wT) + y cos(wT)
Z = z z = Z
Differentiate them all.
dt = dT
dx = dX cos(wT) + dY sin(wT) + dT[-Xw sin(wT) + Yw cos(wT)]
dy = -dX sin(wT) + dY cos(wT) + dT[-Xw cos(wT) - Yw sin(wT)]
dz = dZ
Then plug into the Minkowski metric and group terms to find the same space
described with a different coordinate system.
ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
= (-1+w^2(X^2+Y^2))dT^2 + dX^2 + dY^2 + dZ^2 + 2YwdXdT - 2XwdYdT
Pick off terms to write g_uv as a matrix and stick into the geodesic
equation and you'll get the centrifugal and Coriolis force terms without
introducing any postulates besides those of special relativity and without
explicitly using Christoffel symbols. But the information in the
Christoffel symbols is, of course, in there because the basis vectors are
not Cartesian.
Now if you apply the principle of equivalence and relate curvature to
sources with G=8piT, *then* you're doing general relativity.
Ken S. Tucker
(( and all these points will be donated to my local
whorehouse, so they may teach me the sin of my ways))
> > In 1911, in the same Dover book, Einstein published,
> > "ON THE INFLUENCE OF GRAVITATION ON THE
> > PROPAGATION OF LIGHT", where he attempted to use
> > Special Relativity to predict the bending of light in accelerating
> > frames, and got the wrong answer.
> > This indicates one needs to know GR before one can
> > simplify acceleration problems to be *handled* by SR.
> > In effect an acceleration problem *handled* by SR is a
> > knowledgeable simplification and specialization of general
> > procedures prescribed by GR, as Einstein discovered.
>
> A fascinating observation. But since this was not a claim that SR could
> only be used in inertial frames, no points.
What if I beg, begging should count for something.
(pity points????)
> greywolf42
> ubi dubium ibi libertas
No disrespect intended...just a bit of fun
Ken
Dirk Van de moortel
"greywolf42" <min...@sim-ss.com> wrote in message news:vkhq065...@corp.supernews.com...
> shuba <tim....@eudoramail.com> wrote in message
> news:tim.shuba-7A6AF...@corp.supernews.com...
> > Gregory L. Hansen wrote:
> >
> > > In article <vkf8ifq...@corp.supernews.com>,
> > > greywolf42 <min...@sim-ss.com> wrote:
> >
> > > >The following exchange occured on sci.physics.relativity:
> >
> > Typical of the bogus scholarship of the Bastard greywolf, this is
> > incorrect. The exchange did *not* take place in s.p.relativity,
> > though follow-ups were set there by Baez. I have again set the
> > follow-ups there. By the way, the "Tim S" referred to below was
> > not me.
>
> Timmy seems to have a personal problem. The post that I read *was* in SPR.
> However, you have no standing to 'change' followups. Go play in your own
> sandbox.
It was in sci.physics.research. Not in sci.physics.relativity.
What kind of personal problem can *you* possibly have
to lie about this???
Dirk Vdm
greywolf42
Perhaps it does where you want your points donated. :)
> No disrespect intended...just a bit of fun
Nothing wrong with that.
Bill Hobba
> The essential difference that GR recognizes that SR fails to
> respect is the positional variation as a result of the spatial
> metrics differnces within a field that are provided by
> g11 ....g33. These metrics are physically proven by the
> experimental results derived by the study of EM radiation
> bent by the sun.
> At this point, we should call into jury, or... respect those
> reported results.
I would not call that the essential difference. In GR SR still applies
locally - as I said. The essential difference between GR and SR lies in the
principle of general invariance. The POR applies only to inertial reference
frames but the principle of general invariance applies to all coordinate
systems. From this principle we see that the metric must be a dynamical
variable to have 'no prior geometry' and the EFE's follow. An argument
could even be mounted that the difference is that in GR only local inertial
coordinate systems exist. But when you combine the two - local inertial
coordinate systems and the principle of general invariance then GR naturally
results.
Thanks
Bill
Bill Hobba
> Sigh. SR *does* deal with acceleration. Acceleration relative to
> inertial frame is acceleration.
Sorry Mati I must disagree with you there. SR can used to deal with
coordinate systems accelerated WRT to inertial reference frames. It does
this by noting that during an infinitesimal time period the accelerated
frame can be considered inertial so SR can be used. Dirk in another post
provided the full mathematical detail for this and I fully agree and retract
any statement I may have made that SR can not be used to deal with
acceleration. But by the very definition of inertial reference frames and
the POR SR does not deal directly with accelerated reference frames - it is
only by noting the above trick or others similar that it can be done.
Thanks
Bill
Bill Hobba
> SR deals with the constant velocity of *light*.
> All other sci.relativity sci.fiction about
> math concepts about gravity is bullshit.
And your view of the POR would be bullshit I suppose. Go read a good book
on Relativity like Rindler - Introduction to Special Relativity especially
page 19 where the role of the second axiom (ie the speed of light postulate)
is discussed is detail. After reading that; post where you believe
Rindler's error is when he states the second axiom is only required to fix
the value of speed that naturally occurs from the POR alone.
Thanks
Bill
Bill Hobba
> Mati Meron wrote:
> > Sigh. SR *does* deal with acceleration. Acceleration relative to
> > inertial frame is acceleration.
>
Bill Hobba replied:
> Sorry Mati I must disagree with you there. SR can used to deal with
> coordinate systems accelerated WRT to inertial reference frames. It does
> this by noting that during an infinitesimal time period the accelerated
> frame can be considered inertial so SR can be used. Dirk in another post
> provided the full mathematical detail for this and I fully agree and
retract
> any statement I may have made that SR can not be used to deal with
> acceleration. But by the very definition of inertial reference frames and
> the POR SR does not deal directly with accelerated reference frames - it
is
> only by noting the above trick or others similar that it can be done.
>
On further reflection my response may be degenerating this discussion into a
semantic issue which I am loath to do as it gets no one nowhere. If 'SR
*does* deal with acceleration' means it can be used to analyze accelerated
reference frames then my response indicates I am logically forced to agree.
If however it is meant to indicate that SR directly deals with acceleration
then I disagree.
Thanks
Bill
me...@cars3.uchicago.edu
> Mati Meron wrote:
>> Sigh. SR *does* deal with acceleration. Acceleration relative to
>> inertial frame is acceleration.
>
>Sorry Mati I must disagree with you there. SR can used to deal with
>coordinate systems accelerated WRT to inertial reference frames. It does
>this by noting that during an infinitesimal time period the accelerated
>frame can be considered inertial so SR can be used.
Yes, for sure.
> Dirk in another post
>provided the full mathematical detail for this and I fully agree and retract
>any statement I may have made that SR can not be used to deal with
>acceleration. But by the very definition of inertial reference frames and
>the POR SR does not deal directly with accelerated reference frames - it is
>only by noting the above trick or others similar that it can be done.
>
Certainly, no argument (though said "trick" is really no different
from what we're doing in Newtonian mechanics, there too inertial
frames are the base reference). Note, however, what I wrote i.e.
"acceleration relative to inertial frame..." No mention of "accelerated
frame". What I'm talking about is the treatment of accelerated motion
relative to inertial frame. And SR has no problem with this.
acceleration relative to inertial frame =! motion relative to
accelerated frame.
Bill Hobba
> > Mati Meron wrote:
> >> Sigh. SR *does* deal with acceleration. Acceleration relative to
> >> inertial frame is acceleration.
> >
Bill Hobba repleid
> >Sorry Mati I must disagree with you there. SR can used to deal with
> >coordinate systems accelerated WRT to inertial reference frames. It does
> >this by noting that during an infinitesimal time period the accelerated
> >frame can be considered inertial so SR can be used.
>
Mati Meron wrote
> Yes, for sure.
>
Bill Hobba wrote:
> > Dirk in another post
> >provided the full mathematical detail for this and I fully agree and
retract
> >any statement I may have made that SR can not be used to deal with
> >acceleration. But by the very definition of inertial reference frames
and
> >the POR SR does not deal directly with accelerated reference frames - it
is
> >only by noting the above trick or others similar that it can be done.
> >
Mati Meron wrote
> Certainly, no argument (though said "trick" is really no different
> from what we're doing in Newtonian mechanics, there too inertial
> frames are the base reference). Note, however, what I wrote i.e.
> "acceleration relative to inertial frame..." No mention of "accelerated
> frame". What I'm talking about is the treatment of accelerated motion
> relative to inertial frame. And SR has no problem with this.
>
> acceleration relative to inertial frame =! motion relative to
> accelerated frame.
>
It is always a pleasure to discuss an issue when both sides end up in
complete agreement. This seems to be the case here. I appreciated the
discussion and feel my understanding in the area is now a bit better.
Thanks
Bill
ZZBunker
I've already finished reading all the books on Relativity,
I'm ever going to read, 20 years ago.
But, since the Equivalence Posulate follows from
bullshit and vapor, it's just as naturally irrelevent.
> Thanks
> Bill
Ken S. Tucker
>In article <2202379a.03082...@posting.google.com>,
>Ken S. Tucker <dyna...@vianet.on.ca> wrote:
>>glha...@indiana.edu (Gregory L. Hansen) wrote in message
>>news:<8ce5c97e.03082...@posting.google.com>...
>>
>>>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03082...@posting.google.com>...
>>>> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
>>news:<bi8c8u$56h$1...@hood.uits.indiana.edu>...
>>
>>Hi Gregory
>>Read your note several times, thank you, and posted
>>comments within context. But first I'd like to have some
>>cooperation on dilineating the boundary of GR and SR.
>> In my view, the covariant acceleration is described by
>>GR 's geodesic equation given by the familiar form,
>>
>>dU^u/ds = GAMMA^u_ab U^a U^b
>>
>>(U^u = dx^u/ds),
>>
>>and this equation is GR, and not SR, and delineates
>>the difference, because it is generally covariant.
>>
>>This equation corelates with the *absolute derivative*
>>
>>DU^u = 0 ,
>
>Why would you make that the distinction between SR and GR?
In SR, Eq. GAMMA^u_ab=0 always, hence dU^u/ds=0.
In GR, Eq. GAMMA^u_ab =/=0 always.
> There's
>nothing about it that conflicts with the postulates of special relativity,
Of course that's true.
>and using it does not require additional postulates.
I think it does, GR postulates *no absolute acceleration
exists*, (in the equation DU^u=0), SR (also known as
Restricted Relativity) is restricted to *absolute uniform
motion does not exist*.
>It doesn't even describe acceleration unless you're in an
>accelerated frame.
What? How about freefall...
>The geodesic
>equation is also used in *Newtonian* mechanics! The terms on the right
>disappear in Cartesian coordinates, but try finding the geodesic in polar
>coordinates. The geodesic in a rotating frame will give you the Coriolis
>force terms. It's just geometry. What is the shortest path between two
>points on a sphere? That's not a general relativity problem, it's just
>the geometry of a surface in Euclidean space.
>
>For that matter, I can plug a metric like
>
> ds^2 = dt^2 + dx^2 - dy^2 - dz^2
>
>into the geodesic equation and it won't be special relativity or general
>relativity, it won't be anything.
>
>The student typically doesn't see that math until he starts learning
>general relativity. But general relativity is one of many applications of
>differential geometry.
So what, Plancks E=hf is an algebra application. I getting
worried that you are dismissing the physical meaning of the
content of the equations.
>Some authors do seem to *define* general relativity as relativity with
>acceleration. For instance, R. H. Good in "Uniformly accelerated
>reference frame and the twin paradox" in Am. J. Phys. 50, 232 (1982). He
>starts with the Lorentz transformations to find a composition of
>velocities, takes a derivative of that to find acceleration, integrates
>along paths... and at the end he says "When general relativity is
>included in analysis of the traveling twin's trip..." But he never
>brought any postulate of general relativity into the analysis. He didn't
>use the equivalence principle, he didn't use G=8piT. All he did was
>manipulate the Lorentz transformations in a space with a flat metric. So
>calling that general relativity is just confused and spurious. The theory
>is defined by the postulates, not by which problem you apply the
>postulates to.
Agreed.
>>(DU^u = U^u;v U^v where ';' is covariant differentiation)
>>It is always possible to find a Coordinate System (CS)
>>where U^u =0 therefore DU^u =0 is always true.
Essentially SR postulates U^u=0 in some FoR and GR
postulates DU^u=0 in this FoR, which is really SR with
the application of differential calculus. However,
Because the integral of DU^u=0 given by,
U^u = $ (DU^u = 0) ds =/=0 always,
because of the integration constant, then
U^u =0 is not true in general. A case in point
is when the constant of integration = c. SR fails here
while GR's DU^u=0 holds. SR's postulates provide
the wrong answers regarding the motion of light.
>>What I just wrote is the General Theory of Relativity,
>>(gravitation is an application).
[snip]
>>Not just gravity, electrodynamics fails when accelerating
>>charges in Special Relativity. (See Dovers P of R,
>>ELECTRODYNAMICS, chp 10, entitled 'Dynamics of
>>the Slowly Accelerated Electron".)
>> In view of the Quantum Theory, the listed accelerations
>>derived by SR are wrong. However in GR, in view of
>>DU^u=0 these are correct, and are QT compatible.
>>(invariant mass m*DU^u=0 = DE^u=0, meaning NO
>>continuous change in energy is measureable).
>
>I have no idea what to think of the preceeding two paragraphs.
I provided a reference where SR failed to properly treat
accelerations, and showed how GR's DU^u=0 does do it
correctly.
>But if
>you're using GR in this sense as relativity with acceleration, then OF
>COURSE electrodynamics would fail when accelerating charges in special
>relativity because accelerating charges require acceleration.
I think we agree here.
[snip]
>>>Working with accelerated frames in special relativity may bring up
>>>some mathematical apparatus like Christoffel symbols, and that will
>>>cause some people to immediately shout "Curvature! General
>>>relativity!" But you get Christoffel symbols when you work in
>>>accelerated frames, or non-Cartesian coordinates, in Newtonian
>>>mechanics, too.
>>In view of your statement, can you provide the subsitutions into
>>the Christoffel symbol that account for centrifugal force without
>>using the geodesical equation DU^u =0?
>
>Can you find the shortest path between two points in plane polar
>coordinates without using the geodesic equation? Does that make it GR?
>
>Do you remember where the geodesic equation came from?
As I explained above from DU^u =0.
>The path length between two points is given by
> s = int sqrt(g_uv dx^u dx^v)
>That's not general relativity or special relativity or anything, it's the
>rule for finding a distance along a path in any metric space.
>Parameterize the path.
> s = int sqrt(g_uv dx^u/d(tau) dx^v/d(tau)) d(tau)
>It's still not any kind of relativity since the only purpose tau serves so
>far is to parameterize the path. Ultimately we'll chose a metric that
>gives tau as an invariant length, and call it the proper time.
When you call it proper time you are giving this physical content,
by equating a heretofore mathematical theory physical significance.
>Now use the variational principle, integrate by parts, just think of s as
>the Lagrangian L when you derive the Euler-Lagrange equations. Minimizing
>the action is done the same way that minimizing the path length is done.
>Then stick s (or L) into the Euler-Lagrange equations and solve. Strictly
>speaking, it's still not any kind of relativity until we choose a metric
>and a physical interpretation.
>
>>In other words, derive the Christoffel metrics from SR
>>for centrifugal force.
These next answers of yours are very cool...nice, thanks...
>Define rotating coordinate system
>
> T = t t = T
> X = x cos(wt) - y sin(wt) <=> x = X cos(wT) + Y sin(wT)
> Y = x sin(wt) + y cos(wt) y = -X sin(wT) + y cos(wT)
> Z = z z = Z
>
>Differentiate them all.
>
> dt = dT
> dx = dX cos(wT) + dY sin(wT) + dT[-Xw sin(wT) + Yw cos(wT)]
> dy = -dX sin(wT) + dY cos(wT) + dT[-Xw cos(wT) - Yw sin(wT)]
> dz = dZ
>
>Then plug into the Minkowski metric and group terms to find the same space
>described with a different coordinate system.
>
> ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
>
> = (-1+w^2(X^2+Y^2))dT^2 + dX^2 + dY^2 + dZ^2 + 2YwdXdT - 2XwdYdT
>
>Pick off terms to write g_uv as a matrix and stick into the geodesic
>equation and you'll get the centrifugal and Coriolis force terms without
>introducing any postulates besides those of special relativity and without
>explicitly using Christoffel symbols.
Well I can't put the metrics in the geodesic without using Christoffel
symbols.
Ok, this is intersting, patiently correct me if I'm wrong, assume
the metrics components are covariant, (saves writing g_11),
and Z and dZ are zilch, then,
R^2 = X^2 + Y^2,
g00 = -(1- w^2 R^2), g11 = g22 =1, g10 =Yw, g20=-Xw
(assuming g10 and g20 are symmetrical ie g10 = g01)
Right?
>Now if you apply the principle of equivalence and relate curvature to
>sources with G=8piT, *then* you're doing general relativity.
I can do that with a rocket engine slightly increasing the
acceleration on an elevator. The elevator frame can't
distinguish between an increasing rocket thrust OR if
the elevator FoR is in uniform motion moving downward
in a g-field, and thus experincing increasing inertia.
While these are the same measurements, SR has only
one conclusion, ie, thrust, while GR accepts either.
Given that every point in the universe is in free-fall,
SR applications have very little application.
Regards
Ken S. Tucker
me...@cars3.uchicago.edu
>
>> > Mati Meron wrote:
>> >> Sigh. SR *does* deal with acceleration. Acceleration relative to
>> >> inertial frame is acceleration.
>> >
>
>> >Sorry Mati I must disagree with you there. SR can used to deal with
>> >coordinate systems accelerated WRT to inertial reference frames. It does
>> >this by noting that during an infinitesimal time period the accelerated
>> >frame can be considered inertial so SR can be used.
