Conservation of Energy[Schwarzschild Metric]

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Anamitra Palit

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Jan 3, 2015, 2:40:05 PM1/3/15
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Conservation of Energy wih Schwarzschild metric[by a simple technique]:
https://drive.google.com/file/d/0BymT8iD6LY1nZ3A3TW44X2d1Zjg/view?usp=sharing

Jack...@hotmail.com

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Jan 30, 2015, 10:00:02 AM1/30/15
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Not so fast.
>From Eq. 1, by inspection, dtau/dt = sqrt(-1+2m/r) which conflicts
with the 'usual result'. It seems necessary to account for the
imaginaries. They have not gone away.
John Polasek

Roland Franzius

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Jan 30, 2015, 2:10:03 PM1/30/15
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The length parameter dtau^2 is negative, meaning it is a metric suitable
for spacelike distances.

But geodesics in spacelike directions have no known physical relevance.

He just has switch to the metric for timelike curves by changing the
sign in front of dtau so that it fits to dt in a state of spatial rest.

Since the geodesic equation is nothing else but the Euler-Langrange
equations derived from Hamiltons action principle for timelike curves
with a free Langrangian L = G(dx/dtau,dx/dtau) and G the Schwarzschild
bilinear form on the tangent vector,

we have the general result:

Lagrangian L(dx/dt,x) and the Hamiltonian

H(p,x) = G^-1(dL/dx,dL/x)

are independent of the coordinates tau, t, and phi. Their corresponding
canocal moments are constant along the worldlines, which are solutions
of the geododesic equations.

Consequently the values of of

the Hamiltonian

H = G^-1(p,p) = m^2 c^2 is constant because dH/dtau=0

the time component of the momentum 4-vector, is constant because

d/dtau dH/dp_t = d/dtau m u_t = -d H/dt =0,

and finally any component of the angular momentum are constant, because
its true for the 3-axis momentum

d/dtau p_phi = - dH/dphi = 0

and the axis can be choosen to be 3-orthogonal to the plane of the
3-trajectory.

Of course one can identify these constants of motion by separation of
variables in the geodesic equations to. But even there one uses tacitly
the assumption that the proper time used as curve parameter is metrical
constant along the timelike geodesic.

The Lagrangian method in Schwarzschild geometry you will find
demonstrated perfectly eg in MTW's phonebook.

--

Roland Franzius

Roland Franzius

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Jan 31, 2015, 1:30:03 PM1/31/15
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Am 30.01.2015 um 20:05 schrieb Roland Franzius:
>
> the time component of the momentum 4-vector, is constant because
>
> d/dtau dH/dp_t = d/dtau m u_t = -d H/dt =0,

Sorry, this is a typo.

The canonical proper time derivative of the energy
p_t = dL/d(dt/dtau)i is of course

d/dtau p_t = d/dtau (m dt/dtau) = - d H/dt = 0.

--

Roland Franzius

Jack...@hotmail.com

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Jan 31, 2015, 1:30:06 PM1/31/15
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Priority: Normal Date: Friday, January 30, 2015 5:22 PM Size: 3 KB

Thank you for your well thought out and elaborate answer, none of
which I would try to dispute, but I have an idea that general
relativity has a serious problem.
You can get a suggestion of it in the MTW gravity
'phone book' on page 51 under the title "Farewell to ict". They say
"Since the imaginary cannot be used… It will not be used".
After this accomplishment, we see nothing but Square terms, ict having
been 'gentrified'.
Also (in Wikipedia, not in MTW) you can find a square diagonal matrix,
the Schwarzschild metric, all of its entries squared. It is a faux
matrix that might just as well be collapsed into a vector. it is not
possible to perform a similarity transformation which is a necessary
property of Cartesian tensors. The vectors on which the matrix would
be expected to act, are not defined.
A metric tensor written in matrix form will look like XGX'
The quadratic product hasG, as a tensor, that does the stretching and
pulling, and, following good mathematical form, X' is first processed
by G, and this is post-multiplied into X (or the reverse order).
GX' is the distorted frame.
A tensor G is dX'/dX, whose entries would never be squared.
Again, g00 is set down as -1+ 2m/r and there really is no graceful
way to change the sign.
Furthermore the determinant is negative, converting
positive volumes into negative ones.
General relativity starts out on
very shaky grounds and is really not very useful even for orbiting
satellites in our solar system.
At least that's the way I see it, with a magnifying glass.
John Polasek



