Geodesic Definition
Dan Becker
& Nightingale. They introduce the idea of a geodesic by first talking
about Euclidean space and describing a geodesic as a straight line,
saying "What makes a straight line straight is that it's tangent
vectors all point in the same direction". They then go on to derive a
differential equation (involving the connection coefficients and
metric) which a straight line must satisfy.
Later, they use the same equation to describe a geodesic on a surface
in Euclidean space, and also to describe a geodesic in a manifold.
I don't understand this idea that "tangent vectors all point in the
same direction". In Euclidean space this makes sense. But for, the
surface of a sphere, it doesn't make sense to me. If your curve is
the equator, the tangent vectors don't all point in the same
direction, right?
The more I think about it, the less the less I understand. Doesn't
each point in a manifold have a *different* tangent space? So how can
it even be meaningful to talk about tangent vectors at different
points "pointing in the same direction".
Confused,
Dan
Robert Low
> I'm reading the book "A Short Course In General Relativity" by Foster
> & Nightingale. ... They then go on to derive a
> differential equation (involving the connection coefficients and
> metric) which a straight line must satisfy.
> ...
> The more I think about it, the less the less I understand. Doesn't
> each point in a manifold have a *different* tangent space? So how can
> it even be meaningful to talk about tangent vectors at different
> points "pointing in the same direction".
This is exactly correct. What a connection does is give
you a way of carrying a vector along a curve. If the tangent
vector is carried into itself by this process, then you
have a geodesic. Note that whether a curve is a geodesic
depends on the connection you choose.
--
Rob
Gordon D. Pusch
mtx...@coventry.ac.uk (Robert Low) writes:
Note, however, that this definition assumes the connection is
torsion-free. If the connection has torsion, then a "geodesic"
defined as an extremal path is in general different from an
"autoparallel" defined vis parallel transport.
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
Daniel Grubb
Yes, the tangent spaces at different points of a manifold are different.
This is overcome by the idea of a connection. Essentially, a connection
allows you to compare vectors at different points *given a path between
these points*. In other words, given a path between points x and y, there
is an isomorphism of the tangent spaces at x and at y. This isomorphism
depends on the path chosen between the points. It is by means of this
isomorphism that we talk about 'parallel transport' of vectors along
paths.
Now, a geodesic is a path such that the tangent vectors are parallel
transported to tangent vectors through parallel transport *along the
path*. This is what your book was trying to say when it said that the
tangent vectors point in the same direction. Every Riemannian manifold
has a natural connection, the Levi-Civita connection, that is used to
determine the 'natural' parallel transport of vectors. The connection
coefficients that you mentioned are those from the Levi-Civita
connection.
There is also another way of deriving the the differential equation
of a geodesic. Write down the integral that describes 'arc length'
of a curve using the metric. When you use calculaus of variations
to find paths of extremal arc length, you get the geodesic formula.
In the case of Reimannian manifolds, 'extremal' means 'minimal', so
geodesics are paths of (locally) minimal arc length. For general
relativity, the metric is not positive definite (it has negative signs
along the diagonal), so the extremal paths are actually *maximal* arc
length. Also important is that the 'arc length' of a path is just the
proper time along that curve. Thus, the geodesics in GR are paths with
(locally) maximal proper time.
I hope this helps!
--Dan Grubb
Oz
>In the case of Reimannian manifolds, 'extremal' means 'minimal', so
>geodesics are paths of (locally) minimal arc length. For general
>relativity, the metric is not positive definite (it has negative signs
>along the diagonal), so the extremal paths are actually *maximal* arc
>length. Also important is that the 'arc length' of a path is just the
>proper time along that curve. Thus, the geodesics in GR are paths with
>(locally) maximal proper time.
Does this have any relationship with minimisation of action?
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Note: soon (maybe already) only posts via despammed.com will be accepted.
Paul Reilly
Oz <aco...@btopenworld.com> wrote in message news:<b74ma1$t1g$1...@lfa222122.richmond.edu>...
> Daniel Grubb <gr...@lola.math.niu.edu> writes
>
> >In the case of Reimannian manifolds, 'extremal' means 'minimal', so
> >geodesics are paths of (locally) minimal arc length. For general
> >relativity, the metric is not positive definite (it has negative signs
> >along the diagonal), so the extremal paths are actually *maximal* arc
> >length. Also important is that the 'arc length' of a path is just the
> >proper time along that curve. Thus, the geodesics in GR are paths with
> >(locally) maximal proper time.
