Grokking the Einstein Equation
Edward Green
We only have to do one simple thing: understand the significance of
the Ricci tensor minus 1/2 the metric tensor times the scalar
curvature. This of course is interpreted as the... ?
What the heck is this thing, anyway? :-/
Eric Gisse
Einstein tensor, a tensor that is guaranteed to be divergence free.
Basically the desire was to have second derivatives of the metric
equal to matter terms. The Einstein tensor is the simplest way to do
it.
Edward Green
> On Oct 23, 5:36 pm, Edward Green <spamspamsp...@netzero.com> wrote:
>
> > I feel it is time to grok the Einstein equation.
>
> > We only have to do one simple thing: understand the significance of
> > the Ricci tensor minus 1/2 the metric tensor times the scalar
> > curvature. This of course is interpreted as the... ?
>
> > What the heck is this thing, anyway? :-/
>
> Einstein tensor,
What's in a name?
> a tensor that is guaranteed to be divergence free.
Well, that's something.
> Basically the desire was to have second derivatives of the metric
> equal to matter terms.
Then how is is at least part of this tensor consists of the metric
multiplied by a contraction of RIEMANN (yeah, I know RICCI is also a
contraction of RIEMANN, to a lesser degree). Is that obviously
something involving second derivatives of the metric?
I was hoping for some geometric insight into the terms, as they
directly relate the local geometry of spacetime to momentum and
energy.
Koobee Wublee
> On Oct 23, 10:29 pm, Eric Gisse wrote:
> > Einstein tensor,
>
> What's in a name?
>
> > a tensor that is guaranteed to be divergence free.
>
> Well, that's something.
The Riemann tensor was derived by Ricci and named after Riemann.
The Ricci tensor was derived by Ricci’s student Levi-Civita and named
after Ricci.
The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
had since then championed to be named after their messiah Einstein,
the nitwit, the plagiarist, and the liar. <shrug>
> > Basically the desire was to have second derivatives of the metric
> > equal to matter terms.
>
> Then how is is at least part of this tensor consists of the metric
> multiplied by a contraction of RIEMANN (yeah, I know RICCI is also a
> contraction of RIEMANN, to a lesser degree). Is that obviously
> something involving second derivatives of the metric?
You need to understand how the Einstein tensor is derived to fully
understand that. <shrug>
> I was hoping for some geometric insight into the terms, as they
> directly relate the local geometry of spacetime to momentum and
> energy.
Any geometric insight is merely an interpretation to the mathematics
involved.
> > The Einstein tensor is the simplest way to do it.
For those who are still struggling to understand the Einstein tensor,
here is a derivation. Please pay attention. It is actually very
simple.
Hilbert pulled out the following out of nowhere which is now called
the Lagrangian to the Einstein-Hilbert action. It does not even
satisfy as a Langrangian according to the Lagrangian methods.
L = (H R + rho c^2) sqrt(- det[g])
Where
** H = Constant
** R = [g]^ij [R]_ij = Ricci Scalar
** rho = Mass density (kg/m^3 in MKS system)
** det[g] = determinant of [g]
** [g] = The metric, a symmetric 4-by-4 matrix
To get the field equations, all you have to do is to take the partial
derivative of the Lagrangian L with respect to [g]^ij where they are
elements to the inverse of the matrix [g] with elements [g]_ij.
Setting that to null, you get
@L/@[g]^ij = 0
Or
H @R/@[g]^ij sqrt(- det[g]) = (H R + rho c^2) @det[g]/@[g]^ij / sqrt(-
det[g]) / 2
Well,
** @R/@[g]^ij = [R]_ij
** @det[g]/@[g]^ij = - det[g] / [g]^ij
Thus, you get
[R]_ij – R / [g]^ij / 2 = rho c^2 / [g]^ij / H / 2
Just flip the raising and lowering of the indices several times, you
can mathemagically transform the above into
[R]_ij – R / [g]_ij / 2 = (rho c^2 / H / 2) [g]_ij
While the rest of the world is still bedazzled by your mathemagic
trick, you quickly mystify the right-hand side of the equations above
by calling it the energy momentum tensor [T]_ij in which it possess
all the mathemagical properties. So,
[R]_ij – R / [g]_ij / 2 = Constant [T]_ij
There is nothing to the highly worshipped field equations if you can
identify the mathemagical steps. Deriving that constant is another
chapter in the nonsense of the field equations. I will save it for
next time.
Jim Black
> Hilbert pulled out the following out of nowhere which is now called
> the Lagrangian to the Einstein-Hilbert action. It does not even
> satisfy as a Langrangian according to the Lagrangian methods.
>
> L = (H R + rho c^2) sqrt(- det[g])
>
> Where
>
> ** H = Constant
> ** R = [g]^ij [R]_ij = Ricci Scalar
> ** rho = Mass density (kg/m^3 in MKS system)
> ** det[g] = determinant of [g]
> ** [g] = The metric, a symmetric 4-by-4 matrix
[snip calculations]
> [R]_ij - R / [g]_ij / 2 = (rho c^2 / H / 2) [g]_ij
>
> While the rest of the world is still bedazzled by your mathemagic
> trick, you quickly mystify the right-hand side of the equations above
> by calling it the energy momentum tensor [T]_ij in which it possess
> all the mathemagical properties. So,
>
> [R]_ij - R / [g]_ij / 2 = Constant [T]_ij
Or you could try adding an actual matter term to the Lagrangian.
<shrug>
Poor Koobee Wublee, writing down all these equations but unable to
tell a cosmological constant from a rock. This is not grokking.
Koobee Wublee
> Koobee Wublee wrote:
> > Hilbert pulled out the following out of nowhere which is now called
> > the Lagrangian to the Einstein-Hilbert action. It does not even
> > satisfy as a Langrangian according to the Lagrangian methods.
>
> > L = (H R + rho c^2) sqrt(- det[g])
>
> > Where
>
> > ** H = Constant
> > ** R = [g]^ij [R]_ij = Ricci Scalar
> > ** rho = Mass density (kg/m^3 in MKS system)
> > ** det[g] = determinant of [g]
> > ** [g] = The metric, a symmetric 4-by-4 matrix
>
> [snip calculations]
It suits some dimwits who are unable to follow such simple mathematics
hyped up as the most complicated mathemagic bullsh*t. <shrug>
> > [R]_ij - R / [g]_ij / 2 = (rho c^2 / H / 2) [g]_ij
>
> > While the rest of the world is still bedazzled by your mathemagic
> > trick, you quickly mystify the right-hand side of the equations above
> > by calling it the energy momentum tensor [T]_ij in which it possess
> > all the mathemagical properties. So,
>
> > [R]_ij - R / [g]_ij / 2 = Constant [T]_ij
>
> Or you could try adding an actual matter term to the Lagrangian.
> <shrug>
I did while your ignorance is abound. <shrug>
> Poor Koobee Wublee, writing down all these equations but unable to
> tell a cosmological constant from a rock.
Yes, I am poor all right with all these kids I have to support. Hey,
that is worth it. Some folks choose to go through life collecting as
much busslsh*t, wealth, lovers, or whatever as possible. I choose to
maximize on true knowledge and offsprings. Please don’t be jealous of
me. <shrug>
> This is not grokking.
There is no logical addition of the Cosmological constant into the
field equations. The field equations were thought to result in
Newtonian type solutions where gravitation is asymptotically flat.
Einstein the nitwit, the plagiarist, and the liar chose to toss
everything on the ground and declared the existence of negative mass
density in vacuum to fudge into the observed static universe. The
result was the Cosmological constant. For once, Einstein the nitwit,
the plagiarist, and the liar was right. The Cosmological constant was
the biggest and the only blunder of his life. <shrug>
Happy slurping up Einstein’s diarrhea. Please don’t make me vomit and
keep that “wonderful” and sickly experience to yourself. <shrug>
Edward Green
> On Oct 23, 8:09 pm, Edward Green wrote:
>
> > On Oct 23, 10:29 pm, Eric Gisse wrote:
> > > Einstein tensor,
>
> > What's in a name?
>
> > > a tensor that is guaranteed to be divergence free.
>
> > Well, that's something.
>
> The Riemann tensor was derived by Ricci and named after Riemann.
>
> The Ricci tensor was derived by Ricci’s student Levi-Civita and named
> after Ricci.
>
> The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
> had since then championed to be named after their messiah Einstein,
> the nitwit, the plagiarist, and the liar. <shrug>
Look at this from my point of view: I don't know how much you
understand because I understand so little. Now, I realize perhaps you
don't care, but which way will your sobriquet "the nitwit, the
plagiarist, and the liar" likely drive my Bayesian priors? You may
not care, but I am unlikely to be the only one affected this way.
> > > Basically the desire was to have second derivatives of the metric
> > > equal to matter terms.
>
> > Then how is is at least part of this tensor consists of the metric
> > multiplied by a contraction of RIEMANN (yeah, I know RICCI is also a
> > contraction of RIEMANN, to a lesser degree). Is that obviously
> > something involving second derivatives of the metric?
>
> You need to understand how the Einstein tensor is derived to fully
> understand that. <shrug>
Possibly. The Einstein equation cries out for direct apprehension,
however: I'm not sure this requires us to repeat the
"derivation" (heuristic derivation). The equation says that some
geometric property of spacetime, expressed as a sum of two parts, is
equal to some matter terms. I have some idea about the matter terms.
I have little apprehension of the geometric terms.
Since I am led to believe the Riemann tensor, and hence, of course,
the Ricci tensor and the scalar curvature, can be derived from he
metric, I wonder if the expression is not more complicated than need
be. It could be written as some operator applied to the metric:
E(g) = T
> > I was hoping for some geometric insight into the terms, as they
> > directly relate the local geometry of spacetime to momentum and
> > energy.
>
> Any geometric insight is merely an interpretation to the mathematics
> involved.
Right. Well, suppose we tried to write the equation in a local Lorentz
frame, which we are guaranteed to be able to do. Then we have a
direct component wise decomposition/interpretation: the ij_th
component of the Ricci tensor - 1/2 the ij_th component of the metric
times the scalar curvature = the flow of i-momentum in the j-direction
(or vice verse -- we can be sloppy because of symmetry).
We ought to be able to develop a direct appreciation of _that_.
_What_ local stretching of spacetime is equal to the ij_th momentum
flux. Whatever it is, what could be simpler? At this level, the
relation is completely linear.
> > > The Einstein tensor is the simplest way to do it.
>
> For those who are still struggling to understand the Einstein tensor,
> here is a derivation. Please pay attention. It is actually very
> simple.
>
> Hilbert pulled out the following out of nowhere which is now called
> the Lagrangian to the Einstein-Hilbert action. It does not even
> satisfy as a Langrangian according to the Lagrangian methods.
>
> L = (H R + rho c^2) sqrt(- det[g])
>
> Where
>
> ** H = Constant
> ** R = [g]^ij [R]_ij = Ricci Scalar
hmm... is that true? Interesting.
Well, let us say you "derived" them. Any such derivation must be
heuristic, since they constitute a physical theory/postulate. The
result is much simpler looking than the derivation, however, which
should enable a more direct "aha, that's just..." experience.
Daryl McCullough
>Look at this from my point of view: I don't know how much you
>understand because I understand so little. Now, I realize perhaps you
>don't care, but which way will your sobriquet "the nitwit, the
>plagiarist, and the liar" likely drive my Bayesian priors? You may
>not care, but I am unlikely to be the only one affected this way.
It seems to be the case that Koobee understands nothing at all about
General Relativity or tensors or differential geometry. In particular,
he doesn't seem to understand how the metric tensor is relevant to
computing areas and volumes.
--
Daryl McCullough
Ithaca, NY
Eric Gisse
> On Oct 23, 10:29 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Oct 23, 5:36 pm, Edward Green <spamspamsp...@netzero.com> wrote:
>
> > > I feel it is time to grok the Einstein equation.
>
> > > We only have to do one simple thing: understand the significance of
> > > the Ricci tensor minus 1/2 the metric tensor times the scalar
> > > curvature. This of course is interpreted as the... ?
>
> > > What the heck is this thing, anyway? :-/
>
> > Einstein tensor,
>
> What's in a name?
>
> > a tensor that is guaranteed to be divergence free.
>
> Well, that's something.
>
> > Basically the desire was to have second derivatives of the metric
> > equal to matter terms.
>
> Then how is is at least part of this tensor consists of the metric
> multiplied by a contraction of RIEMANN (yeah, I know RICCI is also a
> contraction of RIEMANN, to a lesser degree). Is that obviously
> something involving second derivatives of the metric?
The Einstein tensor is simply a function of Riemann.
R_uv - 1/2 R g_uv
Both the Ricci scalar and the Ricci tensor are contractions of
Riemann. The information encoded in the field equations is enough to
completely determine Riemann, which satisfies your previous request.
>
> I was hoping for some geometric insight into the terms, as they
> directly relate the local geometry of spacetime to momentum and
> energy.
The Einstein tensor is also automatically conserved to preserve the
Bianchi identities, which are statements about Riemann.
xxein
xxein: It is the attempted explanation of the Eddington measurement
where it is shown that light is bent doubly as per Newton in gravity.
It relates to what we can measure with different scales of measurement
and choose to accept as a theory.
Do you want more?
Androcles
"xxein" <xxe...@bellsouth.net> wrote in message
news:36b4f77a-7fad-4978...@v30g2000hsa.googlegroups.com...
On Oct 23, 9:36 pm, Edward Green <spamspamsp...@netzero.com> wrote:
> I feel it is time to grok the Einstein equation.
>
> We only have to do one simple thing: understand the significance of
> the Ricci tensor minus 1/2 the metric tensor times the scalar
> curvature. This of course is interpreted as the... ?
>
> What the heck is this thing, anyway? :-/
xxein: It is the attempted explanation of the Eddington measurement
where it is shown that light is bent doubly as per Newton in gravity.
============================================
Liar.
Eric Gisse
> On Oct 23, 8:09 pm, Edward Green wrote:
>
> > On Oct 23, 10:29 pm, Eric Gisse wrote:
> > > Einstein tensor,
>
> > What's in a name?
>
> > > a tensor that is guaranteed to be divergence free.
>
> > Well, that's something.
>
> The Riemann tensor was derived by Ricci and named after Riemann.
>
> The Ricci tensor was derived by Ricci’s student Levi-Civita and named
> after Ricci.
>
> The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
> had since then championed to be named after their messiah Einstein,
> the nitwit, the plagiarist, and the liar. <shrug>
You only think it was fudged because you don't understand any of this.
>
> > > Basically the desire was to have second derivatives of the metric
> > > equal to matter terms.
>
> > Then how is is at least part of this tensor consists of the metric
> > multiplied by a contraction of RIEMANN (yeah, I know RICCI is also a
> > contraction of RIEMANN, to a lesser degree). Is that obviously
> > something involving second derivatives of the metric?
>
> You need to understand how the Einstein tensor is derived to fully
> understand that. <shrug>
True, he does. He will learn though, but you never will.
>
> > I was hoping for some geometric insight into the terms, as they
> > directly relate the local geometry of spacetime to momentum and
> > energy.
>
> Any geometric insight is merely an interpretation to the mathematics
> involved.
Now reconcile this usage of 'interpretation' to your oft-used claim
that SR is an 'interpretation' to the Lorentz transform.
>
> > > The Einstein tensor is the simplest way to do it.
>
> For those who are still struggling to understand the Einstein tensor,
> here is a derivation. Please pay attention. It is actually very
> simple.
I once again ask the question: Where did you learn this material?
>
> Hilbert pulled out the following out of nowhere which is now called
> the Lagrangian to the Einstein-Hilbert action. It does not even
> satisfy as a Langrangian according to the Lagrangian methods.
>
> L = (H R + rho c^2) sqrt(- det[g])
No, *YOU* pulled it out of nowhere. The actual Einstein-Hilbert
Lagrangian is R sqrt(-|g|). You can include an _arbitrary_ matter
term, however what you wrote is simply wrong.
I don't even see how you can claim something 'does not even satisfy as
a Lagrangian'. Not just in the sense of your tortured English, but the
thesis itself. ANYTHING can be a Lagrangian, as there is no
requirement past being mathematically well-posed.
Just because you don't know how to vary the Lagrangian doesn't mean
there is anything wrong with it, just that you are an idiot.
>
> Where
>
> ** H = Constant
> ** R = [g]^ij [R]_ij = Ricci Scalar
> ** rho = Mass density (kg/m^3 in MKS system)
It isn't even covariant. Little concepts like 'covariance' just slip
by you, don't they?
> ** det[g] = determinant of [g]
> ** [g] = The metric, a symmetric 4-by-4 matrix
The metric isn't a matrix, as a matrix is just an ordered arrangement
of objects. The metric, however, is a tensor which has additional
properties to that of a matrix.
Try to learn the difference before you die.
>
> To get the field equations, all you have to do is to take the partial
> derivative of the Lagrangian L with respect to [g]^ij where they are
> elements to the inverse of the matrix [g] with elements [g]_ij.
> Setting that to null, you get
>
> @L/@[g]^ij = 0
Nope. You fucked up step 0 *and* step 1 of working with a Lagrangian.
You couldn't even write the correct Lagrangian, or the correct field
equations.
You need to vary the Lagrangian not only with respect to the metric
components g_ij but the _derivatives_ of the metric components. Which
you aren't doing.
>
> Or
>
> H @R/@[g]^ij sqrt(- det[g]) = (H R + rho c^2) @det[g]/@[g]^ij / sqrt(-
> det[g]) / 2
Wrong, chuckles. Remember the other half of the Euler-Lagrange
equations you consistently told me to learn years after I did? They
involve _derivatives of the metric_. Any particular reason you feel
the metric derivatives don't count?
>
> Well,
>
> ** @R/@[g]^ij = [R]_ij
This isn't even right.
Do you actually think there is no functional dependence on the metric
past that?
> ** @det[g]/@[g]^ij = - det[g] / [g]^ij
Seriously? You actually think that?
You can't divide by a tensor. Try again.
>
> Thus, you get
>
> [R]_ij – R / [g]^ij / 2 = rho c^2 / [g]^ij / H / 2
This is ridiculous, even for you. You have division by tensors, the
wrong matter contribution, wrong attempt at varying the
Lagrangian...what a clusterfuck.
>
> Just flip the raising and lowering of the indices several times, you
> can mathemagically transform the above into
>
> [R]_ij – R / [g]_ij / 2 = (rho c^2 / H / 2) [g]_ij
You are still dividing by a tensor. How on Earth did you hallucinate
the ability to understand this?
>
> While the rest of the world is still bedazzled by your mathemagic
> trick, you quickly mystify the right-hand side of the equations above
> by calling it the energy momentum tensor [T]_ij in which it possess
> all the mathemagical properties. So,
>
> [R]_ij – R / [g]_ij / 2 = Constant [T]_ij
That's not the stress-energy tensor.
>
> There is nothing to the highly worshipped field equations if you can
> identify the mathemagical steps. Deriving that constant is another
> chapter in the nonsense of the field equations. I will save it for
> next time.
Since you didn't even do it right the first time, would you like to
see how it is _actually_ done?
Eric Gisse
[...]
He is an idiot. Everything he wrote was wrong, so don't even try to
learn from him.
Igor
If you let Wooblee guide you, you'll both end up in the ditch.
Don Stockbauer
Edward Green wrote:
> I feel it is time to grok the Einstein equation.
>
Who the hell says "grok" anymore???????
Tom Roberts
People who still like Robert Heinlein's novels, and who recognize that
English has no word that captures the fullness of this one.
Tom Roberts
Koobee Wublee
> On Oct 23, 7:09 pm, Edward Green wrote:
> > Then how is is at least part of this tensor consists of the metric
> > multiplied by a contraction of RIEMANN (yeah, I know RICCI is also a
> > contraction of RIEMANN, to a lesser degree). Is that obviously
> > something involving second derivatives of the metric?
>
> The Einstein tensor is simply a function of Riemann.
>
> R_uv - 1/2 R g_uv
>
> Both the Ricci scalar and the Ricci tensor are contractions of
> Riemann.
Yeah, the original poster knows that. <shrug> The question is why?
Just tell him that Levi-Civita chose to do so. Given the Riemann
curvature tensor R^n_ijk, why did Levi-Civita define the Ricci tensor
to be the following?
R_ij = R^j_ijk
And not
R_ij = R^k_ijk
Or
R_ij = R^i_ijk
> The information encoded in the field equations is enough to
> completely determine Riemann, which satisfies your previous request.
Get real. Given the Ricci tensor R_ij, you cannot recover the Riemann
tensor R^n_ijk.
> > I was hoping for some geometric insight into the terms, as they
> > directly relate the local geometry of spacetime to momentum and
> > energy.
>
> The Einstein tensor is also automatically conserved to preserve the
> Bianchi identities, which are statements about Riemann.
There is no big deal about the Bianchi identity. You are spreading
mysticism in differential geometry. <shrug>
Koobee Wublee
> On Oct 23, 9:19 pm, Koobee Wublee wrote:
> > The Riemann tensor was derived by Ricci and named after Riemann.
>
> > The Ricci tensor was derived by Ricci’s student Levi-Civita and named
> > after Ricci.
>
> > The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
> > had since then championed to be named after their messiah Einstein,
> > the nitwit, the plagiarist, and the liar. <shrug>
>
> [...]
More whining crap from a multi-year super-senior without a college
degree is mercifully snipped. I guess with mommy’s financial support,
you can sleep well pass noon every single day. <shrug>
> > Hilbert pulled out the following out of nowhere which is now called
> > the Lagrangian to the Einstein-Hilbert action. It does not even
> > satisfy as a Langrangian according to the Lagrangian methods.
>
> > L = (H R + rho c^2) sqrt(- det[g])
>
> No, *YOU* pulled it out of nowhere. The actual Einstein-Hilbert
> Lagrangian is R sqrt(-|g|).
Yes, but Hilbert really meant what I wrote above. With just (R sqrt(-
deg[g]), you don’t get the energy momentum tensor. <shrug>
> You can include an _arbitrary_ matter
> term, however what you wrote is simply wrong.
>
> I don't even see how you can claim something 'does not even satisfy as
> a Lagrangian'.
You have to understand the calculus of variations first. <shrug>
> [...]
More whining, yapping crap is mercifully snipped.
> > @L/@[g]^ij = 0
>
> Nope. You fucked up step 0 *and* step 1 of working with a Lagrangian.
> You couldn't even write the correct Lagrangian, or the correct field
> equations.
>
> You need to vary the Lagrangian not only with respect to the metric
> components g_ij but the _derivatives_ of the metric components. Which
> you aren't doing.
Oh, this is a classic. I promise you. If you ever were to learn the
calculus of variations, your statement will even disgust yourself.
<shrug>
> > H @R/@[g]^ij sqrt(- det[g]) = (H R + rho c^2) @det[g]/@[g]^ij / sqrt(-
> > det[g]) / 2
>
> Wrong, chuckles. Remember the other half of the Euler-Lagrange
> equations you consistently told me to learn years after I did? They
> involve _derivatives of the metric_. Any particular reason you feel
> the metric derivatives don't count?
As I said, you still don’t get it. You only know the calculus of
variation in skin depth. I have been trying to tell you all all along
that the derivation of the field equations does not involve any Euler-
Lagrange equations. <shrug>
> > Well,
>
> > ** @R/@[g]^ij = [R]_ij
>
> This isn't even right.
>
> Do you actually think there is no functional dependence on the metric
> past that?
[R]_ij is a function of the metric [g]. So, you are just whining
about things you do not understand as well. <shrug>
> > ** @det[g]/@[g]^ij = - det[g] / [g]^ij
>
> Seriously? You actually think that?
Yes. <shrug>
> You can't divide by a tensor. Try again.
Try again. [g]^ij is an element (a scalar) to the matrix of the
inverse of the matrix [g] the metric. <shrug>
> [...]
I have had enough of the whining crap from the multi-year super-senior
without any college degree. <shrug>
Koobee Wublee
> On Oct 24, 1:19 am, Koobee Wublee wrote:
> > The Riemann tensor was derived by Ricci and named after Riemann.
>
> > The Ricci tensor was derived by Ricci’s student Levi-Civita and named
> > after Ricci.
>
> > The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
> > had since then championed to be named after their messiah Einstein,
> > the nitwit, the plagiarist, and the liar. <shrug>
>
> Look at this from my point of view: I don't know how much you
> understand because I understand so little.
Hey, I was once in your shoes. I did not understand but tried and
still could not understand until concentrated from the pure
mathematical points of view. Mathematics is the truth while the
interpretations are mostly bullsh*t. Then, I understand it crystal-
clear. <shrug>
> Now, I realize perhaps you don't care,
You are very correct. Whether you understand it or not has no impact
on my life. <shrug> No hard feelings.
> but which way will your sobriquet "the nitwit, the
> plagiarist, and the liar" likely drive my Bayesian priors?
