Can two vectors define the Kerr metric?

[chemguy]

Any point in space-time surrounding a massive object is identified by a position vector. The position vector is based at the geometric center of the object.

A displacement is associated with the identified point. The displacement is represented by an incremental displacement vector acting on the identified point. Both vectors may be called “the Kerr vectors”.

The Kerr vectors must relate to each other, to the dimensions of space-time, and to the characteristics of the object (length scales).

Can the Kerr metric be defined by these two vectors?

Reference; http://newstuff77.weebly.com 30 The Kerr Vectors

[JanPB]

On Tuesday, September 25, 2018 at 1:29:23 PM UTC-7, chemguy wrote:
> Any point in space-time surrounding a massive object is identified by a position vector.

That works only in flat (affine) spaces/spacetimes. In general a point
is typically defined by its coordinate values (which presumes the point
is inside a coordinate domain, obviously).

>
> Can the Kerr metric be defined by these two vectors?

No. A metric in general is a dot product on each tangent space which
can vary from point to point (that's why metric coefficients are
usually non-constant functions of position).

So you need to specify that, either directly or indirectly, and nothing
else will do.

--
Jan

[Thomas 'PointedEars' Lahn]

chemguy wrote:
> Any point in space-time surrounding a massive object is identified by a position vector.

Rather, it *can* be identified by a position vector.

> The position vector is based at the geometric center of the object.

Not necessarily, and usually it is not.

> A displacement is associated with the identified point.

No, it is not.

> [not even wrong]

--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

[Hilton Packett]

JanPB wrote:

> On Tuesday, September 25, 2018 at 1:29:23 PM UTC-7, chemguy wrote:
>> Any point in space-time surrounding a massive object is identified by a
>> position vector.
>
> That works only in flat (affine) spaces/spacetimes. In general a point
> is typically defined by its coordinate values (which presumes the point
> is inside a coordinate domain, obviously).

Absolutely, but I see many wannabe homepage programmers around here,
saying that the shortest shortest path is the geodesic of light. Which
reveals they dident understood what is going on in relativity.

[Tom Roberts]

On 9/25/18 3:29 PM, chemguy wrote:
> [...]

No. A metric has MANY more degrees of freedom than two vectors.

Tom Roberts

[Tom Roberts]

On 9/25/18 4:58 PM, Thomas 'PointedEars' Lahn wrote:
> chemguy wrote:
>> Any point in space-time surrounding a massive object is identified by a position vector.
>
> Rather, it *can* be identified by a position vector.

Hmmm. In general the position of an object is NOT a vector of any sort.
Differential displacements in a manifold are vectors, but over
finite/macroscopic distances the result depends on the path over which you
integrate, and is thus not well defined, and CERTAINLY not a vector.

Historically, in EUCLIDEAN SPACE, positions were indeed represented as vectors.
That does not generalize to other manifolds, and is a PUN on "vector".

Tom Roberts

[mitchr...@gmail.com]

On Tuesday, September 25, 2018 at 1:29:23 PM UTC-7, chemguy wrote:

How would we know if any even horizon is lopsided?

Mitchell Raemsch

[Thomas 'PointedEars' Lahn]

Tom Roberts wrote:
> On 9/25/18 4:58 PM, Thomas 'PointedEars' Lahn wrote:
>> chemguy wrote:
>>> Any point in space-time surrounding a massive object is identified by a position vector.
>> Rather, it *can* be identified by a position vector.
>
> Hmmm. In general the position of an object is NOT a vector of any sort.

Learn to read. Nobody said that the position would *be* a vector.

However, it is often formulated this way (for example, in the mathematics
and physics lectures that I heard at the university this week), and that is
certainly not wrong, no matter the number of dimensions.

The defining feature of a vector is that it is a mathematical object whose
sign/direction changes under a parity transformation. This is certainly
true for objects in 3-d space called “position vectors”.

[Tom Roberts]

On 9/27/18 7:51 PM, Thomas 'PointedEars' Lahn wrote:
> Tom Roberts wrote:
>> On 9/25/18 4:58 PM, Thomas 'PointedEars' Lahn wrote:
>>> chemguy wrote:
>>>> Any point in space-time surrounding a massive object is identified by a position vector.
>>> Rather, it *can* be identified by a position vector.
>>
>> Hmmm. In general the position of an object is NOT a vector of any sort.
>
> Learn to read. Nobody said that the position would *be* a vector.

