Theory of Relativity
Ken S. Tucker
extraordinary beauty and simplicity I find
in the Theory of Relativity.
It employs the ingenious logic developed
by mathematicians called tensor analysis.
The fundamental assumption of relativity is
*absolute _spatial_ motion does not exist*,
however placing that principle into a succint
mathematical form seems to be difficult. So
that's what I'll try to do.
Beginning with the well known
ds^2 = g_uv dx^u dx^v , u,v,w={0,1,2,3}.
We can use association provided the covariant
derivative,
g_uv;w = 0.
Then by association,
ds^2 = dx_u dx^u.
Expanding to time and space gives,
ds^2 = dx_0 dx^0 + dx_i dx^i , i,j={1,2,3}.
The absolute spatial motion I'll define by
dx_i dx^i = Absolute spatial motion.
Absolute spatial motion cannot exist, IOW's
it vanishes, hence,
dx_i dx^i =0.
However, relative spatial motion cannot vanish,
it is always possible, so I'll select dx^i to
be relative spatial motion, so that dx^i >0 generally
and then require
dx_i =0 always,
to insure
dx_i dx^i=0 always,
and is the mathematical description of the
Principle of Relativity. More formally it's
expressed using the covariant 3-velocity,
U_i = dx_i/ds =0.
By using tensor algebra we obtain from that,
g_0i = - g_ij dx^j/dx^0,
and generally,
ds^2 = g_00 dx^0 dx^0 - g_ij dx^i dx^j , (always).
For an SR application, sub the metric values,
g_00 = g_11 = g_22 = g_33 =1,
g_ij =0 when i =/= j and
g_0i = -dx^i/dx^0,
and find by algebra,
ds^2 = g_00 dx^0 dx^0 - g_ij dx^i dx^j
== dt^2 - dx^2 - dy^2 - dz^2.
The succinct U_i=0 provides Minkowski spacetime,
which embodies the Lorentz transformation, but
done Generally Covariantly.
Moving to General Relativity, the following absolute
derivative vanishes (because U_i=0),
DU_i = U_i;w dx^w =0 .
Using association,
U_i = g_iu U^u =0
therefore,
DU_i = g_iu DU^u =0
and thus,
DU^u =0, aka the geodesic equation.
We have arrived at the equation for the geodesic
(for ref see Weinberg's, Grav&Cosmo, Eq.(5.1.6))
using the Principle of Relativity given by U_i =0
and g_uv;w =0.
The g_uv;w=0 is the mathematical expression for
the Principle of Equivalence, and as is obvious,
is required to get DU^u=0 from U_i=0.
The geodesic equation is expanded to,
DU^u/ds = dU^u/ds + GAMMA^u_ab U^a U^b = 0
(ref, see Weinberg's Eq. (5.1.7)), and is the equation of
motion in General Relativity.
Up to this point we've used two assumptions
1) U_i=0
2) g_uv;w=0
where (1) is a statement of the law of Relativity that excludes
"absolute motion", and (2) is a statement that excludes absolute
acceleration, and at the same time is the Principle of Equivalence,
used to derive the General Relativity geodesic.
I use (1) and (2) in relativity mathematics, is there a reason not to?
TIA
Ken S. Tucker
Koobee Wublee
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:1114717067.2...@f14g2000cwb.googlegroups.com...
>
> The fundamental assumption of relativity is
> *absolute _spatial_ motion does not exist*,
I think this issue is blown way over board. Who cares if the absolute rest
frame exists or not. The laws of physics should apply to all inertial
frames which every frame should be inertial.
> The absolute spatial motion I'll define by
>
> dx_i dx^i = Absolute spatial motion.
>
> Absolute spatial motion cannot exist, IOW's
> it vanishes, hence,
>
> dx_i dx^i =0.
If (i) also includes the time dimension, your equation only applies to
photons. If (i) does not include time, then it is very ludicrous to say any
volume of space is zero.
> [...]
>
> U_i = dx_i/ds =0.
(dx_i/ds) is a by-product of applying the principle of least action. It
works for non-photons becuase
ds = time +/- space
In reality, space has nothing to do if the action of an event is minimized
or not. It is the passage of time only that represents the minimal action
of an event.
How do you deal with photons with the equation above where a photon has (ds
= 0). Are you going to be as creative as Ciufolini on his derivation of a
gravitationally deflected photon?
