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Topic: Science > Physics
User: ""
Date: 19 Jan 2006 10:26:42 PM
Object: resolve to perpendicular components, because they are independent
Hi Physicers,
Any help will be greatly appreciated.
It is my understanding that we often resolve vectors (like force,
velocity) into perpendicular components because they are so called
"independent." If one component changes value, it doesn't affect the
value of the other pependicular components (I guess this is where
"Independent comes from").
Yet I have been recently shown how to resolve vectors (force) into
components that aren't perpendicular using a reverse-parallelogram
rule. Say a force is acting on a structure like < at the left point, I
have to resolve it into two forces alone the two branches (not
orthogonal). I encountered this in my self study of certain questions,
so I don't really have anyone reliable to ask why this is viable in
light of my previous knowledge of "independent vectors need to be
perpendicular."
Can someone shed light on any of this?
.

User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 02:23:52 AM
wrote:

Hi Physicers,
Any help will be greatly appreciated.
It is my understanding that we often resolve vectors (like force,
velocity) into perpendicular components because they are so called
"independent." If one component changes value, it doesn't affect the
value of the other pependicular components (I guess this is where
"Independent comes from").
Yet I have been recently shown how to resolve vectors (force) into
components that aren't perpendicular using a reverse-parallelogram
rule. Say a force is acting on a structure like < at the left point, I
have to resolve it into two forces alone the two branches (not
orthogonal). I encountered this in my self study of certain questions,
so I don't really have anyone reliable to ask why this is viable in
light of my previous knowledge of "independent vectors need to be
perpendicular."
Can someone shed light on any of this?

I think you want to study the *Kronecker Delta*. I accept that
a number of axes x^1, x^2...x^n are *independant* if
((& is partial))
&x^a / &x^b = 1 if a=b, and 0 if a=/=b.
Is this a structural engineering question?
Ken
.
User: ""

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 02:55:20 AM
In article <1137745432.546723.23260@f14g2000cwb.googlegroups.com>, "Ken S. Tucker" <dynamics@vianet.on.ca> writes:


kenneth.bull@gmail.com wrote:

Hi Physicers,
Any help will be greatly appreciated.
It is my understanding that we often resolve vectors (like force,
velocity) into perpendicular components because they are so called
"independent." If one component changes value, it doesn't affect the
value of the other pependicular components (I guess this is where
"Independent comes from").
Yet I have been recently shown how to resolve vectors (force) into
components that aren't perpendicular using a reverse-parallelogram
rule. Say a force is acting on a structure like < at the left point, I
have to resolve it into two forces alone the two branches (not
orthogonal). I encountered this in my self study of certain questions,
so I don't really have anyone reliable to ask why this is viable in
light of my previous knowledge of "independent vectors need to be
perpendicular."
Can someone shed light on any of this?


I think you want to study the *Kronecker Delta*. I accept that
a number of axes x^1, x^2...x^n are *independant* if
((& is partial))

&x^a / &x^b = 1 if a=b, and 0 if a=/=b.

Everybody (well, nearly everybody) is confusing the poor guy. Let me
try to set the record straight.
Mathematically, a set of vectors is independent if non of them can be
expressed as a linear combination of the other ones. For the 3D case
you care about it simply means that 3 vectors are independent if
they're not lying in the same plane. Translating to axes it means the
same, you cannot have 3 axes lying in the same plane, otherwise the
angles between them are arbitrary. So, no, independent vectors do not
need to be orthogonal.
But, but... non orthogonal axes are a pain. Many of the formulas you
use implicitly assume orthogonal vector base. Once it is not
orthogonal then, to begin with the scalar product isn't given by the
simple formula (a1,a2,a3) /dot ((b1,b2,b3) = a1*b1 + a2*b2 + a3*b3.
You've a more complex formula instead, one that is taking the angles
betwee the axes into account. Vector product is even worse. And when
you get to vector operators like grad or curl, all hell breaks out.
So, I'll second Timo's advice, i.e. unless you've a good enough reson
to go non-orthogonal, don't. But, if you choose to do, it is legit
(though painful).
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 03:29:39 AM
Mati and Timo are probably right, but...
mmeron@cars3.uchicago.edu wrote:

In article <1137745432.546723.23260@f14g2000cwb.googlegroups.com>, "Ken S. Tucker" <dynamics@vianet.on.ca> writes:


kenneth.bull@gmail.com wrote:

Hi Physicers,
Any help will be greatly appreciated.
It is my understanding that we often resolve vectors (like force,
velocity) into perpendicular components because they are so called
"independent." If one component changes value, it doesn't affect the
value of the other pependicular components (I guess this is where
"Independent comes from").
Yet I have been recently shown how to resolve vectors (force) into
components that aren't perpendicular using a reverse-parallelogram
rule. Say a force is acting on a structure like < at the left point, I
have to resolve it into two forces alone the two branches (not
orthogonal). I encountered this in my self study of certain questions,
so I don't really have anyone reliable to ask why this is viable in
light of my previous knowledge of "independent vectors need to be
perpendicular."
Can someone shed light on any of this?


I think you want to study the *Kronecker Delta*. I accept that
a number of axes x^1, x^2...x^n are *independant* if
((& is partial))

&x^a / &x^b = 1 if a=b, and 0 if a=/=b.


Everybody (well, nearly everybody) is confusing the poor guy. Let me
try to set the record straight.

Mathematically, a set of vectors is independent if non of them can be
expressed as a linear combination of the other ones. For the 3D case
you care about it simply means that 3 vectors are independent if
they're not lying in the same plane. Translating to axes it means the
same, you cannot have 3 axes lying in the same plane, otherwise the
angles between them are arbitrary. So, no, independent vectors do not
need to be orthogonal.

But, but... non orthogonal axes are a pain. Many of the formulas you
use implicitly assume orthogonal vector base. Once it is not
orthogonal then, to begin with the scalar product isn't given by the
simple formula (a1,a2,a3) /dot ((b1,b2,b3) = a1*b1 + a2*b2 + a3*b3.
You've a more complex formula instead, one that is taking the angles
betwee the axes into account. Vector product is even worse. And when
you get to vector operators like grad or curl, all hell breaks out.

I find nonorthogonal axes easier than orthogonal, indeed a Curl
becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
manipulating equations in tensors is streamlined by notation.
For me, orthogonal metrics are challenging, specifically defining
a metric that can be transformed to
X_uv = 1 or 0 (u=v or u=/=v)
in a more general metric like,
g_uv = X_uv + A_u B_v .
Ken

So, I'll second Timo's advice, i.e. unless you've a good enough reson
to go non-orthogonal, don't. But, if you choose to do, it is legit
(though painful).

.
User: "Timo Nieminen"

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 04:36:31 AM
On Fri, 20 Jan 2006, Ken S. Tucker wrote:

I find nonorthogonal axes easier than orthogonal,

Then you must be some kind of bizarre freak of nature!!!

indeed a Curl
becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
manipulating equations in tensors is streamlined by notation.

Can't you just do that with orthogonal metrics too? (Mixing covariant and
contravariant is just a naughty little trick to hide the metric tensor!)
--
Timo
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 05:47:34 PM
Timo Nieminen wrote:

On Fri, 20 Jan 2006, Ken S. Tucker wrote:

I find nonorthogonal axes easier than orthogonal,


Then you must be some kind of bizarre freak of nature!!!

Not really, as in Chess, solving problems in mathematical
physics consists of keeping your options open, to be
closed by physical principle, and certainly not by an aprior
preceived convenience. It's well known "orthogonality" is
at best an approximation in a g-field, but Reimann and his
"gang" evolved quite a nice "tensor" analysis notation that
is easier to use than clunky "ijk" unit vectors.

indeed a Curl
becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
manipulating equations in tensors is streamlined by notation.


Can't you just do that with orthogonal metrics too? (Mixing covariant and
contravariant is just a naughty little trick to hide the metric tensor!)