>>
>
>> Yes, for sure.
>>
>
>Bill Hobba wrote:
>> >provided the full mathematical detail for this and I fully agree and
>retract
>> >any statement I may have made that SR can not be used to deal with
>> >acceleration. But by the very definition of inertial reference frames
>and
>> >the POR SR does not deal directly with accelerated reference frames - it
>is
>> >only by noting the above trick or others similar that it can be done.
>> >
>
>> Certainly, no argument (though said "trick" is really no different
>> from what we're doing in Newtonian mechanics, there too inertial
>> frames are the base reference). Note, however, what I wrote i.e.
>> "acceleration relative to inertial frame..." No mention of "accelerated
>> frame". What I'm talking about is the treatment of accelerated motion
>> relative to inertial frame. And SR has no problem with this.
>>
>> acceleration relative to inertial frame =! motion relative to
>> accelerated frame.
>>
>
>complete agreement. This seems to be the case here. I appreciated the
>discussion and feel my understanding in the area is now a bit better.
>
sacred traditions of this newsgroup, right?:-)
Jeffery
> In article <BB68376D...@timsilverman.demon.co.uk>,
> Tim S <T...@timsilverman.demon.co.uk> wrote:
>
> >I'm puzzled.
> >
> >Why do so many people say that special relativity can't handle non-inertial
> >reference frames? Or, even more amazingly, that it can't handle acceleration
> >at all (hence presumably involves only free particles and/or fields, with no
> >forces)?
>
> upon a time, and the error found its way into a bunch
> of popularizations and maybe even some textbooks. A
> this mistake, but this appears to be impossible - it's
> one of those memes that won't die. It would be interesting
> (but depressing) to study the history of this mistake and
> its spread.
>
> One way people spread this mistake is by telling a superficially
> plausible story like this:
>
> [TOTAL BALONEY FOLLOWS:]
>
> "Einstein came up with special relativity, which handles
> only unaccelerated frames of reference [or worse: unaccelerated
> motion]. Then he realized that an accelerated frame of reference
> is indistinguishable from a gravitational field - his famous
> `elevator' thought experiment. This led him to invent general
> relativity, which handles general frames of reference and gravity."
>
> This is superficially plausible, in part because even Einstein
> was a bit mixed up about this stuff at first: he made up
> the term "general relativity" in part to emphasize the
> "general covariance" of the theory, i.e. the invariance
> of the equation of motion under arbitrary coordinate
> transformations. But in fact, special relativity is
> already "generally covariant" in this sense! What matters
> is that special relativity is not a "background-free" theory.
> This distinction has taken a long time to sink in. See
>
> http://math.ucr.edu/home/baez/background.html
>
> for more.
>
> It's mildly annoying when people say special relativity
> can only handle unaccelerated (= inertial) frames of
> reference. But it goes beyond annoying into hilarious
> when people say it's unable to handle accelerated motion.
> If that were true, it couldn't handle the concept of force!
> It would be pretty much useless.
>
> (I'll set followups to a more accelerated newsgroup.)
In your following article
http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#14
you write
"Indeed, the concept of velocity is not a very useful one in general
relativity and this makes it difficult to define what 'faster than
light' means."
and this is your explanation for why the moon is not going faster than
light if you spin around, because if you can't strictly define
velocity then you can't say the Moon is going faster than light, but
then now you're saying that special relativity can easily handle
acceleration and non-inertial reference frames. If special relativity
can handle acceleration, then general relativity could handle
velocity, so what is the explanation for the so-called Moon paradox?
Jeffery Winkler
Gregory L. Hansen
> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<bib6ss$40h$1...@hood.uits.indiana.edu>...
> >In article <2202379a.03082...@posting.google.com>,
> >Ken S. Tucker <dyna...@vianet.on.ca> wrote:
> >>glha...@indiana.edu (Gregory L. Hansen) wrote in message
> >>news:<8ce5c97e.03082...@posting.google.com>...
> >>
> >>>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03082...@posting.google.com>...
> >>>> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
> >>news:<bi8c8u$56h$1...@hood.uits.indiana.edu>...
> >>
> >>Hi Gregory
> >>Read your note several times, thank you, and posted
> >>comments within context. But first I'd like to have some
> >>cooperation on dilineating the boundary of GR and SR.
> >> In my view, the covariant acceleration is described by
> >>GR 's geodesic equation given by the familiar form,
> >>
> >>dU^u/ds = GAMMA^u_ab U^a U^b
> >>
> >>(U^u = dx^u/ds),
> >>
> >>and this equation is GR, and not SR, and delineates
> >>the difference, because it is generally covariant.
> >>
> >>This equation corelates with the *absolute derivative*
> >>
> >>DU^u = 0 ,
> >
> >Why would you make that the distinction between SR and GR?
>
> In SR, Eq. GAMMA^u_ab=0 always, hence dU^u/ds=0.
No, it's not.
> In GR, Eq. GAMMA^u_ab =/=0 always.
In SR, the laws of physics have the same form in any inertial
reference frame and the speed of light is the same in any inertial
reference frame. Anything that uses those postulates, doesn't
conflict with them, and doesn't require additional postulates is SR.
Otherwise you're redefining the theory.
In SR, Gamma^u_ab=0 if and *only* if you're using Cartesian
coordinates. In Newtonian mechanics, Gamma^u_ab=0 if and *only* if
you're using Cartesian coordinates. Non-zero Christoffel symbols mean
the magnitude and/or direction of your basis vectors aren't constant
in all space, so you have to take derivatives of your basis vectors
along with your vector components. That's true, for instance, of
curvilinear coordinates in any system of mechanics.
Do these exercises.
Calculate the Christoffel symbols for plane polar coordinates, with
the metric
ds^2 = dr^2 + r^2 d(theta)^2
(Answer: Gamma^r_tt=-r, Gamma^t_rt=Gamma^t_tr=1/r, others zero, where
t stands in for theta.)
There's no time in there because it's not mechanics, it's just plane
geometry expressed in polar coordinates.
Now use the geodesic equation to find the equations describing a
straight line in plane polar coordinates.
Now write out the free particle Lagrangian in plane polar coordinates,
L = 1/2 m ((dr/dt)^2 + r^2(d theta/dt)^2)
crank it through the Euler-Lagrange equations, and see what you get.
You should get exactly the same differential equations as the geodesic
equation gave you, from a Newtonian problem. And it's not
coincidence. A particle, in the absence of applied forces, goes in a
straight line. When you plug the Lagrangian for a free particle into
the Euler-Lagrange equations, that *is* the geodesic equation! It's
not just similar to, it's the same darn thing with identical
derivation, scaled by m/2 which cancels out because the right-hand
side equals zero.
For your next exercise, if you're not tired of this yet, find the
curvature tensor of the surface of a sphere in three dimensional
Euclidean space, and think of a Newtonian problem to apply it to.
Confine a marble to the surface and let it fall under gravity or
something.
What I hope to make clear by that is there's nothing uniquely general
relativistic or Einsteinian at all about Christoffel symbols or
geodesic equations or even curvature tensors. They are mathematical
methods that can be applied to many things, general relativity is only
one of them. They tend to be introduced to the student only when he
studies general relativity because in any study besides general
relativity the motion of a free particle is trivially resolved with
Cartesian coordinates, but you can't use Cartesian coordinates in GR.
But that doesn't make any of that stuff unique to GR, it just means
the problems you've been given in other classes have involved simple
geometry.
Can you think of any problem in special relativity that you've done or
seen in a problem set that involved non-Cartesian coordinates?
Probably not, it's just not done. That doesn't mean you can't do
special relativity in curvilinear coordinates. In fact, it's very
straight-forward. It just means that most students haven't seen it.
So I think some students can confuse parts of pure geometry with
physical theory when they're seen for the first time introduced
together.
Simple example of curvilinear coordinates in special relativity: a
clock in a centrifuge. I made this one up because I haven't seen
curvilinear coordinates used in SR, either. The metric is
ds^2 = c^2 dt^2 - dr^2 - r^2 d(theta)^2
r=R=constant, and the centrifuge spins with angular velocity
w=d(theta)/dt in the lab frame. Find the rate of the spinning clock
versus a lab clock. With velocity v=wR, relate it to the traditional
formula for time dilation.
I took a class in classical many-body physics. I didn't get much out
of it, but the prof spent the first half semester developing (in
modern form) the differential geometry that deals with curved
surfaces, and then we spent a lot of time using Liouville's theorem on
the constant-energy manifold in phase space. That's a Newtonian
application on a manifold that's not even supposed to directly
represent space and time.
>
> > There's
> >nothing about it that conflicts with the postulates of special relativity,
>
> Of course that's true.
>
> >and using it does not require additional postulates.
>
> I think it does, GR postulates *no absolute acceleration
> exists*,
That's a derived result.
> (in the equation DU^u=0), SR (also known as
> Restricted Relativity) is restricted to *absolute uniform
> motion does not exist*.
>
> >It doesn't even describe acceleration unless you're in an
> >accelerated frame.
>
> What? How about freefall...
How about a metric that doesn't describe freefall? The geodesic
equation tells you all kinds of wonderful general relativistic things
when you put a general relativistic metric into it. You can put a
special relativistic or Newtonian or plane geometry metric into it and
it will tell you things that have nothing at all to do with general
relativity. It applies to all Riemannian spaces. You get general
relativistic metrics from Einstein's field equations, but you can get
metrics from lots of other places, too.
In special relativity, any metric you use is a coordinate transform
away from
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
If you're boosting or rotating you may have t=(x,t;a,v),
x=x(y,z,t;a,v), etc.,
or something like that, but in the end you can get your new metric by
expressing the new coordinate system in terms of x,y,z,t and plugging
them into the relation above.
And I would say that distinguishes SR from GR.
Stupid relativity trick: Use the Lorentz transforms to boost in the x
direction and calculate the metric of that coordinate system. It
wasn't what I expected, but made perfect sense in retrospect, given
the invariance of the interval.
[...]
> >Do you remember where the geodesic equation came from?
>
> As I explained above from DU^u =0.
>
> >The path length between two points is given by
> > s = int sqrt(g_uv dx^u dx^v)
> >That's not general relativity or special relativity or anything, it's the
> >rule for finding a distance along a path in any metric space.
> >Parameterize the path.
> > s = int sqrt(g_uv dx^u/d(tau) dx^v/d(tau)) d(tau)
> >It's still not any kind of relativity since the only purpose tau serves so
> >far is to parameterize the path. Ultimately we'll chose a metric that
> >gives tau as an invariant length, and call it the proper time.
>
> When you call it proper time you are giving this physical content,
> by equating a heretofore mathematical theory physical significance.
I haven't called it proper time yet. tau is simply a parameter. It
could be a path length, if you like. Which is really what proper time
is in Minkowski's geometry, but that's beside the point.
>
> >Now use the variational principle, integrate by parts, just think of s as
> >the Lagrangian L when you derive the Euler-Lagrange equations. Minimizing
> >the action is done the same way that minimizing the path length is done.
> >Then stick s (or L) into the Euler-Lagrange equations and solve. Strictly
> >speaking, it's still not any kind of relativity until we choose a metric
> >and a physical interpretation.
> >
> >>In other words, derive the Christoffel metrics from SR
> >>for centrifugal force.
>
> These next answers of yours are very cool...nice, thanks...
It was a homework problem that I found particularly delightful.
Let me take the lazy way out and look it up... Yes, that's right.
>
> >Now if you apply the principle of equivalence and relate curvature to
> >sources with G=8piT, *then* you're doing general relativity.
>
> I can do that with a rocket engine slightly increasing the
> acceleration on an elevator. The elevator frame can't
> distinguish between an increasing rocket thrust OR if
> the elevator FoR is in uniform motion moving downward
> in a g-field, and thus experincing increasing inertia.
> While these are the same measurements, SR has only
> one conclusion, ie, thrust, while GR accepts either.
They are, after all, different theories. Or rather one a subset of
the other.
> Given that every point in the universe is in free-fall,
> SR applications have very little application.
Eh... SR is reduced to engineering practice. The only actual
application of GR so far, besides science and more science (testing GR
and applications in astronomy and cosmology), is the GPS system.
That's why GR remains an esoteric field that some physics departments
only teach every other year or even not at all. At both the
University of Minnesota and Indiana University my schedule was phased
such that the class was never offered when I'd have been able to take
it, although I wanted to. I got the homework problem from problem
sets and solutions posted on the web at IU after I'd finished classes
and moved out of state to do research.
Ken S. Tucker
>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03082...@posting.google.com>...
>> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<bib6ss$40h$1...@hood.uits.indiana.edu>...
>> >In article <2202379a.03082...@posting.google.com>,
>> >Ken S. Tucker <dyna...@vianet.on.ca> wrote:
[snip]
>> >Why would you make that the distinction between SR and GR?
>> In SR, Eq. GAMMA^u_ab=0 always, hence dU^u/ds=0.
>
>No, it's not.
Read (as in red) through your post several times, and think
I understand the basis of the differences in our points of view.
I'm using metrics measured and defined by the velocity of light,
which of course is a constant velocity c in SR. In Relativity,
this is the only acceptable means of defining metrics. This
leads to GAMMA^u_ab=0 in SR. Of course c is not a
constant in GR, (light deflects).
Your arguments define the metric by other means, in
applications that employ assumptions about the metric,
which is fine, if this is known.
>> In GR, Eq. GAMMA^u_ab =/=0 always.
>
>In SR, the laws of physics have the same form in any inertial
>reference frame and the speed of light is the same in any inertial
>reference frame. Anything that uses those postulates, doesn't
>conflict with them, and doesn't require additional postulates is SR.
>Otherwise you're redefining the theory.
>In SR, Gamma^u_ab=0 if and *only* if you're using Cartesian
>coordinates. In Newtonian mechanics, Gamma^u_ab=0 if and *only* if
>you're using Cartesian coordinates. Non-zero Christoffel symbols mean
>the magnitude and/or direction of your basis vectors aren't constant
>in all space, so you have to take derivatives of your basis vectors
>along with your vector components. That's true, for instance, of
>curvilinear coordinates in any system of mechanics.
Your slowly beginning to use 3 dimensional thinking in a
Minkowski 4D Coordinate System (CS) required by SR.
>Do these exercises.
>Calculate the Christoffel symbols for plane polar coordinates, with
>the metric
>
> ds^2 = dr^2 + r^2 d(theta)^2
>
>(Answer: Gamma^r_tt=-r, Gamma^t_rt=Gamma^t_tr=1/r, others zero, where
>t stands in for theta.)
>There's no time in there because it's not mechanics, it's just plane
>geometry expressed in polar coordinates.
Right, of course, but this is not relativity, it's 3D tensor analysis,
and does not apply.
>Now use the geodesic equation to find the equations describing a
>straight line in plane polar coordinates.
>Now write out the free particle Lagrangian in plane polar coordinates,
> L = 1/2 m ((dr/dt)^2 + r^2(d theta/dt)^2)
>crank it through the Euler-Lagrange equations, and see what you get.
>You should get exactly the same differential equations as the geodesic
>equation gave you, from a Newtonian problem. And it's not
>coincidence. A particle, in the absence of applied forces, goes in a
>straight line. When you plug the Lagrangian for a free particle into
>the Euler-Lagrange equations, that *is* the geodesic equation! It's
>not just similar to, it's the same darn thing with identical
>derivation, scaled by m/2 which cancels out because the right-hand
>side equals zero.
>For your next exercise, if you're not tired of this yet, find the
>curvature tensor of the surface of a sphere in three dimensional
>Euclidean space, and think of a Newtonian problem to apply it to.
>Confine a marble to the surface and let it fall under gravity or
>something.
This is getting far from relativity.
>What I hope to make clear by that is there's nothing uniquely general
>relativistic or Einsteinian at all about Christoffel symbols or
>geodesic equations or even curvature tensors. They are mathematical
>methods that can be applied to many things, general relativity is only
>one of them. They tend to be introduced to the student only when he
>studies general relativity because in any study besides general
>relativity the motion of a free particle is trivially resolved with
>Cartesian coordinates, but you can't use Cartesian coordinates in GR.
>But that doesn't make any of that stuff unique to GR, it just means
>the problems you've been given in other classes have involved simple
>geometry.
>Can you think of any problem in special relativity that you've done or
>seen in a problem set that involved non-Cartesian coordinates?
Yes, I habitually use nonorthogonal geometry (g01 =/=0) as you
have used below.
>Probably not, it's just not done. That doesn't mean you can't do
>special relativity in curvilinear coordinates. In fact, it's very
>straight-forward. It just means that most students haven't seen it.
>So I think some students can confuse parts of pure geometry with
>physical theory when they're seen for the first time introduced
>together.
Right, as I mentioned above, metrics correlate to the velocity
of light.
>Simple example of curvilinear coordinates in special relativity: a
>clock in a centrifuge. I made this one up because I haven't seen
>curvilinear coordinates used in SR, either. The metric is
>
> ds^2 = c^2 dt^2 - dr^2 - r^2 d(theta)^2
>
>r=R=constant, and the centrifuge spins with angular velocity
>w=d(theta)/dt in the lab frame. Find the rate of the spinning clock
>versus a lab clock. With velocity v=wR, relate it to the traditional
>formula for time dilation.
Ok, let me try...
r^2 d(theta)^2 = r^2 d(theta/dt)^2 dt^2 = r^2 w^2 dt^2
= v^2 dt^2, and dr^2=0 so,
ds^2 = c^2 dt^2 - v^2 dt^2.
Barring math error on my part, the metric is erroneous,
because a spinning clock clearly requires a non-euclidean
CS and the derived result is Minkowskian.
