Metric for a Uniform Gravitational Field
Pmb
the metric for a uniformly accelerating frame of reference. The derivation
is
---------------------------------------------------------
As is well known (the derivation has been given in this newsgroup)
if a rocket ship undergoes constant proper acceleration, then its
path will be given (as a function of proper time aboard the ship)
by (choosing units where c=1)
x = 1/g cosh(g tau)
t = 1/g sinh(g tau)
Now, we want to change coordinates to a system in which the
coordinate velocity of the rocket is zero. An obvious choice
is X and T defined by
X = square-root(x^2 - t^2)
T = arctanh(t/square-root(x^2 - t^2))
or the inverse transformations
x = X cosh(gT)
t = X sinh(gT)
To find the metric in these new coordinates, we use the
invariance of the interval ds^2
ds^2 = dt^2 - dx^2
= g'_00 dT^2 + 2 g'_01 dX dT + g'_11 dX^2
Now, we use the transformation equations to relate the
differentials dt and dx to dT and dX:
dt = dX sinh(gT) + gX cosh(gT) dT
dx = dX cosh(gT) + gX sinh(gT) dT
dt^2 = dX^2 sinh^2(gT) + 2 gX dX dT sinh(gT) cosh(gT)
+ g^2 X^2 cosh^2(gT) dT^2
dx^2 = dX^2 cosh^2(gT) + 2 gX dX dT sinh(gT) cosh(gT)
+ g^2 X^2 sinh^2(gT) dT^2
dt^2 - dx^2 = g^2 X^2 dT^2 - dX^2
So, the metric components in terms of X,T are
g'_00 = g^2 X^2
g'_11 = -1
g'_01 = 0
For a clock at "rest" in the accelerated coordinate system,
dX = 0, so the proper time is given by
d tau = square-root(ds^2) = square-root(g^2 X^2 dT^2)
= g X dT
So, for clocks at rest, the clock "rate" increases linearly with X.
(Higher clocks run faster).
There is no equivalence principle used here---this is all
just Special Relativity plus calculus.
---------------------------------------------------------
I would like the opinions of others in this newsgroup on this derivation. I
don't think its valid but have not gone through it completely. However it is
incompatible with the metric for a uniformly accelerating frame of
reference/uniform gravitational field as found everywhere else in the
literature i.e.
ds^2 = c^2(1+gz/c^2)dt^2 - dx^2 - dy^2 - dz^2
In particular the above metric is not valid for X = 0. The transformation
has imaginarty values for all x such that
x^2 - t^2 < 0
Comments please?
Pmb
tes...@tum.bot
> A derivation was provided in sci.physics.relativity by Daryl McCullough for
> the metric for a uniformly accelerating frame of reference. The derivation
> is
[snip quote]
>From the bit you quoted it does not appear that Darryl ever claimed to
derive "a line element for a uniform gravitational field". Rather, he
gave a standard derivation of the standard Rindler line element
ds^2 = -(ax)^2 dt^2 + dx^2 + dy^2 + dz^2,
x > 0
for a halfspace in -Minkowski- spacetime. This does admit an obvious
family of observers who can be interpreted as realizing Einstein's famous
"elevator experiment", but of course the gravitational field (curvature
tensor) of this spacetime vanishes everywhere.
We have often discussed the Rindler chart here in the past. For example,
you can look for a past post studying in detail the trajectories of null
geodesics drawn in the spatial slice t = 0, x > 0; these turn out to have
an interesting and perhaps surprising characterization which is easily
found by elementary reasoning. This observation is related to Darryl's
comment about "gravitational frequency shifting" for the Rindler family of
uniformly accelerating observers.
> I would like the opinions of others in this newsgroup on this
> derivation. I don't think its valid
Are you sure you understand correctly what he was trying to accomplish?
The bit you quoted is completely standard (and valid!)--- in fact I think
you can find it in Taylor & Wheeler.
> it is incompatible with the metric for a uniformly accelerating frame of
> reference/uniform gravitational field as found everywhere else in the
> literature i.e.
>
> ds^2 = c^2(1+gz/c^2)dt^2 - dx^2 - dy^2 - dz^2
No, this is simply a minor variation of the standard Rindler chart! To
see this, consider the chart
ds^2 = -[A(t) x + B(t) y + C(t) z] dt^2 + dx^2 + dy^2 + dz^2,
A(t) x + B(t) y + C(t) z > 0
where A,B,C can be "almost any" smooth functions. As you can see for one
choice of A,B,C this chart specializes to the one given by yourself and
for another choice of A,B,C, it specializes to the chart given by Darryl!
To understand the intuitive meaning of this chart, note that you can
easily read off the frame
e_1 = 1/[A(t) x + B(t) y + C(t) z] @/@t
e_2 = @/@x
e_3 = @/@y
e_4 = @/@z
Here, the timelike unit vector field X = e_1 corresponds to a family of
ideal observers having possibly time-varying acceleration
1
D_X X = ------------------------- [ A(t) e_2 + B(t) e_3 + C(t) e_4 ]
A(t) x + B(t) y + C(t) z
but with vanishing expansion and vorticity tensors. You can easily
compute the curvature tensor and verify that it vanishes, so that the
spacetime in question is locally flat, and indeed it is easy to see that
for reasonable choices of A,B,C, this chart covers a neighborhood
diffeomorphic to a half space of Minkowski spacetime--- just like the
original chart. This should help you understand the relationship between
your version and his.
BTW, if your real question is: "which exact -vacuum- solution in gtr has
constant and uniform -nonzero- gravitational field?", you can look for
past posts (some quite recent!) discussing the Weyl vacuum
ds^2 = -exp(2 m z) dt^2
+ exp(-2 m z) [ exp(-m^2 r^2) (dz^2 + dr^2) + r^2 du^2 ],
-infty < t,z < infty, 0 < r < infty, -pi < u < pi
Under the well-known correspondence between Weyl vacuums and axisymmetric
harmonic functions, it is easy to see that this solution corresponds to
the Newtonian uniform field potential
phi = m z
Furthermore, wrt to the obvious family of static observers, the tidal
tensor is, to second order in the field strength parameter m:
[ 2 m^2 0 0 ]
E_(ab) ~ [ 0 -m^2 0 ]
[ 0 0 -m^2 ]
(To answer a question someone recently asked regarding this notation,
which is fairly standard, the electrogravitic or tidal tensor is one part
of a bipartite decomposition [valid in a Ricci flat spacetime], wrt the
timelike vector field X = e_1, of the fourth rank Riemann tensor into two
-second rank- tensors. The tidal tensor can be defined via
E_(ab) = R_(ambn) X^m X^n
where the indices 1,2,3,4 refer to the basis above. But it is easy to see
that this tensor always "lives" in the spatial hyperplane element
orthogonal to X at each point in the chart, so, always working in the
given frame, we can think of it as indexed by 2,3,4. Furthermore, in the
Newtonian limit it reduces to the Newtonian tidal tensor. This
decomposition, due to Luis Bel, is purely mathematical. It is analogous
to the familiar decomposition, wrt X = e_1, of the EM field tensor (or any
other antisymmetric second rank tensor) into two "spatial" vector fields.
The tidal tensor E(X) is of course analogous to the electric field; the
other tensor appearing in the decomposition is the magnetogravitic tensor
B(X), which is analogous to the magnetic field, and is another "spatial"
tensor. Interested readers can look for many past posts here explaining
all this in great detail, including the reconstruction of the Riemann
tensor from X, E(X), and B(X). You can also search the ArXiV for "Bel
decomposition".)
Thus, to first order in m, in accordance with Einstein's elevator
experiment, the field -vanishes-. IOW, in the weak field limit, the given
solution reduces to Minkowski vacuum. To second order it is nonzero,
constant, and (in the obvious sense for a "diagonal" tensor in Coulomb
form) "uniform". We have discussed in the past the question of whether,
physically speaking, it makes sense to put this solution forward as "the"
solution we asked for.
"T. Essel" (hiding somewhere in cyberspace)
Daryl McCullough
Pmb says...
>
>A derivation was provided in sci.physics.relativity by Daryl McCullough for
>the metric for a uniformly accelerating frame of reference. The derivation
>is
[Skipped]
> ds^2 = ...
> ... = g^2 X^2 dT^2 - dX^2
>I would like the opinions of others in this newsgroup on this derivation. I
>don't think its valid but have not gone through it completely. However it is
>incompatible with the metric for a uniformly accelerating frame of
>reference/uniform gravitational field as found everywhere else in the
>literature i.e.
>
>ds^2 = c^2(1+gz/c^2)dt^2 - dx^2 - dy^2 - dz^2
I think you made a mistake there. The first term should
be (setting c=1)
(1+gz)^2 dt^2
(see for example, page 7 of
http://aether.lbl.gov/www/classes/p139/homework/eight.pdf
or equation 10 of
http://panda.unm.edu/courses/finley/p570/handouts/accelobserv.pdf)
With this correction, the metric above is the same as mine,
except for the substitution X = 1/g + z.
The correct version leads to the equation for proper time
in the case dx = dy = dz = 0
d tau = square-root(ds^2) = (1+gz) dt
which agrees with equation <1> of Peter Brown's
paper http://xxx.lanl.gov/ftp/physics/papers/0204/0204044.pdf
--
Daryl McCullough
Ithaca, NY
Pmb
> >From the bit you quoted it does not appear that Darryl ever claimed to
> derive "a line element for a uniform gravitational field".
According to the equivalence principle the metric for a uniform
gravitational field is identical for the metric for a uniformly accelerating
frame of reference. This was how the subject came up. This was being
discussed in another thread and he did not claim that the equivalence
principle was wrong. In fact his arguements seemed to suggest he agreed with
it.
However his description/arguement for his so-called "derivation" was based
solely on the assumption of a particle, which was accelerating uniformly,
was at rest in the acceleratring frame of referene. He gave no physical
arguements to interpret the meaning of his chosen coordinates. However his
demand that the particle be at rest in that frame is insuficcient
information to obtain the metric for a uniformly accelerating frame of
reference. In this case the form of the metric was correct but the physical
interpretation is wrong. E.g. he claimed in another thread that this metric
is not supposed to work for a clock at the origin of the coordinate system.
However it is clear to me now what is wrong. The X is not a spatial
displacement from the origin. However from his comments it appears to me
that this was what he thought since he seemed unable to understand my
question regarding the rate at which a clock is running at the origin
> Rather, he gave a standard derivation ..
If this derivation is standard then please give me a reference in the GR
literature where it is derived. From how Rindler describes it in his text
darryl did not describe the coordinates in the same way and is giving this a
different physical meaning. According to his and your metric Either X = 0 at
the orgin or, if you're speaking of the metric that Rinlder derives, then X
= 1 at the origin. In the case of the later d(tau) = adt which means that a
clock at the origin of coordinates does not run at the same rate as a clock
at the origin and that cannot be coorrect of course.