>
> Does this have any relationship with minimisation of action?
Yes, very much so.
Take for example a null geodesic in some spacetime. Consider the spacetime
geometry to be fixed and vary the path of low-intensity (low enough not
to disturb the geometry) electromagnetic wave. The wave takes an extremal
path because slight variations from the extremal path are in phase,
whereas on a non extremal path the variations change the phase and cause
destructive interference.
You can view this as finding the extreme of the path length, but you can
also view it as extremizing the action. Same thing with a timelike
geodesic - we can do this problem classically or (in a fixed background)
quantum mechanically. The quantum mechanical version is very similar to
the result for EM waves.
The classical version: write down the normal action for your matter or
energy (being careful with volume elements) plus a term equal to the
Ricci scalar times the square root of minus the determinant of the metric.
Moderator, please fix this if I said it wrong - I should pop out Dirac's
little GR book to get it right for sure.
When you vary the R*sqrt(-g) term, you get the Einstein equations, but
when you vary the action for the matter you get geodesic paths. It's
explained very nicely in Dirac's tiny book on general relativity, better
IMO than it is in Wald or MTW, and it's a book that's safe to drop on your
foot.
Sorry for not posting equations in detail but I am tired and fear making
mistakes of capitalization or whatever and getting pounced on :) I can
post a little derivation of geodesics from action later if you like.
The main point I want to make is that your intuition is correct - it's
not just that you have two ways to find the geodesic via path length vs.
action extremization, it is that the whole idea of what "path length"
means is tied up with the Einstein-Hilbert action, in GR.
Iain
<snip>
> Take for example a null geodesic in some spacetime. Consider the spacetime
> geometry to be fixed and vary the path of low-intensity (low enough not
> to disturb the geometry) electromagnetic wave. The wave takes an extremal
> path because slight variations from the extremal path are in phase,
> whereas on a non extremal path the variations change the phase and cause
> destructive interference.
I am rather skeptical about this phase cancellation argument. I sounds
very much like hand-waving to me - can you back up your statement that
the change in phase causes destructive interference?
-I
Tim S
on 9/4/03 12:33 am, Dan Becker at dt...@yahoo.com wrote:
> I'm reading the book "A Short Course In General Relativity" by Foster
> & Nightingale.
My goodness me, does that still exist? It was the first GR book I ever read,
when I was 15 or so. From it, I learned that I should have read a book on
differential geometry first :-).
> They introduce the idea of a geodesic by first talking
> about Euclidean space and describing a geodesic as a straight line,
> saying "What makes a straight line straight is that it's tangent
> vectors all point in the same direction". They then go on to derive a
> differential equation (involving the connection coefficients and
> metric) which a straight line must satisfy.
OK...
>
> Later, they use the same equation to describe a geodesic on a surface
> in Euclidean space, and also to describe a geodesic in a manifold.
OK...
>
> I don't understand this idea that "tangent vectors all point in the
> same direction". In Euclidean space this makes sense. But for, the
> surface of a sphere, it doesn't make sense to me. If your curve is
> the equator, the tangent vectors don't all point in the same
> direction, right?
Yes, but we don't need something that strong.
>
> The more I think about it, the less the less I understand. Doesn't
> each point in a manifold have a *different* tangent space?
Indeed.
> So how can
> it even be meaningful to talk about tangent vectors at different
> points "pointing in the same direction".
What we need is something to connect the tangent spaces together. This is
what a 'connection' does -- the thing whose coefficients are the 'connection
coefficients'. What we want is to be able to take a path in the manifold,
with a vector chosen at each point along the path, and say how, and how
quickly, this field of vectors varies as we move along the path. This is the
covariant derivative, and it is defined using the connecion. If the rate of
change of the vector is zero, we are 'parallel transporting' it along the
path. If parallel transporting the tangent to the path from one point gives
us the tangents at the other points, then the path is a geodesic.
For instance, if we take a tangent vector on the equator of a sphere, and
push it along the equator, always keeping it tangent to the sphere and not
rotating it to the north or the south, then we are parallel transporting it.