>I took a class in classical many-body physics. I didn't get much out
>of it, but the prof spent the first half semester developing (in
>modern form) the differential geometry that deals with curved
>surfaces, and then we spent a lot of time using Liouville's theorem on
>the constant-energy manifold in phase space. That's a Newtonian
>application on a manifold that's not even supposed to directly
>represent space and time.
[snip]
>> I think it does, GR postulates *no absolute acceleration
>> exists*,
>That's a derived result.
What's your point, arguably every theory is a derived
result.
[snip]
>> >It doesn't even describe acceleration unless you're in an
>> >accelerated frame.
>>
>> What? How about freefall...
>
>How about a metric that doesn't describe freefall? The geodesic
>equation tells you all kinds of wonderful general relativistic things
>when you put a general relativistic metric into it. You can put a
>special relativistic or Newtonian or plane geometry metric into it and
>it will tell you things that have nothing at all to do with general
>relativity. It applies to all Riemannian spaces. You get general
>relativistic metrics from Einstein's field equations, but you can get
>metrics from lots of other places, too.
Ok, that's were we agree.
[snip]
>> These next answers of yours are very cool...nice, thanks...
>
>It was a homework problem that I found particularly delightful.
Yes, we share this delight, perhaps I can entertain you.
Check out these last two metrics and see that,
g10/Y = w, g20/X = -w and
(g10/Y = - g20/X) == (g10/x2 = - g20/x1)
which is antisymmetrical in x and y.
I'm quite aware that nothing is proven in tensor calculus,
after a specialization of a CS, so allow an illustration.
In Eq. g10/x2 = - g20/x1 the indices 10 and 20 are
symmetrical. ie, g10 = g01 and g20 = g02, provided
each is performed on both sides. So,
g12/x0 = - g21/x0, by illustration.
It follows g12 = - g21 is suggested by illustration.
This gets a bit tough but you seem qualified...
Suppose the earth is in a circular orbit around the sun, (xy plane),
with an orbital speed V, then we should expect an aberration of
V/c.
Assume axis x connects the Sun and Earth and does NOT
rotate. This is important because we are taking a CS that
we usually assume to be rotating (because of our notion
that earth revolves about the Sun) and specify this CS to
be at rest. This is an application of General Covariance.
The aberration of light (and gravity) is given by,
g12 = (y/r)*(dx/cdt) - (x/r)*(dy/cdt).
to first order as measured by light-ray paths.
With x=r and y=0, g12 = -dy/cdt, and describes clockwise
motion.
If x=-r then g12 = dy/cdt and is also clockwise motion,
(anti-clockwise motion is given by g21=-g12).
The term -dy/cdt is equal to (orbital velocity)/c, = -V/c,
and accounts for aberration of light and gravity, and
g12 (and g21) account for revolution in either direction.
The quantity g12 is defined by the relation (aberration)
of light signals from the sun to the earth, and so can be
regarded as a component.defining a relating spacetime
metric.
The Earth and Sun remain the same distance apart therefore,
du^i/ds = 0, where u^i is the relative 4-velocity.
There is no specific need to consider covariancy or
contravariancy details for the accuracy required, the
coordinate acceleration from the usual geodesic equation
in direction x can be written (without using "^" and "_"),
du1/ds = 0 = -1/2*g11*(g14,4 + g41,4 - g44,1)*u4*u4
Let g11~1 and u4~1, which provides,
du1/ds = 0 = 1/2*g44,1 - g14,4
(g14 is considered symmetrical herein.)
where
1/2*g44,1 = gravitational acceleration
g14,4 = inertial (centrifugal) acceleration.
Now lets calculate the force in the direction y,
(these are the aberration terms when x=r and y=0),
du2/ds = - 1/2*g21*(2*g14,4 - g44,1)
and use
g12 = (y/r)*(dx/dt ) - (x/r)*(dy/dt), (c=1) .
(since x,y is the orbital plane, for this set-up,
g12= - dy/dt = -V and g21 = V).
Break out the following,
Aberrated gravitation acceleration = 1/2*V*g44,1
(an acceleration in direction y),
This is the term that *presumeably* causes the Earth to
respond to the aberrated position of the sun, except it is
exactly balanced by inertial acceleration.
Aberrated centrifugal acceleration = - V*g14,4
(an acceleration equal in magnitude to gravitational
acceleration in direction -y)
Recall x=r, and the sum of accelerations in the
direction 'y' are,
du2/ds = 1/2*V*g44,1 - V*g14,4 = 0
These are the equal and opposite terms that nullify
the aberrated gravitational acceleration.
The above is a transparent example of the Principle of
Equivalence demonstrated using the geodesic equation
and including the aberrating effects of the rotational
metric (g12).
Frankly, I think it's a tribute to Einstein's GR that such a
complex problem can be so simplified and easily applied.
Ie. applying generally covariant equations (the geodesic)
involving gravity, inertia and rotation be expressed more
clearly, and applied so directly to find speed of gravity=c.
>> >Now if you apply the principle of equivalence and relate curvature to
>> >sources with G=8piT, *then* you're doing general relativity.
>>
>> I can do that with a rocket engine slightly increasing the
>> acceleration on an elevator. The elevator frame can't
>> distinguish between an increasing rocket thrust OR if
>> the elevator FoR is in uniform motion moving downward
>> in a g-field, and thus experincing increasing inertia.
>> While these are the same measurements, SR has only
>> one conclusion, ie, thrust, while GR accepts either.
>
>They are, after all, different theories. Or rather one a subset of
>the other.
Again agree.
>> Given that every point in the universe is in free-fall,
>> SR applications have very little application.
>
>Eh... SR is reduced to engineering practice. The only actual
>application of GR so far, besides science and more science (testing GR
>and applications in astronomy and cosmology), is the GPS system.
>That's why GR remains an esoteric field that some physics departments
>only teach every other year or even not at all. At both the
>University of Minnesota and Indiana University my schedule was phased
>such that the class was never offered when I'd have been able to take
>it, although I wanted to. I got the homework problem from problem
>sets and solutions posted on the web at IU after I'd finished classes
>and moved out of state to do research.
True, in ordinary engineering GR is certainly rarely required.
I agree that GR is esoteric, but I remain impressed that every
FoR is equal, this sounds like a mathematical solution for
democracy.
I hope this NG provides you with opportunities to discuss
GR issues, if you ever visit Ontario, stop in for a few weeks,
and, as I have found (such in our discourse), there are a few
hundred people seeking understanding (aside from those who
need to get their grants renewed, by any invention possible).
Most of how I learned was by *flash tutorial*, be prepared,
make an appointment with a local prof, and discuss like you're
smart. The prof takes you seriously, enjoys your company,
and wants more stimulation, and arranges for another meeting.
Thanks Again Greg, hope I'm not too boring....
Ken S. Tucker
Sam Wormley
Roshard Davis wrote:
>
> Greetings everyone. I have a question concerning Albert Einstein's
> Theory of Relativity. I heard that the theory is split up into 2
> sections call Special Relativity and General Relativity and I was
> wondering, what is difference between these two and what compelled
> Einstein to create to different sections on this Theory of
> Relativity?Thanks.
Read These:
http://scienceworld.wolfram.com/biography/Einstein.html
http://scienceworld.wolfram.com/physics/SpecialRelativity.html
http://scienceworld.wolfram.com/physics/GeneralRelativity.html
Are There Any Good Books on Relativity Theory?
http://math.ucr.edu/home/baez/physics/Administrivia/rel_booklist.html
Gregory L. Hansen
> glha...@indiana.edu (Gregory L. Hansen) wrote in message news:<8ce5c97e.0308...@posting.google.com>...
>
> >dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03082...@posting.google.com>...
> >> glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<bib6ss$40h$1...@hood.uits.indiana.edu>...
> >> >In article <2202379a.03082...@posting.google.com>,
> >> >Ken S. Tucker <dyna...@vianet.on.ca> wrote:
> [snip]
> >> >Why would you make that the distinction between SR and GR?
>
> >> In SR, Eq. GAMMA^u_ab=0 always, hence dU^u/ds=0.
> >
> >No, it's not.
>
> Read (as in red) through your post several times, and think
> I understand the basis of the differences in our points of view.
> I'm using metrics measured and defined by the velocity of light,
> which of course is a constant velocity c in SR. In Relativity,
> this is the only acceptable means of defining metrics. This
> leads to GAMMA^u_ab=0 in SR.
I feel like you haven't read my posts. Gamma^u_ab is not, in general,
0 in SR. Not in Newtonian mechanics. Not in plane geometry. Not in
anything whatsoever, *except* in the special case that you're using
Cartesian coordinates. I demonstrated that. The Christoffel symbols
are derivatives of the basis vectors, there'll be non-zero Christoffel
symbols whenever you use non-Cartesian coordinates. That has nothing
to do with relativity, it has to do with your coordinate system.
> Of course c is not a
> constant in GR, (light deflects).
Not a constant in SR, either, when viewed from an accelerated frame.
> Your arguments define the metric by other means, in
> applications that employ assumptions about the metric,
> which is fine, if this is known.
No, they don't. The speed of light is constant in an inertial
reference frame. If you transform to an accelerated frame the speed
of light doesn't need to remain constant, but the accelerated frame is
related to inertial frames. What makes it still special relativity is
that you've used the Lorentz transformations to find an accelerated
frame that's equivalent to the inertial frame. Lorentz
transformations form a group, that means they can't get you out of
special relativity no matter how many of them you do.
You know the difference between a vector and the components of a
vector, right? You know that (1,1) and (sqrt(2), pi/4) are the same
vector, the first expressed in Cartesian coordinates and the second
expressed in polar coordinates, right? Changing your coordinate
system doesn't change the physical model, doesn't change the physics,
it's just a different representation of the same thing. If we had
special relativity in one coordinate system, then we still have
special relativity in any coordinate system. I think you're confusing
the representation with the physics.
So your criterion above that in special relativity Gamma^u_ab=0
amounts to an additional and spurious postulate.
1. The laws of physics have the same form in any inertial reference
frame.
2. The speed of light is the same in all inertial reference frames.
3. Transformations and coordinate systems shall be restricted to
inertial and Cartesian.
Number 3 is just not required by the first two, it's something you've
made up. The first two refer specifically to inertial frames, but
those two postulates also determine what happens in accelerated
frames.
My argument distinguishes special from general relativity by the
postulates, because that's what a theory is. And the postulates of
special relativity lead to a metric
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
And you can plug any coordinate system you like into there,
x = x(u,v,w,s)
y = y(u,v,w,s)
z = z(u,v,w,s)
t = t(u,v,w,s)
and you'll still have special relativity because it's just
re-expressing the same metric. And you can use Lorentz
transformations to go to any other coordinate system you like and it's
the same structure of space and time from another point of view (first
postulate).
You can certainly watch, in the context of special relativity, the
acceleration of a rocket from the lab frame. Transforming to the
rocket frame is just a different point of view; if the problem from
the lab frame was special relativity, then the problem from the rocket
frame is also special relativity.
And you can never get a theory of gravity from all that. Not without
additional postulates like those of GR. Even the rocket frame only
looks like gravity locally, not over extended distances.
>
> >> In GR, Eq. GAMMA^u_ab =/=0 always.
> >
> >In SR, the laws of physics have the same form in any inertial
> >reference frame and the speed of light is the same in any inertial
> >reference frame. Anything that uses those postulates, doesn't
> >conflict with them, and doesn't require additional postulates is SR.
> >Otherwise you're redefining the theory.
> >In SR, Gamma^u_ab=0 if and *only* if you're using Cartesian
> >coordinates. In Newtonian mechanics, Gamma^u_ab=0 if and *only* if
> >you're using Cartesian coordinates. Non-zero Christoffel symbols mean
> >the magnitude and/or direction of your basis vectors aren't constant
> >in all space, so you have to take derivatives of your basis vectors
> >along with your vector components. That's true, for instance, of
> >curvilinear coordinates in any system of mechanics.
>
> Your slowly beginning to use 3 dimensional thinking in a
> Minkowski 4D Coordinate System (CS) required by SR.
I've deliberately negelcted time because g_00 is constant, so any
derivative of it goes to zero. But I was trying to show that the
distinction you made above would call a lot of things general
relativity that aren't general relativity.
And doing a Lorentz boost in curvilinear coordinates is a bear.
>
> >Do these exercises.
> >Calculate the Christoffel symbols for plane polar coordinates, with
> >the metric
> >
> > ds^2 = dr^2 + r^2 d(theta)^2
> >
> >(Answer: Gamma^r_tt=-r, Gamma^t_rt=Gamma^t_tr=1/r, others zero, where
> >t stands in for theta.)
> >There's no time in there because it's not mechanics, it's just plane
> >geometry expressed in polar coordinates.
>
> Right, of course, but this is not relativity, it's 3D tensor analysis,
> and does not apply.
Special relativity has three spatial coordinates, so it applies. Time
is trivially added, and doesn't eliminate those non-zero Christoffel
symbols.
>
> >Now use the geodesic equation to find the equations describing a
> >straight line in plane polar coordinates.
> >Now write out the free particle Lagrangian in plane polar coordinates,
> > L = 1/2 m ((dr/dt)^2 + r^2(d theta/dt)^2)
> >crank it through the Euler-Lagrange equations, and see what you get.
> >You should get exactly the same differential equations as the geodesic
> >equation gave you, from a Newtonian problem. And it's not
> >coincidence. A particle, in the absence of applied forces, goes in a
> >straight line. When you plug the Lagrangian for a free particle into
> >the Euler-Lagrange equations, that *is* the geodesic equation! It's
> >not just similar to, it's the same darn thing with identical
> >derivation, scaled by m/2 which cancels out because the right-hand
> >side equals zero.
> >For your next exercise, if you're not tired of this yet, find the
> >curvature tensor of the surface of a sphere in three dimensional
> >Euclidean space, and think of a Newtonian problem to apply it to.
> >Confine a marble to the surface and let it fall under gravity or
> >something.
>
> This is getting far from relativity.
That's the point. It's the same math that you've used to demarcate
special from general relativity, but I'm trying to show that your
criterion says nothing. If I can do the above with geodesic equations
and Christoffel symbols, then I don't see how you can make conditions
with them to distinguish the relativities. I could give you any
number of problems where Gamma^u_ab!=0 that have nothing at all to do
with general or any relativity.
And again, it's not hard to stick t terms into that metric and turn
the problem into a relativistic one.
What makes the difference is the metric you stick in there. And the
interpretation of it.
>
> >What I hope to make clear by that is there's nothing uniquely general
> >relativistic or Einsteinian at all about Christoffel symbols or
> >geodesic equations or even curvature tensors. They are mathematical
> >methods that can be applied to many things, general relativity is only
> >one of them. They tend to be introduced to the student only when he
> >studies general relativity because in any study besides general
> >relativity the motion of a free particle is trivially resolved with
> >Cartesian coordinates, but you can't use Cartesian coordinates in GR.
> >But that doesn't make any of that stuff unique to GR, it just means
> >the problems you've been given in other classes have involved simple
> >geometry.
> >Can you think of any problem in special relativity that you've done or
> >seen in a problem set that involved non-Cartesian coordinates?
>
> Yes, I habitually use nonorthogonal geometry (g01 =/=0) as you
> have used below.
Ah, well, it's not often seen in the classroom.
>
> >Probably not, it's just not done. That doesn't mean you can't do
> >special relativity in curvilinear coordinates. In fact, it's very
> >straight-forward. It just means that most students haven't seen it.
> >So I think some students can confuse parts of pure geometry with
> >physical theory when they're seen for the first time introduced
> >together.
>
> Right, as I mentioned above, metrics correlate to the velocity
> of light.
>
> >Simple example of curvilinear coordinates in special relativity: a
> >clock in a centrifuge. I made this one up because I haven't seen
> >curvilinear coordinates used in SR, either. The metric is
> >
> > ds^2 = c^2 dt^2 - dr^2 - r^2 d(theta)^2
> >
> >r=R=constant, and the centrifuge spins with angular velocity
> >w=d(theta)/dt in the lab frame. Find the rate of the spinning clock
> >versus a lab clock. With velocity v=wR, relate it to the traditional
> >formula for time dilation.
>
> Ok, let me try...
>
> r^2 d(theta)^2 = r^2 d(theta/dt)^2 dt^2 = r^2 w^2 dt^2
> = v^2 dt^2, and dr^2=0 so,
c^2 d(tau)^2 = c^2 dt^2 - R^2 w^2 dt^2
d(tau) = sqrt(1 - R^2 w^2 / c^2) dt
t = tau / sqrt(1 - R^2 w^2 / c^2)
Since v=wR that can be turned into
t = tau / sqrt(1 - v^2/c^2)
which is the traditional time dilation formula, which is how most
students probably would have related the moving clock to the lab
clock.
>
> ds^2 = c^2 dt^2 - v^2 dt^2.
>
> Barring math error on my part, the metric is erroneous,
> because a spinning clock clearly requires a non-euclidean
> CS and the derived result is Minkowskian.
Just stay in the lab frame and it's easy. Special relativity is
always non-Euclidean, anyway.
Most amusing. It will take me a while longer to really figure out
what you did there.
(Feel free to snip mercilessly in any reply.)
Mark Palenik
news:325dbaf1.03082...@posting.google.com...
In SR we can distinguish between inertial and accelerated reference frames,
so we should be able to say with certainty that it is you that's spinning
around and not the moon.
Paul Cardinale
rpo...@yahoo.com (Randy Poe) wrote in message news:<585ab5d8.03081...@posting.google.com>...
> [Moderator's note: I have set followups to sci.physics.relativity. -P.H.]
>
> numberedbea...@webtv.net (Roshard Davis) wrote in message
> news:<f6a3a34f.0308...@posting.google.com>...