> of the standard Rindler line element
>
> ds^2 = -(ax)^2 dt^2 + dx^2 + dy^2 + dz^2,
>
> x > 0
Please define the meaning of these symbols. The X in that metric can't be
spatial distance from the origin. If X was a spatial distance from the
origion then note that it predicts that the proper time interval, d(tau)
measured by the coordinate clock at the origin, dt, is zero since
d(tau) = 0*dt
The metric for a uniform gravitational field is (in your conventions)
ds^2 = -(1+gx)^2dt + dx^2 + dy^2 + dz^2
where x,y,z are spatial displacements. How are these two metrics related?
> Are you sure you understand correctly what he was trying to accomplish?
Of course. But he didn't describe the physical meaning of the X is that
equation.
> The bit you quoted is completely standard (and valid!)--- in fact I think
> you can find it in Taylor & Wheeler.
I'm not saying that its invalid. I'm saying that his results are not what he
thinks they are. However since he didn't explicitly state what he thought
the coordinates meant so I had to guess. He did seem to think that X is a
spatial displacement from the origin though and that is not what X means if
this is supposed to be the Rindler metric.
Taylor and Wheeler do not touch the accelerting frame.
> > it is incompatible with the metric for a uniformly accelerating frame of
> > reference/uniform gravitational field as found everywhere else in the
> > literature i.e.
> >
> > ds^2 = c^2(1+gz/c^2)dt^2 - dx^2 - dy^2 - dz^2
>
> No, this is simply a minor variation of the standard Rindler chart!
Your comment "No." is wrong. The answer is "Yes." This is most certainly the
metric for a uniformly accelerating frame of referance. The metric you gave
was in different coordinates and X is does not represent a spatial
displacement. It's the coordinates/physical meaning, that I'm obviously
inquiring about. I guess I didn't make that clear though - sorry.
> BTW, if your real question is: "which exact -vacuum- solution in gtr has
> constant and uniform -nonzero- gravitational field?",
Nope. I already know that solution. It's in several GR texts, e.g. Mould,
MTW. It was also derived in the literature in several places, e.g.
"Principle of Equivalence," F. Rohrlich, Ann. Phys. 22, 169-191, (1963),
page 173
Taking a look at Rindler's text/derivation it is clear that X = exp(2*Phi)
where Phi is the gravitational potential. Therefore X is *not* a spatial
displacement from the origin since at the origin Phi = 0, X = 1 and
therefore d(tau) = dt as expected. But that's using the metic as Rindler
gives it. Not as the one you provided above.
Expand exp(2*Phi) and ignore second order terms in Phi
exp(z ) ~ 1 + z
z = 2*Phi
exp(2*Phi) ~ 1 + 2*Phi
And that is quite consistent with the metric I gave above and it is
consistent with experiment of course.
Pmb
Daryl McCullough
[My equation]
> ds^2 = ... g^2 X^2 dT^2 - dX^2
>I would like the opinions of others in this newsgroup on this derivation. I
>don't think its valid but have not gone through it completely. However it is
>incompatible with the metric for a uniformly accelerating frame of
>reference/uniform gravitational field as found everywhere else in the
>literature i.e.
>
>ds^2 = c^2(1+gz/c^2)dt^2 - dx^2 - dy^2 - dz^2
Okay, I looked it up in Misner, Thorne, and Wheeler's "Gravitation"
and it agrees with me: page 173, equation 6.18
ds^2 = -(1 + gE1')^2(dE0')^2 + (dE1')^2 + (dE2')^2 + (dE3')^2
(where the Es are actually some squiggly Greek letter)
[Shortest Moderator's Note Ever: xi. -TB]
This is the same as my metric except for (1) I drop all but one spatial
coordinate, (2) I chose the opposite sign (arbitrary convention),
and (3) you have to use the substitutions X = 1/g + E1', and
T = E0'.
So, to the extent that MTW is part of the standard literature,
I think you're wrong that my equation is incompatible with the
ones found everywhere else in the literature.
Daryl McCullough
Pmb says...
>
><tes...@tum.bot> wrote in message news:4035e4db$1...@news.sentex.net...
>> of the standard Rindler line element
>>
>> ds^2 = -(ax)^2 dt^2 + dx^2 + dy^2 + dz^2,
>>
>> x > 0
>
>Please define the meaning of these symbols. The X in that metric can't be
>spatial distance from the origin.
Yes, it is.
>If X was a spatial distance from the
>origion then note that it predicts that the proper time interval, d(tau)
>measured by the coordinate clock at the origin, dt, is zero since
>
>d(tau) = 0*dt
>
>The metric for a uniform gravitational field is (in your conventions)
>
>ds^2 = -(1+gx)^2dt + dx^2 + dy^2 + dz^2
>
>where x,y,z are spatial displacements. How are these two metrics related?
Isn't it pretty clear that if you perform the substitution
x' = x + 1/g,
then your metric is converted into the one above? So for your
metric, the problem arises as x = -1/g, rather than x'=0. Physically,
the choice of the origin doesn't make any difference.
As to the physical meaning of the coordinate x in
ds^2 = -(ax)^2 dt^2 + dx^2 + dy^2 + dz^2
it's this:
Start with a ruler marked off in millimeters at rest in some
inertial reference frame. Then let each point on the ruler undergo
uniform accelerated motion with the proper acceleration of the point
marked x given by
g_x = 1/x
Then the proper length of any infinitesimal segment of the ruler
will remain constant. This fact allows this accelerated ruler to
be used as a standard of length for an observer at rest relative
to the ruler (that is, at rest relative to the closest point on
the ruler). This behavior on the part of the ruler is the closest
thing to a "rigid ruler" possible in an accelerated coordinate
system.
The point marked x on the accelerated ruler will follow
the trajectory (as described in the initial inertial coordinate system)
x(t) = square-root(x^2 + t^2)
>From this, it is immediately seen that the point x=0 corresponds
to the path of a light signal. That's why it's not a physically
meaningful point: No massive physical object can follow that trajectory,
so no massive object can be at the point x=0 in the accelerated
coordinate system.
>However since he didn't explicitly state what he thought
>the coordinates meant so I had to guess. He did seem to think that X is a
>spatial displacement from the origin though and that is not what X means if
>this is supposed to be the Rindler metric.
I think that the confusion might be about the choice of "origin".
The accelerated observer accelerating at uniform proper acceleration
g is *not* at the origin in my coordinates, he is at the point
X = 1/g
But the meaning of X is certainly an ordinary spatial distance. It
is the *initial* distance from the origin (in my coordinates---if
you prefer to let the accelerated observer with proper acceleration g
be at the origin, then my X is 1/g + the distance to the origin).
But X is certainly an ordinary spatial coordinate in this sense:
If you have two nearby observers both at "rest" in the accelerated
coordinate system, one at position X1, and the other at X2, then the distance
between those observers, as measured using comoving metersticks, will
be given by |X2 - X1|.
Pmb
<tes...@tum.bot> wrote in message news:4035e4db$1...@news.sentex.net...
> >From the bit you quoted it does not appear that Darryl ever claimed to
> derive "a line element for a uniform gravitational field".
According to the equivalence principle the metric for a uniform
> of the standard Rindler line element
>
> ds^2 = -(ax)^2 dt^2 + dx^2 + dy^2 + dz^2,
>
> x > 0
Please define the meaning of these symbols. The X in that metric can't be
spatial distance from the origin. If X was a spatial distance from the
origion then note that it predicts that the proper time interval, d(tau)
measured by the coordinate clock at the origin, dt, is zero since
d(tau) = 0*dt
The metric for a uniform gravitational field is (in your conventions)
ds^2 = -(1+gx)^2dt + dx^2 + dy^2 + dz^2
where x,y,z are spatial displacements. How are these two metrics related?
> Are you sure you understand correctly what he was trying to accomplish?
Of course. But he didn't describe the physical meaning of the X is that
equation.
> The bit you quoted is completely standard (and valid!)--- in fact I think
> you can find it in Taylor & Wheeler.
I'm not saying that its invalid. I'm saying that his results are not what he
thinks they are. However since he didn't explicitly state what he thought
the coordinates meant so I had to guess. He did seem to think that X is a
spatial displacement from the origin though and that is not what X means if
this is supposed to be the Rindler metric.
Taylor and Wheeler do not touch the accelerting frame.
> > it is incompatible with the metric for a uniformly accelerating frame of
> > reference/uniform gravitational field as found everywhere else in the
> > literature i.e.
> >
> > ds^2 = c^2(1+gz/c^2)dt^2 - dx^2 - dy^2 - dz^2
>
> No, this is simply a minor variation of the standard Rindler chart!
Your comment "No." is wrong. The answer is "Yes." This is most certainly the
metric for a uniformly accelerating frame of referance. The metric you gave
was in different coordinates and X is does not represent a spatial
displacement. It's the coordinates/physical meaning, that I'm obviously
inquiring about. I guess I didn't make that clear though - sorry.
> BTW, if your real question is: "which exact -vacuum- solution in gtr has
> constant and uniform -nonzero- gravitational field?",
Nope. I already know that solution. It's in several GR texts, e.g. Mould,
tes...@tum.bot
[snip numerous erroneous claims, all based on elementary misconceptions]
> The X in that metric can't be spatial distance from the origin.
It seems that some poster (possibly me) told you something like this: "the
geometric meaning of the coordinate x in the Rindler chart
ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2, x > 0
is that it measures spatial distance from the locus x = 0". This claim is
a bit too vague to be perfectly well-defined as stated, but there is only
one reasonable interpretation of the intended meaning!
Verifying this kind of claim in your head is a pretty basic skill, but
here is one way to do it, in absolutely excruciating detail:
Start by reading off from the given line element the coframe (four
one-forms)
o^1 = x dt
o^2 = dx
o^3 = dy
o^4 = dz
and its dual frame (four orthonormal vectors)
e_1 = 1/x @/@t
e_2 = @/@x
e_3 = @/@y
e_4 = @/@z
Note these are well-defined on x > 0. Now consider the following integral
curve of the unit spacelike vector field Y = e_2:
f(q) = (0,q,0,0), 0 < q < infty
As you can see, this is a coordinate ray from the locus x = 0, which we
have parameterized by q. It is also a spacelike geodesic, since
D_Y Y = 0
Choose 0 < x1 < x2 and consider the events
(0,x1,0,0) (0,x2,0,0)
Both events lie on our ray. Using the metric we easily obtain the
invariant interval between these events as -measured along our ray-.