Jack...@hotmail.com

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Jan 31, 2015, 1:30:09 PM1/31/15
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On Sat, 3 Jan 2015 13:37:55 CST, Anamitra Palit
<palit.a...@gmail.com> wrote:

Please disregard my last diatribe re general relativity. It will just
cause hard feelings. I didnt realize I had hit the Send button.
John Polasek

Jack...@hotmail.com

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Feb 1, 2015, 1:20:03 PM2/1/15
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On Fri, 30 Jan 2015 13:05:53 CST, Roland Franzius
<roland....@uos.de> wrote:

>Am 30.01.2015 um 15:50 schrieb Jack...@hotmail.com:
>> On Sat, 3 Jan 2015 13:37:55 CST, Anamitra Palit
>> <palit.a...@gmail.com> wrote:
>>
>>> Conservation of Energy wih Schwarzschild metric[by a simple technique]:
>>> https://drive.google.com/file/d/0BymT8iD6LY1nZ3A3TW44X2d1Zjg/view?usp=sharing
>> Not so fast.
>>>From Eq. 1, by inspection, dtau/dt = sqrt(-1+2m/r) which conflicts
>> with the 'usual result'. It seems necessary to account for the
>> imaginaries. They have not gone away.
>> John Polasek
>>
>
>The length parameter c meaning it is a metric suitable
>for spacelike distances.

dtau^2 is negative?
What is the prohibition against taking the square root here?
It hardly requires any prompting. In fact, it is screaming for someone
to do it.
The result will, of course, be imaginary, yet, in the 'official'
version there are no imaginaries to be found.
Serious obfuscation is suspected.
In the end, GR does not explain gravity.

>But geodesics in spacelike directions have no known physical relevance.

Well, it's a start,-but they must have relevance. There is no room for
ornaments.
I dont think the Lagrangian was part of Einsteins method.
As I recall generalized forces came from d/dt(dL/dQdot). Where is L
defined in the empty space of relativity?

The original complaint, about changing the sign of g00, requres a
quite specific and transparent mathematical operation. It is not like
the choice of covariant and contravariant which may be freely done for
convenience.

I don't care to start an endless series of arguments, I just thought
I'd point a few things out.

John Polasek

Roland Franzius

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Feb 2, 2015, 9:50:02 AM2/2/15
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For a timelike worldline

tau -> { x^0(tau),x(tau),y(tau),z(tau) }

L = 1/2 (dct/dtau)^2- 1/2 ( dx/dtau)^2 + (dy/dtau) + (dz/dtau)^2

> The original complaint, about changing the sign of g00, requres a
> quite specific and transparent mathematical operation. It is not like
> the choice of covariant and contravariant which may be freely done for
> convenience.


There is no problem because changing the sign of g or L changes the
signs of the energy-momentum vector d(+-L)/d(dx/dtau)and the forces
d(+-L)/dx together. Thats the art of invariant tangent vector definitions.

Just for convenience, one takes L>0 for quantum physics and point
mechanics because g00>0 and positive energy are an important concept.

Cosmologists and geometers prefer g00<0 and space geometry g_ik at fixed
time positive definite in order to translate the formulae of Riemannian
geometry 1-1.

This goes under the titles est coast - west coast metrics.

And do not hesitate to accept that velocities and momenta of the
negative part in L point into opposite directions. Thats inevitable for
the Ricci/Cartan forms and vectors machinery to work.