> > Greetings everyone. I have a question concerning Albert Einstein's
> > Theory of Relativity. I heard that the theory is split up into 2
> > sections call Special Relativity and General Relativity and I was
> > wondering, what is difference between these two and what compelled
> > Einstein to create to different sections on this Theory of
> > Relativity?Thanks.
>
> what happens when two things are moving at some velocity
> relative to each other, but it doesn't handle acceleration.
> It is limited to what are called "inertial frames".
That is not correct. It handles acceleration just fine.
Paul Cardinale
Cesar Sirvent
> note that not everyone agrees (to say the least) with the notion that
> special relativity can't handle accelerations. The FAQ entry
I disagree. Special relativity can handle accelerations. What SR
cannot do is hadle accelerated reference systems, which is what is
usually meant.
Tim S
> ba...@galaxy.ucr.edu (John Baez) wrote in message
> news:<bi75vd$qbm$1...@glue.ucr.edu>...
>> In article <BB68376D...@timsilverman.demon.co.uk>,
>> Tim S <T...@timsilverman.demon.co.uk> wrote:
>>
>>> I'm puzzled.
>>>
>>> Why do so many people say that special relativity can't handle non-inertial
>>> reference frames? Or, even more amazingly, that it can't handle acceleration
>>> at all (hence presumably involves only free particles and/or fields, with no
>>> forces)?
>>
>> It's because
<snip the sad but true reasons why>
>>
>> It's mildly annoying when people say special relativity
>> can only handle unaccelerated (= inertial) frames of
>> reference. But it goes beyond annoying into hilarious
>> when people say it's unable to handle accelerated motion.
>> If that were true, it couldn't handle the concept of force!
>> It would be pretty much useless.
>>
>> (I'll set followups to a more accelerated newsgroup.)
>
> In your following article
>
> http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#14
>
> you write
>
> "Indeed, the concept of velocity is not a very useful one in general
> relativity and this makes it difficult to define what 'faster than
> light' means."
>
> and this is your explanation for why the moon is not going faster than
> light if you spin around, because if you can't strictly define
> velocity then you can't say the Moon is going faster than light, but
> then now you're saying that special relativity can easily handle
> acceleration and non-inertial reference frames. If special relativity
> can handle acceleration, then general relativity could handle
> velocity,
Nope! Special relativity has trouble with relative velocity too. You have to
pick an inertial reference frame in order to define it sensibly; worse,
different inertial frames disagree with each other about what the relative
velocities actually are (unlike in the Newtonian/Galilean case). In GR,
there usually _aren't_ any inertial frames, so GR has problems which are
even worse.
> so what is the explanation for the so-called Moon paradox?
SR allows you define relative velocity given an inertial frame; it also
handles non-inertial frames; but it doesn't claim that non-inertial frames
are inertial! You can't define relative velocity in a non-inertial frame.
(Or at least not in any way that's at all enlightening or useful.)
Tim
Ken S. Tucker
>(Feel free to snip mercilessly in any reply.)
I'll try editing, but certainly leave all of your previous reply,
and it's motive, this is pretty interesting stuff...
Do you have a copy of Dover's Principle of Relativity?
(I'll try to stay with this single reference)
>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03082...@posting.google.com>...
>> glha...@indiana.edu (Gregory L. Hansen) wrote in message news:<8ce5c97e.0308...@posting.google.com>...
[snip]...
>>This leads to GAMMA^u_ab=0 in SR.
>
>I feel like you haven't read my posts. Gamma^u_ab is not, in general,
>0 in SR. Not in Newtonian mechanics. Not in plane geometry. Not in
>anything whatsoever, *except* in the special case that you're using
>Cartesian coordinates. I demonstrated that. The Christoffel symbols
>are derivatives of the basis vectors, there'll be non-zero Christoffel
>symbols whenever you use non-Cartesian coordinates. That has nothing
>to do with relativity, it has to do with your coordinate system.
Yes I do understand this. In the Ref, see GR Eq. (46), and
onwards, I quote, "If the GAMMA vanish, then the point
moves uniformly in a straight line."....
>> Of course c is not a constant in GR, (light deflects).
>
>Not a constant in SR, either, when viewed from an accelerated frame.
The 2nd postulate of SR defines c to be a constant velocity.
By removing this condition, you are in effect, changing the
definition of SR.
>> Your arguments define the metric by other means, in
>> applications that employ assumptions about the metric,
>> which is fine, if this is known.
>
>No, they don't. The speed of light is constant in an inertial
>reference frame. If you transform to an accelerated frame the speed
>of light doesn't need to remain constant,
But the Lorentz Transform requires c be a constant, so
you are no longer using the LT, that you have described
below.
>but the accelerated frame is related to inertial frames.
>What makes it still special relativity is
>that you've used the Lorentz transformations to find an accelerated
>frame that's equivalent to the inertial frame. Lorentz
>transformations form a group, that means they can't get you out of
>special relativity no matter how many of them you do.
>
>You know the difference between a vector and the components of a
>vector, right? You know that (1,1) and (sqrt(2), pi/4) are the same
>vector, the first expressed in Cartesian coordinates and the second
>expressed in polar coordinates, right? Changing your coordinate
>system doesn't change the physical model, doesn't change the physics,
>it's just a different representation of the same thing. If we had
>special relativity in one coordinate system, then we still have
>special relativity in any coordinate system. I think you're confusing
>the representation with the physics.
>So your criterion above that in special relativity Gamma^u_ab=0
>amounts to an additional and spurious postulate.
>1. The laws of physics have the same form in any inertial reference
>frame.
>2. The speed of light is the same in all inertial reference frames.
>3. Transformations and coordinate systems shall be restricted to
>inertial and Cartesian.
>Number 3 is just not required by the first two, it's something you've
>made up.
Is this true: Implicit in SR is the requirement the metric
transformation,
g'_ab = (&x^u/&x'^a) (&x^v/&x'^b) delta_uv, (KST1)
(summed over 0,1,2,3 delta_uv is Kroneckers,
& is a partial diff). is true globally. It is only defined to be
true locally in GR, this corrects the use of Cartesian in
postulate 3, and then I agree with it. If your metric cannot
comply with Eq.(KST1) globally, (g00 =1, g11...g33=-1
for example) then your making Eq.(KST1) local and this
is GR.
>The first two refer specifically to inertial frames, but
>those two postulates also determine what happens in accelerated
>frames.
No because c is not a constant in an accelerated frame
(as you pointed out) using only LT as I'll explain below...
>My argument distinguishes special from general relativity by the
>postulates, because that's what a theory is. And the postulates of
>special relativity lead to a metric
>
> ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
>
>And you can plug any coordinate system you like into there,
> x = x(u,v,w,s)
> y = y(u,v,w,s)
> z = z(u,v,w,s)
> t = t(u,v,w,s)
>and you'll still have special relativity because it's just
>re-expressing the same metric. And you can use Lorentz
>transformations to go to any other coordinate system you like and it's
>the same structure of space and time from another point of view (first
>postulate).
>You can certainly watch, in the context of special relativity, the
>acceleration of a rocket from the lab frame. Transforming to the
>rocket frame is just a different point of view;
LT permits and is based on,
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
IMO there is no way one can transform this to
contain acceleration components like d^2x/dt^2.
within the requirement of Eq. (KST1).
>if the problem from
>the lab frame was special relativity, then the problem from the rocket
>frame is also special relativity.
I think otherwise, using your centrifugal example below,
"g00 = -(1- w^2 R^2)" is certainly not in accord with
Eq. (KST1).
>And you can never get a theory of gravity from all that. Not without
>additional postulates like those of GR. Even the rocket frame only
>looks like gravity locally, not over extended distances.
>> Your slowly beginning to use 3 dimensional thinking in a
>> Minkowski 4D Coordinate System (CS) required by SR.
>
>I've deliberately negelcted time because g_00 is constant, so any
>derivative of it goes to zero. But I was trying to show that the
>distinction you made above would call a lot of things general
>relativity that aren't general relativity.
>And doing a Lorentz boost in curvilinear coordinates is a bear.
...
>> >(Answer: Gamma^r_tt=-r, Gamma^t_rt=Gamma^t_tr=1/r, others zero, where
>> >t stands in for theta.)
>> >There's no time in there because it's not mechanics, it's just plane
>> >geometry expressed in polar coordinates.
>>
>> Right, of course, but this is not relativity, it's 3D tensor analysis,
>> and does not apply.
>
>Special relativity has three spatial coordinates, so it applies. Time
>is trivially added, and doesn't eliminate those non-zero Christoffel
>symbols.
Time cannot be added trivially, it must be included so that
c is a constant for everyone, everywhere in SR for LT to
be valid.
....
>> This is getting far from relativity.
>
>That's the point. It's the same math that you've used to demarcate
>special from general relativity, but I'm trying to show that your
>criterion says nothing. If I can do the above with geodesic equations
>and Christoffel symbols, then I don't see how you can make conditions
>with them to distinguish the relativities. I could give you any
>number of problems where Gamma^u_ab!=0 that have nothing at all to do
>with general or any relativity.
>And again, it's not hard to stick t terms into that metric and turn
>the problem into a relativistic one.
>What makes the difference is the metric you stick in there. And the
>interpretation of it.
.....
>> Yes, I habitually use nonorthogonal geometry (g01 =/=0) as you
>> have used below.
>Ah, well, it's not often seen in the classroom.
Should be though, as you demonstrated.
....
>> Ok, let me try...
>>
>> r^2 d(theta)^2 = r^2 d(theta/dt)^2 dt^2 = r^2 w^2 dt^2
>> = v^2 dt^2, and dr^2=0 so,
>
> c^2 d(tau)^2 = c^2 dt^2 - R^2 w^2 dt^2
>
> d(tau) = sqrt(1 - R^2 w^2 / c^2) dt
>
> t = tau / sqrt(1 - R^2 w^2 / c^2)
>
>Since v=wR that can be turned into
>
> t = tau / sqrt(1 - v^2/c^2)
>
>which is the traditional time dilation formula, which is how most
>students probably would have related the moving clock to the lab
>clock.
>> Barring math error on my part, the metric is erroneous,
>> because a spinning clock clearly requires a non-euclidean
>> CS and the derived result is Minkowskian.
>
>Just stay in the lab frame and it's easy. Special relativity is
>always non-Euclidean, anyway.
[snop = big old snip]
>> >> R^2 = X^2 + Y^2,
>> >> g00 = -(1- w^2 R^2), g11 = g22 =1, g10 =Yw, g20=-Xw
>> >> (assuming g10 and g20 are symmetrical ie g10 = g01)
>> >Let me take the lazy way out and look it up... Yes, that's right.
>> Check out these last two metrics and see that,
>> g10/Y = w, g20/X = -w and
[snip post on g12=-g21 until Greg comments]
>Most amusing. It will take me a while longer to really figure out
>what you did there.
Feel free to ask and criticize...
>(Feel free to snip mercilessly in any reply.)
Same to you, please retain my most recent reply,
Thanks
Ken S. Tucker
Starblade Darksquall
Uncle Al <Uncl...@hate.spam.net> wrote in message
news:<bhti2j$ce1$1...@lfa222122.richmond.edu>...
> One tacitly assumes c=c, G=0, h=0 and goes on from there. Newton is
> made to fit Maxwell's equations. Galilean transforms fall to
> Lorentzian transforms. SR does not handle gravitation or accelerated
> reference frames (one can tap dance a bit in the latter case).
>
> General Relativity is SR plus the Equivalence Principle - all local
> masses fall identically in vacuum. GR is gravitation. One tacitly
> assumes c=c, G=G, h=0 and goes on from there. Everything else is
> details (!).
>
> Note that one can formulate equally predictive valid gravitation
> theories by not assuming the Equivalence Principle (e.g.,
> Weitzenboeck).
What is relativity when h is not equal to zero?
I take it from here, that there is no longer any conformal invariance,
and the universe depends on the scale used, I.e., how much larger or
smaller they are than the planck units.
(...Starblade Riven Darksquall...)
Gregory L. Hansen
>
> >(Feel free to snip mercilessly in any reply.)
> I'll try editing, but certainly leave all of your previous reply,
> and it's motive, this is pretty interesting stuff...
> Do you have a copy of Dover's Principle of Relativity?
> (I'll try to stay with this single reference)
I have it, but haven't read it. I'm more familiar with Schutz and
Bergmann.
>
> >dyna...@vianet.on.ca (Ken S. Tucker) wrote in message
news:<2202379a.03082...@posting.google.com>...
> >> glha...@indiana.edu (Gregory L. Hansen) wrote in message
news:<8ce5c97e.0308...@posting.google.com>...
> [snip]...
>
> >>This leads to GAMMA^u_ab=0 in SR.
> >
> >I feel like you haven't read my posts. Gamma^u_ab is not, in general,
> >0 in SR. Not in Newtonian mechanics. Not in plane geometry. Not in
> >anything whatsoever, *except* in the special case that you're using
> >Cartesian coordinates. I demonstrated that. The Christoffel symbols
> >are derivatives of the basis vectors, there'll be non-zero Christoffel
> >symbols whenever you use non-Cartesian coordinates. That has nothing
> >to do with relativity, it has to do with your coordinate system.
>
> Yes I do understand this. In the Ref, see GR Eq. (46), and
> onwards, I quote, "If the GAMMA vanish, then the point
> moves uniformly in a straight line."....
If you're using e.g. polar coordinates, uniform motion in a straight
line
means the Gammas don't vanish.
>
> >> Of course c is not a constant in GR, (light deflects).
> >
> >Not a constant in SR, either, when viewed from an accelerated frame.
>
> The 2nd postulate of SR defines c to be a constant velocity.
> By removing this condition, you are in effect, changing the
> definition of SR.
The 2nd postulate of SR defines c to be a constant velocity in an
inertial
reference frame. Going to an accelerated frame doesn't remove that
postulate, more on that below.
>
> >> Your arguments define the metric by other means, in
> >> applications that employ assumptions about the metric,
> >> which is fine, if this is known.
> >
> >No, they don't. The speed of light is constant in an inertial
> >reference frame. If you transform to an accelerated frame the speed
> >of light doesn't need to remain constant,
>
> But the Lorentz Transform requires c be a constant, so
> you are no longer using the LT, that you have described
> below.
Going into an accelerated frame is effectively doing a boost during a
measurement. Suppose you have two buoys fixed in some inertial frame,
and measure the speed of light by the distance traveled divided by the
elapsed time,
c = (x2 - x1)/(t2 - t1)
Find that measurement in a boosted frame, still inertial. You've seen
this calculation before.
(x2'-x1')/(t2'-t1') = ((x2-x1) + v(t2-t1))/((t2-t1)+v(x2-x1)/c^2)
= ( (x2-x1)/(t2-t1) + v) / ( 1 + (x2-x1)/(t2-t1)v/c^2)
= (c + v) / (1 + cv/c^2)
= c
But suppose right after measurement (t1,x1) you boost by v to a new
frame. Then the speed you measure will be (and watch the primes, I
didn't forget any)
(x2' - x1)/(t2' - t1)
= ( gamma(x2 - v t2) - x1 ) / ( gamma(t2 - v x2/c^2) - t1 )
And I'm not sure right now what's a pretty way to present that, but it
works out to c+stuff. The reason is that by boosting, you effectively
move the marker before the light gets to it. And to 2nd order there's
that stuff with gamma factors and (v/c) going on.
And this is also what happens with the Sagnac effect view from the
rotating frame, although it's criticized by cranks that say it
contradicts relativity because they think the second postulate applies
to all frames and not just inertial frames. (Not that I mean to seem
to imply you're a crank, I just mention it because it came up in an
unrelated discussion I was involved with not so long ago.)
The usual recipie for an accelerated frame is to start somewhere, then
boost by some small v to frame 2, stay there for some small proper
time t.
Then boost by some small v to frame 3, stay there for some small
proper
time t. Then boost by some small v to frame 4, stay there for some
small
proper time t. Lorentz boost, Lorentz boost, Lorentz boost.
Can we agree that a boost, and that a succession of boosts as
described
above, are still special relativity? (Assuming yes...) Then can we
say
there's some point where special relativity ends if we just let t and
v
get small, but in a constant ratio a=v/t?
In simpler problems we'd assume one measurement is completed before
the boost, and a second measurement completed after the boost, etc.
But when t gets very small there is no measurement that's done
entirely in one frame; the measuring apparatus moves to a different
inertial frame before the measurement ends (Schutz calls it the
Momentary Comoving Reference Frame, MCRF.) But what's measured in an
accelerated frame is still absolutely determined by what happens in
the inertial frames since it is, after all, the Lorentz
transformations that construct the accelerated frame for us.
There is actually one more postulate that needs to be introduced: that
acceleration doesn't do anything special to clocks, aside from
mechanical
effects. I think that's suggested just by the fact that time is
continuous through a normal boost in SR, although that may not be
rigorous. But normally a boost is treated as instantaneous, or at
least much faster than some time scale of interest.
The relation above is true in an orthogonal coordinate system, but we
don't have to use orthogonal coordinates. Oblique coordinates are a
simple counter-example. E.g. if you have
x' = x
y' = y + ax
z' = z
t' = t
the metric will look something like
g_ij = ( 1 -a 0 0 )
|-a 1 0 0 |
| 0 0 1 0 |
( 0 0 0 -c^2)
and I wrote out the full metric because you've seemed uncomfortable
before with spatial fragments.
But changing your reference frame doesn't change the physics. Whether
you watch it rocket away or hop on board shouldn't matter, it's just a
different point of view, that doesn't introduce any new physical
phenomena.
Metric without time:
ds^2 = dr^2 + r^2 d(theta)^2
Metric with time.
ds^2 = dr^2 + r^2 d(theta)^2 - c^2 dt^2
I'd call that pretty trivial.