> ds^2 = g_00 dx^0 dx^0 - g_ij dx^i dx^j
>
> == dt^2 - dx^2 - dy^2 - dz^2.
>
> The succinct U_i=0 provides Minkowski spacetime,
> which embodies the Lorentz transformation, but
> done Generally Covariantly.
You have pointed out Noether's Theorem which indicates there are 4 conserved
quantities in Minkowski spacetime. They are
** Energy
** speed along x-axis
** speed along y-axis
** speed along z-axis
Which are all observed parameters.
> Moving to General Relativity, the following absolute
> derivative vanishes (because U_i=0),
>
> DU_i = U_i;w dx^w =0 .
>
> Using association,
>
> U_i = g_iu U^u =0
>
> therefore,
>
> DU_i = g_iu DU^u =0
>
> and thus,
>
> DU^u =0, aka the geodesic equation.
Geodesic equations are the representatives of the principle of least action
where not all equations vanish. In polar coordinate, at best two equations
vanish. For example, using the Schwarzschild metric, the vanished equations
give us the following conserved events:
** energy
** angular momentum
Which are all observed parameters.
Harry
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:1114717067.2...@f14g2000cwb.googlegroups.com...
> extraordinary beauty and simplicity I find
> in the Theory of Relativity.
> It employs the ingenious logic developed
> by mathematicians called tensor analysis.
>
> The fundamental assumption of relativity is
> *absolute _spatial_ motion does not exist*,
That assumption has been definitely disproved by Sagnac and Michelson-Gale,
as they detect a component of absolute spatial motion.
.
Harald
Spoonfed
> In this post, it's my intention to show the
> extraordinary beauty and simplicity I find
> in the Theory of Relativity.
> It employs the ingenious logic developed
> by mathematicians called tensor analysis.
>
> The fundamental assumption of relativity is
> *absolute _spatial_ motion does not exist*,
> however placing that principle into a succint
> mathematical form seems to be difficult. So
> that's what I'll try to do.
>
> Beginning with the well known
>
> ds^2 = g_uv dx^u dx^v , u,v,w={0,1,2,3}.
There doesn't appear to be a w in the equation.
But otherwise, I think your saying this:
g00 g01 g02 g03 dx0
g10 g11 g12 g13 dx1
ds^2=(dx0,dx1,dx2,dx3) ( g20 g21 g22 g23 ) ( dx2 )
g30 g31 g32 g33 dx3
>
> We can use association provided the covariant
> derivative,
>
> g_uv;w = 0.
What about g_uv? Could you explain what the words and equation above
mean?
>
> Then by association,
>
> ds^2 = dx_u dx^u.
>
You seem to have set
g_uv = {u==v ? 1 : 0 } //Computer programmer-speak.
But I think that g_00 is going to be -1/c^2 if dx0 is time.
> Expanding to time and space gives,
>
> ds^2 = dx_0 dx^0 + dx_i dx^i , i,j={1,2,3}.
>
Could you explain your nomenclature? It looks to me like you're going
back and forth between subscripts and superscripts, and you've got rid
of one dimension. Also, there's no j in your equation.
> The absolute spatial motion I'll define by
>
> dx_i dx^i = Absolute spatial motion.
>
That's an implicit sum i={1,2,3}, right?
Motion generally involves dx/dt, not just dx.
But I still don't know your meaning with
the superscripts and subscripts.
So I'll get feedback on your terminology before I get too much more
confused.
Ken S. Tucker
> Ken S. Tucker wrote:
> > In this post, it's my intention to show the
> > extraordinary beauty and simplicity I find
> > in the Theory of Relativity.
> > It employs the ingenious logic developed
> > by mathematicians called tensor analysis.
> >
> > The fundamental assumption of relativity is
> > *absolute _spatial_ motion does not exist*,
> > however placing that principle into a succint
> > mathematical form seems to be difficult. So
> > that's what I'll try to do.
> >
> > Beginning with the well known
> >
> > ds^2 = g_uv dx^u dx^v , u,v,w={0,1,2,3}.
>
> There doesn't appear to be a w in the equation.
> But otherwise, I think your saying this:
>
> g00 g01 g02 g03 dx0
> g10 g11 g12 g13 dx1
> ds^2=(dx0,dx1,dx2,dx3) ( g20 g21 g22 g23 ) ( dx2 )
> g30 g31 g32 g33 dx3
Been awhile since I put things into matrix products
written out. That's superseded by tensors for obvious
reasons of brevity, but it looks good.