--

Roland Franzius

Jack...@hotmail.com

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Feb 5, 2015, 12:00:03 AM2/5/15
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On Mon, 2 Feb 2015 08:45:16 CST, Roland Franzius
It seems to me that general relativity is not really physics. It has
been brought to such a fine degree of sophistication, that by now we
know very considerably more about gravity then Einstein did. But, on
the other hand, it is unfortunate that we also now know more about it
than God.
(A bell should be going off here).
John Polasek

Tom Roberts

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Feb 8, 2015, 1:20:06 AM2/8/15
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On 1/31/15 1/31/15 12:27 PM, Jack...@hotmail.com wrote:
> I have an idea that general relativity has a serious problem. [...] Again,
> g00 is set down as -1+ 2m/r and there really is no graceful way to change
> the sign. Furthermore the determinant is negative, converting positive
> volumes into negative ones.

> dtau^2 is negative? What is the prohibition against taking the square root
> here?

> Serious obfuscation is suspected.

> The original complaint, about changing the sign of g00, requres a quite
> specific and transparent mathematical operation.

All of these refer to a single "problem" you think you see in the mathematics of
GR. The underlying problem is that the authors you are reading do not treat
mathematics with the respect and rigor it deserves. This is, of course, quite
common in physics.

In all the cases you cite, a more careful derivation of the equations would
introduce either absolute-value signs around the determinant of the metric, or a
minus sign in front of it. The "problem" you think you see is not a problem.
Most physicists using GR are aware of this and simply ignore it, always taking
positive values for quantities that are naturally positive (volumes, areas, path
lengths, etc.).

The key point is that semi-Riemannian geometry is NOT Riemannian
geometry. While many theorems of the latter apply to the former,
not all do, and often they get a minus sign or absolute-value.
And for many purposes, one must treat spacelike and timelike
intervals separately (signs often differ).


> In the end, GR does not explain gravity.

Of course not! NO physical theory can ever "explain" natural phenomena. One of
the major lessons of the last century is that science is not able to generate
"explanations", it is only able to provide DESCRIPTIONS. For instance, this
recognition underlies the change in nomenclature from "theories" to "models".

Any "explanations" one might think one has are only "one level
deep". One can ALWAYS keep asking "why that?" or demand to
"explain that"....


> It seems to me that general relativity is not really physics.

It seems to me you don't know enough about it to have a sensible opinion.


Tom Roberts

Jack...@hotmail.com

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Feb 9, 2015, 12:40:02 PM2/9/15
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I'm saying that I can't believe in general relativity, because even
just the mathematical basis is defective. If it was valid mathematics,
you would have a "metric tensor" that would bend space-time, in which
dtau/dt = g00 = sqrt(1-2MG/rc^2). instead you use the negative square
of that term.
implicit, in such a case, would be an obligation to specifically list
the basis vectors for that space, but that can't be done without
listing ict.
Instead, for no good reason except stealth, all terms already
initially squared.
I point out the only case known to me, of an overt attempt at such
expurgation is in "Gravitation" page 51, by MTW, titled "Farewell to
ICT". It is a pathetic explanation, something like not wanting X4,
with its ict, to confuse people and therefore selecting X0 as the time
axis.
Included is some twaddle about vectors being entirely different
creatures from one forms, when you really only need covariant and
contravariant forms of vector corresponding to row and column
matrices.
I'm not sure of the exact impact of making that discrimination.
Quadratic form should be of the form s^2 = X'GX or x^i*x_jGij but
definitely not G*(x^i)^2.
I bow to your superior knowledge of general relativity. I'm just
telling you that there's so much that is hard to believe in, given the
slovenly treatment of mathematics.
I know enough real physics to know that time is not dilated; the
effects merely describe the change in the velocity of light in
gravity.
John Polasek

Tom Roberts

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Feb 17, 2015, 1:20:03 PM2/17/15
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On 2/9/15 2/9/15 - 11:33 AM, Jack...@hotmail.com wrote:
> On Sun, 8 Feb 2015 00:14:56 CST, Tom Roberts
> <tjrobe...@sbcglobal.net> wrote:
>> The key point is that semi-Riemannian geometry is NOT Riemannian
>> geometry. While many theorems of the latter apply to the former,
>> not all do, and often they get a minus sign or absolute-value.
>> And for many purposes, one must treat spacelike and timelike
>> intervals separately (signs often differ).