[okay, I'll snip some stuff here]
> >> >> R^2 = X^2 + Y^2,
> >> >> g00 = -(1- w^2 R^2), g11 = g22 =1, g10 =Yw, g20=-Xw
> >> >> (assuming g10 and g20 are symmetrical ie g10 = g01)
> >> >Let me take the lazy way out and look it up... Yes, that's right.
>
> >> Check out these last two metrics and see that,
> >> g10/Y = w, g20/X = -w and
> [snip post on g12=-g21 until Greg comments]
>
> >Most amusing. It will take me a while longer to really figure out
> >what you did there.
>
> Feel free to ask and criticize...
Actually, I don't know if you've seen the thread I've started "Where's
my div?", but I'm trying to work on what seems like an embarassingly
basic problem considering the level of discussion we've been having.
That is, when I try to work out divergence in curvilinear coordinates,
and I've tried two methods and got the same answer, I'm missing some
multiplying factors in front of the terms. What I'm missing,
actually, is sqrt(g^ii), and I presume there'd be some sqrt(g^ij) in
non-orthogonal coordinates. And I know the official word, I can just
follow some formulas and stick those terms in there, but I can't prove
that it should be so, or find the generalization to non-orthogonal
coordinates. In particular, I can't figure out how to find it from
the fancy-pants definition V^i_;i. It's like the terms I find are
related to but not the same as the terms in the inside cover of
Jackson. I'd be trying to figure that out now if I weren't writing to
you.
If you can shed light on that, I'd be much appreciative and ready to
move on.
Pmb
"Tim S" <T...@timsilverman.demon.co.uk> wrote
> Nope! Special relativity has trouble with relative velocity too.
That's an incorrect statement.
> You have to
> pick an inertial reference frame in order to define it sensibly; .
There's nothing wrong with that.
> ..worse,
> different inertial frames disagree with each other about what the relative
> velocities actually are (unlike in the Newtonian/Galilean case).
There's nothing wrong with that either. But in SR each observer agrees with
the magnitude of the relative velocities of their own inertial frames.
>In GR,
> there usually _aren't_ any inertial frames, .
There's always a locally inertial frame,
Pmb
Pmb
"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.03081...@posting.google.com...
That's incorrect by definition - Special relativity is *defined* to be
"physics in an inertial frame of referance". And you can't really say what
physics is like in an accelerating frame unless you do what Einstien did -
assert the equivalence principle
Tom Roberts
> Special relativity is *defined* to be
> "physics in an inertial frame of referance".
Hmmm. Say rather that SR is physics in Minkowski spacetime (i.e. the
flat 4-d Lorentzian manifold with topology R^4). Your "definition" is
too restrictive.
While traditionally SR used inertial corodinates, there's no physical
reason to restrict it to such coordinates. This has been discussed
around here several times over the past few years, and the consensus
among knowledgeable people is that the definition I gave above more
acurately characterizes SR than what you say.
> And you can't really say what
> physics is like in an accelerating frame unless you do what Einstien did -
> assert the equivalence principle
The equivalence principle relates gravitation to acceleration, but no
gravitation is involved here. SR is adequate -- simply use calculus to
relate what happens wrt an inertial frame to the accelerated frame. One
does not need the equivalence principle for this, one merely needs to
assume that in a generally-accelerated frame both clocks and rulers
behave at each instant as if they were at rest in their
instantaneously-comoving local inertial frame. This is a DIFFERENT
assumption from the equivalence principle, and is essentially Einstein's
"hidden" hypothesis that clocks and rulers have no memory.
For clocks there is ample experimental evidence that this is valid (it's
called the "clock hypothesis"; see the FAQ for references to several
experiments that validate it up to accelerations of ~10^18 g). For
rulers, a little thought indicates it had better be valid, as
inter-atomic bonds must behave to make it so (but AFAIK there are no
direct measurmeents, due to the difficulty of doing so -- we can measure
time intervals vastly better than distances; there are no direct
measurements of basic length contraction for inertial motion, either).
Tom Roberts tjroo...@lucent.com
Ken S. Tucker
>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message
>news:<2202379a.0308...@posting.google.com>...
>> glha...@indiana.edu (Gregory L. Hansen) wrote in message news:<8ce5c97e.03082...@posting.google.com>...
>>
>> Do you have a copy of Dover's Principle of Relativity?
>> (I'll try to stay with this single reference)
>
>I have it, but haven't read it. I'm more familiar with Schutz and
>Bergmann.
Bergmann's "Introduction to the Theory of Relativity" is fine.
>> Yes I do understand this. In the Ref, see GR Eq. (46), and
>> onwards, I quote, "If the GAMMA vanish, then the point
>> moves uniformly in a straight line."....
>
>If you're using e.g. polar coordinates, uniform motion in a straight
>line
>means the Gammas don't vanish.
Ok, at the risk of going circular, would you say that the
geodesic equation, (Ref Eq. 46) ie,
dU^u = GAMMA^u_ab U^a U^b (==DU^u=0)
may be non-zero in SR?
...
>> The 2nd postulate of SR defines c to be a constant velocity.
>> By removing this condition, you are in effect, changing the
>> definition of SR.
>
>The 2nd postulate of SR defines c to be a constant velocity in an
>inertial reference frame. Going to an accelerated frame doesn't
>remove that postulate, more on that below.
...
Thinking,
>In simpler problems we'd assume one measurement is completed before
>the boost, and a second measurement completed after the boost, etc.
>But when t gets very small there is no measurement that's done
>entirely in one frame; the measuring apparatus moves to a different
>inertial frame before the measurement ends (Schutz calls it the
>Momentary Comoving Reference Frame, MCRF.) But what's measured in an
>accelerated frame is still absolutely determined by what happens in
>the inertial frames since it is, after all, the Lorentz
>transformations that construct the accelerated frame for us.
>
>There is actually one more postulate that needs to be introduced: that
>acceleration doesn't do anything special to clocks, aside from
>mechanical
>effects. I think that's suggested just by the fact that time is
>continuous through a normal boost in SR, although that may not be
>rigorous. But normally a boost is treated as instantaneous, or at
>least much faster than some time scale of interest.
Hope you have S. Weinberg's "Grav and Cosmo", if so,
how does his Eq.(2.1.8) fair in your argument?
My own 2D view begins with ds^2 = dt^2 - dx^2
ds^2 = (1-v^2) dt^2 = (1 - 2 a r)dt^2 = g00 dt^2
a =acceleration, r = distance, (v^2 = 2 a s)
2 a r = GM/r is some arbituary accelerating potential
(I chose this form of acceleration as a familiar example)
and g00 = 1 - 2 GM/r, the point is &g00/&r =/= 0.
Your example (I think) has &g00/&r discontinuous,
(boost series), that can be accomplished by discrete
instanteous energy inputs, like quanta. It may be sematics
if this is SR or GR.
At this point my question to all is,
does the Lorentz Transform allow the second derivative
d^2 s^2 =/=0 , ie is ds^2 a variable in LT's, or must it be
a scalar invariant?
Thanks,
I think there's a technical problem here. Tensors like g_ij
relate CS's to a single location but the preamble y' = y + ax
is a location difference, I think that's a Galilean Transform.
If I recall recall correctly(?) proper transformations (tensor)
are among CS's at the same location, so I hesitate to
comment, pending your review.
Off hand, quantity "a" above, relates relative motion
between CS's, and is therefore a non-covariant artifact,
but is very real in CS's with relative motion.
...
>> I think otherwise, using your centrifugal example below,
>> "g00 = -(1- w^2 R^2)" is certainly not in accord with
>> Eq. (KST1).
>
>But changing your reference frame doesn't change the physics. Whether
>you watch it rocket away or hop on board shouldn't matter, it's just a
>different point of view, that doesn't introduce any new physical
>phenomena.
Of course, but the relative FoR's are accelerating. Using SR
alone, how are to relate and compare the measurements of
these relatively accelerating FoR's? (Specifically in view of
Weinberg's Eq. 2.1.8)
...
>> Time cannot be added trivially, it must be included so that
>> c is a constant for everyone, everywhere in SR for LT to
>> be valid.
>
>Metric without time:
>
> ds^2 = dr^2 + r^2 d(theta)^2
>
>Metric with time.
>
> ds^2 = dr^2 + r^2 d(theta)^2 - c^2 dt^2
>
>I'd call that pretty trivial.
As you previously pointed out, (and I agree) the
incestuous nonorthogothonal spacetime dynamic
metrics g01, g02, g03 need to be included.
>[okay, I'll snip some stuff here]
...
>> Feel free to ask and criticize...
>
>Actually, I don't know if you've seen the thread I've started "Where's
>my div?", but I'm trying to work on what seems like an embarassingly
>basic problem considering the level of discussion we've been having.
>That is, when I try to work out divergence in curvilinear coordinates,
>and I've tried two methods and got the same answer, I'm missing some
>multiplying factors in front of the terms. What I'm missing,
>actually, is sqrt(g^ii), and I presume there'd be some sqrt(g^ij) in
>non-orthogonal coordinates. And I know the official word, I can just
>follow some formulas and stick those terms in there, but I can't prove
>that it should be so, or find the generalization to non-orthogonal
>coordinates. In particular, I can't figure out how to find it from
>the fancy-pants definition V^i_;i. It's like the terms I find are
>related to but not the same as the terms in the inside cover of
>Jackson. I'd be trying to figure that out now if I weren't writing to
>you.
(Sorry to distract you :-)
>If you can shed light on that, I'd be much appreciative and ready to
>move on.
Ok, will look,
IMVHO (V = Very), Div in tensors looks simple at first,
ie, div A^u = A^u_u = 1/sqrt(g) &/&x^v (sqrt(g) A^v)
But the understanding of the factor sqrt(g) begins to involve
relative tensors. H. Weyl uses them alot in his book,
"Space-Time-Matter". OTOH Einstein avoided them at first
by simply setting sqrt(g) =-1, (see Eq. 18 in Dover's P of R),
but at the expense of General Covariance. In math, proper
transformations are ok provided g=/=0.
Regards Ken S. Tucker
Pmb
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:biqbl2$g...@netnews.proxy.lucent.com...
> Pmb wrote:
> > Special relativity is *defined* to be
> > "physics in an inertial frame of referance".
>
> Hmmm. Say rather that SR is physics in Minkowski spacetime (i.e. the
> flat 4-d Lorentzian manifold with topology R^4). Your "definition" is
> too restrictive.
Tell me something tom. Why do you keep refering to these definitions as
"yours" refering to me? It is *the* definition. I was defined by Einstein
and used throughout most of relativity as are many other definitions you've
refered to as mine
>
> While traditionally SR used inertial corodinates, there's no physical
> reason to restrict it to such coordinates.
Not really - You keep thinking of "special" as refering to flat spacetime.
And it seems that's all you think about when it comes to this term. SR is
defined according to inertial frames since the postulates of SR refer only
to inertial frames. GR is what states that the laws of physics are the same
in all coordinate systems - nor SR.
> This has been discussed
> around here several times over the past few years, and the consensus
> among knowledgeable people is that the definition I gave above more
> acurately characterizes SR than what you say.
That's also incorrect. One simply has to look the term up.
>
> > And you can't really say what
> > physics is like in an accelerating frame unless you do what Einstien
did -
> > assert the equivalence principle
>
> The equivalence principle relates gravitation to acceleration, but no
> gravitation is involved here.
Again you're arguing about definitions again - you forget that Einstein
defined "gravitational field" differently then you use the term . And nobody
has proved Einstein wrong. At best some just choose to define things
differently - but I haven't see that in almost all the literature I've seen
on this.
Look it up = Browse a representative collection
Pmb
Pmb
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:biqbl2$g...@netnews.proxy.lucent.com...
>>Pmb wrote:
>>>Special relativity is *defined* to be
>>>"physics in an inertial frame of referance".
>>
>>Hmmm. Say rather that SR is physics in Minkowski spacetime (i.e. the
>>flat 4-d Lorentzian manifold with topology R^4). Your "definition" is
>>too restrictive.
> Tell me something tom. Why do you keep refering to these definitions as
> "yours" refering to me?
Because you are the one writing. While other people have made such
definitions LONG AGO, today they are outmoded, except to you -- those
other people are not writing TODAY; you are.
> It is *the* definition.
No, it is not. Not as we use the term "SR" today. Accelerated motion was
analyzed in SR many decades ago, including accelerated coordinates.
>>While traditionally SR used inertial corodinates, there's no physical
>>reason to restrict it to such coordinates.
> Not really - You keep thinking of "special" as refering to flat spacetime.
> And it seems that's all you think about when it comes to this term. SR is
> defined according to inertial frames since the postulates of SR refer only
> to inertial frames.
Yes, the postulates of SR refer only to inertial frames. That does not
ipso facto limit the theory to such frames. SR is a PHYSICAL theory, and
as such physicists consider mathematics to be "free", in the sense that
applying mathematics to the postulates of the theory does not yield a
new theory[#]. As I said before, one can simply apply calculus to
inertial frames and determine anything of interest using accelerated
coordinates. As no additional postulates are needed to do that, this
remains within the bounds of SR.
[#} Think about that, as without mathematics there would
be no physical theories....
> GR is what states that the laws of physics are the same
> in all coordinate systems - nor SR.
Right. Apply SR in non-inertial coordinates and the "laws of physics"
are QUITE different than they are in inertial coordinates. That's
because in SR we drop the "complicated" terms related to the connection....
Actually not, because we apply SR in spherical coordinates,
for which the connection is nonzero. Note that accelerated
coordinates are in principle no different from spherical
coordinates in this....
If you prohibit accelerated coordinates, then how can you admit
spherical coordinates?
Useful exercise: write down the Lorentz transform in terms
of spherical coordinates. Hint: this should take no more than
2 minutes. Second hint: this is not a test, and you get no
credit for lengthy algebra.
Useful exercise: describe the range of validity of the
transforms of the previous exercise. Why don't they
cover the manifold?
[Accelerated coordinates (aka Rindler coords.) do not
cover the manifold, which is why I present this second
exercise.]
>>This has been discussed
>>around here several times over the past few years, and the consensus
>>among knowledgeable people is that the definition I gave above more
>>acurately characterizes SR than what you say.
> That's also incorrect. One simply has to look the term up.
It is certainly correct that this has been discussed around here, and
the consensus AMONG KNOWLEDGEABLE PEOPLE was as I said.
Be careful where you "look it up", as elementary books tend to take the
simplistic way out....
Look at MTW section 6, and its relationship to the introduction
of both SR and GR. Why do you think they titled part 2 "Physics
in Flat Spacetime" rather than "SR"?
[my answer: because they did not want to get embroiled in the
dead horse you are flogging.]
>>>And you can't really say what
>>>physics is like in an accelerating frame unless you do what Einstien
>>> did - assert the equivalence principle
>>
>>The equivalence principle relates gravitation to acceleration, but no
>>gravitation is involved here.
>
> Again you're arguing about definitions again - you forget that Einstein
> defined "gravitational field" differently then you use the term .
Not really. But that's irrelvant -- in the Minkowski spacetime of SR
there is no "gravitation" of any sort. So the equivalence principle is
not needed, as I said.
You do need the "hidden" postulate that clocks and rulers have no
memory, as I said.
Useful exercise: How does GR avoid that "hidden" postulate?
Of course the precise "location" of the boundary between SR and GR is
not of major importance -- GR is the real theory and SR is merely a
local approximation to it. So don't expect me to keep beating this....
Tom Roberts tjro...@lucent.com
Gregory L. Hansen
>news:<8ce5c97e.03082...@posting.google.com>...
>>> Yes I do understand this. In the Ref, see GR Eq. (46), and
>>> onwards, I quote, "If the GAMMA vanish, then the point
>>> moves uniformly in a straight line."....
>>
>>If you're using e.g. polar coordinates, uniform motion in a straight
>>line
>>means the Gammas don't vanish.
>
>Ok, at the risk of going circular, would you say that the
>geodesic equation, (Ref Eq. 46) ie,
>
>dU^u = GAMMA^u_ab U^a U^b (==DU^u=0)
>
>may be non-zero in SR?
Sure, it depends on the coordinate system. That's also true of inertial
coordinate systems, I might add. It can be non-zero in Newtonian
mechanics, too. You'll have one fewer dimension, then, but nothing
changes with the math.
>>In simpler problems we'd assume one measurement is completed before
>>the boost, and a second measurement completed after the boost, etc.
>>But when t gets very small there is no measurement that's done
>>entirely in one frame; the measuring apparatus moves to a different
>>inertial frame before the measurement ends (Schutz calls it the
>>Momentary Comoving Reference Frame, MCRF.) But what's measured in an
>>accelerated frame is still absolutely determined by what happens in
>>the inertial frames since it is, after all, the Lorentz
>>transformations that construct the accelerated frame for us.
>>
>>There is actually one more postulate that needs to be introduced: that
>>acceleration doesn't do anything special to clocks, aside from
>>mechanical
>>effects. I think that's suggested just by the fact that time is
>>continuous through a normal boost in SR, although that may not be
>>rigorous. But normally a boost is treated as instantaneous, or at
>>least much faster than some time scale of interest.
>
>Hope you have S. Weinberg's "Grav and Cosmo", if so,
>how does his Eq.(2.1.8) fair in your argument?
I do not have that one.
>
>My own 2D view begins with ds^2 = dt^2 - dx^2
>ds^2 = (1-v^2) dt^2 = (1 - 2 a r)dt^2 = g00 dt^2
>a =acceleration, r = distance, (v^2 = 2 a s)
>2 a r = GM/r is some arbituary accelerating potential
>(I chose this form of acceleration as a familiar example)
>and g00 = 1 - 2 GM/r, the point is &g00/&r =/= 0.
>
>Your example (I think) has &g00/&r discontinuous,
>(boost series), that can be accomplished by discrete
>instanteous energy inputs, like quanta. It may be sematics
>if this is SR or GR.