If it rotated to the north or the south, it would have a non-zero covariant
derivative.
Since the parallel-transported versions of the tangent vector we've picked
are always tangent to the equator, the equator is a geodesic.
Tim
Oz
Paul Reilly <pa...@info-ren.org> writes
>The main point I want to make is that your intuition is correct - it's
>not just that you have two ways to find the geodesic via path length vs.
>action extremization, it is that the whole idea of what "path length"
>means is tied up with the Einstein-Hilbert action, in GR.
OK.
Now the next step.
If, in a gravitational system, least action is simply an expression of
paths following a geodesic, then what happens when we add other forces.
I've asked this before without getting a real answer.
For example imagine a charged particle traversing crossed gravitational
and electric fields.
Presumably one can generate a legrangian for such a system. I presume
(since I certainly do not grok lagrangian formulism) that now we are not
following a geodesic.
This would be fine in itself but I am somewhat uncomfortable with it.
I would prefer, for reasons I can't explain, that the path still follows
a geodesic.
I know, because I have been told (and it's very reasonable), that an
accelerating charged particle emits photons. I fondly imagine (not
really understanding the maths and probably the physics) that the
particle is still following a geodesic but in some space where there is
some "propensity for asymmetric photon in photon emission" (IYSWIM). OK,
that's unclear, I know.
Hmm, put another way if you had a photon-powered spaceship this would be
in some sense indistinguishable from a charged spaceship traversing an
electric field?
Bruce Bowen
dt...@yahoo.com (Dan Becker) wrote in message news:<c935aa8b.03040...@posting.google.com>...
When standing on a hilly surface start walking in any random
direction. Don't turn left or right (if you do you're adding your own
"geodesic" curvature) and your path will be a geodesic.
-Bruce bbo...@pppppppppppppacbell.nettttttttttttttt
Iain
Oz <aco...@btopenworld.com> wrote in message news:<kLr2aNC$Fom+...@btopenworld.com>...
<snip>
> OK.
>
> Now the next step.
>
> If, in a gravitational system, least action is simply an expression of
> paths following a geodesic, then what happens when we add other forces.
>
> I've asked this before without getting a real answer.
>
> For example imagine a charged particle traversing crossed gravitational
> and electric fields.
>
> Presumably one can generate a legrangian for such a system. I presume
> (since I certainly do not grok lagrangian formulism) that now we are not
> following a geodesic.
Yes
> This would be fine in itself but I am somewhat uncomfortable with it.
> I would prefer, for reasons I can't explain, that the path still follows
> a geodesic.
To make that work, you would have to define the other forces as
geometrical, so that instead of a separate EM field, the field itself
would be part of the spatial geometry. I think that this is related to
Kaluza-Klein theories, and possibly superstring theory.
-I
John Baez
In article <b74ma1$t1g$1...@lfa222122.richmond.edu>,
Oz <ozac...@despammed.com> wrote:
>Daniel Grubb <gr...@lola.math.niu.edu> writes
>>Thus, the geodesics in GR are paths with
>>(locally) maximal proper time.
>Does this have any relationship with minimisation of action?
[the Wizard flickers into view]
YES! It's just a special case!
Don't worry too much about the minus signs which
cause the switch from "minimization" to "maximization":
everyone finds them a bit confusing, but it's ultimately
no big deal.
Ken S. Tucker
news:<b7nujm$rpa$1...@glue.ucr.edu>...
Well something has puzzled me since High School about
geodesics. Referencing to Dover's, "Principle of Relativity"
and therein Einstein's Foundation of GR, Eq. (20) where he
employs a basis for a developement of a geodesic, based
on mini-max variations of Integral(ds) and finally in Eq.(22)
derives the geodesic
OTOH one may take the 4-velocity (given by U^u) and
by covariant differentiation, under the condition that
absolute acceleration does not exist, (which is the General
Principle of Relativty) find DU^u = 0 where
DU^u =0, produces Eq. (22) in the above reference,
In other words, the statement DU^u =0 is also a description
of a geodesic, equivalent to that one found using Eq.(20).
I am at a loss to understand if the different means to
arrive at the geodesic is meaningful. (?)
Thanks in advance
Ken S. Tucker
John Baez
In article <2202379a.03041...@posting.google.com>,
Ken S. Tucker <dyna...@vianet.on.ca> wrote something like:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<b7nujm$rpa$1...@glue.ucr.edu>...
