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Space Gets Curved!
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Anamitra Palit
Jan 13, 2012, 8:17:32 PM1/13/12
to
Let’s consider a rectangular (t,x,y,z) in the flat spacetime context.
Two planes z=k and z=-k are considered. Gravity is now turned on—and
we assume the new metric to be a complicated one. The relative
physical distances between the points on each plane as well as the
distance between points lying pairwise on the two different planes
change.
The planes should become physically curved even from the spatial
point of view.
Suppose we are standing on the ground and gravity grows stronger. Is
the curvature of the ground we are standing on going to change?
To understand the situation we may consider a light ,thin rubber sheet
stretched horizontally over four poles on the earth’s surface. If
gravity grows stronger the rubber sheet should curve downwards in
response to gravity, the inertial resistance of its material being
weak. In case of the earth the inertial resistance would be much
stronger. If it was as weak as the of the rubber sheet ,the curvature
of the ground should change. In fact if the earth receives a large
amount of high density mass which gets evenly spread over its surface
a collapse could start despite the inertial resistance of the
earth ,changing is curvature. The collapse might be of an asymmetric
nature if the mass received is not distributed uniformly over the
surface
Conclusion:If a gravitational change occurs the space between the
coordinate labels should[in general] get physically curved instead of
being stretched or compressed.
Two planes z=k and z=-k are considered. Gravity is now turned on—and
we assume the new metric to be a complicated one. The relative
physical distances between the points on each plane as well as the
distance between points lying pairwise on the two different planes
change.
The planes should become physically curved even from the spatial
point of view.
Suppose we are standing on the ground and gravity grows stronger. Is
the curvature of the ground we are standing on going to change?
To understand the situation we may consider a light ,thin rubber sheet
stretched horizontally over four poles on the earth’s surface. If
gravity grows stronger the rubber sheet should curve downwards in
response to gravity, the inertial resistance of its material being
weak. In case of the earth the inertial resistance would be much
stronger. If it was as weak as the of the rubber sheet ,the curvature
of the ground should change. In fact if the earth receives a large
amount of high density mass which gets evenly spread over its surface
a collapse could start despite the inertial resistance of the
earth ,changing is curvature. The collapse might be of an asymmetric
nature if the mass received is not distributed uniformly over the
surface
Conclusion:If a gravitational change occurs the space between the
coordinate labels should[in general] get physically curved instead of
being stretched or compressed.
Anamitra Palit
Jan 13, 2012, 10:23:18 PM1/13/12
to
We may consider the equation: y=x of a straight line in the first
quadrant. Let it represent the path of a light ray in the flat
spacetime context.
We use the following transformations:
x=x’
y=Ay’^2
z=z’
t=t’
A is a constant having the dimension:1/Length and y’ has the dimension
of length.
Initial metric:ds^2=dt^2-dx^2-dy^2-dz^2
Final metric:ds^2=dt’^2-dx’^2-(2Ay’dy')^2-dz’^2
=dt’^2-dx’^2-4A^2 y’^2dy'^2-dz’^2
Incidentally Ay’ is a dimensionless quantity.
Transformed equation of the path of the light ray:
x’=Ay’^2
The above equation is coordinate equation of the light path
representing a parabola. The physical equation remains a straight
line. Incidentally flat spacetime is characterized by straight line
geodesics, both in the coordinate and the physical sense.We have moved
to a new manifold by our transformation!
Initially we had a flat x-y surface. The new x’-y’ surface is an
undulating one which might incorporate stretching/compression of
physical distances . The light ray in the transformed frame moves
through set of coordinate points on the x’-y’ plane whose
projection on the old Euclidean surface is a parabola.
Conclusion:If a gravitational change occurs the space between the
undulation might include stretching or compression when observed
against a background Euclidean surface.
Tom Roberts
Jan 14, 2012, 4:23:28 PM1/14/12
to
On 1/13/12 1/13/12 9:23 PM, Anamitra Palit wrote:
> On Jan 14, 6:17 am, Anamitra Palit<palit.anami...@gmail.com> wrote:
>> Let’s consider a rectangular (t,x,y,z) in the flat spacetime context.