> At this point my question to all is,
>does the Lorentz Transform allow the second derivative
>d^2 s^2 =/=0 , ie is ds^2 a variable in LT's, or must it be
>a scalar invariant?
(ds)^2, the spacetime interval, is a constant. That's sort of one way to
define Lorentz transforms. The left hand side is essentially a
particle's proper time, and a particle only has one proper time. So I
suppose any derivatives, d(ds)/dq have to vanish. It's preserved in the
uniformly accelerating frame, at least. E.g. if sigma is a distance
measured by rigid rods at rest in the accelerating frame, and (x,t) are
inertial frame coordinates, we'd have sigma=sqrt(x^2-t^2) (if all
parameters of interest equal 1).
I guess I can't confidently say it's always constant in the accelerated
frame, but I think it must be. In particular, when it's expressed in the
frame coordinates and integrated along whatever frame-specific path the
particle takes.
>>The relation above is true in an orthogonal coordinate system, but we
>>don't have to use orthogonal coordinates. Oblique coordinates are a
>>simple counter-example. E.g. if you have
>>
>> x' = x
>> y' = y + ax
>> z' = z
>> t' = t
>>
>>the metric will look something like
>>
>> g_ij = ( 1 -a 0 0 )
>> |-a 1 0 0 |
>> | 0 0 1 0 |
>> ( 0 0 0 -c^2)
>>
>>and I wrote out the full metric because you've seemed uncomfortable
>>before with spatial fragments.
>
>Thanks,
>I think there's a technical problem here. Tensors like g_ij
>relate CS's to a single location but the preamble y' = y + ax
>is a location difference, I think that's a Galilean Transform.
> If I recall recall correctly(?) proper transformations (tensor)
>are among CS's at the same location, so I hesitate to
>comment, pending your review.
> Off hand, quantity "a" above, relates relative motion
>between CS's, and is therefore a non-covariant artifact,
>but is very real in CS's with relative motion.
Huh? There's no motion there. In this case, x, y, x', y' are position
coordinates, a is a unitless scaling factor. It's different coordinates
in the same reference frame. Crystallographers use oblique coordinate
systems all the time, but that's not usually in a relativistic context.
> ...
>>> I think otherwise, using your centrifugal example below,
>>> "g00 = -(1- w^2 R^2)" is certainly not in accord with
>>> Eq. (KST1).
>>
>>But changing your reference frame doesn't change the physics. Whether
>>you watch it rocket away or hop on board shouldn't matter, it's just a
>>different point of view, that doesn't introduce any new physical
>>phenomena.
>
>Of course, but the relative FoR's are accelerating. Using SR
>alone, how are to relate and compare the measurements of
>these relatively accelerating FoR's? (Specifically in view of
>Weinberg's Eq. 2.1.8)
I can give you a suggested reading list.
Desloge and Philpott, "Uniformly accelerated reference frames in special
relativity", Am. J. Phys. 55 (3), 252, March 1987. Using elementary
methods they carefully develop the uniformly accelerated reference frame
in one direction, introduce two coordinate systems to describe it and
related it to inertial coordinates, defined by two different measurement
processes, and discuss its properties.
Good, "Uniformly accelerated reference frame and twin paradox", Am. J.
Phys. 50 (3), 232, March 1982. He analyzes the twin paradox from the
perspective of the travelling twin where the traveller starts at rest on
Earth, accelerates, then coasts, then accelerates again to turn around,
coasts, then accelerates to a stop.
Rizzi and Tartaglia, "Speed of light on Rotating Platforms", Foundations
of Physics 28 (11), page 1663, 1998. They analyze the Sagnac effect from
the rotating frame in the context of special relativity, and discuss some
misconceptions about it.
MTW, chapter 6. A whole chapter on acceleration in special relativity,
using non-elementary methods. (Section 6.1 titled "Accelerated Observers
Can Be Analyzed Using Special Relativity".) Chapter 7 is on the
inadequacy of special relativity in a theory of gravity. I need to put a
bit of work into this one.
>>Actually, I don't know if you've seen the thread I've started "Where's
>>my div?", but I'm trying to work on what seems like an embarassingly
>>basic problem considering the level of discussion we've been having.
>>That is, when I try to work out divergence in curvilinear coordinates,
>>and I've tried two methods and got the same answer, I'm missing some
>>multiplying factors in front of the terms. What I'm missing,
>>actually, is sqrt(g^ii), and I presume there'd be some sqrt(g^ij) in
>>non-orthogonal coordinates. And I know the official word, I can just
>>follow some formulas and stick those terms in there, but I can't prove
>>that it should be so, or find the generalization to non-orthogonal
>>coordinates. In particular, I can't figure out how to find it from
>>the fancy-pants definition V^i_;i. It's like the terms I find are
>>related to but not the same as the terms in the inside cover of
>>Jackson. I'd be trying to figure that out now if I weren't writing to
>>you.
>
>(Sorry to distract you :-)
>
>>If you can shed light on that, I'd be much appreciative and ready to
>>move on.
>
>Ok, will look,
>IMVHO (V = Very), Div in tensors looks simple at first,
>ie, div A^u = A^u_u = 1/sqrt(g) &/&x^v (sqrt(g) A^v)
I figured it out. The prescription for finding vector operations in the
relativity texts are always for vectors in a coordinate basis. The
formulas in Jackson and other E&M texts work with unit vectors. The unit
vector u_i is related to the coordinate vector e_i by
u_i = e_i / sqrt(e_i dot e_i) = e_i / sqrt(g_ii)
and the vector components by
f^i e_i = F_i u_i, F_i = sqrt(g_ii) f^i
If I make that correction I get what's on the inside cover of Jackson.
--
"A nice adaptation of conditions will make almost any hypothesis agree
with the phenomena. This will please the imagination but does not advance
our knowledge." -- J. Black, 1803.
Gauge
> Pmb wrote:
> > "Tom Roberts" <tjro...@lucent.com> wrote in message
> > news:biqbl2$g...@netnews.proxy.lucent.com...
> >>Pmb wrote:
> >>>Special relativity is *defined* to be
> >>>"physics in an inertial frame of referance".
> >>
> >>Hmmm. Say rather that SR is physics in Minkowski spacetime (i.e. the
> >>flat 4-d Lorentzian manifold with topology R^4). Your "definition" is
> >>too restrictive.
> > Tell me something tom. Why do you keep refering to these definitions as
> > "yours" refering to me?
>
> Because you are the one writing.
So if I use the term "general relativity" it's supposed to mean that
it's mine?
> While other people have made such
> definitions LONG AGO, today they are outmoded, except to you -- those
> other people are not writing TODAY; you are.
That's totally wrong. Just look in most texts today and you'll see.
We've been through all of this before already so I'm not going to
repeat myself again.
[snipped same old comments from these past years]
Pmb
John Baez
Jeffery <jeffery...@mail.com> wrote:
>In your following article
>
>http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#14
>
>you write:
>
>"Indeed, the concept of velocity is not a very useful one in general
>relativity and this makes it difficult to define what 'faster than
>light' means."
I didn't write most of the articles in the Physics FAQ;
my site is just one of many hosts for this collection of
articles. In particular, if you look at the top of this
particular page, you'll see it was written by Phil Gibbs.
So, maybe *he* can explain what he meant by this, and
how it's related to the so-called Moon Paradox.
Ken S. Tucker
>In article <2202379a.03083...@posting.google.com>,
>Ken S. Tucker <dyna...@vianet.on.ca> wrote:
>>glha...@indiana.edu (Gregory L. Hansen) wrote in message
>>news:<8ce5c97e.03082...@posting.google.com>...
...
>>Ok, at the risk of going circular, would you say that the
>>geodesic equation, (Ref Eq. 46) ie,
>>
>>dU^u = GAMMA^u_ab U^a U^b (==DU^u=0)
>>
>>may be non-zero in SR?
>
>Sure, it depends on the coordinate system. That's also true of inertial
>coordinate systems, I might add. It can be non-zero in Newtonian
>mechanics, too.
We should stay with accelerations in SR, that's the topic,
(big enough for me!).
>You'll have one fewer dimension, then, but nothing
>changes with the math.
To clarify, you figure a non-zero dU^u/ds will occur
as a result of the choice of spatial coordinates?
Let's try a simple example beginning with
dU^u/ds = GAMMA^u_ab U^a U^b
(I'll include the work for reference)
= (1/2) g^uv [g_av,b + g_vb,a - g_ba,v] U^a U^b
Figure to find acceleration in direction x,
dU1/ds = (g11/2) [g_a1,b + g_1b,a - g_ba,1] U^a U^b.
Operating on the box [],
[]= g_a1,b U^a U^b+ g_1b,a U^a U^b- g_ba,1 U^a U^b
gives some differentials,
=dg_a1/ds U^a + dg_1b/ds U^b - g_ba,1 U^a U^b
Letting g11 vary as a function of x we can sum over 0,1 gives,
= 2 dg11/ds U1 + 2dg10/ds U0 - g11,1 U1 U1 - g00,1 U0 U0
in SR g00,1 = 0, and at low speeds U0~1, (ds=dt) so,
[] = 2 dg11/dt U1 + 2dg10/dt - g11,1 U1 U1
At this point, I need to show this is generally zero, or you
need to provide a non-zero example.
For my part, IMO, since this is a tensor equation, I may
select a CS where U1=0, leaving only
[]=2dg10/dt
to produce a non-zero Christoffel symbol in the event
of acceleration in SR. In view of your previous use
of g10 and g20 in your centrifugal example a few posts
ago, I know you understand g10 and thus dg10/dt,
so your comments will be quite relevant.
...
>>My own 2D view begins with ds^2 = dt^2 - dx^2
>>ds^2 = (1-v^2) dt^2 = (1 - 2 a r)dt^2 = g00 dt^2
>>a =acceleration, r = distance, (v^2 = 2 a s)
>>2 a r = GM/r is some arbituary accelerating potential
>>(I chose this form of acceleration as a familiar example)
>>and g00 = 1 - 2 GM/r, the point is &g00/&r =/= 0.
>>
>>Your example (I think) has &g00/&r discontinuous,
>>(boost series), that can be accomplished by discrete
>>instanteous energy inputs, like quanta. It may be sematics
>>if this is SR or GR.
>> At this point my question to all is,
>>does the Lorentz Transform allow the second derivative
>>d^2 s^2 =/=0 , ie is ds^2 a variable in LT's, or must it be
>>a scalar invariant?
>
>(ds)^2, the spacetime interval, is a constant. That's sort of one way to
>define Lorentz transforms. The left hand side is essentially a
>particle's proper time, and a particle only has one proper time. So I
>suppose any derivatives, d(ds)/dq have to vanish.
So I think we can agree at the instant of acceleration,
ds varies, and therefore LT does not apply?
>It's preserved in the
>uniformly accelerating frame, at least. E.g. if sigma is a distance
>measured by rigid rods at rest in the accelerating frame, and (x,t) are
>inertial frame coordinates, we'd have sigma=sqrt(x^2-t^2) (if all
>parameters of interest equal 1).
When d^2 t'/dt^2 <>0 as this is varing between relatively
accelerating systems, your statement implies sigma = sigma',
that's difficult to confirm, (sorry if I'm dense).
>I guess I can't confidently say it's always constant in the accelerated
>frame, but I think it must be. In particular, when it's expressed in the
>frame coordinates and integrated along whatever frame-specific path the
>particle takes.
Well if your neat expression above "d(ds)/dq" varies wrt q
in one system K, but not the other K', ie. ds' is constant in the
system K' where acceleration is zero, but ds varies at the
instant of acceleration in K then a puzzlement results.
Of course you've employed oblique coordinates in g12,
the problem I pointed out is that in relating tensors
only partial derivatives relate CS's.
...
Thank you Sir, I'm swayed by *your* arguments alone if
the above equation,
[]=2dg10/dt
can be non zero in SR. Maybe if this is true, then LT may
be preserved in the acceleration instant, which I'll look
hard at.
...
>>>If you can shed light on that, I'd be much appreciative and ready to
>>>move on.
>
>>Ok, will look,
>>IMVHO (V = Very), Div in tensors looks simple at first,
>>ie, div A^u = A^u_u = 1/sqrt(g) &/&x^v (sqrt(g) A^v)
>
>I figured it out. The prescription for finding vector operations in the
>relativity texts are always for vectors in a coordinate basis. The
>formulas in Jackson and other E&M texts work with unit vectors. The unit
>vector u_i is related to the coordinate vector e_i by
>
> u_i = e_i / sqrt(e_i dot e_i) = e_i / sqrt(g_ii)
Ah yes, in curvilinear coordinates the usual notation
h_i = sqrt(g_ii) is employed, where h_i are scale factors.
But IMO there is no difference between covariant and
contravariant components in curvilinear coordinates (?),
so the use of g_ii or g^ii is meaningless (?). The reason
is nonorthogonal systems where metrics like g_ij are
real, can't be handled by scale factors.
>and the vector components by
>
> f^i e_i = F_i u_i, F_i = sqrt(g_ii) f^i
>
>If I make that correction I get what's on the inside cover of Jackson.
A guy called Alfred Einstead in this NG is super smart
on basis vectors, he bores easily though, or maybe I'm
boring.
Well your further comments would be appreciated,
Ken S. Tucker
Gregory L. Hansen
>news:<bisrgv$622$1...@hood.uits.indiana.edu>...
>
>>In article <2202379a.03083...@posting.google.com>,
>>Ken S. Tucker <dyna...@vianet.on.ca> wrote:
>>>glha...@indiana.edu (Gregory L. Hansen) wrote in message
>>>news:<8ce5c97e.03082...@posting.google.com>...
>>>Ok, at the risk of going circular, would you say that the
>>>geodesic equation, (Ref Eq. 46) ie,
>>>
>>>dU^u = GAMMA^u_ab U^a U^b (==DU^u=0)
>>>
>>>may be non-zero in SR?
>>
>>Sure, it depends on the coordinate system. That's also true of
inertial
>>coordinate systems, I might add. It can be non-zero in Newtonian
>>mechanics, too.
>
>(big enough for me!).
In that case, I think acceleration mainly means the metric has
off-diagonal terms with the time. In the other examples like polar
coordinates the off-diagonal components were all spatial.
Desloge and Philpott give, for a particular measurement procedure in
the
accelerated frame in one dimension,
x = sigma cosh(theta/sigma)
t = sigma sinh(theta/sigma)
And y=y and z=z for the same reasons they do for a Lorentz boost.
sigma
is distance measured by rigid rods at rest in the accelerating frame,
tau
is the time on local clocks at rest in the accelerated frame. The
other
coordinate system they consider is based on bouncing light signals to
measure distances, and a central clock plus half the light reflection
time to determine time.
And clearly that metric will give a d(sigma)d(tau) term. I haven't
worked
out Christoffel symbols for it.
>[] = 2 dg11/dt U1 + 2dg10/dt - g11,1 U1 U1
>
>At this point, I need to show this is generally zero, or you
>need to provide a non-zero example.
>
>For my part, IMO, since this is a tensor equation, I may
>select a CS where U1=0, leaving only
>
>[]=2dg10/dt
>
>to produce a non-zero Christoffel symbol in the event
>of acceleration in SR. In view of your previous use
>of g10 and g20 in your centrifugal example a few posts
>ago, I know you understand g10 and thus dg10/dt,
>so your comments will be quite relevant.
You can do the calculation. I started it, but it really doesn't
simplify
very well, so it started to seem like a lot to do at the keyboard when
I
have duties in the real world demanding my attention. But I got
g_10 = 2 sinh(tau/sigma) cosh(tau/sigma)
- 2 (tau/sigma) (sinh^2(tau/sigma) + cosh^2(tau/sigma))
So whatever the tau derivative of that would be.
In the other coordinate system I mentioned the speed of light would be
constant in the accelerated frame, but trivially so.
>...
>>>My own 2D view begins with ds^2 = dt^2 - dx^2
>>>ds^2 = (1-v^2) dt^2 = (1 - 2 a r)dt^2 = g00 dt^2
>>>a =acceleration, r = distance, (v^2 = 2 a s)
>>>2 a r = GM/r is some arbituary accelerating potential
>>>(I chose this form of acceleration as a familiar example)
>>>and g00 = 1 - 2 GM/r, the point is &g00/&r =/= 0.
>>>
>>>Your example (I think) has &g00/&r discontinuous,
>>>(boost series), that can be accomplished by discrete
>>>instanteous energy inputs, like quanta. It may be sematics
>>>if this is SR or GR.
>>> At this point my question to all is,
>>>does the Lorentz Transform allow the second derivative
>>>d^2 s^2 =/=0 , ie is ds^2 a variable in LT's, or must it be
>>>a scalar invariant?
>>
>>(ds)^2, the spacetime interval, is a constant. That's sort of one
way to
>>define Lorentz transforms. The left hand side is essentially a
>>particle's proper time, and a particle only has one proper time. So
I
>>suppose any derivatives, d(ds)/dq have to vanish.
>
>So I think we can agree at the instant of acceleration,
>ds varies, and therefore LT does not apply?
If it doesn't vary after a boost, I don't see how it can vary for an
acceleration because that's just a lot of boosts.
>Thank you Sir, I'm swayed by *your* arguments alone if
>the above equation,
>
>[]=2dg10/dt
>
>can be non zero in SR. Maybe if this is true, then LT may
>be preserved in the acceleration instant, which I'll look
>hard at.
You're only getting the Usenet version from me. There are some subtle
issues that have to be carefully handled. With the rotating frame in
particular, as many people have made assumptions about it that just
weren't so.
Paul Cardinale
You can easily be shown to be wrong.
IF SR couldn't handle acceleration,
then it should be easy for you to come up with a problem involving
acceleration that can't be solved using SR.
You won't be able to do that.