>Well something has puzzled me since High School about
>geodesics. Referencing to Dover's, "Principle of Relativity"
>and therein Einstein's Foundation of GR, Eq. (20) where he
>employs a basis for a development of a geodesic, based
>on mini-max variations of Integral(ds) and finally in Eq.(22)
>derives the geodesic.
Yup.
> OTOH one may take the 4-velocity (given by U^u) and
>by covariant differentiation, under the condition that
>absolute acceleration does not exist, (which is the General
>Principle of Relativity) find DU^u = 0 where
>DU^u =0, produces Eq. (22) in the above reference,
>In other words, the statement DU^u =0 is also a description
>of a geodesic, equivalent to that one found using Eq.(20).
Yup.
> I am at a loss to understand if the different means to
>arrive at the geodesic are meaningful. (?)
It's not quite clear what you want to know, so I'll just tell
you some stuff and hope it helps.
Whenever you find any sort of path that minimizes any sort of
quantity, the path that minimizes this quantity usually satisfies
a differential equation. This is a standard routine in classical
mechanics, where the differential equation often goes by the name of
"F = ma". You're describing a slightly fancier special case, namely
the general relativity description of a particle in free fall.
If you don't know how the standard routine works, you might find
this special case puzzling. If so, you'd probably enjoy learning
about "Lagrangians", "action", and the "calculus of variations".
You can read about these in any good book on classical mechanics, e.g.:
Herbert Goldstein, Charles Poole, and John Safko, Classical
Mechanics, Addison Wesley, San Francisco, 2002.
Oz
Oz asked:
|This would be fine in itself but I am somewhat uncomfortable with it.
|I would prefer, for reasons I can't explain, that the path still follows
|a geodesic.
Iain <iainm...@yahoo.com> writes:
>To make that work, you would have to define the other forces as
>geometrical, so that instead of a separate EM field, the field itself
>would be part of the spatial geometry. I think that this is related to
>Kaluza-Klein theories, and possibly superstring theory.
Indeed, but it's clearly more complex than that because accelerating
charged particles emit photons. OTOH co-accelerating (in a gravitational
field) observers do not observe (if I have it right) emitted photons so
there is clearly some relationship there.
========= Oz also asked:=====
Hmm, put another way if you had a photon-powered spaceship this would be
in some sense indistinguishable from a charged spaceship traversing an
electric field?
============
Is this 'not even wrong'?
-----------------------------------
NB The kets for brainless bears thread seems to have exhausted the
teachers <sigh>. This is sad, but I offer great thanks to them for what
was achieved. Thank you all very much.
Ken S. Tucker
Oz asked:
|This would be fine in itself but I am somewhat uncomfortable with it.
|I would prefer, for reasons I can't explain, that the path still follows
|a geodesic.
I agree with Oz, if I may venture an explanation, (pardon my
archaic ref, to R.C. Tolman's Relativity, Thermodynamics...)
His Eq. (83.1) is the 'conventional geodesic' derived from
DU^{\sigma} = 0 where D is the *absolute derivative* and
U^{\sigma} is the 4-velocity.
Since DU^{\sigma} = 0 is a tensor, it is a true law of
physics for all observers using any proper coordinate system
of reference. In words, if this is true in one system then,
acceleration =0 in one system, hence acceleration is relative,
and no absolute acceleration exists.
IMO, this is usually undisputed in g-fields, so the geodesic
equation is acceptable in g-fields.
In juxtaposition to this geodesic is one found in the same ref,
in Eq. (103.1). Therein a Lorentz force term is an additive to
Eq. (83.1) to produce,
DU^{\mu} = Lorentz Acceleration
(Lorentz Acceleration is explanatory by the ref.)
This requires an abandonment of the Principle of General
Relativity, by enabling absolute acceleration.
In view of Eq. (103.1), how is it possible to recognize Oz's
intuition, "that the path still follows a geodesic. "
Are there defects in Eq. (103.1)?
A weak argument is , the original geodesic solution
requires from Einstein's Law, G_uv = kT_uv that
G_uv =0 and T_uv=0, and so adding charges and
mass into the solution of the Einstein Law directly
onto the geodesic, may still preserve the geodesic,
if the field equation G_uv=kT_uv was properly solved.