<shrug> There is nothing that can be attributed to Einstein as
original. Einstein was a nitwit, a plagiarist, and a liar. <shrug>
This is the truth if you still further in either history and the
relevant mathematics. The forensic evidence in history lies in the
mathematics itself ironically.
> You may
> not care, but I am unlikely to be the only one affected this way.
<shrug>
> > You need to understand how the Einstein tensor is derived to fully
> > understand that. <shrug>
>
> Possibly. The Einstein equation cries out for direct apprehension,
This is not true. The Einstein field equations are merely the
differential equations that allow you to solve for the metric and
identify what the metric is. The field equations are cooked up in the
pot of alchemy. It is a pure mathematical madness. <shrug>
> however: I'm not sure this requires us to repeat the
> "derivation" (heuristic derivation).
Hmmm... The best way to understand something is to go to where the
beginning is. <shrug>
> The equation says that some
> geometric property of spacetime, expressed as a sum of two parts, is
> equal to some matter terms.
Yes, it can be interpreted that way. <shrug>
> I have some idea about the matter terms.
Hmmm... Matter means mass. <shrug> If you don’t have any idea on
what the matter term is, you will be called an idiot. <shrug>
> I have little apprehension of the geometric terms.
The keystone to understand this subject in differential geometry is
the set of the Christoffel symbols of the second kind. In your
textbooks, they just magically appear. However, if you understand how
Christoffel derived the geodesic equations, there is no more
mysticism. Again, the geodesic equations are Euler-Lagrange equations
with the Lagrangian
L = g_ij dq^i/ds dq^j/ds = 1
Derived from,
ds^2 = g_ij dq^i dq^j
You just cannot get any simpler than that. <shrug>
> Since I am led to believe the Riemann tensor, and hence, of course,
> the Ricci tensor and the scalar curvature, can be derived from he
> metric,
Hmmm... The Riemann tensor, the Ricci tensor, and the Ricci scalar
are all functions of the metric [g] the matrix with elements g_ij or
[g]_ij (my nomenclature).
> I wonder if the expression is not more complicated than need
> be. It could be written as some operator applied to the metric:
>
> E(g) = T
I don’t know what your confusion is. <shrug>
> > Any geometric insight is merely an interpretation to the mathematics
> > involved.
>
> Right. Well, suppose we tried to write the equation in a local Lorentz
> frame, which we are guaranteed to be able to do. Then we have a
> direct component wise decomposition/interpretation: the ij_th
> component of the Ricci tensor - 1/2 the ij_th component of the metric
> times the scalar curvature = the flow of i-momentum in the j-direction
> (or vice verse -- we can be sloppy because of symmetry).
>
> We ought to be able to develop a direct appreciation of _that_.
> _What_ local stretching of spacetime is equal to the ij_th momentum
> flux. Whatever it is, what could be simpler? At this level, the
> relation is completely linear.
You can interpret any piece of mathematical equations anyway you
choose it to be. You can keep it as simple and as mundane as
possible, or you can do so with fairies, unicorns, elves, dwarves,
etc. <shrug> The bottom line is that the Holy Grail of GR is the set
of the field equations which are merely ordinary differential
equations when properly solved would either the metric.
> > For those who are still struggling to understand the Einstein tensor,
> > here is a derivation. Please pay attention. It is actually very
> > simple.
>
> > Hilbert pulled out the following out of nowhere which is now called
> > the Lagrangian to the Einstein-Hilbert action. It does not even
> > satisfy as a Langrangian according to the Lagrangian methods.
>
> > L = (H R + rho c^2) sqrt(- det[g])
>
> > Where
>
> > ** H = Constant
> > ** R = [g]^ij [R]_ij = Ricci Scalar
>
> hmm... is that true? Interesting.
Yes. (rho c^2) represent the “Lagrangian of mass”.
Why Hilbert came up with that is still a mystery, and the academics
tried to keep a very low profile of that because of obvious reasons.
<shrug>
You are wrong on that account. They do not have to be heuristic.
Trying to make something that is not heuristic to be soundly heuristic
is where you are struggling to understand the whole nonsense of GR.
<shrug>
> The
> result is much simpler looking than the derivation, however, which
> should enable a more direct "aha, that's just..." experience.
That is no “ahah” in madness but “OK!”. You cannot reason with
madness or nonsense. <shrug>
GR is purely a mathematical nonsense. <shrug>
Edward Green
I do.
It's a good word, whose meaning seems to match its sound.
Next question?
Edward Green
> On Oct 23, 9:19 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
<...>
> > You need to understand how the Einstein tensor is derived to fully
> > understand that. <shrug>
>
> True, he does. He will learn though, but you never will.
Thanks. Sometimes, one plows through a complicated derivation, only
at the end to see "Oh... that's just...".
I think I may be repeating myself. Sorry.
Anyway, I was just hoping remotely it might be possible to short-
circuit the process in this case, to see, beyond all the intermediate
connections, covariant derivatives, fiber bundles, and what have you,
just what geometric property of spacetime is encoded by the stress
energy tensor, with a kind of "Oh... of course!".
Truth is, it will take some more work on my part, but I think the
required degree of work lies somewhere between zero, and "First, learn
all of differential geometry". I feel something of the frustration of
the pathological anti-relativists, because the apparent simplicity of
the final equation is kind of mocking. It _must_ be possible to
understand this in kind of an immediate way.
The cottage industry of explaining this equation in this way lives in
a very small cottage, though.
Hmm... who added "sci.physics"? Not that it makes much difference.
Daryl McCullough
>Anyway, I was just hoping remotely it might be possible to short-
>circuit the process in this case, to see, beyond all the intermediate
>connections, covariant derivatives, fiber bundles, and what have you,
>just what geometric property of spacetime is encoded by the stress
>energy tensor, with a kind of "Oh... of course!".
Well, John Baez gave the following formulation of Einstein's
field equations in ordinary English (valid for situations in
which the pressure is isotropic; the same in all directions):
Given a small ball of freely falling test particles initially
at rest with respect to each other, the rate at which it begins
to shrink is proportional to its volume times: the energy
density at the center of the ball plus three times the pressure
at that point.
This formulation uses a very special coordinate system:
freefalling coordinates in which all particles are initially
at rest. The business about Riemann tensors and stress-energy
tensors is the way to state things in a coordinate-independent
way.
Daryl McCullough
Edward Green says...
>Anyway, I was just hoping remotely it might be possible to short-
>circuit the process in this case, to see, beyond all the intermediate
>connections, covariant derivatives, fiber bundles, and what have you,
>just what geometric property of spacetime is encoded by the stress
>energy tensor, with a kind of "Oh... of course!".
John Baez' discussion of the intuitive meaning of GR is given here:
http://math.ucr.edu/home/baez/einstein/node3.html
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 24, 5:51 pm, Eric Gisse wrote:
>> On Oct 23, 9:19 pm, Koobee Wublee wrote:
>
>> > The Riemann tensor was derived by Ricci and named after Riemann.
>>
>> > The Ricci tensor was derived by Ricci’s student Levi-Civita and named
>> > after Ricci.
>>
>> > The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
>> > had since then championed to be named after their messiah Einstein,
>> > the nitwit, the plagiarist, and the liar. <shrug>
>>
>> [...]
>
>More whining crap from a multi-year super-senior without a college
>degree is mercifully snipped. I guess with mommy’s financial support,
>you can sleep well pass noon every single day. <shrug>
The standard refrain for when you can't endure the criticism or answer
any of the questions. Why does someone with no shame have such a thin
skin?
>
>> > Hilbert pulled out the following out of nowhere which is now called
>> > the Lagrangian to the Einstein-Hilbert action. It does not even
>> > satisfy as a Langrangian according to the Lagrangian methods.
>>
>> > L = (H R + rho c^2) sqrt(- det[g])
>>
>> No, *YOU* pulled it out of nowhere. The actual Einstein-Hilbert
>> Lagrangian is R sqrt(-|g|).
>
>Yes, but Hilbert really meant what I wrote above. With just (R sqrt(-
>deg[g]), you don’t get the energy momentum tensor. <shrug>
How do you know Hilbert "really meant" that when the inclusion of a
matter term is fucking trivial and _not done that way_? Please answer
in the form of an insult which poorly hides your stupidity.
>
>> You can include an _arbitrary_ matter
>> term, however what you wrote is simply wrong.
>>
>> I don't even see how you can claim something 'does not even satisfy as
>> a Lagrangian'.
>
>You have to understand the calculus of variations first. <shrug>
I already do - and far better from you, judging from your responses
below.
>
>> [...]
>
>More whining, yapping crap is mercifully snipped.
>
>> > @L/@[g]^ij = 0
>>
>> Nope. You fucked up step 0 *and* step 1 of working with a Lagrangian.
>> You couldn't even write the correct Lagrangian, or the correct field
>> equations.
>>
>> You need to vary the Lagrangian not only with respect to the metric
>> components g_ij but the _derivatives_ of the metric components. Which
>> you aren't doing.
>
>Oh, this is a classic. I promise you. If you ever were to learn the
>calculus of variations, your statement will even disgust yourself.
><shrug>
I notice that you don't disagree.
>
>> > H @R/@[g]^ij sqrt(- det[g]) = (H R + rho c^2) @det[g]/@[g]^ij / sqrt(-
>> > det[g]) / 2
>>
>> Wrong, chuckles. Remember the other half of the Euler-Lagrange
>> equations you consistently told me to learn years after I did? They
>> involve _derivatives of the metric_. Any particular reason you feel
>> the metric derivatives don't count?
>
>As I said, you still don’t get it. You only know the calculus of
>variation in skin depth. I have been trying to tell you all all along
>that the derivation of the field equations does not involve any Euler-
>Lagrange equations. <shrug>
Then why are you writing Lagrangians and poorly trying to invoke the
calculus of variations?
The Lagrangian is explicit, as is the variation since it is a textbook
exercise anyone who professes knowledge of the subject should be able
to do. Is there a particular reason you don't understand this?
>
>> > Well,
>>
>> > ** @R/@[g]^ij = [R]_ij
>>
>> This isn't even right.
>>
>> Do you actually think there is no functional dependence on the metric
>> past that?
>
>[R]_ij is a function of the metric [g]. So, you are just whining
>about things you do not understand as well. <shrug>
So why did you not take the derivative of R_ij with respect to the
metric?
The argument is largely irrelevant because you screwed up the
derivation totally, but watching you flail because you can't do simple
derivatives is rather amusing.
>
>> > ** @det[g]/@[g]^ij = - det[g] / [g]^ij
>>
>> Seriously? You actually think that?
>
>Yes. <shrug>
>
>> You can't divide by a tensor. Try again.
>
>Try again. [g]^ij is an element (a scalar) to the matrix of the
>inverse of the matrix [g] the metric. <shrug>
Always delightful to watch you argue something that'd fail a regular
person from the relevant course.
The division operation does not exist. Period. That you wish it does
is irrelevant to the mathematics as they actually exist. Arguing why
the operation does not exist or once again pointing out your
misunderstandings of tensors and matrices would only be useful were
you capable of learning.
>
>> [...]
>
>I have had enough of the whining crap from the multi-year super-senior
>without any college degree. <shrug>
There is no way you act like this in person. Anyone who is this
disgusting would get popped in the face real fucking quick.
Would you like to see how to _actually_ derive the field equations
from the Lagrangian, or are you allergic to learning?
Androcles
"eric gisse" <jowr.pi...@gmail.com> wrote in message
news:vvo9g4tt1s6m4m8kf...@4ax.com...
> On Sat, 25 Oct 2008 22:51:00 -0700 (PDT), Koobee Wublee
> <koobee...@gmail.com> wrote:
>
>>On Oct 24, 5:51 pm, Eric Gisse wrote:
>>> On Oct 23, 9:19 pm, Koobee Wublee wrote:
>>
>>> > The Riemann tensor was derived by Ricci and named after Riemann.
>>>
>>> > The Ricci tensor was derived by Ricci's student Levi-Civita and named
>>> > after Ricci.
>>>
>>> > The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
>>> > had since then championed to be named after their messiah Einstein,
>>> > the nitwit, the plagiarist, and the liar. <shrug>
>>>
>>> [...]
>>
>>More whining crap from a multi-year super-senior without a college
>>degree is mercifully snipped. I guess with mommy's financial support,
>>you can sleep well pass noon every single day. <shrug>
>
> The standard refrain for when you can't endure the criticism or answer
> any of the questions. Why does someone with no shame have such a thin
> skin?
New name not accepted, you are still an ignorant bigot and a cunt and
always will be.
*plonk*
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 24, 7:52 am, Edward Green wrote:
>> On Oct 24, 1:19 am, Koobee Wublee wrote:
>
>> > The Riemann tensor was derived by Ricci and named after Riemann.
>>
>> > The Ricci tensor was derived by Ricci’s student Levi-Civita and named
>> > after Ricci.
>>
>> > The Einstein tensor was fudged by Hilbert. The Einstein Dingleberries
>> > had since then championed to be named after their messiah Einstein,
>> > the nitwit, the plagiarist, and the liar. <shrug>
>>
>> Look at this from my point of view: I don't know how much you
>> understand because I understand so little.
>
>Hey, I was once in your shoes. I did not understand but tried and
>still could not understand until concentrated from the pure
>mathematical points of view. Mathematics is the truth while the
>interpretations are mostly bullsh*t. Then, I understand it crystal-
>clear. <shrug>
Really, where did you learn the truth?
If you aren't interested in sharing the truth, then by all means shut
the fuck up and go away. There's no point in blubbering about how you
know the truth when you refuse to give a source that isn't in your
control.
>
>> Now, I realize perhaps you don't care,
>
>You are very correct. Whether you understand it or not has no impact
>on my life. <shrug> No hard feelings.
>
>> but which way will your sobriquet "the nitwit, the
>> plagiarist, and the liar" likely drive my Bayesian priors?
>
><shrug> There is nothing that can be attributed to Einstein as
>original. Einstein was a nitwit, a plagiarist, and a liar. <shrug>
>This is the truth if you still further in either history and the
>relevant mathematics. The forensic evidence in history lies in the
>mathematics itself ironically.
The mathematics that you do not understand.
You still don't understand how to compute volume of surface area from
the metric.
>
>> You may
>> not care, but I am unlikely to be the only one affected this way.
>
><shrug>
>
>> > You need to understand how the Einstein tensor is derived to fully
>> > understand that. <shrug>
>>
>> Possibly. The Einstein equation cries out for direct apprehension,
>
>This is not true. The Einstein field equations are merely the
>differential equations that allow you to solve for the metric and
>identify what the metric is. The field equations are cooked up in the
>pot of alchemy. It is a pure mathematical madness. <shrug>
Your inability to understand does't make it 'madness'.
Again, would you like to go through the derivation of the field
equations from the Einstein-Hilbert action step by step?
>
>> however: I'm not sure this requires us to repeat the
>> "derivation" (heuristic derivation).
>
>Hmmm... The best way to understand something is to go to where the
>beginning is. <shrug>
>
>> The equation says that some
>> geometric property of spacetime, expressed as a sum of two parts, is
>> equal to some matter terms.
>
>Yes, it can be interpreted that way. <shrug>
>
>> I have some idea about the matter terms.
>
>Hmmm... Matter means mass. <shrug> If you don’t have any idea on
>what the matter term is, you will be called an idiot. <shrug>
Since when does matter mean mass, persistent idiot? Where's the mass
in the source-free Maxwell equations?
>
>> I have little apprehension of the geometric terms.
>
>The keystone to understand this subject in differential geometry is
Funny how you keep talking about the subject like you understand it,
but refuse to give either literature references or even simple
calculations.
>the set of the Christoffel symbols of the second kind. In your
>textbooks, they just magically appear. However, if you understand how
Name one textbook where they magically appear and are never derived.
>Christoffel derived the geodesic equations, there is no more
>mysticism. Again, the geodesic equations are Euler-Lagrange equations
>with the Lagrangian
>
>L = g_ij dq^i/ds dq^j/ds = 1
>
>Derived from,
>
>ds^2 = g_ij dq^i dq^j
>
>You just cannot get any simpler than that. <shrug>
Now why is the Lagrangian equal to 1? What assumptions are required to
make that particular statement?
>
>> Since I am led to believe the Riemann tensor, and hence, of course,
>> the Ricci tensor and the scalar curvature, can be derived from he
>> metric,
>
>Hmmm... The Riemann tensor, the Ricci tensor, and the Ricci scalar
>are all functions of the metric [g] the matrix with elements g_ij or
>[g]_ij (my nomenclature).
Nice to see you still don't know the difference between a matrix
(ordered collection of numbers) and a tensor (ordered collection of
numbers which obeys certain properties under coordinate
transformations).
>
>> I wonder if the expression is not more complicated than need
>> be. It could be written as some operator applied to the metric:
>>
>> E(g) = T
>
>I don’t know what your confusion is. <shrug>
That's because you don't understand general relativity at any level.
>
>> > Any geometric insight is merely an interpretation to the mathematics
>> > involved.
>>
>> Right. Well, suppose we tried to write the equation in a local Lorentz
>> frame, which we are guaranteed to be able to do. Then we have a
>> direct component wise decomposition/interpretation: the ij_th
>> component of the Ricci tensor - 1/2 the ij_th component of the metric
>> times the scalar curvature = the flow of i-momentum in the j-direction
>> (or vice verse -- we can be sloppy because of symmetry).
>>
>> We ought to be able to develop a direct appreciation of _that_.
>> _What_ local stretching of spacetime is equal to the ij_th momentum
>> flux. Whatever it is, what could be simpler? At this level, the
>> relation is completely linear.
>
>You can interpret any piece of mathematical equations anyway you
>choose it to be. You can keep it as simple and as mundane as
>possible, or you can do so with fairies, unicorns, elves, dwarves,
>etc. <shrug> The bottom line is that the Holy Grail of GR is the set
>of the field equations which are merely ordinary differential
>equations when properly solved would either the metric.
My, you don't even know the difference between ordinary and partial
differential equations. Isn't that special?
>
>> > For those who are still struggling to understand the Einstein tensor,
>> > here is a derivation. Please pay attention. It is actually very
>> > simple.
>>
>> > Hilbert pulled out the following out of nowhere which is now called
>> > the Lagrangian to the Einstein-Hilbert action. It does not even
>> > satisfy as a Langrangian according to the Lagrangian methods.
>>
>> > L = (H R + rho c^2) sqrt(- det[g])
>>
>> > Where
>>
>> > ** H = Constant
>> > ** R = [g]^ij [R]_ij = Ricci Scalar
>>
>> hmm... is that true? Interesting.
>
>Yes. (rho c^2) represent the “Lagrangian of mass”.
>
>Why Hilbert came up with that is still a mystery, and the academics
>tried to keep a very low profile of that because of obvious reasons.
><shrug>
Simple answer to a stupid question: Hilbert didn't come pu with it,
you pulled it out of your ass and claimed Hilbert 'meant' to come up
with it.
[...]
>
>GR is purely a mathematical nonsense. <shrug>
Only the way you do it, as you refuse to listen to those better
educated than you.
carlip...@physics.ucdavis.edu
> I feel it is time to grok the Einstein equation.
> We only have to do one simple thing: understand the significance of
> the Ricci tensor minus 1/2 the metric tensor times the scalar
> curvature. This of course is interpreted as the... ?
Try http://arxiv.org/pdf/gr-qc/0103044v5 and
http://arxiv.org/abs/gr-qc/0203100.
Steve Carlip
eric gisse
> On Oct 24, 3:26 pm, Eric Gisse wrote:
>
> > On Oct 23, 7:09 pm, Edward Green wrote:
> > > Then how is is at least part of this tensor consists of the metric
> > > multiplied by a contraction of RIEMANN (yeah, I know RICCI is also a
> > > contraction of RIEMANN, to a lesser degree). Is that obviously
> > > something involving second derivatives of the metric?
>
> > The Einstein tensor is simply a function of Riemann.
>
> > R_uv - 1/2 R g_uv
>
> > Both the Ricci scalar and the Ricci tensor are contractions of
> > Riemann.
>
> Yeah, the original poster knows that. <shrug> The question is why?
> Just tell him that Levi-Civita chose to do so. Given the Riemann
> curvature tensor R^n_ijk, why did Levi-Civita define the Ricci tensor
> to be the following?
>
> R_ij = R^j_ijk
Try again. That evaluates to R_ik. Funny how you fuck up every
calculation, no matter how simple.
>
> And not
>
> R_ij = R^k_ijk
...the one you got right...
>
> Or
>
> R_ij = R^i_ijk
Try again. That evaluates to R_jk. Funny how you fuck up every
calculation, no matter how simple.
The symmetries of Riemann makes the choice of contraction irrelevant,
so I'm not sure why you are making a big deal out of it. Proving they
are the same is a simple exercise - why don't you show me where you
are having difficulties?
>
> > The information encoded in the field equations is enough to
> > completely determine Riemann, which satisfies your previous request.
>
> Get real. Given the Ricci tensor R_ij, you cannot recover the Riemann
> tensor R^n_ijk.
Did I _say_ the words "Ricci tensor" ? No? Then why the fuck are you
beating on a strawman?
>
> > > I was hoping for some geometric insight into the terms, as they
> > > directly relate the local geometry of spacetime to momentum and
> > > energy.
>
> > The Einstein tensor is also automatically conserved to preserve the
> > Bianchi identities, which are statements about Riemann.
>
> There is no big deal about the Bianchi identity. You are spreading
> mysticism in differential geometry. <shrug>
Its' just symmetries of the Riemann tensor, why are you whining about
'mysticism' ?
Edward Green
wrote:
> Edward Green says...
>
> >Anyway, I was just hoping remotely it might be possible to short-
> >circuit the process in this case, to see, beyond all the intermediate
> >connections, covariant derivatives, fiber bundles, and what have you,
> >just what geometric property of spacetime is encoded by the stress
> >energy tensor, with a kind of "Oh... of course!".
>
> Well, John Baez gave the following formulation of Einstein's
> field equations in ordinary English (valid for situations in
> which the pressure is isotropic; the same in all directions):
>
> Given a small ball of freely falling test particles initially
> at rest with respect to each other, the rate at which it begins
> to shrink is proportional to its volume times: the energy
> density at the center of the ball plus three times the pressure
> at that point.
Interesting.... I guess that is the "GR" effect of pressure, to which
would presumably be added the ordinary effect of pressure, causing
clouds of particles to expand. Or these "test particles"... well,
heck, you said they were initially at rest wrt one another, that rules
out their having an intrinsic pressure. Or maybe I'm thinking of the
ordinary effect of pressure gradient: a pressure without a gradient
hardly causes things to expand, as the contents of any intact gas
bottle illustrate.
I wasn't thinking so far as test particles -- maybe I should have
been. But what I was thinking of was the geometry of spacetime. The
more you look at the field equation, the more exceedingly simple the
relation of momentum/energy flux to this geometry looks. If one
immediately apprehended the Ricci tensor and the product of the metric
tensor with the scalar curvature, then one would pretty much
immediately apprehend the relation (assuming one "got" T).
Maybe the key is to retreat to our old idea of tensors, and think of
both T and E as things with slots, and ask just what their output is:
I think if you feed T two tangent vectors, for example, you will get
the mass-energy flux in a particular direction across a particular
normal surface (where the "surface" may have a time like normal).
Now, you plug that into g_mn, and you get the inner product of the
flux direction and the surface normal, times a factor of 1/2 the
scalar curvature, which you subtract from... from what? How do we
interpret the result of feeding RICCI two tangent vectors? Something
to do with spacetime volumes...
Almost there (he naively thinks).
So close: the flux of m-momentum across an n-surface (T) is equal to
the (something something something) in the m and n directions, of the
local geometry of spacetime. The "non-linear" equations are perfectly
linear, in this sense, at a point. Or is that an illusion, from
mixing the metric tensor and byproducts of RIEMANN in the same
equation (including a product of the metric tensor and a byproduct of
RIEMANN).
I wonder if we can always pick a local coordinate system which
diagonalizes E = T. That would be a _big_ help! Then the m-th
component of "4-pressure" is equal to... the m-th component of some
extremely simplified (yet without loss of generality) local summary of
the spacetime geometry.
I want to work out my own idiot's guide before consulting Baez. Then
I'll see how close he got. :-)
Koobee Wublee
> John Baez' discussion of the intuitive meaning of GR is given here:
>
> http://math.ucr.edu/home/baez/einstein/node3.html
Hmmm... How do you like his explanation of the Noether’s theorem. I
hate to tell you that Professor Baez exposes himself to be the biggest
nitwit through his promotion of GR. I thought Professor Carlip has at
least some intelligent, but how can Professor Carlip ever graduate
Professor Baez? I wonder if that claim of direct student-teacher
lineage to Gauss helps his cause. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >More whining crap from a multi-year super-senior without a college
> >degree is mercifully snipped. I guess with mommy’s financial support,
> >you can sleep well pass noon every single day. <shrug>
>
> The standard refrain for when you can't endure the criticism or answer
> any of the questions. Why does someone with no shame have such a thin
> skin?