If your intrinsic dimensionality differs from an integer, i.e
let n= intrinsic dimensionality =2.9, then how the heck do
you expect to squeeze 3 orthogonals into that?
Ken
.
User: "FrediFizzx"

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 06:32:30 PM
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
|
| Timo Nieminen wrote:
| > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| >
| > > I find nonorthogonal axes easier than orthogonal,
| >
| > Then you must be some kind of bizarre freak of nature!!!
|
| Not really, as in Chess, solving problems in mathematical
| physics consists of keeping your options open, to be
| closed by physical principle, and certainly not by an aprior
| preceived convenience. It's well known "orthogonality" is
| at best an approximation in a g-field, but Reimann and his
| "gang" evolved quite a nice "tensor" analysis notation that
| is easier to use than clunky "ijk" unit vectors.
|
| > > indeed a Curl
| > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
| > > manipulating equations in tensors is streamlined by notation.
| >
| > Can't you just do that with orthogonal metrics too? (Mixing
covariant and
| > contravariant is just a naughty little trick to hide the metric
tensor!)
|
| If your intrinsic dimensionality differs from an integer, i.e
| let n= intrinsic dimensionality =2.9, then how the heck do
| you expect to squeeze 3 orthogonals into that?
Hmm... I wonder if that would apply to what Lisa Randall is calling
"Warped Passages"?
FrediFizzx
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 06:53:19 PM
FrediFizzx wrote:

"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
|
| Timo Nieminen wrote:
| > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| >
| > > I find nonorthogonal axes easier than orthogonal,
| >
| > Then you must be some kind of bizarre freak of nature!!!
|
| Not really, as in Chess, solving problems in mathematical
| physics consists of keeping your options open, to be
| closed by physical principle, and certainly not by an aprior
| preceived convenience. It's well known "orthogonality" is
| at best an approximation in a g-field, but Reimann and his
| "gang" evolved quite a nice "tensor" analysis notation that
| is easier to use than clunky "ijk" unit vectors.
|
| > > indeed a Curl
| > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
| > > manipulating equations in tensors is streamlined by notation.
| >
| > Can't you just do that with orthogonal metrics too? (Mixing
covariant and
| > contravariant is just a naughty little trick to hide the metric
tensor!)
|
| If your intrinsic dimensionality differs from an integer, i.e
| let n= intrinsic dimensionality =2.9, then how the heck do
| you expect to squeeze 3 orthogonals into that?

Hmm... I wonder if that would apply to what Lisa Randall is calling
"Warped Passages"?

LOL, ok, how about a link, Randall is super-pop, so
I know your not jokin...

FrediFizzx

Ken
.
User: "FrediFizzx"

Title: Re: resolve to perpendicular components, because they are independent 20 Jan 2006 07:18:08 PM
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137804799.742330.45120@o13g2000cwo.googlegroups.com...
|
| FrediFizzx wrote:
| > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
| > |
| > | Timo Nieminen wrote:
| > | > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| > | >
| > | > > I find nonorthogonal axes easier than orthogonal,
| > | >
| > | > Then you must be some kind of bizarre freak of nature!!!
| > |
| > | Not really, as in Chess, solving problems in mathematical
| > | physics consists of keeping your options open, to be
| > | closed by physical principle, and certainly not by an aprior
| > | preceived convenience. It's well known "orthogonality" is
| > | at best an approximation in a g-field, but Reimann and his
| > | "gang" evolved quite a nice "tensor" analysis notation that
| > | is easier to use than clunky "ijk" unit vectors.
| > |
| > | > > indeed a Curl
| > | > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
| > | > > manipulating equations in tensors is streamlined by notation.
| > | >
| > | > Can't you just do that with orthogonal metrics too? (Mixing
| > covariant and
| > | > contravariant is just a naughty little trick to hide the metric
| > tensor!)
| > |
| > | If your intrinsic dimensionality differs from an integer, i.e
| > | let n= intrinsic dimensionality =2.9, then how the heck do
| > | you expect to squeeze 3 orthogonals into that?
| >
| > Hmm... I wonder if that would apply to what Lisa Randall is calling
| > "Warped Passages"?
|
| LOL, ok, how about a link, Randall is super-pop, so
| I know you're not jokin...
"Discretizing Gravity in Warped Spacetime"
http://www.arxiv.org/abs/hep-th/0507102
I haven't read this yet but maybe it has something. I was mainly
referring to something she was saying in her new book (did you get it
yet? ;-) ). I didn't make the connection at the time I was reading it
until you brought this up (forgot what you call it) again.
FrediFizzx
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 21 Jan 2006 01:29:46 PM
FrediFizzx wrote:

"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137804799.742330.45120@o13g2000cwo.googlegroups.com...
|
| FrediFizzx wrote:
| > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
| > |
| > | Timo Nieminen wrote:
| > | > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| > | >
| > | > > I find nonorthogonal axes easier than orthogonal,
| > | >
| > | > Then you must be some kind of bizarre freak of nature!!!
| > |
| > | Not really, as in Chess, solving problems in mathematical
| > | physics consists of keeping your options open, to be
| > | closed by physical principle, and certainly not by an aprior
| > | preceived convenience. It's well known "orthogonality" is
| > | at best an approximation in a g-field, but Reimann and his
| > | "gang" evolved quite a nice "tensor" analysis notation that
| > | is easier to use than clunky "ijk" unit vectors.
| > |
| > | > > indeed a Curl
| > | > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
| > | > > manipulating equations in tensors is streamlined by notation.
| > | >
| > | > Can't you just do that with orthogonal metrics too? (Mixing
| > covariant and
| > | > contravariant is just a naughty little trick to hide the metric
| > tensor!)
| > |
| > | If your intrinsic dimensionality differs from an integer, i.e
| > | let n= intrinsic dimensionality =2.9, then how the heck do
| > | you expect to squeeze 3 orthogonals into that?
| >
| > Hmm... I wonder if that would apply to what Lisa Randall is calling
| > "Warped Passages"?
|
| LOL, ok, how about a link, Randall is super-pop, so
| I know you're not jokin...

"Discretizing Gravity in Warped Spacetime"
http://www.arxiv.org/abs/hep-th/0507102

I haven't read this yet but maybe it has something. I was mainly
referring to something she was saying in her new book (did you get it
yet? ;-) ). I didn't make the connection at the time I was reading it
until you brought this up (forgot what you call it) again.
FrediFizzx

Here's an interesting quicky...
http://en.wikipedia.org/wiki/Fractional_calculus
that demo's a departure from our usual "integer" thinking, we
commonly apply to both calculus and so to dimensionality.
Recall that when we integrate a line like "x" by
$ x dx = x^2/2 == area
we go from 1D "x" to 2D "x^2" , but what the link above shows
is that integration (and differentiation) can be a continuous thing,
and so can dimensionality.
Is that where we're going?
Ken
.
User: "FrediFizzx"