Paul Cardinale
Gauge
> gau...@hotmail.com (Gauge) wrote in message news:<e7203033.03083...@posting.google.com>...
> > Tom Roberts <tjro...@lucent.com> wrote in message news:<birhat$l...@netnews.proxy.lucent.com>...
> > > Pmb wrote:
> > > > "Tom Roberts" <tjro...@lucent.com> wrote in message
> > > > news:biqbl2$g...@netnews.proxy.lucent.com...
> > > >>Pmb wrote:
> > > >>>Special relativity is *defined* to be
> > > >>>"physics in an inertial frame of referance".
> > > >>
> > > >>Hmmm. Say rather that SR is physics in Minkowski spacetime (i.e. the
> > > >>flat 4-d Lorentzian manifold with topology R^4). Your "definition" is
> > > >>too restrictive.
> > > > Tell me something tom. Why do you keep refering to these definitions as
> > > > "yours" refering to me?
> > >
> > > Because you are the one writing.
> >
> > So if I use the term "general relativity" it's supposed to mean that
> > it's mine?
> >
> > > While other people have made such
> > > definitions LONG AGO, today they are outmoded, except to you -- those
> > > other people are not writing TODAY; you are.
> >
> > That's totally wrong. Just look in most texts today and you'll see.
> >
> > We've been through all of this before already so I'm not going to
> > repeat myself again.
> >
> > [snipped same old comments from these past years]
> >
>
> You can easily be shown to be wrong.
Nope. That is incorrect.
> IF SR couldn't handle acceleration, ...
And there's your misunderstanding right here. You're trying to use the
term to define it instead of the correct sequence wher you define it
and then show what it implies.
Do you think Einstein was so stupid as to not undersand what he was
implying when he defined these terms? Of course not.
Pmb
Ken S. Tucker
>In article <2202379a.03090...@posting.google.com>,
>Ken S. Tucker <dyna...@vianet.on.ca> wrote:
>>glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
>>news:<bisrgv$622$1...@hood.uits.indiana.edu>...
I read posts a few times before commenting, so it might
appear that I comment out of order.
>>We should stay with accelerations in SR, that's the topic,
>>(big enough for me!).
>
>In that case, I think acceleration mainly means the metric has
>off-diagonal terms with the time. In the other examples like polar
>coordinates the off-diagonal components were all spatial.
Yes I agree, please see my recent post entitled
"LT for dummies", (pardon the levity). Anyway I find,
g_0i = - g_ij dx^j/dx^0 ~ - dx^i/cdt = - velocity,
(i,j=1,2,3 only)
which appears consistent with views expressed below.
The post makes use of the idea that a Coordinate System
may be found where the covariant displacement
components dx_i will be zero, ie. dx_i =0, however the
contravariant components dx^i are non-zero.
>keyboard when , I have duties in the real world demanding
>my attention. But I got
> g_10 = 2 sinh(tau/sigma) cosh(tau/sigma)
> - 2 (tau/sigma) (sinh^2(tau/sigma) + cosh^2(tau/sigma))
>So whatever the tau derivative of that would be.
>In the other coordinate system I mentioned the speed of light would be
>constant in the accelerated frame, but trivially so.
Thanks, I avoid hyperbolic trig like the plague, and didn't
mean to distract you from the real world. As explained
above, I think that Lorentz Transforms can be preserved
through the acceleration instant if the CS used sets dx_i =0.
>>So I think we can agree at the instant of acceleration,
>>ds varies, and therefore LT does not apply?
>
>If it doesn't vary after a boost, I don't see how it can vary for an
>acceleration because that's just a lot of boosts.
Ok, I agree dg_i0/dt is LT valid, iff dx_i =0, so far.
[snipped refs, I read MTW, and decided to order it]
>>Thank you Sir, I'm swayed by *your* arguments alone if
>>the above equation,
>>[]=2dg10/dt
>>can be non zero in SR. Maybe if this is true, then LT may
>>be preserved in the acceleration instant, which I'll look
>>hard at.
>
>You're only getting the Usenet version from me. There are some subtle
>issues that have to be carefully handled. With the rotating frame in
>particular, as many people have made assumptions about it that just
>weren't so.
Thanks and agreed. I'm trying to be realistic, essentially
the g_i0 is equivalent to aberration (v/c) so dg_i0/dt is
a change in aberration angle caused by relative acceleration,
and causes a relative rotation of the apparent directions
of two systems.
Thanks and Regards
Ken S. Tucker
Pmb
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:2202379a.03090...@posting.google.com...
> glha...@indiana.edu (Gregory L. Hansen) wrote in message
news:<8ce5c97e.03090...@posting.google.com>...
>
> >In article <2202379a.03090...@posting.google.com>,
> >Ken S. Tucker <dyna...@vianet.on.ca> wrote:
> >>glha...@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
> >>news:<bisrgv$622$1...@hood.uits.indiana.edu>...
>
> I read posts a few times before commenting, so it might
> appear that I comment out of order.
>
> >>We should stay with accelerations in SR, that's the topic,
> >>(big enough for me!).
> >
> >In that case, I think acceleration mainly means the metric has
> >off-diagonal terms with the time. In the other examples like polar
> >coordinates the off-diagonal components were all spatial.
>
> Yes I agree, please see my recent post entitled
> "LT for dummies", (pardon the levity). Anyway I find,
> g_0i = - g_ij dx^j/dx^0 ~ - dx^i/cdt = - velocity,
> (i,j=1,2,3 only)
> which appears consistent with views expressed below.
I don't see that as true as a general rule. In the case of a rotating frame
there is a cross term, g_01, term since this is a non-time orthogonal
spacetime. However there are no diagonal terms for a uniformly accelerated
frame of referance. However if you're in a uniformly accelerated frame and
you change to a frame moving at constant velocity relative to this frame
then there will be a cross term - these cross terms indicate the presence of
frame dragging. As such you can have frame dragging in flat spacetime.
> The post makes use of the idea that a Coordinate System
> may be found where the covariant displacement
> components dx_i will be zero, ie. dx_i =0, however the
> contravariant components dx^i are non-zero.
This makes no sense. The displacement dx_i represents a movement of a
particle in the dx_i direction. Why can't the particle move in the dx_i
direction? And I don't see why you refer to dx_i as a "covariant
displacement" - What do you mean by this?
Let me address the LT for dummies post first. Then we'll come back to this
Pmb
Ken S. Tucker
Just a bit of background here...
Studying Greg Hansen's posts in this thread he suggested
a non-zero Christoffel (acceleration in SR) is compatible
with SR and Lorentz Transforms, and above he mentions,
"off diagonal terms with time". In "LT for dummies", I show
the spacetime metric (g_01) required for a moving clock
is, (in 2D = 0,1)
g_01 = - dx^1/cdt
and also showed
&x'^0/&x^1 =0.
To Pete's question, "Why can't the particle move in the
dx_i direction?".
Consider the association dx_i = g_iu dx^u
{i =1,2,3 u =0,1,2,3}
dx_i = g_ij dx^j + g_i0 dx^0
Using a simple cartesian CS g_ij dx^j = dx^i, so
dx_i = dx^i + g_i0 dx^0
Assuming (from LT for dummies)
g_i0 = - g_ij dx^j/dx^0 then
g_i0 dx^0 = -dx^i so that
dx_i = dx^i + [g_i0 dx^0 = -dx^i] = 0
Covariant (x_u) and contravariant (x^u) axes become
different in non-orthogonal CS's. In SR this is physically
due to effects of aberration, because light is deflected.
Since the direction of light is used to define axes in
relativity, axis directions are different for relatively moving
observers.
As a result of the term in [], dx_i =0 is equivalent to the LT,
when g_00 ...g_33 = 1, which simplifies things.
The contravariant spatial displacement dx^i, is given by
dx^i = g^iu dx_u = g^i0 dx_0, because dx_i=0, so
g^i0 = + dx^i/dx_0
Also ds^2 = dx_u dx^u = dx_0 dx^0 because
dx_i dx^i =0. If I had to define *proper time*
using coordinate time only,
ds^2 =dx_0 dx^0
looks reasonable.
If dx_i dx^i =/=0 this would suggest (to me) an
absolute velocity, so it makes sense that dx_i dx^i =0,
but of course relative velocity dx^i exists, so dx^i =/=0,
and so dx_i =0.
Another interesting result of dx_i =0 (2D) is,
g = |g_uv| = |+1 -v/c| = 1 - v^2/c^2
|-v/c +1|
and shows g=0 when v=c, and is reasonable,
since a proper transformation requires g=/=0.
Therefore a proper transformation is impossible
between CS's moving at relative v=c.
Sorry to run on, regards
Ken S. Tucker
Big Bird
> John Baez:
> "It's because some people mistakenly thought this once upon a time, and the
> error found its way into a bunch of popularizations and maybe even some
> textbooks."
>
>
> And hence, this is a good time for another quest for the origin of a claim.
> Is it myth? Or is it science?
>
The term "origin" here is ambiguous.
I understand from the context that you mean it in the historical
sense, i.e. "who said it first". I have little to add to the things
others have said.
But another way of looking at it is "why do so many people believe in
it" i.e. "what is at the root of the wide distribution of this
notion". And it seems that nobody has mentioned the following yet:
when you introduce relativity to a novice, you'd be out of your mind
to throw accelerated coordinate systems at them. Instead, you derive
everything up to and including Lorentz transforms for inertial
coordinate systems. And then the semester is over and the pupil will
never have to look at MTW ch 6.1.
That pupil then goes on to write some pop-sci book. Which will end
exactly where their own education ended. Or teach science in
highschool. How many science teachers have gone beyond basic Lorentz
transforms? And yet they are the ones who'll have to answer to the
bright highschooler's questions.
Let me propose this as the "origin" of this misconception as a
wide-spread one.
Pmb
"As such you can have frame dragging in flat spacetime"
when I should have said
"As such you can have frame dragging in flat spacetime in an inertial
frame."
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote
> Just a bit of background here...
> Studying Greg Hansen's posts in this thread he suggested
> a non-zero Christoffel (acceleration in SR) is compatible
> with SR and Lorentz Transforms, and above he mentions,
> "off diagonal terms with time". In "LT for dummies", I show
> the spacetime metric (g_01) required for a moving clock
> is, (in 2D = 0,1)
The metric is indenpendant of whether a clock is moving or not. If you find
a metric which depends on the motion of a clock then you've done something
wrong. Typically the coordinate clock is not moving by definition of
"coordinate clock"
> To Pete's question, "Why can't the particle move in the
> dx_i direction?".
>
> Consider the association dx_i = g_iu dx^u
> {i =1,2,3 u =0,1,2,3}
>
> dx_i = g_ij dx^j + g_i0 dx^0
>
> Using a simple cartesian CS g_ij dx^j = dx^i, so
This is incorrect. First off the quantity g_ij dx^j is only part of a
covariant 4-vector. You have it as part of a contravariant 4-vector. And
note that g_ij dx^j **does not** equal dx_i at all.
>
> dx_i = dx^i + g_i0 dx^0
See what you have here? you're doing an invalid addition here. You can't add
covariant components of a vector of a vector and top the contravariant
components of a vector. Either "i" is a superscript of a subscript - not
both.
Pmb
Ken S. Tucker
>Note - I made an error in my post. I said
>"As such you can have frame dragging in flat spacetime"
>when I should have said
>"As such you can have frame dragging in flat spacetime in an inertial
>frame."
Thanks Pete, great questions...
Pete when you previously mentioned "frame-dragging", I
deliberately did not respond. Frame-dragging is usually an
exotic GR reaction caused by rotating masses, and has no
application in this simple massless coordinate transformation.
>"Ken S. Tucker" <dyna...@vianet.on.ca> wrote
>
>> Just a bit of background here...
>> Studying Greg Hansen's posts in this thread he suggested
>> a non-zero Christoffel (acceleration in SR) is compatible
>> with SR and Lorentz Transforms, and above he mentions,
>> "off diagonal terms with time". In "LT for dummies", I show
>> the spacetime metric (g_01) required for a moving clock
>> is, (in 2D = 0,1)
Pete says,
>The metric is independant of whether a clock is moving or not.
Pete has raised an extremely important point here, suggesting,
1) the metric is independant of the phenomena being
measured.
This view is fairly conventional.
While I have suggested,
2) the metric is subject to the phenomena being measured.
Herein I take the view that the metric is defined by a relation
(relativity) and has no absolute significance. The metric needs
two points to be defined.
>If you find a metric which depends on the motion of a clock
>then you've done something wrong.
Not necessarily, there is no law against everyone deciding
to use one single arbituarily moving clock as a reference,
and define ones metrics relatively to that clock's period.
That's the whole point of General Covariance.
Let me put this in Human terms. Little kenny tucker
is sitting at the origin of a CS, kst. At this location,
(the origin) are lots of clocks moving with various
uniform velocities, speeds and accelerations.
Little kenny tucker has no clock (sniff) of his own.
Can little kenny tucker decide one clock is better
than any other clock?
Well no, every clock is correct, all little kenny
tucker needs to do is relate the clocks to his
location.
For example, replace the clocks by stars, lots of
stars move relative to kst, and if you suppose each
has an intrinsic equal brightness and frequency, then
kst will observe many different frequencies and
brightness and variations of location of these stars
from the point of view of kst, when Doppler effects,
aberration and relativistic time dilation is taken into
account.
Now, how the hell does little kenny tucker decide which
clock is more truthful?
Well, the're all truthful, but since little kenny tucker
has no clock in his frame, he's compelled to rely on
other clocks and develope his time base in a manner
that relates all clocks.
>Typically the coordinate clock is not moving by definition of
>"coordinate clock"
Agreed, but, there is no such thing as a coordinate clock in GR.
Any clock is as good as every other clock. There is no preferred
clock, (see what little kenny had to do).
>> To Pete's question, "Why can't the particle move in the
>> dx_i direction?".
>> Consider the association dx_i = g_iu dx^u
>> {i =1,2,3 u =0,1,2,3}
>> dx_i = g_ij dx^j + g_i0 dx^0
>> Using a simple cartesian CS g_ij dx^j = dx^i, so
>
>This is incorrect. First off the quantity g_ij dx^j is only part of a
>covariant 4-vector. You have it as part of a contravariant 4-vector. And
>note that g_ij dx^j **does not** equal dx_i at all.
Pete, I specified [i,j] are summed over 1,2,3 only,
so of course there a 3-velocity only (not a 4-velocity as you
thought)
>> dx_i = dx^i + g_i0 dx^0
>
>See what you have here? you're doing an invalid addition here. You can't add
>covariant components of a vector of a vector and top the contravariant
>components of a vector.
Pete, you need to recognize the condition, I stated,
"Using a simple cartesian CS g_ij dx^j = dx^i, "
where i,j = 1,2,3. Cartesian means g_ij =0 if i=/=j.
>Either "i" is a superscript of a subscript - not both.
>Pmb
After specializing CS subs into a tensor's component's,
things like that show up, they're not generally covariant,
just an artifact of the chosen CS.
Regards Ken S. Tucker
Gauge
> "Pmb" <som...@somewhere.com> wrote in message news:<O4_5b.24286$zL5....@nwrdny02.gnilink.net>...
> >Note - I made an error in my post. I said
> >"As such you can have frame dragging in flat spacetime"
> >when I should have said
> >"As such you can have frame dragging in flat spacetime in an inertial
> >frame."
>
> Thanks Pete, great questions...
> Pete when you previously mentioned "frame-dragging", I
> deliberately did not respond. Frame-dragging is usually an
> exotic GR reaction caused by rotating masses, and has no
> application in this simple massless coordinate transformation.
Simply put - The phenomena of frame dragging is that when a particle
falls its swept away. This is due to the g_0k component of the metric
tensor. Now recall that a uniformly accelerating frame of referance is
equivalent to uniform gravitational field. A uniform gravitational
field can be generated by an infinite sheet of matter of uniform mass
density. If that sheet is in motion (i.e. there is a relative motion
between the material the sheet of and the observer) then this flow of
matter will cause a frame dragging effect.
Think of frame dragging as the analog of a magnetic effect - hence the
term gravitomagnetic field. Now think of the sheet of matter in the
same way you'd think of a sheet with a uniform charge density. Then
such a moving sheet will give rise to a magnetic field. However this
holds when the electric field is uniform as in the case of a sheet of
charge. So too with the sheet of matter. Since the effect will be
there for the moving matter and since the uniform field is equivalent
to a uniformly accelerating frame of referance then it follows that
there will be a frame dragging effect when the observer is moving at
constant velocity perpendicular to the direction of acceleration -
e.g. like a car driving in the cabin of a large accelerating rocket
ship.
To give you a better idea see
http://www.geocities.com/physics_world/gr/uniform_move_trans.htm
Note equation 30. This is the frame dragging effect.
Pmb
Ken S. Tucker
>dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03090...@posting.google.com>...
>> Thanks Pete, great questions...
>> Pete when you previously mentioned "frame-dragging", I
>> deliberately did not respond. Frame-dragging is usually an
>> exotic GR reaction caused by rotating masses, and has no
>> application in this simple massless coordinate transformation.
>
>Simply put - The phenomena of frame dragging is that when a particle
>falls its swept away. This is due to the g_0k component of the metric
>tensor.
Well, *frame dragging* is an unproven theory that Gravity
Probe B intends to research.
>Now recall that a uniformly accelerating frame of referance is
>equivalent to uniform gravitational field. A uniform gravitational
>field can be generated by an infinite sheet of matter of uniform mass
>density. If that sheet is in motion (i.e. there is a relative motion
>between the material the sheet of and the observer) then this flow of
>matter will cause a frame dragging effect.
That's only a theory. You speak about "frame-dragging" as
if it was verified. No one can do that yet.
>Think of frame dragging as the analog of a magnetic effect - hence the
>term gravitomagnetic field.