Talking nonsense again, eh? Where are you now? Back with mommy?
With lowercase associated with your name, your mommy really makes you
look small. <shrug>
>
>
>
> >> > Hilbert pulled out the following out of nowhere which is now called
> >> > the Lagrangian to the Einstein-Hilbert action. It does not even
> >> > satisfy as a Langrangian according to the Lagrangian methods.
>
> >> > L = (H R + rho c^2) sqrt(- det[g])
>
> >> No, *YOU* pulled it out of nowhere. The actual Einstein-Hilbert
> >> Lagrangian is R sqrt(-|g|).
>
> > Yes, but Hilbert [‘s L is] really ... what I [meant to write] above.
> > With just (R sqrt(- deg[g]), you don’t get the energy momentum
> > tensor. <shrug>
>
> How do you know Hilbert "really meant" that when the inclusion of a
> matter term is fucking trivial and _not done that way_? Please answer
> in the form of an insult which poorly hides your stupidity.
This is a simple forensic mathematical study. The field equations are
the evidence. You are indeed shallow-minded. <shrug>
> >You have to understand the calculus of variations first. <shrug>
>
> I already do - and far better from you, judging from your responses
> below.
It does not show. <shrug>
> >Oh, this is a classic. I promise you. If you ever were to learn the
> >calculus of variations, your statement will even disgust yourself.
> ><shrug>
>
> I notice that you don't disagree.
You read it wrong. I have strongly disagreed that the field equations
are the results of the calculus of variations. <shrug>
> >As I said, you still don’t get it. You only know the calculus of
> >variation in skin depth. I have been trying to tell you all all along
> >that the derivation of the field equations does not involve any Euler-
> >Lagrange equations. <shrug>
>
> Then why are you writing Lagrangians and poorly trying to invoke the
> calculus of variations?
Because that is what you misunderstand it to be. <shrug>
> The Lagrangian is explicit, as is the variation since it is a textbook
> exercise anyone who professes knowledge of the subject should be able
> to do. Is there a particular reason you don't understand this?
Then, why are you challenging me that Hilbert’s Lagrangian is indeed a
Lagrangian?
> >[R]_ij is a function of the metric [g]. So, you are just whining
> >about things you do not understand as well. <shrug>
>
> So why did you not take the derivative of R_ij with respect to the
> metric?
No, not the metric but each element of the inverse of the metric.
<shrug>
> The argument is largely irrelevant because you screwed up the
> derivation totally, but watching you flail because you can't do simple
> derivatives is rather amusing.
It is not irrelevant but your lack of understanding. <shrug>
> >Try again. [g]^ij is an element (a scalar) to the matrix of the
> >inverse of the matrix [g] the metric. <shrug>
>
> Always delightful to watch you argue something that'd fail a regular
> person from the relevant course.
>
> The division operation does not exist. Period. [...]
Says who? It actually goes back to the Einstein-Hilbert action
concerning with the integral of (dL dq^0 dq^1 dq^2 dq^3) being zero.
If dL is already zero, the associated field equation does not exist.
<shrug>
> [snipped more whining crap]
Koobee Wublee
> Koobee Wublee wrote:
> >Hey, I was once in your shoes. I did not understand but tried and
> >still could not understand until concentrated from the pure
> >mathematical points of view. Mathematics is the truth while the
> >interpretations are mostly bullsh*t. Then, I understand it crystal-
> >clear. <shrug>
>
> Really, where did you learn the truth?
Does it concern you?
> If you aren't interested in sharing the truth, then by all means shut
> the fuck up and go away. There's no point in blubbering about how you
> know the truth when you refuse to give a source that isn't in your
> control.
Where I learn it is not important, multi-year super-senior without a
college degree. <shrug>
> ><shrug> There is nothing that can be attributed to Einstein as
> >original. Einstein was a nitwit, a plagiarist, and a liar. <shrug>
> >This is the truth if you still further in either history and the
> >relevant mathematics. The forensic evidence in history lies in the
> >mathematics itself ironically.
>
> The mathematics that you do not understand.
As judged from a multi-year super-senior without a college degree and
who also does not understand the calculus of variations. <shrug>
> You still don't understand how to compute volume of surface area from
> the metric.
Your perception is badly flawed since you cannot even recognize the
answer. <shrug>
> >This is not true. The Einstein field equations are merely the
> >differential equations that allow you to solve for the metric and
> >identify what the metric is. The field equations are cooked up in the
> >pot of alchemy. It is a pure mathematical madness. <shrug>
>
> Your inability to understand does't make it 'madness'.
The argument with you is always centered around your lack of
understanding that you erroneously accused me of not to. It is really
getting old. <shrug>
> Again, would you like to go through the derivation of the field
> equations from the Einstein-Hilbert action step by step?
Well, allow me to call your bluff. Yes.
> >Hmmm... Matter means mass. <shrug> If you don’t have any idea on
> >what the matter term is, you will be called an idiot. <shrug>
>
> Since when does matter mean mass, persistent idiot? Where's the mass
> in the source-free Maxwell equations?
So, when does matter not mean mass even in Maxwell’s equations?
> >The keystone to understand this subject in differential geometry is
>
> Funny how you keep talking about the subject like you understand it,
> but refuse to give either literature references or even simple
> calculations.
More yapping from gisse who all of a sudden chooses to be more obscure
by using stupid lowercase as his name. <shrug>
> >the set of the Christoffel symbols of the second kind. In your
> >textbooks, they just magically appear. However, if you understand how
>
> Name one textbook where they magically appear and are never derived.
You tell me. You are the one who sits on piles of textbooks. <shrug>
> >Christoffel derived the geodesic equations, there is no more
> >mysticism. Again, the geodesic equations are Euler-Lagrange equations
> >with the Lagrangian
>
> >L = g_ij dq^i/ds dq^j/ds = 1
>
> >Derived from,
>
> >ds^2 = g_ij dq^i dq^j
>
> >You just cannot get any simpler than that. <shrug>
>
> Now why is the Lagrangian equal to 1?
You are really stupid. <shrug>
> What assumptions are required to
> make that particular statement?
None. Just 9’th grade algebra. <shrug>
> >Hmmm... The Riemann tensor, the Ricci tensor, and the Ricci scalar
> >are all functions of the metric [g] the matrix with elements g_ij or
> >[g]_ij (my nomenclature).
>
> Nice to see you still don't know the difference between a matrix
> (ordered collection of numbers) and a tensor (ordered collection of
> numbers which obeys certain properties under coordinate
> transformations).
Would you ever understand what a tensor really is as meant by the
founding fathers of differential geometry?
> >I don’t know what your confusion is. <shrug>
>
> That's because you don't understand general relativity at any level.
Cheap shot is always ignored. <shrug>
> >You can interpret any piece of mathematical equations anyway you
> >choose it to be. You can keep it as simple and as mundane as
> >possible, or you can do so with fairies, unicorns, elves, dwarves,
> >etc. <shrug> The bottom line is that the Holy Grail of GR is the set
> >of the field equations which are merely ordinary differential
> >equations when properly solved would either the metric.
>
> My, you don't even know the difference between ordinary and partial
> differential equations. Isn't that special?
Hmmm... You still do not understand the subtlety. <shrug>
> >Yes. (rho c^2) represent the “Lagrangian of mass”.
>
> >Why Hilbert came up with that is still a mystery, and the academics
> >tried to keep a very low profile of that because of obvious reasons.
> ><shrug>
>
> Simple answer to a stupid question: Hilbert didn't come pu with it,
> you pulled it out of your ass and claimed Hilbert 'meant' to come up
> with it.
That is because you don’t understand how the field equations are
actually derived. <shrug>
> >GR is purely a mathematical nonsense. <shrug>
>
> Only the way you do it, as you refuse to listen to those better
> educated than you.
Since there is no other way, I am very safe saying that “GR is purely
a mathematical nonsense”. <shrug>
eric gisse
<koobee...@gmail.com> wrote:
[...]
>You read it wrong. I have strongly disagreed that the field equations
>are the results of the calculus of variations. <shrug>
Since the derivation is a textbook exercise, would you like to go
through it?
Koobee Wublee
> On Oct 25, 9:36 pm, Koobee Wublee wrote:
> > Yeah, the original poster knows that. <shrug> The question is why?
> > Just tell him that Levi-Civita chose to do so. Given the Riemann
> > curvature tensor R^n_ijk, why did Levi-Civita define the Ricci tensor
> > to be the following?
>
> > R_ij = R^j_ijk
>
> Try again. That evaluates to R_ik. Funny how you fuck up every
> calculation, no matter how simple.
>
> > And not
>
> > R_ij = R^k_ijk
>
> ...the one you got right...
>
> > Or
>
> > R_ij = R^i_ijk
>
> Try again. That evaluates to R_jk. Funny how you fuck up every
> calculation, no matter how simple.
So, that was just a typo on my part. Allow me to correct that.
** R1_ij = R^k_ijk
** R2_jk = R^i_ijk
** R3_ik = R^j_ijk
Where
** R^n_ijk = Riemann curvature tensor
R1_ij, R2_jk, and R3_ik are all different. However, they can all be
written in R1_ij, R2_ij, R3_ij respectively with new organizations of
indices. <shrug>
> The symmetries of Riemann makes the choice of contraction irrelevant,
> so I'm not sure why you are making a big deal out of it. Proving they
> are the same is a simple exercise - why don't you show me where you
> are having difficulties?
Nonsense.
R^n_ijk = - R^n_ikj
Where
** R^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_km C^m_ij – C^n_jm
C^m_ik
** C^n_ij = Christoffel symbols of the 2nd kind
> > Get real. Given the Ricci tensor R_ij, you cannot recover the Riemann
> > tensor R^n_ijk.
>
> Did I _say_ the words "Ricci tensor" ? No? Then why the fuck are you
> beating on a strawman?
You said “the field equations” which are basically the Ricci tensor.
<shrug>
> > There is no big deal about the Bianchi identity. You are spreading
> > mysticism in differential geometry. <shrug>
>
> Its' just symmetries of the Riemann tensor, why are you whining about
> 'mysticism' ?
The symmetry is subtle. Since you have brought it up, Ricci while
making his Riemann curvature tensor can easily settle on another
version by grouping the connection coefficients differently in the
geodesic variation equations such as:
S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
In doing so, the field equations would look very different if Ricci
chose S^n_ijk as the Riemann curvature instead of R^n_ijk.
Curvature of spacetime is all man-made. Man is not God. <shrug>
Koobee Wublee
> On Oct 26, 11:52 am, Daryl McCullough wrote:
> > Given a small ball of freely falling test particles initially
> > at rest with respect to each other, the rate at which it begins
> > to shrink is proportional to its volume times: the energy
> > density at the center of the ball plus three times the pressure
> > at that point.
Crackpot sighted.
> Maybe the key is to retreat to our old idea of tensors, and think of
> both T and E as things with slots, and ask just what their output is:
> I think if you feed T two tangent vectors, for example, you will get
> the mass-energy flux in a particular direction across a particular
> normal surface (where the "surface" may have a time like normal).
> Now, you plug that into g_mn, and you get the inner product of the
> flux direction and the surface normal, times a factor of 1/2 the
> scalar curvature, which you subtract from... from what? How do we
> interpret the result of feeding RICCI two tangent vectors? Something
> to do with spacetime volumes...
The key to the curvature business is the arrangement of the connection
coefficients. One such arrangement results in the most symmetric
rank-3 tensor which are the Christoffel symbols of the 2nd kind.
Those who cannot understand the mathematics of the derivations of the
geodesic equations came up with the nonsense of teleporting, parallel
porting, tangent whatever. It is all a smoke screen to hide their
lack of understanding in mathematics. <shrug>
Think rationally for a moment. Is there any possible way to derive
such complicated Christoffel symbols through vertical, parallel,
tangential porting of a vector?
GR is mysticism where Orwellian education flourishes:
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
** IGNORANCE IS TRUTH
** STUPIDITY IS INTELLECT
<shrug>
> Almost there (he naively thinks).
Yes, very naïve indeed.
> So close: the flux of m-momentum across an n-surface (T) is equal to
> the (something something something) in the m and n directions, of the
> local geometry of spacetime. The "non-linear" equations are perfectly
> linear, in this sense, at a point. Or is that an illusion, from
> mixing the metric tensor and byproducts of RIEMANN in the same
> equation (including a product of the metric tensor and a byproduct of
> RIEMANN).
>
> I wonder if we can always pick a local coordinate system which
> diagonalizes E = T. That would be a _big_ help! Then the m-th
> component of "4-pressure" is equal to... the m-th component of some
> extremely simplified (yet without loss of generality) local summary of
> the spacetime geometry.
Now, show me how you obtain the Christoffel symbols of the 2nd kind
from all that.
> I want to work out my own idiot's guide before consulting Baez. Then
> I'll see how close he got. :-)
You still would not find how the Christoffel symbols of the 2nd kind
are derived in there. <shrug>
> > This formulation uses a very special coordinate system:
> > freefalling coordinates in which all particles are initially
> > at rest.
After Einstein the nitwit, the plagiarist, and the liar finally
understood the “very difficult concept” of Newtonian law of gravity,
he re-discovered the principle of equivalence. Of course, the nitwit,
the plagiarist, and the liar did not realize Galileo had already done
just that 400 years ago. Through the equivalence principle, Newton
was able to come up with the law of gravity.
Instead of observing how an object behaves in gravity just like Newton
saw that falling apple, the nitwit, the plagiarist, and the liar
decided to picture himself as that falling apple. Unable to do the
math, Einstein enlisted his classmate Grossmann’s help. The “Entwurf”
theory was heavily centered around the mathematics of coordinate
transformation. It got nowhere. Disgusted, Einstein the nitwit, the
plagiarist, and the liar thus dumped Grossmann and went to Goettingen
to solicit Hilbert’s help. The field equations exhibit no such
fruitless effort in coordinate transformation.
> > The business about Riemann tensors and stress-energy
> > tensors is the way to state things in a coordinate-independent
> > way.
Nonsense. Since it is impossible to write the Riemann tensor without
using a particular set of coordinate system, Riemann tensor must be
coordinate dependent while the geometry remains independent of any
coordinate system. <shrug>
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 26, 3:04 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> >Hey, I was once in your shoes. I did not understand but tried and
>> >still could not understand until concentrated from the pure
>> >mathematical points of view. Mathematics is the truth while the
>> >interpretations are mostly bullsh*t. Then, I understand it crystal-
>> >clear. <shrug>
>>
>> Really, where did you learn the truth?
>
>Does it concern you?
If you are dishonest, refusing to disclose your sources is an
excellent way to hide your ignorance.
>
>> If you aren't interested in sharing the truth, then by all means shut
>> the fuck up and go away. There's no point in blubbering about how you
>> know the truth when you refuse to give a source that isn't in your
>> control.
>
>Where I learn it is not important, multi-year super-senior without a
>college degree. <shrug>
Do you have that written down in your copy-paste text file, or is it
reflex at this point?
>
>> ><shrug> There is nothing that can be attributed to Einstein as
>> >original. Einstein was a nitwit, a plagiarist, and a liar. <shrug>
>> >This is the truth if you still further in either history and the
>> >relevant mathematics. The forensic evidence in history lies in the
>> >mathematics itself ironically.
>>
>> The mathematics that you do not understand.
>
>As judged from a multi-year super-senior without a college degree and
>who also does not understand the calculus of variations. <shrug>
Your basis for judgement is what, again?
[...]
>
>> Again, would you like to go through the derivation of the field
>> equations from the Einstein-Hilbert action step by step?
>
>Well, allow me to call your bluff. Yes.
This will be done my way, as it'll take most of an hour to do in terms
of typing and writing, and I anticipate you will get mad at the
mathematics ten lines in and refuse to discuss the rest.
The Einstein-Hilbert action S is equal to int( R sqrt(-g) d^4x).
Recall that the Lagrange equations are obtained by varying the action
with respect to the field variables, and demanding that the variation
is equal to zero at the boundary points.
So the variation considered will be in the field variables g_ij in the
form g^ij --> g^ij + delta(g^ij).
S = int( R_ij g^ij sqrt(-g) d^4x).
delta(S) = delta( int( R_ij g^ij sqrt(-g) d^4x) ) +
The variation seems to behave like a differential operator, so the
terms are expanded as such.
delta(S) = int( delta(R_ij) g^ij sqrt(-g) d^4x) + int( R_ij
delta(g^ij) sqrt(-g) d^4x) + int( R_ij g^ij delta(sqrt(-g)) d^4x)
delta(S) = A + B + C
The B term is already varied with respect to the field variables, so
that chunk of the variation is done.
The C term will now be considered. Normally I would just march through
term by term, but I have my doubts that you'll accept the derivation.
As before, the goal is to get the entire variation in terms of the
varied metric, as opposed to varied functions of the metric.
int( R_ij g^ij delta(sqrt(-g)) d^4x)
There is a more rigorous way of doing this, but since the result is
the same I won't bother. So if I continue to treat it with the rules
of ordinary calculus, the variation ends up like this:
delta(sqrt(-g)) = - delta(g) / ( 2 sqrt(-g) )
delta(sqrt(-g)) = g_ij g delta(g^ij) / ( 2 sqrt(-g) )
delta(sqrt(-g)) = - g_ij sqrt(-g) delta(g^ij) / 2
Now if you look what the delta(sqrt(-g)) term is attached to, you'll
notice a -Rg_ij /2 term. That's half the field equations right there.
Do you wish to argue about any of this, or do you accept the results
so far?
>
>> >Hmmm... Matter means mass. <shrug> If you don’t have any idea on
>> >what the matter term is, you will be called an idiot. <shrug>
>>
>> Since when does matter mean mass, persistent idiot? Where's the mass
>> in the source-free Maxwell equations?
>
>So, when does matter not mean mass even in Maxwell’s equations?
Matter is an umbrella term that includes mass, electromagnetic energy,
or anything else.
The source free Maxwells equations for j^a = (0,0,0,0) have nontrivial
solutions with nonzero energies with no matter sources present. No
mass, but there's still energy.
>
>> >The keystone to understand this subject in differential geometry is
>>
>> Funny how you keep talking about the subject like you understand it,
>> but refuse to give either literature references or even simple
>> calculations.
>
>More yapping from gisse who all of a sudden chooses to be more obscure
>by using stupid lowercase as his name. <shrug>
How many WORDS have you written about an utterly trivial change?
Different newsreader because google groups was being an epic piece of
shit, forgot to capitalize. I see the place to change it not 6" from
these words but I'm not going to because it obviously annoys you.
Why does someone who cowardly hides behind a pseudonym get so butthurt
about a different presentation of a real name?
>
>> >the set of the Christoffel symbols of the second kind. In your
>> >textbooks, they just magically appear. However, if you understand how
>>
>> Name one textbook where they magically appear and are never derived.
>
>You tell me. You are the one who sits on piles of textbooks. <shrug>
Either you know or you don't, and are just making it up. Which is it?
>
>> >Christoffel derived the geodesic equations, there is no more
>> >mysticism. Again, the geodesic equations are Euler-Lagrange equations
>> >with the Lagrangian
>>
>> >L = g_ij dq^i/ds dq^j/ds = 1
>>
>> >Derived from,
>>
>> >ds^2 = g_ij dq^i dq^j
>>
>> >You just cannot get any simpler than that. <shrug>
>>
>> Now why is the Lagrangian equal to 1?
>
>You are really stupid. <shrug>
I want to see if you can explain. Can you explain, or is this just
another fact you memorized without context or understanding?
How about if I said the effective Lagrangian was equal to -1, would I
be wrong?
>
>> What assumptions are required to
>> make that particular statement?
>
>None. Just 9’th grade algebra. <shrug>
Show the algebra then.
>
>> >Hmmm... The Riemann tensor, the Ricci tensor, and the Ricci scalar
>> >are all functions of the metric [g] the matrix with elements g_ij or
>> >[g]_ij (my nomenclature).
>>
>> Nice to see you still don't know the difference between a matrix
>> (ordered collection of numbers) and a tensor (ordered collection of
>> numbers which obeys certain properties under coordinate
>> transformations).
>
>Would you ever understand what a tensor really is as meant by the
>founding fathers of differential geometry?
How can we possibly know when you refuse to give a literature
reference?
Why don't you show me how the 'founding fathers of differential
geometry' defined a tensor?
By the way, since you insist that the metric g_ij is a matrix why
don't you explain why you are trying to divide by them even though the
division operation does not exist for matrices.
>
>> >I don’t know what your confusion is. <shrug>
>>
>> That's because you don't understand general relativity at any level.
>
>Cheap shot is always ignored. <shrug>
Why? You babble incessantly about 'multi year super senior' and other
personal subjects you have no actual knowledge about.
You still don't understand how the metric is relevant to computing
areas or volumes of given constructs in a manifold.
>
>> >You can interpret any piece of mathematical equations anyway you
>> >choose it to be. You can keep it as simple and as mundane as
>> >possible, or you can do so with fairies, unicorns, elves, dwarves,
>> >etc. <shrug> The bottom line is that the Holy Grail of GR is the set
>> >of the field equations which are merely ordinary differential
>> >equations when properly solved would either the metric.
>>
>> My, you don't even know the difference between ordinary and partial
>> differential equations. Isn't that special?
>
>Hmmm... You still do not understand the subtlety. <shrug>
Why don't you explain the subtlety then?
>
>> >Yes. (rho c^2) represent the “Lagrangian of mass”.
>>
>> >Why Hilbert came up with that is still a mystery, and the academics
>> >tried to keep a very low profile of that because of obvious reasons.
>> ><shrug>
>>
>> Simple answer to a stupid question: Hilbert didn't come pu with it,
>> you pulled it out of your ass and claimed Hilbert 'meant' to come up
>> with it.
>
>That is because you don’t understand how the field equations are
>actually derived. <shrug>
Why don't you give a literature reference showing how the field
equations are actually derived?
>
>> >GR is purely a mathematical nonsense. <shrug>
>>
>> Only the way you do it, as you refuse to listen to those better
>> educated than you.
>
>Since there is no other way, I am very safe saying that “GR is purely
>a mathematical nonsense”. <shrug>
How come the only people who agree with you can't even do elementary
calculus?
Edward Green
Thank you.
(I just want to express some acknowledgment, lest you think when you
send me to a dense reference and I never come back it is through
ingratitude!)
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 27, 4:04 pm, eric gisse wrote:
>> On Oct 25, 9:36 pm, Koobee Wublee wrote:
>
>> > Yeah, the original poster knows that. <shrug> The question is why?
>> > Just tell him that Levi-Civita chose to do so. Given the Riemann
>> > curvature tensor R^n_ijk, why did Levi-Civita define the Ricci tensor
>> > to be the following?
>>
>> > R_ij = R^j_ijk
>>
>> Try again. That evaluates to R_ik. Funny how you fuck up every
>> calculation, no matter how simple.
>>
>> > And not
>>
>> > R_ij = R^k_ijk
>>
>> ...the one you got right...
>>
>> > Or
>>
>> > R_ij = R^i_ijk
>>
>> Try again. That evaluates to R_jk. Funny how you fuck up every
>> calculation, no matter how simple.
>
>So, that was just a typo on my part. Allow me to correct that.
>
>** R1_ij = R^k_ijk
>
>** R2_jk = R^i_ijk
>
>** R3_ik = R^j_ijk
So nice to see you can at least get that right when I take the time to
correct it for you.
>
>Where
>
>** R^n_ijk = Riemann curvature tensor
>
>R1_ij, R2_jk, and R3_ik are all different. However, they can all be
>written in R1_ij, R2_ij, R3_ij respectively with new organizations of
>indices. <shrug>
Are they really?
Bianchi identity #1: R_a[bcd] = 0
The Riemann tensor is completely antisymmetric in the last three
indices, so it obeys the Jacobi identity for permutations.
R_abcd + R_acdb + R_adbc = 0
Raise the first index, and remember everything is equal to zero.
R^a_bcd + R^a_dbc + R^a_cdb = 0
Contract on an index - doesn't matter which - to form the Ricci
tensor.
R^z_zcd + R^z_dzc + R^z_cdz = 0
Riemann is antisymmetric in the first two indices, so R^z_zcd = 0 by
antisymmetry. So much for that choice for Ricci.
Riemann is antisymmetric in the last two indices, so R^z_cdz = -
R^z_czd. This satisfies the Bianchi identity and runs out the last two
possible choices for forming Ricci. As I said, the choice is
irrelevant.
>
>> The symmetries of Riemann makes the choice of contraction irrelevant,
>> so I'm not sure why you are making a big deal out of it. Proving they
>> are the same is a simple exercise - why don't you show me where you
>> are having difficulties?
>
>Nonsense.