> I'm saying that I can't believe in general relativity, because even
> just the mathematical basis is defective.

It is YOU who is mistaken, not GR which is "defective".

This is not really about "belief", it is about UNDERSTANDING, which is more
subtle than you appreciate.


> If it was valid mathematics,
> you would have a "metric tensor" that would bend space-time, in which
> dtau/dt = g00 = sqrt(1-2MG/rc^2). instead you use the negative square
> of that term.

Look at the quote up above -- Semi-Riemannian geometry is NOT Riemannian
geometry. You are attempting to force GR into the latter, and that is utterly
hopeless.

[I ignore the difference betwen metric signatures -+++ and +---.]


> implicit, in such a case, would be an obligation to specifically list
> the basis vectors for that space, but that can't be done without
> listing ict.

Sure it can. Indeed "ict" is long gone from GR. As you mention, there is even a
box in MTW that explains why -- "ict" was an attempt to force GR into Riemannian
geometry; like your attempt above, it failed (the introduction of complex
quantities causes various theorems to fail, just as they fail in semi-Riemannian
geometry).

Select a locally inertial frame at a given point in the
spacetime manifold of interest. The basis vectors are:
(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)
The metric is diag(-1,1,1,1). There's not an "i" involved.

Again, your complaint is really about your personal ignorance, not about GR itself.


> I point out the only case known to me, of an overt attempt at such
> expurgation is in "Gravitation" page 51, by MTW, titled "Farewell to
> ICT". It is a pathetic explanation, something like not wanting X4,
> with its ict, to confuse people and therefore selecting X0 as the time
> axis.
> Included is some twaddle about vectors being entirely different
> creatures from one forms, when you really only need covariant and
> contravariant forms of vector corresponding to row and column
> matrices.
> I'm not sure of the exact impact of making that discrimination.

Hmmmm. Vectors _ARE_ different from one-forms; vectors and one-forms _ARE_ the
same as contravariant vectors and covariant vectors, and they _ARE_ also the
same as vectors and covectors.

(Beware the meaning of "co-", as it can mean either "with" or
"against/inverse". In some contexts the meanings of "covariant"
and "contravariant" are inverted. Remember also that a covariant
vector has a set of contravariant components; the confusion
multiplies when tensors are confused with their components.
But still, a co-mathematician would be a system for converting
theorems into coffee.)

Note that vectors do NOT "correspond" to column matrices. You are confusing the
COMPONENTS of the vector with the vector itself. That's invalid (even though
many ancient books confuse them). A vector has a magnitude and a direction. To
turn a column matrix into a vector requires applying the corresponding set of
basis vectors -- these last are indeed VECTORS, and not merely another column
matrix.


> Quadratic form should be of the form s^2 = X'GX or x^i*x_jGij but
> definitely not G*(x^i)^2.

I have no idea what you are trying to say. Every quadratic form I know of in GR
(and in differential geometry) can be written as A(X,X), for a rank-2 tensor
A(.,.) and a vector X; in components that would be A_ij X^i X^j.

Note all three of your formulas are wrong, in one detail or
another. For someone complaining about lack of rigor that is
rather suggestive....


> I bow to your superior knowledge of general relativity. I'm just
> telling you that there's so much that is hard to believe in, given the
> slovenly treatment of mathematics.