>> Two planes z=k and z=-k are considered. Gravity is now turned on
But you cannot "turn on gravity" without violating the known laws of physics,
> On Jan 14, 6:17 am, Anamitra Palit<palit.anami...@gmail.com> wrote:
>> Let’s consider a rectangular (t,x,y,z) in the flat spacetime context.
>> Two planes z=k and z=-k are considered. Gravity is now turned on
and thus negating anything you attempt to say about the physical situation.
Moreover, you seem completely unaware of the arbitrary nature of coordinates,
and how in general (i.e. in a curved manifold) you cannot discuss "straight
lines" or "planes", because those concepts lose their meanings.
>> The planes should become physically curved even from the spatial
>> point of view.
attempt to compare two different manifolds, one flat and one curved (with
gravitating objects), than you cannot uniquely identify points between them --
you can ARBITRARILY choose which points in the curved manifold to identify with
the planes of the flat manifold, but it is clear that such identification has no
physical basis, as it is an ARBITRARY choice you made.
Or you could start in a region of the universe far from any massive object, so
spacetime is approximately flat to excellent accuracy, and mark out your two
planes with a bunch of rice grains all at rest in your coordinates. Then if a
massive planet approaches, and if you are willing to speak quite loosely, you
could say "gravity has been turned on" (or better, "gravity has greatly
increased"). But it should be obvious that the rice grains will all fall to the
planet's surface, destroying any illusion you might have had that they mark out
"two planes".
>> Suppose we are standing on the ground and gravity grows stronger.
There's no point in attempting to discuss a "physical situation" that violates
the known laws of physics.
>> Conclusion:If a gravitational change occurs the space between the
>> coordinate labels should[in general] get physically curved instead of
>> being stretched or compressed.
words you use do not fit together in the way you use them and still retain their
original meanings -- the result is word salad with no meaning.
> We may consider the equation: y=x of a straight line in the first
> quadrant. Let it represent the path of a light ray in the flat
> spacetime context.
y=x is NOT the path of a light ray, because a light ray MOVES. Light follows a
null geodesic, and y=x is clearly not such a geodesic path.
So for the remainder of this discussion, let us use the word "path" to mean the
spatial locus occupied by the light ray, independent of time. This is not an
unusual meaning, but it must be recognized that it is not the trajectory of any
light ray (which is necessarily a null geodesic).
Then y=x is not sufficient to define the path of a light ray, you must specify
y=x,z=0,t=0 to define a line. Note that t=0 shows this "path" is not the usual
time-dependent trajectory of light.
> We use the following transformations:
> x=x’
> y=Ay’^2
> z=z’
> t=t’
> A is a constant having the dimension:1/Length and y’ has the dimension
> of length.
> Initial metric:ds^2=dt^2-dx^2-dy^2-dz^2
OK.
> Final metric:ds^2=dt’^2-dx’^2-(2Ay’dy')^2-dz’^2
> =dt’^2-dx’^2-4A^2 y’^2dy'^2-dz’^2
> Incidentally Ay’ is a dimensionless quantity.
> Transformed equation of the path of the light ray:
> x’=Ay’^2
equation is not linear.
[#] because the manifold is flat this is well defined with its
usual meaning. Remember our meaning of "path" above.
> The above equation is coordinate equation of the light path
> representing a parabola. The physical equation remains a straight
> line. Incidentally flat spacetime is characterized by straight line
> geodesics, both in the coordinate and the physical sense.
the path itself is a straight line.
So why do you then go one to make a blatantly incorrect statement?
> We have moved
> to a new manifold by our transformation!
That's what a coordinate transformation does. These are not Minkowski
coordinates, and straight lines are not necessarily linear in these coordinates,
but they are applied to the same Minkowski manifold as the original (unprimed)
coordinates in terms of which they are defined.
> Initially we had a flat x-y surface. The new x’-y’ surface is an
> undulating one which might incorporate stretching/compression of
> physical distances .
defined by z=0,t=0. The surface defined by z'=0,t'=0 is EXACTLY THE SAME
SURFACE, and it remains a flat plane.
Consider instead the surface x=y,t=0; as y and z vary the surface is seen to be
a flat plane (it is not normal to the x-axis, the y-axis, or the z-axis).
