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Jack Sarfatti
science forum Guru

Joined: 29 Apr 2005
Posts: 487

Posted: Fri May 13, 2005 3:28 pm    Post subject: "Weyl" type hidden gauge invariance in General Relativity

OK we hit a temporary snag solved below. In elementary physics first
rule is to check your units and physical dimensions. Don't mix apples
with oranges etc. Yet GR theorists do that nonchalantly and sloppily
even in text books.

For example, the SSS metric is typically written as

gtt = -(1 - 2GM/c^2r) [dimensionless]

grr = (1 - 2GM/c^2r)^-1 [dimensionless]

But hold on

gthetatheta = r^2 [area]

gphiphi = r^2sin^2theta [area]

Where we have the incommensurate basis set of Cartan 1-forms

dx^0 = cdt
dx^1 = dr
dx^2 = dtheta
dx^3 = dphi

With the Grassmann basis sort of "Clifford" algebra" of 2^4 = 16
p-forms, p = 0,1,2,3,4

*p-form = (4 - p)-form, when N = 4.

1

dx^u

dx^u/\dx^v

dx^u/\dx^v/\dx^w

dx^u/\dx^v/\dx^w/\dx^l

This gives an incommensurate set of Levi-Civita connection field
components in the hovering LNIFs

(LC)^001 = [2(1 - 2GM/c^2r)]^-1 (1 - 2GM/c^2r),r [1/length]

(LC)^122 = -r(1 - 2GM/c^2r) [length]

(LC)^233 = -sinthetacostheta [dimensionless]

(LC)^100 = (1/2)(1 - 2GM/c^2r),r(1 - 2GM/c^2r) [1/length]

(LC)^133 = -(rsin^2theta)(1 - 2GM/c^2r) [length]

(LC)^313 = (LC)^212 = 1/r [1/length]

(LC)^111 = (1/2)(1 - 2GM/c^2r)(1 - 2GM/c^2r)^-1,r [1/length]

(LC)^323 = cottheta [dimensionless]

all other (LC) identically & globally zero in this FRAME BUNDLE of
hovering LNIFs all over this toy model 4D space-time

My original suggestion gthetatheta = gphiphi = 1 will not work here
because physically we have a stretch-squeeze tidal curvature that

Nevertheless we MUST use commensurate infinitesimal basis sets for our
local frames and the (LC) components MUST all be of the same physical
dimension in order to define consistent Diff(4) covariant derivatives.

For example

Au;v = Au,v - (LC)uv^wAw

The GRAVITY-MATTER MINIMAL COUPLING SUM (LC)uv^wAw must have physically
commensurate (LC) components because Au is arbitrary! For example, Au
can be the Maxwell EM vector potential, and all the components of Au
have same physical dimensions.

Therefore ALL the (LC) MUST obey [LC] = 1/length

So, how to we accomplish this?

Simple, use engineering dimensional analysis and introduce a scale L.

What is L? Is L = Lp = (hG/c^3)^1/2 or is L = GM/c^2 or?

For now let's call it "L".

Therefore the SSS metric is now the physically commensurate
dimensionless array

gtt = -(1 - 2GM/c^2r)

grr = (1 - 2GM/c^2r)^-1

gthetatheta = (r/L)^2

gphiphi = (r/L)^2sin^2theta

Where we NOW have the commensurate set of basic 1-forms

dx^0 = cdt
dx^1 = dr
dx^2 = Ldtheta
dx^3 = Ldphi

Note that

,0 = (1/c),t

,1 = ,r

,2 = (1/L),theta

,3 = (l/L),phi

Therefore, all the (LC) are now [1/length]

LC)^001 = [2(1 - 2GM/c^2r)]^-1 (1 - 2GM/c^2r),r

(LC)^122 = -(r/L^2)(1 - 2GM/c^2r)

(LC)^233 = -(1/L)sinthetacostheta

(LC)^100 = (1/2)(1 - 2GM/c^2r),r(1 - 2GM/c^2r)

(LC)^133 = -(rsin^2theta/L^2)(1 - 2GM/c^2r)

(LC)^313 = (LC)^212 = 1/r

(LC)^111 = (1/2)(1 - 2GM/c^2r)(1 - 2GM/c^2r)^-1,r

(LC)^323 = (1/L)cottheta [dimensionless]

The Riemann-Christoffel tensor is now dimensionally self-consistent,
i.e. 1/Area

Note that L cancels out of the frame invariant

ds^2 = guvdx^udx^v

and it must cancel out of any local physical quantity.