Title: Re: resolve to perpendicular components, because they are independent 22 Jan 2006 01:11:01 AM
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137871786.842707.317150@f14g2000cwb.googlegroups.com...
|
| FrediFizzx wrote:
| > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > news:1137804799.742330.45120@o13g2000cwo.googlegroups.com...
| > |
| > | FrediFizzx wrote:
| > | > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > | > news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
| > | > |
| > | > | Timo Nieminen wrote:
| > | > | > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| > | > | >
| > | > | > > I find nonorthogonal axes easier than orthogonal,
| > | > | >
| > | > | > Then you must be some kind of bizarre freak of nature!!!
| > | > |
| > | > | Not really, as in Chess, solving problems in mathematical
| > | > | physics consists of keeping your options open, to be
| > | > | closed by physical principle, and certainly not by an aprior
| > | > | preceived convenience. It's well known "orthogonality" is
| > | > | at best an approximation in a g-field, but Reimann and his
| > | > | "gang" evolved quite a nice "tensor" analysis notation that
| > | > | is easier to use than clunky "ijk" unit vectors.
| > | > |
| > | > | > > indeed a Curl
| > | > | > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
| > | > | > > manipulating equations in tensors is streamlined by
notation.
| > | > | >
| > | > | > Can't you just do that with orthogonal metrics too? (Mixing
| > | > covariant and
| > | > | > contravariant is just a naughty little trick to hide the
metric
| > | > tensor!)
| > | > |
| > | > | If your intrinsic dimensionality differs from an integer, i.e
| > | > | let n= intrinsic dimensionality =2.9, then how the heck do
| > | > | you expect to squeeze 3 orthogonals into that?
| > | >
| > | > Hmm... I wonder if that would apply to what Lisa Randall is
calling
| > | > "Warped Passages"?
| > |
| > | LOL, ok, how about a link, Randall is super-pop, so
| > | I know you're not jokin...
| >
| > "Discretizing Gravity in Warped Spacetime"
| > http://www.arxiv.org/abs/hep-th/0507102
| >
| > I haven't read this yet but maybe it has something. I was mainly
| > referring to something she was saying in her new book (did you get
it
| > yet? ;-) ). I didn't make the connection at the time I was reading
it
| > until you brought this up (forgot what you call it) again.
| > FrediFizzx
|
| Here's an interesting quicky...
|
| http://en.wikipedia.org/wiki/Fractional_calculus
|
| that demo's a departure from our usual "integer" thinking, we
| commonly apply to both calculus and so to dimensionality.
|
| Recall that when we integrate a line like "x" by
|
| $ x dx = x^2/2 == area
|
| we go from 1D "x" to 2D "x^2" , but what the link above shows
| is that integration (and differentiation) can be a continuous thing,
| and so can dimensionality.
|
| Is that where we're going?
Yep, I am really thinking that this is what she is talking about with
"warped" spacetime. Now what is that particular name you had for this?
Sheesh... I can't believe I forgot it!
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.vacuum-physics.com
.
User: "Hexenmeister"

Title: Re: resolve to perpendicular components, because they are independent 22 Jan 2006 01:31:08 AM
"FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:43gpcpF1nr999U1@individual.net...

"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137871786.842707.317150@f14g2000cwb.googlegroups.com...
|
| FrediFizzx wrote:
| > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > news:1137804799.742330.45120@o13g2000cwo.googlegroups.com...
| > |
| > | FrediFizzx wrote:
| > | > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > | > news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
| > | > |
| > | > | Timo Nieminen wrote:
| > | > | > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| > | > | >
| > | > | > > I find nonorthogonal axes easier than orthogonal,
| > | > | >
| > | > | > Then you must be some kind of bizarre freak of nature!!!
| > | > |
| > | > | Not really, as in Chess, solving problems in mathematical
| > | > | physics consists of keeping your options open, to be
| > | > | closed by physical principle, and certainly not by an aprior
| > | > | preceived convenience. It's well known "orthogonality" is
| > | > | at best an approximation in a g-field, but Reimann and his
| > | > | "gang" evolved quite a nice "tensor" analysis notation that
| > | > | is easier to use than clunky "ijk" unit vectors.
| > | > |
| > | > | > > indeed a Curl
| > | > | > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
| > | > | > > manipulating equations in tensors is streamlined by
notation.
| > | > | >
| > | > | > Can't you just do that with orthogonal metrics too? (Mixing
| > | > covariant and
| > | > | > contravariant is just a naughty little trick to hide the
metric
| > | > tensor!)
| > | > |
| > | > | If your intrinsic dimensionality differs from an integer, i.e
| > | > | let n= intrinsic dimensionality =2.9, then how the heck do
| > | > | you expect to squeeze 3 orthogonals into that?
| > | >
| > | > Hmm... I wonder if that would apply to what Lisa Randall is
calling
| > | > "Warped Passages"?
| > |
| > | LOL, ok, how about a link, Randall is super-pop, so
| > | I know you're not jokin...
| >
| > "Discretizing Gravity in Warped Spacetime"
| > http://www.arxiv.org/abs/hep-th/0507102
| >
| > I haven't read this yet but maybe it has something. I was mainly
| > referring to something she was saying in her new book (did you get
it
| > yet? ;-) ). I didn't make the connection at the time I was reading
it
| > until you brought this up (forgot what you call it) again.
| > FrediFizzx
|
| Here's an interesting quicky...
|
| http://en.wikipedia.org/wiki/Fractional_calculus
|
| that demo's a departure from our usual "integer" thinking, we
| commonly apply to both calculus and so to dimensionality.
|
| Recall that when we integrate a line like "x" by
|
| $ x dx = x^2/2 == area
|
| we go from 1D "x" to 2D "x^2" , but what the link above shows
| is that integration (and differentiation) can be a continuous thing,
| and so can dimensionality.
|
| Is that where we're going?

Yep, I am really thinking that this is what she is talking about with
"warped" spacetime. Now what is that particular name you had for this?
Sheesh... I can't believe I forgot it!

Warped mind, perhaps?
Have some curved spacetime:
http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/crls.rxml
Hexenmeister.
.

User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 22 Jan 2006 03:15:40 AM
Hi Fred
FrediFizzx wrote:

"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137871786.842707.317150@f14g2000cwb.googlegroups.com...
|
| FrediFizzx wrote:
| > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > news:1137804799.742330.45120@o13g2000cwo.googlegroups.com...
| > |
| > | FrediFizzx wrote:
| > | > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > | > news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
| > | > |
| > | > | Timo Nieminen wrote:
| > | > | > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| > | > | >
| > | > | > > I find nonorthogonal axes easier than orthogonal,
| > | > | >
| > | > | > Then you must be some kind of bizarre freak of nature!!!
| > | > |
| > | > | Not really, as in Chess, solving problems in mathematical
| > | > | physics consists of keeping your options open, to be
| > | > | closed by physical principle, and certainly not by an aprior
| > | > | preceived convenience. It's well known "orthogonality" is
| > | > | at best an approximation in a g-field, but Reimann and his
| > | > | "gang" evolved quite a nice "tensor" analysis notation that
| > | > | is easier to use than clunky "ijk" unit vectors.
| > | > |
| > | > | > > indeed a Curl
| > | > | > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u), because
| > | > | > > manipulating equations in tensors is streamlined by
notation.
| > | > | >
| > | > | > Can't you just do that with orthogonal metrics too? (Mixing
| > | > covariant and
| > | > | > contravariant is just a naughty little trick to hide the
metric
| > | > tensor!)
| > | > |
| > | > | If your intrinsic dimensionality differs from an integer, i.e
| > | > | let n= intrinsic dimensionality =2.9, then how the heck do
| > | > | you expect to squeeze 3 orthogonals into that?
| > | >
| > | > Hmm... I wonder if that would apply to what Lisa Randall is
calling
| > | > "Warped Passages"?
| > |
| > | LOL, ok, how about a link, Randall is super-pop, so
| > | I know you're not jokin...
| >
| > "Discretizing Gravity in Warped Spacetime"
| > http://www.arxiv.org/abs/hep-th/0507102
| >
| > I haven't read this yet but maybe it has something. I was mainly
| > referring to something she was saying in her new book (did you get
it
| > yet? ;-) ). I didn't make the connection at the time I was reading
it
| > until you brought this up (forgot what you call it) again.
| > FrediFizzx
|
| Here's an interesting quicky...
|
| http://en.wikipedia.org/wiki/Fractional_calculus
|
| that demo's a departure from our usual "integer" thinking, we
| commonly apply to both calculus and so to dimensionality.
|
| Recall that when we integrate a line like "x" by
|
| $ x dx = x^2/2 == area
|
| we go from 1D "x" to 2D "x^2" , but what the link above shows
| is that integration (and differentiation) can be a continuous thing,
| and so can dimensionality.
|
| Is that where we're going?

Yep, I am really thinking that this is what she is talking about with
"warped" spacetime. Now what is that particular name you had for this?

I refer to that as "partial interdimensional transformations",
basically lifting the requirement for an integer in the tensor
indices.
Recall that conventionally a spacetime tensor would
have components like A_u == A_0, A_1, A_2, A_3 , but
when we do the tensor calculus the number u=4 is not
required until we substitute a 4D CS into a specific
physical application.
We also know that the "nonorthogonality" (warp) of spacetime
depends on the strength of the g-field.
An example is the hypothetical "event horizon" where both
time and 1 spatial dimension vanish altogether in that extreme
circumstance.