Consider the surface x'=y',t=0; the intersection of this surface with any
surface z'=const is a parabola [@]. This is probably the curved surface you
wanted to discuss. Note this is a COMPLETELY DIFFERENT SURFACE from x=y,t=0.
[@] I mean the plane geometrical figure is a parabola, INDEPENDENT
of its representation in the primed coordinates.
Note that a point requires 4 values of the coordinates, a line
requires three relations among coordinate values, and a plane
requires two relations among coordinate values.
> The light ray in the transformed frame moves
> through set of coordinate points on the x’-y’ plane whose
> projection on the old Euclidean surface is a parabola.
line in the manifold, is linear in the unprimed coordinates, and is a parabola
in the primed coordinates. This is a single path in the (single) manifold,
described via two different coordinate systems.
> Conclusion:[... more unphysical nonsense]
Your attempt at a conclusion is both unphysical and wrong.
As I have said before, you have some very basic misunderstandings
about geometry and GR. You really need to take a course at a
major university on this, so you can discuss the concepts with an
instructor who understands them. the internet is not a suitable
medium for this.
Tom Roberts
Anamitra Palit
Jan 14, 2012, 4:21:26 PM1/14/12
to
represent the path of a light ray in the flat spacetime context.
We use the following transformations[many-to –one]:
x’=x
y’=Ay^2
z’=Bz+Cx^2+Dy^10
t’=t
A is a constant having the dimension:1/Length
C has the dimension :1/Length
D has the dimension: Length^(-9)
Transformed equation projected on the x’-y’ coordinate plane:
x’=Sqrt[y’/A]
The “coordinate equation” of the projected path of the light ray is
not a straight line in the second frame. It is a parabola. The
physical variables will produce a straight line IF physical time is
taken to be:dT_p=sqrt[g_(00)dt. Incidentally flat spacetime is
characterized by straight line geodesics, both in the coordinate and
the physical sense.We have moved to a new manifold by our
transformation!
Initially we had a flat x-y surface. The new x’-y’ surface is an
undulating one which might incorporate stretching/compression of
physical distances . The light ray in the transformed frame moves
through set of coordinate points whose projection on the old
the physical sense.We have moved to a new manifold by our
transformation!
Initially we had a flat x-y surface. The new x’-y’ surface is an
undulating one which might incorporate stretching/compression of
physical distances . The light ray in the transformed frame moves
Euclidean surface is a parabola.
[In my second posting the x'-y' surface is not an undulating one.The
current posting considers an undulating x'-y' surface]
X-Phy
Jan 14, 2012, 5:39:56 PM1/14/12
to
On 14 jan, 02:17, Anamitra Palit <palit.anami...@gmail.com> wrote:
> Let’s consider a rectangular (t,x,y,z) in the flat spacetime context.
> Two planes z=k and z=-k are considered. Gravity is now turned on—and
> we assume the new metric to be a complicated one. The relative
> physical distances between the points on each plane as well as the
> distance between points lying pairwise on the two different planes
> change.
>
> The planes should become physically curved even from the spatial
> point of view.
The planes get curved, but one can't see it!
> Let’s consider a rectangular (t,x,y,z) in the flat spacetime context.
> Two planes z=k and z=-k are considered. Gravity is now turned on—and
> we assume the new metric to be a complicated one. The relative
> physical distances between the points on each plane as well as the
> distance between points lying pairwise on the two different planes
> change.
>
> The planes should become physically curved even from the spatial
> point of view.
How does one verify that a surface is plane? One looks at it from
within it.
But the trajectories of the photons get curved too.
--
X-Phy
Anamitra Palit
Jan 14, 2012, 10:41:57 PM1/14/12
to
On Jan 15, 2:23 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> OK. Note the PATH remains a straight line [#], but in these coordinates its
> equation is not linear.
>
>
>
> OK. Note the PATH remains a straight line [#], but in these coordinates its
> equation is not linear.
>
>
> > The above equation is coordinate equation of the light path
> > representing a parabola. The physical equation remains a straight
> > line. Incidentally flat spacetime is characterized by straight line
> > geodesics, both in the coordinate and the physical sense.
>
> Yes to all this. Remember the parabola is purely in the primed coordinates, and
> the path itself is a straight line.
