Schwarzschild's paper and the uniqueness of the solution

[JanPB]

This post is a continuation of the thread "Lagranian Method Revisited" [sic]
which discussed Karl Schwarzschild's 1916 paper "Ueber das Gravitationsfeld
eines Massenpunktes nach der Einstein'schen Theorie" (1916) (English
translation "On the gravitational field of a mass point according to
Einstein's theory" available at arvix.org in physics/9905030v1).

It is sometimes claimed (by the above English translators, for example) that
the paper describes a different solution of the Einstein field equations than
the one described currently in textbooks. Specifically, it is asserted that
his original solution does not exhibit a black hole but merely a static
spherically symmetric vacuum region with a pointlike mass in the centre.

In this note I'd like to explain why this conclusion is incorrect and also to
go in some detail through the relevant parts of Schwarzschild's paper in
order to find the subtle error he made.

This is in no way to diminish his achievement, in those days hardly anybody
knew differential geometry anyway, and among physicists it was practically
unknown.

To appreciate the subtlety of the problem we need to begin with simple
examples illustrating potential complications.


Section I.
METRICS AND GEOMETRIES IN MANIFOLDS WITH FIXED COORDINATE SYSTEMS

In this section I'll use exclusively manifolds with the _identity map_ as
their coordinate system. I won't discuss _other_ coordinate systems on
manifolds until Section II.

First we need to state precisely what we mean by geometry. It's important to
keep in mind that metric tensor determines geometry but that relation is NOT
one-to-one. For example, the following tensor describes the flat (Minkowski)
geometry in 2D:

ds1^2 = -dt^2 + dx^2

But the same geometry is also described by a _different_ tensor:

ds2^2 = 4 dt dx

(Again, I'm stressing here that in this section the coordinate system never
changes, so ds2^2 is a _different tensor_ than ds1^2.)

The geometry is the same because the Riemann curvature of the second metric
is identically zero (skipping the calculation here).

So what determines whether two given metrics define the same geometry?

Answer:
Two metrics ds1^2 and ds^2 determine the same geometry if and only if there
exists a _diffeomorphism_ F (of the manifold) which induces a transformation
of one metric tensor into the other, say ds1^2 into ds2^2:

F^*(ds1^2) = ds2^2

The "F with the superscript star" notation stands for the transformation of
covariant tensors induced by F. I'll define this mapping in a moment.

"Diffeomorphism" here simply means "smooth map which is one-to-one and whose
inverse is also smooth". "Smooth" here means "infinitely differentiable".
(A bit later I'll state a useful criterion for checking whether a given map
is a diffeomorphism that's easier to use than this definition.)

For example:
(1) f(x) = sinh(x) is a diffeomorphism of the entire real line (all x) since
it's 1-1, infinitely differentiable for all x, and so is its inverse
f^(-1)(x) = arsinh(x) (recall the derivative of arsinh(x) is 1/sqrt(1+x^2),
and similar for higher derivatives, so the denominator never causes any
trouble).

(2) g(x) = x^3. This is NOT a diffeomorphism of the entire real line.
Although it is 1-1, smooth, and invertible, its inverse is NOT smooth:
g^(-1)(x) = x^(1/3), so its derivative is 1/(3x^(2/3)) - and the denominator
is zero when x = 0. Nevertheless, g(x) _is_ a diffeomorphism of on the
subset of the real line consisting of _all nonzero_ real numbers. So the
domain is important for being a diffeomorphism or not.

OK, so what does this "F^*" mean. Let's say that ds1^2 is:

ds1^2 = g_ij dx^i dx^j

Then F^*(ds1^2) (called the _pullback_ of ds1^2 under F) is defined as
follows:

F^*(ds1^2) = (g_ij . F) d(x^i . F) d(x^j . F)

...where by the dot "." I mean composing functions. So "g_ij.F" is F followed
by g_ij, and "x^i.F" means "the i-th component of F" or "F followed by the
i-th coordinate function".

We then also say that the manifold with the metric ds1^2 is _isometric_ to
the same manifold with the metric ds2^2. The map F is called the _isometry_.
Isometric manifolds are indistinguishable geometrically and physically. The
reason is that isometries preserve the dot product, hence the distances and
the angles, hence the whole geometry. That's why pullbacks enter the picture
here. I should also mention (although this is getting ahead of myself a bit)
that curvature tensors of metrics related by a pullback are themselves
related by pullback which in our case means that if a ds^2 satisfies the
Einstein equation, so does F^*(ds^2) for ANY diffeomorphism F.

All this pullback business should become clear by examining the Minkowski
example from a minute ago:

ds1^2 = -dt^2 + dx^2
ds2^2 = 4 dt dx

In this case they are isometric because ds2^2 is the pullback of ds1^2 under
the following diffeomorphism F of the (t,x)-plane:

F(t, x) = (x - t, x + t)

Let's check it using the definition of the pullback (note that the
compositions are: -1.F = -1 and 1.F = 1):

F^*(ds1^2) =

= F^*(-dt^2 + dx^2) =

= -d("t-component of F")^2 + d("x-component of F")^2 =

= -d(x - t)^2 + d(x + t)^2 =

= -(dx - dt)^2 + (dx + dt)^2 =

= -dx^2 - dt^2 + 2 dt dx + dx^2 + dt^2 + 2 dt dx =

= 4 dt dx =

= ds2^2

The silly factor 4 can be removed by one more pullback via the following
diffeomorphism G:

G(t, x) = (t/2, x/2)


G^*(ds2^2) =

= G^*(4 dt dx) =

= 4 d(t/2) d(x/2) =

= 4 1/2 dt 1/2 dx =

= dt dx

So ds3^2 = dt dx is _also_ the flat Minkowski space (this too can be verified
directly by computing the curvature of ds3^2 by hand).

OK, so this was the easy part. Now the subtle part begins. Let's imagine a
slightly trickier diffeomorphism:

H(t, x) = (t, 1/x)

Notice that its domain isn't the whole (t,x)-plane, it's the plane with the
set of points where x = 0 (i.e., the t-axis) _removed_.

H a diffeomorphism there because it's clearly 1-1 with both F and its inverse
infinitely differentiable at each x =/= 0.

BY THE WAY: there exists a very good criterion for verifying whether a given
map is a diffeomorphism that does NOT require computing the inverse of the
map: it's called "the inverse function theorem" and is one of the two pillars
of real analysis. (The other pillar of real analysis, incidentally, is
"Lebesgue dominated convergence theorem" - but I digress.)

That theorem goes like this: in order to check F is a diffeomorphism all you
need to verify is:

1. F is 1-1,
2. F is smooth,
3. the Jacobian of F is nonzero _at every point of the domain of interest_.

No need to invert F, very handy-dandy. It's easy to check that all our
diffeomorphisms so far satisfy this criterion [exercise].

Let's now calculate the pullback (call it ds4^2) of the Minkowski ds1^2 under
the above H:

ds4^2 =

= H^*(ds1^2) =

= H^*(-dt^2 + dx^2) =

= -dt^2 + d(1/x)^2 =

= -dt^2 + [ (-1/x^2) dx ]^2 =

= -dt^2 + dx^2/x^4

So this is also the flat Minkowski space (and yet again, one can verify this
directly by calculating the curvature by hand).

BUT there is now a kink: the metric ds4^2 is undefined at x = 0! To be sure,
that new tensor is flat wherever it's defined but it's _not defined_ at x = 0!

BINGO.

Here we've just encountered the standard phenomenon of the trade: the _same
geometry_ represented by two tensors _whose domains do not coincide_. In this
case ds1^2 is defined everywhere in the (t,x)-plane but ds4^2 isn't: one of
its coefficients (1/x^4) blows up.

In other words: ds4^2 has a singularity at x = 0 but the geometry (and the
manifold) does NOT. (This ought to sound familiar...)

AGAIN:

The tensor ds4^2 _DOES NOT_ extend over x = 0 (the t-axis), it has a
_genuine_ singularity there, but it is the pullback of a tensor that DOES
extend. (Namely, the ds1^2 one.) So the geometry is well-defined, including
at x = 0. The singular behaviour of ds4^2 is only due to choosing a slightly
misbehaving isometry to transform one into the other.

So:
A geometry does not have to stop where a metric tensor defining it stops. It
may be possible the geometry is simply a restriction of a larger geometry
which can be seen by selecting an appropriate pullback.

In our case, this geometry extending process would go something like this:
someone hands you over the metric ds4^2, tells you that it's been written
with respect to the canonical coordinates, you see it's not defined at x = 0
but for some reason or other you suspect the singularity is not really
present in the geometry (say, your reason is that you've calculated the
curvature of ds4^2 and obtained a value (namely, zero!) which does NOT blow
up at x = 0). To prove it we need to produce an explicit pullback that
extends over x = 0. There is no general recipe for doing this other than
educated guesswork. In our case we know what to do, of course, given that we
know where ds4^2 came from. We set the following diffeomorphism (the inverse
of H, which happens to look the same as H):

K(t, x) = (t, 1/x).

Then:

K^*(ds4^2) =

= K^*(-dt^2 + dx^2/x^4) =

= -dt^2 + d(1/x)^2 / (1/x)^4 =

= -dt^2 + (-1/x^2)^2 dx^2 x^4 =

= -dt^2 + dx^2

...and this is defined on the whole plane, so is the geometry.

Time for two more examples illustrating certain notable patterns.

Example A:

The location of tensor singularity is not fixed, it can change in a
pullback. Let's look at ds4^2 one more time:

ds4^2 = -dt^2 + dx^2/x^4

It's singular at x = 0. But consider a really basic diffeomorphism:

L(t, x) = (t, x + 1)

Pulling back:

L^*(ds4^2) =

= -dt^2 + [d(x + 1)]^2/(x + 1)^4 =

= -dt^2 + dx^2/(x + 1)^4

Now we got a tensor describing the Minkowski geometry which is singular at
x = -1. This is borderline trivial but worth remembering: the coordinate
value of the singularity is not a fixed property of the tensors defining
the geometry.

Example B:

Minkowski ds1^2 again:

ds1^2 = -dt^2 + dx^2

Now let's look at this easy diffeomorphism:

M(t, x) = (t, 2x)

Therefore,

M^*(ds1^2) =

= -dt^2 + d(2x)^2 =

= -dt^2 + 4 dx^2

Well, nothing deep here, it would work the samy way for any nonzero number
instead of 2 just the same:

N(t, x) = (t, Cx) where C is any nonzero constant (you should know by
now why we insist on C =/= 0 here)

Therefore,

N^*(ds1^2) =

= -dt^2 + d(Cx)^2 =

= -dt^2 + C^2 dx^2

Now we got a tensor describing the Minkowski geometry which apparently
depends on an _extra constant_. This is again borderline trivial but worth
remembering: a constant appearing in metric tensor coefficients does NOT
necessarily have ANY physical significance! It can be simply an artifact of
the method used to calculate the given tensor and in such cases can be
"pulled back away". Here it can be naturally "pulled back away" by the
inverse diffeomorphism:

N^(-1)(t, x) = (t, x/C)


It would be nice if we could now start discussing Schwarzschild's paper but
"unfortunately" we need to cover one more degree of freedom: coordinate
changes. This is necessary because Einstein, Schwarzschild, and modern
textbooks all use this concept. I said "unfortunately" because the "single
coordinate system + pullbacks" approach used in the previous section is IMHO
clearer to understand and to use, esp. when tricky subjects like
singularities and geometry extensions arise. OTOH coordinate changes
introduce a certain extra degree of topological complication, you'll see what
I mean in a minute.


Section II.
METRICS AND GEOMETRIES IN DIFFERENT COORDINATES

This is in a sense a "dual" degree of freedom: given a metric tensor written
with respect to some coordinates, we can keep it _unchanged_ but instead
change the linear bases of the tangent spaces of the manifold and expand the
tensor at each point in those new bases.

Warning: it turns out BOTH approaches produce FAPP same-looking formulas, so
one must be specific and always state clearly what a given formula
represents: _two different tensors_ (one before, one after the pullback) in a
_fixed_ coordinate system OR _one tensor_ written with respect to _two
different coordinate systems_.

From the point of view of geometry or physics, both approaches are equivalent
since they produce different but _isometric_ manifolds.

NB: The pullback approach of Section I is sometimes referred to as "active
view" and the coordinate change approach of this section as "passive view".
The terminology suggests the former view changes the object in question (the
tensor) while the latter one does not.

Let's again consider one of the metrics from Section I to see how this
approach works:

ds1^2 = -dt^2 + dx^2

...written, as before, in the canonical coordinates (t, x). Let's now
introduce, for the first time in this posting, _new_ coordinates, call them
(T, X), defined as:

T = (x - t)/2 (*)
X = (x + t)/2

i.e.:

t = X - T
x = X + T

Hence:

dt = dX - dT
dx = dX + dT

So we obtain:

ds1^2 =

= -(dX - dT)^2 + (dX + dT)^2 =

= -dX^2 -dT^2 + 2 dT dX + dX^2 + dT^2 + 2 dT dX =

= 4 dT dX

This looks almost identical to the calculation in Section I, the mathematical
transformation process is the same (except for using the upper-case letters)
but the meaning and the objects involved are very different:

--> in Section I we had _two different tensors_, ds1^2 and F^*(ds1^2), both
written in terms of _the same coordinate system_. Here we have _one tensor_,
ds1^2, written in terms of _two different coordinates_: the first system,
(t,x), being the canonical one, and the second system, (T,X), being defined
by the formulas (*) above.

In all cases we have ONE geometry: in Section I that's because ds1^2 and
F^*(ds1^2) are related by a pullback, hence they are isometric, and in this
section we have just one tensor, so one geometry by definition.

Let's do one more example, again similar to something in Section I, to show
how tensor domains can change in different coordinates. This is where it gets
a bit hairy. Start with ds1^2 again:

ds1^2 = -dt^2 + dx^2

Change the coordinates like so:

T = t
X = 1/x (obviously defined for x =/= 0 only)

This means:

t = T
x = 1/X

therefore (this should look annoyingly familiar by now):

ds1^2 =

= -dT^2 + d(1/X)^2 =

= -dT^2 + [ -1/x^2 dX ]^2 =

= -dT^2 + dX^2/X^4

Again:

1. the mathematical manipulation looks practically the same as in Section I
except for the upper-/lower-case distinguishing the coordinates,

2. the same geometry throughout (the flat Minkowski).

But here comes a subtle point which I never see properly discussed in
textbooks, even the unusually careful Robert Wald glosses over it. So let's
look at another, deceptively similar scenario: someone gives you a metric
like the one above but written instead in the _standard_ coordinates (t, x):

ds^2 = -dt^2 + dx^2/x^4

Remember, these coordinates are well-defined everywhere, so the blowing up of
the coefficient 1/x^4 at x = 0 means the tensor ds^2 _itself_ blows up there
too.

No matter, we can change the coordinates anyway, why not try the (T, X)
again. We get, by the same type of calculation as before:

ds^2 = -dT^2 + dX^2

...and this expression does extend to X = 0. Wait a minute, didn't we just
say that ds^2 blows up? We haven't changed ds^2, only the coordinates. So
how come it suddenly doesn't blow up at zero?

The answer is that we extend it over X = 0 which in the original coordinates
does NOT correspond to _any point_ in the (t,x)-plane! (since 0 cannot equal
1/x) So where on earth does this ds^2 extend _to_? Where _is_ this point X =
0 if it's nowhere in the (t,x)-plane?

What really happens is that we are extending the original expression (-dt^2 +
dx^2/x^4) over _an extra set of manifold points which wasn't there before_.
Namely we extend -dt^2 + dx^2/x^4 over the extra "line at x = +/- infinity"
while at the same time removing the "line at x = 0" (the t-axis) over which
-dt^2 + dx^2/x^4 _does not_ extend.

The resulting manifold consists of a slightly different set of points but
it's _isometric_ to the Minkowski plane.

You can imagine the original (t,x)-plane wrapped around a cylinder (the
t-axis pointing in the direction of the cylinder's axis) so that the two
extremities of the plane where x approaches plus and minus infinity end up
_just_ shy of the vertical line on the other side of the cylinder, opposite
the t-axis x = 0. It is THAT vertical line which we add while removing the
line x = 0. The result is clearly an isometric copy of the entire Minkowski
plane but sort of "turned inside out".

This sort of thing is precisely what is involved in extending the standard
modern textbook Schwarzschild metric (typically written in the spherical
coordinates space) over the horizon by switching to, say, the
Eddington-Finkelstein coordinates: the metric does not extend over the space
r = 2M but instead extends over a _copy_ of that space attached _differently_
(namely, attached as limit points of the infalling (say) geodesics on the
usual (t,r)-diagram).

(BTW, this type of cutting and pasting parts of manifolds is called "surgery"
and there is a HUGE field of topology devoted to it. GR textbooks avoid
discussing this, presumably because this operation isn't much done in GR
otherwise and ignoring it is of no consequence 99.9% of the time.
Historically this sort of manifold manipulations and manifold extension arose
for the first time in the context of complex analysis in the study of Riemann
surfaces.)