From the point of view of experimental mathematics there

is a physical basis to consider fractional dimensionality just
as we may consider,
http://en.wikipedia.org/wiki/Fractional_calculus
to show fractional derivatives and integrals is rational
mathematics.
So if we start with some vector (or tensor) "A" we can
manipulate the components "A_n" without specifing n
to be an integer. But why would we want to?
My reasoning is based on a sort of General Covariance,
where the laws of nature are independent of preconceived
dimensionality. If I was an electron, the laws of nature
would apply to me, but I would not need a wrist watch,
because an electron is totally stable in time, and wouldn't
work :-).
Suppose for example we have preconceived laws of nature
cast in 4D and find difficulties of applying GR inside a sub-
atomic particle. Would we dismiss GR or question the idea
that our macroscopic 4D view of dimensionality is true at all
scales?
So formulating the laws of physics independent of fixed
dimensionality is reasonable.
Regards
Ken S. Tucker


FrediFizzx

http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps

http://www.vacuum-physics.com

.
User: "FrediFizzx"

Title: Re: resolve to perpendicular components, because they are independent 22 Jan 2006 03:19:04 PM
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137921340.012633.216030@z14g2000cwz.googlegroups.com...
| Hi Fred
|
| FrediFizzx wrote:
| > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > news:1137871786.842707.317150@f14g2000cwb.googlegroups.com...
| > |
| > | FrediFizzx wrote:
| > | > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > | > news:1137804799.742330.45120@o13g2000cwo.googlegroups.com...
| > | > |
| > | > | FrediFizzx wrote:
| > | > | > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > | > | > news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
| > | > | > |
| > | > | > | Timo Nieminen wrote:
| > | > | > | > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| > | > | > | >
| > | > | > | > > I find nonorthogonal axes easier than orthogonal,
| > | > | > | >
| > | > | > | > Then you must be some kind of bizarre freak of nature!!!
| > | > | > |
| > | > | > | Not really, as in Chess, solving problems in mathematical
| > | > | > | physics consists of keeping your options open, to be
| > | > | > | closed by physical principle, and certainly not by an
aprior
| > | > | > | preceived convenience. It's well known "orthogonality" is
| > | > | > | at best an approximation in a g-field, but Reimann and his
| > | > | > | "gang" evolved quite a nice "tensor" analysis notation
that
| > | > | > | is easier to use than clunky "ijk" unit vectors.
| > | > | > |
| > | > | > | > > indeed a Curl
| > | > | > | > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u),
because
| > | > | > | > > manipulating equations in tensors is streamlined by
| > notation.
| > | > | > | >
| > | > | > | > Can't you just do that with orthogonal metrics too?
(Mixing
| > | > | > covariant and
| > | > | > | > contravariant is just a naughty little trick to hide the
| > metric
| > | > | > tensor!)
| > | > | > |
| > | > | > | If your intrinsic dimensionality differs from an integer,
i.e
| > | > | > | let n= intrinsic dimensionality =2.9, then how the heck do
| > | > | > | you expect to squeeze 3 orthogonals into that?
| > | > | >
| > | > | > Hmm... I wonder if that would apply to what Lisa Randall is
| > calling
| > | > | > "Warped Passages"?
| > | > |
| > | > | LOL, ok, how about a link, Randall is super-pop, so
| > | > | I know you're not jokin...
| > | >
| > | > "Discretizing Gravity in Warped Spacetime"
| > | > http://www.arxiv.org/abs/hep-th/0507102
| > | >
| > | > I haven't read this yet but maybe it has something. I was
mainly
| > | > referring to something she was saying in her new book (did you
get
| > it
| > | > yet? ;-) ). I didn't make the connection at the time I was
reading
| > it
| > | > until you brought this up (forgot what you call it) again.
| > | > FrediFizzx
| > |
| > | Here's an interesting quicky...
| > |
| > | http://en.wikipedia.org/wiki/Fractional_calculus
| > |
| > | that demo's a departure from our usual "integer" thinking, we
| > | commonly apply to both calculus and so to dimensionality.
| > |
| > | Recall that when we integrate a line like "x" by
| > |
| > | $ x dx = x^2/2 == area
| > |
| > | we go from 1D "x" to 2D "x^2" , but what the link above shows
| > | is that integration (and differentiation) can be a continuous
thing,
| > | and so can dimensionality.
| > |
| > | Is that where we're going?
| >
| > Yep, I am really thinking that this is what she is talking about
with
| > "warped" spacetime. Now what is that particular name you had for
this?
|
| I refer to that as "partial interdimensional transformations",
| basically lifting the requirement for an integer in the tensor
| indices.
Ok thanks. Maybe we can call this PIT? ;-) As a way to remember it.
| Recall that conventionally a spacetime tensor would
| have components like A_u == A_0, A_1, A_2, A_3 , but
| when we do the tensor calculus the number u=4 is not
| required until we substitute a 4D CS into a specific
| physical application.
| We also know that the "nonorthogonality" (warp) of spacetime
| depends on the strength of the g-field.
| An example is the hypothetical "event horizon" where both
| time and 1 spatial dimension vanish altogether in that extreme
| circumstance.
Sure. I think Randall is applying the warp factor to a 5th dimension.
However she is using branes that are separated. I wonder if what she is
doing would work with the branes intersecting somewhat? IOW, the 5th
dimension "distance" instead of a separation distance would be an
intersection distance.
| >From the point of view of experimental mathematics there
| is a physical basis to consider fractional dimensionality just
| as we may consider,
|
| http://en.wikipedia.org/wiki/Fractional_calculus
|
| to show fractional derivatives and integrals is rational
| mathematics.
|
| So if we start with some vector (or tensor) "A" we can
| manipulate the components "A_n" without specifing n
| to be an integer. But why would we want to?
Do you mean; why would we want to specify an integer? If so, I would
imagine that it makes calculations easier? ;-)
| My reasoning is based on a sort of General Covariance,
| where the laws of nature are independent of preconceived
| dimensionality. If I was an electron, the laws of nature
| would apply to me, but I would not need a wrist watch,
| because an electron is totally stable in time, and wouldn't
| work :-).
|
| Suppose for example we have preconceived laws of nature
| cast in 4D and find difficulties of applying GR inside a sub-
| atomic particle. Would we dismiss GR or question the idea
| that our macroscopic 4D view of dimensionality is true at all
| scales?
| So formulating the laws of physics independent of fixed
| dimensionality is reasonable.
Hmm... I thought that using tensors already took care of that. So you
are saying that using integer indices in tensor notation doesn't really
"match" nature? I guess it is due to the fact that we are imposing a
Lorentzian signature.
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.vacuum-physics.com
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 22 Jan 2006 06:57:01 PM
FrediFizzx wrote:

"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1137921340.012633.216030@z14g2000cwz.googlegroups.com...
| Hi Fred
|
| FrediFizzx wrote:
| > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > news:1137871786.842707.317150@f14g2000cwb.googlegroups.com...
| > |
| > | FrediFizzx wrote:
| > | > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > | > news:1137804799.742330.45120@o13g2000cwo.googlegroups.com...
| > | > |
| > | > | FrediFizzx wrote:
| > | > | > "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
| > | > | > news:1137800854.404588.75250@z14g2000cwz.googlegroups.com...
| > | > | > |
| > | > | > | Timo Nieminen wrote:
| > | > | > | > On Fri, 20 Jan 2006, Ken S. Tucker wrote:
| > | > | > | >
| > | > | > | > > I find nonorthogonal axes easier than orthogonal,
| > | > | > | >
| > | > | > | > Then you must be some kind of bizarre freak of nature!!!
| > | > | > |
| > | > | > | Not really, as in Chess, solving problems in mathematical
| > | > | > | physics consists of keeping your options open, to be
| > | > | > | closed by physical principle, and certainly not by an
aprior
| > | > | > | preceived convenience. It's well known "orthogonality" is
| > | > | > | at best an approximation in a g-field, but Reimann and his
| > | > | > | "gang" evolved quite a nice "tensor" analysis notation
that
| > | > | > | is easier to use than clunky "ijk" unit vectors.
| > | > | > |
| > | > | > | > > indeed a Curl
| > | > | > | > > becomes A_u,v - A_v,u (== &A_u/&x^v - &A_v/&x^u),
because
| > | > | > | > > manipulating equations in tensors is streamlined by
| > notation.
| > | > | > | >
| > | > | > | > Can't you just do that with orthogonal metrics too?
(Mixing
| > | > | > covariant and
| > | > | > | > contravariant is just a naughty little trick to hide the
| > metric
| > | > | > tensor!)
| > | > | > |
| > | > | > | If your intrinsic dimensionality differs from an integer,
i.e
| > | > | > | let n= intrinsic dimensionality =2.9, then how the heck do

| > | > | > | you expect to squeeze 3 orthogonals into that?
| > | > | >
| > | > | > Hmm... I wonder if that would apply to what Lisa Randall is
| > calling
| > | > | > "Warped Passages"?
| > | > |
| > | > | LOL, ok, how about a link, Randall is super-pop, so
| > | > | I know you're not jokin...
| > | >
| > | > "Discretizing Gravity in Warped Spacetime"
| > | > http://www.arxiv.org/abs/hep-th/0507102
| > | >
| > | > I haven't read this yet but maybe it has something. I was
mainly
| > | > referring to something she was saying in her new book (did you
get
| > it
| > | > yet? ;-) ). I didn't make the connection at the time I was
reading
| > it
| > | > until you brought this up (forgot what you call it) again.
| > | > FrediFizzx
| > |
| > | Here's an interesting quicky...
| > |
| > | http://en.wikipedia.org/wiki/Fractional_calculus
| > |
| > | that demo's a departure from our usual "integer" thinking, we
| > | commonly apply to both calculus and so to dimensionality.
| > |
| > | Recall that when we integrate a line like "x" by
| > |
| > | $ x dx = x^2/2 == area
| > |
| > | we go from 1D "x" to 2D "x^2" , but what the link above shows
| > | is that integration (and differentiation) can be a continuous
thing,
| > | and so can dimensionality.
| > |
| > | Is that where we're going?
| >
| > Yep, I am really thinking that this is what she is talking about
with
| > "warped" spacetime. Now what is that particular name you had for
this?
|
| I refer to that as "partial interdimensional transformations",
| basically lifting the requirement for an integer in the tensor
| indices.

Ok thanks. Maybe we can call this PIT? ;-) As a way to remember it.

Ok with PIT.

| Recall that conventionally a spacetime tensor would
| have components like A_u == A_0, A_1, A_2, A_3 , but
| when we do the tensor calculus the number u=4 is not
| required until we substitute a 4D CS into a specific
| physical application.
| We also know that the "nonorthogonality" (warp) of spacetime
| depends on the strength of the g-field.
| An example is the hypothetical "event horizon" where both
| time and 1 spatial dimension vanish altogether in that extreme
| circumstance.

Sure. I think Randall is applying the warp factor to a 5th dimension.
However she is using branes that are separated. I wonder if what she is
doing would work with the branes intersecting somewhat? IOW, the 5th
dimension "distance" instead of a separation distance would be an
intersection distance.

It's reasonable to consider 5D, just look around your office and
see that your desk is denser than the air above it, so far so good.
OTOH, the density of the desk may be described by a 4D warp.
Some physicists perfer to impose a 5th D variable on an orthogonal
4D, but then that becomes a nonorthogonal 4D, as in GR.

| >From the point of view of experimental mathematics there
| is a physical basis to consider fractional dimensionality just
| as we may consider,
|
| http://en.wikipedia.org/wiki/Fractional_calculus
|
| to show fractional derivatives and integrals is rational
| mathematics.
|
| So if we start with some vector (or tensor) "A" we can
| manipulate the components "A_n" without specifing n
| to be an integer. But why would we want to?

Do you mean; why would we want to specify an integer? If so, I would
imagine that it makes calculations easier? ;-)
| My reasoning is based on a sort of General Covariance,
| where the laws of nature are independent of preconceived
| dimensionality. If I was an electron, the laws of nature
| would apply to me, but I would not need a wrist watch,
| because an electron is totally stable in time, and wouldn't
| work :-).
|
| Suppose for example we have preconceived laws of nature
| cast in 4D and find difficulties of applying GR inside a sub-
| atomic particle. Would we dismiss GR or question the idea
| that our macroscopic 4D view of dimensionality is true at all
| scales?
| So formulating the laws of physics independent of fixed
| dimensionality is reasonable.

Hmm... I thought that using tensors already took care of that. So you
are saying that using integer indices in tensor notation doesn't really
"match" nature? I guess it is due to the fact that we are imposing a
Lorentzian signature.

I would say the imposition of the Lorentz signature is required
by specializating a spacetime co-ordinated measure that is
fundamental to coordinating experiments, but not necessarily
the required dimensionality to express the physics laws in.

FrediFizzx

http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps

http://www.vacuum-physics.com

Best Regards
Ken
.
User: ""

Title: Re: resolve to perpendicular components, because they are independent 23 Jan 2006 02:40:45 AM
Ken S. Tucker wrote:

It's reasonable to consider 5D, just look around your office and
see that your desk is denser than the air above it, so far so good.
OTOH, the density of the desk may be described by a 4D warp.
Some physicists perfer to impose a 5th D variable on an orthogonal
4D, but then that becomes a nonorthogonal 4D, as in GR.

Can You explain, how a fourth dimension can be orthogonal to the three
of space, measured in cm=B3?
My results so far: Density ( Volume of particles /( Volume of space
swept out by particles in movement) ) can be described in assigning a
value of density [ cm=B3/ cm=B3] to every point of Your desk and the air
above it. This makes a scalar field.
Regards
Hero
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 23 Jan 2006 08:48:30 AM
wrote:

Ken S. Tucker wrote:

It's reasonable to consider 5D, just look around your office and
see that your desk is denser than the air above it, so far so good.
OTOH, the density of the desk may be described by a 4D warp.
Some physicists perfer to impose a 5th D variable on an orthogonal
4D, but then that becomes a nonorthogonal 4D, as in GR.

Can You explain, how a fourth dimension can be orthogonal to the three
of space, measured in cm=B3?

One of the best explanations I've seen is in the movie "Time Machine"
(Original with Rod Taylor). Orthogonality can't always be assumed.

My results so far: Density ( Volume of particles /( Volume of space
swept out by particles in movement) ) can be described in assigning a
value of density [ cm=B3/ cm=B3] to every point of Your desk and the air
above it. This makes a scalar field.

Yes I think with care you can think of it that way. In GR a socalled
"curvature scalar" is proportional to a density scalar.
This is rather like the density of air changing the amount it refracts
light, the refraction being the curvature.

Regards
Hero

Regards
Ken
.
User: ""

Title: Re: resolve to perpendicular components, because they are independent 23 Jan 2006 05:21:37 PM
Ken S. Tucker wrote about

....an orthogonal 4D....