Such legalistic nit-picking is unwarranted. Using the phrases "position vector"
and "can be identified" give essentially the same meaning -- you attempt to make
a distinction without a meaningful difference.

Acting as if you were the arbiter of correct English usage
is rather silly for a non-native English speaker in a
PHYSICS newsgroup. Though I understand the motivation.

> However, it is often formulated this way (for example, in the mathematics
> and physics lectures that I heard at the university this week), and that is
> certainly not wrong, no matter the number of dimensions.

Yes, as I said, this ONLY applies in Euclidean space. This simply does not work
for any manifold with curvature.

Moreover, this is a PUN on "vector", and is related to physicists' traditional
disdain for mathematical rigor. That disdain is disappearing as we learn that
GR, string theory, and loop quantum gravity all require careful attention to the
details of the math.

> The defining feature of a vector is that it is a mathematical object whose
> sign/direction changes under a parity transformation.

No. The defining characteristic of a vector is that it is a member of a vector
space [#]. This includes requirements about addition and multiplication by a
scalar, which fail for "position vectors" in curved manifolds.

[#] Despite the names, this is not circular.

Note the importance of curved manifolds in theoretical physics cannot be
ignored, as the Lagrangian density for EVERY current physical theory is the
scalar curvature of an appropriate manifold.

Neither string theory nor loop quantum gravity are current
physical theories. Yet....

Tom Roberts

[Thomas 'PointedEars' Lahn]

Tom Roberts wrote:
> On 9/27/18 7:51 PM, Thomas 'PointedEars' Lahn wrote:
>> Tom Roberts wrote:
>>> On 9/25/18 4:58 PM, Thomas 'PointedEars' Lahn wrote:
>>>> chemguy wrote:
>>>>> Any point in space-time surrounding a massive object is identified by a position vector.
>>>> Rather, it *can* be identified by a position vector.
>>>
>>> Hmmm. In general the position of an object is NOT a vector of any sort.
>>
>> Learn to read. Nobody said that the position would *be* a vector.
>
> Such legalistic nit-picking is unwarranted.

No, it is not. There is a difference between the identification of a
position and the equivalence with a position..

Using the phrases "position vector"
> and "can be identified" give essentially the same meaning --

No.

> you attempt to make a distinction without a meaningful difference.

You are wrong.

>> However, it is often formulated this way (for example, in the mathematics
>> and physics lectures that I heard at the university this week), and that is
>> certainly not wrong, no matter the number of dimensions.
>
> Yes, as I said, this ONLY applies in Euclidean space. This simply does not work
> for any manifold with curvature.

Example?

> Moreover, this is a PUN on "vector", and is related to physicists' traditional
> disdain for mathematical rigor.

Doubtful.

>> The defining feature of a vector is that it is a mathematical object whose
>> sign/direction changes under a parity transformation.
>
> No. The defining characteristic of a vector is that it is a member of a vector
> space [#].

You have it backwards.

[JanPB]

On Tuesday, September 25, 2018 at 2:58:49 PM UTC-7, Thomas 'PointedEars' Lahn wrote:
> chemguy wrote:
> > Any point in space-time surrounding a massive object is identified by a position vector.
>
> Rather, it *can* be identified by a position vector.

In a fixed coordinate system, yes, because of the identification of
the relevant piece of the manifold with a linear space. But that's not
what the original poster meant: he wrote "Any point in space-time..."
which, while imprecise, implies no specific coordinates chosen since e.g.
it's not true in general every manifold possesses a coordinate system
covering it completely.

--
Jan

[Tom Roberts]

On 9/28/18 11:55 AM, Thomas 'PointedEars' Lahn wrote:
> Tom Roberts wrote:
>> Yes, as I said, this ONLY applies in Euclidean space. This simply does not work
>> for any manifold with curvature.
>
> Example?