You can see now why I prefer the "active" approach of Section I where
typically there is no need for surgery, just transform tensors by pullbacks
as needed which keeps the geometry unchanged and the coordinate systems
fixed, so less confusion.

Finally:


Section III.
SCHWARZSCHILD'S PAPER

Here we'll find out why the solution described in this paper is identical to
the one described in all textbooks, i.e. why it describes the usual black
hole geometry, including the horizon and the inner, non-static, region.

Let's begin with a quick summary of his method. Then we'll stop at the
"smoking gun moment" where he jumps to a conclusion too fast.

He begins by writing down the vacuum field equations, Ricci = 0, for any
coordinate system in which the determinant of the metric is identically -1.
Under this restriction the field equations simplify quite a bit (to half
their "native" size), he writes this as his equation (4). He then proceeds to
construct such coordinate system.

First he quotes Einstein's 1915 paper "Erklaerung der Perihelbewegung des
Merkur aus der allgemeinen Relativitaetstheorie" ("Explanation of the
Perihelion Motion of Mercury from General Relativity Theory", a rather bad
English translation is available here:
http://www.gsjournal.net/old/eeuro/vankov.pdf) by stating that "according to
Mr. Einstein's list" the following conditions must be satisfied:

1. all metric components are independent of the time x_4 (he denotes the time
coordinate by x_4),

2. g_i4 = g_4i = 0 for i = 1, 2, 3

3. the solution is spatially spherically symmetric wrt to the origin,

4. boundary conditions at infinity: g_44 = 1, g_11 = g_22 = g_33 = -1.

So right away we are faced with the question: what exactly is the basis for
claims 1 and 2? The field equations are a complicated system of 10 coupled
PDEs for 10 unknown functions of four variables, why would merely assuming
spherical symmetry and staticity imply 1 and 2 just like that? This
conclusion seems wrong and it's easy to concoct a counterexample in fact:

Consider the 4D vacuum with the following metric written in spherical
coordinates:

ds^2 = -cosh^2(t) dt^2 - 2 cosh(t) dt dr + r^2 dO^2

...where "dO^2" is the shorthand for the line element of the unit sphere:
dtheta^2 + sin^2(theta) dphi^2.

This ds^2 has BOTH time-dependent coefficients AND a cross-term (dt dr), so
it violates conditions 1 and 2 in Schwarzschild's paper, yet it is the
Minkowski geometry AGAIN, in particular a spherically symmetric static
vacuum.

To see this, just calculate its curvature or pull ds^2 back via the following
map (easily seen to be a diffeomorphism by the inverse function theorem
quoted earlier):

F(t, r, theta, phi) = (arsinh(t - r), r, theta, phi)

(Or use the equivalent "passive" change of coordinates: T = r + sinh(t), R =
r, THETA = theta, PHI = phi.)

So just saying "the solution is spherically symmetric and static" does not
seem to _guarantee_ 1 and 2.

Unfortunately many texbooks continue to sloppily insist that it is so, for
example Ohanian & Ruffini, 2nd edition, p.392: "The demand for a static
solution forbids terms of the type dr dt, dtheta dt, dphi dt in the
expression for the spacetime interval because those terms change sign when we
perform the time reversal t -> -t".

But of course _nothing_ changes when we perform the time reversal, in fact we
can "perform" ANY diffeomorphism whatsoever without affecting the geometry at
all! What the authors really have in mind is that the _tensor itself_ should
be invariant under such transformations. But this too is fishy because
spherical symmetry refers to the 3D _spatial_ slices of spacetime. QUESTION:
which ones? OK, so let's say "the spatial slices t = const. where "t" is one
of the canonical (identity) coordinates on R^4. Well, actually, not quite,
because due to spherical symmetry we want to use spherical coordinates in
place of the canonical (x,y,z). So we really mean the spherical coordinate
system on each spatial slice t = const.

Well, it's flaky but it kind of works. Now the "static" part. A metric tensor
in the above coordinates is static if its coefficients are t-independent.

As you can see, the above definition is sloppy. But it's good enough
for us here. Good modern texts like Wald or d'Inverno or Hawking & Ellis
define these things properly, in a way that captures the physics and the
geometry in a way that does not presume any coordinates. I won't dwell
more on that point.

So now Schwarzschild begins solving the Einstein equation by positing his
condition 1 and 2 as an Ansatz: seek the solution in the following form,
written in the above system consisting of spherical coordinates on slices
t = const (he sometimes denotes time by "t", sometimes by "x_4"). This is
Schwarzschild's equation (6):

ds^2 = A(r) dt^2 - B(r) dr^2 - C(r) (dtheta^2 + sin^2(theta) dphi^2)

...where I took the liberty of denoting by A(r), B(r), C(r) certain functions
of r he momentarily keeps written out explicitly and which arise from his
Cartesian-to-spherical switch (their exact form is unimportant here).

He finally makes one more coordinate change: from (t, r, theta, phi) to
(x_1, x_2, x_3, x_4) [he reuses the letters which at the beginning of his
paper denote the Cartesians] like so:

x_1 = r^3/3, x_2 = -cos(theta), x_3 = phi, x_4 = t

...with the obvious ranges inherited from the spherical coordinates:

x_1 > 0, -1 < x_2 < 1, 0 < x_3 < 2pi, -infinity < x_4 < +infinity

In these new coordinates he writes the final general form of the desired
solution and gets on with the actual solving the equation. That final general
form is (his equation (9)):

ds^2 =

= f_4 dx_4^2 -

- f_1 dx_1^2 -

- f_2/(1 - x_2^2) dx_2^2 -

- f_3 (1 - x_2^2) dx_3^2

f_1, f_2, f_3, f_4 are the unknowns to solve for.

All this means that _I'm retracting_ my previous claim that an extra
coordinate change is needed beyond those mentioned above _in the static
case_. When I was discussing this with Koobee, I had in mind the general case
of the spherically symmetric geometry (i.e., no assumption of static). In
that case one cannot posit that _all_ cross-terms disappear: the dt dr term
has to stay BUT one can then construct an ADDITIONAL change of coordinates
which removes it. This is the coordinate change I was talking about.
Nevertheless, this additional coordinate change is not relevant anyway
because as it turns out it is a _further_ change down the road that's
responsible for moving the points surrounding the r = 0 set _away_
from the origin.

Remember this is the reason for this post: to show r = 0 ceases to be
"pointlike" in Schwarzschild.

OK, so so far we still have: r is the polar coordinate with r = 0 denoting
the origin, and x_1 = r^3/3, so 1_4 has the same property.

It looks like we're stuck with the "central pointlike mass at r = 0"
but hang on.

Let's continue following Schwarzschild. He plugs the above ds^2 in the field
equations and solves them for the above unknown metric coefficients f_1, f_2,
f_3, f_4. These solutions are his equations (10), (11), (12).


THE SMOKING GUN
And at this point, right after his equation (12), he stumbles. He notes that
one of the metric coefficients (f_1) is discontinuous and he _immediately_
assumes this discontinuity _is_ the pointlike mass at the origin! Well, in
1916 it must have seemed perfectly sensible: what OTHER singularity IS there??

But now we know, and we've seen examples of this earlier, that _a singularity
of a metric tensor does not necessarily imply the singularity of the
geometry_! There may exist a pullback which will change the tensor to
another one with a different domain which may well cover the supposed
geometric singularity. (Equivalently: there may exist a coordinate change...
etc. etc., you know the rest).

In fact, his "stumble" appears for the first time a bit earlier, on page 3
[English translation] where he lists conditions his f-coefficients should
fulfil, specifically his condition 4 which insists on:

"Continuity of the f, except for x_1 = 0".

This is incorrect as matric tensor singularities do not necessarily arise
from geometry singularities and therefore are NOT restricted to these
locations.

So his conclusion that rho = alpha^3 is likewise jumping the gun. Instead, he
should have noticed that one of the two constants (rho) was _redundant_, just
like the constant C in our Example B at the end of Section I. In other words,
with the right diffeomorphism the constant rho could be absorbed by the
pullback, with the tensor domain adjusted accordingly. This is the freedom
allowed by the fact that Einstein's equation is a PDE for _unknown tensors_,
not just a regular PDE for _unknown functions_. This would have left him with
only one constant (alpha) which then could have been shown tied to the mass
of the central "body". Of course Schwarzschild was hampered in his
pullback/coordinate freedom by that constraint of the unit determinant.

Let's see how this constant absorbing would've worked had Schwarzschild done
it in the paper. Schwarzschild's metric ds^2 is hard to write down in ASCII,
so I'm going to provide here a link to a properly typeset PDF page, but the
point is that one can transform rho away by using the following simple
diffeomorphism:

F(x_1, x_2, x_3, x_4) = ((x_1 - rho)/3, x_2, x_3, x_4)

Here is the link to the PDF which is much easier on the eyes:
https://drive.google.com/file/d/0B2N1X7SgQnLRY1ExWUczZldYZXM

Summing up in ASCII: pulling Schwarzschild's ds^2 back by F yields:

F^*(ds^2) =

= (1 - alpha*x_1^(-1/3)) dx_4^2 -

x_1^(-4/3)/(1 - alpha*x_1^(-1/3)) * (1/3)^2 dx_1^2 -

(x_1^(2/3))/(1 - x_2^2) dx_2^2 -

(x_1^(2/3))*(1 - x_2^2) dx_3^2

...with the x_1 coordinate of the pullback satisfying: x_1 > rho (recall
Schwarzschild's original x_1 was defined as the polar r^3/3, hence his x_1
satisfied x_1 > 0).

His original singularity of ds^2 was at (his) x_1 = (alpha^3 - rho)/3.

Thus the above pullback F^*(ds^2) has the same singularity (unless it happens
to extend, which it does not) at:

x_1 = 3((the old singularity location) + rho =

= 3((alpha^3 - rho)/3 + rho) =

= alpha^3

...which is a strictly positive number. THIS is the coordinate change that
moves the "r = 0" away from the origin.

Since x_1 > rho, where rho is any constant, we are free to extend the metric
as far "down" as we please but another glance at F^*(ds^2) shows that the
metric has another singularity at x_1 = 0 (because of the cube roots of x_1
in the denominators). A curvature calculation then shows it's actually a
genuine singularity of the geometry (and you know the rest).

As usual, a completely analogous reasoning can be made using the "passive"
(coordinate change) viewpoint.

PS:
Another reason why Schwarzschild's set r = 0 is not a point is that if one
surrounds it by a sphere, it's surface area does NOT go to zero as r
approaches 0 (it approaches 4pi alpha^2 which is a nonzero number since alpha
is nonzero whenever the mass is nonzero). For a proof check this link:
https://drive.google.com/open?id=0B2N1X7SgQnLRV1otWnE2T2F6aEk

--
Jan

[Koobee Wublee]

On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:

> This post is a continuation of the thread "Lagranian Method Revisited" [sic]
> which discussed Karl Schwarzschild's 1916 paper "Ueber das Gravitationsfeld
> eines Massenpunktes nach der Einstein'schen Theorie" (1916) (English
> translation "On the gravitational field of a mass point according to
> Einstein's theory" available at arvix.org in physics/9905030v1).

Koobee Wublee has read the entire post and found it rather disappointingly that Jan is supposed to have come up with the differential equations that are coordinate free and to have shown how the solutions can only the Schwarzschild metric. It is full of wild claims and errors that junior high schools kids interested in geometry would not have made. <shrug>

> In this note I'd like to explain why this conclusion is incorrect and also to
> go in some detail through the relevant parts of Schwarzschild's paper in
> order to find the subtle error he made.

Schwarzschild did not make any errors as Jan the relativistic moron has claimed. <shrug>

> First we need to state precisely what we mean by geometry.

Unbelievably stupid. The geometry represents reality where no one can possibly change it. Since reality is invariant, the geometry is invariant. <shrug>

> It's important to keep in mind that metric tensor determines geometry but
> that relation is NOT one-to-one.

No, metric reflects geometry through the chosen coordinate system. <shrug>

> For example, the following tensor describes the flat (Minkowski)
> geometry in 2D:
>
> ds1^2 = -dt^2 + dx^2

That is flat spacetime in 2D but a unique geometry. The following is such a flat Minkowski geometry in 2D. <shrug>

> dsN^2 = c^2 dt^2 - dx^2

> But the same geometry is also described by a _different_ tensor:
>
> ds2^2 = 4 dt dx

No, it is not even flat and another very unique geometry. <shrug>

> (Again, I'm stressing here that in this section the coordinate system never
> changes, so ds2^2 is a _different tensor_ than ds1^2.)

Glad that Jan can tell the metric tensors for both are different. Does Jan realize the coordinate system is the same. <shrug>

> The geometry is the same because the Riemann curvature of the second metric
> is identically zero (skipping the calculation here).

The Riemann tensor does not come into the field equations. So, who gives a damn about the Riemann tensor? The Riemann tensor does not play any roles in the measurement of the curvature thing. <shrug>

> [rest of this section and the next snipped due to these errors]

> [Schwarzschild] begins by writing down the vacuum field equations,

> Ricci = 0, for any coordinate system in which the determinant of
> the metric is identically -1.

No, he did not. He started with the Cartesian coordinate system and soon realized the metric is not diagonal. So, he changed the coordinate system to the polar coordinate system. At this moment, the determinant of the metric did not have to be -1. <shrug>

The null Ricci tensor was suggested to be the spacetime equivalence of the Laplace equation by Nordstrom years before 1915. Schwarzschild should have plenty of time solving the null Ricci tensor and would have found the Schwarzschild metric to the null Ricci tensor years beforehand. <shrug>

** ds^2 = c^2 (1 - K / r) - dr^2 / (1 - K / r) - r^2 dO^2

Where

** K = constant
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

After Hilbert, then plagiarized by Einstein the nitwit, the plagiarist, and the liar, had published the field equations, Schwarzschild had to deal with the null Einstein tensor instead of the null Ricci tensor. <shrug>

Schwarzschild did not realize the null Einstein and the null Ricci tensors are the same thing where one should be able to trivially derive the null Ricci tensor from the null Einstein tensor. However, after understanding how the field equations are derived, Schwarzschild realized if the determinant of the metric is -1, the trace term vanishes. Thus, the null Einstein tensor becomes the null Ricci tensor. <shrug>

Knowing what the solution is in the null Ricci tensor, Schwarzschild had great motivation to transform the polar coordinate system into one that yields a determinant of -1. After the transform into this new coordinate system, the solution becomes so simple. He, then, had to transform the coordinate back to the polar coordinate system to get the final answer where his final answer is nowhere close to the Schwarzschild metric. <shrug>

> 1. all metric components are independent of the time x_4 (he denotes the time
> coordinate by x_4),

Newtonian solution is static. Thus, it is not a bad idea to guess that the solution would be static which mean independent of time. <shrug>

> 2. g_i4 = g_4i = 0 for i = 1, 2, 3

If Schwarzschild did not make this assumption, he will have to deal with a lot more differential equations. Thus, the assumption is a good starting point to solve for the differential equations. Again, Jan the relativistic moron has never solved any differential equations before. Solving them requires a lot trial and errors. What Schwarzschild did was very reasonable. <shrug>

> 3. the solution is spatially spherically symmetric wrt to the origin,

Since the gravitating body is a sphere, given that the Newtonian solution is spherically symmetric, it is no brainer to assume that the solution would also be symmetrically symmetric. Again, Jan the relativistic moron has never solved any differential equations before. It is always wise to think about what the possible solution would be beforehand. <shrug>

> 4. boundary conditions at infinity: g_44 = 1, g_11 = g_22 = g_33 = -1.

This was not Schwarzschild's assumption. It just happened so. Jan the relativistic moron does not understand Schwarzschild. <shrug>

> [rest of purely Jan's misunderstandings snipped]

It is very disappointing to see Jan the relativistic moron to come up with garbage like that after more than a month to come up with such a lengthy but bullshit answer. <shrug>

[JanPB]

On Tuesday, February 23, 2016 at 10:28:49 PM UTC-8, Koobee Wublee wrote:
>
> It is very disappointing to see Jan the relativistic moron to come up with garbage like that after more than a month to come up with such a lengthy but bullshit answer. <shrug>

You cannot count: it was 2 weeks and 2 days. The rest of your post is nonsense.

--
Jan

[Koobee Wublee]

On Tuesday, February 23, 2016, Jan the relativistic moron wrote:

> On Tuesday, February 23, 2016 at 10:28:49 PM, Koobee Wublee wrote:

> > It is very disappointing to see Jan the relativistic moron to
> > come up with garbage like that after more than a month to come
> > up with such a lengthy but bullshit answer. <shrug>
>
> You cannot count: it was 2 weeks and 2 days.