>
>R^n_ijk = - R^n_ikj
>
>Where
>
>** R^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_km C^m_ij – C^n_jm
>C^m_ik
>
>** C^n_ij = Christoffel symbols of the 2nd kind
Are you reminding me or yourself?
>
>> > Get real. Given the Ricci tensor R_ij, you cannot recover the Riemann
>> > tensor R^n_ijk.
>>
>> Did I _say_ the words "Ricci tensor" ? No? Then why the fuck are you
>> beating on a strawman?
>
>You said “the field equations” which are basically the Ricci tensor.
><shrug>
Except it isn't "basically the Ricci tensor". The divergence-free
condition of the Einstein tensor encodes the contributions from the
ENTIRE Riemann tensor, not just the trace or trace-free components.
>
>> > There is no big deal about the Bianchi identity. You are spreading
>> > mysticism in differential geometry. <shrug>
>>
>> Its' just symmetries of the Riemann tensor, why are you whining about
>> 'mysticism' ?
>
>The symmetry is subtle. Since you have brought it up, Ricci while
>making his Riemann curvature tensor can easily settle on another
>version by grouping the connection coefficients differently in the
>geodesic variation equations such as:
>
>S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
>C^m_ik
Note that you have a supposed Riemann tensor that is third order in
the Christoffel symbols. Try again, or show how your definition
follows from attempting to derive the geodesic equation.
Regardless, the grouping is irrelevant as any 'different' definition
for the Riemann tensor that is defined in this way can be shown to be
equivalent because the coefficients are _always_ symmetric in their
lower indices.
>
>In doing so, the field equations would look very different if Ricci
>chose S^n_ijk as the Riemann curvature instead of R^n_ijk.
Try writing a definition for S^n_ijk that not only has the correct
amount of terms but isn't related to the other 'choices' by index
permutations.
Koobee Wublee
> Koobee Wublee wrote:
> >So, that was just a typo on my part. Allow me to correct that.
>
> >** R1_ij = R^k_ijk
>
> >** R2_jk = R^i_ijk
>
> >** R3_ik = R^j_ijk
>
> So nice to see you can at least get that right when I take the time to
> correct it for you.
Well, I only have a few minutes in the day that I can post while you
have hours with mommy’s support. <shrug>
> >Where
>
> >** R^n_ijk = Riemann curvature tensor
>
> >R1_ij, R2_jk, and R3_ik are all different. However, they can all be
> >written in R1_ij, R2_ij, R3_ij respectively with new organizations of
> >indices. <shrug>
>
> Are they really?
Yes.
> Bianchi identity #1: R_a[bcd] = 0
>
> The Riemann tensor is completely antisymmetric in the last three
> indices, so it obeys the Jacobi identity for permutations.
>
> R_abcd + R_acdb + R_adbc = 0
>
> Raise the first index, and remember everything is equal to zero.
>
> R^a_bcd + R^a_dbc + R^a_cdb = 0
>
> Contract on an index - doesn't matter which - to form the Ricci
> tensor.
>
> R^z_zcd + R^z_dzc + R^z_cdz = 0
>
> Riemann is antisymmetric in the first two indices, so R^z_zcd = 0 by
> antisymmetry. So much for that choice for Ricci.
>
> Riemann is antisymmetric in the last two indices, so R^z_cdz = -
> R^z_czd. This satisfies the Bianchi identity and runs out the last two
> possible choices for forming Ricci. As I said, the choice is
> irrelevant.
That does not refute what I wrote on the Ricci tensor above. <shrug>
> >R^n_ijk = - R^n_ikj
>
> >Where
>
> >** R^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_km C^m_ij – C^n_jm
> >C^m_ik
>
> >** C^n_ij = Christoffel symbols of the 2nd kind
>
> Are you reminding me or yourself?
You. Are you the one claiming “symmetries of Riemann makes the choice
of contraction irrelevant”?
> >You said “the field equations” which are basically the Ricci tensor.
> ><shrug>
>
> Except it isn't "basically the Ricci tensor". The divergence-free
> condition of the Einstein tensor encodes the contributions from the
> ENTIRE Riemann tensor, not just the trace or trace-free components.
Well, the only verifiable condition against the Newtonian limit is in
vacuum. Guess what? The Ricci tensor is the set of the field
equations. <shrug>
> >The symmetry is subtle. Since you have brought it up, Ricci while
> >making his Riemann curvature tensor can easily settle on another
> >version by grouping the connection coefficients differently in the
> >geodesic variation equations such as:
>
> >S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
> >C^m_ik
>
> Note that you have a supposed Riemann tensor that is third order in
> the Christoffel symbols.
No, I have not. S^n_ijk has the same order as R^n_ijk. <shrug>
> Try again, or show how your definition
> follows from attempting to derive the geodesic equation.
S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
The extra term did not appear in my original post. You dishonestly
added it in. <shrug>
> Regardless, the grouping is irrelevant as any 'different' definition
> for the Riemann tensor that is defined in this way can be shown to be
> equivalent because the coefficients are _always_ symmetric in their
> lower indices.
There is no such requirement mathematical. Maybe a consensus, but
that is not physics but legal laws. <shrug>
> >In doing so, the field equations would look very different if Ricci
> >chose S^n_ijk as the Riemann curvature instead of R^n_ijk.
>
> Try writing a definition for S^n_ijk that not only has the correct
> amount of terms but isn't related to the other 'choices' by index
> permutations.
S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
> >Curvature of spacetime is all man-made. Man is not God. <shrug>
Man is still no God. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >You read it wrong. I have strongly disagreed that the field equations
> >are the results of the calculus of variations. <shrug>
>
> Since the derivation is a textbook exercise, would you like to go
> through it?
Oh, yes. Once again, I call your bluff. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >Does it concern you?
>
> If you are dishonest, refusing to disclose your sources is an
> excellent way to hide your ignorance.
What is so dishonest in not disclosing my education background?
> >Where I learn it is not important, multi-year super-senior without a
> >college degree. <shrug>
>
> Do you have that written down in your copy-paste text file, or is it
> reflex at this point?
Actually, no, but thanks for the great suggestion.
> >As judged from a multi-year super-senior without a college degree and
> >who also does not understand the calculus of variations. <shrug>
>
> Your basis for judgement is what, again?
Your gross ignorance. Einstein was still a nitwit, a plagiarist, and
a liar. <shrug>
> >Well, allow me to call your bluff. Yes.
>
> This will be done my way, as it'll take most of an hour to do in terms
> of typing and writing, and I anticipate you will get mad at the
> mathematics ten lines in and refuse to discuss the rest.
Here we go. As long as my wife has not started yelling at me to do
the household chores before going to bed, I would love to go through
that.
> The Einstein-Hilbert action S is equal to int( R sqrt(-g) d^4x).
Correction: integral(R sqrt(- det[g]) d^4q)
> Recall that the Lagrange equations are obtained by varying the action
> with respect to the field variables, and demanding that the variation
> is equal to zero at the boundary points.
Don’t forget the change of the field variables.
> So the variation considered will be in the field variables g_ij in the
> form g^ij --> g^ij + delta(g^ij).
Or just d[g]^ij where [g]^ij = g^ij.
> S = int( R_ij g^ij sqrt(-g) d^4x).
Or just
S = integral(R sqrt(- [g]) d^4[q])
> delta(S) = delta( int( R_ij g^ij sqrt(-g) d^4x) ) +
Or
dS = integral(@(R sqrt(- [g]) d^4[q]))/@[g]^ij d[g]^ij)
> The variation seems to behave like a differential operator, so the
> terms are expanded as such.
>
> delta(S) = int( delta(R_ij) g^ij sqrt(-g) d^4x) + int( R_ij
> delta(g^ij) sqrt(-g) d^4x) + int( R_ij g^ij delta(sqrt(-g)) d^4x)
Why not
dS = integral(sqrt(- det[g]) @R/@[g]^ij d[g]^ij + R @(sqrt(- det[g]))/
@[g]^ij d[g]^ij)
> [...]
>
> As before, the goal is to get the entire variation in terms of the
> varied metric, as opposed to varied functions of the metric.
>
> int( R_ij g^ij delta(sqrt(-g)) d^4x)
>
> There is a more rigorous way of doing this, but since the result is
> the same I won't bother.
So, you can only continue by applying mysticism. <shrug>
> So if I continue to treat it with the rules
> of ordinary calculus,
The whole thing is ordinary. There are no divine signs in these
equations. <shrug>
> the variation ends up like this:
>
> delta(sqrt(-g)) = - delta(g) / ( 2 sqrt(-g) )
Since g or my [g] is a matrix, it makes no sense. However, I have
shown that it is the determinate of the matrix [g] that is applied.
Then, it looks like the following.
@sqrt(- det[g]) = - @det[g]/@[g]^ij / sqrt(- det[g]) / 2
> delta(sqrt(-g)) = g_ij g delta(g^ij) / ( 2 sqrt(-g) )
> delta(sqrt(-g)) = - g_ij sqrt(-g) delta(g^ij) / 2
These are utterly nonsense.
> Now if you look what the delta(sqrt(-g)) term is attached to, you'll
> notice a -Rg_ij /2 term. That's half the field equations right there.
No, the above equation becomes
@sqrt(- det[g]) = det[g] / [g]^ij / sqrt(- det[g]) / 2
Because
@det[g]/@[g]^ij = - det[g] / [g]^ij
> Do you wish to argue about any of this, or do you accept the results
> so far?
Just did. <shrug>
Well, why stop now? Tired of copying a textbook?
> >So, when does matter not mean mass even in Maxwell’s equations?
>
> Matter is an umbrella term that includes mass, electromagnetic energy,
> or anything else.
Says who? The committee of GR worshippers?
> The source free Maxwells equations for j^a = (0,0,0,0) have nontrivial
> solutions with nonzero energies with no matter sources present. No
> mass, but there's still energy.
You can deny energy as not mass, but mathematically energy is the same
as mass. <shrug>
> >More yapping from gisse who all of a sudden chooses to be more obscure
> >by using stupid lowercase as his name. <shrug>
>
> How many WORDS have you written about an utterly trivial change?
What trivial change?
> Different newsreader because google groups was being an epic piece of
> shit, forgot to capitalize. I see the place to change it not 6" from
> these words but I'm not going to because it obviously annoys you.
Ranting against Google is no concern of mine. <shrug>
> Why does someone who cowardly hides behind a pseudonym get so butthurt
> about a different presentation of a real name?
Here we go again. Hormone imbalance?
> >You tell me. You are the one who sits on piles of textbooks. <shrug>
>
> Either you know or you don't, and are just making it up. Which is it?
We are talking about you. Remember?
> >Christoffel derived the geodesic equations, there is no more
> >mysticism. Again, the geodesic equations are Euler-Lagrange equations
> >with the Lagrangian
>
> >L = g_ij dq^i/ds dq^j/ds = 1
>
> >Derived from,
>
> >ds^2 = g_ij dq^i dq^j
>
> I want to see if you can explain. Can you explain, or is this just
> another fact you memorized without context or understanding?
It is all in the mathematics above. It only takes a 7th grade algebra
to understand all that. <shrug>
> How about if I said the effective Lagrangian was equal to -1, would I
> be wrong?
Yes.
> >> What assumptions are required to
> >> make that particular statement?
>
> >None. Just 9’th grade algebra. <shrug>
>
> Show the algebra then.
I am not sinking to 7th grade level to show you the 7th grade algebra.
> >Would you ever understand what a tensor really is as meant by the
> >founding fathers of differential geometry?
>
> How can we possibly know when you refuse to give a literature
> reference?
>
> Why don't you show me how the 'founding fathers of differential
> geometry' defined a tensor?
Are you not sitting on piles of textbooks? Without these books, you
have no arguments. <shrug>
> By the way, since you insist that the metric g_ij is a matrix why
> don't you explain why you are trying to divide by them even though the
> division operation does not exist for matrices.
I am dividing the non-zero elements of the matrix [g]. <shrug>
> >Cheap shot is always ignored. <shrug>
>
> Why? You babble incessantly about 'multi year super senior' and other
> personal subjects you have no actual knowledge about.
Is it not true that you remain a multi-year super-senior without a
college degree?
> You still don't understand how the metric is relevant to computing
> areas or volumes of given constructs in a manifold.
I can see why. You do not even understand 7th grade algebra. <shrug>
> >Hmmm... You still do not understand the subtlety. <shrug>
>
> Why don't you explain the subtlety then?
I am not your professor. <shrug>
> >That is because you don’t understand how the field equations are
> >actually derived. <shrug>
>
> Why don't you give a literature reference showing how the field
> equations are actually derived?
I gave you the derivation already. <shrug>
> >Since there is no other way, I am very safe saying that “GR is purely
> >a mathematical nonsense”. <shrug>
>
> How come the only people who agree with you can't even do elementary
> calculus?
That is because it was judged by someone who remains a multi-year
super-senior and cannot do rudimentary 7th grade algebra. <shrug>
eric gisse
<koobee...@gmail.com> wrote:
[...]
>> Riemann is antisymmetric in the last two indices, so R^z_cdz = -
>> R^z_czd. This satisfies the Bianchi identity and runs out the last two
>> possible choices for forming Ricci. As I said, the choice is
>> irrelevant.
>
>That does not refute what I wrote on the Ricci tensor above. <shrug>
Of course it does.
You only have three choices for contracting on R^a_bcd, and I have
shown that one is equal to zero and the other two are the same up to a
minus sign which has no effect on the field equations.
Now tell me again how the choice is relevant when you are choosing
between "obviously zero" and "same up to a sign" when the choice has
no relevance for dynamics.
Do disagree on any particular point, or are you just unwilling to
admit you are wrong?
[...]
>
>> >You said “the field equations” which are basically the Ricci tensor.
>> ><shrug>
>>
>> Except it isn't "basically the Ricci tensor". The divergence-free
>> condition of the Einstein tensor encodes the contributions from the
>> ENTIRE Riemann tensor, not just the trace or trace-free components.
>
>Well, the only verifiable condition against the Newtonian limit is in
>vacuum. Guess what? The Ricci tensor is the set of the field
>equations. <shrug>
You have no argument, as you simply repeated your claim and tossed in
a non-sequitur for good measure. The _set_ of field equations are G_uv
= k T_uv, not R_uv.
You know that R_uv != G_uv, right?
Do you not understand why the Einstein tensor is distinct from the
Ricci tensor? Do you need help understanding?
>
>> >The symmetry is subtle. Since you have brought it up, Ricci while
>> >making his Riemann curvature tensor can easily settle on another
>> >version by grouping the connection coefficients differently in the
>> >geodesic variation equations such as:
>>
>> >S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
>> >C^m_ik
>>
>> Note that you have a supposed Riemann tensor that is third order in
>> the Christoffel symbols.
>
>No, I have not. S^n_ijk has the same order as R^n_ijk. <shrug>
>
>> Try again, or show how your definition
>> follows from attempting to derive the geodesic equation.
>
>S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
>
>The extra term did not appear in my original post. You dishonestly
>added it in. <shrug>
True ... not true. A mistake from copying and pasting that should not
have followed through the actual posting.
My point stands. Take the difference S^n_ijk - R^n_ijk. The partial
terms disappear, and the last terms from each disappear leaving the
following:
S^n_ijk - R^n_ijk = C^n_km C^m_ij - C^n_im C^m_jk
Swap i and j in the first term, then swap k and i. Both operations are
permitted as per the definition of the connection coeffecients. The
difference disappears, which proves my claim.
Do disagree on any particular point, or are you just unwilling to
admit you are wrong?
>
>> Regardless, the grouping is irrelevant as any 'different' definition
>> for the Riemann tensor that is defined in this way can be shown to be
>> equivalent because the coefficients are _always_ symmetric in their
>> lower indices.
>
>There is no such requirement mathematical. Maybe a consensus, but
>that is not physics but legal laws. <shrug>
Actually there is. I'm surprised the one true guru of differential
geometry doesn't know this!
Riemannian manifolds have a symmetric metric by definition. You can
argue about non-symmetric metrics but they are non-physical and are
not relevant to general relativity.
[...]
eric gisse
<koobee...@gmail.com> wrote:
[...]
>> This will be done my way, as it'll take most of an hour to do in terms
>> of typing and writing, and I anticipate you will get mad at the
>> mathematics ten lines in and refuse to discuss the rest.
>
>Here we go. As long as my wife has not started yelling at me to do
>the household chores before going to bed, I would love to go through
>that.
>
>> The Einstein-Hilbert action S is equal to int( R sqrt(-g) d^4x).
>
>Correction: integral(R sqrt(- det[g]) d^4q)
The subscripts weren't dropped because I forgot. My usage is
consistent with the literature and easy to write - I suggest you catch
up.
For the purposes of this section, the g_ij is a four by four matrix,
as the tensorial properties of the metric aren't relevant here. All
the rules of matrix algebra apply - up to and INCLUDING the
non-existence of the division operator.
>
>> Recall that the Lagrange equations are obtained by varying the action
>> with respect to the field variables, and demanding that the variation
>> is equal to zero at the boundary points.
>
>Don’t forget the change of the field variables.
There is no change of field variables. The g_ij are the only relevant
fields, as the derivatives don't make a contribution when varying this
way.
>
>> So the variation considered will be in the field variables g_ij in the
>> form g^ij --> g^ij + delta(g^ij).
>
>Or just d[g]^ij where [g]^ij = g^ij.
I use d(bla) for differentials. Since I would have d(bla) representing
two different things repeatedly, I'm not going to do that. Trivial to
write on paper, easy for LaTeX, hard for ACSII.
[...]
>
>> the variation ends up like this:
>>
>> delta(sqrt(-g)) = - delta(g) / ( 2 sqrt(-g) )
>
>Since g or my [g] is a matrix, it makes no sense. However, I have
>shown that it is the determinate of the matrix [g] that is applied.
>Then, it looks like the following.
It makes plenty of sense. Is there a particular part you don't
understand? I'm using the standard rules of matrix algebra and matrix
calculus.
Regardless, if you are going to whine and cry about something
simple...
For any square matrix A with nonzero determinant, the following is
true:
ln( det(A) ) = tr( ln(A) )
Vary with respect to A, and you have:
1/det(A) * delta(det(A)) = tr(A^-1 delta(A))
Do you need your hand held further by expanding in a power series, or
can you accept this very simple thing? Once again _try_ think in terms
of the differential operator.
Now, exchange A for g_ij.
1/det(g_ij) * delta(det(g_ij)) = g^ij delta(g_ij)
delta(g) = g g^ij delta(g_ij)
Using delta(g_ij) = - g_ia g_jb delta(g^ab), you have:
(i) delta(g) = -g g_ij delta(g^ij)
>
>@sqrt(- det[g]) = - @det[g]/@[g]^ij / sqrt(- det[g]) / 2
Good lord, did you even pass linear algebra when you first took it?
There is no division operation _nor does this even make any sense_. No
free indices on the left, two free indices on the right.
>
>> delta(sqrt(-g)) = g_ij g delta(g^ij) / ( 2 sqrt(-g) )
>> delta(sqrt(-g)) = - g_ij sqrt(-g) delta(g^ij) / 2
>
>These are utterly nonsense.
What part of this is confusing you?
I am varying the sqrt(-g) term.
Straight application of the delta yields:
delta(sqrt(-g)) = -delta(g) / ( 2 sqrt( -g )
Do I need to expand in a power series, or can you accept the Obvious?
Substitute (i) in and you get:
delta(sqrt(-g)) = - 1/2 sqrt(-g) g_ij delta(g^ij)
Exactly as I claimed. Do you disagree on a particular point, or are
you disagreeing just for the hell of it?
>
>> Now if you look what the delta(sqrt(-g)) term is attached to, you'll
>> notice a -Rg_ij /2 term. That's half the field equations right there.
>
>No, the above equation becomes
>
>@sqrt(- det[g]) = det[g] / [g]^ij / sqrt(- det[g]) / 2
>
>Because
>
>@det[g]/@[g]^ij = - det[g] / [g]^ij
There is no division operator in matrix algebra. Try again.
>
>> Do you wish to argue about any of this, or do you accept the results
>> so far?
>
>Just did. <shrug>
All I have seen are assertions. I have given proofs, and I'm
_explicit_ in what I'm assuming.
Go back to your previous assertion:
"However, I have shown that it is the determinate of the matrix [g]
that is applied."
You forgot the 'show' part of "I have shown", and you ran head first
into nonsense.
>
>Well, why stop now? Tired of copying a textbook?
Two reasons.
a) If you cannot agree with the simple variation of sqrt(-g), how can
you possibly understand the variation of R_ij?
b) The variation of R_ij takes a lot of writing, and playing fallout 3
is slightly more intresting.
Taking potshots at me is simply a sign that you have no argument
against the mathematics.
>
>> >So, when does matter not mean mass even in Maxwell’s equations?
>>
>> Matter is an umbrella term that includes mass, electromagnetic energy,
>> or anything else.
>
>Says who? The committee of GR worshippers?
It is the terminology. Use it or don't but attempting to redefine it
is a fools errand, and arguing about the terminology in the hopes of
winning is even stupider.
[...]
>> >You tell me. You are the one who sits on piles of textbooks. <shrug>
>>
>> Either you know or you don't, and are just making it up. Which is it?
>
>We are talking about you. Remember?
Looks like you don't know - thanks for playing.
>
>> >Christoffel derived the geodesic equations, there is no more
>> >mysticism. Again, the geodesic equations are Euler-Lagrange equations
>> >with the Lagrangian
>>
>> >L = g_ij dq^i/ds dq^j/ds = 1
>>
>> >Derived from,
>>
>> >ds^2 = g_ij dq^i dq^j
>>
>> I want to see if you can explain. Can you explain, or is this just
>> another fact you memorized without context or understanding?
>
>It is all in the mathematics above. It only takes a 7th grade algebra
>to understand all that. <shrug>
Let us see if you can do 7th grade algebra then.
>
>> How about if I said the effective Lagrangian was equal to -1, would I
>> be wrong?
>
>Yes.
Except I'm not, as you didn't specify either the type of motion
(time/space-like, null) *or* the metric signature.
>
>> >> What assumptions are required to
>> >> make that particular statement?
>>
>> >None. Just 9’th grade algebra. <shrug>
>>
>> Show the algebra then.
>
>I am not sinking to 7th grade level to show you the 7th grade algebra.
Looks like you don't even know 7th grade algebra, otherwise you'd
prove me wrong.
Isn't it funny to see how this always goes? You assert I'm wrong. I
ask you to prove it. You refuse. I assert you are wrong. I prove it.
You refuse to acknowledge the proof.
Is this a fulfilling use of your retirement years? Least you could do
is use your time to learn something new instead of being belligerant.
>
>> >Would you ever understand what a tensor really is as meant by the
>> >founding fathers of differential geometry?
>>
>> How can we possibly know when you refuse to give a literature
>> reference?
>>
>> Why don't you show me how the 'founding fathers of differential
>> geometry' defined a tensor?
>
>Are you not sitting on piles of textbooks? Without these books, you
>have no arguments. <shrug>
I'm asking you. Do you know?
>
>> By the way, since you insist that the metric g_ij is a matrix why
>> don't you explain why you are trying to divide by them even though the
>> division operation does not exist for matrices.
>
>I am dividing the non-zero elements of the matrix [g]. <shrug>
Nope, and even if you thought you were you are still wrong since there
are usually nonzero components of the metric that you can't just
ignore. The field equations aren't true for "most" of the components -
they are true for _all_ of the components, even the zero ones.
Back to square one since you are clearly confused: Write the action S
= int(R sqrt(-g) d^4x).
Look at the Ricci scalar component:
R = R_ij g^ij
The i and j are _summed_ over 1 to 4, as per the Einstein summation
convention. You are operating on _ALL_ of the i and j, not just
individual elements.
You have clearly forgotten this.
>
>> >Cheap shot is always ignored. <shrug>
>>
>> Why? You babble incessantly about 'multi year super senior' and other
>> personal subjects you have no actual knowledge about.
>
>Is it not true that you remain a multi-year super-senior without a
>college degree?
I entered the physics degree program at UAF in 2003. Minus the year I
was away, I've been here for 4 years as of this September. So it isn't
true.
I'm comfortable with my education and my ability to further it. Are
you?
>
>> You still don't understand how the metric is relevant to computing
>> areas or volumes of given constructs in a manifold.
>
>I can see why. You do not even understand 7th grade algebra. <shrug>
Asking you to justify your claims isn't ignorance on my part as I
understand the subject far better than you.
>
>> >Hmmm... You still do not understand the subtlety. <shrug>
>>
>> Why don't you explain the subtlety then?
>
>I am not your professor. <shrug>
This is the, what, fifth time now in this very thread where I have
asked you to explain something you consider simple and you have
refused?
>
>> >That is because you don’t understand how the field equations are
>> >actually derived. <shrug>
>>
>> Why don't you give a literature reference showing how the field
>> equations are actually derived?