To be honest, i feel like a woman, being told by her husband "Yes dear,
You are right." while he is thinking "'...and i got my peace."', so my
words got labelled and stored away in the back part of the wrong
drawer.
I'm not a physicist, having just basic knowledge, but i notice, that
You avoid a direct answer to:
"Can You explain, how a fourth dimension can be orthogonal to the three
of space, measured in cm=B3?,
which arose from Your mentioning of "an orthogonal 4D".
The mathematical ideas and speculations of Riemann, which can be found
here
http://www.maths.tcd.ie/pub/HistMath
led to best science fiction and some interesting math and he could
differ between both. This is testified and described in the last pages
of
James Clark Maxwell's "A treatise of electricity and magnetism".
(Actually i always thought the "Preliminary" of the first 31 pages
would be all i could achieve understanding in my lifetime, but reading
here so much about aether, so i wanted to know, what Maxwell thought
about this. Just read for Yourself the last sentences of his work and
what was his "constant aim in this treatise" of about a thousand pages.
That intrigues me very much, so i hope, i can achieve some
understanding of this too in my life.)
Now, one hears a lot of loud talking about Riemann, but i never met
even one, who didn't mix the math with the math-fiction to an
undistinguishing mud, something Riemann always tried to avoid. All what
they are doing, is:
Expanding a 2D-space twice. Some do this on purpose, like Hilbert, and
some out of not-knowing better - but all skip the third dimension.
Take a simple sphere and consider a diameter - that are three points on
a straight line, two are on the surface and one is the center. The line
is the shortest connection between the two outer points with extension
( the two points mark an intervall on this line). Where do You find
this with Hilbert? How can one proceed to "an orthogonal 4D" and
skipping this 3D-shortest-path? I don't grasp.
Just one step further is a step back to Thales: From every point of a
circle a diameter appears under a right angle. This is true for every
point on a sphere too ( the proof is left to the reader).
And that's it in 3D. Now please do this in 4D and try to explain.
Or- if You preferr analytical structure over geometrical - please
explain, how the square of a density can be on equal footing with the
square of a length of a distance and at the same time doesn't coincide
somehow with three orthogonal extensions (and this for every set of
two points possible)? To illustrate density in 3D i only know the use
of different shades of grey, or "translated" into different colours or
into "frames"
(the discret form of displaying time with films and computers). Please
show me "an orthogonal 4D".
Regards
Hero
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 24 Jan 2006 09:54:12 AM
wrote:

Ken S. Tucker wrote about

....an orthogonal 4D....

To be honest, i feel like a woman, being told by her husband "Yes dear,
You are right." while he is thinking "'...and i got my peace."', so my
words got labelled and stored away in the back part of the wrong
drawer.
I'm not a physicist, having just basic knowledge, but i notice, that
You avoid a direct answer to:
"Can You explain, how a fourth dimension can be orthogonal to the three
of space, measured in cm=B3?,
which arose from Your mentioning of "an orthogonal 4D".

3 spatial dimensions and 1 time dimension, is commonly
called 4D. An *orthogonal* 4D implies 3 orthogonal spatial
axes x,y,z, and 1 time axis perpendicular to x,y,z.

The mathematical ideas and speculations of Riemann, which can be found
here
http://www.maths.tcd.ie/pub/HistMath
led to best science fiction and some interesting math and he could
differ between both. This is testified and described in the last pages
of
James Clark Maxwell's "A treatise of electricity and magnetism".
(Actually i always thought the "Preliminary" of the first 31 pages
would be all i could achieve understanding in my lifetime, but reading
here so much about aether, so i wanted to know, what Maxwell thought
about this. Just read for Yourself the last sentences of his work and
what was his "constant aim in this treatise" of about a thousand pages.
That intrigues me very much, so i hope, i can achieve some
understanding of this too in my life.)
Now, one hears a lot of loud talking about Riemann, but i never met
even one, who didn't mix the math with the math-fiction to an
undistinguishing mud, something Riemann always tried to avoid. All what
they are doing, is:
Expanding a 2D-space twice. Some do this on purpose, like Hilbert, and
some out of not-knowing better - but all skip the third dimension.
Take a simple sphere and consider a diameter - that are three points on
a straight line, two are on the surface and one is the center. The line
is the shortest connection between the two outer points with extension
( the two points mark an intervall on this line). Where do You find
this with Hilbert? How can one proceed to "an orthogonal 4D" and
skipping this 3D-shortest-path? I don't grasp.
Just one step further is a step back to Thales: From every point of a
circle a diameter appears under a right angle. This is true for every
point on a sphere too ( the proof is left to the reader).
And that's it in 3D. Now please do this in 4D and try to explain.
Or- if You preferr analytical structure over geometrical - please
explain, how the square of a density can be on equal footing with the
square of a length of a distance and at the same time doesn't coincide
somehow with three orthogonal extensions (and this for every set of
two points possible)? To illustrate density in 3D i only know the use
of different shades of grey, or "translated" into different colours or
into "frames"
(the discret form of displaying time with films and computers). Please
show me "an orthogonal 4D".

Ok, maybe you could inform me/us of your background
in math & physics, especially geometry. A term like
"orthogonal" can be easy to explain but difficult to define.
Ken

Regards
Hero

.
User: ""

Title: Re: resolve to perpendicular components, because they are independent 25 Jan 2006 06:00:01 AM
Ken S. Tucker wrote:

Hero.van.Jindelt@gmx.de wrote:

Ken S. Tucker wrote about

....an orthogonal 4D....

"Can You explain, how a fourth dimension can be orthogonal to the
three of space, measured in cm=B3?,

3 spatial dimensions and 1 time dimension, is commonly
called 4D. An *orthogonal* 4D implies 3 orthogonal spatial
axes x,y,z, and 1 time axis perpendicular to x,y,z.

Ken, again avoiding "body contact", replacing "orthogonal" by
"perpendicular".
Your are thinking too much coordinate-minded, 3D with a point of
origin. That's how we all are brought up, years of looking at 2D
blackboards, Book-illustrations and monitors. With the monitor came
"frames", discret time. Continous time displayed one has in math only
with life teaching: the movement of the hand can be repeated, different
from static looking at the trace, it left to blackboard or in
Alexandria-times in the sand.
Nowadays i go this way:
Going from one point to a different point, gives a point and a
direction (1D=3D2P=3Dtwo points) {and a distance too}.
3 Points, that is -most often - two directions from one point, the
difference of directions is called an angle, an oriented one, if we
assign order to the directions (f.e. one principal or
reference-direction and - in math standard - a left or clockwise turn
to the other direction). And 3 points, that is - most often - a plane
too (2D=3D3P).
There are several ways to an right angle.
I'm adding a fourth point (each two on a seperate line without the
other two points). So we have another direction, from the plane to the
fourth point, a normal. This normal is perpendicular and orthogonal to
the plane and also to the two directions, as given before.
Standard orientation in math is left, but this time left-screw (which
can include left-clockwise). Illustrated by a right hand, naming the
thumb 1, the pointer 2 and the middle finger 3 -this gives a
left-clockwize (left on a plane) orientation. Naming the root of the
three fingers in the palm 0, one gets a left screw, going from 0 to 1
to 2 to 3. (3D=3D4P).
Now, not to forget about all the 4D-talk, most basic:
0D, a point. (0D=3D1P) !!
In here we all saw lots of 2D pictures of what is named "4D-space" -
but that's not what we are talking about. We both talk about a 3D-space
and we both can proceed to dynamic geometry: ( 3D, time).But time being
"orthogonal" or "perpendicular" to space or something of it ?
A three-hours-walk can be a distance (with speed known) and ten minutes
is displayed by a finger of the hand of a clock changing direction and
the tip moving along a distance of 1/6 th of 2*pi* (length of finger).
A coordinate system can display two movements, one - most regular -
when one draws the "time"-axis (mostly the x-axis), t being here the
independent variable. And when one draws the the graph of a function,
the set { ( (t, f(t) ) | f ....} as a line. Is this movement or the
time in any way "perp"? Consider the traces left, the projection of any
point onto the axes gives a point on the t-axis and one on the y-axis.
Both axes perp to each other- but that's space, a picture of time, nit
time..... and so forth. so

Please show me "an orthogonal 4D".


Ok, ...

That would be great.

...maybe you could inform me/us of your background
in math & physics, especially geometry.

Look into the PS-attachment below.

A term like
"orthogonal" can be easy to explain but difficult to define.
Ken

I put my trust in You.
So after Your answer ( and may be a few more posts) at the end, we
could ] otagai ni rei [.
I hope kenneth.h... is enjoying reading this, and some others too. Any
word of You is welcome too.
Regards
Hero
PS

...maybe you could inform me/us of your background
in math & physics, especially geometry.