So, this time limit is hurting Jan's pride, eh? <shrug>

Koobee Wublee forgot to point out when Jan the relativistic moron utter below:

> ... a transformation of one metric tensor into the other, say ds1^2

> into ds2^2:
>
> F^*(ds1^2) = ds2^2

Both ds1 and ds2 represent geometry as Jan the relativistic moron had already stated. They are not tensors. Mathematically, it does not make any sense to transform one reality to another reality. Reality is whatever it is regardless how many transformations are involved. Jan the relativistic moron certainly does not know what it is babbling about. <shrug>

[JanPB]

On Tuesday, February 23, 2016 at 11:01:46 PM UTC-8, Koobee Wublee wrote:
> On Tuesday, February 23, 2016, Jan the relativistic moron wrote:
> > On Tuesday, February 23, 2016 at 10:28:49 PM, Koobee Wublee wrote:
>
> > > It is very disappointing to see Jan the relativistic moron to
> > > come up with garbage like that after more than a month to come
> > > up with such a lengthy but bullshit answer. <shrug>
> >
> > You cannot count: it was 2 weeks and 2 days.
>
> So, this time limit is hurting Jan's pride, eh? <shrug>

No, you simply lied.

> Koobee Wublee forgot to point out when Jan the relativistic moron utter below:
>
> > ... a transformation of one metric tensor into the other, say ds1^2
> > into ds2^2:
> >
> > F^*(ds1^2) = ds2^2
>
> Both ds1 and ds2 represent geometry as Jan the relativistic moron had already stated. They are not tensors. Mathematically, it does not make any sense to transform one reality to another reality. Reality is whatever it is regardless how many transformations are involved. Jan the relativistic moron certainly does not know what it is babbling about. <shrug>

Nonsense.

--
Jan

[Koobee Wublee]

On Tuesday, February 23, 2016 at 11:04:50 PM UTC-8, JanPB wrote:

> On Tuesday, February 23, 2016 at 11:01:46 PM, Koobee Wublee wrote:

> > So, this time limit is hurting Jan's pride, eh? <shrug>
>
> No, you simply lied.

Jan the relativistic moron is behaving like a 3 year-old. Jan the relativistic moron has claimed all solutions to the field equations would miraculously become the Schwarzschild metric without itself laying out and solving for these differential equations. On top of that, Jan the relativistic moron had bragged about coming up with these differential equations to the field equations in the form without any coordinate system but is unable to deliver. This is no brainer. Among these thousands of textbooks, there is not a single one bragging about laying out these differential equations in the form without any coordinate system. It is a task that is impossible to achieve as Koobee Wublee has explained so many, fvcking many times over. <shrug>

Yes, Jan the relativistic moron has no fvcking idea what the subject is insisted so. Of course, this is the internet where no one can possibly win an argument over cyberspace. Jan the relativistic moron can pretend to still living in that fat castle in the air. In the meantime, eventually GR will be found stupid, invalid, and false. GR does not represent an evolution to Newtonian physics but a false bud full of stupidity and self contradictions. <shrug>

[JanPB]

Nonsense. Get help.

--
Jan

[Koobee Wublee]

Well, Jan with the last resort, you can join PD's (aka Odd Bodkin aka Sylvia Else aka others) quest to go to Caltech where the institution had rewarded Koobee Wublee with these degrees and demand them to revoke them all. That will make Koobee Wublee's day, and don't forget to tell them how Koobee Wublee has driven you idiots to this last resort through debates on the internet of this subject where no one can possibly win an argument in. Ahahahaha... <ahahaha...>

[JanPB]

I don't care about winning any argument. I enjoy watching the zoo.

--
Jan

[David Waite]

On Wednesday, February 24, 2016 at 12:27:27 AM UTC-7, Koobee Wublee wrote:
>Jan the relativistic moron has claimed all solutions to the field equations would miraculously become the Schwarzschild metric...

I know you're too stupid to realize that you don't understand what Jan is saying, so plainly, that's not whats being claimed. Any spherically symmetric exact vacuum solution to the field equations is just the Schwarzschild solution whether you recognize it expressed in whatever coordinates are chosen to express it or not. In Schwarzschild's original paper coordinates were used to express it that are no longer commonly used, so you just don't recognize that its the same solution expressed in today's Schwarzschild coordinates as
ds^2 = (1-R/r)dct^2 - dr^2/(1-R/r) - r^2*dOmega^2
Now if the matter is collapsed beneath the horizon of these coordinates, it describes a black hole and it doesn't matter if you recognize it as a black hole because you may transform the expression of it to other coordinates, you don't change what it physically describes, that you still have a black hole. This is even the case if you were to translate the r coordinate to put the horizon at a new "r" coordinate position of 0 and hide where the matter physically collapsed to at some negative value of the new coordinate like you've tried to do.
As for another example, you can do a coordinate transformation to a new time coordinate like
ds^2 = dct'^2-dr^2-r^2*dOmega^2-(R/r)[(dr+/-dct')^2]
and for constant R this is still just the Schwarzschild solution just expressed in a different time coordinate so you may not recognize it, and if the matter was collapsed beneath r=R still describes a black hole, even though there is no coordinate singularity at the horizon.
And by the way letting R be a function of r+/-ct' in this last expression becomes the Vaidya solution where the difference is you no longer have vacuum, but rather EM radiation either forming or radiating from a shrinking hole in a spherically symmetric fashion depending on the +/- choice.

[Maciej Woźniak]

Użytkownik "JanPB" napisał w wiadomości grup
dyskusyjnych:444001cc-c1c0-4433...@googlegroups.com...

A lie. As expected from a relativistic trash.

[David Waite]

For those interested in the math the time transformation is
ct'=ct -/+ Rln|(r/R)-1| + k

[Maciej Woźniak]

Użytkownik "David Waite" napisał w wiadomości grup
dyskusyjnych:adaf181e-c4ef-4ad2...@googlegroups.com...

|For those interested in the math the time transformation is
|ct'=ct -/+ Rln|(r/R)-1| + k

And for those interested in facts, REAL clocks indicate t'=t.

[Odd Bodkin]

On 2/24/2016 1:27 AM, Koobee Wublee wrote:
> Of course, this is the internet where no one can possibly win an argument over cyberspace.

Which is of course what Koobee Wublee is counting on. Because he needs a
space where he does not lose arguments.

--
Odd Bodkin --- maker of fine toys, tools, tables

[Odd Bodkin]

On 2/24/2016 1:40 AM, Koobee Wublee wrote:
> Well, Jan with the last resort, you can join PD's (aka Odd Bodkin aka Sylvia Else
> aka others) quest to go to Caltech where the institution had rewarded Koobee Wublee
> with these degrees and demand them to revoke them all. That will make Koobee Wublee's
> day, and don't forget to tell them how Koobee Wublee has driven you idiots to this
> last resort through debates on the internet of this subject where no one can possibly
> win an argument in. Ahahahaha... <ahahaha...>

Koobee, three comments:
1. It's a sign of paranoia when you consolidate your critics into a
unified persona. Your mind does it to reduce the threat, by making it a
one-on-one battle, rather than a one-on-many.

2. Congratulations on your one-time accomplishment of CalTech degrees
from a long time ago. This does not change the unfortunate fact that you
have severely deteriorated since that accomplishment. You can keep your
degrees, but you do need to address your PRESENT state.

3. You lose arguments even if you don't acknowledge that you've lost
them. The way you lose them is loss of reputation and rise of general
ridicule. Now, I understand that you might be confused about your loss
of reputation and the tide of ridicule if you tell yourself, "But...
but... I never lost an argument. Not once!"

[Dono,]

On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:

This is a very nice work. Might be a good idea to make into a paper and submit it to AmJourPhys. I liked the style.

[Paul B. Andersen]

On 24.02.2016 08:40, Koobee Wublee wrote:
>
> Well, Jan with the last resort, you can join PD's
> (aka Odd Bodkin aka Sylvia Else aka others) quest
> to go to Caltech where the institution had rewarded
> Koobee Wublee with these degrees and demand them to revoke them all.
> That will make Koobee Wublee's day,

Give us your real name, and I will make your day.

--
Paul

https://paulba.no/

[Koobee Wublee]

On February 24, 2016, Jan the relativistic moron wrote:

> Koobee Wublee wrote:

> > Jan the relativistic moron has claimed all solutions to the field
> > equations would miraculously become the Schwarzschild metric
> > without itself laying out and solving for these differential
> > equations. On top of that, Jan the relativistic moron had bragged
> > about coming up with these differential equations to the field
> > equations in the form without any coordinate system but is unable
> > to deliver. This is no brainer. Among these thousands of
> > textbooks, there is not a single one bragging about laying out
> > these differential equations in the form without any coordinate
> > system. It is a task that is impossible to achieve as Koobee
> > Wublee has explained so many, fvcking many times over. <shrug>
>

> > Jan the relativistic moron can pretend to still living in that fat
> > castle in the air. In the meantime, eventually GR will be found
> > stupid, invalid, and false. GR does not represent an evolution to
> > Newtonian physics but a false bud full of stupidity and self
> > contradictions. <shrug>
>

> I don't care about winning any argument. I enjoy watching the zoo.

After all these years, Koobee Wublee has finally understood what the myth plaguing the Einstein dingleberries is. Koobee Wublee would never have dreamt about such a infantile confusion. This confusion is like saying Jan Bielawski is of Polish decent, and so is Wozniak. Thus, they are of the same person. Oh, man! What an idiot Jan the relativistic moron really is! <shaking head>

[David Waite]

No dumb nut, the analogy would be like one person expresses Jan Bielawski as Jan B, a second expresses Jan Bielawski as J.B., a third expresses Jan Bielawski as J. Bielawski, but they are all referring to the same person no matter how they expressed it.

[Koobee Wublee]

On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
> On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:

> > Two metrics ds1^2 and ds^2 determine the same geometry if and
> > only if there exists a _diffeomorphism_ F (of the manifold)
> > which induces a transformation of one metric tensor into the
> > other, say ds1^2 into ds2^2:
> >
> > F^*(ds1^2) = ds2^2
> >
> > The "F with the superscript star" notation stands for the
> > transformation of covariant tensors induced by F.
>

> This is a very nice work. Might be a good idea to make into a paper
> and submit it to AmJourPhys. I liked the style.

Well, ds1 and ds2 can be two different geometries/realties. Transforming one reality to another is in the realm of alchemists and the occult. <shrug>

The transformation that differential geometry has been doing is transforming from one coordinate system to another but never from one invariant geometry (reality) to another reality. It is like transforming a piece of shit to a delicious hot dog. In reality, the piece of shit can be observed differently thus transformed. One person such as Jan may observe it to be a piece of shit, but Dono the tampon head may think otherwise (by transforming Jan's observation in a delicious hot dog). Can you dig it? <shrug>

[Dono,]

On Wednesday, February 24, 2016 at 10:31:28 PM UTC-8, Koobee Wublee wrote:
> On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
> > On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:
>
> > > Two metrics ds1^2 and ds^2 determine the same geometry if and
> > > only if there exists a _diffeomorphism_ F (of the manifold)
> > > which induces a transformation of one metric tensor into the
> > > other, say ds1^2 into ds2^2:
> > >
> > > F^*(ds1^2) = ds2^2
> > >
> > > The "F with the superscript star" notation stands for the
> > > transformation of covariant tensors induced by F.
> >
> > This is a very nice work. Might be a good idea to make into a paper
> > and submit it to AmJourPhys. I liked the style.
>
> Well, ds1 and ds2 can be two different geometries/realties.

Not if the condition F^*(ds1^2) = ds2^2 is fulfilled.
Cretin. Incurable.

[JanPB]

On Wednesday, February 24, 2016 at 10:31:28 PM UTC-8, Koobee Wublee wrote:

> On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
> > On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:
>
> > > Two metrics ds1^2 and ds^2 determine the same geometry if and
> > > only if there exists a _diffeomorphism_ F (of the manifold)
> > > which induces a transformation of one metric tensor into the
> > > other, say ds1^2 into ds2^2:
> > >
> > > F^*(ds1^2) = ds2^2
> > >
> > > The "F with the superscript star" notation stands for the
> > > transformation of covariant tensors induced by F.
> >
> > This is a very nice work. Might be a good idea to make into a paper
> > and submit it to AmJourPhys. I liked the style.
>
> Well, ds1 and ds2 can be two different geometries/realties.

They cannot. What F^*(ds1^2) means is that F transforms manifold points and tangent vectors
in such a way that dot products between the latter are unchanged.

> Transforming one reality to another is in the realm of alchemists and the occult. <shrug>

Reality is not transformed. Think of it like the electrostatic potential: it can be transformed
by adding a constant, yet the electrostatic force it defines stays the same.

--
Jan

[PC]

David Waite wrote:

>> The transformation that differential geometry has been doing is
>> transforming from one coordinate system to another but never from one

>> invariant geometry (reality) to another reality...
>
> I know this will be over your head, but for the sake of others
> interested

What others? Nobody is interested.

[David Waite]

On Wednesday, February 24, 2016 at 11:31:28 PM UTC-7, Koobee Wublee wrote:
> On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
> > On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:
>
> > > Two metrics ds1^2 and ds^2 determine the same geometry if and
> > > only if there exists a _diffeomorphism_ F (of the manifold)
> > > which induces a transformation of one metric tensor into the
> > > other, say ds1^2 into ds2^2:
> > >
> > > F^*(ds1^2) = ds2^2
> > >
> > > The "F with the superscript star" notation stands for the
> > > transformation of covariant tensors induced by F.
> >
> > This is a very nice work. Might be a good idea to make into a paper
> > and submit it to AmJourPhys. I liked the style.
>
> Well, ds1 and ds2 can be two different geometries/realties. Transforming one reality to another is in the realm of alchemists and the occult. <shrug>
>

> The transformation that differential geometry has been doing is transforming from one coordinate system to another...

I know this will be over your head, but for the sake of others interested here's an example
Let
ds1a^2 = (1-R/r)dct^2-dr^2/(1-R/r)-r^2*dOmega^2
ds1b^2 = dct'^2-dr^2-r^2dOmega^2-(R/r)[(r +/- dct')^2]
ds2a^2 = dct^2-dx^2-dy^2-dz^2
ds2b^2 = [(1+Az'/c^2)^2]dct'^2-dx^2-dy^2-dz'^2
Now it turns out that ds1a and ds1b are two different coordinate expressions for the SAME spacetime. They are both the Schwarzschild solution. You literally have
ds1a=ds1b.
It also turns out that ds2a and ds2b are different coordinate expressions for the SAME spacetime. You literally have
ds2a=ds2b
However, the ds1 spacetime is fundamentally different from the ds2 spacetime. There exists NO coordinate transformation that *globally* transforms either of ds1 to either of ds2. This is what makes the ds1 expressions a different vacuum solution to the field equations than the ds2 expressions.

[David Waite]

Then feel free to killfile me.

[Odd Bodkin]

Heaven knows I killfile him, with one click, each and every time he
spends a while creating a new identity.

[David Waite]

So your trolling with lies, fine, post here the length of time since I've posted here with a different name and then explain why you lied about it being too often to killfile me.

[PC]

Odd Bodkin wrote:

>>>> I know this will be over your head, but for the sake of others
>>>> interested
>>>
>>> What others? Nobody is interested.
>>
>> Then feel free to killfile me.
>
> Heaven knows I killfile him, with one click, each and every time he
> spends a while creating a new identity.

Don't know what you are talking about. You are doing a big mistake. The
thing is that detection of "gravity waves" overbably speaking is something
absurd. See it for yourself. Modern Science is frauduously taxing
nutrients to force the developing countries not to develop.

The Great Global Warming Swindle Full Movie.
https://www.youtube.com/watch?v=D-m09lKtYT4

[Odd Bodkin]

Slow down, hoss. I was referring to my killfilling PC, not you. I
responded to YOU, and the "him" was the 3rd person in that conversation.
Sorry for the confusion.

[PC]

David Waite wrote:

>> > Then feel free to killfile me.
>> >
>> >
>> Heaven knows I killfile him, with one click, each and every time he
>> spends a while creating a new identity.
>

> So your trolling with lies, fine, post here the length of time since
> I've posted here with a different name and then explain why you lied
> about it being too often to killfile me.

http://yournewswire.com/the-vatican-own-a-time-travel-device-insiders-claim/

[David Waite]

Ok, sorry.

[David Waite]

Since they also believe they are given authority of some sort from a magic flying zombie with three personalities, why do we care what they claim?