The strongest argument for Oz's intuition exists by
a violation of Quantum Theory. The equation
DU^{\mu} = Lorentz Acceleration
enables a continuous variation of energy, and this has
been ruled out by experiment.
We must accept that continuous variation of invariant
energy is impossible, in accord with quantum principles.
And this in turn renders, Lorentz Acceleration=0
Regards and Thanks
Ken S. Tucker
PS: Snippable...
By vectors, in support of Oz, kindly consider E dot R,
where E is the Electric field, a charge q is in, and R is q's
distance from E.
E dot R > 0 means charge q is actually spiralling into an E
field in a classical sense. Quantum Reality describes this
as impossible, because E dot R means a continuous variation
of energy.
KST
Iain
Oz <aco...@btopenworld.com> wrote in message news:<b8478r$dqm$1...@panther.uwo.ca>...
> Oz asked:
>
> |This would be fine in itself but I am somewhat uncomfortable with it.
> |I would prefer, for reasons I can't explain, that the path still follows
> |a geodesic.
>
> Iain <iainm...@yahoo.com> writes:
>
> >To make that work, you would have to define the other forces as
> >geometrical, so that instead of a separate EM field, the field itself
> >would be part of the spatial geometry. I think that this is related to
> >Kaluza-Klein theories, and possibly superstring theory.
>
> Indeed, but it's clearly more complex than that because accelerating
> charged particles emit photons. OTOH co-accelerating (in a gravitational
> field) observers do not observe (if I have it right) emitted photons so
> there is clearly some relationship there.
Suppose the two freely falling observers are in the same local
inertial frame, and so can be considered from the standpoint of SR.
Therefore, the EM field causes the emission of photons, but the
gravitational one doesn't, because there is locally, no detectable
gravitational field. When you treat EM and gravity together, one
considers the EM field as arising from a "curled-up" dimension (AFAIK
it is a circle), and the laws of Maxwell's electromagnetism are
included in this framework. This includes the emission of radiation by
charged particles. I am unaware of the details of how the geometry
causes the difference in the behaviour of particles on different types
of geodesics.
> ========= Oz also asked:=====
> Hmm, put another way if you had a photon-powered spaceship this would be
> in some sense indistinguishable from a charged spaceship traversing an
> electric field?
> ============
>
> Is this 'not even wrong'?
Are you thinking of some kind of equivalence principle applied to EM
fields, so that the effect of such a field is locally
indistinguishable from acceleration? I don't know if there is in
Kaluza-Klein an extension to the principle of equivalence relating
charge as well as gravitational mass to inertial mass.
> -----------------------------------
>
> NB The kets for brainless bears thread seems to have exhausted the
> teachers <sigh>. This is sad, but I offer great thanks to them for what
> was achieved. Thank you all very much.
I don't think there are any _brainless_ bears.
-I
Oz
Iain <iainm...@yahoo.com> writes
>
>Suppose the two freely falling observers are in the same local
>inertial frame, and so can be considered from the standpoint of SR.
OK.
>Therefore, the EM field causes the emission of photons, but the
>gravitational one doesn't, because there is locally, no detectable
>gravitational field.
OK. I must have been confused when reading the posts between mighty
wizards. This is what regularly happens when you haven't a clue what
they are on about.
However as I understand it (ie have gleaned from the crumbs off wizardly
tables) we have certain 'odd' things happening when geodesics are not
followed.
1) An uncharged accelerating observer sees himself in a 'hotter'
environment due to unruh radiation. This is a splendidly weak effect as
befits weak gravity.
2) I get the impression that this radiation is NOT uniform but comes
from (probably) the direction the observer is accelerating in.
3) I imagine non-accelerating observers will see the accelerating
observer as hotter if he is accelerating towards them.
4) This effect is orders too small to be a gravitational equivalent to
cyclotron radiation.
5) A charged particle in an electric field will typically accelerate and
emit photons.
6) I am unclear whether such photons would be observed by comoving
identical particles.
7) They presumably also emit/see unruh radiation.
Hmm. I just had a thought. What if gravitons were in fact supremely
unstable (I guess virtual ones might not be) and rapidly decayed in to
unruh radiation so free ones were never seen? Ooops, I have a feeling
this fits into the fireball-worthy 'not even wrong' category.