Computing this number involves evaluating a very easy definite integral;
as you can verify, the result is
x2 - x1 = delta x
We obtained this number via a coordinate computation, but of course it has
geometric meaning independent of the particular chart we used; if we had
integrated along the same interval represented in some other chart, we
would have obtained the same number. Hence the meaning of the phrase
"invariant interval along a curve".
Finally, recall that our ray is a spacelike geodesic segment. It is
standard to call an invariant interval measured along a spacelike curve a
"distance" measured along that curve. After introducing the limit x1 ->
0, we can summarize our results by saying that "the x coordinate measures
spatial distance to the locus x = 0".
> If X was a spatial distance from the origion then note that it predicts
> that the proper time interval, d(tau) measured by the coordinate clock
> at the origin, dt, is zero since
>
> d(tau) = 0*dt
More excruciating detail:
The locus x = 0 is -not- covered by the Rindler chart--- that is precisely
why I wrote the condition "x > 0" above! Understanding the reason why the
chart doesn't cover x = 0 is a crucial part of understanding the geometric
meaning of the coordinates, but this is standard material, so I'll refer
you to standard textbooks.
You can't ask about ideal clocks sitting on x = 0, but you -can- ask about
ideal clocks sitting on some coordinate hyperplane x = x0, where x0 > 0.
A typical world line of such a clock is:
f(q) = (q,x0,0,0)
This curve is a coordinate line and also an integral curve of X = e_1,
i.e. a timelike curve, but it is not a -geodesic-, since
D_X X = 1/x e_2
(Physically, of course, this vector is just the acceleration of the
Rindler observer carrying our clock.) Nonetheless, you can pick two
events lying on this curve, and by integrating you can obtain the
invariant interval measured along the curve between those events. It
would be standard practice to call this number something like "the elapsed
time measured by an ideal clock whose world line coincides with the given
curve". Again, even though we computed this number using a coordinate
chart, the answer we obtain has a well-defined geometric meaning
-independent- of our chart.
Again choose 0 < x0 < x1, but this time consider two clocks with world
lines
(q,x0,0,0) (q',x1,0,0)
where q,q' are parameters. Pick suitable pairs of events connected by
suitable null geodesics running between the two curves. Further easy
computations and further standard physical interpretation of all this
Lorentzian geometry eventually leads to a precise prediction of the
"frequency shift" which would be observed, in our thought experiment, by a
Rindler observer who measures radar pulses emitted by a second Rindler
observer. This frequency shift is clearly analogous, via the EP, to a
"gravitational frequency shift". Exhibiting this analogy is a major
pedagogical reason for introducing the Rindler observers in the first
place.
> > of the standard Rindler line element
> >
> > ds^2 = -(ax)^2 dt^2 + dx^2 + dy^2 + dz^2,
> >
> > x > 0
>
> Please define the meaning of these symbols.
Because the Rindler chart and the geometric meaning of the coordinates is
standard material, and not particularly subtle or hard to understand--- if
you have a good picture to look at!--- I will simply refer you to the very
clear discussion in MTW, which features several excellent figures. If
reading MTW doesn't lead you to retract everything you wrote, I'll give
up.
Pmb
<tes...@tum.bot> wrote in message
news:c1t2ln$147q$1...@fiasco.xenopsyche.net...
> "Pmb" wrote:
>
> [snip numerous erroneous claims, all based on elementary misconceptions]
Why did you post a claim that there as a misconception and then not explain
why you think there is a misconception?
> > If X was a spatial distance from the origion then note that it predicts
> > that the proper time interval, d(tau) measured by the coordinate clock
> > at the origin, dt, is zero since
> >
> > d(tau) = 0*dt
>
> More excruciating detail:
>
> The locus x = 0 is -not- covered by the Rindler chart--- that is precisely
> why I wrote the condition "x > 0" above!
I understand what you wrote. I disagree with what you wrote. Rindler shows
that X = exp(Phi) where Phi = gravitational potential - exactly what I
suspected - X does not measure spatial distance.
> Because the Rindler chart and the geometric meaning of the coordinates is
> standard material, and not particularly subtle or hard to understand--- if
> you have a good picture to look at!--- I will simply refer you to the very
> clear discussion in MTW, which features several excellent figures. If
> reading MTW doesn't lead you to retract everything you wrote, I'll give
> up.
Thanks anyway but Rindler explains the meaning of the Rinler line element
nicely - as he should! :-)
Daryl McCullough
>Rindler shows that X = exp(Phi) where Phi = gravitational potential
>- exactly what I suspected - X does not measure spatial distance.
In the metric
ds^2 = (aX)^2 dT^2 - dX^2 - dY^2 - dZ^2
the quantity dX is indeed a spatial distance. If you have two
objects A and B that are both "at rest" in the coordinate
system (X,Y,Z,T), A at coordinates (X=X_A, Y=0, Z=0) and B
at coordinates (X=X_B, Y=0, Z=0)
dist = X_B - X_A
as measured using comoving rulers.
For reference, see
http://www.theorie.physik.uni-muenchen.de/~serge/T6/SS03-T6-LN8.pdf
In Section 8.1.2, the author writes (about uniformly accelerated
observer)
"Now we need to find a suitable coordinate system (tau,xi) in which
the accelerated observer is at rest. The time coordinate tau and the
spatial coordinate xi should give the proper time and proper distance
as measured by the accelerated observer..."
He gives the Rindler metric in equation 8.8, which is the same as
the above, except for the choice of the origin:
T --> tau
X --> xi + 1/a
He also explains the alternative form of the Rindler metric: If you
change coordinates once again to xi~ = 1/a log(1 + a xi) then
the metric is converted into the form
ds^2 = exp(2a xi~) (d tau^2 - d xi~^2)
tes...@um.bot
> <tes...@tum.bot> wrote
>
> > "Pmb" wrote:
> >
> > [snip numerous erroneous claims, all based on elementary misconceptions]
>
> Why did you post a claim that there as a misconception and then not explain
> why you think there is a misconception?
Because this thread seemed to have degenerated to the level of
A: "Is!"
B: "Is -not-!"
A: "Is -so-!"
I thought the problem might be that you misunderstood the conventional but
ambiguous statement "the x coordinate in the Rindler chart measures
distance from the locus x=0". That's why I exhibited the unambiguous (but
boring) computation which that statement abbreviates. But it seems you
disagree:
> I understand what you wrote. I disagree with what you wrote.
Hmm... I probably should have left in the magnitude a of the constant
uniform acceleration of our Rindler observers, since setting a = 0
(instead of setting a = 1, as in my computation) gives
ds^2 = -dt^2 + (dx^2 + dy^2 + dz^2)
The conventional but ambiguous statement at issue now becomes "the x
coordinate measures distance to the locus x = 0, as measured by the
observers with world lines parallel to the t axis".
> Rindler shows that X = exp(Phi) where Phi = gravitational potential -
> exactly what I suspected
Fine, but did you perhaps overlook the fact that
exp(phi) ~ 1 + phi
to first order in phi?
> - X does not measure spatial distance.
But there is no contradiction here with what I wrote!
I think Rindler must somehow be confusing you--- maybe you should try
another book? If you don't like MTW, another very nice discussion of
weak-field theory can be found here:
author = {Ohanian, Hans C and Ruffini, Remo},
title = {Gravitation and Spacetime},
edition = {second},
publisher = {Norton},
year = 1994}
Many people find this book much more readable than MTW.
> > If reading MTW doesn't lead you to retract everything you wrote, I'll
> > give up.
>
> Thanks anyway but Rindler explains the meaning of the Rinler line element
> nicely - as he should! :-)
I certainly do not believe that Rindler and Darryl or I are in
disagreement about anything, but since you declined to retract, I'll give
up here, as I promised to do.
Gauge
> I thought the problem might be that you misunderstood the conventional but
> ambiguous statement "the x coordinate in the Rindler chart measures
> distance from the locus x=0".
There is a difference between understanding a claim and agreeing witht
that claim. Because I question a claim does not mean that I'm stating
that the claim is wrong - it means only that I question that claim.
> That's why I exhibited the unambiguous (but
> boring) computation which that statement abbreviates. But it seems you
> disagree:
>
> > I understand what you wrote. I disagree with what you wrote.
>
> Hmm... I probably should have left in the magnitude a of the constant
> uniform acceleration of our Rindler observers, since setting a = 0
> (instead of setting a = 1, as in my computation) gives
>
> ds^2 = -dt^2 + (dx^2 + dy^2 + dz^2)
>
> The conventional but ambiguous statement at issue now becomes "the x
> coordinate measures distance to the locus x = 0, as measured by the
> observers with world lines parallel to the t axis".
I should have said that I *question* what you wrote rather than
disagree with it. Sorry.
> > - X does not measure spatial distance.
Seems that Rindler contradicts my statement here. Puzzling!
>
> But there is no contradiction here with what I wrote!
>
> I think Rindler must somehow be confusing you--- ..
Did you see me say that I was confused? No. You didn't. Because
someone questions something doesn't mean that they are confused. It
means that something is missing. I understand precisely what Rindler
states. But the metric that Rindler arrives at is different than that
obtained by others. i.e. that found in
"Principle of Equivalence," F. Rohrlich, Ann. Phys. 22, 169-191,
(1963), page 173
"Basic Relativity," Richard A. Mould, page 232
"Gravitation," Misner, Thorne and Wheeler, page 173
Eg MTW derive the metric
ds^2 = c^2 (1 + gx/c^2)^2 dt^2 - dx^2 - dy^2 - dz^2
This is different from Rindler and yet x,y,z are supposed to have the
exact same physical meaning. What I wanted to know is why there are
two metrics whose coordinates have the same physical meaning and yet
are different.
> maybe you should try another book?
I did. five years ago when I first started studying the accelerating
frame. Mould was clear and was consistent with Rohrlich and MTW.
Rindler was different - I put off learning why there is a difference
until I had more time and was more interested.
> If you don't like MTW, another very nice discussion of
> weak-field theory can be found here:
>
> author = {Ohanian, Hans C and Ruffini, Remo},
> title = {Gravitation and Spacetime},
> edition = {second},
Ohanian does not discuss this and MTW give a different answer.
> Many people find this book much more readable than MTW.
Why are you refering to MTW when they give
ds^2 = c^2 (1 + gx/c^2)^2 dt^2 - dx^2 - dy^2 - dz^2
I assume that you're referencing them due to the fact that you think
the answers are identical. If so then please explain the reason for
the difference in metric when the terms have the same physical
meaning. Unless you think that they don't have the same physical
meaning?
>
> > > If reading MTW doesn't lead you to retract everything you wrote, I'll
> > > give up.