[Jack...@hotmail.com]

On Wed, 24 Feb 2016 22:31:23 -0800 (PST), Koobee Wublee
<koobee...@gmail.com> wrote:

>On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
>> On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:
>
>> > Two metrics ds1^2 and ds^2 determine the same geometry if and
>> > only if there exists a _diffeomorphism_ F (of the manifold)
>> > which induces a transformation of one metric tensor into the
>> > other, say ds1^2 into ds2^2:
>> >
>> > F^*(ds1^2) = ds2^2

ds1^2 and ds2^2 are not tensors, they are called "metrics"and they
are pure scalars. Once ds's are squared, they become emasculated and
they have buried important information and are no longer candidates
for transformation, i.e. are the same in any coordinate system, just a
number. (Only their cross product would be a vector).
There is no evidenceof a tensor in any of this discussion. if you are
trying to use the Schwartzschild metric as a square diagonal 4 x 4
matrix as a tensor F to bend space-time then you need to state what
the base coordinate set, X, consists of.
I think what you will find is you cannot get along without "ict" which
is to be abhorred but which is concealed by all these squared terms.
So you need to deduce a 4 x 4 matrix, the metric tensor, G, such that
X' = GX that would show the bending of space time by gravity.
G = dX'/dX
The metric must be calculated thus: S^2 = X'GX (= 0 for light Ray
travels straight line).

>> > The "F with the superscript star" notation stands for the
>> > transformation of covariant tensors induced by F.

John Polasek

[JanPB]

On Thursday, February 25, 2016 at 11:06:18 AM UTC-8, Jack...@hotmail.com wrote:
> On Wed, 24 Feb 2016 22:31:23 -0800 (PST), Koobee Wublee
> <koobee...@gmail.com> wrote:
>
> >On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
> >> On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:
> >
> >> > Two metrics ds1^2 and ds^2 determine the same geometry if and
> >> > only if there exists a _diffeomorphism_ F (of the manifold)
> >> > which induces a transformation of one metric tensor into the
> >> > other, say ds1^2 into ds2^2:
> >> >
> >> > F^*(ds1^2) = ds2^2
>
> ds1^2 and ds2^2 are not tensors, they are called "metrics"and they
> are pure scalars.

No, they are tensors. To be exact: ds^2 is a rank-2 symmetric covariant tensor.

The notation like:

dt dr (say)

...is really defined as:

1/2 * (dt X dr + dr X dt)

...where "X" denotes the tensor product of the covectors (i.e. rank-1
covariant tensors).

It is traditional to omit the symmetric-tensor-product multiplication sign,
so rather than writing, say:

dt . dr or dt \/ dr or dt * dr or whatever, people just write:

dt dr

Sometimes the "\/"-notation or a dot-in-a-circle is used, but it's rare
and most people think it's unnecessary.

As an aside: in the somewhat analogous case of the rank-2 ANTI-symmetric
covariant tensors, the relevant notation is "/\", like so:

dt /\ dr

...which is a shorthand for:

dt X dr - dr X dt

...where "X" again is the tensor product. Notice the minus sign instead
of the plus, and the absence of the 1/2 factor (it's a convention that's
least cumbersome in calculations).

Note that the letter "d" used all over the place has different meanings
in different places (some authors stress it by putting an overbar over
some "d" and not the others): the "d" in front of "dt", "dx^i", and the
like is an actual operation: the exterior derivative of the relevant
coordinate function ("t" or "x^i" in this case).

OTOH the "d" in "ds^2" is merely a convenient metaphor, it's neither a
derivative nor a square of anything. The reason for this metaphoric
symbol is that the line element in path integrals (denoted by "ds"
usually, that "d" again is metaphoric) is equal to the square root
of the "ds^2" tensor evaluated at the tangent vector V to the path:

"ds in line integrals" = sqrt( ds^2(V,V) )

People tend to re-use the notation as much as possible, this makes
things easy to manipulate but creates confusion for some students
initially.

> Once ds's are squared,

They are not "squared" in this sense, they are "squares" out of the box.
In general there exists NO tensor T such that ds^2 = T X T. The square
only is relevant in the above sense (i.e. with tangent vectors plugged in).

--
Jan

[PC]

JanPB wrote:

>> >> > F^*(ds1^2) = ds2^2
>>
>> ds1^2 and ds2^2 are not tensors, they are called "metrics"and they
>> are pure scalars.
>
> No, they are tensors. To be exact: ds^2 is a rank-2 symmetric covariant
> tensor.

The Development of Our Views on the Composition and Essence of Radiation
(1909) by Albert Einstein


"This experiment demonstrated that matter does not completely carry along
its ether but, in general, the ether is
moving relative to matter. Now the Earth is a material object, which moves
in different directions over the course of
a year relative to the solar system. The ether in our laboratories was
assumed to not participate in the Earth's motion
completely, just as the ether did not participate in the water's motion
completely in Fizeau's experiment. Thus, the
conclusion was that the ether was moving relative to our instruments, and
that this relative motion changed over the
course of a day and of a year. This relative motion was expected to
produce a visible anisotropy of space, i.e., optical
phenomena were expected to depend on the orientation of the apparatus. The
most diverse experiments were
performed without detecting the expected dependence of phenomena on
orientation.
This contradiction was chiefly eliminated by the pioneering work of H. A.
Lorentz"

[Jack...@hotmail.com]

willOn Thu, 25 Feb 2016 12:35:22 -0800 (PST), JanPB

<fil...@gmail.com> wrote:

>On Thursday, February 25, 2016 at 11:06:18 AM UTC-8, Jack...@hotmail.com wrote:
>> On Wed, 24 Feb 2016 22:31:23 -0800 (PST), Koobee Wublee
>> <koobee...@gmail.com> wrote:
>>
>> >On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
>> >> On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:
>> >
>> >> > Two metrics ds1^2 and ds^2 determine the same geometry if and
>> >> > only if there exists a _diffeomorphism_ F (of the manifold)
>> >> > which induces a transformation of one metric tensor into the
>> >> > other, say ds1^2 into ds2^2:
>> >> >
>> >> > F^*(ds1^2) = ds2^2
>>
>> ds1^2 and ds2^2 are not tensors, they are called "metrics"and they
>> are pure scalars.
>
>No, they are tensors. To be exact: ds^2 is a rank-2 symmetric covariant tensor.

No, ds^2 is a scaler product formed from ds^2 = dX^i*dX_j delta ij,
Einstein summation. The dX's are covariant and contravariant vectors
(components of dX).

>The notation like:
>
> dt dr (say)
>
>...is really defined as:
>
> 1/2 * (dt X dr + dr X dt)
>
>...where "X" denotes the tensor product of the covectors (i.e. rank-1
>covariant tensors).

X is not a product, it has to be an operator, it seems to me. If X
denotes the cross product, the quantity in your brackets is
identically 0. How do you explain that?

If r and t are basis vectors, they must be orthogonal to each other.
What is the object in trying to marry these two?

>It is traditional to omit the symmetric-tensor-product multiplication sign,
>so rather than writing, say:
>
> dt . dr or dt \/ dr or dt * dr or whatever, people just write:
>
> dt dr

Please show me an example of where they use that form dt dr.

>Sometimes the "\/"-notation or a dot-in-a-circle is used, but it's rare
>and most people think it's unnecessary.
>
>As an aside: in the somewhat analogous case of the rank-2 ANTI-symmetric
>covariant tensors, the relevant notation is "/\", like so:
>
> dt /\ dr
>
>...which is a shorthand for:
>
> dt X dr - dr X dt
>
>...where "X" again is the tensor product. Notice the minus sign instead
>of the plus, and the absence of the 1/2 factor (it's a convention that's
>least cumbersome in calculations).
>
>Note that the letter "d" used all over the place has different meanings
>in different places (some authors stress it by putting an overbar over
>some "d" and not the others): the "d" in front of "dt", "dx^i", and the
>like is an actual operation: the exterior derivative of the relevant
>coordinate function ("t" or "x^i" in this case).
>
>OTOH the "d" in "ds^2" is merely a convenient metaphor,

"d" is a convenient metaphor? (this canard seems to set the tone for
the rest of the discussion.)

>it's neither a
>derivative nor a square of anything. The reason for this metaphoric
>symbol is that the line element in path integrals (denoted by "ds"
>usually, that "d" again is metaphoric) is equal to the square root
>of the "ds^2" tensor evaluated at the tangent vector V to the path:
>
> "ds in line integrals" = sqrt( ds^2(V,V) )
>
>People tend to re-use the notation as much as possible, this makes
>things easy to manipulate but creates confusion for some students
>initially.
>
>> Once ds's are squared,
>
>They are not "squared" in this sense, they are "squares" out of the box.

No they are squares out of the box only if you are trying to make the
Schwarzschild metric into the metric tensor, and they comprise the
field, which is not but it's the only way it will "work".

>In general there exists NO tensor T such that ds^2 = T X T. The square
>only is relevant in the above sense (i.e. with tangent vectors plugged in).

No, you have it backwards, I said it had to be ds^2 = XTX not TXT
where X is the collection of dx_i's.
I think you're working out of some specialized branch of algebra which
is purely mathematical and does not conform to the tensors we use in
cosmological physics
John Polasek.

[JanPB]

On Thursday, February 25, 2016 at 2:37:49 PM UTC-8, Jack...@hotmail.com wrote:
> willOn Thu, 25 Feb 2016 12:35:22 -0800 (PST), JanPB
> <fil...@gmail.com> wrote:
>
> >On Thursday, February 25, 2016 at 11:06:18 AM UTC-8, Jack...@hotmail.com wrote:
> >> On Wed, 24 Feb 2016 22:31:23 -0800 (PST), Koobee Wublee
> >> <koobee...@gmail.com> wrote:
> >>
> >> >On Wednesday, February 24, 2016 at 7:16:13 AM UTC-8, Dono, wrote:
> >> >> On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:
> >> >
> >> >> > Two metrics ds1^2 and ds^2 determine the same geometry if and
> >> >> > only if there exists a _diffeomorphism_ F (of the manifold)
> >> >> > which induces a transformation of one metric tensor into the
> >> >> > other, say ds1^2 into ds2^2:
> >> >> >
> >> >> > F^*(ds1^2) = ds2^2
> >>
> >> ds1^2 and ds2^2 are not tensors, they are called "metrics"and they
> >> are pure scalars.
> >
> >No, they are tensors. To be exact: ds^2 is a rank-2 symmetric covariant tensor.
> No, ds^2 is a scaler product formed from ds^2 = dX^i*dX_j delta ij,
> Einstein summation. The dX's are covariant and contravariant vectors
> (components of dX).

No, that's incorrect. Your ds^2 is equal to dx^i dx_i which is not true.
Also "The dX's are covariant and contravariant vectors (components of dX)"
is a nonsensical sequence of words.

> >The notation like:
> >
> > dt dr (say)
> >
> >...is really defined as:
> >
> > 1/2 * (dt X dr + dr X dt)
> >
> >...where "X" denotes the tensor product of the covectors (i.e. rank-1
> >covariant tensors).
> X is not a product, it has to be an operator, it seems to me. If X
> denotes the cross product, the quantity in your brackets is
> identically 0. How do you explain that?

I wrote "X" is a tensor product (in print it's denoted by an "x" multiplication
sign in a circle). This product is NOT commutative, so a X b is not equal
to b X a.

> If r and t are basis vectors, they must be orthogonal to each other.

No. t and r are coordinate functions.

> >It is traditional to omit the symmetric-tensor-product multiplication sign,
> >so rather than writing, say:
> >
> > dt . dr or dt \/ dr or dt * dr or whatever, people just write:
> >
> > dt dr
> Please show me an example of where they use that form dt dr.

Everywhere. This is the _definition_ of "ds^2". I suggest e.g. Michael Spivak's
"Comprehensive Introduction to Differential Geometry", vol. I. But any book
on general Riemannian geometry should have it.

Oh give me a break. Ask or hire someone to teach you this, at the stage
you are in right now the Internet won't do you much good in that department
unless you have 200 years to spare.

--
Jan

[Koobee Wublee]

On Thursday, February 25, 2016, JanPB wrote:
> On February 25, 2016 at 11:06:18 AM, Jack...@hotmail.com wrote:

> > Jan Bielawski wrote:

> > > Two metrics ds1^2 and ds^2 determine the same geometry if and
> > > only if there exists a _diffeomorphism_ F (of the manifold)
> > > which induces a transformation of one metric tensor into the
> > > other, say ds1^2 into ds2^2:
>
> > > F^*(ds1^2) = ds2^2
>
> > ds1^2 and ds2^2 are not tensors, they are called "metrics" and
> > they are pure scalars.
>
> No, they are tensors.

No, ds^2 is just a scalar that represents the geometry. In the following description, you can find what the metric [g] and the coordinate system are. <shrug>

** ds^2 = [g]_ij d[q]^i d[q]^j

Where

** ds = spacetime
** [g] = matrix that represents the metric
** [q] = matrix that represents the coordinate system
** [g]^i, [g]^j, [q]^i, [q]^j = elements to these matrices

> [matheMAGICS snipped]

>
> OTOH the "d" in "ds^2" is merely a convenient metaphor, it's neither a
> derivative nor a square of anything. The reason for this metaphoric
> symbol is that the line element in path integrals (denoted by "ds"
> usually, that "d" again is metaphoric) is equal to the square root
> of the "ds^2" tensor evaluated at the tangent vector V to the path:
>
> "ds in line integrals" = sqrt( ds^2(V,V) )

Not even wrong! <shrug>

> [more nonsense snipped]

[JanPB]

On Thursday, February 25, 2016 at 3:47:09 PM UTC-8, Koobee Wublee wrote:
> On Thursday, February 25, 2016, JanPB wrote:
> > On February 25, 2016 at 11:06:18 AM, Jack...@hotmail.com wrote:
> > > Jan Bielawski wrote:
>
> > > > Two metrics ds1^2 and ds^2 determine the same geometry if and
> > > > only if there exists a _diffeomorphism_ F (of the manifold)
> > > > which induces a transformation of one metric tensor into the
> > > > other, say ds1^2 into ds2^2:
> >
> > > > F^*(ds1^2) = ds2^2
> >
> > > ds1^2 and ds2^2 are not tensors, they are called "metrics" and
> > > they are pure scalars.
> >
> > No, they are tensors.
>
> No, ds^2 is just a scalar that represents the geometry.

Incorrect (also, the phrase "scalar that represents the geometry" is
a meaningless string of words).

You don't understand the rest at all, so I won't bother any further.

--
Jan

[Koobee Wublee]

On Thursday, February 25, 2016, Jan the relativistic moron wrote:
> Koobee Wublee wrote:

> > ds^2 is just a scalar that represents the geometry. In the
> > following description, you can find what the metric [g] and
> > the coordinate system are. <shrug>
>
> > ** ds^2 = [g]_ij d[q]^i d[q]^j
>
> > Where
>
> > ** ds = spacetime
> > ** [g] = matrix that represents the metric
> > ** [q] = matrix that represents the coordinate system
> > ** [g]^i, [g]^j, [q]^i, [q]^j = elements to these matrices

> Incorrect (also, the phrase "scalar that represents the geometry"
> is a meaningless string of words).

It is meaningless to relativistic morons. <shrug>

> You don't understand the rest at all, so I won't bother any
> further.

For example, the Minkowski spacetime can be described as the following in the Cartesian coordinate system. <shrug>

** ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2

Notice that it is a number with a unit of length^2. It is not a tensor nor a metric. When you are that fvcking dumb, it is best not to broadcast your ignorance. <shrug>

[Koobee Wublee]

On Thursday, February 25, 2016 at 9:12:01 AM UTC-8, JanPB wrote:

> On February 24, 2016 at 10:31:28 PM UTC-8, Koobee Wublee wrote:
> > On Tuesday, February 23, 2016 at 5:05:13 PM UTC-8, JanPB wrote:

> > > Two metrics ds1^2 and ds^2 determine the same geometry if and
> > > only if there exists a _diffeomorphism_ F (of the manifold)
> > > which induces a transformation of one metric tensor into the
> > > other, say ds1^2 into ds2^2:
>
> > > F^*(ds1^2) = ds2^2
>

> > Well, ds1 and ds2 can be two different geometries/realties.

> They cannot. What F^*(ds1^2) means is that F transforms manifold
> points and tangent vectors in such a way that dot products between
> the latter are unchanged.

You cannot play God transforming one reality to another. You can transform reality into your observation or transform that same reality to someone else's observation. That is what the change of coordinate system is all about under differential geometry. <shrug>

[Odd Bodkin]

s^2 = c^2 * t^2 - x^2 - y^2 - z^2 is a number with units of length^2.
Notice the OPERATOR d in front in the expression above.

I'm not surprised that you don't know the difference. You obviously
didn't know much about calculus anymore. Pity you've forgotten so much.

To illustrate your ignorance, consider a function y = x^2, where x has
units meters. So y has a scalar VALUE of 4 m^2 when x=2 m. Now, Koobee,
kindly tell me the VALUE of dy at that value of x.

PS. You can keep your CalTech diplomas, if you like. They are the only
remaining evidence that you ever learned anything.

[JanPB]

Pretty much everything you say here is incorrect.