Actually Pete I spent a couple of days at NASA's Marshall
Spaceflight Center evaluating, Dr. Ning Li's applications of
theoretical physics to the problem of gravitational attenuation
by gravitomagnetic effects induced by spinning super-conducting
discs, at the invitation of Lott Brantley.
>Now think of the sheet of matter in the
>same way you'd think of a sheet with a uniform charge density. Then
>such a moving sheet will give rise to a magnetic field.
You can't use this simplistic analogy. An object in free-fall
has an accelometer reading that is always zero. However,
an object subject to EM-force has the accelometer reading
varying in accord with the photonic energy deviating this
otherwise inertial system into a non-inertial (accelerated)
system, and path.
>However this
>holds when the electric field is uniform as in the case of a sheet of
>charge. So too with the sheet of matter.
I think this is a forced analogy.
>Since the effect will be there for the moving matter and since the
>uniform field is equivalent
>to a uniformly accelerating frame of referance then it follows that
>there will be a frame dragging effect when the observer is moving at
>constant velocity perpendicular to the direction of acceleration -
Pete, I think the term frame dragging applies in the presence of
matter, Hansen and I were trying to figure out if it's reasonable
to have non-zero Christoffel's in SR. Hansen suggested, terms
like g_0i exist and vary by relative CS acceleration without
gravity. That's what this thread is about.
>e.g. like a car driving in the cabin of a large accelerating rocket
>ship.
>
>To give you a better idea see
>http://www.geocities.com/physics_world/gr/uniform_move_trans.htm
>Note equation 30. This is the frame dragging effect.
>Pmb
Pete, I find the *frame dragging* effect does not exist, and
if you want we should start a new thread on that subject.
Pmb
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:2202379a.03090...@posting.google.com...
> gau...@hotmail.com (Gauge) wrote in message
news:<e7203033.03090...@posting.google.com>...
>
> >dyna...@vianet.on.ca (Ken S. Tucker) wrote in message
news:<2202379a.03090...@posting.google.com>...
> >> Thanks Pete, great questions...
> >> Pete when you previously mentioned "frame-dragging", I
> >> deliberately did not respond. Frame-dragging is usually an
> >> exotic GR reaction caused by rotating masses, and has no
> >> application in this simple massless coordinate transformation.
> >
> >Simply put - The phenomena of frame dragging is that when a particle
> >falls its swept away. This is due to the g_0k component of the metric
> >tensor.
>
> Well, *frame dragging* is an unproven theory that Gravity
> Probe B intends to research.
The meaning is well defined. Whether nature exhibits frame dragging is
another thing.
>
> >Now recall that a uniformly accelerating frame of referance is
> >equivalent to uniform gravitational field. A uniform gravitational
> >field can be generated by an infinite sheet of matter of uniform mass
> >density. If that sheet is in motion (i.e. there is a relative motion
> >between the material the sheet of and the observer) then this flow of
> >matter will cause a frame dragging effect.
>
> That's only a theory. You speak about "frame-dragging" as
> if it was verified. No one can do that yet.
We're speaking about relativity and what relativity predicts right?
>
> >Now think of the sheet of matter in the
> >same way you'd think of a sheet with a uniform charge density. Then
> >such a moving sheet will give rise to a magnetic field.
>
> You can't use this simplistic analogy. An object in free-fall
> has an accelometer reading that is always zero. However,
> an object subject to EM-force has the accelometer reading
> varying in accord with the photonic energy deviating this
> otherwise inertial system into a non-inertial (accelerated)
> system, and path.
That's beside the point Ken. We're talking about the motion of a particle
and not the operation of an accelerometer. The equations are nearly
identical. And the topic here is the frame dragging which is predicted using
gravitomagnetism.
>
> >However this
> >holds when the electric field is uniform as in the case of a sheet of
> >charge. So too with the sheet of matter.
>
> I think this is a forced analogy.
Have you ever studied gravitomagnetism and the equations of motion?
> Pete, I think the term frame dragging applies in the presence of
> matter, ...
Tell you what - If you find an error in the derivations and then we'll
continue.
Pmb
Tom Roberts
> Well, *frame dragging* is an unproven theory that Gravity
> Probe B intends to research.
Not true. Frame dragging is a direct consequence of GR, and is not any
sort of independent "theory". It has been measured using the LAGEOS
satellites (with error bars ~50%); I believe it has also been measured
in binary pulsars, but am not certain.
Note that no theory of physics can ever be proven. The best one can do
is have a theory that has not been refuted by experiments (within its
domain of applicability). Such as GR.
> That's only a theory. You speak about "frame-dragging" as
> if it was verified. No one can do that yet.
Look up the LAGEOS satellite measurements.
Tom Roberts tjro...@lucent.com
Ken S. Tucker
>"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
>news:2202379a.03090...@posting.google.com...
>> gau...@hotmail.com (Gauge) wrote in message news:<e7203033.03090...@posting.google.com>...
>>
>> >dyna...@vianet.on.ca (Ken S. Tucker) wrote in message news:<2202379a.03090...@posting.google.com>...
...
>> >Simply put - The phenomena of frame dragging is that when a particle
>> >falls its swept away. This is due to the g_0k component of the metric
>> >tensor.
>>
>> Well, *frame dragging* is an unproven theory that Gravity
>> Probe B intends to research.
>
>The meaning is well defined. Whether nature exhibits frame dragging is
>another thing.
Agreed.
>> >Now recall that a uniformly accelerating frame of referance is
>> >equivalent to uniform gravitational field. A uniform gravitational
>> >field can be generated by an infinite sheet of matter of uniform mass
>> >density. If that sheet is in motion (i.e. there is a relative motion
>> >between the material the sheet of and the observer) then this flow of
>> >matter will cause a frame dragging effect.
>>
>> That's only a theory. You speak about "frame-dragging" as
>> if it was verified. No one can do that yet.
>
>We're speaking about relativity and what relativity predicts right?
Ok, if you say so, more below.
>> >Now think of the sheet of matter in the
>> >same way you'd think of a sheet with a uniform charge density. Then
>> >such a moving sheet will give rise to a magnetic field.
>>
>> You can't use this simplistic analogy. An object in free-fall
>> has an accelometer reading that is always zero. However,
>> an object subject to EM-force has the accelometer reading
>> varying in accord with the photonic energy deviating this
>> otherwise inertial system into a non-inertial (accelerated)
>> system, and path.
>
>That's beside the point Ken. We're talking about the motion of a particle
>and not the operation of an accelerometer. The equations are nearly
>identical.
Nearly indentical? The difference is acceleration due to gravitation
and acceleration due to quantum events. IMO the source of
acceleration is important and distinct.
>And the topic here is the frame dragging which is predicted using
>gravitomagnetism.
Right, but (and hear me out because this statement is well
referenced), 'gravitomagnetism is not a prediction of relativity,
it is a prediction of Einstein's based on Mach's principle hypo-
thetically applied to relavity'. Please see, A. Einstein's "The
Meaning of Relativity", (I have the 5th Edition). Turn to pg.
100, and read from, "But in the second place...probable that
Mach...", and please note the word *probable*.
The *frame-dragging* effect is found in Eq. (117) of that ref.
(gamma _4a = ).
For a second reference, that involves a modern treatment
of frame dragging, (using the Kerr metric), please see,
S. Weinberg's, "Grav and Cosmo", pg 240. Read down
a bit and see that Weinberg says " *looks* like the metric",
(his emphasis), in reference to the Kerr metric.
Neither of these authors says relativity *predicts* frame-
dragging, on the contrary, it is only a suggestion.
>> >However this
>> >holds when the electric field is uniform as in the case of a sheet of
>> >charge. So too with the sheet of matter.
>>
>> I think this is a forced analogy.
>
>Have you ever studied gravitomagnetism and the equations of motion?
Yes, I have not yet been able to substantiate the *prediction*.
>> Pete, I think the term frame dragging applies in the presence of
>> matter, ...
>
>Tell you what - If you find an error in the derivations and then we'll
>continue.
Well, it's not a matter of finding errors in calculations, it's a
matter of the principles supporting the assumptions of those
calculations, as my references show. If you would re-read
your web page on this, and detail the assumptions and
hypothesis, as Einstein and Weinberg have, it would be
easier to understand.
>Pmb
Regards
Ken S. Tucker
Ken S. Tucker
>Ken S. Tucker wrote:
>> Well, *frame dragging* is an unproven theory that Gravity
>> Probe B intends to research.
>
>Not true. Frame dragging is a direct consequence of GR, and is not any
>sort of independent "theory".
Tom, I disagree for reasons that I posted to Pmb today,
please have a glance at that post.
Einstein and Weinberg do not agree with you, (and I can't
either).
IMO, Mach's Principle is ill defined.
>It has been measured using the LAGEOS
>satellites (with error bars ~50%); I believe it has also been measured
>in binary pulsars, but am not certain.
Yeah, experimentally the issue is clearly undecided, as
it appears to be theoretically. That's why GP-B is such
a good experiment.
>Note that no theory of physics can ever be proven. The best one can do
>is have a theory that has not been refuted by experiments (within its
>domain of applicability). Such as GR.
Yes, sofar the data is inconclusive.
>> That's only a theory. You speak about "frame-dragging" as
>> if it was verified. No one can do that yet.
>
>Look up the LAGEOS satellite measurements.
Thanks Tom. I do understand Mach's Principle and
where Einstein's coming from, but the theory is weak
in principle, and data determining a probable decision
is foggy.
((On a personal note, I've postponed further theoretical
research into frame-dragging phenomena pending the
data return from GP-b, ie. launch in 1999....2003?)).
>Tom Roberts tjro...@lucent.com
Pmb
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote
> Nearly indentical? The difference is acceleration due to gravitation
> and acceleration due to quantum events. IMO the source of
> acceleration is important and distinct.
Nope. The difference in the EM equations and the equations of
gravitomagnetism is but a mere factor of 2 - period.
>
> >And the topic here is the frame dragging which is predicted using
> >gravitomagnetism.
>
> Right, but (and hear me out because this statement is well
> referenced), 'gravitomagnetism is not a prediction of relativity,
> it is a prediction of Einstein's based on Mach's principle hypo-
> thetically applied to relavity'.
Do you know what gravitomagnetism is?
Pmb
Ken S. Tucker
Hi Pete
>"Ken S. Tucker" <dyna...@vianet.on.ca> wrote
>
>> Nearly indentical? The difference is acceleration due to gravitation
>> and acceleration due to quantum events. IMO the source of
>> acceleration is important and distinct.
>
>Nope. The difference in the EM equations and the equations of
>gravitomagnetism is but a mere factor of 2 - period.
Ok, it's best I review more recent work prior to commenting
further.
>> >And the topic here is the frame dragging which is predicted using
>> >gravitomagnetism.
>>
>> Right, but (and hear me out because this statement is well
>> referenced), 'gravitomagnetism is not a prediction of relativity,
>> it is a prediction of Einstein's based on Mach's principle hypo-
>> thetically applied to relavity'.
((Pete, you could have indicated that you snipped the opinions
of two Nobel prize winners, you over snip sometimes)).
>Do you know what gravitomagnetism is?
>Pmb
((GRAVITOMAGNETISM - PLO))
Yes, it's suspiciously simple. Let me explain my view...
Place a spinning flywheel in a box, find an experiment
to determine which way the flywheel is spinning using
gravity alone.
An answer: Have two objects of unit mass revolve
equally but in opposite directions around the box, in
the plane of spin. The rate of rotation of the flywheel
will be different relatively to the objects, hence the
energy of the flywheel (and it's gravitational mass) is
different for the contrary revolving objects.
So the object contra-revolving senses the spinning
wheel to rotate faster than the co-revolving object,
and thus the contra-revolving object senses the kinetic
and gravitational energy to be greater than the
co-revolving object measures.
Slight differences in the orbits of the contra-revolving
and co-revolving objects reveals the direction of spin.
Sound simple enough?
Chapter 2
Sister Kenny strolls in carrying a purse, with GR in it.
|
***********
* G_uv=0 *
***********
And throws her purse at any point beyond the rim
of the spinning flywheel and asks, how can my purse
find direction to occur from my symmetrical G_uv=0
at every point in space outside of the box?
She asks, how can frame-dragging exist if G_i4 is
symmetrical? (she's not too bright, but he's is in drag).
My point is: if you intend to have frame-dragging
consistent with G_uv=0, outside of the box then
G_i4 = - G_4i =0 is the solution, and I haven't been
able to solve this.
If you want frame-dragging then we need a non-
symmetrical G_uv, that is found from an non-
symmetrical T_uv. This requires anti-symmetrical
components within T_uv that depend on direction
of matter rotation, (in the spinning flywheel), so
that beyond the flywheel T_uv <>0, and it follows
G_uv <>0.
That's a characteristic of EM-fields when circular
currents exist, but hard to create with neutral matter
in GR.
Is that gravitomagnetism?
Regards
Ken S. Tucker
Pmb
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:2202379a.03090...@posting.google.com...
All gravitomagnetism is is general relativity - they are one in the same
thing - the difference is zero. Its called gravitomagnetism when you're
using the weak field limit and using the analogy with electromagnetism since
they have nearly identical equations. If you think GR is right then there's
zero reason to think that gravitomagnetism is wrong.
Pmb
Pmb
> > Well, *frame dragging* is an unproven theory that Gravity
> > Probe B intends to research.
>
> Not true. Frame dragging is a direct consequence of GR, and is not any
> sort of independent "theory". It has been measured using the LAGEOS
> satellites (with error bars ~50%); I believe it has also been measured
> in binary pulsars, but am not certain.
I assume that you're referring to the binary pulsar PSR 1913+16?
Pmb
Ken S. Tucker
>"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
>news:2202379a.03090...@posting.google.com...
Pete, no insults are intended, I'm a bit of a joker...
That statement is imprecise.
>Its called gravitomagnetism when you're
>using the weak field limit and using the analogy with electromagnetism
We all know how dangerous analogies can be in complex
problems.
>since they have nearly identical equations.
"nearly identical", how near? Are we playing horse shoes?
(pardon the jest).
For example, we can shield electromagnetism, but can
we expect to shield gravitomagnetism? That's a big diff.
>If you think GR is right then there's
>zero reason to think that gravitomagnetism is wrong.
Pete, you're forcing a conclusion. Let me ask you a
direct question: Does G_uv=0 describe a field that
is spherically symmetrical about a gravitating mass
point?
I think Pmb (and all) agree it's yes.
And so (IMO) this cannot provide a solution the causes
a deflection orthogonal to the radial direction, ie. the
frame-dragging.
Chapter 3 ((PLO))
So to find a gravitomagnetic field we'll need to resort
to G_uv = k T_uv. In this case, I introduced above
the counter revolving satellites. These satellites have
different revolutionary velocities relatively to the
spinning gravitating body they orbit, in other words,
differing energy density currents.
Awhile back Pmb wrote quite a bit on the stress-
energy tensor, so his opinion should be considered
quite valid.
Let me keep this simple, we have two satellites, one
orbiting CW (clockwise) and the other (CCW), and the
gravitating mass is spinning CW or CCW as preferred.
This produces two separate field equation solutions, (k=1),
G_uv (CW) = T_uv (CW)
and
G_uv (CCW) = T_uv (CCW)
for the satellites.
This is not simple, but we can try by taking the differences,
dG_uv = G_uv (CW) - G_uv (CCW)
dT_uv = T_uv (CW) - T_uv (CCW)
If the gravitating body is NOT spinning then,
dG_uv =0 and dT_uv=0, therefore the quantity
dT_uv <>0 could be caused by the spinning body and
potentially cause a gravitomagnetic effect in a direction
related to the direction of spin.
Here's my problem, dT_uv must somehow encode the
direction of the spin of the gravitating body.
Supposing the plane of spin and the satellites orbital
revolution are in X and Y plane, then we must find
anti-symmetrical components to describe the bodies
spin using T_uv like,
dT_12 = - dT_21 =/=0.
that encodes the spin of the body and creates the
gravitomagnetic deflection in the field difference dG_uv,
that varies the satellites geodesics from being symmetrical.
Are anti-symmetrical components generated by
differentiating T_uv?
Regards
Ken S. Tucker
Pmb
Why?
>
> >Its called gravitomagnetism when you're
> >using the weak field limit and using the analogy with electromagnetism
>
> We all know how dangerous analogies can be in complex
> problems.
'Then back up your claim that its dangerous in the case of GR.
>
> >since they have nearly identical equations.
>
> "nearly identical", how near?
A factor of 2 in one equation as I recall. I thought you said you were
familiar with it?
> For example, we can shield electromagnetism, but can
> we expect to shield gravitomagnetism? That's a big diff.
Ken - I said the equations are nearly identical. I did not say that GR is
nearly identical to EM. That's so obviously not true as to not be worthy of
mention.
>
> >If you think GR is right then there's
> >zero reason to think that gravitomagnetism is wrong.
>
> Pete, you're forcing a conclusion. Let me ask you a
> direct question: Does G_uv=0 describe a field that
> is spherically symmetrical about a gravitating mass
> point?
That's not a meaningfull question Ken. All that says is that there is no
matter at the point you're describing. Stating "There is no matter at point
R" does not tell us anything except that there is no matter there.
Have you ever seen the equations of gravitomagnetism?
Pmb
Bill Hobba
Tom Roberts wrote:
> Be careful where you "look it up", as elementary books tend to take the
> simplistic way out....
>
Aren't that the truth. On page 5 of Introduction to Special Relativity
Rindler defines SR as:
'special relativity is the theory of an ideal physics refereed to an ideal
set of infinitely extended gravity-free inertial reference frames'
Now of course Rindler is a genuine expert so all you can conclude is that
what Tom said is true.
BTW he also used rigid scales in his definition of an inertial reference
frame. Rigid in SR?
Thanks
Bill