Consider the surface of the unit sphere S^2. Pick any point on it and
lay down a vector of length pi in some direction. Add to it another
vector of length pi parallel to the first. You of course come back to
the original point. So you have added two non-zero vectors and obtain a
zero vector. That violates the requirements of a vector space.

Tom Roberts

[Thomas 'PointedEars' Lahn]

Pardon? Of course not. That the addition of two non-zero vectors gives
a zero vector happens all the time in vector spaces. For example, in the
conservation of total linear momentum.

[pass...@domain.invalid]

:)

>
> Tom Roberts

[Tom Roberts]

On 9/29/18 8:56 AM, Thomas 'PointedEars' Lahn wrote:
> Tom Roberts wrote:
>> On 9/28/18 11:55 AM, Thomas 'PointedEars' Lahn wrote:
>>> Tom Roberts wrote:
>>>> Yes, as I said, this ONLY applies in Euclidean space. This simply does not work
>>>> for any manifold with curvature.
>>> Example?
>>
>> Consider the surface of the unit sphere S^2. Pick any point on it and
>> lay down a vector of length pi in some direction. Add to it another
>> vector of length pi parallel to the first. You of course come back to
>> the original point. So you have added two non-zero vectors and obtain a
>> zero vector. That violates the requirements of a vector space.
>
> Pardon? Of course not. That the addition of two non-zero vectors gives
> a zero vector happens all the time in vector spaces. For example, in the
> conservation of total linear momentum.

Yep. My mistake.

Tom Roberts

[Hilton Packett]

No is not. Is his.

[Koobee Wublee]

On Tuesday, September 25, 2018 at 1:29:23 PM UTC-7, chemguy wrote:

> Any point in space-time is identified by a position vector. The position

> vector is based at the geometric center of the object.

Yes. This is true for any geometry of spacetime. Which idiot says otherwise? <shrug>

> A displacement is associated with the identified point.

Yes, again. <shrug>

Say <s> is a position vector. Then, d<s> or <ds> is a point vector that the self-styled physicists call tangent vector or something. <shrug>

> Can the Kerr metric be defined by these two vectors?

Any geometry is the inner product of two d<s>. For example,

** ds² = d<s> * d<s> = [g]_ij d[q]^i d[q]^j

Where

** ds = geometry of spacetime
** [g] = matrix that represents the metric
** [g]_ij = elements to [g]
** [q] = matrix that represents the coordinate system
** [q]^i, [q]^j = elements to [q]
** <a> * <b> = inner product of two vectors <a> and <b>

[Paul Gish]

Koobee Wublee wrote:

>> Can the Kerr metric be defined by these two vectors?
>
> Any geometry is the inner product of two d<s>. For example,
> ** ds² = d<s> * d<s> = [g]_ij d[q]^i d[q]^j
> Where
> ** ds = geometry of spacetime ** [g] = matrix that represents the
> metric ** [g]_ij = elements to [g]
> ** [q] = matrix that represents the coordinate system ** [q]^i, [q]^j
> = elements to [q]

Probably not, but you have to express the units, ie what units is that
"geometry of spacetime" using? I am a serbian.

[JanPB]

On Wednesday, October 3, 2018 at 12:53:22 PM UTC-7, Koobee Wublee wrote:
> On Tuesday, September 25, 2018 at 1:29:23 PM UTC-7, chemguy wrote:
>
> > Any point in space-time is identified by a position vector. The position
> > vector is based at the geometric center of the object.
>
> Yes. This is true for any geometry of spacetime.

No, this is false. In general "position vector" applies only to vector
spaces. Manifold points OTOH are identified by coordinate values. On
certain subsets, namely on coordinate domains on the manifold, one can
use the resulting identification (which is non-canonical) of manifold
points with the coordinate space points. The coordinate space is linear,
hence one can use its vector and affine structures interchangeably:
either its points or its vectors (the canonical identification of affine
and vector structures arises as soon as a point is selected as a
reference, or "origin").

But it's false to say that ANY point in space time can be universally
identified by a vector because in general manifolds are not covered
by a _single_ coordinate domain.

Bottom line is invoking vetcors in the manifold context is very misleading
as they only have a canonical existence as elements of tangent spaces, NOT
as manifold points.