--
Jan

[Koobee Wublee]

On Thursday, February 25, 2016 at 4:30:07 PM UTC-8, Odd Bodkin wrote:

> On 2/25/2016, Koobee Wublee wrote:

> > ds^2 is just a scalar that represents the geometry. In the
> > following description, you can find what the metric [g] and
> > the coordinate system are. <shrug>
>
> > ** ds^2 = [g]_ij d[q]^i d[q]^j
>
> > Where
>
> > ** ds = spacetime
> > ** [g] = matrix that represents the metric
> > ** [q] = matrix that represents the coordinate system
> > ** [g]^i, [g]^j, [q]^i, [q]^j = elements to these matrices
>

> > For example, the Minkowski spacetime can be described as the
> > following in the Cartesian coordinate system. <shrug>
> >
> > ** ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
> >
> > Notice that it is a number with a unit of length^2. It is not a

> > tensor or a metric.

> s^2 = c^2 * t^2 - x^2 - y^2 - z^2 is a number with units of length^2.
> Notice the OPERATOR d in front in the expression above.

<shaking head>

> I'm not surprised that you don't know the difference. You obviously
> didn't know much about calculus anymore. Pity you've forgotten so much.

The 'd' in ds^2 above is not an operator. Technically, d of dANYTHING is not an operator. Leibniz was the one who founded the set of rules to reduce it. You are an idiot, PD. <shrug>

> [rest of garbage snipped]

[John Gogo]

On Tuesday, February 23, 2016 at 7:05:13 PM UTC-6, JanPB wrote:
> This post is a continuation of the thread "Lagranian Method Revisited" [sic]
> which discussed Karl Schwarzschild's 1916 paper "Ueber das Gravitationsfeld
> eines Massenpunktes nach der Einstein'schen Theorie" (1916) (English
> translation "On the gravitational field of a mass point according to
> Einstein's theory" available at arvix.org in physics/9905030v1).
>
> It is sometimes claimed (by the above English translators, for example) that
> the paper describes a different solution of the Einstein field equations than
> the one described currently in textbooks. Specifically, it is asserted that
> his original solution does not exhibit a black hole but merely a static
> spherically symmetric vacuum region with a pointlike mass in the centre.
>
> In this note I'd like to explain why this conclusion is incorrect and also to
> go in some detail through the relevant parts of Schwarzschild's paper in
> order to find the subtle error he made.
>
> This is in no way to diminish his achievement, in those days hardly anybody
> knew differential geometry anyway, and among physicists it was practically
> unknown.
>
> To appreciate the subtlety of the problem we need to begin with simple
> examples illustrating potential complications.
>
>
> Section I.
> METRICS AND GEOMETRIES IN MANIFOLDS WITH FIXED COORDINATE SYSTEMS
>
> In this section I'll use exclusively manifolds with the _identity map_ as
> their coordinate system. I won't discuss _other_ coordinate systems on
> manifolds until Section II.
>
> First we need to state precisely what we mean by geometry. It's important to
> keep in mind that metric tensor determines geometry but that relation is NOT
> one-to-one. For example, the following tensor describes the flat (Minkowski)
> geometry in 2D:
>
> ds1^2 = -dt^2 + dx^2
>
> But the same geometry is also described by a _different_ tensor:
>
> ds2^2 = 4 dt dx
>
> (Again, I'm stressing here that in this section the coordinate system never
> changes, so ds2^2 is a _different tensor_ than ds1^2.)
>
> The geometry is the same because the Riemann curvature of the second metric
> is identically zero (skipping the calculation here).
>
> So what determines whether two given metrics define the same geometry?
>
> Answer:

> Two metrics ds1^2 and ds^2 determine the same geometry if and only if there
> exists a _diffeomorphism_ F (of the manifold) which induces a transformation
> of one metric tensor into the other, say ds1^2 into ds2^2:
>
> F^*(ds1^2) = ds2^2
>