>
> So why do you then go one to make a blatantly incorrect statement?
>
> > representing a parabola. The physical equation remains a straight
> > line. Incidentally flat spacetime is characterized by straight line
> > geodesics, both in the coordinate and the physical sense.
>
> Yes to all this. Remember the parabola is purely in the primed coordinates, and
> the path itself is a straight line.
>
> So why do you then go one to make a blatantly incorrect statement?
>
> You mixed up the meanings of words, but yes, the path y=x,z=0,t=0 is a straight
> line in the manifold, is linear in the unprimed coordinates, and is a parabola
> in the primed coordinates. This is a single path in the (single) manifold,
> described via two different coordinate systems.
>
> Tom Roberts
> line in the manifold, is linear in the unprimed coordinates, and is a parabola
> in the primed coordinates. This is a single path in the (single) manifold,
> described via two different coordinate systems.
>
The path of a light ray in the flat spacetime context is a straight
line. By some suitable transformation you can always convert this path
into a curved line in a different frame. No harm if you treat the
second frame as "Mathematical WorkSpace".
But you could always find some true physical manifold where the light
ray follows exactly the SAME curved path as it follows wrt to the new
coordinates.
I have a circular metal ring in front of me. By working out
transformations on a piece of paper I can get get an oval shaped
object for the ring.That does not distort the ring in the physical
sense. But I can always get an oval ring[of exactly the same shape and
size as I have got out the transformation] in the physical world.
Now lets consider the metrics referred to in my last second posting:
Initial metric:ds^2=dt^2-dx^2-dy^2-dz^2 ----------------(1)
Final metric:ds^2=dt’^2-dx’^2-(2Ay’dy')^2-dz’^2
=dt’^2-dx’^2-4A^2 y’^2dy'^2-dz’^2
Incidentally Ay’ is a dimensionless quantity.
Transformations used:
x=x'
y=Ay'^2 [Dimension of A:1/Length]]
You may derive the equation of the null geodesic [and other
geodesics]directly from the second metric. The spatial part of the
null geodesic is not a straight line.
Point to be considered:
A transformation can create a new workspace--a mathematical one where
objects from the first one get distorted.It is quite possible that we
can have such distorted objects in the physical world.
Tom Roberts
Jan 16, 2012, 1:11:14 AM1/16/12
to
On 1/14/12 1/14/12 9:41 PM, Anamitra Palit wrote:
> The path of a light ray in the flat spacetime context is a straight
> line.
Yes.
> The path of a light ray in the flat spacetime context is a straight
> line.
> By some suitable transformation you can always convert this path
> into a curved line in a different frame.
You MUST be more careful about what you mean. If by "path" you apply the usual
meaning of a 1-d locus in spacetime (see my previous post in this thread), then
your statement is just plain wrong -- the locus is a straight line and no
coordinate transformation can possibly change that. If by "path" you apply an
ABNORMAL meaning of the equation in some coordinate system, then certainly the
description of the straight-line path (normal meaning) can appear to be "curved".
> But you could always find some true physical manifold where the light
> ray follows exactly the SAME curved path as it follows wrt to the new
> coordinates.
context.
I don't know what it means, as there is no definitive and unique
method to compare paths in different manifolds. For instance, one
could say "the same" means there is a diffeomorphism between the
manifolds that takes the path in one to the path in the other --
that is useless as almost every pair of paths would be "the same".
BTW here you seem to be applying the usual meaning of "path", making your
confusion evident.
> I have a circular metal ring in front of me. By working out
> transformations on a piece of paper I can get get an oval shaped
> object for the ring.
possibly change THE RING ITSELF. What has changed in transforming to these new
coordinates is the DESCRIPTION of the ring -- there is ONE AND ONLY ONE RING.
One of the most important lessons of General Relativity is that
coordinate-dependent descriptions are so fungible that they are almost useless.
One must discuss INVARIANTS.
This is remarkably close to the dictum from Quantum Mechanics:
discuss only observables. Surely there is a deep connection here....
> That does not distort the ring in the physical
> sense. But I can always get an oval ring[of exactly the same shape and
> size as I have got out the transformation] in the physical world.
coordinate transform did NOTHING WHATSOEVER to the shape and size of the ring;
it merely changed the DESCRIPTION of the ring in terms of the new coordinates.