>When you treat EM and gravity together, one
>considers the EM field as arising from a "curled-up" dimension (AFAIK
>it is a circle), and the laws of Maxwell's electromagnetism are
>included in this framework. This includes the emission of radiation by
>charged particles. I am unaware of the details of how the geometry
>causes the difference in the behaviour of particles on different types
>of geodesics.
I'm certainly unaware, too .....
>> ========= Oz also asked:=====
>> Hmm, put another way if you had a photon-powered spaceship this would be
>> in some sense indistinguishable from a charged spaceship traversing an
>> electric field?
>> ============
>>
>> Is this 'not even wrong'?
>
>Are you thinking of some kind of equivalence principle applied to EM
>fields, so that the effect of such a field is locally
>indistinguishable from acceleration? I don't know if there is in
>Kaluza-Klein an extension to the principle of equivalence relating
>charge as well as gravitational mass to inertial mass.
Actually it was a simple question with a simple answer not requiring
super-sophisticated physics. It does, however, show the depth of my
ignorance.
Let me put it another way.
If we take an electron with some momentum m0 and then accelerate it in
an electric field to some momentum m1 emitting photons of total momentum
mp can we say that m0 = m1 + mp?
Or is some momentum transferred directly to the charging apparatus?
>> -----------------------------------
>>
>> NB The kets for brainless bears thread seems to have exhausted the
>> teachers <sigh>. This is sad, but I offer great thanks to them for what
>> was achieved. Thank you all very much.
>
>I don't think there are any _brainless_ bears.
This refers to a stuffed child's doll, which is a close approximation to
yours truly and is indeed brainless. Fortunately it is also pretty
harmless. It is generally given pats on the head and spoken to in baby
talk, occasionally thrown across the room for entertainment, too.
Hmmmm, an irritatingly good description .....
Ken S. Tucker
Oz <aco...@btopenworld.com> wrote in message news:<tx3UUSB8...@btopenworld.com>...
Hi Oz et al...
I would venture (cautiously) an explanation this way.
Start with a hydrogen atom (proton + electron), and then
let the electron and proton move closer and emit a photon.
In context of Oz's question we may call the proton, the
"charging apparatus".
After this has happened, the proton and electron had
an equal relative change in potential, electrostatically and
magnetically, based on two assumptions,
1) The electron moved the same distance to the proton as
the proton is to the electron, (electrostatically equal change).
2) The rate of rotation of the electron around the proton
is the same as the proton around the electron (magnetic =).
So an electromagnetic diagram describing the changes of the
relatively emitting particles looks like,
p+ ===> . <====e-
But these are opposite charges in opposite directions,
in terms of same charges in same directions becomes,
P+===>. ====>e+
or
P-<====.<====e-
and these vectors are the relatively incremented
electromagnetic potentials of the proton and electron.
A diagram that summarizes this effect should look
like this,
e-......
|......e-
~~~~~~~~> photon
|......p+
p+.....
1st state.......>.2nd state
My reason for these diagrams is to show that the emission
of radiation is due to the relative motion of the apparatus (p+)
and the electron.
Referring to Oz's question, "m0 = m1 + mp?"
The term, "momentum" alone carries relativistic components.
Perhaps the invariant energy would be easier to start with.
In 1st state, Invariant Energy = Invariant (p1) + Invariant (e1).
In 2nd state, Invariant Energy = Invariant (p2) + Invariant (e2)
+ Invariant (photon)
The invariant energies of p2 and e2 are less than p1 and e1
because each surrenders equal electromagnetic potential to
create the photon's invariant energy.
And to Oz's question..." can we say that m0 = m1 + mp?"
I would say yes (?) if these quantities are invariants.
Regards and thanx...
Ken S. Tucker
>>> NB The kets for brainless bears thread seems to have exhausted the
>>> teachers <sigh>. This is sad, but I offer great thanks to them for what
>>> was achieved. Thank you all very much.
>>
>>I don't think there are any _brainless_ bears.
>
>This refers to a stuffed child's doll, which is a close approximation to
>yours truly and is indeed brainless. Fortunately it is also pretty
>harmless. It is generally given pats on the head and spoken to in baby
>talk, occasionally thrown across the room for entertainment, too.