You need a better textbook. Physicists are well known for their casual attitude
about mathematical rigor. You also need to improve the precision of your own
thoughts and words, as YOU are excessively casual, too.


> I know enough real physics to know that time is not dilated; the
> effects merely describe the change in the velocity of light in
> gravity.

What you think you know is wrong. The local vacuum speed of light DOES NOT
CHANGE, in gravity or anywhere else. When you omit the qualifier "local" you
merely exhibit your personal ignorance of its importance, demonstrating that you
fail to understand the geometrical structure of GR. Omitting the qualifier
"vacuum" indicates you do not understand the underlying physics.

As I have said so often, it simply is not possible to understand subtle
concepts, such as relativity, without precision in thought and word. Your
statements here fall short of what is needed. Relativity is not only more subtle
than you think, it is more subtle than it is possible for you to think, until
you clean up your act. Complaining about lack of rigor in other peoples'
presentations is the least of your problems.


Tom Roberts

Jack...@hotmail.com

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Feb 18, 2015, 11:20:04 AM2/18/15
to
On Tue, 17 Feb 2015 12:18:31 CST, Tom Roberts
<tjrobe...@sbcglobal.net> wrote:

>On 2/9/15 2/9/15 - 11:33 AM, Jack...@hotmail.com wrote:
>> On Sun, 8 Feb 2015 00:14:56 CST, Tom Roberts
>> <tjrobe...@sbcglobal.net> wrote:
>>> The key point is that semi-Riemannian geometry is NOT Riemannian
>>> geometry. While many theorems of the latter apply to the former,
>>> not all do, and often they get a minus sign or absolute-value.
>>> And for many purposes, one must treat spacelike and timelike
>>> intervals separately (signs often differ).
>
>> I'm saying that I can't believe in general relativity, because even
>> just the mathematical basis is defective.
>
>It is YOU who is mistaken, not GR which is "defective".
>
>This is not really about "belief", it is about UNDERSTANDING, which is more
>subtle than you appreciate.
>
>
>> If it was valid mathematics,
>> you would have a "metric tensor" that would bend space-time, in which
>> dtau/dt = g00 = sqrt(1-2MG/rc^2). instead you use the negative square
>> of that term.
>
>Look at the quote up above -- Semi-Riemannian geometry is NOT Riemannian
>geometry. You are attempting to force GR into the latter, and that is utterly
>hopeless.
This talk about semiRiemannian is a cover for the fact that the
"tensor" cannot be rotated, which disqualifies it. Rotating it would
entail changing x with y or z, etc. but time is "stuck".
> [I ignore the difference betwen metric signatures -+++ and +---.]
>
>
>> implicit, in such a case, would be an obligation to specifically list
>> the basis vectors for that space, but that can't be done without
>> listing ict.
>
>Sure it can. Indeed "ict" is long gone from GR. As you mention, there is even a
>box in MTW that explains why -- "ict" was an attempt to force GR into Riemannian
>geometry; like your attempt above, it failed (the introduction of complex
>quantities causes various theorems to fail, just as they fail in semi-Riemannian
>geometry).
>
> Select a locally inertial frame at a given point in the
> spacetime manifold of interest. The basis vectors are:
> (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)
> The metric is diag(-1,1,1,1). There's not an "i" involved.
>as below, and it is plotted as a residual in the lower panel of Figure 2.
>Again, your complaint is really about your personal ignorance, not about GR itself.
Then, as a simple test, where x0 =(1 0 0 0), we can write the
transformation x0 * g0j = (-1 0 0 0 ). WE should be surprised to find
that
the vector has been arbitrarily reversed, which I think you'll agree
is totally unacceptable.
So this ' tensor ' fails the first test, the
simple transformation of an input vector.