All of the above is global analysis 101.

--
Jan

[Koobee Wublee]

On Wednesday, October 3, 2018 at 1:51:48 PM UTC-7, JanPB wrote:
> On Wednesday, October 3, 2018 at 12:53:22 PM UTC-7, Koobee Wublee wrote:

> > Say <s> is a position vector. Then, d<s> or <ds> is a point vector
> > that the self-styled physicists call tangent vector or something.
> > <shrug>
>

> > Any geometry is the inner product of two d<s>. For example,
>
> > ** ds² = d<s> * d<s> = [g]_ij d[q]^i d[q]^j
>
> > Where
>
> > ** ds = geometry of spacetime
> > ** [g] = matrix that represents the metric
> > ** [g]_ij = elements to [g]
> > ** [q] = matrix that represents the coordinate system
> > ** [q]^i, [q]^j = elements to [q]

> > ** <?> * <?> = inner product of two vectors

> No, this is false. All of the above is global analysis 101.
>
> [Garbage snipped]

All this is geometry 1. Any point is space or spacetime regardless how curved up it is can be described by a position vector. This is a very simple concept of mapping the geometry of reality into observed space using the position vector <s> where in turn <s> can be described by the observer’s choice of coordinate system. <shrug>

[Paul Gish]

Mapping is the oldest trick in the book. As for instance using addition
instead of multiplication.

[JanPB]

On Wednesday, October 3, 2018 at 2:24:09 PM UTC-7, Koobee Wublee wrote:
> On Wednesday, October 3, 2018 at 1:51:48 PM UTC-7, JanPB wrote:
> > On Wednesday, October 3, 2018 at 12:53:22 PM UTC-7, Koobee Wublee wrote:
>
> > > Say <s> is a position vector. Then, d<s> or <ds> is a point vector
> > > that the self-styled physicists call tangent vector or something.
> > > <shrug>
> >
> > > Any geometry is the inner product of two d<s>. For example,
> >
> > > ** ds² = d<s> * d<s> = [g]_ij d[q]^i d[q]^j
> >
> > > Where
> >
> > > ** ds = geometry of spacetime
> > > ** [g] = matrix that represents the metric
> > > ** [g]_ij = elements to [g]
> > > ** [q] = matrix that represents the coordinate system
> > > ** [q]^i, [q]^j = elements to [q]
> > > ** <?> * <?> = inner product of two vectors
>
> > No, this is false. All of the above is global analysis 101.
> >
> > [Garbage snipped]
>
> All this is geometry 1. Any point is space or spacetime regardless how curved up it is can be described by a position vector.

Incorrect.

--
Jan

[Koobee Wublee]

On Wednesday, October 3, 2018 at 2:53:08 PM UTC-7, JanPB wrote:
> On Wednesday, October 3, 2018 at 2:24:09 PM UTC-7, Koobee Wublee wrote:

> > Any point is space or spacetime regardless how curved up it is can be

> > described by a position vector. This is a very simple concept of
> > mapping the geometry of reality into observed space using the position
> > vector <s> where in turn <s> can be described by the observer’s choice
> > of coordinate system. <shrug>
>

> > d<s> or <ds> is a point vector that the self-styled physicists call

> > tangent vector or something. Any geometry is the inner product of two

> > d<s>. For example,
>
> > ** ds² = d<s> * d<s> = [g]_ij d[q]^i d[q]^j
>
> > Where
>
> > ** ds = geometry of spacetime
> > ** [g] = matrix that represents the metric
> > ** [g]_ij = elements to [g]
> > ** [q] = matrix that represents the coordinate system
> > ** [q]^i, [q]^j = elements to [q]
> > ** <?> * <?> = inner product of two vectors

> Incorrect.

That is the opinion of the queer of England who never has comprehended geometry 1. <shrug>

[Paul Gish]

JanPB wrote:

>> All this is geometry 1. Any point is space or spacetime regardless how
>> curved up it is can be described by a position vector.
>
> Incorrect.

Combined with the gradient, it can. Go back under your bridge.

[JanPB]

Not even wrong.

--
Jan

[Paul Gish]

Piece of cake, packed into a tensor combined with the curvature deficit.