>
>I gave you the derivation already. <shrug>
It was wrong.
>
>> >Since there is no other way, I am very safe saying that “GR is purely
>> >a mathematical nonsense”. <shrug>
>>
>> How come the only people who agree with you can't even do elementary
>> calculus?
>
>That is because it was judged by someone who remains a multi-year
>super-senior and cannot do rudimentary 7th grade algebra. <shrug>
You must have been a pleasure to work with in your previous job with
your blatant dishonesty. All I asked you to do was prove your claim
and you are leaping to dishonest conclusions about my mathematical
ability.
Daryl McCullough
>
>On Tue, 28 Oct 2008 23:00:45 -0700 (PDT), Koobee Wublee
><koobee...@gmail.com> wrote:
>[...]
>
>>> This will be done my way, as it'll take most of an hour to do in terms
>>> of typing and writing, and I anticipate you will get mad at the
>>> mathematics ten lines in and refuse to discuss the rest.
Why in the world are you going through this? We've already established
that Koobee doesn't know how to use the metric to compute areas and
volumes. In other words, he knows nothing at all about the subject
of differential geometry. Why in the world would you expect him to
follow a derivation involving the Ricci tensor?
Koobee doesn't understand *anything* about physics. He's a fraud.
eric gisse
McCullough) wrote:
It amuses me because I want to watch him flail, and more importantly
it is a useful exercise.
Edward Green
R is the trace, more or less, of R_ij (or should I write R_mn).
So the Einstein tensor is possibly, more or less, the traceless part
of R_ij.
That is, if we ignore factors of 2, and the g_ij terms.
Anyway, the traceless part of R_ij measures, more or less, the
relative volumetric distortion as a function of direction(s), but
without the "bulk" part.
Or at least it would, except for the mysterious inclusion of the g_ij
term, which is quite obfuscating, because it really hides all the
information of R_ijkl in it, doesn't it!
Groping, not groking. Somebody chime in, please.
Possibly this groping is all wrong, even as groping.
I'm also missing the possibility of the activity even when T = 0,
clearly: I mean, you can grok all you want how T couples to spacetime,
but even in vacuum, a lot of interesting stuff is going on. So
grokking the Einstein equation is at best half the... err... grok.
Kind of a gr or an ok. Hmm...
At least we see why GR is OK is exactly the correct term for
understanding, here. :-)
Interesting how none of the stuff that goes on in vacuum, including,
apparently, gravity waves, involves energy or momentum. Incredible,
actually.
eric gisse
<spamsp...@netzero.com> wrote:
>A baby step...
>
>R is the trace, more or less, of R_ij (or should I write R_mn).
Write whatever.
R = g^ij R_ij
The indices don't care what they are called.
>
>So the Einstein tensor is possibly, more or less, the traceless part
>of R_ij.
No. The Einstein tensor is derived from symmetries and contractions of
the Weyl tensor, which relate to the traceless parts of the Riemann
tensor.
>
>That is, if we ignore factors of 2, and the g_ij terms.
Same mistake kooby was making, but this is in good faith.
>
>Anyway, the traceless part of R_ij measures, more or less, the
>relative volumetric distortion as a function of direction(s), but
>without the "bulk" part.
No.
http://math.ucr.edu/home/baez/einstein/einstein.html
>
>Or at least it would, except for the mysterious inclusion of the g_ij
>term, which is quite obfuscating, because it really hides all the
>information of R_ijkl in it, doesn't it!
No.
Same thing I told kooby: the trace-free parts of Riemann come into
play through the Einstein tensor, and the trace parts of Riemann work
through direct inclusion. All the information is there, but it is
rather compact and mashed together through the all the symmetries.
>
>Groping, not groking. Somebody chime in, please.
>
>Possibly this groping is all wrong, even as groping.
You NEED to get a textbook on the subject. You won't learn much of
anything this way.
>
>I'm also missing the possibility of the activity even when T = 0,
>clearly: I mean, you can grok all you want how T couples to spacetime,
>but even in vacuum, a lot of interesting stuff is going on. So
>grokking the Einstein equation is at best half the... err... grok.
>Kind of a gr or an ok. Hmm...
I treat Einstein's equation as analogus to F = ma. You have the
dynamical behavior of the system (a in Newton, metric in GR) that is
proportional to contributions from matter.
There is a more direct physical (and correct) interpretation on Baez's
page, but it is far less complicated than you think.
>
>At least we see why GR is OK is exactly the correct term for
>understanding, here. :-)
>
>Interesting how none of the stuff that goes on in vacuum, including,
>apparently, gravity waves, involves energy or momentum. Incredible,
>actually.
They do involve energy and momentum - gravitational radiation
transports energy and angular momentum from a system. Just like
regular ol' electromagnetic dipole radiation.
Koobee Wublee
> Koobee Wublee wrote:
> >That does not refute what I wrote on the Ricci tensor above. <shrug>
>
> Of course it does.
<shrug>
> You only have three choices for contracting on R^a_bcd, and I have
> shown that one is equal to zero and the other two are the same up to a
> minus sign which has no effect on the field equations.
Still does not refute what I wrote. <shg>
> Now tell me again how the choice is relevant when you are choosing
> between "obviously zero" and "same up to a sign" when the choice has
> no relevance for dynamics.
Just numbers. <shrug>
> Do disagree on any particular point, or are you just unwilling to
> admit you are wrong?
This part of mathematics has no significance in real life. <shrug>
You came up all these possible mathematical outcomes. You just choose
one of your liking. While they should be considered nonsense just
like Hilbert had done and walked away from all there mess he created.
<shrug>
> >Well, the only verifiable condition against the Newtonian limit is in
> >vacuum. Guess what? The Ricci tensor is the set of the field
> >equations. <shrug>
>
> You have no argument, as you simply repeated your claim and tossed in
> a non-sequitur for good measure. The _set_ of field equations are G_uv
> = k T_uv, not R_uv.
If T is zero, R_ij = G_ij. <shrug>
> You know that R_uv != G_uv, right?
Yes, in general, but I specifically say in vacuum. <shrug>
> Do you not understand why the Einstein tensor is distinct from the
> Ricci tensor? Do you need help understanding?
Keep yapping.
> >S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
>
> >The extra term did not appear in my original post. You dishonestly
> >added it in. <shrug>
>
> True ... not true. A mistake from copying and pasting that should not
> have followed through the actual posting.
Then, you should not accuse me of error on this part. It is your own
gross mistake. <shrug>
> My point stands. Take the difference S^n_ijk - R^n_ijk. The partial
> terms disappear, and the last terms from each disappear leaving the
> following:
>
> S^n_ijk - R^n_ijk = C^n_km C^m_ij - C^n_im C^m_jk
Correction:
S^n_ijk - R^n_ijk = C^n_im C^m_jk - C^n_km C^m_ij
Where
** S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm
C^m_ik
** R^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_km C^m_ij – C^n_jm
C^m_ik
** C^n_ij = Christoffel symbols of the 2nd kind
> Swap i and j in the first term, then swap k and i. Both operations are
> permitted as per the definition of the connection coeffecients. The
> difference disappears, which proves my claim.
That is some matheMAGIC trick!
Swapping i and j and then k and i from R^n_ijk, you get
T^n_ijk = @C^n_jk/@q^i - @C^n_ij/@q^k + C^n_im C^m_jk – C^n_km C^m_ik
Then, tell me what you get for (S^n_ijk – T^n_ijk).
> Do disagree on any particular point, or are you just unwilling to
> admit you are wrong?
For a no-degree multi-year super-senior, you are indeed very cocky.
<shrug>
> >There is no such requirement mathematical. Maybe a consensus, but
> >that is not physics but legal laws. <shrug>
>
> Actually there is. I'm surprised the one true guru of differential
> geometry doesn't know this!
Who would this guru be?
> Riemannian manifolds have a symmetric metric by definition. You can
> argue about non-symmetric metrics but they are non-physical and are
> not relevant to general relativity.
With a rank-4 tensor, I am not convinced that the Riemann curvature
tensor needs to be symmetric. It is only a pure mathematical
construct. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >Correction: integral(R sqrt(- det[g]) d^4q)
>
> The subscripts weren't dropped because I forgot. My usage is
> consistent with the literature and easy to write - I suggest you catch
> up.
What subscripts? I corrected the determinant part. <shrug>
> For the purposes of this section, the g_ij is a four by four matrix,
> as the tensorial properties of the metric aren't relevant here.
No, g_ij must be elements to the metric or the matrix g. That is why
I like to write g as [g] instead. With [g]^ij or [g]_ij as elements
to the matrix [g^-1] and [g] respectively, then there is no confusion.
> All
> the rules of matrix algebra apply - up to and INCLUDING the
> non-existence of the division operator.
The following field equations apply only to the elements of the
tensors involved.
R_ij – R g_ij / 2 = constant T_ij
Writing it in matrix form, you get the following instead
[R] – R [g] / 2 = constant [T]
Where
** [R], [g], [T] = 4x4 matrices
Learn the difference and avoid more matheMAGICAL embarrassments.
> >Don’t forget the change of the field variables.
>
> There is no change of field variables. The g_ij are the only relevant
> fields, as the derivatives don't make a contribution when varying this
> way.
That is correct. Thus, the field equations do not qualify as the
Euler-Lagrange equations, do they?
> >Or just d[g]^ij where [g]^ij = g^ij.
>
> I use d(bla) for differentials. Since I would have d(bla) representing
> two different things repeatedly, I'm not going to do that. Trivial to
> write on paper, easy for LaTeX, hard for ACSII.
OK, whatever you say. <shrug>
My terminology is simpler and more clear. If you don’t want to adopt
it, that is fine with me. <shrug>
>
> [...]
>
>
>
> >> the variation ends up like this:
>
> >> delta(sqrt(-g)) = - delta(g) / ( 2 sqrt(-g) )
>
> >Since g or my [g] is a matrix, it makes no sense. However, I have
> >shown that it is the determinate of the matrix [g] that is applied.
> >Then, it looks like the following.
>
> It makes plenty of sense. Is there a particular part you don't
> understand? I'm using the standard rules of matrix algebra and matrix
> calculus.
You cannot trivially divide a matrix by another matrix. You need to
find the inverse first. <shrug>
> Regardless, if you are going to whine and cry about something
> simple...
That is why g_ij represents the elements to the matrix g. <shrug>
> For any square matrix A with nonzero determinant, the following is
> true:
>
> ln( det(A) ) = tr( ln(A) )
>
> Vary with respect to A, and you have:
>
> 1/det(A) * delta(det(A)) = tr(A^-1 delta(A))
>
> Do you need your hand held further by expanding in a power series, or
> can you accept this very simple thing?
Well, you need to prove that.
> [...]
> >@sqrt(- det[g]) = - @det[g]/@[g]^ij / sqrt(- det[g]) / 2
>
> Good lord, did you even pass linear algebra when you first took it?
> There is no division operation _nor does this even make any sense_. No
> free indices on the left, two free indices on the right.
Relax. It is only a typo with the following correction. <shrug>
@sqrt(- det[g])/@[g]^ij = - @det[g]/@[g]^ij / sqrt(- det[g]) / 2
> >> delta(sqrt(-g)) = g_ij g delta(g^ij) / ( 2 sqrt(-g) )
> >> delta(sqrt(-g)) = - g_ij sqrt(-g) delta(g^ij) / 2
>
> >These are utterly nonsense.
>
> What part of this is confusing you?
All of them. <shrug>
> I am varying the sqrt(-g) term.
<shrug>
> Straight application of the delta yields:
>
> delta(sqrt(-g)) = -delta(g) / ( 2 sqrt( -g )
>
> Do I need to expand in a power series, or can you accept the Obvious?
>
> Substitute (i) in and you get:
>
> delta(sqrt(-g)) = - 1/2 sqrt(-g) g_ij delta(g^ij)
That is wrong.
> Exactly as I claimed. Do you disagree on a particular point, or are
> you disagreeing just for the hell of it?
This is just wrong.
>
>
>
> >> Now if you look what the delta(sqrt(-g)) term is attached to, you'll
> >> notice a -Rg_ij /2 term. That's half the field equations right there.
>
> >No, the above equation becomes
>
> >@sqrt(- det[g]) = det[g] / [g]^ij / sqrt(- det[g]) / 2
>
> >Because
>
> >@det[g]/@[g]^ij = - det[g] / [g]^ij
>
> There is no division operator in matrix algebra. Try again.
I told you already. [g]^ij = g^ij are elements to the inverse of the
matrix [g]. <shrug>
> >Just did. <shrug>
>
> All I have seen are assertions. I have given proofs, and I'm
> _explicit_ in what I'm assuming.
MatheMAGIC tricks on your part. <shrug>
> Go back to your previous assertion:
>
> "However, I have shown that it is the determinate of the matrix [g]
> that is applied."
>
> You forgot the 'show' part of "I have shown", and you ran head first
> into nonsense.
Say what?
> >Well, why stop now? Tired of copying a textbook?
>
> Two reasons.
>
> a) If you cannot agree with the simple variation of sqrt(-g), how can
> you possibly understand the variation of R_ij?
>
> b) The variation of R_ij takes a lot of writing, and playing fallout 3
> is slightly more intresting.
>
> Taking potshots at me is simply a sign that you have no argument
> against the mathematics.
Well, I know why stop now. That is because you cannot get the energy
momentum part out of just that Einstein-Hilbert action. <shrug>
> >Says who? The committee of GR worshippers?
>
> It is the terminology. Use it or don't but attempting to redefine it
> is a fools errand, and arguing about the terminology in the hopes of
> winning is even stupider.
There are really no golden set of rules on terminology usage. <shrug>
> >We are talking about you. Remember?
>
> Looks like you don't know - thanks for playing.
We are still talking about you. <shrug>
> >It is all in the mathematics above. It only takes a 7th grade algebra
> >to understand all that. <shrug>
>
> Let us see if you can do 7th grade algebra then.
Yes, I can, but you can’t. Did you even graduate from junior high?
> >Yes.
>
> Except I'm not, as you didn't specify either the type of motion
> (time/space-like, null) *or* the metric signature.
The Lagrangian is not and cannot be -1. <shrug>
> >I am not sinking to 7th grade level to show you the 7th grade algebra.
>
> [...]
More yapping to hide eric gisse’s weak skills in algebra snipped.
> >Are you not sitting on piles of textbooks? Without these books, you
> >have no arguments. <shrug>
>
> I'm asking you. Do you know?
You are the one sitting on all these books. <shrug>
> >I am [not] dividing the non-zero elements of the matrix [g]. <shrug>
>
> Nope, and even if you thought you were you are still wrong since there
> are usually nonzero components of the metric that you can't just
> ignore.
That is correct. Writing the Einstein-Hilbert action below,
A = Integral(L d^4[q])
Where
** A = Einstein-Hilbert action
** L = Lagrangian
** d^4[q] = Correlation of all coordinate components
A stationary action demands the following.
dA = Integral(dL d^4[q]) = 0
Or
dA = Integral(@L/@[g]^ij d[g]^ij d^4[q]) = 0
Where
** dL = @L/@[g]^ij d[g]^ij
If d[g]^ij != 0, then @L/@[g]^ij does not have to be zero.
If d[g]^ij == 0, then @L/@[g]^ij must be zero.
That is why if d[g]^ij == 0, you have to ignore the results of @L/
@[g]^ij. <shrug>
> The field equations aren't true for "most" of the components -
> they are true for _all_ of the components, even the zero ones.
For that reason I have shown you above, what you said above is just
not true. <shrug>
> Back to square one since you are clearly confused: Write the action S
> = int(R sqrt(-g) d^4x).
>
> Look at the Ricci scalar component:
>
> R = R_ij g^ij
>
> The i and j are _summed_ over 1 to 4, as per the Einstein summation
> convention. You are operating on _ALL_ of the i and j, not just
> individual elements.
That is correct.
> You have clearly forgotten this.
When did I forget that? I have always been saying the Ricci scalar R
is a scalar. <shrug>
> >Is it not true that you remain a multi-year super-senior without a
> >college degree?
I have to snip the rest of yapping crap from you.
[snip, snip, snip]
Tom Roberts
In GR, "Vacuum" is defined as T=0 (meaning the tensor, not just its
trace). So in vacuum the energy density is zero, as are the momentum
density and the stress density.
> They do involve energy and momentum - gravitational radiation
> transports energy and angular momentum from a system. Just like
> regular ol' electromagnetic dipole radiation.
One must be careful, as in your statement "energy" and "momentum" have
different meanings from the usual ones, and different meanings for the
first and second times you use them.
Consider a localized system contained within a closed spatial boundary B
in a spatially-infinite manifold that is vacuum (T=0) outside B. Use the
obvious foliation in which B is at rest, and let the system be emitting
gravitational radiation. The "total energy inside B" is not conserved
when that means the integral of T_tt over the volume inside B. But yet
T=0 outside B, so there is no energy outside B, either (note carefully
the two terms with and without quotation marks). What gives? -- the
manifold is not flat, and integration over the volume inside B is not
well defined, so the "total energy inside B" is not well defined,
either. That is, the value you obtain depends in detail how you perform
the integral, and the result will change over time even though you do
your best to do it the same way at each time [#]. IOW: that "obvious
foliation" is not really so obvious inside B (where things are moving
around), and it is not possible to "do it the same way" at different times.
So yes, the gravitational radiation is transporting energy
away from the system inside B, as long as one is a bit
loose about what that problematical word means. But the
gravitational radiation outside B does not have any energy
(T=0).
[#] An exception is when T=0 throughout, in which
case the integral is well defined to be zero. Such as
outside B. Hence no quotation marks there.
If one defines 'energy' as the Noether current conjugate to time
translation, it is not conserved. We know this because there is no
timelike Killing vector (the presence of gravitational radiation
prevents it, as does the motion inside B).
For this physical situation one can define an energy-momentum
pseudo-tensor, and if one integrates it over the volume inside a closed
3-surface K that is everywhere outside B, then one obtains a
well-defined value. In the limit K -> spatial infinity the total
pseudo-energy is conserved.
So that is four different meanings of a single word -- be careful not to
confuse them.
All this applies to "momentum"/'momentum'/momentum/pseudo-momentum also.
[I am not an expert on this, but that's how I understand it.]
Tom Roberts
eric gisse
>> >> the variation ends up like this:
>>
>> >> delta(sqrt(-g)) = - delta(g) / ( 2 sqrt(-g) )
>>
>> >Since g or my [g] is a matrix, it makes no sense. However, I have
>> >shown that it is the determinate of the matrix [g] that is applied.
>> >Then, it looks like the following.
>>
>> It makes plenty of sense. Is there a particular part you don't
>> understand? I'm using the standard rules of matrix algebra and matrix
>> calculus.
>
>You cannot trivially divide a matrix by another matrix. You need to
>find the inverse first. <shrug>
I'm not. The objects being divided by are scalar quantities. Or did
you forget that g is the determinant of the metric?
>
>> Regardless, if you are going to whine and cry about something
>> simple...
>
>That is why g_ij represents the elements to the matrix g. <shrug>
It represents _all_ of the elements. The rules of manipulating
matrices apply - as in, you can't divide by g_ij and other silly shit.
This is all linear algebra and calculus. The only difference is that
I'm using the Einstein summation convention to represent the objects,
which is a perfectly valid method.
>
>> For any square matrix A with nonzero determinant, the following is
>> true:
>>
>> ln( det(A) ) = tr( ln(A) )
>>
>> Vary with respect to A, and you have:
>>
>> 1/det(A) * delta(det(A)) = tr(A^-1 delta(A))
>>
>> Do you need your hand held further by expanding in a power series, or
>> can you accept this very simple thing?
>
>Well, you need to prove that.
The infintesimal variation obeys the _exact same properties_ as the
ordinary differential operator. Go back to the variational methods you
have told me to study, and bone the fuck up so we can skip these
little diversions.
Given that the determinant and trace are linear functions of their
arguments, do I really _honestly_ need to hold your hand? Will you ask
me to prove the chain rule next?
>
>> [...]
It looks like you agree with what I wrote.
It seems to be more that you don't think I can prove any of this, and
are making me prove more and more trivial results in a feeble attempt
to catch me clueless.
>
>> >@sqrt(- det[g]) = - @det[g]/@[g]^ij / sqrt(- det[g]) / 2
>>
>> Good lord, did you even pass linear algebra when you first took it?
>> There is no division operation _nor does this even make any sense_. No
>> free indices on the left, two free indices on the right.
>
>Relax. It is only a typo with the following correction. <shrug>
>
>@sqrt(- det[g])/@[g]^ij = - @det[g]/@[g]^ij / sqrt(- det[g]) / 2
Still wrong - there is no division operator, and the operation is on
the _entire_ matrix of elements g_ij. Notice how I'm only using
properties of linear algebra and calculus?
>
>> >> delta(sqrt(-g)) = g_ij g delta(g^ij) / ( 2 sqrt(-g) )
>> >> delta(sqrt(-g)) = - g_ij sqrt(-g) delta(g^ij) / 2
>>
>> >These are utterly nonsense.
>>
>> What part of this is confusing you?
>
>All of them. <shrug>
Probably because you are operating on the notion that you can operate
only on a particular element of the metric then the entire metric at a
whim, and are getting confused when the inevitible conflicts arise.
>
>> I am varying the sqrt(-g) term.
>
><shrug>
Since I have to explain rules derived from introductory calculus and
linear algebra, it seems like I have to explain these things.
Now do you agree that the sqrt(-g) term is being varied, or do I need
to go back to the beginning and explain the concept of calculus of
variations?
>
>> Straight application of the delta yields:
>>
>> delta(sqrt(-g)) = -delta(g) / ( 2 sqrt( -g )
>>
>> Do I need to expand in a power series, or can you accept the Obvious?
>>
>> Substitute (i) in and you get:
>>
>> delta(sqrt(-g)) = - 1/2 sqrt(-g) g_ij delta(g^ij)
>
>That is wrong.
Why? I wrote out every step - point to the first part you think is
wrong.
>
>> Exactly as I claimed. Do you disagree on a particular point, or are
>> you disagreeing just for the hell of it?
>
>This is just wrong.
You thought I was bluffing when I made the offer, and now you are
pissed off because I followed through and you don't understand a
single fucking word I'm writing. Ain't that a bitch?
[...]
>> All I have seen are assertions. I have given proofs, and I'm
>> _explicit_ in what I'm assuming.
>
>MatheMAGIC tricks on your part. <shrug>
It stops being magic when you learn how it is done, as it is with real
magic.
[...]
>Well, I know why stop now. That is because you cannot get the energy
>momentum part out of just that Einstein-Hilbert action. <shrug>
These are the vacuum field equations - I thought that was obvious from
the complete non-inclusion of the matter term.
Dealing with the matter term and proving it is symmetric and conserved
is just a few lines of mathematics. I'll save that for when were done
with varying the sqrt(-g) term, which you seem to be having difficulty
in understanding for some reason.
[...]
>> >I am [not] dividing the non-zero elements of the matrix [g]. <shrug>
>>
>> Nope, and even if you thought you were you are still wrong since there
>> are usually nonzero components of the metric that you can't just
>> ignore.
>
>That is correct. Writing the Einstein-Hilbert action below,
>
>A = Integral(L d^4[q])
>
>Where
>
>** A = Einstein-Hilbert action
>** L = Lagrangian
>** d^4[q] = Correlation of all coordinate components
The correct term is "volume element", as the integral is done over all
of space and time.
>
>A stationary action demands the following.
>
>dA = Integral(dL d^4[q]) = 0
>
>Or
>
>dA = Integral(@L/@[g]^ij d[g]^ij d^4[q]) = 0
No. You are ignoring the contributions from the derivatives of the
field variables - again. You don't even know how to include the
contributions from the derivatives of the field variables!
Varying the field under a small deflection _IS_ how the Euler-Lagrange
equations are derived in case you forgot.
>
>Where
>
>** dL = @L/@[g]^ij d[g]^ij
>
>If d[g]^ij != 0, then @L/@[g]^ij does not have to be zero.
>
>If d[g]^ij == 0, then @L/@[g]^ij must be zero.
Once again you make the same mistake as before. You are summing over
_all_ the components. Not just the ones you want to.
[...]
>> >Is it not true that you remain a multi-year super-senior without a
>> >college degree?
>
>I have to snip the rest of yapping crap from you.
>
>[snip, snip, snip]
Once again you refuse to hear what you don't want to believe.
Don Stockbauer
> Don Stockbauer wrote:
> > Edward Green wrote:
> >> I feel it is time to grok the Einstein equation.
>
>
> People who still like Robert Heinlein's novels, and who recognize that
> English has no word that captures the fullness of this one.
>
> Tom Roberts
I grok that.
Don Stockbauer
> On Oct 25, 8:36 am, Don Stockbauer <donstockba...@hotmail.com> wrote:
>
> > Edward Green wrote:
> > > I feel it is time to grok the Einstein equation.