>
>> I point out the only case known to me, of an overt attempt at such
>> expurgation is in "Gravitation" page 51, by MTW, titled "Farewell to
>> ICT". It is a pathetic explanation, something like not wanting X4,
>> with its ict, to confuse people and therefore selecting X0 as the time
>> axis.
>> Included is some twaddle about vectors being entirely different
>> creatures from one forms, when you really only need covariant and
>> contravariant forms of vector corresponding to row and column
>> matrices.
>> I'm not sure of the exact impact of making that discrimination.
>
>Hmmmm. Vectors _ARE_ different from one-forms; vectors and one-forms _ARE_ the
>same as contravariant vectors and covariant vectors, and they _ARE_ also the
>same as vectors and covectors.
>
> (Beware the meaning of "co-", as it can mean either "with" or
> "against/inverse". In some contexts the meanings of "covariant"
> and "contravariant" are inverted. Remember also that a covariant
> vector has a set of contravariant components; the confusion
> multiplies when tensors are confused with their components.
> But still, a co-mathematician would be a system for converting
> theorems into coffee.)
>
>Note that vectors do NOT "correspond" to column matrices. You are confusing the
>COMPONENTS of the vector with the vector itself.
I am very conversant that nothing is done in mathematics with lines
and arrows and angles but only with the components (X Y Z).
You failed to grasp what I was trying to tell you, covariant and
contravariant types of vectors(yes yes the components) have as their
counterparts, row vectors and column vectors. There is no place I can
see for one forms.
It just occurred to me, the the authors of Gravitation possibly had
never heard of row vectors, (few have) but they are absolutely
essential, like contravariant vectors.

>That's invalid (even though
>many ancient books confuse them). A vector has a magnitude and a direction. To
>turn a column matrix into a vector requires applying the corresponding set of
>basis vectors -- these last are indeed VECTORS, and not merely another column
>matrix.
>
>
>> Quadratic form should be of the form s^2 = X'GX or x^i*x_jGij but
>> definitely not G*(x^i)^2.
>
>I have no idea what you are trying to say. Every quadratic form I know of in GR
>(and in differential geometry) can be written as A(X,X), for a rank-2 tensor
>A(.,.) and a vector X; in components that would be A_ij X^i X^j.
By quadratic form you must mean that polynomial that issues from a
proper XGX, an operation that must be done in sequence: XG = X' and
then X'X = polynomial. But just quoting the polynomial covers up the
fact that the underlying matrix G is unsustainable.
This nonsense about A(XX) for a 2-rank tensor probably comes from the
glib remarks in Gravitation about the tensor being a machine that
takes
in 2 vectors and delivers a third! How puerile!


> Note all three of your formulas are wrong, in one detail or
> another. For someone complaining about lack of rigor that is
> rather suggestive....
>
>
>> I bow to your superior knowledge of general relativity. I'm just
>> telling you that there's so much that is hard to believe in, given the
>> slovenly treatment of mathematics.
>
>You need a better textbook. Physicists are well known for their casual attitude
>about mathematical rigor. You also need to improve the precision of your own
>thoughts and words, as YOU are excessively casual, too.
I have to remind you again that in a paper in 1967 in the Journal of
American optical society I made up my own tensors from scratch for
mirrors and also for the rotating gimbles that held them and also for
the rotation matrices that transformed them. I have hands-on
experience
with tensors.
>
>> I know enough real physics to know that time is not dilated; the
>> effects merely describe the change in the velocity of light in
>> gravity.
>
>What you think you know is wrong. The local vacuum speed of light DOES NOT
>CHANGE, in gravity or anywhere else. When you omit the qualifier "local" you
>merely exhibit your personal ignorance of its importance, demonstrating that you
>fail to understand the geometrical structure of GR. Omitting the qualifier
>"vacuum" indicates you do not understand the underlying physics.
>
>As I have said so often, it simply is not possible to understand subtle
>concepts, such as relativity, without precision in thought and word. Your
>statements here fall short of what is needed. Relativity is not only more subtle
>than you think, it is more subtle than it is possible for you to think, until
>you clean up your act. Complaining about lack of rigor in other peoples'
>presentations is the least of your problems.
>
>
>Tom Roberts
GR is a failed concept, having little to do with physics.
John Polasek