> > We can use association provided the covariant
> > derivative,
> >
> > g_uv;w = 0.
>
> What about g_uv? Could you explain what the words and equation above
> mean?
The g_uv are very nearly one, but vary slightly from
one, and slight variation produces the gravitational
effects evident by our bums stuck to our chairs:-)
> > Then by association,
> >
> > ds^2 = dx_u dx^u.
> >
>
> You seem to have set
> g_uv = {u==v ? 1 : 0 } //Computer programmer-speak.
>
> But I think that g_00 is going to be -1/c^2 if dx0 is time.
>
> > Expanding to time and space gives,
> >
> > ds^2 = dx_0 dx^0 + dx_i dx^i , i,j={1,2,3}.
> >
>
> Could you explain your nomenclature? It looks to me like you're
going
> back and forth between subscripts and superscripts, and you've got
rid
> of one dimension. Also, there's no j in your equation.
I defined the "j" and "w" indice for brevity, because
they're used later.
> > The absolute spatial motion I'll define by
> >
> > dx_i dx^i = Absolute spatial motion.
> >
>
> That's an implicit sum i={1,2,3}, right?
Yes
> Motion generally involves dx/dt, not just dx.
> But I still don't know your meaning with
> the superscripts and subscripts.
> So I'll get feedback on your terminology before I get too much more
> confused.
Ok...:-)...
To a Spoonfed Harry Kooblee :-)...
As prefaced in my post, a reasonable familiarity
with tensors is required. I use covariant "A_u",
and contravariant "A^u" notation, also the summation
convention, association, and the Absolute derivative,
as is standard, and well founded by mathematicians.
When mathematicians develope new branches of
mathematics they prefer to generalize and reduce
the number restictions.
However relativity excludes things like *absolute
motion* that mathematicians will include in a general
mathemematical way, after-all physics is only one
application.
So for physical applications we form *constraints*
on what's allowable, i.e. it was deduced,
U_i =0 (1st constraint)
is a physical law that permits only relative motion.
The second *constraint* is,
g_uv;w=0 (2nd constraint)
meaning if any CS can be found where the g_uv are
constant then g_uv;w=0 in all CS's, which is a
reasonable definition of the Principle of Equivalence.
That 2nd constraint is frequently questioned, by the
motion of charges in EM-fields.
The arguments I understand go like this: a CS attached
to a charged particle *cannot* transform to zero the
induced acceleration of the EM-field, hence the g_uv;w
cannot be transformed away by selecting the charged
particle as a reference.
Nevertheless I would argue the laws of physics are as
true to a CS attached to the charged particle as any
other CS, so I'm obligated to find Lorentz force,
f_u = q*F_uv U^v =0,
so that f_u=0 relative to a CS attached to "q".
The point of my OP was to summarize in a consistent
and succint way the Theory of Relativity.
Regards
Ken S. Tucker
Ken S. Tucker
On the basis on the presented OP understanding
of relativity the GP-b experiment should null,
as I'll explain below...
Below...
Here the metric under the condition U_i=0
has collapsed so that g_0i terms are excluded.
For ref to the basis of GP-b, please see Weinbergs,
"Grav & Cosmo" pg 239-241, and note the Kerr metric.
The Kerr metric depends upon the use of g_0i, that
can be excluded by the vanishing of *absolute motion*
demo'd by U_i=0.
And further note, the need for Mach's principle.
It can found in AE's book, "The Meaning of Relativity",
on pg. 100, AE writes, *a dependancy on Mach's Principle*
that I find tenuous.
Tucker predicts, on the basis of his posts, the expected
result from GP-b, will not be measured, and shows why.
The GP-b is testing Kerr's metric or Mach's Principle,
depending on interpretation, but neither works if
U_i=0 is true.
I'll get my ask kicked when I'm wrong, but that's
where I come down to a decision.
Regards
Ken S. Tucker
[...good stuff]
Tom Roberts
> The fundamental assumption of relativity is
> *absolute _spatial_ motion does not exist*,
Well, that's a new approach AFAIK....
It has a fundamental problem: how does one define "motion" so this even
makes sense?
For this approach to be fruitful, I'm pretty sure you'll have to define
"motion" as _timelike_ -- then your assertion "Absolute spatial motion
does not exist" is trivial. In any case, we'll see below that your
attempt here is hopeless.