> The "F with the superscript star" notation stands for the transformation of

> covariant tensors induced by F. I'll define this mapping in a moment.
>
> "Diffeomorphism" here simply means "smooth map which is one-to-one and whose
> inverse is also smooth". "Smooth" here means "infinitely differentiable".
> (A bit later I'll state a useful criterion for checking whether a given map
> is a diffeomorphism that's easier to use than this definition.)
>
> For example:
> (1) f(x) = sinh(x) is a diffeomorphism of the entire real line (all x) since
> it's 1-1, infinitely differentiable for all x, and so is its inverse
> f^(-1)(x) = arsinh(x) (recall the derivative of arsinh(x) is 1/sqrt(1+x^2),
> and similar for higher derivatives, so the denominator never causes any
> trouble).
>
> (2) g(x) = x^3. This is NOT a diffeomorphism of the entire real line.
> Although it is 1-1, smooth, and invertible, its inverse is NOT smooth:
> g^(-1)(x) = x^(1/3), so its derivative is 1/(3x^(2/3)) - and the denominator
> is zero when x = 0. Nevertheless, g(x) _is_ a diffeomorphism of on the
> subset of the real line consisting of _all nonzero_ real numbers. So the
> domain is important for being a diffeomorphism or not.
>
> OK, so what does this "F^*" mean. Let's say that ds1^2 is:
>
> ds1^2 = g_ij dx^i dx^j
>
> Then F^*(ds1^2) (called the _pullback_ of ds1^2 under F) is defined as
> follows:
>
> F^*(ds1^2) = (g_ij . F) d(x^i . F) d(x^j . F)
>
> ...where by the dot "." I mean composing functions. So "g_ij.F" is F followed
> by g_ij, and "x^i.F" means "the i-th component of F" or "F followed by the
> i-th coordinate function".
>
> We then also say that the manifold with the metric ds1^2 is _isometric_ to
> the same manifold with the metric ds2^2. The map F is called the _isometry_.
> Isometric manifolds are indistinguishable geometrically and physically. The
> reason is that isometries preserve the dot product, hence the distances and
> the angles, hence the whole geometry. That's why pullbacks enter the picture
> here. I should also mention (although this is getting ahead of myself a bit)
> that curvature tensors of metrics related by a pullback are themselves
> related by pullback which in our case means that if a ds^2 satisfies the
> Einstein equation, so does F^*(ds^2) for ANY diffeomorphism F.
>
> All this pullback business should become clear by examining the Minkowski
> example from a minute ago:
>
> ds1^2 = -dt^2 + dx^2
> ds2^2 = 4 dt dx
>
> In this case they are isometric because ds2^2 is the pullback of ds1^2 under
> the following diffeomorphism F of the (t,x)-plane:
>
> F(t, x) = (x - t, x + t)
>
> Let's check it using the definition of the pullback (note that the
> compositions are: -1.F = -1 and 1.F = 1):
>
> F^*(ds1^2) =
>
> = F^*(-dt^2 + dx^2) =
>
> = -d("t-component of F")^2 + d("x-component of F")^2 =
>
> = -d(x - t)^2 + d(x + t)^2 =
>
> = -(dx - dt)^2 + (dx + dt)^2 =
>
> = -dx^2 - dt^2 + 2 dt dx + dx^2 + dt^2 + 2 dt dx =
>
> = 4 dt dx =
>
> = ds2^2
>
> The silly factor 4 can be removed by one more pullback via the following
> diffeomorphism G:
>
> G(t, x) = (t/2, x/2)
>
>
> G^*(ds2^2) =
>
> = G^*(4 dt dx) =
>
> = 4 d(t/2) d(x/2) =
>
> = 4 1/2 dt 1/2 dx =
>
> = dt dx
>
> So ds3^2 = dt dx is _also_ the flat Minkowski space (this too can be verified
> directly by computing the curvature of ds3^2 by hand).
>
> OK, so this was the easy part. Now the subtle part begins. Let's imagine a
> slightly trickier diffeomorphism:
>
> H(t, x) = (t, 1/x)
>
> Notice that its domain isn't the whole (t,x)-plane, it's the plane with the
> set of points where x = 0 (i.e., the t-axis) _removed_.
>
> H a diffeomorphism there because it's clearly 1-1 with both F and its inverse
> infinitely differentiable at each x =/= 0.
>
> BY THE WAY: there exists a very good criterion for verifying whether a given
> map is a diffeomorphism that does NOT require computing the inverse of the
> map: it's called "the inverse function theorem" and is one of the two pillars
> of real analysis. (The other pillar of real analysis, incidentally, is
> "Lebesgue dominated convergence theorem" - but I digress.)
>
> That theorem goes like this: in order to check F is a diffeomorphism all you
> need to verify is:
>
> 1. F is 1-1,
> 2. F is smooth,
> 3. the Jacobian of F is nonzero _at every point of the domain of interest_.
>
> No need to invert F, very handy-dandy. It's easy to check that all our
> diffeomorphisms so far satisfy this criterion [exercise].
>
> Let's now calculate the pullback (call it ds4^2) of the Minkowski ds1^2 under
> the above H:
>
> ds4^2 =
>
> = H^*(ds1^2) =
>
> = H^*(-dt^2 + dx^2) =
>
> = -dt^2 + d(1/x)^2 =
>
> = -dt^2 + [ (-1/x^2) dx ]^2 =
>
> = -dt^2 + dx^2/x^4
>
> So this is also the flat Minkowski space (and yet again, one can verify this
> directly by calculating the curvature by hand).
>
> BUT there is now a kink: the metric ds4^2 is undefined at x = 0! To be sure,
> that new tensor is flat wherever it's defined but it's _not defined_ at x = 0!
>
> BINGO.
>
> Here we've just encountered the standard phenomenon of the trade: the _same
> geometry_ represented by two tensors _whose domains do not coincide_. In this
> case ds1^2 is defined everywhere in the (t,x)-plane but ds4^2 isn't: one of
> its coefficients (1/x^4) blows up.
>
> In other words: ds4^2 has a singularity at x = 0 but the geometry (and the
> manifold) does NOT. (This ought to sound familiar...)
>
> AGAIN:
>
> The tensor ds4^2 _DOES NOT_ extend over x = 0 (the t-axis), it has a
> _genuine_ singularity there, but it is the pullback of a tensor that DOES
> extend. (Namely, the ds1^2 one.) So the geometry is well-defined, including
> at x = 0. The singular behaviour of ds4^2 is only due to choosing a slightly
> misbehaving isometry to transform one into the other.
>
> So:
> A geometry does not have to stop where a metric tensor defining it stops. It
> may be possible the geometry is simply a restriction of a larger geometry
> which can be seen by selecting an appropriate pullback.
>
> In our case, this geometry extending process would go something like this:
> someone hands you over the metric ds4^2, tells you that it's been written
> with respect to the canonical coordinates, you see it's not defined at x = 0
> but for some reason or other you suspect the singularity is not really
> present in the geometry (say, your reason is that you've calculated the
> curvature of ds4^2 and obtained a value (namely, zero!) which does NOT blow
> up at x = 0). To prove it we need to produce an explicit pullback that
> extends over x = 0. There is no general recipe for doing this other than
> educated guesswork. In our case we know what to do, of course, given that we
> know where ds4^2 came from. We set the following diffeomorphism (the inverse
> of H, which happens to look the same as H):
>
> K(t, x) = (t, 1/x).
>
> Then:
>
> K^*(ds4^2) =
>
> = K^*(-dt^2 + dx^2/x^4) =
>
> = -dt^2 + d(1/x)^2 / (1/x)^4 =
>
> = -dt^2 + (-1/x^2)^2 dx^2 x^4 =
>
> = -dt^2 + dx^2
>
> ...and this is defined on the whole plane, so is the geometry.
>
> Time for two more examples illustrating certain notable patterns.
>
> Example A:
>
> The location of tensor singularity is not fixed, it can change in a
> pullback. Let's look at ds4^2 one more time:
>
> ds4^2 = -dt^2 + dx^2/x^4
>
> It's singular at x = 0. But consider a really basic diffeomorphism:
>
> L(t, x) = (t, x + 1)
>
> Pulling back:
>
> L^*(ds4^2) =
>
> = -dt^2 + [d(x + 1)]^2/(x + 1)^4 =
>
> = -dt^2 + dx^2/(x + 1)^4
>
> Now we got a tensor describing the Minkowski geometry which is singular at
> x = -1. This is borderline trivial but worth remembering: the coordinate
> value of the singularity is not a fixed property of the tensors defining
> the geometry.
>
> Example B:
>
> Minkowski ds1^2 again:
>
> ds1^2 = -dt^2 + dx^2
>
> Now let's look at this easy diffeomorphism:
>
> M(t, x) = (t, 2x)
>
> Therefore,
>
> M^*(ds1^2) =
>
> = -dt^2 + d(2x)^2 =
>
> = -dt^2 + 4 dx^2
>
> Well, nothing deep here, it would work the samy way for any nonzero number
> instead of 2 just the same:
>
> N(t, x) = (t, Cx) where C is any nonzero constant (you should know by
> now why we insist on C =/= 0 here)
>
> Therefore,
>
> N^*(ds1^2) =
>
> = -dt^2 + d(Cx)^2 =
>
> = -dt^2 + C^2 dx^2
>
> Now we got a tensor describing the Minkowski geometry which apparently
> depends on an _extra constant_. This is again borderline trivial but worth
> remembering: a constant appearing in metric tensor coefficients does NOT
> necessarily have ANY physical significance! It can be simply an artifact of
> the method used to calculate the given tensor and in such cases can be
> "pulled back away". Here it can be naturally "pulled back away" by the
> inverse diffeomorphism:
>
> N^(-1)(t, x) = (t, x/C)
>
>
> It would be nice if we could now start discussing Schwarzschild's paper but
> "unfortunately" we need to cover one more degree of freedom: coordinate
> changes. This is necessary because Einstein, Schwarzschild, and modern
> textbooks all use this concept. I said "unfortunately" because the "single
> coordinate system + pullbacks" approach used in the previous section is IMHO
> clearer to understand and to use, esp. when tricky subjects like
> singularities and geometry extensions arise. OTOH coordinate changes
> introduce a certain extra degree of topological complication, you'll see what
> I mean in a minute.
>
>
> Section II.
> METRICS AND GEOMETRIES IN DIFFERENT COORDINATES
>
> This is in a sense a "dual" degree of freedom: given a metric tensor written
> with respect to some coordinates, we can keep it _unchanged_ but instead
> change the linear bases of the tangent spaces of the manifold and expand the
> tensor at each point in those new bases.
>
> Warning: it turns out BOTH approaches produce FAPP same-looking formulas, so
> one must be specific and always state clearly what a given formula
> represents: _two different tensors_ (one before, one after the pullback) in a
> _fixed_ coordinate system OR _one tensor_ written with respect to _two
> different coordinate systems_.
>
> From the point of view of geometry or physics, both approaches are equivalent
> since they produce different but _isometric_ manifolds.
>
> NB: The pullback approach of Section I is sometimes referred to as "active
> view" and the coordinate change approach of this section as "passive view".
> The terminology suggests the former view changes the object in question (the
> tensor) while the latter one does not.
>
> Let's again consider one of the metrics from Section I to see how this
> approach works:
>
> ds1^2 = -dt^2 + dx^2
>
> ...written, as before, in the canonical coordinates (t, x). Let's now
> introduce, for the first time in this posting, _new_ coordinates, call them
> (T, X), defined as:
>
> T = (x - t)/2 (*)
> X = (x + t)/2
>
> i.e.:
>
> t = X - T
> x = X + T
>
> Hence:
>
> dt = dX - dT
> dx = dX + dT
>
> So we obtain:
>
> ds1^2 =
>
> = -(dX - dT)^2 + (dX + dT)^2 =
>
> = -dX^2 -dT^2 + 2 dT dX + dX^2 + dT^2 + 2 dT dX =
>
> = 4 dT dX
>
> This looks almost identical to the calculation in Section I, the mathematical
> transformation process is the same (except for using the upper-case letters)
> but the meaning and the objects involved are very different:
>
> --> in Section I we had _two different tensors_, ds1^2 and F^*(ds1^2), both
> written in terms of _the same coordinate system_. Here we have _one tensor_,
> ds1^2, written in terms of _two different coordinates_: the first system,
> (t,x), being the canonical one, and the second system, (T,X), being defined
> by the formulas (*) above.
>
> In all cases we have ONE geometry: in Section I that's because ds1^2 and
> F^*(ds1^2) are related by a pullback, hence they are isometric, and in this
> section we have just one tensor, so one geometry by definition.
>
> Let's do one more example, again similar to something in Section I, to show
> how tensor domains can change in different coordinates. This is where it gets
> a bit hairy. Start with ds1^2 again:
>
> ds1^2 = -dt^2 + dx^2
>
> Change the coordinates like so:
>
> T = t
> X = 1/x (obviously defined for x =/= 0 only)
>
> This means:
>
> t = T
> x = 1/X
>
> therefore (this should look annoyingly familiar by now):
>
> ds1^2 =
>
> = -dT^2 + d(1/X)^2 =
>
> = -dT^2 + [ -1/x^2 dX ]^2 =
>
> = -dT^2 + dX^2/X^4
>
> Again:
>
> 1. the mathematical manipulation looks practically the same as in Section I
> except for the upper-/lower-case distinguishing the coordinates,
>
> 2. the same geometry throughout (the flat Minkowski).
>
> But here comes a subtle point which I never see properly discussed in
> textbooks, even the unusually careful Robert Wald glosses over it. So let's
> look at another, deceptively similar scenario: someone gives you a metric
> like the one above but written instead in the _standard_ coordinates (t, x):
>
> ds^2 = -dt^2 + dx^2/x^4
>
> Remember, these coordinates are well-defined everywhere, so the blowing up of
> the coefficient 1/x^4 at x = 0 means the tensor ds^2 _itself_ blows up there
> too.
>
> No matter, we can change the coordinates anyway, why not try the (T, X)
> again. We get, by the same type of calculation as before:
>
> ds^2 = -dT^2 + dX^2
>
> ...and this expression does extend to X = 0. Wait a minute, didn't we just
> say that ds^2 blows up? We haven't changed ds^2, only the coordinates. So
> how come it suddenly doesn't blow up at zero?
>
> The answer is that we extend it over X = 0 which in the original coordinates
> does NOT correspond to _any point_ in the (t,x)-plane! (since 0 cannot equal
> 1/x) So where on earth does this ds^2 extend _to_? Where _is_ this point X =
> 0 if it's nowhere in the (t,x)-plane?
>
> What really happens is that we are extending the original expression (-dt^2 +
> dx^2/x^4) over _an extra set of manifold points which wasn't there before_.
> Namely we extend -dt^2 + dx^2/x^4 over the extra "line at x = +/- infinity"
> while at the same time removing the "line at x = 0" (the t-axis) over which
> -dt^2 + dx^2/x^4 _does not_ extend.
>
> The resulting manifold consists of a slightly different set of points but
> it's _isometric_ to the Minkowski plane.
>
> You can imagine the original (t,x)-plane wrapped around a cylinder (the
> t-axis pointing in the direction of the cylinder's axis) so that the two
> extremities of the plane where x approaches plus and minus infinity end up
> _just_ shy of the vertical line on the other side of the cylinder, opposite
> the t-axis x = 0. It is THAT vertical line which we add while removing the
> line x = 0. The result is clearly an isometric copy of the entire Minkowski
> plane but sort of "turned inside out".
>
> This sort of thing is precisely what is involved in extending the standard
> modern textbook Schwarzschild metric (typically written in the spherical
> coordinates space) over the horizon by switching to, say, the
> Eddington-Finkelstein coordinates: the metric does not extend over the space
> r = 2M but instead extends over a _copy_ of that space attached _differently_
> (namely, attached as limit points of the infalling (say) geodesics on the
> usual (t,r)-diagram).
>
> (BTW, this type of cutting and pasting parts of manifolds is called "surgery"
> and there is a HUGE field of topology devoted to it. GR textbooks avoid
> discussing this, presumably because this operation isn't much done in GR
> otherwise and ignoring it is of no consequence 99.9% of the time.
> Historically this sort of manifold manipulations and manifold extension arose
> for the first time in the context of complex analysis in the study of Riemann
> surfaces.)
>
> You can see now why I prefer the "active" approach of Section I where
> typically there is no need for surgery, just transform tensors by pullbacks
> as needed which keeps the geometry unchanged and the coordinate systems
> fixed, so less confusion.
>
> Finally:
>
>
> Section III.
> SCHWARZSCHILD'S PAPER
>
> Here we'll find out why the solution described in this paper is identical to
> the one described in all textbooks, i.e. why it describes the usual black
> hole geometry, including the horizon and the inner, non-static, region.
>
> Let's begin with a quick summary of his method. Then we'll stop at the
> "smoking gun moment" where he jumps to a conclusion too fast.
>
> He begins by writing down the vacuum field equations, Ricci = 0, for any
> coordinate system in which the determinant of the metric is identically -1.
> Under this restriction the field equations simplify quite a bit (to half
> their "native" size), he writes this as his equation (4). He then proceeds to
> construct such coordinate system.
>
> First he quotes Einstein's 1915 paper "Erklaerung der Perihelbewegung des
> Merkur aus der allgemeinen Relativitaetstheorie" ("Explanation of the
> Perihelion Motion of Mercury from General Relativity Theory", a rather bad
> English translation is available here:
> http://www.gsjournal.net/old/eeuro/vankov.pdf) by stating that "according to
> Mr. Einstein's list" the following conditions must be satisfied:
>
> 1. all metric components are independent of the time x_4 (he denotes the time
> coordinate by x_4),
>
> 2. g_i4 = g_4i = 0 for i = 1, 2, 3
>
> 3. the solution is spatially spherically symmetric wrt to the origin,
>
> 4. boundary conditions at infinity: g_44 = 1, g_11 = g_22 = g_33 = -1.
>
> So right away we are faced with the question: what exactly is the basis for
> claims 1 and 2? The field equations are a complicated system of 10 coupled
> PDEs for 10 unknown functions of four variables, why would merely assuming
> spherical symmetry and staticity imply 1 and 2 just like that? This
> conclusion seems wrong and it's easy to concoct a counterexample in fact:
>
> Consider the 4D vacuum with the following metric written in spherical
> coordinates:
>
> ds^2 = -cosh^2(t) dt^2 - 2 cosh(t) dt dr + r^2 dO^2
>
> ...where "dO^2" is the shorthand for the line element of the unit sphere:
> dtheta^2 + sin^2(theta) dphi^2.
>
> This ds^2 has BOTH time-dependent coefficients AND a cross-term (dt dr), so
> it violates conditions 1 and 2 in Schwarzschild's paper, yet it is the
> Minkowski geometry AGAIN, in particular a spherically symmetric static
> vacuum.
>
> To see this, just calculate its curvature or pull ds^2 back via the following
> map (easily seen to be a diffeomorphism by the inverse function theorem
> quoted earlier):
>
> F(t, r, theta, phi) = (arsinh(t - r), r, theta, phi)
>
> (Or use the equivalent "passive" change of coordinates: T = r + sinh(t), R =
> r, THETA = theta, PHI = phi.)
>
> So just saying "the solution is spherically symmetric and static" does not
> seem to _guarantee_ 1 and 2.
>
> Unfortunately many texbooks continue to sloppily insist that it is so, for
> example Ohanian & Ruffini, 2nd edition, p.392: "The demand for a static
> solution forbids terms of the type dr dt, dtheta dt, dphi dt in the
> expression for the spacetime interval because those terms change sign when we
> perform the time reversal t -> -t".
>
> But of course _nothing_ changes when we perform the time reversal, in fact we
> can "perform" ANY diffeomorphism whatsoever without affecting the geometry at
> all! What the authors really have in mind is that the _tensor itself_ should
> be invariant under such transformations. But this too is fishy because
> spherical symmetry refers to the 3D _spatial_ slices of spacetime. QUESTION:
> which ones? OK, so let's say "the spatial slices t = const. where "t" is one
> of the canonical (identity) coordinates on R^4. Well, actually, not quite,
> because due to spherical symmetry we want to use spherical coordinates in
> place of the canonical (x,y,z). So we really mean the spherical coordinate
> system on each spatial slice t = const.
>
> Well, it's flaky but it kind of works. Now the "static" part. A metric tensor
> in the above coordinates is static if its coefficients are t-independent.
>
> As you can see, the above definition is sloppy. But it's good enough
> for us here. Good modern texts like Wald or d'Inverno or Hawking & Ellis
> define these things properly, in a way that captures the physics and the
> geometry in a way that does not presume any coordinates. I won't dwell
> more on that point.
>
> So now Schwarzschild begins solving the Einstein equation by positing his
> condition 1 and 2 as an Ansatz: seek the solution in the following form,
> written in the above system consisting of spherical coordinates on slices
> t = const (he sometimes denotes time by "t", sometimes by "x_4"). This is
> Schwarzschild's equation (6):
>
> ds^2 = A(r) dt^2 - B(r) dr^2 - C(r) (dtheta^2 + sin^2(theta) dphi^2)
>
> ...where I took the liberty of denoting by A(r), B(r), C(r) certain functions
> of r he momentarily keeps written out explicitly and which arise from his
> Cartesian-to-spherical switch (their exact form is unimportant here).
>
> He finally makes one more coordinate change: from (t, r, theta, phi) to
> (x_1, x_2, x_3, x_4) [he reuses the letters which at the beginning of his
> paper denote the Cartesians] like so:
>
> x_1 = r^3/3, x_2 = -cos(theta), x_3 = phi, x_4 = t
>
> ...with the obvious ranges inherited from the spherical coordinates:
>
> x_1 > 0, -1 < x_2 < 1, 0 < x_3 < 2pi, -infinity < x_4 < +infinity
>
> In these new coordinates he writes the final general form of the desired
> solution and gets on with the actual solving the equation. That final general
> form is (his equation (9)):
>
> ds^2 =
>
> = f_4 dx_4^2 -
>
> - f_1 dx_1^2 -
>
> - f_2/(1 - x_2^2) dx_2^2 -
>
> - f_3 (1 - x_2^2) dx_3^2
>
> f_1, f_2, f_3, f_4 are the unknowns to solve for.
>
> All this means that _I'm retracting_ my previous claim that an extra
> coordinate change is needed beyond those mentioned above _in the static
> case_. When I was discussing this with Koobee, I had in mind the general case
> of the spherically symmetric geometry (i.e., no assumption of static). In
> that case one cannot posit that _all_ cross-terms disappear: the dt dr term
> has to stay BUT one can then construct an ADDITIONAL change of coordinates
> which removes it. This is the coordinate change I was talking about.
> Nevertheless, this additional coordinate change is not relevant anyway
> because as it turns out it is a _further_ change down the road that's
> responsible for moving the points surrounding the r = 0 set _away_
> from the origin.
>
> Remember this is the reason for this post: to show r = 0 ceases to be
> "pointlike" in Schwarzschild.
>
> OK, so so far we still have: r is the polar coordinate with r = 0 denoting
> the origin, and x_1 = r^3/3, so 1_4 has the same property.
>
> It looks like we're stuck with the "central pointlike mass at r = 0"
> but hang on.
>
> Let's continue following Schwarzschild. He plugs the above ds^2 in the field
> equations and solves them for the above unknown metric coefficients f_1, f_2,
> f_3, f_4. These solutions are his equations (10), (11), (12).
>
>
> THE SMOKING GUN
> And at this point, right after his equation (12), he stumbles. He notes that
> one of the metric coefficients (f_1) is discontinuous and he _immediately_
> assumes this discontinuity _is_ the pointlike mass at the origin! Well, in
> 1916 it must have seemed perfectly sensible: what OTHER singularity IS there??
>
> But now we know, and we've seen examples of this earlier, that _a singularity
> of a metric tensor does not necessarily imply the singularity of the
> geometry_! There may exist a pullback which will change the tensor to
> another one with a different domain which may well cover the supposed
> geometric singularity. (Equivalently: there may exist a coordinate change...
> etc. etc., you know the rest).
>
> In fact, his "stumble" appears for the first time a bit earlier, on page 3
> [English translation] where he lists conditions his f-coefficients should
> fulfil, specifically his condition 4 which insists on:
>
> "Continuity of the f, except for x_1 = 0".
>
> This is incorrect as matric tensor singularities do not necessarily arise
> from geometry singularities and therefore are NOT restricted to these
> locations.
>
> So his conclusion that rho = alpha^3 is likewise jumping the gun. Instead, he
> should have noticed that one of the two constants (rho) was _redundant_, just
> like the constant C in our Example B at the end of Section I. In other words,
> with the right diffeomorphism the constant rho could be absorbed by the
> pullback, with the tensor domain adjusted accordingly. This is the freedom
> allowed by the fact that Einstein's equation is a PDE for _unknown tensors_,
> not just a regular PDE for _unknown functions_. This would have left him with
> only one constant (alpha) which then could have been shown tied to the mass
> of the central "body". Of course Schwarzschild was hampered in his
> pullback/coordinate freedom by that constraint of the unit determinant.
>
> Let's see how this constant absorbing would've worked had Schwarzschild done
> it in the paper. Schwarzschild's metric ds^2 is hard to write down in ASCII,
> so I'm going to provide here a link to a properly typeset PDF page, but the
> point is that one can transform rho away by using the following simple
> diffeomorphism:
>
> F(x_1, x_2, x_3, x_4) = ((x_1 - rho)/3, x_2, x_3, x_4)
>
> Here is the link to the PDF which is much easier on the eyes:
> https://drive.google.com/file/d/0B2N1X7SgQnLRY1ExWUczZldYZXM
>
> Summing up in ASCII: pulling Schwarzschild's ds^2 back by F yields:
>
> F^*(ds^2) =
>
> = (1 - alpha*x_1^(-1/3)) dx_4^2 -
>
> x_1^(-4/3)/(1 - alpha*x_1^(-1/3)) * (1/3)^2 dx_1^2 -
>
> (x_1^(2/3))/(1 - x_2^2) dx_2^2 -
>
> (x_1^(2/3))*(1 - x_2^2) dx_3^2
>
> ...with the x_1 coordinate of the pullback satisfying: x_1 > rho (recall
> Schwarzschild's original x_1 was defined as the polar r^3/3, hence his x_1
> satisfied x_1 > 0).
>
> His original singularity of ds^2 was at (his) x_1 = (alpha^3 - rho)/3.
>
> Thus the above pullback F^*(ds^2) has the same singularity (unless it happens
> to extend, which it does not) at:
>
> x_1 = 3((the old singularity location) + rho =
>
> = 3((alpha^3 - rho)/3 + rho) =
>
> = alpha^3
>
> ...which is a strictly positive number. THIS is the coordinate change that
> moves the "r = 0" away from the origin.
>
> Since x_1 > rho, where rho is any constant, we are free to extend the metric
> as far "down" as we please but another glance at F^*(ds^2) shows that the
> metric has another singularity at x_1 = 0 (because of the cube roots of x_1
> in the denominators). A curvature calculation then shows it's actually a
> genuine singularity of the geometry (and you know the rest).
>
> As usual, a completely analogous reasoning can be made using the "passive"
> (coordinate change) viewpoint.
>
> PS:
> Another reason why Schwarzschild's set r = 0 is not a point is that if one
> surrounds it by a sphere, it's surface area does NOT go to zero as r
> approaches 0 (it approaches 4pi alpha^2 which is a nonzero number since alpha
> is nonzero whenever the mass is nonzero). For a proof check this link:
> https://drive.google.com/open?id=0B2N1X7SgQnLRV1otWnE2T2F6aEk
>
> --
> Jan

MHPF is called the isometry.
Isometric manifolds are indistinguishable geometrically and physically.
Pullbacks, dot products, diffeomorphisms, inverse function theorem. 4pi alpha ^2 which is a nonzero number- which is 10,000.

[PC]

JanPB wrote:

>> No, ds^2 is a scaler product formed from ds^2 = dX^i*dX_j delta ij,
>> Einstein summation. The dX's are covariant and contravariant vectors
>> (components of dX).
>
> No, that's incorrect. Your ds^2 is equal to dx^i dx_i which is not true.
> Also "The dX's are covariant and contravariant vectors (components of
> dX)" is a nonsensical sequence of words.

That's what happen when you give same thing ten meaning, Janny. You must
admit Koobee Woblee is stronger than you when it comes to tensors.

[David Waite]

Oh please, I doubt koobee could even calculate g^mu^nu from g_mu_nu for
ds^2=dx0^2-dr^2-r^2dOmega^2-(R/r)[(dr-dx0)^2]
if his life depended on it. He probably couldn't even write down a matrix representation of g_mu_nu for it.

[PC]

Economic Update: Capitalism is the Problem
http://www.rdwolff.com/download/sites/default/files/Econ_Update_Truthout_2016.02.18.WEB_.mp3

[Gary Harnagel]

On Friday, February 26, 2016 at 7:59:51 AM UTC-7, PC wrote:
>
> David Waite wrote:
> >

> > Oh please, I doubt koobee could even calculate g^mu^nu from g_mu_nu for
> > ds^2=dx0^2-dr^2-r^2dOmega^2-(R/r)[(dr-dx0)^2]
> > if his life depended on it. He probably couldn't even write down a matrix
> > representation of g_mu_nu for it.
>
> Economic Update: Capitalism is the Problem
> http://www.rdwolff.com/download/sites/default/files/Econ_Update_Truthout_2016.02.18.WEB_.mp3

Non sequitur.

Dishonest person's scheme: When out of his depth, change the subject.

[David Waite]

On Friday, February 26, 2016 at 7:59:51 AM UTC-7, PC wrote:

And what "ism" isn't in some way a problem? Not relevant anyway.

[David Waite]

Hell, I'd bet money he doesn't even know what g^mu_nu is, which is a no brainer.

[al...@interia.pl]

You are a stupid amateur, not any mathematician.
And you proved that very well - just writing this idiotic text.