> >
> > Thanks anyway but Rindler explains the meaning of the Rinler line element
> > nicely - as he should! :-)
>
> I certainly do not believe that Rindler and Darryl or I are in
> disagreement about anything, but since you declined to retract, I'll give
> up here, as I promised to do.
Then explain
ds^2 = c^2 (1 + gx/c^2)^2 dt^2 - dx^2 - dy^2 - dz^2
That is found in
"Principle of Equivalence," F. Rohrlich, Ann. Phys. 22, 169-191,
(1963), page 173
"Basic Relativity," Richard A. Mould, page 232
"Gravitation," Misner, Thorne and Wheeler, page 173
Thanks
Please note that when someone disagrees with something it doesn't mean
that they haven't thought it through very carefully.
Pmb
Moderator - please send comments to peter....@verizon.net and
snipt this e-mail address out before posting. Thanks
Gauge
> Fine, but did you perhaps overlook the fact that
>
> exp(phi) ~ 1 + phi
>
> to first order in phi?
No. I stated this fact right off the bat in another thread. But for
some reason that thread didn't get posted for a very long time.
Pmb
Gauge
----------------------------------------------
We were discussing the Rinler metric
ds^2 = c^2 X^2 dT^2 - dX^2 - dY^2 - dZ^2
First off I wouldn't call that the Rindler metric.
----------------------------------------------
That was a mistake. I was speaking of something before that metric and
I inserted a comment later and this got all messed up./
I meant that if you do a coordinate transformation on
ds^2 = c^2 X^2 dT^2 - dX^2 - dY^2 - dZ^2
Then its no longer the Rindler metric.
Pmb
Gauge
with an acquantance of mine who is somewhat knowledgeable on the
Rindler metric. He made it all quite clear. I feel like a dummy for
missing the obvious. Doh! :-D
da...@atc-nycorp.com (Daryl McCullough) wrote
> "Now we need to find a suitable coordinate system (tau,xi) in which
> the accelerated observer is at rest. The time coordinate tau and the
> spatial coordinate xi should give the proper time and proper distance
> as measured by the accelerated observer..."
>
> He gives the Rindler metric in equation 8.8, which is the same as
> the above, except for the choice of the origin:
>
> T --> tau
> X --> xi + 1/a
We were discussing the Rinler metric
ds^2 = c^2 X^2 dT^2 - dX^2 - dY^2 - dZ^2
First off I wouldn't call that the Rindler metric. What you just did
here is a coordinate transforamtion and a coordinate transformation
gives another metric. The coordinate transformation you started with
took you to the Rindler metric from the Minkowski metric. Doing
another coordinate transformation brings you to another metric. The
one I spoke of in another post, i.e. the one on my website
ds^2 = c^2(1+gx/c^2)^dt^2 - dx^2 - dy^2 - dz^2
(I usual choose z as the direction of acceleration but we'll use x)
(sorry about missing the square before)
This metric is what I consider the metric for a unifrom gravitational
field. I.e. it pertains to an observer whose clock is located at x = 0
and which has a proper acceleration of "g". That's like a rocket which
is accelerating such that each point of his rocket has a constant
proper acceleration and, of course, points of different z have
different proper accelerations. This metric has an intrinsic value "g"
which is tied to this particular frame of referance. Other frames of
reference are either higher in the "field" or lowere in the "field" -
at least according to what I'd call "frame of reference".
The metric you gave, i.e. Rindler metric, pertains to a uniformly
accelerating lattice. The lattice not having an intrinsic value of
"g". Recall my earlier comments about setting X = 0 and as such a
clock sitting there there stops. Doh! I missed that one too. That's
the Rindler horizon! The coordinate clock, i.e. the clock which
measures "T" is sitting at X = 1. And yes - X is proportional to
spatial distance and by taking a particular rocket acceleration we
choose a particular frame corresponding to a translations in z
somewhere along the lattice.
So I can perform the coordinate transformation
X = 1 + gx
and that translates me from the Rindler Horizon to the place where a
lattice point is accelerating with proper acceleration "g".
Of course we could have just started from the inertial frame and
transformed straight to this metric as MTW et al did.
Had you explained that X = 0 was the Rinlder Horizon then I would have
picked this up a bit faster.
I don't see where you get X --> xi + 1/a though?
In any case this is still GR since we're using the postulate from GR
that the equations of physics are valid in all coordinate systems.
That's a GR postulate and not an SR postulate. SR restricts to
inertial frames. It was not obvious before GR that the equations of
physics are valid for all spacetime coordinates/frames of reference
Thanks
Pmb
Daryl McCullough
>First off I wouldn't call that the Rindler metric. What you just did
>here is a coordinate transforamtion and a coordinate transformation
>gives another metric.
Picky, picky. A translation of the origin is a coordinate
transformation that's hardly worth mentioning. I didn't start
from the Rindler metric and do a coordinate transformation for
no good reason, I started from the physical situation, namely
a uniformly accelerated observer, and derived the metric from
that.
>The coordinate transformation you started with
>took you to the Rindler metric from the Minkowski metric. Doing
>another coordinate transformation brings you to another metric. The
>one I spoke of in another post, i.e. the one on my website
>
>ds^2 = c^2(1+gx/c^2)^dt^2 - dx^2 - dy^2 - dz^2
>
>(I usual choose z as the direction of acceleration but we'll use x)
>
>(sorry about missing the square before)
>
>This metric is what I consider the metric for a unifrom gravitational
>field. I.e. it pertains to an observer whose clock is located at x = 0
>and which has a proper acceleration of "g".
Yes, that's how I derived my metric.
>The metric you gave, i.e. Rindler metric, pertains to a uniformly
>accelerating lattice. The lattice not having an intrinsic value of
>"g".
The metric I was using was this
ds^2 = (gX)^2 dt^2 - dX^2
That's the same as your metric, except for shifting the origin by
1/g. That is, my metric is the metric for a clock undergoing uniform
acceleration at proper acceleration g, starting at a location X=1/g.
The physical situation isn't changed by starting at X=1/g as opposed
to x=0.
>So I can perform the coordinate transformation
>
>X = 1 + gx
>
>I don't see where you get X --> xi + 1/a though?
My metric was
(aX)^2 dT^2 - dX^2
if you perform the substitution X = xi + 1/a, then this
is converted to
(1 + a xi)^2 dT^2 - (dxi)^2
which is the same as your metric.
>In any case this is still GR since we're using the postulate from GR
>that the equations of physics are valid in all coordinate systems.
>That's a GR postulate and not an SR postulate.
GR doesn't have anything to do with it. If you have a map between two
sets of coordinates, in my case, between the Minkowsky coordinates x,t
and the Rindler coordinates X,T, and you have a description of physics
for the (x,t) coordinates, then you can apply your map to find out what
physics is like in the (X,T) coordinates. That's just mathematics. There
is no physical postulate involved. If the proper time in (x,t) coordinate
is given by
ds = square-root(dt^2 - dx^2)
then proper time in (X,T) coordinates is given by
ds = square-root((aX)^2 dT^2 - dX^2)
That's just simple substitution, using
dt = dX sinh(aT) + aX cosh(aT)dT
dx = dX cosh(aT) + aX sinh(aT)dT
>SR restricts to inertial frames.
No, it doesn't. It is *simplest* in inertial frames. The form SR
takes in other frames is derivable from the case for inertial frames
by using calculus.
>It was not obvious before GR that the equations of
>physics are valid for all spacetime coordinates/frames
>of reference
That's true, it wasn't obvious. But in hindsight, it *should*
have been obvious. Physics can't depend on the coordinates used
to describe it.
Pmb
"Daryl McCullough" <da...@atc-nycorp.com> wrote in message
news:c3ddt...@drn.newsguy.com...
> Gauge says...
>
> >First off I wouldn't call that the Rindler metric. What you just did
> >here is a coordinate transforamtion and a coordinate transformation
> >gives another metric.
>
> Picky, picky. A translation of the origin is a coordinate
> transformation that's hardly worth mentioning.
Not if it changes the strength of the gravitational field. You accomplished
a transformation such that the coordinate clock is now ticking at a
different rate.
> >The metric you gave, i.e. Rindler metric, pertains to a uniformly
> >accelerating lattice. The lattice not having an intrinsic value of
> >"g".
>
> The metric I was using was this
>
> ds^2 = (gX)^2 dt^2 - dX^2
>
> That's the same as your metric, except for shifting the origin by
> 1/g.
That's called a "coordinate transformation" and its a coordinate
trasformation which changes the rate at which the coordinate clock tick.
Hardly a trivial notion physically, although mathematically it is trivial. I
can just as well say that the Rinder metric is just the Minkowski metric in
a different coordinate system. After all consider one definition of the
metric tensor - it maps vectors to scalars and that mapping does not depend
of the coordinates. If you're going to go by the convention you've chosen
(which I agree with) then the name of the tensor is a function of the
coordinate you represent the components in. However I would wager that if
Rindler himself were here that he'd call the metric ds^2 = c^2(1+gz/c^2)^2
dt^2 - dx^2 - dy^2 - dz^2 a silly metric since "g" has no intrinsic
significance.
> >In any case this is still GR since we're using the postulate from GR
> >that the equations of physics are valid in all coordinate systems.
> >That's a GR postulate and not an SR postulate.
>
> GR doesn't have anything to do with it.
Sure it does. One of the basic postulates of GR is that the laws of physics
are the same in all coordinate systems. SR pertains only to the laws in
inertial frames. Let me state this as Classical Mechanics - 3rd Ed.,"
Goldstein, Safko and Poole, pp 325
--------------------------------------------------------------------
Principle of covariance - in the special theory, all inertial observers are
equivalent. The general theory extends this idea by postulating the
principle of covariance. This principle states that all observers, inertial
or not, observer the same laws of physics. That means taht the laws can be
expressed in terms of tensors, since tensors are geometric objects defined
independant of any coordinate system.
--------------------------------------------------------------------
All you're doing is renaming SR and the principle of covariance.
Pmb
Daryl McCullough
>"Daryl McCullough" <da...@atc-nycorp.com> wrote in message
>> >The metric you gave, i.e. Rindler metric, pertains to a uniformly
>> >accelerating lattice. The lattice not having an intrinsic value of
>> >"g".
>>
>> The metric I was using was this
>>
>> ds^2 = (gX)^2 dt^2 - dX^2
>>
>> That's the same as your metric, except for shifting the origin by
>> 1/g.
>
>That's called a "coordinate transformation" and its a coordinate
>trasformation which changes the rate at which the coordinate clock tick.
>Hardly a trivial notion physically, although mathematically it is trivial. I
>can just as well say that the Rinder metric is just the Minkowski metric in
>a different coordinate system.