Actually i'm living in this space and time, i'm talking about. My most
prominent teacher was my father, f.e. showing me from a bridge over the
railway, how the rails ( proper prolonged) meet at infinity, so the
train driver doesn't has to worry, that he can't pass the - so to be
seen - diminishing distance between the two rails. No border - in
finis, what a wonderful universe.
The other teachers, aside from the persons at the blackboard are the
same as Yours, Archimedes, Hamilton - You tell me. In physics i told
You already about how far i am with Maxwell - and that's symptomatic
for me. For a better picture of what i learned, google
sp,sm and de.sci.mathematik for my name.
And a photo is displayed at my website
http://1iz.de
or, the same:
http://i-is-no-longer-imaginary.gmxhome.de
It's most about my biggest achievement so far: there is no imaginary
axis, (even Riemann just talks about the y-axis, link was given) or a
complex or Gauss-plane!
Actually i'm still hunting for this wo-man, who introduced this
notation into math and physics, as s-he pested some deennia of my life.
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 26 Jan 2006 12:38:24 PM
wrote:

Ken S. Tucker wrote:

wrote:

Ken S. Tucker wrote about

....an orthogonal 4D....

"Can You explain, how a fourth dimension can be orthogonal to the
three of space, measured in cm=B3?,

3 spatial dimensions and 1 time dimension, is commonly
called 4D. An *orthogonal* 4D implies 3 orthogonal spatial
axes x,y,z, and 1 time axis perpendicular to x,y,z.

Ken, again avoiding "body contact", replacing "orthogonal" by
"perpendicular".
Your are thinking too much coordinate-minded, 3D with a point of
origin. That's how we all are brought up, years of looking at 2D
blackboards, Book-illustrations and monitors. With the monitor came
"frames", discret time. Continous time displayed one has in math only
with life teaching: the movement of the hand can be repeated, different
from static looking at the trace, it left to blackboard or in

The above meanders much to much for me without focusing on
any particular problem.

Alexandria-times in the sand.
Nowadays i go this way:
Going from one point to a different point, gives a point and a
direction (1D=3D2P=3Dtwo points) {and a distance too}.
3 Points, that is -most often - two directions from one point, the
difference of directions is called an angle, an oriented one, if we
assign order to the directions (f.e. one principal or
reference-direction and - in math standard - a left or clockwise turn
to the other direction). And 3 points, that is - most often - a plane
too (2D=3D3P).
There are several ways to an right angle.
I'm adding a fourth point (each two on a seperate line without the
other two points). So we have another direction, from the plane to the
fourth point, a normal. This normal is perpendicular and orthogonal to
the plane and also to the two directions, as given before.
Standard orientation in math is left, but this time left-screw (which
can include left-clockwise). Illustrated by a right hand, naming the
thumb 1, the pointer 2 and the middle finger 3 -this gives a
left-clockwize (left on a plane) orientation. Naming the root of the
three fingers in the palm 0, one gets a left screw, going from 0 to 1
to 2 to 3. (3D=3D4P).

Now, not to forget about all the 4D-talk, most basic:
0D, a point. (0D=3D1P) !!

In here we all saw lots of 2D pictures of what is named "4D-space" -
but that's not what we are talking about. We both talk about a 3D-space
and we both can proceed to dynamic geometry: ( 3D, time).But time being
"orthogonal" or "perpendicular" to space or something of it ?

A three-hours-walk can be a distance (with speed known) and ten minutes
is displayed by a finger of the hand of a clock changing direction and
the tip moving along a distance of 1/6 th of 2*pi* (length of finger).
A coordinate system can display two movements, one - most regular -
when one draws the "time"-axis (mostly the x-axis), t being here the
independent variable. And when one draws the the graph of a function,
the set { ( (t, f(t) ) | f ....} as a line. Is this movement or the
time in any way "perp"? Consider the traces left, the projection of any
point onto the axes gives a point on the t-axis and one on the y-axis.
Both axes perp to each other- but that's space, a picture of time, nit
time..... and so forth. so

Please show me "an orthogonal 4D".


Ok, ...

That would be great.

...maybe you could inform me/us of your background
in math & physics, especially geometry.

Look into the PS-attachment below.

A term like
"orthogonal" can be easy to explain but difficult to define.
Ken

I put my trust in You.
So after Your answer ( and may be a few more posts) at the end, we
could ] otagai ni rei [.
I hope kenneth.h... is enjoying reading this, and some others too. Any
word of You is welcome too.
Regards
Hero
PS

...maybe you could inform me/us of your background
in math & physics, especially geometry.

Actually i'm living in this space and time, i'm talking about. My most
prominent teacher was my father, f.e. showing me from a bridge over the
railway, how the rails ( proper prolonged) meet at infinity, so the
train driver doesn't has to worry, that he can't pass the - so to be
seen - diminishing distance between the two rails. No border - in
finis, what a wonderful universe.
The other teachers, aside from the persons at the blackboard are the
same as Yours, Archimedes, Hamilton - You tell me. In physics i told
You already about how far i am with Maxwell - and that's symptomatic
for me. For a better picture of what i learned, google
sp,sm and de.sci.mathematik for my name.
And a photo is displayed at my website
http://1iz.de
or, the same:
http://i-is-no-longer-imaginary.gmxhome.de
It's most about my biggest achievement so far: there is no imaginary
axis, (even Riemann just talks about the y-axis, link was given) or a
complex or Gauss-plane!
Actually i'm still hunting for this wo-man, who introduced this
notation into math and physics, as s-he pested some deennia of my life.

Perhaps you could make a succint argument/question,
I think you'll get more replies.
Ken
.
User: ""

Title: Re: resolve to perpendicular components, because they are independent 30 Jan 2006 02:31:23 AM

Hero.van.Jindelt@gmx.de wrote:

Ken S. Tucker wrote:

Hero.van.Jindelt@gmx.de wrote:

Ken S. Tucker wrote about

....an orthogonal 4D....

"Can You explain, how a fourth dimension can be orthogonal to the
three of space, measured in cm=B3?,

3 spatial dimensions and 1 time dimension, is commonly
called 4D. An *orthogonal* 4D implies 3 orthogonal spatial
axes x,y,z, and 1 time axis perpendicular to x,y,z.


Ken, again avoiding "body contact", replacing "orthogonal" by
"perpendicular".
Your are thinking too much coordinate-minded, 3D with a point of
origin. That's how we all are brought up,...........


The above meanders much to much for me without focusing on
any particular problem.

Nobody is stopping You to hit the focus: tell us, what You think, when
You talk
about "... 1 time axis perpendicular to x,y,z."
..=2E..............

Nowadays i go this way:
Going from one point to a different point, gives a point and a
direction (1D=3D2P=3Dtwo points)........................

Now, not to forget about all the 4D-talk, most basic:
0D, a point. (0D=3D1P) !!

In here we all saw lots of 2D pictures of what is named "4D-space" -
but that's not what we are talking about. We both talk about a 3D-space
and we both can proceed to dynamic geometry: ( 3D, time).But time being
"orthogonal" or "perpendicular" to space or something of it ?

A three-hours-walk can be a distance (with speed known) ...............=

..=2E..

Please show me "an orthogonal 4D".


Ok, ...

That would be great................

A term like
"orthogonal" can be easy to explain but difficult to define.
Ken

I put my trust in You.

Perhaps you could make a succint argument/question,

Why? Ken, You said "Ok."

I think you'll get more replies.
Ken

I'm afraid, we will not get an explanation from You. May be Time is
different from Space.
If i looked it up correct, then a 'pendiculum' is a balance of weights,
horizontal to earth-surface.
And 'perpendicular" thus means vertical, straight through the
horizontal balance. We can add
to these two directions, which are at a right angle to each other, the
axis of the balance, when it
balances or moves, again at a right angle to the two others.
No place for time left, unless You show us.It seems You cannot, so i've
got the impression,
that You are in connection with these people, who want to revive the
medivial obscure theory,
that the earth is not moving, to which we all know the famous reply: "
..=2E.and yet she moves!"
One can freeze movements, bring a balance to rest - but times...
they area achanging. I like this.
Thanks for Your comments.
Regards
Hero
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 30 Jan 2006 11:09:20 AM
wrote:

wrote:

Ken S. Tucker wrote:

wrote:

Ken S. Tucker wrote about

....an orthogonal 4D....