>
> > Who the hell says "grok" anymore???????
>
>
> It's a good word, whose meaning seems to match its sound.
>
> Next question?
So its meaning is "the sound a frog makes in a pond"?
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 28, 11:42 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> >That does not refute what I wrote on the Ricci tensor above. <shrug>
>>
>> Of course it does.
>
><shrug>
An 8 shrug post! You must be pretty pissed off, since you can't find
anything wrong with what I wrote.
>
>> You only have three choices for contracting on R^a_bcd, and I have
>> shown that one is equal to zero and the other two are the same up to a
>> minus sign which has no effect on the field equations.
>
>Still does not refute what I wrote. <shg>
It most certainly does. You asserted that the choice mattered - I
proved that the choice doesn't, as you either get two of the same
thing or zero.
Your steadfast unwillingness to accept this isn't my problem, though.
>
>> Now tell me again how the choice is relevant when you are choosing
>> between "obviously zero" and "same up to a sign" when the choice has
>> no relevance for dynamics.
>
>Just numbers. <shrug>
>
>> Do disagree on any particular point, or are you just unwilling to
>> admit you are wrong?
>
>This part of mathematics has no significance in real life. <shrug>
/Classic/ sour grapes. You know you lost so you are trying to argue
that the argument didn't really matter in the first place.
>
>You came up all these possible mathematical outcomes. You just choose
>one of your liking. While they should be considered nonsense just
>like Hilbert had done and walked away from all there mess he created.
><shrug>
Since the two nontrivial choices are the same up to a sign change that
has absolutely zero effect on the field equations , its hardly a
'mess'.
Just because _YOU_ do not understand does not mean it is wrong.
>
>> >Well, the only verifiable condition against the Newtonian limit is in
>> >vacuum. Guess what? The Ricci tensor is the set of the field
>> >equations. <shrug>
>>
>> You have no argument, as you simply repeated your claim and tossed in
>> a non-sequitur for good measure. The _set_ of field equations are G_uv
>> = k T_uv, not R_uv.
>
>If T is zero, R_ij = G_ij. <shrug>
Only since both are equal to zero. It isn't true in general, which
destroys your argument.
>
>> You know that R_uv != G_uv, right?
>
>Yes, in general, but I specifically say in vacuum. <shrug>
My argument is true in general which you are unable to refute. You are
like Sue - you don't understand the basic concepts so you refer to
things you [or think that you] know to be true in an effort to avoid
discussing the thing you do not understand.
>
>> Do you not understand why the Einstein tensor is distinct from the
>> Ricci tensor? Do you need help understanding?
>
>Keep yapping.
>
>> >S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm C^m_ik
>>
>> >The extra term did not appear in my original post. You dishonestly
>> >added it in. <shrug>
>>
>> True ... not true. A mistake from copying and pasting that should not
>> have followed through the actual posting.
>
>Then, you should not accuse me of error on this part. It is your own
>gross mistake. <shrug>
>
>> My point stands. Take the difference S^n_ijk - R^n_ijk. The partial
>> terms disappear, and the last terms from each disappear leaving the
>> following:
>>
>> S^n_ijk - R^n_ijk = C^n_km C^m_ij - C^n_im C^m_jk
>
>Correction:
>
>S^n_ijk - R^n_ijk = C^n_im C^m_jk - C^n_km C^m_ij
Distinction without difference - the terms cancel either way.
>
>Where
>
>** S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm
>C^m_ik
>** R^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_km C^m_ij – C^n_jm
>C^m_ik
>** C^n_ij = Christoffel symbols of the 2nd kind
>
>> Swap i and j in the first term, then swap k and i. Both operations are
>> permitted as per the definition of the connection coeffecients. The
>> difference disappears, which proves my claim.
>
>That is some matheMAGIC trick!
Funny how every complaint you have boils down to you believing it is a
trick. You can't find anything wrong with the trick - you just know
there is a trick. But you don't know what it is so I must be wrong.
>
>Swapping i and j and then k and i from R^n_ijk, you get
>
>T^n_ijk = @C^n_jk/@q^i - @C^n_ij/@q^k + C^n_im C^m_jk – C^n_km C^m_ik
>
>Then, tell me what you get for (S^n_ijk – T^n_ijk).
Learn to read. Swap i and j then k and i on the first term in ( C^n_km
C^m_ij - C^n_km C^m_ij).
The operation is permitted by the symmetry of the connection. Learn
the rules or don't play the game.
You are wrong, and proven so. Find a new argument.
>
>> Do disagree on any particular point, or are you just unwilling to
>> admit you are wrong?
>
>For a no-degree multi-year super-senior, you are indeed very cocky.
><shrug>
Funny how you run right back to the insults when you have nowhere else
to go.
Why is it the 'no-degree multi-year super-senior' has to explain
freshman concepts to you?
>
>> >There is no such requirement mathematical. Maybe a consensus, but
>> >that is not physics but legal laws. <shrug>
>>
>> Actually there is. I'm surprised the one true guru of differential
>> geometry doesn't know this!
>
>Who would this guru be?
>
>> Riemannian manifolds have a symmetric metric by definition. You can
>> argue about non-symmetric metrics but they are non-physical and are
>> not relevant to general relativity.
>
>With a rank-4 tensor, I am not convinced that the Riemann curvature
>tensor needs to be symmetric. It is only a pure mathematical
>construct. <shrug>
>
Your inability to read is staggering. I didn't say "Riemann curvature
tensor", I said "metric". Do you not know the difference or are you
that fucking stupid?
So what if GR is a mathematical construct? If you want to argue about
GR, you need to play by the established rules instead of making shit
up as you go along.
Koobee Wublee
> Koobee Wublee wrote:
> An 8 shrug post! You must be pretty pissed off, since you can't find
> anything wrong with what I wrote.
It is very obvious that you have a lot of more bandwidth than I do.
Thus, in these series of exchanges I will try to stick to the on-topic
stuff. In doing so, I will snip all you yapping, whining nonsense.
So, here we go.
> [whining sh*t snipped]
>
> >You came up all these possible mathematical outcomes. You just choose
> >one of your liking. While they should be considered nonsense just
> >like Hilbert had done and walked away from all there mess he created.
> ><shrug>
>
> Since the two nontrivial choices are the same up to a sign change that
> has absolutely zero effect on the field equations , its hardly a
> 'mess'.
Any sign change in the Ricci tensor will affect the field equations in
general. <shrug>
> [whining sh*t snipped]
>
> >Correction:
>
> >S^n_ijk - R^n_ijk = C^n_im C^m_jk - C^n_km C^m_ij
>
> [whining sh*t snipped]
>
> >Where
>
> >** S^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_im C^m_jk – C^n_jm
> >C^m_ik
> >** R^n_ijk = @C^n_ij/@q^k - @C^n_ik/@q^j + C^n_km C^m_ij – C^n_jm
> >C^m_ik
> >** C^n_ij = Christoffel symbols of the 2nd kind
>
> [whining sh*t snipped]
>
> >Swapping i and j and then k and i from R^n_ijk, you get
>
> >T^n_ijk = @C^n_jk/@q^i - @C^n_ij/@q^k + C^n_im C^m_jk – C^n_km C^m_ik
>
> >Then, tell me what you get for (S^n_ijk – T^n_ijk).
>
> [whining sh*t snipped]
>
> The operation is permitted by the symmetry of the connection. Learn
> the rules or don't play the game.
I have pointed out your mathematical mistakes, and that is final.
<shrug>
> [whining sh*t snipped]
Wow! Done! That’s all. My time can be more productive snipping all
your irrelevant whining crap. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >You cannot trivially divide a matrix by another matrix. You need to
> >find the inverse first. <shrug>
>
> I'm not. The objects being divided by are scalar quantities. Or did
> you forget that g is the determinant of the metric?
Oh, my. You said earlier that g is not a determinant. I have to snip
all your whining wishy-washy crap. <shrug>
Koobee Wublee
> R is the trace, more or less, of R_ij (or should I write R_mn).
R is defined as the dot/inner product of two matrices.
R = [g] * [R]
Where
** [g] = the metric, a 4x4 matrix
** [R] = Ricci tensor, also a 4x4 matrix
There are 3 possible operations of two matrices of the same elements.
#1 Inner product
[A] * [B] = C
Or
[A]^ij [B]_ij = C
Where
** [A] = n-by-m matrix
** [B] = n-by-m matrix
** C = scalar
You are most familiar with (m = 1) where [A] and [B] are vectors.
#2 Multiplication product
[A] X [B] = [C]
Or
[A]^ik [B]_kj = [C]^i_j
Where
** [A] = n-by-p matrix
** [B] = p-by-m matrix
** [C] = n-by-m matrix
This is what you have learned in junior high school.
#3 Scalar product
[A] x [B] = [C]
Or
[A]^i_j [B]_ij = [C]_ij
Where
** [A] = n-by-m matrix
** [B] = n-by-m matrix
** [C] = n-by-m matrix
> So the Einstein tensor is possibly, more or less, the traceless part
> of R_ij.
Trace will confuse the heck out of you. <shrug>
> That is, if we ignore factors of 2, and the g_ij terms.
You cannot ignore what are integral parts of the equations. <shrug>
> Anyway, the traceless part of R_ij measures, more or less, the
> relative volumetric distortion as a function of direction(s), but
> without the "bulk" part.
>
> Or at least it would, except for the mysterious inclusion of the g_ij
> term, which is quite obfuscating, because it really hides all the
> information of R_ijkl in it, doesn't it!
No, [g]_ij does not hide any information of [R]_ijkl. [R]_ijkl is a
function of [g]. <shrug>
> Groping, not groking. Somebody chime in, please.
I prefer the word “analyze”. <shrug>
> Possibly this groping is all wrong, even as groping.
<shrug>
> I'm also missing the possibility of the activity even when T = 0,
The field equations only have accountability when T == 0. Otherwise,
you can claim whatever you want and not subject to any experimental
policeman. <shrug>
> clearly: I mean, you can grok all you want how T couples to spacetime,
> but even in vacuum, a lot of interesting stuff is going on. So
> grokking the Einstein equation is at best half the... err... grok.
> Kind of a gr or an ok. Hmm...
Again, the field equations are a set of mathematical construct with no
particular physical insights to justify their derivations. <shrug>
> At least we see why GR is OK is exactly the correct term for
> understanding, here. :-)
GR is OK only under some special mathematical applications. <shrug>
> Interesting how none of the stuff that goes on in vacuum, including,
> apparently, gravity waves, involves energy or momentum. Incredible,
> actually.
Gravity wave is another mathematical construct squeezed out between
the geodesic crack of two adjacent points in spacetime. It is the
same by-product of a particular mathematical method that also yields
the Riemann curvature tensor. <shrug>
eric gisse
You always knew what g was, you just want to flee the discussion
because you lost.
I alawys knew you wouldn't be able to follow the discussion even if I
wrote out every step for you. Thanks for playing.
Koobee Wublee
> Koobee Wublee wrote:
> >Oh, my. You said earlier that g is not a determinant. I have to snip
> >all your whining wishy-washy crap. <shrug>
>
> You always knew what g was,
And I always understand you meant g to be the matrix. It also
reflects in your calculations. <shrug>
> you just want to flee the discussion
> because you lost.
You are the wishy-washy one. I have told you from the very beginning
that g_ij or g^ij represent the elements to the matrix g and g^-1
respectively. You were the one confusing g_ij and g^ij as matrices.
<shrug>
> I alawys knew you wouldn't be able to follow the discussion even if I
> wrote out every step for you. Thanks for playing.
I just don’t have the patience to carry on stupid discussions.
<shrug>
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 29, 8:24 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> An 8 shrug post! You must be pretty pissed off, since you can't find
>> anything wrong with what I wrote.
>
>It is very obvious that you have a lot of more bandwidth than I do.
>Thus, in these series of exchanges I will try to stick to the on-topic
>stuff. In doing so, I will snip all you yapping, whining nonsense.
>So, here we go.
Looks like the discussion is over. You don't know how to vary the
Einstein-Hilbert action. You don't know how to vary the square root of
a determinant. You don't know the rules of freshman calculus, nor do
you know the basic rules of matrix algebra.
I was hoping to _at least_ make it to varying the Ricci tensor, but
you couldn't even handle varying the square root of the determinant!
That's amazing!
You've been arguing about this since 2005, and you couldn't even work
through a textbook exercise with every step laid out for you. Do you
have any idea how ridiculous and stupid you look right now?
>
>> [whining sh*t snipped]
>>
>> >You came up all these possible mathematical outcomes. You just choose
>> >one of your liking. While they should be considered nonsense just
>> >like Hilbert had done and walked away from all there mess he created.
>> ><shrug>
>>
>> Since the two nontrivial choices are the same up to a sign change that
>> has absolutely zero effect on the field equations , its hardly a
>> 'mess'.
>
>Any sign change in the Ricci tensor will affect the field equations in
>general. <shrug>
Nope. The negative will carry through the Ricci tensor and the Ricci
scalar, which is then absorbed into the proportionality constant.
You have no argument other than sheer contrarianism. You whined that
the choice mattered - I proved that it doesnt. You whined that
alterante definitions of the Riemann tensor were available - I proved
they are equivalent.
Thanks for playing.
[...]
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 29, 3:31 pm, Edward Green wrote:
>
>> R is the trace, more or less, of R_ij (or should I write R_mn).
>
>R is defined as the dot/inner product of two matrices.
>
>R = [g] * [R]
No, it isn't. It is defined to be R = g_ab R^ab as tensors ARE NOT
MATRICES.
>
>Where
>
>** [g] = the metric, a 4x4 matrix
>** [R] = Ricci tensor, also a 4x4 matrix
They are tensors, but it is a distinction without difference to you
since you don't really understand either tensor of matrix algebra as
abundantly proved in this very thread.
>
>There are 3 possible operations of two matrices of the same elements.
>
>#1 Inner product
>
>[A] * [B] = C
>
>Or
>
>[A]^ij [B]_ij = C
Same operation in matrix algebra, retard. Explains why you kept trying
to insert the [BLA] matrix notation - you don't even understand that
the summation convention notation works for matrix algebra.
[...]
All that linear algebra and so little understanding. What a waste of
electrons.
>
>> So the Einstein tensor is possibly, more or less, the traceless part
>> of R_ij.
>
>Trace will confuse the heck out of you. <shrug>
Classic projection of one's problems on someone else.
Another 8 <shrug> post - having a bad day on the internets?
>
>> That is, if we ignore factors of 2, and the g_ij terms.
>
>You cannot ignore what are integral parts of the equations. <shrug>
Says mister "the field equations are basically the Ricci tensor".
Don't criticize someone for doing the same thing you do.
[...]
>> I'm also missing the possibility of the activity even when T = 0,
>
>The field equations only have accountability when T == 0. Otherwise,
>you can claim whatever you want and not subject to any experimental
>policeman. <shrug>
Just like F = ma, F = dp/dt, H|psi> = E|psi>, d/dt (@L/@v) - @L / @ q
= F_q right? I mean you can put ANYTHING in H, F, or F_q!
>
>> clearly: I mean, you can grok all you want how T couples to spacetime,
>> but even in vacuum, a lot of interesting stuff is going on. So
>> grokking the Einstein equation is at best half the... err... grok.
>> Kind of a gr or an ok. Hmm...
>
>Again, the field equations are a set of mathematical construct with no
>particular physical insights to justify their derivations. <shrug>
An amusing claim since everyone watching this thread _JUST_ watched
you trip and fall as I laid out the derivation of the field equations
step by step.
>
>> At least we see why GR is OK is exactly the correct term for
>> understanding, here. :-)
>
>GR is OK only under some special mathematical applications. <shrug>
Since you don't understand GR, this is yet another laughable claim.
>
>> Interesting how none of the stuff that goes on in vacuum, including,
>> apparently, gravity waves, involves energy or momentum. Incredible,
>> actually.
>
>Gravity wave is another mathematical construct squeezed out between
>the geodesic crack of two adjacent points in spacetime. It is the
>same by-product that yields the Riemann curvature tensor. <shrug>
This is wrong and provably so, but as you clearly don't know what you
are talking about there's no point.
You don't understand - we get it. Now dry up and blow away.
Koobee Wublee
> Koobee Wublee wrote:
> [more whining crap mercifully snipped]
>
> >Any sign change in the Ricci tensor will affect the field equations in
> >general. <shrug>
>
> Nope. The negative will carry through the Ricci tensor and the Ricci
> scalar, which is then absorbed into the proportionality constant.
This is just no true since there is the energy momentum tensor
involved. Maybe in a few years, you will release your mistake. In
the meantime, enjoy your mommy’s support of being a multi-year super-
senior.
Koobee Wublee
> Koobee Wublee wrote:
> >On Oct 29, 3:31 pm, Edward Green wrote:
> >R is defined as the dot/inner product of two matrices.
>
> >R = [g] * [R]
>
> No, it isn't. It is defined to be R = g_ab R^ab as tensors ARE NOT
> MATRICES.
What an idiot your are. <shrug>
I meant
R = [g^-1] * [R]
Where
** [g^-1] = inverse of [g]
** R = [g]^ij * [R]
** [g^ij] = Elements of [g^-1]
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 29, 10:16 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> >Oh, my. You said earlier that g is not a determinant. I have to snip
>> >all your whining wishy-washy crap. <shrug>
>>
>> You always knew what g was,
>
>And I always understand you meant g to be the matrix. It also
>reflects in your calculations. <shrug>
You are lying. Or stupid - there is no such operation "square root of
a matrix".
http://groups.google.com/group/sci.physics.relativity/msg/e57f2556ad2fa579?dmode=source
You weren't confused there - the determinant was explicitly spelled
out and you weren't complaining. You even read my source and saw the
same notation that I'm using NOW.
You have been arguing about GR under this pseudonym since 2005 and you
expect me to believe that you have been confused all this time? Try
again.
>
>> you just want to flee the discussion
>> because you lost.
>
>You are the wishy-washy one. I have told you from the very beginning
>that g_ij or g^ij represent the elements to the matrix g and g^-1
>respectively. You were the one confusing g_ij and g^ij as matrices.
><shrug>
Except they aren't just elements, the usage is clearly in reference to
ALL OF THEM. Just like actual matrix algebra!
S = SUM OVER i, SUM OVER j int(g^ij R_ij sqrt(-g) d^4x)
Are you that fucking stupid?
>
>> I alawys knew you wouldn't be able to follow the discussion even if I
>> wrote out every step for you. Thanks for playing.
>
>I just don’t have the patience to carry on stupid discussions.
><shrug>
You had plenty of patience until you got cornered. You fool nobody but
yourself.
eric gisse
So what? The proportionality constant is experimentally determined.
Under the alternate definition of the Ricci tensor the constant will
simply pick up a minus sign to counter the other minus sign to
preserve correspondence with Newtonian gravitation. Nothing changes.
Like I said:
"The symmetries of Riemann makes the choice of contraction irrelevant,
so I'm not sure why you are making a big deal out of it. Proving they
are the same is a simple exercise - why don't you show me where you
are having difficulties?"
a) I proved the symmetrys make the choice irrelevant. You either get
zero or the same thing.
b) I still don't know why you are making a big deal out of it beyond
the fact you are being proven wrong.
c) It is a simple exercise, and you couldn't find a single error in
the mathematics so you resort to your boilerplate whines about
'mathemagic' which is code for 'i don't understand what you just did'.
Again, thanks for playing. I can't make you look stupid without your
participation in the exercise.
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 29, 10:50 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>> >On Oct 29, 3:31 pm, Edward Green wrote:
>
>> >R is defined as the dot/inner product of two matrices.
>>
>> >R = [g] * [R]
>>
>> No, it isn't. It is defined to be R = g_ab R^ab as tensors ARE NOT
>> MATRICES.
>
>What an idiot your are. <shrug>
Three years under this name and you repeatedly prove you are clueless.
Is this how you intend to spend your retirement?
>
>I meant
>
>R = [g^-1] * [R]
You have no idea what you meant as you confuse yourself with your poor
notation.
R = g_ij R^ij
R = g^ij R_ij
It is rather simple but nature has to come along and throw in a
retired engineer who can't tell his ass from a hole in the ground but
insists he knows better than everyone else.
Go ahead, give ONE SINGLE REFERENCE that uses your confusion notation.
ONE REFERENCE.
Or <shrug>, insult me, and prove you are the blithering idiot everyone
knows you are.
Daryl McCullough
>
>A baby step...
>
>R is the trace, more or less, of R_ij (or should I write R_mn).
>
>So the Einstein tensor is possibly, more or less, the traceless part
>of R_ij.
But the Einstein tensor *isn't* traceless (I don't think so, anyway)
G_ij = R_ij - 1/2 g_ij R
trace(G_ij) = trace(R_ij) - 1/2 trace(g_ij) R
Since R = trace(R_ij), and trace(g_ij) = 4,
we have
trace(G_ij) = R - 2R = -R
Tom Roberts
> But the Einstein tensor *isn't* traceless
Right. It is divergenceless:
D_i G^ij = 0
Where D_i is the covariant derivative wrt x^i.
You are not expected to recognize this immediately, but this is a direct
consequence of the Bianchi identities.
Tom Roberts
RustyJames
thats because he introduce his tensor with a multi dimesional manifold
to explane the curviture of space time
try looking up local and global generalized Bianchi identities
for classical field theories by resorting to the gauge-natural
invariance of the
Lagrangian and via the application of the Noether Theorems
Koobee Wublee
> Koobee Wublee wrote:
> >And I always understand you meant g to be the matrix. It also
> >reflects in your calculations. <shrug>
>
> You are lying. Or stupid - there is no such operation "square root of
> a matrix".
A square root of a matrix actually makes sense if the matrix is
square. <shrug>
> http://groups.google.com/group/sci.physics.relativity/msg/e57f2556ad2...
http://groups.google.com/group/sci.physics.relativity/msg/3e178ac8395b2984
> You weren't confused there - the determinant was explicitly spelled
> out and you weren't complaining. You even read my source and saw the
> same notation that I'm using NOW.
You meant g to be the matrix itself. <shrug>
> You have been arguing about GR under this pseudonym since 2005 and you
> expect me to believe that you have been confused all this time? Try
> again.
Hormone imbalance again?
> >You are the wishy-washy one. I have told you from the very beginning
> >that g_ij or g^ij represent the elements to the matrix g and g^-1
> >respectively. You were the one confusing g_ij and g^ij as matrices.
> ><shrug>
>
> Except they aren't just elements, the usage is clearly in reference to
> ALL OF THEM. Just like actual matrix algebra!
Again, g_ij and g^ij are elements to the matrices g and g^-1
respectively where g^-1 is the inverse of g. <shrug>
> S = SUM OVER i, SUM OVER j int(g^ij R_ij sqrt(-g) d^4x)
If you indeed mean g to be the determinant of g. <shrug>
> Are you that fucking stupid?
Shouting does not get you anywhere. <shrug>
> >I just don’t have the patience to carry on stupid discussions.
> ><shrug>
>
> You had plenty of patience until you got cornered.
There is no need to behave like a four-year-old when you are pointed
out of your mistakes. <shrug>
> You fool nobody but yourself.
You are wrong. I don’t, won’t, and will not attempt to fool anyone.
<shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >This is just no true since there is the energy momentum tensor
> >involved. Maybe in a few years, you will release your mistake. In
> >the meantime, enjoy your mommy’s support of being a multi-year super-
> >senior.
>
> So what? The proportionality constant is experimentally determined.
Not true. That constant was mathematically fudged to 8 pi something.
<shrug>
> Under the alternate definition of the Ricci tensor the constant will
> simply pick up a minus sign to counter the other minus sign to
> preserve correspondence with Newtonian gravitation. Nothing changes.
Well, that is because you have ignored the Lagrangian of mass term
that will allow you to get the energy momentum tensor. <shrug>
> Like I said:
>
> "The symmetries of Riemann makes the choice of contraction irrelevant,
> so I'm not sure why you are making a big deal out of it. Proving they
> are the same is a simple exercise - why don't you show me where you
> are having difficulties?"
For Chirst’s sake, there is no restriction on if the arrangement of
the connection coefficients has to be symmetric. Other than the
Christoffel symbols of the 2nd kind, I have shown you one that is not
symmetric. <shrug>
> a) I proved the symmetrys make the choice irrelevant. You either get
> zero or the same thing.
You prove jack sh*t. <shrug>
> b) I still don't know why you are making a big deal out of it beyond
> the fact you are being proven wrong.
That is because you are unreasonable. <shrug>
> c) It is a simple exercise, and you couldn't find a single error in
> the mathematics so you resort to your boilerplate whines about
> 'mathemagic' which is code for 'i don't understand what you just did'.
That is because you believe in mathemagics instead of rationalizing
with mathematics. <hrug>
> Again, thanks for playing. I can't make you look stupid without your
> participation in the exercise.