>Hmmmm, an irritatingly good description .....
>Oz
PS: In empathy, I have less lustre than the shiniest book on
the shelf...but something would be lost, if we were burned.
KST
Patrick Massot
> The more I think about it, the less the less I understand. Doesn't
> each point in a manifold have a *different* tangent space? So how can
> it even be meaningful to talk about tangent vectors at different
> points "pointing in the same direction".
>
> Confused,
Hi,
Connections are one of the many things from differential geometry that
you need to know if you want to understand GR and that's not a trivial
concept but you're lucky because there exists one book :
/Tensor geometry/ by Dodson & Poston
which, as far as I know, is much better than any other to explain what
is a connection and it is a very elementary book. There are two
editions, the old one is typed without LaTeX and has some strange
notations but it is still worth reading if you don't have the new
Springer-Verlag edition.
This book begins with the definition of a vector space and ends with
general relativity ! Everything is explained with nice pictures *and*
a rigorous mathematical treatement.
Patrick Massot
Iain
Oz <aco...@btopenworld.com> wrote in message news:<tx3UUSB8...@btopenworld.com>...
This effect is new to me, though I suppose that I should have expected
it, given the existence of Hawking radiation. Unruh radiation is a QFT
effect, experienced by highly accelerated particles, much as Hawking
radiation exists in strong gravitational fields. It is a small effect
- the temperature of the radiation is that of Hawking radiation from a
black hole, but replacing the gravitational field strength with the
acceleration in the equation. The radiation is emitted in the
direction the observer is accelerating in - contrary to what you had.
AFAICS both comoving particles will observe Unruh radiation if they
are accelerated.
> Hmm. I just had a thought. What if gravitons were in fact supremely
> unstable (I guess virtual ones might not be) and rapidly decayed in to
> unruh radiation so free ones were never seen? Ooops, I have a feeling
> this fits into the fireball-worthy 'not even wrong' category.
If gravitons decayed like that, then the gravitational force would be
very short range. For instance, the pion can be regarded as the
carrier of the strong force, and it has an energy ~200MeV. Therefore,
virtual pions can only exist for ~h/E ~2*10^-23 s, and so can travel a
maximum distance of ~ch/E ~6*10^-15 m. The same idea applies to
gravitons (pretending that we have a theory of quantum gravity), so
massive gravitons -> short range virtual gravitons -> no long range
gravity.
<snip>
> >> ========= Oz also asked:=====
> >> Hmm, put another way if you had a photon-powered spaceship this would be
> >> in some sense indistinguishable from a charged spaceship traversing an
> >> electric field?
> >> ============
> >>
> >> Is this 'not even wrong'?
> >
> >Are you thinking of some kind of equivalence principle applied to EM
> >fields, so that the effect of such a field is locally
> >indistinguishable from acceleration? I don't know if there is in
> >Kaluza-Klein an extension to the principle of equivalence relating
> >charge as well as gravitational mass to inertial mass.
>
> Actually it was a simple question with a simple answer not requiring
> super-sophisticated physics. It does, however, show the depth of my
> ignorance.
>
> Let me put it another way.
>
> If we take an electron with some momentum m0 and then accelerate it in
> an electric field to some momentum m1 emitting photons of total momentum
> mp can we say that m0 = m1 + mp?
>
> Or is some momentum transferred directly to the charging apparatus?
One way of thinking about this is to imagine the "apparatus" as
becoming smaller and smaller, with less charge, until there is only a
proton left. It is then easy to see that some momentum is transferred
to the proton. The same applies to the large apparatus. Starting with
the total initial momentum of the system as:
m0 + p0
The final momentum is distributed among the photons, the electron, and
the proton, giving
m1 + mp + p1
As you know these are related by
m0 + p0 = m1 + mp + p1
Giving
m0 = m1 + mp + (p1 - p0)
So yes, some momentum is transferred to the apparatus.
-I
Ken S. Tucker
iainm...@yahoo.com (Iain) wrote in message
news:<68961e7.03050...@posting.google.com>...
>Oz <aco...@btopenworld.com> wrote in message
>news:<tx3UUSB8...@btopenworld.com>...
[truly enormous wads of quoted text deleted by moderator]
>> Let me put it another way.