Tom Roberts

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Mar 11, 2015, 10:20:02 AM3/11/15
to
On 2/18/15 2/18/15 10:15 AM, Jack...@hotmail.com wrote:
> On Tue, 17 Feb 2015 12:18:31 CST, Tom Roberts
>> Semi-Riemannian geometry is NOT Riemannian
>> geometry. You are attempting to force GR into the latter, and that is utterly
>> hopeless.
> This talk about semiRiemannian is a cover for the fact that the
> "tensor" cannot be rotated, which disqualifies it. Rotating it would
> entail changing x with y or z, etc. but time is "stuck".

Nonsense. You are hung up on Riemannian geometry, and need to learn about
semi-Riemannian geometry before you can have any chance of understanding GR and
modern physics.

The rotations you think cannot happen are called boosts. In relativity, relative
motion _IS_ a rotation -- an element of SO(1,3).


> Then, as a simple test, where x0 =(1 0 0 0), we can write the
> transformation x0 * g0j = (-1 0 0 0 ).

That is NOT a transformation.

Sure, you can multiply by -1 and reverse the vector. SO WHAT???


> [... too many repetitions of the same errors; too much nonsense]


Tom Roberts

Jack...@hotmail.com

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Mar 17, 2015, 12:40:02 PM3/17/15
to
On Wed, 11 Mar 2015 09:15:35 CST, Tom Roberts
<tjrobe...@sbcglobal.net> wrote:

>On 2/18/15 2/18/15 10:15 AM, Jack...@hotmail.com wrote:
>> On Tue, 17 Feb 2015 12:18:31 CST, Tom Roberts
>>> Semi-Riemannian geometry is NOT Riemannian
>>> geometry. You are attempting to force GR into the latter, and that is utterly
>>> hopeless.
>> This talk about semiRiemannian is a cover for the fact that the
>> "tensor" cannot be rotated, which disqualifies it. Rotating it would
>> entail changing x with y or z, etc. but time is "stuck".
>
>Nonsense. You are hung up on Riemannian geometry, and need to learn about
>semi-Riemannian geometry before you can have any chance of understanding GR and
>modern physics.
>
>The rotations you think cannot happen are called boosts. In relativity, relative
>motion _IS_ a rotation -- an element of SO(1,3).
>
>
>> Then, as a simple test, where x0 =(1 0 0 0), we can write the
>> transformation x0 * g0j = (-1 0 0 0 ). will

A metric tensor must, by all logic, exist for GR theory to hold up,
but it does not exist. It would be a square array that stretches the
base vectors one way or another. There would be the element G00 =
dt/tau and its value must be close to one but not -1.
>
It is a transformation if gij is a 2nd rank tensor but it turns time
backwards. If you use the other "signature" 1-1-1-1, then all of the
space terms are reversed. It's all an elaborate scheme to cover over
the fact that time has to be imaginary, ICT.
Let's just call it a draw and be done with it.

John Polasek

Jack...@hotmail.com

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Mar 17, 2015, 12:40:05 PM3/17/15
to
On Wed, 11 Mar 2015 09:15:35 CST, Tom Roberts
<tjrobe...@sbcglobal.net> wrote:

>On 2/18/15 2/18/15 10:15 AM, Jack...@hotmail.com wrote:
>> On Tue, 17 Feb It2015 12:18:31 CST, Tom Roberts
>>> Semi-Riemannian geometry is NOT Riemannian
>>> geometry. You are attempting to force GR into the latter, and that is utterly
>>> hopeless.
>> This talk about semiRiemannian is a cover for the fact that the
>> "tensor" cannot be rotated, which disqualifies it. Rotating it would
>> entail changing x with y or z, etc. but time is "stuck".
>
>Nonsense. You are hung up on Riemannian geometry, and need to learn about
>semi-Riemannian geometry before you can have any chance of understanding GR and
>modern physics.
>
>The rotations you think cannot happen are called boosts. In relativity, relative
>motion _IS_ a rotation -- an element of SO(1,3).