I go by the 15 second rule. If it takes less than 15 seconds to perform
the coordinate transformation in your head, then it's not worth mentioning.
>> >In any case this is still GR since we're using the postulate from GR
>> >that the equations of physics are valid in all coordinate systems.
>> >That's a GR postulate and not an SR postulate.
>>
>> GR doesn't have anything to do with it.
>
>Sure it does.
No, it certainly does not. If the usual metric
ds^2 = dt^2 - dx^2 - dy^2 - dz^2
is valid when (x,y,z,t) are inertial coordinates, then
the metric
ds^2 = (aX)^2 dT^2 - dX^2 - dY^2 - dZ^2
is valid when the coordinates (X,Y,Z,T) are used, instead,
where X = square-root(x^2 - t^2), T = 1/a arctanh(t/x),
Y=y, Z=z. You don't need GR to tell you that, it follows
from mathematics alone.
>One of the basic postulates of GR is that the laws of physics
>are the same in all coordinate systems. SR pertains only to the laws in
>inertial frames.
The usual equations of SR (the Lorentz transformations) pertain to
inertial coordinate systems. But the transformation to noninertial
coordinates is no more difficult than transforming from Cartesian
to polar coordinates. You no more need General Relativity to use
noninertial coordinates than you need GR to use polar coordinates.
The difference between GR and SR is that the usual equations
of GR are covariant---they have the same form, regardless of
what coordinate system you use. In contrast, the equations of
SR have a very different form in noninertial coordinate systems.
That doesn't mean you can't use these other coordinate systems,
but it means you have to do a little calculation to figure out
what SR looks like in them.
>Let me state this as Classical Mechanics - 3rd Ed.,"
>Goldstein, Safko and Poole, pp 325
>--------------------------------------------------------------------
>Principle of covariance - in the special theory, all inertial observers are
>equivalent. The general theory extends this idea by postulating the
>principle of covariance. This principle states that all observers, inertial
>or not, observer the same laws of physics. That means taht the laws can be
>expressed in terms of tensors, since tensors are geometric objects defined
>independant of any coordinate system.
>--------------------------------------------------------------------
>
>All you're doing is renaming SR and the principle of covariance.
In GR, all coordinate systems are treated equally. In SR, the
inertial coordinate systems are special, because they are the
only ones that have a nice, constant, diagonal metric. That doesn't
mean you can't use other coordinate systems, it's just that in SR,
other coordinate systems are harder to work with than inertial
coordinates.
Pmb
"Daryl McCullough" <da...@atc-nycorp.com> wrote
> I go by the 15 second rule. If it takes less than 15 seconds to perform
> the coordinate transformation in your head, then it's not worth
mentioning.
So what you're saying is that whether a metric is the Rindler metric depends
on the person who is transforming it from the Minkowski metric?
> No, it certainly does not. If the usual metric
>
> ds^2 = dt^2 - dx^2 - dy^2 - dz^2
>
> is valid when (x,y,z,t) are inertial coordinates, then
> the metric
>
> ds^2 = (aX)^2 dT^2 - dX^2 - dY^2 - dZ^2
>
> is valid when the coordinates (X,Y,Z,T) are used, ...
That is incorrect. You simply *cannot* assume that to be true unlesss you
invoke the principle of general covariance.
>That doesn't
> mean you can't use other coordinate systems, it's just that in SR,
> other coordinate systems are harder to work with than inertial
> coordinates.
That is incorrect. SR only states that the laws of physics are the same in
all **inertial** frames of reference. It is the principle of general
covariance which states that the laws of physics are the same in **all**
frames of reference.
You simply cannot state that the laws of physics are the same in all
coordinate systems without invoking the principle of general covariance
since **that is** the principle of general covariance. Einstein wasn't a
dummy you know. He have very good reasons for defining SR as he did. He knew
that to claim that the laws of physics are valid in all frames that he had
to create a new postulate.
Don't you ever wonder why Einstein didn't state in 1905 that the laws of
physics are the same in *all* frames of reference?
I believe Einstein explains all this in his little book "The Meaning of
Relativity"
Pmb
Michael Varney
"Pmb" <peter.bro...@verizon.net> wrote in message
news:6QL7c.151$7F....@nwrdny03.gnilink.net...
>
> "Daryl McCullough" <da...@atc-nycorp.com> wrote
>
> > I go by the 15 second rule. If it takes less than 15 seconds to perform
> > the coordinate transformation in your head, then it's not worth
> mentioning.
>
> So what you're saying is that whether a metric is the Rindler metric
depends
> on the person who is transforming it from the Minkowski metric?
No Peter. Daryl stated that if it takes less than 15 seconds to perform the
coordinate transformation in your head, then it is not worth mentioning.
Of course Daryl might have meant somthing else, but from his posting history
he seems quite clear and logical in his statments.
> > No, it certainly does not. If the usual metric
> >
> > ds^2 = dt^2 - dx^2 - dy^2 - dz^2
> >
> > is valid when (x,y,z,t) are inertial coordinates, then
> > the metric
> >
> > ds^2 = (aX)^2 dT^2 - dX^2 - dY^2 - dZ^2
> >
> > is valid when the coordinates (X,Y,Z,T) are used, ...
>
> That is incorrect. You simply *cannot* assume that to be true unlesss you
> invoke the principle of general covariance.
>
> >That doesn't
> > mean you can't use other coordinate systems, it's just that in SR,
> > other coordinate systems are harder to work with than inertial
> > coordinates.
>
> That is incorrect.
Actually, this is correct. Check out Rindler's Introduction to Special
Relativity (second Edition) for some examples of SR dealing with
non-inertial reference frames.
> SR only states that the laws of physics are the same in
> all **inertial** frames of reference. It is the principle of general
> covariance which states that the laws of physics are the same in **all**
> frames of reference.
That is a postulate of SR. This does not mean that you cannot apply SR to
non-inertial frames of reference. See examples on four-acceleration and the
trick of considering instantaneously inertial frames, or the angular
momentum four-tensor.
> You simply cannot state that the laws of physics are the same in all
> coordinate systems without invoking the principle of general covariance
> since **that is** the principle of general covariance. Einstein wasn't a
> dummy you know.
No one said he was, Peter.
> He have very good reasons for defining SR as he did. He knew
> that to claim that the laws of physics are valid in all frames
He did not claim that the laws of physics were valid in _all_ frames.
In any event, Rindlers test is very nice. You should check out chapter 3
and chapter 5.
eb...@lfa221051.richmond.edu
In article <e7203033.04031...@posting.google.com>,
Gauge <gau...@hotmail.com> wrote:
>First off I wouldn't call that the Rindler metric. What you just did
>here is a coordinate transforamtion and a coordinate transformation
>gives another metric.
This is just a terminological point, but I think it's worthwhile.
It seems to me quite unwise to say that different coordinate systems
give different metrics. The metric is a tensor, and a tensor
is something that exists independent of what coordinates one chooses.
So personally, I wouldn't talk about "the Rindler metric" as
something distinct from the Minkowski metric; I'd talk about
Rindler coordinates.
Am I alone in this?
-Ted
--
[E-mail me at na...@domain.edu, as opposed to na...@machine.domain.edu.]
mandro
Rindler coordinates, whatever they are
physically, simply represent a chart on
the minkowsky manifold. Now anyone who knows
about metrics and about manifolds, knows that
when you have a manifold with a metric, then
for each given chart, you have a matrix valued
function g_{i,j} that is called the components
of the metric w.r.t. that chart. If it helps,
here's a picture
g_{ij} g*_{ij}
_______ ________
^ ^
\ f /h
___________ M
f is the Minkowski chart, and
h is the rindler chart. A technical note here,
the Minkowsky chart covers all of M, but the
Rindler chart covers only part of M. Now How
you can see all of this in action, in a concrete
fashion is find the transition map T between
the Rindler and Minkowski charts, and using
the Minkowski g ij's find the rindler g* i j's.
Gauge
card on this discussion. Let's move on. -TB]
"Michael Varney" <varney@colorado_no_spam.edu> wrote in message news:<m408c.201$Qg4....@news.uswest.net>...
> "Pmb" <peter.bro...@verizon.net> wrote in message
> news:6QL7c.151$7F....@nwrdny03.gnilink.net...
> >
> > "Daryl McCullough" <da...@atc-nycorp.com> wrote
> >
> > > I go by the 15 second rule. If it takes less than 15 seconds to perform
> > > the coordinate transformation in your head, then it's not worth
> > mentioning.
> >
> > So what you're saying is that whether a metric is the Rindler metric
> depends
> > on the person who is transforming it from the Minkowski metric?
>
>
> No Peter. Daryl stated that if it takes less than 15 seconds to perform the
> coordinate transformation in your head, then it is not worth mentioning.
No michael. Since daryl is using a criteria of 15 seconds to perform a
transformation then that must of neccesity be dependant on the person
doing the transformation since no two people calculate in their heads
at the same speed. What one person does in 15 seconds another does in
5 seconds. What another person does in 5 seconds could take someone
else 45 seconds.
> Actually, this is correct. Check out Rindler's Introduction to Special
> Relativity (second Edition) for some examples of SR dealing with
> non-inertial reference frames.
Actually that is incorrect. And Rindler is incorrect too but he's
certainly entitled to his opinion.
To be able to make such a transformation requires that one knows that
the transformation yields a correct law of physics and that requires
the principle of general covariance and that is part of GR, not SR.
> > SR only states that the laws of physics are the same in
> > all **inertial** frames of reference. It is the principle of general
> > covariance which states that the laws of physics are the same in **all**
> > frames of reference.
>
> That is a postulate of SR.
That is incorrect. The postulates of SR are
(1) Principle of Relativity - The laws of physics are the same in all
inertial frames of reference
(2) Light Postulate - The speed of light in an inertial frame of
reference is independent of the source
The postulates of GR are different. The principle of general
covariance lies in GR. Look it up in Goldstein as I referenced.
> This does not mean that you cannot apply SR to
> non-inertial frames of reference.
It sure does.
> See examples on four-acceleration and the
> trick of considering instantaneously inertial frames, or the angular
> momentum four-tensor.
You can do all the frame jumping you like in SR. That won't tell you
that the laws of GR must be tensor equations and it won't tell you how
to obtain Maxwell
s equations in all coordinate systems etc. That is GR.
> > He have very good reasons for defining SR as he did. He knew
> > that to claim that the laws of physics are valid in all frames
>
> He did not claim that the laws of physics were valid in _all_ frames.
That is incorrect. From "The Foundation of the General Theory of
Relativity," Albert Einstein (1916) (Annalen der Physik 49)
---------------------------------------
The laws of physics must be of such nature that they apply to systems
of reference in any kind of motion. Along this road we arrive at an
extension of the postulate of relativity.