> Beginning with the well known
> ds^2 = g_uv dx^u dx^v , u,v,w={0,1,2,3}.
OK. But please remember what you seem to have forgotten: this defines
the invariant interval ds BETWEEN TWO GIVEN INFINITESIMALLY-SEPARATED
POINTS IN THE MANIFOLD in terms of the metric components {g_uv} and the
COORDINATE DIFFERENCES BETWEEN THE TWO POINTS {dx^u}.
> We can use association provided the covariant
> derivative,
> g_uv;w = 0.
That merely states that your choice of covariant derivative is
compatible with the metric. But no matter, you don't use the covariant
derivative anywhere.
> Then by association,
> ds^2 = dx_u dx^u.
I'm not sure what you mean by "association". And to do that you don't
need to use g_uv;w=0, just _define_ dx_u = d_uv dx^v. Of course we
normally define those symbols just that way for components of vectors,
but differentials are a rather different kettle of fish.... But ignore
that as there are bigger problems here:
> Expanding to time and space gives,
> ds2 = dx_0 dx0 + dx_i dx^i , i,j={1,2,3}.
Wrong! You forgot all the terms involving g_0i and g_i0. But ignore that
as there are bigger problems here:
> Expanding to time and space gives,
> ds^2 = dx_0 dx^0 + dx_i dx^i , i,j={1,2,3}.
> The absolute spatial motion I'll define by
> dx_i dx^i = Absolute spatial motion.
Definining it so does not make it so:
1-1 = 0
and I'll define "absolute spatial motion" to be 1-1. There is no
difference in principle between your definition and mine -- both are
merely words and symbols -- to justify the choice of words "absolute
spatial motion" you need to relate your symbols (dx_i dx^i) to the usual
meaning of "absolute spatial motion" -- that seems impossible to me, as
both the {dx_i} and the {dx^i} are manifestly coordinate dependent, so
there is nothing "absolute" here at all.
And your expression is not invariant over coordinate
transforms (if it were, you could claim it is "absolute" in
a restricted sense).
> Absolute spatial motion cannot exist, IOW's
> it vanishes, hence,
> dx_i dx^i =0.
You have gone completely off the rails by mindlessly manipulating
symbols without understanding their meaning.
Imagine using the {dx^i} and the {dx_i} to represent the differential
between two corners of my desk, measured simultaneously in the inertial
rest frame of the desk. Then dx_i dx^i is simply the distance between
them -- manifestly not zero.
I repeat: you really need to learn the basics.
Tom Roberts tjro...@lucent.com
Spoonfed
Ken S. Tucker wrote:
> As prefaced in my post, a reasonable familiarity
> with tensors is required. I use covariant "A_u",
> and contravariant "A^u" notation, also the summation
> convention, association, and the Absolute derivative,
> as is standard, and well founded by mathematicians.
Well I'm up to page 83 of Introduction to Vector and Tensor Analysis,
by Robert Wrede, although I'm planning to reread 65 through 83 tomorrow
(where the concepts of covariant and contravariant are introduced). I
was frustrated with it at first, since I found it impossible to skip
through it. There's no way to look up mathematical conventions and
semantics in the index. For instance, lower-case letters are from the
original matrix, upper case letters are cofactors. a_i^k is a rotation
matrix. b_i^k is something else, and X, unfortunately seems to have a
different version of a very vague definition every time it comes up.
Eventually, I'll get immersed in it and be even harder to understand.
;)
Ken S. Tucker
Spoonfed wrote:
> Ken S. Tucker wrote:
> > As prefaced in my post, a reasonable familiarity
> > with tensors is required. I use covariant "A_u",
> > and contravariant "A^u" notation, also the summation
> > convention, association, and the Absolute derivative,
> > as is standard, and well founded by mathematicians.
>
> Well I'm up to page 83 of Introduction to Vector and Tensor Analysis,
> by Robert Wrede, although I'm planning to reread 65 through 83
tomorrow
> (where the concepts of covariant and contravariant are introduced).
I
> was frustrated with it at first, since I found it impossible to skip
> through it.
Sorry, I don't have that book. Learning Tensors by oneself
is difficult (I had the benefit of Profs), for even a good
math student. It would be good if could speak to a local prof
for 15 minutes a week, just to get by some the sticklers.