The stupidity is actually on you, mutli-year super-senior who cannot
even do 7th grade algebra. <shrug>
Koobee Wublee
> Edward Green says...
> >R is the trace, more or less, of R_ij (or should I write R_mn).
>
> >So the Einstein tensor is possibly, more or less, the traceless part
> >of R_ij.
>
> But the Einstein tensor *isn't* traceless (I don't think so, anyway)
>
> G_ij = R_ij - 1/2 g_ij R
>
> trace(G_ij) = trace(R_ij) - 1/2 trace(g_ij) R
What is so big deal about this trace anyway?
> Since R = trace(R_ij), and trace(g_ij) = 4,
> we have
According to the following definition,
http://en.wikipedia.org/wiki/Trace_(linear_algebra)
trace([R]) = sum(i=0..3) of [R]_ii
Where
** [R] = Ricci tensor, a 4x4 matrix
And
trace([g]) = sum(i=0..3) of [g]_ii
Where
** [g] = metric, a 4x4 matrix
Given the spherically symmetric polar coordinate system, it does not
even make any sense to talk about this trace because of different
units in the elements of the metric.
Thus, you actually mean the trace to be
trace([R]) = [g^-1] * [R]
Where
** [g^-1] * [R] = sum(i=0..3,j=0..3) of ([g^-1]^ij [R]_ij)
And
trace([g]) = [g^-1] * [g]
Where
** [g^-1] * [g] = sum(i=0..3,j=0..3) of ([g^-1]^ij [R]_ij)
Also, as you have pointed out,
[g^-1]^ij [g] = 4
> trace(G_ij) = R - 2R = -R
Given the field equations,
[G] = k [g]
Where
** [G] = Einstein tensor, a 4x4 matrix
** k [g] = Energy momentum tensor, a 4x4 matrix
** k = 8 pi G rho something
[g^-1] * [G] = k [g^-1] * [g]
Or
trace([G]) = 4 k
A constant, not – R
<shrug>
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 29, 11:27 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> >And I always understand you meant g to be the matrix. It also
>> >reflects in your calculations. <shrug>
>>
>> You are lying. Or stupid - there is no such operation "square root of
>> a matrix".
>
>A square root of a matrix actually makes sense if the matrix is
>square. <shrug>
You should be able to define the square root of a matrix then.
My usage was obvious - you are just an idiot. But since you know now,
go back and read the derivation again.
>
>> http://groups.google.com/group/sci.physics.relativity/msg/e57f2556ad2...
>
>http://groups.google.com/group/sci.physics.relativity/msg/3e178ac8395b2984
>
>> You weren't confused there - the determinant was explicitly spelled
>> out and you weren't complaining. You even read my source and saw the
>> same notation that I'm using NOW.
>
>You meant g to be the matrix itself. <shrug>
Nope - g is and always was the _determinant_ of the metric. You have
no idea what I meant, so don't presume.
[...]
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 29, 11:33 pm, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> >This is just no true since there is the energy momentum tensor
>> >involved. Maybe in a few years, you will release your mistake. In
>> >the meantime, enjoy your mommy’s support of being a multi-year super-
>> >senior.
>>
>> So what? The proportionality constant is experimentally determined.
>
>Not true. That constant was mathematically fudged to 8 pi something.
><shrug>
I cannot think of a stronger claim for you not understanding general
relativity than this.
groups.google.com author:wublee
"mathemagical" : 35 hits
"mathemagic" : 52 hits
"mathemagics" : 8 hits
"magic" : 41 hits
"fudged" : 60 hits
None of this stuff is magical - YOU JUST DO NOT UNDERSTAND.
The fixing of the proportionality constant in G_uv = k T_uv is a
_textbook exercise_. Funny how you just never seem to get those.
>
>> Under the alternate definition of the Ricci tensor the constant will
>> simply pick up a minus sign to counter the other minus sign to
>> preserve correspondence with Newtonian gravitation. Nothing changes.
>
>Well, that is because you have ignored the Lagrangian of mass term
>that will allow you to get the energy momentum tensor. <shrug>
Not ignored - put to the side. You couldn't get through the variation
of the square root of the metric determinant, so there was no point in
discussing that.
Regardless, as I said before the extra negative sign can be absorbed
into the proportionality constant and will be negated anyway when the
comparison against Newton is made.
>
>> Like I said:
>>
>> "The symmetries of Riemann makes the choice of contraction irrelevant,
>> so I'm not sure why you are making a big deal out of it. Proving they
>> are the same is a simple exercise - why don't you show me where you
>> are having difficulties?"
>
>For Chirst’s sake, there is no restriction on if the arrangement of
>the connection coefficients has to be symmetric. Other than the
>Christoffel symbols of the 2nd kind, I have shown you one that is not
>symmetric. <shrug>
Yes, there is. The symmetry of the metric and the torsion-free
condition forces the symmetry of the connection.
This is all BY DEFINITION in a Riemannian manifold. If you want to
argue about non-symmetric connections you are arguing about /something
else/. Why do I have to explain basic definitions to someone who
thinks he understands the theory?
>
>> a) I proved the symmetrys make the choice irrelevant. You either get
>> zero or the same thing.
>
>You prove jack sh*t. <shrug>
Yet you can't find error in the proof. Why?
>
>> b) I still don't know why you are making a big deal out of it beyond
>> the fact you are being proven wrong.
>
>That is because you are unreasonable. <shrug>
Oh yes, so unreasonable. Asking you to support your assertions?
UNREASONABLE. Asking for references? UNREASONABLE. Asking for you to
explain why you think I'm wrong? UNREASONABLE.
>
>> c) It is a simple exercise, and you couldn't find a single error in
>> the mathematics so you resort to your boilerplate whines about
>> 'mathemagic' which is code for 'i don't understand what you just did'.
>
>That is because you believe in mathemagics instead of rationalizing
>with mathematics. <hrug>
I'm yet to see you make one actual _argument_ against my mathematics -
why is that?
You can't negate a proof just because I know stuff that you don't.
>
>> Again, thanks for playing. I can't make you look stupid without your
>> participation in the exercise.
>
>The stupidity is actually on you, mutli-year super-senior who cannot
>even do 7th grade algebra. <shrug>
Try to remember who wrote practically all the math in this thread.
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 30, 9:42 am, Daryl McCullough wrote:
>> Edward Green says...
>
>> >R is the trace, more or less, of R_ij (or should I write R_mn).
>>
>> >So the Einstein tensor is possibly, more or less, the traceless part
>> >of R_ij.
>>
>> But the Einstein tensor *isn't* traceless (I don't think so, anyway)
>>
>> G_ij = R_ij - 1/2 g_ij R
>>
>> trace(G_ij) = trace(R_ij) - 1/2 trace(g_ij) R
>
>What is so big deal about this trace anyway?
Nothing - it is simply one tool among many, which you don't understand
since you think any invokation of something like this is 'magic'.
>
>> Since R = trace(R_ij), and trace(g_ij) = 4,
>> we have
>
>According to the following definition,
>
>http://en.wikipedia.org/wiki/Trace_(linear_algebra)
Gosh - right there on that page:
det(exp(A)) = exp(tr(A))
>
>trace([R]) = sum(i=0..3) of [R]_ii
>
>Where
>
>** [R] = Ricci tensor, a 4x4 matrix
>
>And
>
>trace([g]) = sum(i=0..3) of [g]_ii
>
>Where
>
>** [g] = metric, a 4x4 matrix
Still don't know the difference between a matrix and a tensor? Looks
like you never will.
>
>Given the spherically symmetric polar coordinate system, it does not
>even make any sense to talk about this trace because of different
>units in the elements of the metric.
Fucking 'tard. The trace of a tensor is a coordinate-independent
quantity. It is an easy proof which you - amazingly - still do not
understand.
The trace of the metric is g_ij g^ij = delta(i,j) delta(i,j)
That the trace of this is equal to 4 is Pretty Fucking Obvious(tm).
Except to you, of course. But you are an idiot.
>
>Thus, you actually mean the trace to be
Bwahah. The pattern continues!
First you say something stupid and wrong. Then you notice your stupid
and wrong statement conflicts with what a more educated person says.
Finally, you realize that the educated person MUST be wrong so you
"re-interpret" what was said into something that is more in-line with
your stupid and wrong statement.
>
>trace([R]) = [g^-1] * [R]
>
>Where
>
>** [g^-1] * [R] = sum(i=0..3,j=0..3) of ([g^-1]^ij [R]_ij)
Except you have no idea how to get from here to g_ii. You have no idea
how to manipulate the indices, so you write stuff like this.
>
>And
>
>trace([g]) = [g^-1] * [g]
>
>Where
>
>** [g^-1] * [g] = sum(i=0..3,j=0..3) of ([g^-1]^ij [R]_ij)
>
>Also, as you have pointed out,
>
>[g^-1]^ij [g] = 4
>
>> trace(G_ij) = R - 2R = -R
>
>Given the field equations,
>
>[G] = k [g]
What a moron. You just wrote that the Einstein tensor is proportional
to the metric in your fucked-up, confusing notation.
>
>Where
>
>** [G] = Einstein tensor, a 4x4 matrix
TENSOR. Retard.
>** k [g] = Energy momentum tensor, a 4x4 matrix
TENSOR. Retard.
This new bit of stupidity certainly fits in with your past stupidities
about how the energy-momentum tensor is a 'constant'.
>** k = 8 pi G rho something
>
>[g^-1] * [G] = k [g^-1] * [g]
Oh look the retard attempts to take the trace using matrix mechanics
because he has no fucking idea how to manipulate the tensors.
The stress-energy tensor isn't proportional to the metric, you
blithering idiot. Is there nothing you can't fuck up with your years
of arrogant stupidity?
>
>Or
>
>trace([G]) = 4 k
Wrong again, chuckles.
>
>A constant, not – R
>
><shrug>
"Oh look I just said something ridiculously stupid! Time to shrug and
walk away!"
Why not <fart> ? Same effect - you stink up the place then leave.
Koobee Wublee
> Koobee Wublee wrote:
> >Not true. That constant was mathematically fudged to 8 pi something.
> ><shrug>
>
> I cannot think of a stronger claim for you not understanding general
> relativity than this.
Your opinion to me is very worthless. <shrug>
> groups.google.com author:wublee
>
> "mathemagical" : 35 hits
> "mathemagic" : 52 hits
> "mathemagics" : 8 hits
> "magic" : 41 hits
> "fudged" : 60 hits
Wow, that many hits for me!!!
> None of this stuff is magical - YOU JUST DO NOT UNDERSTAND.
I DO, BUT YOU DON’T. THAT IS WHY YOU DO NOT SEE THE MATEHMAGICS IN
THEM. <SHRUG>
> The fixing of the proportionality constant in G_uv = k T_uv is a
> _textbook exercise_. Funny how you just never seem to get those.
So what if it is fudged in textbooks. <shrug> Which textbooks that
you are sitting on again?
> >Well, that is because you have ignored the Lagrangian of mass term
> >that will allow you to get the energy momentum tensor. <shrug>
>
> Not ignored - put to the side.
I will remember that defense. Somehow, I doubt it that it would go
anywhere. <shrug>
> You couldn't get through the variation
You cannot realize even if I do. Who is it to judge? A multi-year
super-senior without any college degree?
> of the square root of the metric determinant, so there was no point in
> discussing that.
Apparently, your aptitude is very low. That explains why you are a
multi-year super-senior. Your bully-like attitude needs to mitigate.
Or else, you are not getting anywhere. <shrug>
> Regardless, as I said before the extra negative sign can be absorbed
> into the proportionality constant and will be negated anyway when the
> comparison against Newton is made.
I have told you many times that your wishful thinking cannot happen if
the Lagrangian mass term is present. <shrug>
> >For Chirst’s sake, there is no restriction on if the arrangement of
> >the connection coefficients has to be symmetric. Other than the
> >Christoffel symbols of the 2nd kind, I have shown you one that is not
> >symmetric. <shrug>
>
> Yes, there is. The symmetry of the metric and the torsion-free
> condition forces the symmetry of the connection.
That is total nonsense. <shrug>
> This is all BY DEFINITION in a Riemannian manifold. If you want to
> argue about non-symmetric connections you are arguing about /something
> else/. Why do I have to explain basic definitions to someone who
> thinks he understands the theory?
I have shown you that the arrangement of the connection coefficients
does not have to be symmetrical. There is no rational argument to
counter that. <shrug>
> >You prove jack sh*t. <shrug>
>
> Yet you can't find error in the proof. Why?
You are not listening. <shrug>
> >That is because you are unreasonable. <shrug>
>
> Oh yes, so unreasonable. Asking you to support your assertions?
> UNREASONABLE. Asking for references? UNREASONABLE. Asking for you to
> explain why you think I'm wrong? UNREASONABLE.
Yes, multi-year super-senior with no college degree. Mommy’s
financial support into your late 20’s is very rare in this society any
more. <shrug>
> >That is because you believe in mathemagics instead of rationalizing
> >with mathematics. <hrug>
>
> [...]
More yapping crap snipped. Yes, mercifully.
> >The stupidity is actually on you, mutli-year super-senior who cannot
> >even do 7th grade algebra. <shrug>
>
> Try to remember who wrote practically all the math in this thread.
Koobee Wublee did. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >A square root of a matrix actually makes sense if the matrix is
> >square. <shrug>
>
> You should be able to define the square root of a matrix then.
Yes, through Taylor series expansion, but that is irrelevant. <shrug>
> My usage was obvious - you are just an idiot. But since you know now,
> go back and read the derivation again.
You are a devoted worshipper of Einstein the nitwit, the plagiarist,
and the liar. <shrug>
> >You meant g to be the matrix itself. <shrug>
>
> Nope - g is and always was the _determinant_ of the metric. You have
> no idea what I meant, so don't presume.
This is nonsense. g usually means the matrix that is the metric
itself. To express it as the determinant, it is more clear to say
det[g] instead. That way there is no confusion. <shrug>
You are really something else. It is so obvious that you are trying
to hide your own ignorance by blowing your screw-up in terminology
like that. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >What is so big deal about this trace anyway?
>
> Nothing - it is simply one tool among many, which you don't understand
> since you think any invokation of something like this is 'magic'.
Oh, you are elevating yourself again from the status of that multi-
year super-senior. <shrug>
> >According to the following definition,
>
> >http://en.wikipedia.org/wiki/Trace_(linear_algebra)
>
> Gosh - right there on that page:
>
> det(exp(A)) = exp(tr(A))
Yeah? <shrug>
> >trace([R]) = sum(i=0..3) of [R]_ii
>
> >Where
>
> >** [R] = Ricci tensor, a 4x4 matrix
>
> >And
>
> >trace([g]) = sum(i=0..3) of [g]_ii
>
> >Where
>
> >** [g] = metric, a 4x4 matrix
>
> Still don't know the difference between a matrix and a tensor? Looks
> like you never will.
A matrix is merely a rank-n tensor where n is less than or equal to
2. <shrug>
> >Given the spherically symmetric polar coordinate system, it does not
> >even make any sense to talk about this trace because of different
> >units in the elements of the metric.
>
> Fucking 'tard. The trace of a tensor is a coordinate-independent
> quantity. It is an easy proof which you - amazingly - still do not
> understand.
Well, you are the f*cking retard since the following clearly defined
what the trace of a matrix is. Your comprehension skill really
reflects that you remain a multi-year super-senior today. <shrug>
> The trace of the metric is g_ij g^ij = delta(i,j) delta(i,j)
Hmmm... That is merely a definition. <shrug>
> That the trace of this is equal to 4 is Pretty Fucking Obvious(tm).
> Except to you, of course. But you are an idiot.
It depends on how you want to define that operator ‘trace’. <shrug>
> >Thus, you actually mean the trace to be
>
> Bwahah. The pattern continues!
Your whining crap is showing all over now as usual.
> First you say something stupid and wrong. Then you notice your stupid
> and wrong statement conflicts with what a more educated person says.
> Finally, you realize that the educated person MUST be wrong so you
> "re-interpret" what was said into something that is more in-line with
> your stupid and wrong statement.
You are just way too shallow minded. Trace is a user-defined
operator. You can define it whatever you want it to be. <shrug>
Your lack of understand once again reflects your status as a multi-
year super-senior today without any college degree. Is mommy proud of
that?
> >trace([R]) = [g^-1] * [R]
>
> >Where
>
> >** [g^-1] * [R] = sum(i=0..3,j=0..3) of ([g^-1]^ij [R]_ij)
>
> Except you have no idea how to get from here to g_ii. You have no idea
> how to manipulate the indices, so you write stuff like this.
<shrug>
> >And
>
> >trace([g]) = [g^-1] * [g]
>
> >Where
>
> >** [g^-1] * [g] = sum(i=0..3,j=0..3) of ([g^-1]^ij [R]_ij)
>
> >Also, as you have pointed out,
>
> >[g^-1]^ij [g] = 4
>
> >> trace(G_ij) = R - 2R = -R
>
> >Given the field equations,
>
> >[G] = k [g]
>
> What a moron. You just wrote that the Einstein tensor is proportional
> to the metric in your fucked-up, confusing notation.
Einstein tensor is proportional to the metric, yes, according to the
original derivation of the field equations. <shrug>
You don’t see that because you have refused to add the Lagrangian of
mass term into that so-call Lagrangian. <shrug>
> >Where
>
> >** [G] = Einstein tensor, a 4x4 matrix
>
> TENSOR. Retard.
A matrix is rank-0, 1, or 2 tensor. <shrug>
> >** k [g] = Energy momentum tensor, a 4x4 matrix
>
> TENSOR. Retard.
A matrix is rank-0, 1, or 2 tensor. <shrug>
> This new bit of stupidity certainly fits in with your past stupidities
> about how the energy-momentum tensor is a 'constant'.
Energy momentum tensor according to the traditional form of the field
equations cannot be a constant because of the metric term. <shrug>
> >** k = 8 pi G rho something
>
> >[g^-1] * [G] = k [g^-1] * [g]
>
> Oh look the retard attempts to take the trace using matrix mechanics
> because he has no fucking idea how to manipulate the tensors.
Yes, the multi-year super-senior is jealous. <shrug>
> The stress-energy tensor isn't proportional to the metric, you
> blithering idiot. Is there nothing you can't fuck up with your years
> of arrogant stupidity?
You are just wrong about this. Once again, if you include the
Lagrangian of mass term into the overall Lagrangian, you will arrive
at the same conclusion as I do. That is call mathematics not
matheMAGICS. <shrug>
> >Or
>
> >trace([G]) = 4 k
>
> Wrong again, chuckles.
Keep giggling. The fact remains that you are a multi-year super-
senior without any college degree. <shrug>
> >A constant, not – R
>
> ><shrug>
>
> "Oh look I just said something ridiculously stupid! Time to shrug and
> walk away!"
That is the best quote of Eric Gisse on himself. I will remember
that.
> Why not <fart> ? Same effect - you stink up the place then leave.
<shrug>
eric gisse
<koobee...@gmail.com> wrote:
[...]
>> None of this stuff is magical - YOU JUST DO NOT UNDERSTAND.
>
>I DO, BUT YOU DON’T. THAT IS WHY YOU DO NOT SEE THE MATEHMAGICS IN
>THEM. <SHRUG>
Then start participating and stop sniping. I wrote the math - point
out the error.
>
>> The fixing of the proportionality constant in G_uv = k T_uv is a
>> _textbook exercise_. Funny how you just never seem to get those.
>
>So what if it is fudged in textbooks. <shrug> Which textbooks that
>you are sitting on again?
Prove it is fudged. I'll go through the derivation top to bottom if
you'll participate.
>
>> >Well, that is because you have ignored the Lagrangian of mass term
>> >that will allow you to get the energy momentum tensor. <shrug>
>>
>> Not ignored - put to the side.
>
>I will remember that defense. Somehow, I doubt it that it would go
>anywhere. <shrug>
>
>> You couldn't get through the variation
>
>You cannot realize even if I do. Who is it to judge? A multi-year
>super-senior without any college degree?
Who decided to rationalize deleting my entire derivation with a
balatant lie? Hint: it was you.
>
>> of the square root of the metric determinant, so there was no point in
>> discussing that.
>
>Apparently, your aptitude is very low. That explains why you are a
>multi-year super-senior. Your bully-like attitude needs to mitigate.
>Or else, you are not getting anywhere. <shrug>
>
>> Regardless, as I said before the extra negative sign can be absorbed
>> into the proportionality constant and will be negated anyway when the
>> comparison against Newton is made.
>
>I have told you many times that your wishful thinking cannot happen if
>the Lagrangian mass term is present. <shrug>
Sure - the way you do it.
>
>> >For Chirst’s sake, there is no restriction on if the arrangement of
>> >the connection coefficients has to be symmetric. Other than the
>> >Christoffel symbols of the 2nd kind, I have shown you one that is not
>> >symmetric. <shrug>
>>
>> Yes, there is. The symmetry of the metric and the torsion-free
>> condition forces the symmetry of the connection.
>
>That is total nonsense. <shrug>
Prove it.
>
>> This is all BY DEFINITION in a Riemannian manifold. If you want to
>> argue about non-symmetric connections you are arguing about /something
>> else/. Why do I have to explain basic definitions to someone who
>> thinks he understands the theory?
>
>I have shown you that the arrangement of the connection coefficients
>does not have to be symmetrical. There is no rational argument to
>counter that. <shrug>
Yeah there is.
a) The connection is torsion-free. That means - in your notation -
C^a_[cd] = 0. That's fundamental to general relativity, and the
assumption is _always_ made explicit.
This isn't a condition that has to be true - allowing for torsion
obtains Einstein-Cartan theory. This ground has been traveled - look
some time.
b) The metric is symmetric. That one is pretty fucking simple - do you
disagree that the metric has to be symmetric?
[...]
eric gisse
<koobee...@gmail.com> wrote:
[snip]
>
>You are really something else. It is so obvious that you are trying
>to hide your own ignorance by blowing your screw-up in terminology
>like that. <shrug>
My usage is consistent with references on the subject going back to
Misner, Thorne, and Wheeler which is now 30+ years old, and is fucking
/obvious/ from context. I have no idea why you'd think the square root
of g is a matrix quantity, since the entire entity under consideration
is a SCALAR.
Furthermore you should fucking KNOW the Einstein-Hilbert action. You
should KNOW what the terms mean, and you should DAMN well better know
that every integral will be weighted by a factor of sqrt(-g) because
the volume element is only a density.
I'm reminded of a certain 'tard in one of my upper division classes
who argued that 1/r^2 in an integral was 'confusing' because it 'could
have been' not just r.r but (POSSIBLY!) the cross product r x r or the
dyad product rr.
Like there, the usage is obvious from context. Like there, the person
arguing had more mouth than sense.
Now, once again here is the derivation. If you think something is
wrong, say why you think it is wrong. Whining about 'mathematic',
'fudging', or simply asserting something is 'wrong' doesn't make what
I wrote wrong. Find the error or accept what is written so I can move
on to the variation of R_ij.
The Einstein-Hilbert action S is equal to int( R sqrt(-g) d^4x). My
usage of terminology is consistent with the literature - stick with
it, or shut up.
I'm not altering the derivation to suit your personal vocabulary.
Ever.
Recall that the Lagrange equations are obtained by varying the action
with respect to the field variables, and demanding that the variation
is equal to zero at the boundary points.
So the variation considered will be in the field variables g_ij in the
form g^ij --> g^ij + delta(g^ij).
S = int( R_ij g^ij sqrt(-g) d^4x) with g = det(g_ij).
delta(S) = delta( int( R_ij g^ij sqrt(-g) d^4x) ) +
The variation behaves like a differential operator, so the
terms are expanded as such.
delta(S) = int( delta(R_ij) g^ij sqrt(-g) d^4x) + int( R_ij
delta(g^ij) sqrt(-g) d^4x) + int( R_ij g^ij delta(sqrt(-g)) d^4x)
delta(S) = A + B + C
The B term is already varied with respect to the field variables, so
that chunk of the variation is done.
The C term will now be considered. Normally I would just march through
term by term, but I have my doubts that you'll accept the derivation.
As before, the goal is to get the entire variation in terms of the
varied metric, as opposed to varied functions of the metric.
C: int( R_ij g^ij delta(sqrt(-g)) d^4x)
Now to treat the delta(sqrt(-g) term:
Basic fact about square matrices: ln( det(A) ) = tr( ln(A) )
I'm not proving it - open a linear algebra textbook or look at the
Wikipedia page you cited earlier today.
Vary with respect to A, and you have:
1/det(A) * delta(det(A)) = tr(A^-1 delta(A))
The infintesimal variation obeys the same rules as the differential
operator.
Now, exchange A for g_ij.