Besides, one can imagine mathematical coordinate transforms that
distort the equations of the ring in unphysical ways.
> Now lets consider the metrics referred to in my last second posting:
> Initial metric:ds^2=dt^2-dx^2-dy^2-dz^2 ----------------(1)
> Final metric:ds^2=dt’^2-dx’^2-(2Ay’dy')^2-dz’^2
> =dt’^2-dx’^2-4A^2 y’^2dy'^2-dz’^2
> --------------- (2)
> Incidentally Ay’ is a dimensionless quantity.
> Transformations used:
> x=x'
> y=Ay'^2 [Dimension of A:1/Length]]
>
> You may derive the equation of the null geodesic [and other
> geodesics]directly from the second metric. The spatial part of the
> null geodesic is not a straight line.
flat, and the path of the light ray is a straight line (using either our earlier
meaning of "path" or the usual meaning). Coordinate transformations do not
change ANY geometrical or physical aspect of anything, they merely change the
DESCRIPTIONS.
One could describe some object in English, or in German, or in
French, or in any other language. Clearly such translations do
not affect the object. Coordinates are like the different
languages and coordinate transformations are like translations
between languages. The analogy is quite good, insofar as both
languages and coordinates are used for descriptions of objects.
A major difference is that mathematical transformations do not
have the ambiguities inherent in translating human languages.
> Point to be considered:
> A transformation can create a new workspace--a mathematical one where
> objects from the first one get distorted.It is quite possible that we
> can have such distorted objects in the physical world.
the standard vocabulary of mathematics and physics, and then use it. This is a
well-traveled road -- making up your own vocabulary is wasted effort, and is
useless for communicating with others.
In particular, you need to learn what a diffeomorphism is, what
coordinates are, what coordinate transformations are, and how
a diffeomorphism between manifolds differs from a coordinate
transformation. You are VERY confused, and seem to not understand
the very basic aspects of this, including the very important
distinction between model and world (the essence of the difference
between mathematics and physics).
In any case, it is clear to me that any mathematical manipulations you make have
NO EFFECT WHATSOEVER on the real world, so your last sentence is nonsense.
Another aspect of your confusion is that whenever one describes
something using a given set of coordinates, the description cannot
be separated from the coordinates -- BOTH are needed to know the
object. You repeatedly attempt to discuss them disjointed.
Moreover, the coordinates cannot be separated from the underlying
manifold.
Tom Roberts
Anamitra Palit
Feb 7, 2012, 10:46:15 AM2/7/12
to
On Jan 16, 11:11 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> On 1/14/12 1/14/12 9:41 PM, Anamitra Palit wrote:
>
>
> > I have a circular metal ring in front of me. By working out
> > transformations on a piece of paper I can get get an oval shaped
> > object for the ring.
>
> No! Just THINK about what you are doing -- no mathematical manipulation can
> possibly change THE RING ITSELF. What has changed in transforming to these new
> coordinates is the DESCRIPTION of the ring -- there is ONE AND ONLY ONE RING.
> Tom Roberts
> On 1/14/12 1/14/12 9:41 PM, Anamitra Palit wrote:
>
>
> > I have a circular metal ring in front of me. By working out
> > transformations on a piece of paper I can get get an oval shaped
> > object for the ring.
>
> No! Just THINK about what you are doing -- no mathematical manipulation can
> possibly change THE RING ITSELF. What has changed in transforming to these new
> coordinates is the DESCRIPTION of the ring -- there is ONE AND ONLY ONE RING.
You may again consider a the same circular metal ring. The coordinate
description may be a circle , an ellipse or some irregular shape
depending on the coordinate system.If the ring was NOT in front of
you,how would you decide the nature of the physical object by looking
at the graphs corresponding to the metal ring? Suppose I keep the
metal ring in the cupboard and present before you a hundred
"equivalent " graphs which I drew when the object was in front of me.
In the meantime Harry enters the room with hundred different objects
with shapes [and size] identical with the different graphs.What
conclusion would you draw regarding the physical nature of the object
represented by the graphs?