>> Then, as a simple test, where x0 =(1 0 0 0), we can write the
>> transformation x0 * g0j = (-1 0 0 0 ).
You are saying this is not a transformation, and that is right,
because gij is not a tensor. It does not act on vector quantities ,
only their squares. That makes it convenient for the obvious term ICT,
when squared, to look legitimate; why shouldn't it be negative, what
could it hurt?
A metric tensor must, by all logic, exist for GR theory to hold up,
but it does not exist. It would be a square array that stretches the
base vectors one way or another. There would be the element G00 =
dt/tau and its value must be close to one but not -1.
There is the other problem, that is, getting specific about the name
of vectors. x0. x4 is a copout. x0 is really ICT, but it doesn't look
nice if you just write it down.
..>
>Tom Roberts
John Polasek

Tom Roberts

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Mar 23, 2015, 1:30:02 PM3/23/15
to
On 3/17/15 3/17/15 - 12:36 PM, Jack...@hotmail.com wrote:
> On Wed, 11 Mar 2015 09:15:35 CST, Tom Roberts
> <tjrobe...@sbcglobal.net> wrote:
> [...]
>
> A metric tensor must, by all logic, exist for GR theory to hold up,[...]

You merely display your complete and utter ignorance of GR. The metric tensor is
central to the theory, and to its geometry.

Essentially everything you have said in this thread is just plain wrong. Your
refusal to consider semi-Riemannian geometry is excessively limiting, and your
fixation with "imaginary time" is archaic, silly, and wrong.

There's no point in continuing until you educate yourself on the basics of GR;
this is NOT a suitable medium for that. Get a good textbook. A good starting
place is:

Geroch, _General_Relativity_from_A_to_B_.


Tom Roberts

Jack...@hotmail.com

unread,
Mar 24, 2015, 7:40:02 PM3/24/15
to
On Mon, 23 Mar 2015 12:28:17 CST, Tom Roberts
<tjrobe...@sbcglobal.net> wrote:

>On 3/17/15 3/17/15 - 12:36 PM, Jack...@hotmail.com wrote:
>> On Wed, 11 Mar 2015 09:15:35 CST, Tom Roberts
>> <tjrobe...@sbcglobal.net> wrote:
>> [...]
>>
>> A metric tensor must, by all logic, exist for GR theory to hold up,[...]
In Wikipedia under metric tensor they show the Schwarzschild metric, a
diagonal object with a number of squared terms in it. It's the only
place I have seen the elements displayed. Notice that is really never
used, being simply a diagonal presentation always shown as ds^2 =.....
They justify this primitive treatment by another example on the same
page.
They show a "round metric on a sphere", being a 2 x 2 diagonal matrix
with the prescription that "this is usually written in the form" ds2 =
d02 + sin2dphi2. (Not at all;no 2 x 2 matrix can show that without
appropriate vectors).
Not so fast. The square matrix should be preceded by a row vector and
post multiplied by a column vector, not just quickly multiply all 3
together and dump the result down in a line to be memorized.
You need the latter VGV arrangement where you can see that either the
row vector or the column vector will be 1st affected by the matrix
after which the scalar product is formed with the original vector.
This multiplication must stand on its own and there is no question
that some of the components will be totally reversed, an intolerable
condition.These can only be done one operation at a time. This is also
true of the Smetric If you multiply one of the vectors of the matrix,
you get its counterpart, which one would hardly expect to be reversed
180°. Yet that's what they have done here, sacrificing XYZ as negative
to make CT look good.
I realize you have a lot of time and energy invested in this
relativity but it's a false theory; space cannot be bent and even
Einstein is unable to show how to do it.
John Polasek

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