---------------------------------------
It's a nice paper. You should read it.
Pmb
[moderator: Please send all comments either to
peter....@verizon.com or pm...@hotmail.com - preferably the later
since verizon mail isn't working]
Pmb
<eb...@lfa221051.richmond.edu> wrote in message
news:c3ssmt$fm5$1...@lfa222122.richmond.edu...
>
>
>
> In article <e7203033.04031...@posting.google.com>,
> Gauge <gau...@hotmail.com> wrote:
>
> >First off I wouldn't call that the Rindler metric. What you just did
> >here is a coordinate transforamtion and a coordinate transformation
> >gives another metric.
>
> This is just a terminological point, but I think it's worthwhile.
>
> It seems to me quite unwise to say that different coordinate systems
> give different metrics. The metric is a tensor, and a tensor
> is something that exists independent of what coordinates one chooses.
> So personally, I wouldn't talk about "the Rindler metric" as
> something distinct from the Minkowski metric; I'd talk about
> Rindler coordinates.
>
> Am I alone in this?
No. You're not alone. Many people feel the way that you do.
Pmb (aka "Gauge")
Daryl McCullough
>"Daryl McCullough" <da...@atc-nycorp.com> wrote
>
>> I go by the 15 second rule. If it takes less than 15 seconds to perform
>> the coordinate transformation in your head, then it's not worth
>mentioning.
>
>So what you're saying is that whether a metric is the Rindler metric depends
>on the person who is transforming it from the Minkowski metric?
No, I'm saying that the choice between placing the origin of
your coordinate system at one point rather than another has no
physical significance.
>
>> No, it certainly does not. If the usual metric
>>
>> ds^2 = dt^2 - dx^2 - dy^2 - dz^2
>>
>> is valid when (x,y,z,t) are inertial coordinates, then
>> the metric
>>
>> ds^2 = (aX)^2 dT^2 - dX^2 - dY^2 - dZ^2
>>
>> is valid when the coordinates (X,Y,Z,T) are used, ...
>
>That is incorrect.
Yes, it is correct. It's a pure *mathematical* fact:
If proper time in the (x,t) coordinate system is given by
T = Integral of square-root((dt/ds)^2 - (dx/ds)^2)
for the parameterized path x(s), t(s) then in a different
coordinate system (X,T) (where X = square-root(x^2-t^2),
T=arctanh(t/x)), the same integral is given by
T = Integral of square-root((aX)^2 dT^2 - dX^2)
That's just a change of variables in calculus. I don't need any
"law of covariance" to tell me that I can do variable substitution
inside an integral. Using coordinates that "mix" x and t is no more
problematic for Special Relativity than using coordinates (namely
polar coordinates) that mix x and y. I no more need a principle of
covariance to use noninertial coordinates than I need a principle of
covariance to use polar coordinates. Mathematically, the two cases
are exactly analogous.
>You simply *cannot* assume that to be true unlesss you
>invoke the principle of general covariance.
No, you don't *assume* it is true---you *prove* it true using
calculus.
>That is incorrect. SR only states that the laws of physics are the same in
>all **inertial** frames of reference. It is the principle of general
>covariance which states that the laws of physics are the same in **all**
>frames of reference.
I didn't assume that the laws of physics are the same (in form)
in noninertial coordinate systems. As a matter of fact, they *aren't*,
if you use SR as your physics. In nice inertial coordinates, the
law for proper time is simply ds^2 = dt^2 - dx^2. In noninertial
coordinates, the law is ds^2 = (aX)^2 dT^2 - dX^2. I didn't use
the same law in the new coordinate system, I *derived* the form
the laws take in the new coordinate system.
>You simply cannot state that the laws of physics are the same in all
>coordinate systems
I didn't state such a thing. What I stated was that if you know the
laws in one coordinate system, then you can *derive* the corresponding
laws in any other coordinate system using calculus. For example, if
you know that Newton's law for force-free motion looks like
(d/dt)^2 x = 0
(d/dt)^2 y = 0
in Cartesian coordinates, then you can derive that in polar coordinates
(R,A) (with R=square-root(x^2+y^2) A = arctan(y/x)) the law for force-free
motion looks like
(d/dt)^2 R - R (dA/dt)^2 = 0
R (d/dt)^2 A + 2 (dR/dt) (dA/dt) = 0
Newton's laws look simple only in inertial cartesian coordinates,
but you can derive them in any coordinates using mathematics. Same
is true of SR.
In general, you can think of a "law of physics" in a particular
coordinate system as a mathematical function that takes you from
descriptions of experimental setups in that coordinate system to
descriptions of experimental outcomes:
f(D_setup) = D_outcome
A coordinate transformation corresponds to a function T that
maps from descriptions in one coordinate system to descriptions
in another.
T(D) = D'
Then given a law of physics f and a transformation T, we can
find a corresponding law of physics in the new coordinate system
as follows:
f'(D_setup') = T( f( T^{-1}( D_setup' ) ) )
Less mathematically:
1. Take the description of the setup D_setup' in the new coordinate
system.
2. Translate it back into the old coordinate system using T^{-1}
(the inverse transformation).
3. Compute the outcome in the old coordinate system using f.
4. Translate the outcome into the new coordinate system using T.
So if you know the laws of physics in one coordinate system, and
you know how to translate into a new coordinate system, then you
know the laws of physics in the new coordinate system. It's just
mathematics.
>without invoking the principle of general covariance
>since **that is** the principle of general covariance.
The principle of covariance is more of a heuristic tool than a law
of physics. *Any* theory of physics can be recast into an equivalent
form that is generally covariant. Newton's physics can be so rewritten,
Special Relativity can be so rewritten. The claim that the physical
laws are covariant has no physical content. But general covariance is
a useful heuristic tool in discovering laws of physics---you search for
laws that look *simple* in generally covariant form. The laws should
have the fewest ad hoc structures and assumptions.
That's the beauty of General Relativity compared with Special
Relativity or Newtonian physics. Special Relativity and Newtonian
physics when cast into covariant form have additional structure that
seems ad hoc: fields that act upon matter but are not themselves acted
upon. In SR, it is the metric: it affects the motion of matter, but
is not affected by it. In Newtonian physics, it is absolute time
(in addition to a fixed spatial metric) which acts like a field that
acts upon matter but is not affected by it.
In contrast, in General Relativity, there are no fields (scalars, tensors,
etc.) that act upon matter but are not acted upon by it. In that sense of
"no prior geometry", GR is simpler than SR or Newtonian mechanics. But
general covariance by itself doesn't single out GR as opposed to SR or
Newtonian physics.
Tom Snyder
> give different metrics. The metric is a tensor, and a tensor
> is something that exists independent of what coordinates one chooses.
> So personally, I wouldn't talk about "the Rindler metric" as
> something distinct from the Minkowski metric; I'd talk about
> Rindler coordinates.
>
> Am I alone in this?
>
I doubt if you're alone.
However, not everyone uses the word 'metric' to always denote the metric
_tensor_. Some very respectable authors sometimes use the word 'metric' to
refer to the form of the line element ds^2 as written out in a particular
coordinate system. Thus, they speak of the 'Schwarzschild metric', the
'Kerr metric', etc. I'm not particularly bothered by it since the meaning
is usually clear from the context.
Tom S.
Arnold Neumaier
> However, not everyone uses the word 'metric' to always denote the metric
> _tensor_. Some very respectable authors sometimes use the word 'metric' to
> refer to the form of the line element ds^2 as written out in a particular
> coordinate system. Thus, they speak of the 'Schwarzschild metric', the
> 'Kerr metric', etc. I'm not particularly bothered by it since the meaning
> is usually clear from the context.
These are in fact canonically equivalent. Given the metric tensor
one gets the line element
ds^2 = ds_mu g_{mu nu} ds_nu,
and given the line element, one can read off the metric tensor, using
the same formula. The line element is as invariant as the metric tensor,
but to write any of these down explicitly for a particular case, one has
to choose coordinates.
Thus the two ways of using the word are more or less the same,
and it is just standard colloquial sloppiness that is here at work.
Arnold Neumaier
Pmb
"Arnold Neumaier" <Arnold....@univie.ac.at> wrote in message
news:c42a2m$i7j$1...@lfa222122.richmond.edu...
MTW explain all this in Box 3.2 of their text in part D.
Pmb
tes...@um.bot
> not everyone uses the word 'metric' to always denote the metric
> _tensor_. Some very respectable authors sometimes use the word 'metric'
> to refer to the form of the line element ds^2 as written out in a
> particular coordinate system. Thus, they speak of the 'Schwarzschild
> metric', the 'Kerr metric', etc. I'm not particularly bothered by it
> since the meaning is usually clear from the context.
IMO, it is best to regard "the line element" simply as a rough and ready
way to define a metric tensor -in terms of a specific coordinate chart-.
My experience suggests that in the context of gtr or other gravitation
theories:
BAD GOOD
Kerr metric, Kerr solution Kerr vacuum
Vaidya metric, Vaidya solution Vaidya null dust
matter dominatated Friedmann solution FRW dust
radiation dominated Friedmann solution FRW radiation fluid
BLAH metric BLAH model
One reason for this is that we want to avoid confusing -coordinate charts-
with -Lorentzian manifolds- (M,g). In particular, what some authors very
carelessly call "the Rindler metric" is really the good old -Minkowski
metric- rewritten in terms of -the Rindler coordinate chart-! So we
should refer to -the Rindler chart- for (a half space in) -the Minkowksi
vacuum-.
OTH, the Weyl vacuum I mentioned, which in the canonical chart can be
written
ds^2 = exp(2 m z) [-dt^2 + exp(m r^2) (dz^2 + dr^2) ]
+ exp(-2 m z) r^2 du^2,
-infty < t,z < infty, 0 < r < infty, -pi < u < pi
is -not- even locally isometric to the Minkowksi vacuum, yet it can be
regarded as a natural candidate for the closest thing we have in gtr to "a
uniform gravitational field". (I've already explained all this at very
great length; see previous posts in this thread and posts here from a few
years ago.) So this should be called -the Weyl canonical chart- (a
particular type of coordinate chart) for -the plane symmetric Weyl vacuum-
(a particular spacetime model in gtr).
Similarly, we should refer to -the Robertson/Walker chart- for -FRW
models-. (The RW metrics can be formulated for both radiation fluid and
dust FRW models.) We should refer to -the Schwarzschild chart- or -the
Painleve chart- or -the Eddington chart- or -the Costa chart- or... for
-the Schwarzschild vacuum-. And so forth and so on.