1/det(g_ij) * delta(det(g_ij)) = g^ij delta(g_ij)
delta(g) = g g^ij delta(g_ij)
Using delta(g_ij) = - g_ia g_jb delta(g^ab), you have:
(i) delta(g) = -g g_ij delta(g^ij)
Now carry the infintesimal through delta(sqrt(-g)):
(ii) delta(sqrt(-g)) = -delta(g) / ( 2 sqrt( -g )
Substitute (i) into (ii) and you get:
delta(sqrt(-g)) = - 1/2 sqrt(-g) g_ij delta(g^ij)
Variation done. Any questions? Whines? Wholesale snips of material?
eric gisse
<koobee...@gmail.com> wrote:
[...]
>>
>> Still don't know the difference between a matrix and a tensor? Looks
>> like you never will.
>
>A matrix is merely a rank-n tensor where n is less than or equal to
>2. <shrug>
Nope. A tensor obeys a certain transformation rule when transforming
the elements from one coordinate system to another.
You cannot provide one reference that supports your claims, but that
doesn't stop you from repeating them endlessly.
>
>> >Given the spherically symmetric polar coordinate system, it does not
>> >even make any sense to talk about this trace because of different
>> >units in the elements of the metric.
>>
>> Fucking 'tard. The trace of a tensor is a coordinate-independent
>> quantity. It is an easy proof which you - amazingly - still do not
>> understand.
>
>Well, you are the f*cking retard since the following clearly defined
>what the trace of a matrix is. Your comprehension skill really
>reflects that you remain a multi-year super-senior today. <shrug>
>
>> The trace of the metric is g_ij g^ij = delta(i,j) delta(i,j)
>
>Hmmm... That is merely a definition. <shrug>
It is THE definition. Why do you keep invoking your own personal set
of words, symbols, and mathematics for a well defined field?
>
>> That the trace of this is equal to 4 is Pretty Fucking Obvious(tm).
>> Except to you, of course. But you are an idiot.
>
>It depends on how you want to define that operator ‘trace’. <shrug>
>
>> >Thus, you actually mean the trace to be
>>
>> Bwahah. The pattern continues!
>
>Your whining crap is showing all over now as usual.
>
>> First you say something stupid and wrong. Then you notice your stupid
>> and wrong statement conflicts with what a more educated person says.
>> Finally, you realize that the educated person MUST be wrong so you
>> "re-interpret" what was said into something that is more in-line with
>> your stupid and wrong statement.
>
>You are just way too shallow minded. Trace is a user-defined
>operator. You can define it whatever you want it to be. <shrug>
Nope - the trace of a matrix [or a tensor] is the sum of the diagonal
components. There is no 'alternate' definition.
[...]
>> >Given the field equations,
>>
>> >[G] = k [g]
>>
>> What a moron. You just wrote that the Einstein tensor is proportional
>> to the metric in your fucked-up, confusing notation.
>
>Einstein tensor is proportional to the metric, yes, according to the
>original derivation of the field equations. <shrug>
>
You are lying since this is /clearly/ not true. You can read
Einstein's 1916 publication of general relativity /yourself/ and see
that it isn't true. Or you can keep lying because you are shameless in
your dishonesty.
>You don’t see that because you have refused to add the Lagrangian of
>mass term into that so-call Lagrangian. <shrug>
The variation of the matter term is trivial. I simply won't do it
until you can follow through something more relevant.
>
>> >Where
>>
>> >** [G] = Einstein tensor, a 4x4 matrix
>>
>> TENSOR. Retard.
>
>A matrix is rank-0, 1, or 2 tensor. <shrug>
>
>> >** k [g] = Energy momentum tensor, a 4x4 matrix
>>
>> TENSOR. Retard.
>
>A matrix is rank-0, 1, or 2 tensor. <shrug>
>
>> This new bit of stupidity certainly fits in with your past stupidities
>> about how the energy-momentum tensor is a 'constant'.
>
>Energy momentum tensor according to the traditional form of the field
>equations cannot be a constant because of the metric term. <shrug>
There is no 'metric term'.
This is what I'm talking about with your pattern of stupidity. You say
something wrong and stupid - notice the contradiction - and say even
more wrong and stupid things to preserve the original wrong and stupid
assertion.
[snip rest]
RustyJames
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
If the stress-energy tensor is the result of varying the action with
respect to the metric tensor, then it must be symmetric because the
metric tensor is symmetric
no mathmagic just logic
Don Stockbauer
> On Oct 26, 9:48 am, Edward Green <spamspamsp...@netzero.com> wrote:
>
> > On Oct 25, 8:36 am, Don Stockbauer <donstockba...@hotmail.com> wrote:
>
> > > Edward Green wrote:
> > > > I feel it is time to grok the Einstein equation.
>
> > > Who the hell says "grok" anymore???????
>
> > I do.
>
> > It's a good word, whose meaning seems to match its sound.
>
> > Next question?
>
> So its meaning is "the sound a frog makes in a pond"?
All I'm saying is that back when Heinlein wrote "Stranger in a Strange
Land" it was a wonderful word and people used it freely but now it
comes across as dated and younger folks have never heard it but in
reality I wish such were not the case for I agree with everything said
here in defense of it and maybe I wish I had more courage to trot it
out every now and again.
Jeff▲Relf
but listen closely, not for very much longer, I've got to keep control.
I remember doing the Time Warp, drinking those moments when
the blackness would hit me and the void would be calling.
Let's do the time warp again .. Let's do the time warp again !
It's just a jump to the left and then a step to the right;
with your hands on your hips, you bring your knees in tight.
But it's the pelvic thrust that really drives you insane,
Let's do the Time Warp again !
It's so dreamy, oh fantasy free me
so you can't see me,
no not at all
In another dimension,
with voyeuristic intention
Well-secluded, I see all, with a bit of a mind flip;
you're there in the time slip.
And nothing can ever be the same, you're spaced out on sensation,
like you're under sedation
Let's do the Time Warp again !
Well I was walking down the street, just a-having a think ..
when a snake of a guy gave me an evil wink.
He shook me up, he took me by surprise,
He had a pickup truck and the devil's eyes.
He stared at me and I felt a change
Time meant nothing, never would again.
Let's do the Time Warp again !
Hayek
I immediately thought about this book when I saw the subject line, but I
presumed it had other slang use.
I especially liked "Door into Summer" , because of the time travel, and
the tom cat. I have a Tom, and now its turning almost winter and he is
constantly complaining about the lack of this door.
About the Einstein equation : quite impossible to "grok" it.
Start with Clifford Will's "Was Einstein right", with the thought
experiments that inspired Einstein to the theory, which led to the
equations.
In no time, you will outgeodese those who claim to understand the
equations, but absolutely do not, because in contradiction with
Einstein's Original thought experiments.
Only solutions consisting of a black hole are well known, and those are
worth studying in order to get more insight in the matter.
Frame dragging really gets dramatic with black holes, and illustrate the
issue better.
As a result of this frame dragging, the inertia made by the universe
looses against the inertia of the Black hole, and the orbits of objects
around a BH catastrophically decay. Compared to that, the Mercury
perihelion shift is peanuts.
Beware of GR "specialists" that do not grok inertia.
Uwe Hayek.
Koobee Wublee
> Koobee Wublee wrote:
> >A matrix is merely a rank-n tensor where n is less than or equal to
> >2. <shrug>
>
> Nope. A tensor obeys a certain transformation rule when transforming
> the elements from one coordinate system to another.
This is just not true. There is no such restriction on tensors.
<shrug>
> [crap snipped]
>
> >Einstein tensor is proportional to the metric, yes, according to the
> >original derivation of the field equations. <shrug>
>
> You are lying since this is /clearly/ not true. You can read
> Einstein's 1916 publication of general relativity /yourself/ and see
> that it isn't true. Or you can keep lying because you are shameless in
> your dishonesty.
Einstein was a nitwit, a plagiarist, and a liar. <shrug>
> [more crap snipped]
Koobee Wublee
> Koobee Wublee wrote:
> [crap snipped]
> >I have shown you that the arrangement of the connection coefficients
> >does not have to be symmetrical. There is no rational argument to
> >counter that. <shrug>
>
> Yeah there is.
>
> a) The connection is torsion-free. That means - in your notation -
> C^a_[cd] = 0. That's fundamental to general relativity, and the
> assumption is _always_ made explicit.
You are totally f*cked up. There is no requirement on the Christoffel
symbols to be null. <shrug>
> [rest of crap also mercifully snipped]
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 31, 3:37 am, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> >A matrix is merely a rank-n tensor where n is less than or equal to
>> >2. <shrug>
>>
>> Nope. A tensor obeys a certain transformation rule when transforming
>> the elements from one coordinate system to another.
>
>This is just not true. There is no such restriction on tensors.
><shrug>
Sure there is.
http://en.wikipedia.org/wiki/Tensor
http://mathworld.wolfram.com/Tensor.html
>
>> [crap snipped]
>>
>> >Einstein tensor is proportional to the metric, yes, according to the
>> >original derivation of the field equations. <shrug>
>>
>> You are lying since this is /clearly/ not true. You can read
>> Einstein's 1916 publication of general relativity /yourself/ and see
>> that it isn't true. Or you can keep lying because you are shameless in
>> your dishonesty.
>
>Einstein was a nitwit, a plagiarist, and a liar. <shrug>
So you have to lie to make your point?
>
>> [more crap snipped]
eric gisse
Except I didn't say that. Learn to read.
Koobee Wublee
> Koobee Wublee wrote:
> [crap snipped]
>
> Furthermore you should fucking KNOW the Einstein-Hilbert action. You
> should KNOW what the terms mean, and you should DAMN well better know
> that every integral will be weighted by a factor of sqrt(-g) because
> the volume element is only a density.
No, I do not know what the Einstein-Hilbert action really means.
<shrug>
> [more crap snipped]
>
> The Einstein-Hilbert action S is equal to int( R sqrt(-g) d^4x).
What does that mean rationally? I see it as a corpse sewn together
from cadavers scavenged from the neighborhood morgue. <shrug>
> [yapping nonsense snipped]
> Recall that the Lagrange equations are obtained by varying the action
> with respect to the field variables, and demanding that the variation
> is equal to zero at the boundary points.
Why does this action have to be extremized?
> So the variation considered will be in the field variables g_ij in the
> form g^ij --> g^ij + delta(g^ij).
Why is that S has to be extremized while varying g^ij?
> S = int( R_ij g^ij sqrt(-g) d^4x) with g = det(g_ij).
With g = determinant, that is more clear what you are doing now.
Thank you.
> delta(S) = delta( int( R_ij g^ij sqrt(-g) d^4x) ) +
>
> The variation behaves like a differential operator, so the
> terms are expanded as such.
>
> delta(S) = int( delta(R_ij) g^ij sqrt(-g) d^4x) + int( R_ij
> delta(g^ij) sqrt(-g) d^4x) + int( R_ij g^ij delta(sqrt(-g)) d^4x)
You should be writing down the following instead.
dS = d(int(g^ij R_ij sqrt(- det(g)) d^4q)
With variation in g^ij, you get
dS = int(R_ij sqrt(- det(g)) dg^ij+ g^ij R_ij @sqrt(- det(g))/@g^ij
dg^ij d^4q)
> delta(S) = A + B + C
>
> The B term is already varied with respect to the field variables, so
> that chunk of the variation is done.
This is why you have to explain why S has to be extremized under any
variations in g^ij.
> [Snipped suspense part]
>
> delta(g) = g g^ij delta(g_ij)
This is wrong. Just look at the very simple 2-by-2 matrix.
[a c]
[c b]
Where
** g_00 = a
** g_01 = g_10 = c
** g_11 = b
** g^00 = b / (ab – c^2)
** g^01 = g^10 = - c / (ab – c^2)
** g^11 = a / (ab – c^2)
** deg[g] = ab – c^2
You can write
** det(g) = b / g^00 = - c / g^01 = - c / g^10 = a / g^11
So,
** @det(g)/@g^00 = - b / g^00 / g^00 = - det(g) / g^00
** @det(g)/@g^01 = c / g^01 / g^01 = - det(g) / g^01
** @det(g)/@g^10 = c / g^10 / g^10 = - det(g) / g^10
** @det(g)/@g^11 = - a / g^11 / g^11 = - det(g) / g^11
Therefore,
@det(g)/@g^ij = - det(g) / g^ij
Or in your terminology,
Delta det(g) = - det(g) delta(g^ij) / g^ij
> [nonsense snipped]
>
> delta(sqrt(-g)) = - 1/2 sqrt(-g) g_ij delta(g^ij)
@sqrt(- det(g))/@g^ij = - @det(g)/@g^ij / sqrt(- det(g)) / 2
Or
@sqrt(- det(g))/@g^ij = det(g) / g^ij / sqrt(- det(g)) / 2
> Variation done.
Not quite. Then,
dS = int(sqrt(- det(g)) (R_ij – R / g^ij / 2))
Claiming dS to be null, you get
R_ij – R / g^ij / 2 = 0
That is very different from the Einstein tensor of
G_ij = R_ij – R g_ij / 2
Any questions? Whines? Wholesale snips of material?
<CHECKMATE>
eric gisse
<koobee...@gmail.com> wrote:
>On Oct 31, 3:24 am, eric gisse wrote:
>> Koobee Wublee wrote:
>
>> [crap snipped]
>>
>> Furthermore you should fucking KNOW the Einstein-Hilbert action. You
>> should KNOW what the terms mean, and you should DAMN well better know
>> that every integral will be weighted by a factor of sqrt(-g) because
>> the volume element is only a density.
>
>No, I do not know what the Einstein-Hilbert action really means.
><shrug>
Why not? Didn't you learn this stuff from the sources you won't talk
about when you were learning GR?
>
>> [more crap snipped]
>>
>> The Einstein-Hilbert action S is equal to int( R sqrt(-g) d^4x).
>
>What does that mean rationally? I see it as a corpse sewn together
>from cadavers scavenged from the neighborhood morgue. <shrug>
To what degree of simplicity must it be explained to you before you
can understand?
It was assumed that the simplest scalar quantity that contains nothing
higher in order than second derivatives of the metric would be a good
starter for the Lagrangian.
The simplest nontrivial choice is the Ricci scalar.
Except the Lagrangian is must be integrated over the entire volume of
space-time to obtain the action. The volume element is not a scalar,
rather a tensor density. The sqrt(-g) term offsets the weighting of
the tensor density.
This is explained in every reference that adequately deals with the
Lagrangian formulation of general relativity. Is there a reason you
haven't looked at a modern reference on the subject?
>
>> [yapping nonsense snipped]
Of course you snipped something which I had to explain again.
>
>> Recall that the Lagrange equations are obtained by varying the action
>> with respect to the field variables, and demanding that the variation
>> is equal to zero at the boundary points.
>
>Why does this action have to be extremized?
Really? Did you honestly ask me why an action has to be extremized?
You surrendered the right to ever tell me to learn calculus of
variations ever again.
Consult any introductory variational calculus textbook.
>
>> So the variation considered will be in the field variables g_ij in the
>> form g^ij --> g^ij + delta(g^ij).
>
>Why is that S has to be extremized while varying g^ij?
Calculus of variations - the g^ij are the field variables.
>
>> S = int( R_ij g^ij sqrt(-g) d^4x) with g = det(g_ij).
>
>With g = determinant, that is more clear what you are doing now.
>Thank you.
>
>> delta(S) = delta( int( R_ij g^ij sqrt(-g) d^4x) ) +
>>
>> The variation behaves like a differential operator, so the
>> terms are expanded as such.
>>
>> delta(S) = int( delta(R_ij) g^ij sqrt(-g) d^4x) + int( R_ij
>> delta(g^ij) sqrt(-g) d^4x) + int( R_ij g^ij delta(sqrt(-g)) d^4x)
>
>You should be writing down the following instead.
>
>dS = d(int(g^ij R_ij sqrt(- det(g)) d^4q)
I don't because it gets repetitive, the usage is obvious from context,
and the notation is consistent with every textbook on the subject that
I have ever read. Learn the terminology.
>
>
>With variation in g^ij, you get
>
>dS = int(R_ij sqrt(- det(g)) dg^ij+ g^ij R_ij @sqrt(- det(g))/@g^ij
>dg^ij d^4q)
Here is where your notation fucks things up - you are using d to mean
both the infintesimal variation *and* the differential in the same
equation.
You didn't allow for the variation in R_ij, either. Which you seem to
think is OK since there is a R_ij in the field equations.
>
>> delta(S) = A + B + C
>>
>> The B term is already varied with respect to the field variables, so
>> that chunk of the variation is done.
>
>This is why you have to explain why S has to be extremized under any
>variations in g^ij.
Because the g^ij are the field variables that are being perturbed.
Why are you asking me to explain to you the most basic aspects taught
in calculus of variations?
>
>> [Snipped suspense part]
>>
>> delta(g) = g g^ij delta(g_ij)
>
>This is wrong. Just look at the very simple 2-by-2 matrix.
>
>[a c]
>[c b]
>
>Where
>
>** g_00 = a
>** g_01 = g_10 = c
>** g_11 = b
>** g^00 = b / (ab – c^2)
>** g^01 = g^10 = - c / (ab – c^2)
>** g^11 = a / (ab – c^2)
>** deg[g] = ab – c^2
>
>You can write
>
>** det(g) = b / g^00 = - c / g^01 = - c / g^10 = a / g^11
>
>So,
>
>** @det(g)/@g^00 = - b / g^00 / g^00 = - det(g) / g^00
>** @det(g)/@g^01 = c / g^01 / g^01 = - det(g) / g^01
>** @det(g)/@g^10 = c / g^10 / g^10 = - det(g) / g^10
>** @det(g)/@g^11 = - a / g^11 / g^11 = - det(g) / g^11
Wrong. det(g) = det(g_ij). You have to vary with respect to ALL THE
FIELD VARIABLES, not just one.
You seem to be so badly tutored in the subject that you don't even
understand the fundamental consequences of the summation.
>
>Therefore,
>
>@det(g)/@g^ij = - det(g) / g^ij
Again you continue to misunderstand that you _can not_ divide by a
matrix. Why do I have to keep repeating myself? If you would stop
making this error, we could be done with this section by now.
Look at the original equation:
delta(g) = g g^ij delta(g_ij)
This is shorthand for:
delta(det(g_ij)) = (SUM, i = 0..3) (SUM j = 0...3) det(g_ij) g^ij
delta(g_ij)
This is the _definition_ of multiplication by matrices.
http://en.wikipedia.org/wiki/Summation_convention
Or in your bastardized but in this case conceptually useful
terminology:
delta(det(g_ij)) = delta( det([g]) ) = det([g]) * [g]^-1 delta([g])
You can't divide by matrices. It really is as simple as that.
>
>Or in your terminology,
>
>Delta det(g) = - det(g) delta(g^ij) / g^ij
>
>> [nonsense snipped]
>>
>> delta(sqrt(-g)) = - 1/2 sqrt(-g) g_ij delta(g^ij)
>
>@sqrt(- det(g))/@g^ij = - @det(g)/@g^ij / sqrt(- det(g)) / 2
Only true if the @ represents variation with respect to g^ij.
>
>Or
>
>@sqrt(- det(g))/@g^ij = det(g) / g^ij / sqrt(- det(g)) / 2
You are dividing by a matrix. This is wrong.
Your principle fuckup seems to be that you don't really understand
that there is a sum on the i and j, so you then think that you only
operate on one at a time.
>
>> Variation done.
>
>Not quite. Then,
>
>dS = int(sqrt(- det(g)) (R_ij – R / g^ij / 2))
Wrong.
a) dS has to be a scalar - this isnt.
b) You are dividing by the metric. Again. Even though this is
obviously wrong.
c) The total variation is not even close to done yet. R_ij has to be
varied as well - or do you think R_ij remains the same under a small
change in the metric?
>
>Claiming dS to be null, you get
>
>R_ij – R / g^ij / 2 = 0
>
>That is very different from the Einstein tensor of
>
>G_ij = R_ij – R g_ij / 2
You used matrix division and got nonsense, and now you are surprised
you are getting gibberish?!
>
>Any questions? Whines? Wholesale snips of material?
>
><CHECKMATE>
Its' poor form to claim checkmate when you are repeating the same
errors of misunderstanding regarding linear algebra.
Koobee Wublee
> Koobee Wublee wrote:
> [yapping sh*t snipped]
>
> >This is wrong. Just look at the very simple 2-by-2 matrix.
>
> >[a c]
> >[c b]
>
> >Where
>
> >** g_00 = a
> >** g_01 = g_10 = c
> >** g_11 = b
> >** g^00 = b / (ab – c^2)
> >** g^01 = g^10 = - c / (ab – c^2)
> >** g^11 = a / (ab – c^2)
> >** deg[g] = ab – c^2
>
> >You can write
>
> >** det(g) = b / g^00 = - c / g^01 = - c / g^10 = a / g^11
>
> >So,
>
> >** @det(g)/@g^00 = - b / g^00 / g^00 = - det(g) / g^00
> >** @det(g)/@g^01 = c / g^01 / g^01 = - det(g) / g^01
> >** @det(g)/@g^10 = c / g^10 / g^10 = - det(g) / g^10
> >** @det(g)/@g^11 = - a / g^11 / g^11 = - det(g) / g^11
>
> Wrong. det(g) = det(g_ij). You have to vary with respect to ALL THE
> FIELD VARIABLES, not just one.
>
> >Therefore,
>
> >@det(g)/@g^ij = - det(g) / g^ij
>
> Again you continue to misunderstand that you _can not_ divide by a
> matrix.
>
> [more crap snipped]
I give up. You are a fucking idiot. With no college degree, being a
multi-year super-senior would only result in a burden for the
society. In the near future, mommy’s support and Alaska’s socialist
welfare program you seem to be very comfortable, but it won’t last.
<shrug>
Edward Green
<...>
> About the Einstein equation : quite impossible to "grok" it.
Quite hard maybe. I doubt it is impossible. How would you know,
anyway? Did some messenger from an alien race tell you that our
brains are just too small, as you slept?
As I already said, I see it is but half the story: there is lots of
interesting physics in E = 0, so we have to grok exactly what aspects
of spacetime are constrained by T, and which are free.
How constricted some people are in their speech. I had no idea I
sounded old fogeyish using "grok", but if somebody had asked about
"digging" or "being hep to" the Einstein equation, I would, in the
back of my mind, hear the archaic slang, but in the front of my mind I
would hear the question about physics. So let me rephrase my original
question: how can we be hep to the jive in the Einstein equation?
Don Stockbauer
> On Oct 31, 11:13 pm, Hayek <haye...@nospam.xs4all.nl> wrote:
>
> <...>
>
> > About the Einstein equation : quite impossible to "grok" it.
>
> Quite hard maybe. I doubt it is impossible. How would you know,
> anyway? Did some messenger from an alien race tell you that our
> brains are just too small, as you slept?
Brains can be linked via communication channels into meta-brains.
>
> As I already said, I see it is but half the story: there is lots of
> interesting physics in E = 0, so we have to grok exactly what aspects
> of spacetime are constrained by T, and which are free.
>
> How constricted some people are in their speech. I had no idea I
> sounded old fogeyish using "grok", but if somebody had asked about
> "digging" or "being hep to" the Einstein equation, I would, in the
> back of my mind, hear the archaic slang, but in the front of my mind I
> would hear the question about physics. So let me rephrase my original
> question: how can we be hep to the jive in the Einstein equation?
I wish I hadn't said anything. "Grok" is a perfectly all right term.
Hayek
> On Oct 31, 11:13 pm, Hayek <haye...@nospam.xs4all.nl> wrote:
>
> <...>
>
>> About the Einstein equation : quite impossible to "grok" it.
>
> Quite hard maybe. I doubt it is impossible. How would you know,
> anyway? Did some messenger from an alien race tell you that our
> brains are just too small, as you slept?
I just explained that to you, but if you cannot grok normal language, I
doubt you will ever grok the Einstein tensor.
Uwe Hayek.
--
Als ik nu op dit moment geld transfereer [in Belgi隴 naar een
andere rekening staat dat een uur later daar gecrediteerd.
-- Boutros Gali, realiteitsdeskundige.
Hayek
We tried to explain that, but foggy does not understand.
eric gisse
Funny how through several hundred lines of words from you the only
arguments have been how you don't understand what I'm doing. Thanks
for playing.
eric gisse
wrote:
>Edward Green wrote:
>> On Oct 31, 11:13 pm, Hayek <haye...@nospam.xs4all.nl> wrote:
>>
>> <...>
>>
>>> About the Einstein equation : quite impossible to "grok" it.
>>
>> Quite hard maybe. I doubt it is impossible. How would you know,
>> anyway? Did some messenger from an alien race tell you that our
>> brains are just too small, as you slept?
>
>I just explained that to you, but if you cannot grok normal language, I
>doubt you will ever grok the Einstein tensor.
>
>Uwe Hayek.
Except you didn't explain it. None of the words you wrote were even
relevant to explaining the meaning of the Einstein equation. Or tensor
- you know that the Einstein equation is not the same as the Einstein
tensor, right?
Hayek
Whatever. Another one that can't grok language.