[You are not allowed to consider any frame as preferred over the
others]
Anamitra
Tom Roberts
Feb 8, 2012, 12:00:13 AM2/8/12
to
On 2/7/12 2/7/12 9:46 AM, Anamitra Palit wrote:
> On Jan 16, 11:11 am, Tom Roberts<tjroberts...@sbcglobal.net> wrote:
>> On 1/14/12 1/14/12 9:41 PM, Anamitra Palit wrote:
>>> I have a circular metal ring in front of me. By working out
>>> transformations on a piece of paper I can get get an oval shaped
>>> object for the ring.
>> No! Just THINK about what you are doing -- no mathematical manipulation can
>> possibly change THE RING ITSELF. What has changed in transforming to these new
>> coordinates is the DESCRIPTION of the ring -- there is ONE AND ONLY ONE RING.
>
> On Jan 16, 11:11 am, Tom Roberts<tjroberts...@sbcglobal.net> wrote:
>> On 1/14/12 1/14/12 9:41 PM, Anamitra Palit wrote:
>>> I have a circular metal ring in front of me. By working out
>>> transformations on a piece of paper I can get get an oval shaped
>>> object for the ring.
>> No! Just THINK about what you are doing -- no mathematical manipulation can
>> possibly change THE RING ITSELF. What has changed in transforming to these new
>> coordinates is the DESCRIPTION of the ring -- there is ONE AND ONLY ONE RING.
>
> You may again consider a the same circular metal ring. The coordinate
> description may be a circle , an ellipse or some irregular shape
> depending on the coordinate system.If the ring was NOT in front of
> you,how would you decide the nature of the physical object by looking
> at the graphs corresponding to the metal ring? [...]
> description may be a circle , an ellipse or some irregular shape
> depending on the coordinate system.If the ring was NOT in front of
> you,how would you decide the nature of the physical object by looking
> [You are not allowed to consider any frame as preferred over the
> others]
[I will answer in the context of Euclidean geometry, as that seems to
> others]
me to be what you have in mind, and your description uses concepts
that are specific to Euclidean geometry. My answer will generalize
to any N-dimensional (semi-)Riemannian manifold, but I won't attempt
to keep the description general enough for that.]
In addition to the description of the ring in a given set of coordinates, one
also needs the metric projected onto those coordinates, and the coordinates must
be valid in a simply-connected region that encloses the entire ring (you could
not claim to have a coordinate description without this, though you might not
have realized the need for the region to be simply connected -- that ensures
that the integrals below are well defined).
First, consider how to determine if a certain 1-d locus is a circle, given a
description in a given set of coordinates: For any circle, all points lie in
some 2-d plane [@], and in that plane there exists a point P which we call its
center, such that for every point Q of the locus we have
\integral ds = R
where R is a real constant called the radius, the integral is taken over the
unique geodesic path connecting P to Q [#], and ds is the usual line element of
the manifold, and in these coordinates it is written:
ds^2 = g_ij dx^i dx^j
where the {g_ij} are the metric components for these coordinates, and the {dx^i}
are coordinate differentials along the path. Implicitly all these quantities are
a function of position in the manifold, which means functions of the coordinates
{x^i}. The locus is a circle iff such a point P exists, and all points that
satisfy the above are in the locus.
That is a (relatively) convenient approach, because you start with the locus
given in terms of the coordinates. It clearly does not depend on which
coordinates you use.
[@] One difficulty in generalizing this is defining what this means
in a non-Euclidean manifold.
[#] We know this geodesic exists and is unique because the context
is Euclidean geometry and the region is simply connected. One of
the complexities in generalizing this method is replacing this with
a suitable requirement.
An entirely different approach is to start with the coordinates and their metric
components, and construct a set of Cartesian coordinates (aka Riemann normal
coordinates) in the region. Then use the coordinate transform from the original
coordinates to get the description in terms of the Cartesian coordinates. It
will be much easier to recognize standard shapes (such as a circle) in Cartesian
coordinates. But this might be considered to violate your "no preferred frame"
requirement.
That is a start, and the general idea; for your physical ring described in terms
of coordinates, you need to establish that its surface consists only of circles
in the appropriate orientations and positions. The details are too complex, and
too specific to this rather artificial example, for me to be interested in.
Tom Roberts
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