See also my similar terminological comments (submitted today) in another
thread.
tes...@um.bot
> In article <e7203033.04031...@posting.google.com>,
> Gauge <gau...@hotmail.com> wrote:
>
> >First off I wouldn't call that the Rindler metric. What you just did
> >here is a coordinate transforamtion and a coordinate transformation
> >gives another metric.
>
> This is just a terminological point, but I think it's worthwhile.
>
> It seems to me quite unwise to say that different coordinate systems
> give different metrics. The metric is a tensor, and a tensor
> is something that exists independent of what coordinates one chooses.
> So personally, I wouldn't talk about "the Rindler metric" as
> something distinct from the Minkowski metric; I'd talk about
> Rindler coordinates.
>
> Am I alone in this?
Not at all--- I, for one, completely agree! (I commented previously on
pmb's post, but not, IIRC, on this particular point--- this omission was
due to lack of time plus a desire to avoid overwhelming the OP.)
I always suggest using terminology similar to the following examples:
Example 1: a rational prolate spheroidal chart for (a piece of) Minkowski
vacuum:
ds^2 = -dt^2 + A^2 (x^2-y^2)[ dx^2/(x^2-1) + dy^2/(1-y^2) ]
+ A^2 (x^2-1)(1-y^2) du^2 ],
-infty < t < infty, 1 < x < infty, -1 < y < 1, -pi < u < pi
Example 2: a generalized Rindler chart for (a piece of) Mink vacuum:
ds^2 = -[ A(t) x + B(t) y + C(t) z ]^2 dt^2 + dx^2 + dy^2 + dz^2,
A(t) x + B(t) y + C(t) z > 0
Example 3: "slowly falling" frame written in the ingoing Eddington chart
for the Scharzschild vacuum:
e_1 = @/@t - m/r @/@r
e_2 = @/@t + (1-m/r) @/@r
e_3 = 1/r @/@u
e_4 = sin(u)/r @/@v
-infty < t < infty, 0 < r < infty, 0 < u < pi, -pi < v < pi
I'm pretty sure Ted has seen all this before, but for the benefit of
newbies here are a few brief glosses:
1. These examples all define a metric tensor (as classical authors would
say, by "defining the line element") on some domain, which is specified by
naming the legal range of the coordinates. This domain is of course the
domain of the intended coordinate chart in the sense of manifold theory.
2. This terminology/notation is a compromise between ambiguity and
convenience. In my experience, naming the chart, the intended manifold,
-and- the range of coordinates, will clear up the kind of ambiguity which
most often causes confusion.
3. In many cases I advocate going further and describing a "frame"
(timelike vector field and three spacelike vector fields, orthonormal at
each event), as in the third example above. This can be regarded as a
family of idealized observers in the given spacetime. Often this is what
one really wants in setting up (for example) a thought experiment or a
computation. (BTW, in the example, the name "slowfall frame" comes from
the fact that, putting X = e_1, we have
D_X X = m/r^2 e_2
i.e. our observers are accelerating by amount just sufficient to "hover"
at a given radius according to Newton's theory, but are nonetheless
falling in according to Einstein's theory; this illustrates one effect of
the nonlinearity of the EFE.)
4. There are at least three levels of structure exhibited in the last
example (none depending on any field equation or indeed on any physics):
(a) a Lorentzian manifold (M,g),
(b) a coordinate chart (with domain U) on (M,g),
(c) a frame defined on U.
The metric tensor (on domain U) can be determined from any frame defined
on U; in the example it is
ds^2 = -(1-2m/r) dt^2 + 2 dt dr + r^2 (du^2 + sin(u)^2 dv^2),
-infty < t < infty, 0 < r < infty, 0 < u < pi, -pi < v < pi
from which you can see at once that in the Eddington chart "@/@r" is a
null vector, so it's not the same thing as the @/@r in the standard
Schwarzschild chart (which is a spacelike vector).
Further reading for interested newbies: coordinate charts and manifolds
are discussed in many places, but one which physics students may find
particularly useful is
author = {Theodore Frankel},
title = {The Geometry of Physics: An Introduction},
publisher = {Cambridge University Press},
year = 1997}
John Baez
news:c3ssmt$fm5$1...@lfa222122.richmond.edu...
> It seems to me quite unwise to say that different coordinate systems
> give different metrics. The metric is a tensor, and a tensor
> is something that exists independent of what coordinates one chooses.
> So personally, I wouldn't talk about "the Rindler metric" as
> something distinct from the Minkowski metric; I'd talk about
> Rindler coordinates.
>
> Am I alone in this?
No, this is what all mathematicians do, and many physicists - but
not *all* physicists.
And even if we follow Ted Bunn's wise advice, we can easily get ourselves
confused if we're not careful. We agree that changing coordinates doesn't
change the metric on a manifold. This is what some people call a
"passive" coordinate transformation. But, applying a diffeomorphism to
the metric *does* change it - and this is what they call an "active"
coordinate transformation.
Since there can be situations where what Herr Professor Schmidt regards
as a "passive" coordinate transformation is regarded by Herr Professor
Schultz as an "active" one, the situation is ripe for confusion. Schmidt
will say that the metric doesn't change, while Schultz will insist it does!
So, we gotta be careful.
If you want to talk about a metric modulo diffeomorphisms, call it
a "geometry". Both active and passive coordinate transformations leave
a geometry unchanged.
In GR, two metrics giving the same geometry give the same physics.
In short, we've got:
formulas for metrics in terms of coordinates -
changed by both active and passive coordinate transformations
metrics -
changed by active coordinate transformations but not passive ones
geometries -
unchanged by both active and passive coordinate transformations
I leave as a puzzle to figure out what the 4th possibility is like,
and whether people actually talk about this one!
Leonard
>In short, we've got:
> formulas for metrics in terms of coordinates -
>changed by both active and passive coordinate transformations
>metrics -
>changed by active coordinate transformations but not passive ones
>geometries -
>unchanged by both active and passive coordinate transformations
>I leave as a puzzle to figure out what the 4th possibility is like,
>and whether people actually talk about this one!
topologies
unchanged by homeomorphisms
they do talk about it
if there is another better answer to the puzzle it
would be interesting
Morris Carré
> <eb...@lfa221051.richmond.edu> wrote in message
> news:c3ssmt$fm5$1...@lfa222122.richmond.edu...
>
>
>>It seems to me quite unwise to say that different coordinate systems
>>give different metrics. The metric is a tensor, and a tensor
>>is something that exists independent of what coordinates one chooses.
>>So personally, I wouldn't talk about "the Rindler metric" as
>>something distinct from the Minkowski metric; I'd talk about
>>Rindler coordinates.
>>
>>Am I alone in this?
>
>
> No, this is what all mathematicians do, and many physicists - but
> not *all* physicists.
>
> And even if we follow Ted Bunn's wise advice, we can easily get ourselves
> confused if we're not careful. We agree that changing coordinates doesn't
> change the metric on a manifold. This is what some people call a
> "passive" coordinate transformation. But, applying a diffeomorphism to
> the metric *does* change it - and this is what they call an "active"
> coordinate transformation.
>
> Since there can be situations where what Herr Professor Schmidt regards
> as a "passive" coordinate transformation is regarded by Herr Professor
> Schultz as an "active" one, the situation is ripe for confusion. Schmidt
> will say that the metric doesn't change, while Schultz will insist it does!
This reminds me of a fact I heard about in this newsgroup - that the
equivalence of two solutions to Einstein's equations was in all generality an
undecidable problem. Is my memory correct ? Can I read your portrait of
Schmidt and Schultz as a representation of the same fact ?
Regards, Boris Borcic
--
"The book of revelations is to sense as fiscal law is to money"
Stephen Blake
> We agree that changing coordinates doesn't
> change the metric on a manifold. This is what some people call a
> "passive" coordinate transformation. But, applying a diffeomorphism to
> the metric *does* change it - and this is what they call an "active"
> coordinate transformation.
>
> If you want to talk about a metric modulo diffeomorphisms, call it
> a "geometry". Both active and passive coordinate transformations leave
> a geometry unchanged.
>
> In GR, two metrics giving the same geometry give the same physics.
>
Dr. Baez, in your view, is a diffeomorphism/active transformation the same
as the transformation between two observers and is a coordinate
transformation/passive transformation merely a re-labelling of coordinates
by a single observer?
Arkadiusz Jadczyk
wrote:
>If you want to talk about a metric modulo diffeomorphisms, call it
>a "geometry". Both active and passive coordinate transformations leave
>a geometry unchanged.
>
>In GR, two metrics giving the same geometry give the same physics.
Perhaps with one little comments:
Spinors are sometimes considered to be a part of GR. Dirac operator is,
by some, considered as quite important, perhaps even more important than
the metric itself (Alain Connes) Its spectrum may depend on the spin
structure, and there may be inequivalent structures for two equivalent
metrics.
But that is also a side remark, not that relevant to the original
question....
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Danny Ross Lunsford
Arkadiusz Jadczyk wrote:
>>If you want to talk about a metric modulo diffeomorphisms, call it
>>a "geometry". Both active and passive coordinate transformations leave
>>a geometry unchanged.
>>
>>In GR, two metrics giving the same geometry give the same physics.
>
> Perhaps with one little comments:
>
> Spinors are sometimes considered to be a part of GR. Dirac operator is,
> by some, considered as quite important, perhaps even more important than
> the metric itself (Alain Connes) Its spectrum may depend on the spin
> structure, and there may be inequivalent structures for two equivalent
> metrics.
>
> But that is also a side remark, not that relevant to the original
> question....
No no, this is the whole point! You can always cook up a metric given a
frame. Likewise given some coordinate system one can set up frames.
(Just minutes ago I was trying to imagine some tensors written out
explicitly in terms of the frame instead of g. And then I was sort of
struck by the thought that both GR and field theory are about these
frames, this is the one thing they really have in common.)
The real problem with this is - the metric is not gravity, it's just the
metric. The particular odd form of the Christoffel connection is
gravity. It doesn't need spinors for expression.
-drl
tes...@um.bot
> This reminds me of a fact I heard about in this newsgroup - that the
> equivalence of two solutions to Einstein's equations was in all
> generality an undecidable problem. Is my memory correct ? Can I read
> your portrait of Schmidt and Schultz as a representation of the same
> fact ?
Google for old posts to this group with keywords "Karlhede" and "Cartan",
where you will find elaboration of the following point:
Cartan's very general procedure for recognizing locally isometric
manifolds (which was streamlined by Karlhede for the special case of
Lorentzian geometry) ultimately reduces the problem to an algebraic one.
There is no algorithm for solving this algebraic problem in every case.
Nonetheless, in practice this potential defect seems not to have arisen.