In particular it must cancel out of the geodesic equation and the tidal
geodesic deviation.

It's pretty obvious that L will be physically locally unobservable. It's
a bit like the Weyl gauge parameter.

Note that the geodesic equation for a non-spinning point test particle is

D^2x^u/ds^2 = d^2x^u/ds^2 - (LC)^uvw(dx^v/ds)(dx^w/ds) = 0

So the 1/L's in the (LC)s cancel the L's in x2 & x^3

Similarly with geodesic deviation

d(x^u - x'^u)/ds = R^uvwl(x^v - x'^v)(dx^w/ds)(dx^l/ds)

Note that (LC)^uvw and R^uvwl are NEVER MEASURED DIRECTLY in isolation.
What is measured is

D^2x^u/ds^2

and

d(x^u - x'^u)/ds
Ken S. Tucker
science forum Guru

Joined: 30 Apr 2005
Posts: 1230

Posted: Sun May 15, 2005 1:13 am    Post subject: Re: "Weyl" type hidden gauge invariance in General Relativity

Jack Sarfatti wrote:
 Quote: OK we hit a temporary snag solved below. In elementary physics first rule is to check your units and physical dimensions. Don't mix apples with oranges etc. Yet GR theorists do that nonchalantly and sloppily even in text books.

Agreed, I'd rather call this thread *units in tensor
analysis*, that can be ambiguous, let's do a primitive
example.

1)Draw a straight Line on a blank piece of paper.
2)Trace that Line using 1 inch graph paper.
3)Trace using 1 cm graph paper.

All 3 lines are equal, in all CS's (independant of the
graph paper) and so the Line is invariant.

Let's set X^1 = inches and x^1 = cm's then

X^1 = 2.54 x^1 , ie. 1 inch = 2.54 cm's,

is the tranformation.

The invariant *Line* is given by,

Line = E_1 X^1 = e_1 x^1

where

E_1 = 1/inch in direction X^1,

e_1 = 1/cm in direction x^1 .

Please note Line has no units on the blank paper
(1) above. It can only have a measurement in
(2) or (3) or other graph paper.
The "E_1" and "e_1" are refered to as covariant
basis vectors.

We'll follow the usual convention and define the
"metric tensor" by scalar (dot) product like,

g_UV = E_U.E_V and g_uv = e_u.e_v .

It follows g_UV and g_uv have units of 1/area,
i.e. 1/inch^2 and 1/cm^2 respectively, so that
Line is invariant (has no units in any CS).

We then define the usual "norm" by

Line^2 = g_UV X^U X^V = g_uv x^u x^v = invariant.

You can test that using the transformation,
(check out a handy 30cm ~ 12 inch ruler), and
find the invariant to a *unitless* scalar as
all scalars must be.

Since 1 second =~ 3*10^5 km by international
agreement, extending the above to 4D SpaceTime
CS's is straightfoward.