I'm rather poor at math, but I did come to appreciate the
diff between covariant and contravariant tensors in GR, for
example, g^00 = 1/g_00 (approximately) and g^11=1/g_11,
where x1 is parallel to radius.
So the g-field makes spacetime go like g_00 == g^11 and
g^00 == g_11 so what remains is g_00 g^00 == g_11 g^11
which is neat.
> There's no way to look up mathematical conventions and
> semantics in the index. For instance, lower-case letters are from
the
> original matrix, upper case letters are cofactors. a_i^k is a
rotation
> matrix. b_i^k is something else, and X, unfortunately seems to have
a
> different version of a very vague definition every time it comes up.
>
> Eventually, I'll get immersed in it and be even harder to understand.
> ;)
Well, there's lots of people in this group who would like to
help, but doing through ascii is difficult, but try it and see.
Ken
Ken S. Tucker
Tom Roberts wrote:
> Ken S. Tucker wrote:
> > The fundamental assumption of relativity is
> > *absolute _spatial_ motion does not exist*,
>
> Well, that's a new approach AFAIK....
Actually not, it's the juxtaposition of saying
spatial motion is relative.
> It has a fundamental problem: how does one define "motion" so this
even
> makes sense?
Yes, that's a problem.
> For this approach to be fruitful, I'm pretty sure you'll have to
define
> "motion" as _timelike_ -- then your assertion "Absolute spatial
motion
> does not exist" is trivial.
Yes, it needs to be defined to vanish, but the
equation to do that is NOT trivial.
> > Beginning with the well known
> > ds^2 = g_uv dx^u dx^v , u,v,w={0,1,2,3}.
>
> OK. But please remember what you seem to have forgotten:
Why do you claim somethings forgotten? What is it that
you do not understand?
...
> > We can use association provided the covariant
> > derivative,
> > g_uv;w = 0.
>
> That merely states that your choice of covariant derivative is
> compatible with the metric. But no matter, you don't use the
covariant
> derivative anywhere.
No sorry, if one uses g_uv or g^uv in association
then one must stipulate g_uv;w=0, that's often
overlooked. It's a common error for beginners but
now you know.
> > Then by association,
> > ds^2 = dx_u dx^u.
>
> I'm not sure what you mean by "association".
By association use,
dx_u = g_uv dx^v
...
> > Expanding to time and space gives,
> > ds2 = dx_0 dx0 + dx_i dx^i , i,j={1,2,3}.
...
> > Expanding to time and space gives,
> > ds^2 = dx_0 dx^0 + dx_i dx^i , i,j={1,2,3}.
> > The absolute spatial motion I'll define by
> > dx_i dx^i = Absolute spatial motion.
> Defining it so does not make it so:
Agreed, good of you to notice. This is where
physical intuition is required, but you'll
need to follow the math I'll keep simple.
In 3D
dr^2 = dx_i dx^i , {i,j=1,2,3}.
That "dr" is an invariant displacement, aka
an invariant displacement, aka an absolute
displacement. When we go to 4D we can divide
that dr by dt to define "absolute motion",
hence,
dr/dt = invariant motion in 4D.
Relativity is quite clear on this point,
invariant motion cannot be measured, therefore
it vanishes, hence,
dr^2= dx_i dx^i =0
in 4D.
...
> And your expression is not invariant over coordinate
> transforms (if it were, you could claim it is "absolute" in
> a restricted sense).
That's strictly correct mathematically, however not
any or all substitutions for the metric are true
in spacetime. Most begin with the approximation
gii=-1, g00=1
for no other reason except to get from
ds^2 = g_uv dx^u dx^v
to
ds^2 = dt^2 - dx^2
mindlessly, mainly because it's easy to teach.
But that's not how it really should done.
The metrics are defined by a field equations,
to constrain those that a physically viable, so
U_i=0
is the first constraint.
> > Absolute spatial motion cannot exist, IOW's
> > it vanishes, hence,
> > dx_i dx^i =0.
> Imagine using the {dx^i} and the {dx_i} to represent the differential
> between two corners of my desk, measured simultaneously in the
inertial
> rest frame of the desk. Then dx_i dx^i is simply the distance between
> them -- manifestly not zero.
Tom, you aren't going to learn 4D spacetime and
relativity by looking at the corner of your desk,
it's not that simple.
It's easy to draw a 4D diagram on paper just using
x and t with oblique axes.
Regards
Ken