For Cartan's procedure, see
author = {Peter J. Olver},
title = {Equivalence, Invariants, and Symmetry},
publisher = {Cambridge University Press},
year = 1995}
For Karlhede's procedure, see
author = {A. Karlhede and M. A. H. MacCallum},
title = {On Determing the Isometry Group of a Riemannian Space},
journal = {Gen. Rel. Grav.},
volume = 14,
year = 1982,
pages = {673--682}}
author = {D. Pollney and J. E. F. Skea and R. A. D'Inverno},
title = {Classifying geometries in general relativity. {I.} {S}tandard
forms for symmetric spinors},
journal = {Class. Quant. Grav.},
volume = 17,
number = 3,
year = 2000,
pages = {643-663}}
(and succeeding issues for next two parts).
HTH,
"T. Essel"
John Baez
Stephen Blake <ste...@ntlworld.com> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<c4n11l$8i1$1...@glue.ucr.edu>...
>> We agree that changing coordinates doesn't
>> change the metric on a manifold. This is what some people call a
>> "passive" coordinate transformation. But, applying a diffeomorphism to
>> the metric *does* change it - and this is what they call an "active"
>> coordinate transformation.
>Dr. Baez, in your view, is a diffeomorphism/active transformation the same
>as the transformation between two observers and is a coordinate
>transformation/passive transformation merely a re-labelling of coordinates
>by a single observer?
I try to avoid the term "observer" when I'm trying to think about
things precisely. The concept of an observer can be very helpful when
we're fumbling around trying to crack certain physics problems, but
it's devilishly tricky to make precise, except in certain limited contexts.
Since I guess we're trying to be precise here, instead of trying to
crack a physics problem, I'll avoid giving a yes-or-no answer to your
question!
Here's how I think about it. Suppose we have an n-dimensional
space X that admits globally defined coordinates. A "coordinate
system" is then a diffeomorphism
f: X -> R^n
assigning to each point x in X its n-tuple of coordinates, f(x) in R^n.
We can change coordinates in two ways: actively and passively.
A "passive" change of coordinates is a diffeomorphism
g: R^n -> R^n
If we apply this to the coordinate system f we get a new coordinate
system gf, defined by
gf(x) = g(f(x))
In other words, we figure out the coordinates f(x) and then apply
some function to them to get some new coordinates g(f(x)).
An "active" change of coordinates is a diffeomorphism
g: X -> X
If we apply this to the coordinate system f we get a new coordinate
system fg, defined by
fg(x) = f(g(x))
In other words, we move the point x over to the point g(x) and
then figure out its coordinates using f, obtaining f(g(x)).
This is sort of pretty: we see that the difference between
"passive" and "active" changes of coordinates is just the
difference between something like
f |-> gf
and something like
f |-> fg
This should make clear how active and passive coordinate changes
act differently on things like metrics.
Of course the funny thing is that since X is diffeomorphic to R^n,
it could be that X secretly *is* R^n! In this case, coordinates
AND active changes of coordinates AND passive changes of coordinates
are all just diffeomorphisms of R^n - thought of in different ways!
Still, I prefer to think of physical space(time) as some anonymous
manifold X, while our coordinates are lists of numbers, hence R^n.
Also, note that we could replace R^n by some other manifold Y throughout
the entire above discussion (up to but not including this paragraph).
This generalizes the concept of "coordinates" a little bit, in a way
that can be very useful sometimes. We then say:
a "coordinate system on X valued in Y" is a diffeomorphism f: X -> Y
a "passive change of coordinates" is a diffeomorphism g: Y -> Y
an "active change of coordinates" is a diffeomorphism g: X -> X
If I had more time I'd relate all this to the Schroedinger versus
Heisenberg pictures in quantum mechanics and then maybe even tackle
your question about observers, but someone is yelling at me telling me
to set the table for dinner!!!
Sorry to have taken so long to reply. I'll cc this to you but
if you respond, please do so on sci.physics.research.
John Baez
Leonard <leo...@planck.com> wrote:
>John Baez wrote:
There has to be a better answer, since I was asking for something
almost *exactly* like a metric, which is changed by passive
coordinate transformations but not active ones!
A topology is sorta vaguely like a metric, but not almost exactly.
Also, topologies are unchanged by both passive and active coordinate
transformations, where we allow our coordinate transformations to
be homeomorphisms rather than diffeomorphisms. So, a topology is
more like a watered-down "geometry" than the correct answer to my
question. (Watered-down, since any geometry gives a topology, but
the topology is invariant under a bigger group.)
It may help to read my little discussion of active versus
passive coordinate transformations, below.
For people who like math, this can be summarized by saying that
"coordinate systems" form a left torsor of the group of passive
coordinate transformations, and a right torsor of the group of
active coordinate transformations. Thus, the set of coordinate
systems gives an example of what mathematicians call a "bitorsor".
Torsors show up everywhere one discusses "gauge symmetries":
http://math.ucr.edu/home/baez/torsors.html
so it's not surprising that they show up here. BUT: you don't
need to know or care about torsors to solve my puzzle. So, don't
let this paragraph intimidate or distract you if you don't know
what it means!
I could say more, but not without giving away the answer completely.
I'm still hoping someone will solve the puzzle...
..........................................................................
Zig
> In article <mailman.108130...@olympus.het.brown.edu>,
> Leonard <leo...@planck.com> wrote:
>
>
>>John Baez wrote:
>
>
>>>In short, we've got:
>>
>>>formulas for metrics in terms of coordinates -
>>>changed by both active and passive coordinate transformations
>>
>>>metrics -
>>>changed by active coordinate transformations but not passive ones
>>
>>>geometries -
>>>unchanged by both active and passive coordinate transformations
>>
>>>I leave as a puzzle to figure out what the 4th possibility is like,
>>>and whether people actually talk about this one!
>
>
>>topologies
>>unchanged by homeomorphisms
>>they do talk about it
>>
>>if there is another better answer to the puzzle it
>>would be interesting
>
>
> There has to be a better answer, since I was asking for something
> almost *exactly* like a metric, which is changed by passive
> coordinate transformations but not active ones!
>
OK, I would like to try my hand at this puzzle.
We are looking for something that is changed under passive
transformations, which means it is some animal expressed in coordinates.
But something that is left unchanged by active transformations, which
can change the metric.
My guess is that it should be something topological, so that it is
independent of the metric, but it should be built out tensors, so that
its form will change under coordinate changes. So how about something
like the Chern form, expressed in local coordinates?
Stephen Blake
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<c845ea$rtu$1...@glue.ucr.edu>...
>I try to avoid the term "observer" when I'm trying to think about
>things precisely. The concept of an observer can be very helpful
when
>we're fumbling around trying to crack certain physics problems, but
>it's devilishly tricky to make precise, except in certain limited
contexts.
>Since I guess we're trying to be precise here, instead of trying to
>crack a physics problem, I'll avoid giving a yes-or-no answer to your
>question!
I'm puzzled by Dr. Baez's reluctance to identify the
diffeomorphisms/active transformations as the transformations that
connect the different viewpoints of two observers. This was something
I thought I really understood.
I'll argue by the analogy between a general manifold M and de Sitter
space. I choose dS space because it is a manifold with non-zero
curvature which has observer-dependent horizons, so it has the
interesting properties one might encounter in a general manifold, but
the group of congruences of dS space is SO(4,1) which is much easier
to think about than the huge diffeomorphism group Diff(M).
An element g of SO(4,1) is a congruence that acts on the dS manifold
g: dS -> dS. Physically, if a spacetime event appears as point p in
the dS manifold to observer A, then the event appears to observer B as
point g(p) in the dS manifold where the group element g is the
transformation from A's viewpoint to B's viewpoint. A similar
situation exists for a physical quantity. A physical quantity in dS
space must be a realization (or representation) of the congruence
group SO(4,1). So if some physical quantity appears to observer A as
v, the the quantity appears to observer B as T(g)v where T(g) is a rep
of SO(4,1). The congruence group SO(4,1) gives the changes in the
viewpoints of inertial observers in dS space, that is, observers who
are not acted upon by forces.
The above paragraph is presumably uncontroversial, so it seems to me
that nothing goes wrong if I repeat the above paragraph with a few
changes of words to make things more general.
An element g of Diff(M) is a congruence that acts on the manifold g: M
-> M. Physically, if a spacetime event appears as point p in the
manifold M to observer A, then the event appears to observer B as
point g(p) where the group element g is the transformation from A's
viewpoint to B's viewpoint. A similar situation exists for a physical
quantity. A physical quantity must be a realization (or
representation) of Diff(M). For example, a vector field on M is an
element of the Lie algebra of Diff(M) which is a pretty simple rep of
Diff(M). So if some physical quantity appears to observer A as v, the
quantity appears to observer B as T(g)v where T(g) is a rep of
Diff(M). The congruence group Diff(M) gives the changes in the
viewpoints of arbitrary observers in M.
Stephen Blake
--
http://homepage.ntlworld.com/stebla
Alfred Einstead
> We agree that changing coordinates doesn't
> change the metric on a manifold. This is what some people call a
> "passive" coordinate transformation. But, applying a diffeomorphism to
> the metric *does* change it - and this is what they call an "active"
> coordinate transformation.
The more interesting question is: can these considerations be extended
so as to be applicable to connections too? Let W, W' be two
connections with their difference W' - W = K. The structure equations
yield:
d(e^m) + W^m_n ^ e^n = T^m
d(W^m_n) + W^m_p ^ W^p_n = R^m_n
(summation convention used), or more concisely as:
de + We = T; dW + WW = R,
with T and R being the torsion and curvature forms. A
change W -> W' = W + K induces a change:
T' = de + (W+K)e = T + De
R' = d(W+K) + (W+K)(W+K) = R + dK + WK + KW + KK
or in component form:
T'^m = T^m + (K^m_n ^ e^n)
R'^m_n = R^m_n + (DK^m_n + K^m_p ^ K^p_n)
D = covariant exterior differential
which yields the SAME theory, modulo effective spin and
stress tensor terms:
G' = G + delta(G):
delta(G)_{mn} = delta(R)^p_{mpn}) - 1/2 g_{mn} delta(R)^{pq}_{pq}
with
delta(R)^m_{npq} = D_p K^m_{qn} - D_q K^m_{pn}
+ K^m_{pr} K^r_{qn} - K^m_{qr} K^r_{pn}.
This is a more generalized relativity, which equates the two
theories by the duality:
(W,T,R,G,S) --> (W+D,T',R',G+delta(G),S+delta(S))
S = spin tensor,
both yielding the same observational consequences, with all the
differences picked up by the tensors G and S.
Interestingly, one can transform (locally) R to 0 to T to 0,
above, yielding either a curvature-free or torsion-free
formulation of the same theory.