Below are some issues Jack highlights...
Regards
Ken S. Tucker

 Quote: For example, the SSS metric is typically written as gtt = -(1 - 2GM/c^2r) [dimensionless] grr = (1 - 2GM/c^2r)^-1 [dimensionless] But hold on gthetatheta = r^2 [area] gphiphi = r^2sin^2theta [area] Where we have the incommensurate basis set of Cartan 1-forms dx^0 = cdt dx^1 = dr dx^2 = dtheta dx^3 = dphi With the Grassmann basis sort of "Clifford" algebra" of 2^4 = 16 p-forms, p = 0,1,2,3,4 *p-form = (4 - p)-form, when N = 4. 1 dx^u dx^u/\dx^v dx^u/\dx^v/\dx^w dx^u/\dx^v/\dx^w/\dx^l This gives an incommensurate set of Levi-Civita connection field components in the hovering LNIFs (LC)^001 = [2(1 - 2GM/c^2r)]^-1 (1 - 2GM/c^2r),r [1/length] (LC)^122 = -r(1 - 2GM/c^2r) [length] (LC)^233 = -sinthetacostheta [dimensionless] (LC)^100 = (1/2)(1 - 2GM/c^2r),r(1 - 2GM/c^2r) [1/length] (LC)^133 = -(rsin^2theta)(1 - 2GM/c^2r) [length] (LC)^313 = (LC)^212 = 1/r [1/length] (LC)^111 = (1/2)(1 - 2GM/c^2r)(1 - 2GM/c^2r)^-1,r [1/length] (LC)^323 = cottheta [dimensionless] all other (LC) identically & globally zero in this FRAME BUNDLE of hovering LNIFs all over this toy model 4D space-time My original suggestion gthetatheta = gphiphi = 1 will not work here because physically we have a stretch-squeeze tidal curvature that requires the theta dependence in addition to the radial dependence. Nevertheless we MUST use commensurate infinitesimal basis sets for our local frames and the (LC) components MUST all be of the same physical dimension in order to define consistent Diff(4) covariant derivatives. For example Au;v = Au,v - (LC)uv^wAw The GRAVITY-MATTER MINIMAL COUPLING SUM (LC)uv^wAw must have physically commensurate (LC) components because Au is arbitrary! For example, Au can be the Maxwell EM vector potential, and all the components of Au have same physical dimensions. Therefore ALL the (LC) MUST obey [LC] = 1/length So, how to we accomplish this? Simple, use engineering dimensional analysis and introduce a scale L. What is L? Is L = Lp = (hG/c^3)^1/2 or is L = GM/c^2 or? For now let's call it "L". Therefore the SSS metric is now the physically commensurate dimensionless array gtt = -(1 - 2GM/c^2r) grr = (1 - 2GM/c^2r)^-1 gthetatheta = (r/L)^2 gphiphi = (r/L)^2sin^2theta Where we NOW have the commensurate set of basic 1-forms dx^0 = cdt dx^1 = dr dx^2 = Ldtheta dx^3 = Ldphi Note that ,0 = (1/c),t ,1 = ,r ,2 = (1/L),theta ,3 = (l/L),phi Therefore, all the (LC) are now [1/length] LC)^001 = [2(1 - 2GM/c^2r)]^-1 (1 - 2GM/c^2r),r (LC)^122 = -(r/L^2)(1 - 2GM/c^2r) (LC)^233 = -(1/L)sinthetacostheta (LC)^100 = (1/2)(1 - 2GM/c^2r),r(1 - 2GM/c^2r) (LC)^133 = -(rsin^2theta/L^2)(1 - 2GM/c^2r) (LC)^313 = (LC)^212 = 1/r (LC)^111 = (1/2)(1 - 2GM/c^2r)(1 - 2GM/c^2r)^-1,r (LC)^323 = (1/L)cottheta [dimensionless] The Riemann-Christoffel tensor is now dimensionally self-consistent, i.e. 1/Area Note that L cancels out of the frame invariant ds^2 = guvdx^udx^v and it must cancel out of any local physical quantity. In particular it must cancel out of the geodesic equation and the tidal geodesic deviation. It's pretty obvious that L will be physically locally unobservable. It's a bit like the Weyl gauge parameter. Note that the geodesic equation for a non-spinning point test particle is D^2x^u/ds^2 = d^2x^u/ds^2 - (LC)^uvw(dx^v/ds)(dx^w/ds) = 0 So the 1/L's in the (LC)s cancel the L's in x2 & x^3 Similarly with geodesic deviation d(x^u - x'^u)/ds = R^uvwl(x^v - x'^v)(dx^w/ds)(dx^l/ds) Note that (LC)^uvw and R^uvwl are NEVER MEASURED DIRECTLY in isolation. What is measured is D^2x^u/ds^2 and d(x^u - x'^u)/ds

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