[PC]

Gary Harnagel wrote:

>> > Oh please, I doubt koobee could even calculate g^mu^nu from g_mu_nu
>> > for ds^2=dx0^2-dr^2-r^2dOmega^2-(R/r)[(dr-dx0)^2]
>> > if his life depended on it. He probably couldn't even write down a
>> > matrix representation of g_mu_nu for it.
>>
>> Economic Update: Capitalism is the Problem
>> http://www.rdwolff.com/download/sites/default/files/
Econ_Update_Truthout_2016.02.18.WEB_.mp3
>
> Non sequitur.
>
> Dishonest person's scheme: When out of his depth, change the subject.

Shut up pretender. You know nothing about anything.

[PC]

David Waite wrote:

>> > That's what happen when you give same thing ten meaning, Janny. You
>> > must admit Koobee Woblee is stronger than you when it comes to
>> > tensors.
>>
>> Oh please, I doubt koobee could even calculate g^mu^nu from g_mu_nu for
>> ds^2=dx0^2-dr^2-r^2dOmega^2-(R/r)[(dr-dx0)^2]
>> if his life depended on it. He probably couldn't even write down a
>> matrix representation of g_mu_nu for it.
>
> Hell, I'd bet money he doesn't even know what g^mu_nu is, which is a no
> brainer.

Shut up, David Waste. Or answer the question.

[Paul B. Andersen]

On 26.02.2016 01:09, Koobee Wublee broadcast his ignorance:

>
> For example, the Minkowski spacetime can be described as
> the following in the Cartesian coordinate system. <shrug>
>
> ** ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
>
> Notice that it is a number with a unit of length^2.
> It is not a tensor nor a metric. When you are that fvcking dumb,
> it is best not to broadcast your ignorance. <shrug>

A very wise advice! <shrug>

--
Paul

https://paulba.no/

[PC]

Why are you doing it then, exactly that?

[JanPB]

Anything else besides 2 lines of hurt feelings?

--
Jan

[PC]

David Waite wrote:

>> > Oh please, I doubt koobee could even calculate g^mu^nu from g_mu_nu
>> > for ds^2=dx0^2-dr^2-r^2dOmega^2-(R/r)[(dr-dx0)^2]
>> > if his life depended on it. He probably couldn't even write down a
>> > matrix representation of g_mu_nu for it.
>>
>> Economic Update: Capitalism is the Problem
>> http://www.rdwolff.com/download/sites/default/files/Econ_Update_Truthout_2016.02.18.WEB_.mp3
>
> And what "ism" isn't in some way a problem? Not relevant anyway.

You are deluded. Try this then.

Michael Parenti - The U.S. War on Yugoslavia
https://www.youtube.com/watch?v=GEzOgpMWnVs

[JanPB]

You are probably right.

--
Jan

[al...@interia.pl]

W dniu piątek, 26 lutego 2016 20:33:59 UTC+1 użytkownik JanPB napisał:

> > You are a stupid amateur, not any mathematician.
> > And you proved that very well - just writing this idiotic text.
>
> Anything else besides 2 lines of hurt feelings?
>
> --
> Jan

The math is a very strict science... there are
no place for any fanfaronades, like you wrote.

[JanPB]

Still no content.

--
Jan

[Thomas 'PointedEars' Lahn]

_Dimension_, not “units”; _quantity_, not “number”. Even then your
statement is only true if c has the dimension length∕time, t has the
dimension time, and x, y, and z have the dimension length.

> Notice the OPERATOR d in front in the expression above.

That does not change a lot regarding the question at hand. To me there
appears to be as much confusion on the one side as negligence on the others
so far.

According even to Einstein himself, ds *is* “a scalar that represents the
(spacetime) geometry” indeed; that is not at all “a meaningless string of
words”. It is a spacetime differential, the infinitesimal (infinitely
small) change in spacetime position. If ds < 0, then (using the time-like
sign convention), if I do not confuse the first and the last ones, the
spacetime interval is space-like (events thus connected lie outside of their
light cones), if ds = 0 it is light-like (… on their light cones), and if ds
> 0 it is time-like (… within their light cones). The same is true for just
s if t, x, y, and z represent *measured* *differences* in corresponding
coordinates (but it is important that they do).

Compare the Pythagorean theorem in flat 2-space of which this is, in a
manner of speaking, just the calculus version adapted for the peculiarities
of Minkowski (flat)/pseudo-Riemannian (curved) space(time). In the
Pythagorean theorem in its common form, you calculate the length of the
hypotenuse c from the square of the length of the cathetes a and b: c² = a²
+ b²; you *actually* calculate the *distance* between the points on either
cathete that the hypothenuse connects, based on their coordinates. The
distance between points is always a number, a scalar. That does not change
with a different metric or higher dimensions.

Now, by definition, a scalar is a 0th-order tensor (just as a vector is a
1st-order tensor, and a matrix is a 2nd-order tensor), so it is not
incorrect (of Jan) to say of that scalar that it is a tensor, and it is
incorrect (of the other person) to say that it is not a tensor; but it also
does not strike me as overly useful to insist on the term “tensor” for ds
(and, IIUC, for dt, dx, dy, and dz).

It is true that ds is _not_ “spacetime” (that would be the entire manifold
instead); it is the *line element* of the corresponding spacetime instead.
But “g” *is* the spacetime *metric* (not something else that “represents the
metric”).

It certainly *is* useful to know that the “g” and “dq” are also tensors in
the aforementioned equation for *curved* spacetime, because “g” – therefore
also called the metric *tensor* – occurs in the Einstein field equations,
and it matters how tensors transform.

See also:
<http://www.britannica.com/topic/Albert-Einstein-on-Space-Time-1987141>

More technical:
<http://mathworld.wolfram.com/MetricTensor.html>

CMIIW.

> I'm not surprised that you don't know the difference. You obviously
> didn't know much about calculus anymore. Pity you've forgotten so much.

^^^^^^^^^^^
YGCIB.

> To illustrate your ignorance, consider a function y = x^2, where x has
> units meters.

I do not think it is helpful to introduce an physical unit of measurement
here. This problem can be solved without such (more easily).

> So y has a scalar VALUE of 4 m^2 when x=2 m. Now, Koobee,
> kindly tell me the VALUE of dy at that value of x.

I could tell you the value of ∂y(x)∕∂x for x, and at that value of x (it is,
of course, equivalent to the [function of the] slope of the corresponding
curve at a value of x, which is a number/scalar). But “the value of dy (at
that value of x)” – does that even have meaning in and of itself,
considering that dy is a differential, an *infinitesimally small* value?


PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

[al...@interia.pl]

Maybe you should write this fallacy to the sci...fundations or physicsforums - there are many similar impotent imbecilic peple to you;
so your fantastic pseudomath would be probably... very nice there.

BTW: Is the Shapiro delay correct in your math.. at least?
Oh! it's still the same - too short 20us... I'm very sorry.

[JanPB]

On Friday, February 26, 2016 at 2:05:01 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
>
> According even to Einstein himself, ds *is* "a scalar that represents the
> (spacetime) geometry" indeed; that is not at all "a meaningless string of
> words".

This is because professionals don't need to be pedantic and the difference
between ds^2 as a tensor (which is what it really is) and its value on
the relevant curve's tangent vector (which is a number), as well as the
square root of this number (the "ds") is presumed understood and obvious
and not worth elaborating. Everyone knows what's what.

> It is a spacetime differential, the infinitesimal (infinitely
> small) change in spacetime position.

Well, yeah, except "infinitesimal" is an undefined notion (unless you
are explicitly using non-standard analysis which AFAICT no physicist
ever used or uses). "Infinitely small" is not a number: for any real x,
either x < 0 or x = 0 or x > 0. There is no room in this scheme for
"positive but smaller than all positive numbers" other than as a means
of torturing undergraduate students with it.

Again, professionals can use this hand-waving terminology because they
know what this shortcut refers to. It's like that saying about eating
fish: those who know that one ought not to use fork and knife for fish
can use fork and knife for fish.

Finally, keep in mind Einstein's notation is a bit old-fashioned in places,
none of the developments in global analysis due to Georges de Rham (Lausanne
and Geneva!) and Élie Cartan seem to have made any impact on his writing.

> --
> A neutron walks into a bar and inquires how much a drink costs.
> The bartender replies, "For you? No charge."
>
> (from: WolframAlpha)

"That's strange", said the kaon.

--
Jan

[JanPB]

Still no content, just rhetoric. Anyone can say "X is fantastic pseudomath".
So?

--
Jan

[Koobee Wublee]

The ranting above from Jan the relativistic moron does not contain a single correct sentence. <shrug>

[Thomas 'PointedEars' Lahn]

JanPB wrote:

> On Friday, February 26, 2016 at 2:05:01 PM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> According even to Einstein himself, ds *is* "a scalar that represents the
>> (spacetime) geometry" indeed; that is not at all "a meaningless string of
>> words".
>
> This is because professionals don't need to be pedantic and the difference
> between ds^2 as a tensor (which is what it really is) and its value on
> the relevant curve's tangent vector (which is a number), as well as the
> square root of this number (the "ds") is presumed understood and obvious
> and not worth elaborating. Everyone knows what's what.

You have been at least ambiguous about the correctness of the use of
“scalar”. You have claimed that using “scalar” instead of “tensor” would
not be correct, and that something “represents the [spacetime] geometry”
would be nonsense, by saying to this “incorrect” and calling it “a
meaningless string of words”. Neither is true.

It would be appropriate if you admitted your mistake instead of making
further spurious claims (with weasel words) about what “everyone knows”.
If one is explaining, one inherently starts from the assumption that _not_
“everyone knows”. Being ambiguous and imprecise is the failure of the
person explaining, not of the person something is explained to.

>> It is a spacetime differential, the infinitesimal (infinitely
>> small) change in spacetime position.
>
> Well, yeah, except "infinitesimal" is an undefined notion (unless you
> are explicitly using non-standard analysis which AFAICT no physicist
> ever used or uses).

“Infinitesimal *change*” (which I said) and “infinitesimal” (which I did not
say) are neither “undefined notion” nor “non-standard analysis”.

,-<http://mathworld.wolfram.com/Derivative.html>
|
| The derivative of a function represents an infinitesimal change in the
^^^^^^^^^^^^^^^^^^^^
| function with respect to one of its variables. […]
|
| […]
| REFERENCES:
| Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical
| Functions with Formulas, Graphs, and Mathematical Tables, 9th printing.
| New York: Dover, p. 11, 1972.
|
| Amend, B. Camp FoxTrot. Kansas City, MO: Andrews McMeel, p. 19, 1998.
|
| Anton, H. Calculus: A New Horizon, 6th ed. New York: Wiley, 1999.
|
| calc101.com. "Step-by-Step Differentiation."
| http://www.calc101.com/webMathematica/MSP/Calc101/WalkD.
|
| Beyer, W. H. "Derivatives." CRC Standard Mathematical Tables, 28th ed.
| Boca Raton, FL: CRC Press, pp. 229-232, 1987.
|
| Griewank, A. Principles and Techniques of Algorithmic Differentiation.
| Philadelphia, PA: SIAM, 2000.
|
| Mitchell, C. W. Jr. In "Media Clips" (Ed. M. Cibes and J. Greenwood).
| Math. Teacher 100, 339, Dec. 2006/Jan. 2007. Sloane, N. J. A. Sequence
| A021009 in "The On-Line Encyclopedia of Integer Sequences."

,-<http://mathworld.wolfram.com/Infinitesimal.html>
|
| An infinitesimal is some quantity that is explicitly nonzero and yet
| smaller in absolute value than any real quantity.
|
| […]
| REFERENCES:
| Bell, J. L. A Primer of Infinitesimal Analysis. Cambridge, England:
| Cambridge University Press, 1998.
|
| Wolfram, S. A New Kind of Science. Champaign, IL: Wolfram Media, p. 1168,
| 2002.

,-<https://en.wikipedia.org/wiki/Differential_(infinitesimal)>
|
| The term *differential* is used in calculus to refer to an infinitesimal
^^^^^^^^^^^^^^^^
| (infinitely small) change in some varying quantity. For example, if x is a
^^^^^^^^^^^^^^^^^^^^^^^^^
| variable, then a change in the value of x is often denoted Δx (pronounced
| delta x). The differential dx represents an infinitely small change in the
| variable x. […]
|
| Using calculus, it is possible to relate the infinitely small changes of
| various variables to each other mathematically using derivatives. If y is
| a function of x, then the differential dy of y is related to dx by the
| formula
|
| dy = dy∕dx dx
|
| where dy∕dx denotes the derivative of y with respect to x. This formula
| summarizes the intuitive idea that the derivative of y with respect to x
| is the limit of the ratio of differences Δy/Δx as Δx becomes
| infinitesimal.

,-<https://en.wikipedia.org/wiki/Infinitesimal>
|
| In mathematics, *infinitesimals* are things so small that there is no way
| to measure them. The insight with exploiting infinitesimals was that
| entities could still retain certain specific properties, such as angle or
| slope, even though these entities were quantitatively small.[1]
|
| […]
| [1] http://plato.stanford.edu/entries/continuity/#1

By contrast, <http://mathworld.wolfram.com/NonstandardAnalysis.html>.

> "Infinitely small" is not a number:

Yes, it most definitely is.

> for any real x, either x < 0 or x = 0 or x > 0.

Yes.

> There is no room in this scheme for "positive but smaller than all
> positive numbers" other than as a means of torturing undergraduate
> students with it.

Non sequitur.

> Finally, keep in mind Einstein's notation is a bit old-fashioned in
> places, none of the developments in global analysis due to Georges de Rham
> (Lausanne and Geneva!) and Élie Cartan seem to have made any impact on his
> writing.

AISB, you will find the *notion* in other, modern places as well. Now you
can even follow the references.

>> --
>> A neutron walks into a bar and inquires how much a drink costs.
>> The bartender replies, "For you? No charge."
>>
>> (from: WolframAlpha)
>
> "That's strange", said the kaon.

:)


PointedEars
--
“Science is empirical: knowing the answer means nothing;
testing your knowledge means everything.”
—Dr. Lawrence M. Krauss, theoretical physicist,
in “A Universe from Nothing” (2009)

[JanPB]

On Saturday, February 27, 2016 at 1:27:22 AM UTC-8, Thomas 'PointedEars' Lahn wrote:
> JanPB wrote:
>
> > On Friday, February 26, 2016 at 2:05:01 PM UTC-8, Thomas 'PointedEars'
> > Lahn wrote:
> >> According even to Einstein himself, ds *is* "a scalar that represents the
> >> (spacetime) geometry" indeed; that is not at all "a meaningless string of
> >> words".
> >
> > This is because professionals don't need to be pedantic and the difference
> > between ds^2 as a tensor (which is what it really is) and its value on
> > the relevant curve's tangent vector (which is a number), as well as the
> > square root of this number (the "ds") is presumed understood and obvious
> > and not worth elaborating. Everyone knows what's what.
>
> You have been at least ambiguous about the correctness of the use of
> “scalar”. You have claimed that using “scalar” instead of “tensor” would
> not be correct, and that something “represents the [spacetime] geometry”
> would be nonsense, by saying to this “incorrect” and calling it “a
> meaningless string of words”. Neither is true.

It is. Maybe we simply differ in our tolerance to using loose terminology.
The fact is that ds^2 is a tensor, not a number. And I don't approve of statements
like "X represents the geometry" - what does it mean? If X defines the geometry,
that's how it should be phrased.

> It would be appropriate if you admitted your mistake instead of making
> further spurious claims (with weasel words) about what “everyone knows”.

What mistake?

> If one is explaining, one inherently starts from the assumption that _not_
> “everyone knows”. Being ambiguous and imprecise is the failure of the
> person explaining, not of the person something is explained to.

It is me who is trying to be precise.

> >> It is a spacetime differential, the infinitesimal (infinitely
> >> small) change in spacetime position.
> >
> > Well, yeah, except "infinitesimal" is an undefined notion (unless you
> > are explicitly using non-standard analysis which AFAICT no physicist
> > ever used or uses).
>
> “Infinitesimal *change*” (which I said) and “infinitesimal” (which I did not
> say) are neither “undefined notion” nor “non-standard analysis”.

It is an undefined object in mathematics unless you use non-standard analysis
or something similar.

> ,-<http://mathworld.wolfram.com/Derivative.html>
> |
> | The derivative of a function represents an infinitesimal change in the
> ^^^^^^^^^^^^^^^^^^^^
> | function with respect to one of its variables. […]

It's just a pop-sci definition. It doesn't define anything.

Same thing. There all hand wavings.

> | Using calculus, it is possible to relate the infinitely small changes of
> | various variables to each other mathematically using derivatives. If y is
> | a function of x, then the differential dy of y is related to dx by the
> | formula
> |
> | dy = dy∕dx dx
> |
> | where dy∕dx denotes the derivative of y with respect to x. This formula
> | summarizes the intuitive idea that the derivative of y with respect to x
> | is the limit of the ratio of differences Δy/Δx as Δx becomes
> | infinitesimal.