>>
>> If we take an electron with some momentum m0 and then accelerate it in
>> an electric field to some momentum m1 emitting photons of total momentum
>> mp can we say that m0 = m1 + mp?
>>
>> Or is some momentum transferred directly to the charging apparatus?
>One way of thinking about this is to imagine the "apparatus" as
>becoming smaller and smaller, with less charge, until there is only a
>proton left. It is then easy to see that some momentum is transferred
>to the proton. The same applies to the large apparatus. Starting with
>the total initial momentum of the system as:
>
> m0 + p0
>
>The final momentum is distributed among the photons, the electron, and
>the proton, giving
>
> m1 + mp + p1
>
>As you know these are related by
>
> m0 + p0 = m1 + mp + p1
>
>Giving
>
> m0 = m1 + mp + (p1 - p0)
>
>So yes, some momentum is transferred to the apparatus.
At the risk of over using classical thinking in this example,
imagine a rocket firing thrusters to reduce orbital altitude.
This thruster action also reduces angular energy and
momentum of the rocket by expelling exhaust gas.
From what I understand a similar thing would
happen as an electron reduces orbital altitude relative
to a Proton by emittin a photon. But a complication
occurs because a Proton and Electron are really orbiting
a common center of mass, and so their respective angular
momentums about this CM would be,
Electron Angular Momentum = m*r*w
Proton's Angular Momentum = M*R*w
where m,M and r,R are mass and distance from CM
respectively, and w is a constant for both in radians per
second.
Two objects of unequal masses revolving about a
common CM, are like a balance beam, so that
m*r =M*R, and then,
Electron Angular Momentum = Proton's Angular Momentum
Assuming the frame of reference K remains constant before
and after photon emission (constant meaning no accelerations
applied to the measuring reference) we could reasonably
expect the photon emision will increment the angular
momentum of the Proton and the Electron equally, to
maintain the angular momentum equality. In another post
in this thread, an explanation was given accounting for an
equal reduction in Electromagnetic energy by the Proton
and the Electron when emitting a photon. In this posted
classical point of view, suggests equality of increments
of angular momentum as well, and partly addresses
OZ's question on momentum, in agreement with Iain's
momentum conservation.
But, I'm confused about linear momentum...
It looks like, following the emission of the photon,
a linear momentum reaction is imparted to the
Proton/Electron couple relative to K, and following
the photon emision, the linear velocity of the Proton
will have been incremented as much as the *average*
Electron velocity (to maintain a circular orbit), but
because the Protons mass is greater, (~1800x) then
the Proton's linear momentum somehow received
that much more (1800x) linear momentum.
Regards
Ken S. Tucker
Snippable PS: The relation of an Electron and a
Proton feels a lot like rolling a marble on a bowling
ball (like in the Rolling Ball? post). Some *mirror*
symmetries occur.
Oz
>Referring to Oz's question, "m0 = m1 + mp?"
> The term, "momentum" alone carries relativistic components.
>Perhaps the invariant energy would be easier to start with.
>
>In 1st state, Invariant Energy = Invariant (p1) + Invariant (e1).
>
>In 2nd state, Invariant Energy = Invariant (p2) + Invariant (e2)
>+ Invariant (photon)
>
>The invariant energies of p2 and e2 are less than p1 and e1
>because each surrenders equal electromagnetic potential to
>create the photon's invariant energy.
Of course. A beautiful balance, I should have seen this before asking,
although it's not quite the question I asked.
Paul Reilly
> pa...@info-ren.org (Paul Reilly) wrote in message
news:<2dab5ea.03041...@posting.google.com>...
>
> <snip>
> > Take for example a null geodesic in some spacetime. Consider the spacetime
> I am rather skeptical about this phase cancellation argument. I sounds
> very much like hand-waving to me - can you back up your statement that
> the change in phase causes destructive interference?
>
> -I
See for example Feynmann and Hibbs "Quantum Mechanics and Path
Integrals" (McGraw Hill 1965) chapter 3. I'm sure there are more modern
texts out there but this is handy.
Let me try to find a web reference also... googling... aha
http://www.scs.uiuc.edu/~makri/preprints/paper40.pdf
There is also a book which does this in curved spacetime... I have it at
home and can cite it in another post. Something like "Relativistic
Quantum Field Theory In Curved Spacetime" is the title.