"Can You explain, how a fourth dimension can be orthogonal to the
three of space, measured in cm=B3?,

3 spatial dimensions and 1 time dimension, is commonly
called 4D. An *orthogonal* 4D implies 3 orthogonal spatial
axes x,y,z, and 1 time axis perpendicular to x,y,z.


Ken, again avoiding "body contact", replacing "orthogonal" by
"perpendicular".
Your are thinking too much coordinate-minded, 3D with a point of
origin. That's how we all are brought up,...........


The above meanders much to much for me without focusing on
any particular problem.


Nobody is stopping You to hit the focus: tell us, what You think, when
You talk
about "... 1 time axis perpendicular to x,y,z."

That's the old Newtonian space & time, and applies
well enough if velocities and g-fields are small.

................

Nowadays i go this way:
Going from one point to a different point, gives a point and a
direction (1D=3D2P=3Dtwo points)........................

Now, not to forget about all the 4D-talk, most basic:
0D, a point. (0D=3D1P) !!

In here we all saw lots of 2D pictures of what is named "4D-space" -
but that's not what we are talking about. We both talk about a 3D-spa=

ce

and we both can proceed to dynamic geometry: ( 3D, time).But time bei=

ng

"orthogonal" or "perpendicular" to space or something of it ?

A three-hours-walk can be a distance (with speed known) .............=

..=2E....

Please show me "an orthogonal 4D".


Ok, ...

That would be great................

A term like
"orthogonal" can be easy to explain but difficult to define.
Ken

I put my trust in You.

Perhaps you could make a succint argument/question,

Why? Ken, You said "Ok."

I think you'll get more replies.
Ken

I'm afraid, we will not get an explanation from You. May be Time is
different from Space.

Not much in spacetime. In 1983 it was agreed to define
the meter=3Dc*second, I'm fine with that. This means we
can equate spacetime with 4 dimensions.

If i looked it up correct, then a 'pendiculum' is a balance of weights,
horizontal to earth-surface.
And 'perpendicular" thus means vertical, straight through the
horizontal balance. We can add
to these two directions, which are at a right angle to each other, the
axis of the balance, when it
balances or moves, again at a right angle to the two others.
No place for time left, unless You show us.It seems You cannot, so i've
got the impression,
that You are in connection with these people, who want to revive the
medivial obscure theory,
that the earth is not moving, to which we all know the famous reply: "
...and yet she moves!"
One can freeze movements, bring a balance to rest - but times...
they area achanging. I like this.
Thanks for Your comments.

Your Welcome.
Ken
PS:It is currently popular to discuss 5D+, however I guess
I'm a bit old fashioned and find the evidentally experential
4D works fine macroscopically.
.
User: ""

Title: Re: resolve to perpendicular components, because they are independent 30 Jan 2006 01:11:49 PM
Ken S. Tucker wrote about

....an orthogonal 4D....

There was discussion and Hero wrote:

One can freeze movements, bring a balance to rest - but times...
they area achanging. I like this.
Thanks for Your comments.

Your Welcome.
Ken
PS:It is currently popular to discuss 5D+, however I guess
I'm a bit old fashioned and find the evidentally experential
4D works fine macroscopically.

I'm still not so far in my studies, so i still learn about 9D ( 9
mathematical diemnsions), about 3x3-matrices describing frozen
movements in three space dimensions, but i hope to advance to
time-dependent variables in these matrices too.
See You later
Hero
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 30 Jan 2006 01:15:44 PM
wrote:

Ken S. Tucker wrote about

....an orthogonal 4D....

There was discussion and Hero wrote:

One can freeze movements, bring a balance to rest - but times...
they area achanging. I like this.
Thanks for Your comments.


Your Welcome.
Ken
PS:It is currently popular to discuss 5D+, however I guess
I'm a bit old fashioned and find the evidentally experential
4D works fine macroscopically.

I'm still not so far in my studies, so i still learn about 9D ( 9
mathematical diemnsions), about 3x3-matrices describing frozen
movements in three space dimensions, but i hope to advance to
time-dependent variables in these matrices too.
See You later
Hero

Good Luck Ace
.
User: ""

Title: Re: resolve to perpendicular components, because they are independent 30 Jan 2006 06:09:18 PM
Ken S. Tucker wrote:

Good Luck Ace

Thanks - i take this as a compliment.
In german cards the ace is at the beginning and at the end,
the value is lowest (1 point) or highest (11 points).
So i sattle on one point, remember 1P = 0D (zero dimensions),
but not in a coordinate-system, where for example one point, the
origin, can be the turning- and angle point, the universal reference.
I hope You give me of the four colours the heart.
One personal point of reference, an ace of hearts - that's a nice card
for everyone. Who then wants to rise to be a 2, or even a king?
And Shakespeare associates the ace with the *****, one of my favourite
animals.
Have fun or make fun.
Hero
PS Only one card can replace the ace (just as it can do with every
card).
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 30 Jan 2006 11:18:31 PM
wrote:

Ken S. Tucker wrote:

Good Luck Ace

Thanks - i take this as a compliment.

Sure if can make 9D work you deserve the compliment,
otherwise ACE is 1 point.

In german cards the ace is at the beginning and at the end,
the value is lowest (1 point) or highest (11 points).
So i sattle on one point, remember 1P = 0D (zero dimensions),

agreed

but not in a coordinate-system, where for example one point, the
origin, can be the turning- and angle point, the universal reference.
I hope You give me of the four colours the heart.
One personal point of reference, an ace of hearts - that's a nice card
for everyone. Who then wants to rise to be a 2, or even a king?
And Shakespeare associates the ace with the *****, one of my favourite
animals.
Have fun or make fun.

My favorite animal is a tree.

Hero
PS Only one card can replace the ace (just as it can do with every
card).

.




















User: "Timo Nieminen"

Title: Re: resolve to perpendicular components, because they are independent 28 Jan 2006 12:42:01 AM
On Sat, 20 Jan 2006, Ken S. Tucker wrote:

Timo Nieminen wrote:

On Fri, 20 Jan 2006, Ken S. Tucker wrote:

I find nonorthogonal axes easier than orthogonal,


Then you must be some kind of bizarre freak of nature!!!


Not really, as in Chess, solving problems in mathematical
physics consists of keeping your options open, to be
closed by physical principle, and certainly not by an aprior
preceived convenience. It's well known "orthogonality" is
at best an approximation in a g-field, but Reimann and his
"gang" evolved quite a nice "tensor" analysis notation that
is easier to use than clunky "ijk" unit vectors.

But tensor analysis can be used for orthogonal systems of coordinates as
well. If you had written "just as easy as", rather than "easier than", I
wouldn't have responded as above.
When a general method also works for the special case, how can the special
case be more difficult than the general case? Equal difficulty, sure, but
_more_ difficult?
--
Timo
.
User: "Ken S. Tucker"

Title: Re: resolve to perpendicular components, because they are independent 28 Jan 2006 12:40:27 PM
Timo Nieminen wrote:

On Sat, 20 Jan 2006, Ken S. Tucker wrote:

Timo Nieminen wrote:

On Fri, 20 Jan 2006, Ken S. Tucker wrote:

I find nonorthogonal axes easier than orthogonal,


Then you must be some kind of bizarre freak of nature!!!


Not really, as in Chess, solving problems in mathematical
physics consists of keeping your options open, to be
closed by physical principle, and certainly not by an aprior
preceived convenience. It's well known "orthogonality" is
at best an approximation in a g-field, but Reimann and his
"gang" evolved quite a nice "tensor" analysis notation that
is easier to use than clunky "ijk" unit vectors.


But tensor analysis can be used for orthogonal systems of coordinates as
well. If you had written "just as easy as", rather than "easier than", I
wouldn't have responded as above.

When a general method also works for the special case, how can the special
case be more difficult than the general case? Equal difficulty, sure, but
_more_ difficult?

Let me tell you and all my problem, suggestions welcome...
I can work with the general metric tensor "g_uv" without