This defines a limit (which is correct) but it uses incorrect terminology again: "as Δx becomes
infinitesimal." It should read "as Δx approaches zero". (If THAT's what they mean by
"infinitesimal", then they should simply say so).

> ,-<https://en.wikipedia.org/wiki/Infinitesimal>
> |
> | In mathematics, *infinitesimals* are things so small that there is no way
> | to measure them.

This is ancient terminology. Today this sort of thing is used only as an intuitive
conceptual aid (perhaps).

> | The insight with exploiting infinitesimals was that
> | entities could still retain certain specific properties, such as angle or
> | slope, even though these entities were quantitatively small.[1]
> |
> | […]
> | [1] http://plato.stanford.edu/entries/continuity/#1
>
> By contrast, <http://mathworld.wolfram.com/NonstandardAnalysis.html>.
>
> > "Infinitely small" is not a number:
>
> Yes, it most definitely is.

"Infinitely small" is not a number.

> > for any real x, either x < 0 or x = 0 or x > 0.
>
> Yes.

QED.

> > There is no room in this scheme for "positive but smaller than all
> > positive numbers" other than as a means of torturing undergraduate
> > students with it.
>
> Non sequitur.

Come again? Show me the number X which is positive and smaller than all positive
real numbers:

X = ??? <--- fill this space

--
Jan

[Thomas 'PointedEars' Lahn]

JanPB wrote:

> On Saturday, February 27, 2016 at 1:27:22 AM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> JanPB wrote:
>> > On Friday, February 26, 2016 at 2:05:01 PM UTC-8, Thomas 'PointedEars'
>> > Lahn wrote:
>> >> According even to Einstein himself, ds *is* "a scalar that represents
>> >> the (spacetime) geometry" indeed; that is not at all "a meaningless
>> >> string of words".
>> >
>> > This is because professionals don't need to be pedantic and the
>> > difference between ds^2 as a tensor (which is what it really is) and
>> > its value on the relevant curve's tangent vector (which is a number),
>> > as well as the square root of this number (the "ds") is presumed
>> > understood and obvious and not worth elaborating. Everyone knows what's
>> > what.
>>
>> You have been at least ambiguous about the correctness of the use of
>> “scalar”. You have claimed that using “scalar” instead of “tensor” would
>> not be correct, and that something “represents the [spacetime] geometry”
>> would be nonsense, by saying to this “incorrect” and calling it “a
>> meaningless string of words”. Neither is true.
>
> It is. Maybe we simply differ in our tolerance to using loose terminology.
> The fact is that ds^2 is a tensor, not a number.

^^^^^^^^^^^^
How did you get that idea?

> And I don't approve of statements like "X represents the geometry" - what
> does it mean? If X defines the geometry, that's how it should be phrased.

But “ds” alone does _not_ define the geometry; the equation does that (most
notably the metric tensor, which was hidden behind the signs in the equation
given for Minkowski space(time). We have been over this:

ds² = c²dt² − dx² − dy² − dz²
= c²(dx₀)² − (dx₁)² − (dx₂)² − (dx₃)²
= g_µν dx^µ dx^ν

for the metric (tensor)

(c² 0 0 0)
g_µν = (0 −1 0 0)
(0 0 −1 0)
(0 0 0 −1)

(which – surprise! – *is* also a matrix indeed).

“ds” is the line element of the manifold only; it defines how distances in
that manifold, Minkowski spacetime, are measured. And in that sense, it
represents the (spacetime) geometry. Your claim that it does not is simply,
provably and proven wrong.

>> It would be appropriate if you admitted your mistake instead of making
>> further spurious claims (with weasel words) about what “everyone knows”.
>
> What mistake?

,-<news:ab39b83e-4cf4-4245...@googlegroups.com>
|
| > No, ds^2 is just a scalar that represents the geometry.

|
| Incorrect (also, the phrase "scalar that represents the geometry" is
| a meaningless string of words).

*That* mistake.

>> If one is explaining, one inherently starts from the assumption that
>> _not_ “everyone knows”. Being ambiguous and imprecise is the failure
>> of the person explaining, not of the person something is explained to.
>
> It is me who is trying to be precise.

Then you are doing a pretty bad job at it.

>> “Infinitesimal *change*” (which I said) and “infinitesimal” (which I did
>> not say) are neither “undefined notion” nor “non-standard analysis”.
>
> It is an undefined object in mathematics unless you use non-standard
> analysis or something similar.

Already shown by me to be wrong.

>> ,-<http://mathworld.wolfram.com/Derivative.html>
>> |
>> | The derivative of a function represents an infinitesimal change in the
>> ^^^^^^^^^^^^^^^^^^^^
>> | function with respect to one of its variables. […]
>
> It's just a pop-sci definition. It doesn't define anything.

Wolfram MathWorld, with all its references to mathematics textbooks, is not
“pop-sci”. ISTM you just do not like being shown wrong.

>> | […]
>> | REFERENCES:
>> | […]
>>
>> ,-<https://en.wikipedia.org/wiki/Differential_(infinitesimal)>

>> |
>> | […]
>
> Same thing. There all hand wavings.

If you summarily dismiss arguments without given any reasoning yourself, you
do not need to quote them in full. Your irrationality in this is less
wastefully shown by trimming.

>> | Using calculus, it is possible to relate the infinitely small changes
>> | of various variables to each other mathematically using derivatives. If
>> | y is a function of x, then the differential dy of y is related to dx by
>> | the formula
>> |
>> | dy = dy∕dx dx
>> |
>> | where dy∕dx denotes the derivative of y with respect to x. This formula
>> | summarizes the intuitive idea that the derivative of y with respect to
>> | x is the limit of the ratio of differences Δy/Δx as Δx becomes
>> | infinitesimal.
>
> This defines a limit (which is correct) but it uses incorrect terminology
> again:

Utter nonsense.

> "as Δx becomes infinitesimal." It should read "as Δx approaches zero".

I beg your pardon? That is *precisely* what it means in mathematics (see
the *referenced* definitions that I referenced), so there is no need for a
different wording.

dy∕dx = d∕dx dy := lim ∆y∕∆x
∆x → 0

or

d∕dx f(x₀) := lim (f(x₀ + h) − f(x₀))∕h
h → 0

where

∆y = f(x₀ + h) − f(x₀)
∆x = h

if y = f(x) is differentiable at x = x₀.

> (If THAT's what they mean by "infinitesimal", then they should
> simply say so).

Oh for crying out loud.

As I proved to you, “infinitesimal”, both as an adjective and a noun,
*is* a mathematical term with a well-defined meaning *today*.

Particularly (but as I proved to you, not exclusively), in German *today*
even and *especially* in academia, „Infinitesimalrechnung“ refers to the
techniques of both „Differentialrechnung“ (“calculation with differentials”)
and „Integralrechnung“ (“calculation with integrals”). This has to do
with the fact that the German mathematician Gottfried Wilhelm Leibniz,
independent of the Englishman Isaac Newton, invented the basics of
calculus – in English, the term is “calculus” now, from the historic
“calculus of infinitesimals” (but see below).

>> ,-<https://en.wikipedia.org/wiki/Infinitesimal>
>> |
>> | In mathematics, *infinitesimals* are things so small that there is no
>> | way to measure them.
>
> This is ancient terminology. Today this sort of thing is used only as an
> intuitive conceptual aid (perhaps).

Utter nonsense. It is a fallacy to assume that because the term “calculus
of infinitesimals” is historic, “infinitesimal” would be as well.

>> > "Infinitely small" is not a number:
>> Yes, it most definitely is.
>
> "Infinitely small" is not a number.

Argumentum ad nauseam.

>> > for any real x, either x < 0 or x = 0 or x > 0.
>> Yes.
>
> QED.

You have proven nothing so far except your utter ignorance and stubbornness.

[Emigdio Sophronius]

Thomas 'PointedEars' Lahn wrote:

>> It is an undefined object in mathematics unless you use non-standard
>> analysis or something similar.
>
> Already shown by me to be wrong.

Yes, you are wrong. No need to show. We can just assume.

[Thomas 'PointedEars' Lahn]

The ’nym-shifting troll posted as "Emigdio Sophronius":

As we can see here again, the ’nym-shifting troll can neither read nor
write English properly.

*PLONK*


PointedEars
--
Q: Who's on the case when the electricity goes out?
A: Sherlock Ohms.

(from: WolframAlpha)

[Thomas 'PointedEars' Lahn]

David Waite wrote:

>> (c² 0 0 0)
>> g_µν = (0 −1 0 0)
>> (0 0 −1 0)
>> (0 0 0 −1)
>

> If you're going to be pedantic about Jan's semantics than I will be about
> your definitions. It is utterly ridiculous to define your terms in such a
> way that a tensor has different units carried by different elements.

Utter nonsense.


PointedEars
--
Q: How many theoretical physicists specializing in general relativity
does it take to change a light bulb?
A: Two: one to hold the bulb and one to rotate the universe.
(from: WolframAlpha)

[David Waite]

>On Sunday, February 28, 2016 at 4:13:08 AM UTC-7, Thomas 'PointedEars' Lahn wrote:
> (c² 0 0 0)
> g_µν = (0 −1 0 0)
> (0 0 −1 0)
> (0 0 0 −1)

If you're going to be pedantic about Jan's semantics than I will be about your definitions. It is utterly ridiculous to define your terms in such a way that a tensor has different units carried by different elements when its not necessary to do so. Just define x0 by ct rather than t so that it has all elements with the same units. Granted use of spherical coordinates for example would necessitate different units for different elements, but for the rectilinear case your posing there and therefor as a standard for simplicity you should just use x0=ct so your example should be

(1 0 0 0)

[David Waite]

On Sunday, February 28, 2016 at 9:46:06 AM UTC-7, Thomas 'PointedEars' Lahn wrote:
> Utter nonsense.

Argument from incredulity isn't logical Mr.Spock. You were wrong so you didn't have a valid response.

[Emigdio Sophronius]

Thomas 'PointedEars' Lahn wrote:

>>> (c² 0 0 0)
>>> g_µν = (0 −1 0 0)
>>> (0 0 −1 0)
>>> (0 0 0 −1)
>>
>> If you're going to be pedantic about Jan's semantics than I will be
>> about your definitions. It is utterly ridiculous to define your terms
>> in such a way that a tensor has different units carried by different
>> elements.
>
> Utter nonsense. PointedEars

You don't know what a tensors stands for, especially in Physics. Is not a
data storage, idiot. What am I crazy?? I am talking to an almost half of
an engineer.

[Thomas 'PointedEars' Lahn]

David Waite wrote:

> On Sunday, February 28, 2016 at 4:56:17 AM UTC-7, Thomas 'PointedEars'
> Lahn wrote:
>> On Sunday, February 28, 2016 at 4:13:08 AM UTC-7, Thomas 'PointedEars'

>> Lahn wrote:
>> (c² 0 0 0)
>> g_µν = (0 −1 0 0)
>> (0 0 −1 0)
>> (0 0 0 −1)
>

> If you're going to be pedantic about Jan's semantics than I will be about
> your definitions. It is utterly ridiculous to define your terms in such a

> way that a tensor has different units carried by different elements when

> its not necessary to do so. Just get a brain and define x0 by ct rather

> than t so that it has all elements with the same units.

That is the third time now (I can see that you have posted a fourth time as
I was writing this). How more often are you going to post this (what I
think is nonsense as you are confusing the signature with the metric)?

I think I am following convention here. To recapture, I made the following
substitutions to arrive and the summantion notation (now superscript not
following paranthesis meaning the index in a contravariant tensor, not an
exponent):

ds² = (c)² (dt)² − (dx)² − (dy)² − (dz)²
= (c)² (dx⁰)² − (dx¹)² − (dx²)² − (dx³)²
= g₀₀ dx⁰ dx⁰ − g₁₁ dx¹ dx¹ − g₂₂ dx² dx² − g₃₃ dx³ dx³

How is “g” going to look like if I “define x0 by ct rather than t”?

There is no “t” there; there is “dt”.


PointedEars
--
Q: What did the nuclear physicist post on the laboratory door
when he went camping?
A: 'Gone fission'.
(from: WolframAlpha)

[David Fuller]

10:48 AMDavid Waite

Just admit it, we are inside a black hole
Space is (gravitational time dilation)


http://i65.tinypic.com/5ofksg.jpg



That is why your matrix is filled with zeros and (negative one)

(5.60697995045 / (4 * pi * (10^(-7)))) / (376.730313^2) = 31.4382241648 = 10pi

[Thomas 'PointedEars' Lahn]

David Waite wrote:

>>On Sunday, February 28, 2016 at 4:13:08 AM UTC-7, Thomas 'PointedEars'

> has all elements with the same units.

Not a requirement for a metric (tensor).

> Granted use of spherical coordinates for example would necessitate
> different units for different elements, but for the rectilinear case
> your posing there and therefor as a standard for simplicity you should
> just use x0=ct so your example should be
>
> (1 0 0 0)
> g_µν = (0 −1 0 0)
> (0 0 −1 0)
> (0 0 0 −1)

That is the signature of Minkowski space(time), not its metric of it. If
you think otherwise, please write down the equation for ds² for Minkowski
space for me in Einstein summation convention using the metric you
suggested.


PointedEars
--
Two neutrinos go through a bar ...

(from: WolframAlpha)

[Thomas 'PointedEars' Lahn]

David Waite wrote:

>>On Sunday, February 28, 2016 at 4:13:08 AM UTC-7, Thomas 'PointedEars'

> has all elements with the same units.

Not a requirement for a metric (tensor).

> Granted use of spherical coordinates for example would necessitate
> different units for different elements, but for the rectilinear case
> your posing there and therefor as a standard for simplicity you should
> just use x0=ct so your example should be
>
> (1 0 0 0)
> g_µν = (0 −1 0 0)
> (0 0 −1 0)
> (0 0 0 −1)

That is the signature of Minkowski space(time), not its metric. If you

[David Waite]

On Sunday, February 28, 2016 at 10:19:12 AM UTC-7, Thomas 'PointedEars' Lahn wrote:

> David Waite wrote:
> > Granted use of spherical coordinates for example would necessitate
> > different units for different elements, but for the rectilinear case
> > your posing there and therefor as a standard for simplicity you should
> > just use x0=ct so your example should be
> >
> > (1 0 0 0)
> > g_µν = (0 −1 0 0)
> > (0 0 −1 0)
> > (0 0 0 −1)
>
> That is the signature of Minkowski space(time), not its metric.

No, its the full expression for the Minkowski metric tensor, crank. Since you don’t understand general relativity, you really shouldn’t be making assertions about it. In fact I am convinced by that statement and by your defining the Minkowski metric tensor with different units with for elements, that you don’t know any more about it than koobee, but since he still hasn’t responded to his challenge I’ll prove you don’t with one for you. This is the Thomas 'PointedEars' Lahn challenge just for you.
Here is an exact solution to Einstein's field equations(one of mine):
ds^2={dct^2/[(1-(A*|z|/c^2))^2]}-[(1-(A*|z|/c^2))^(2n)]dx^2-[(1-(A*|z|/c^2))^(2n/(n-1))]dy^2-[(1-(A*|z|/c^2))^((2*(n^2-2n+2))/(n-1))]dz^2
which is vacuum for |z| greater than 0. Find the stress-energy tensor for the sheet at z=0.

[Thomas 'PointedEars' Lahn]

Thomas 'PointedEars' Lahn wrote:

> I think I am following convention here. To recapture, I made the
> following substitutions to arrive and the summantion notation (now
> superscript not following paranthesis meaning the index in a contravariant
> tensor, not an exponent):
>
> ds² = (c)² (dt)² − (dx)² − (dy)² − (dz)²
> = (c)² (dx⁰)² − (dx¹)² − (dx²)² − (dx³)²
> = g₀₀ dx⁰ dx⁰ − g₁₁ dx¹ dx¹ − g₂₂ dx² dx² − g₃₃ dx³ dx³

Sorry, I meant of course:

= g₀₀ dx⁰ dx⁰ + g₁₁ dx¹ dx¹ + g₂₂ dx² dx² + g₃₃ dx³ dx³

> How is “g” going to look like if I “define x0 by ct rather than t”?
>
> There is no “t” there; there is “dt”.


PointedEars
--

Q: Where are offenders sentenced for light crimes?
A: To a prism.

(from: WolframAlpha)

[Thomas 'PointedEars' Lahn]

David Waite wrote:

> On Sunday, February 28, 2016 at 10:19:12 AM UTC-7, Thomas 'PointedEars'
> Lahn wrote:
>> David Waite wrote:
>> > Granted use of spherical coordinates for example would necessitate
>> > different units for different elements, but for the rectilinear case
>> > your posing there and therefor as a standard for simplicity you should
>> > just use x0=ct so your example should be
>> >
>> > (1 0 0 0)
>> > g_µν = (0 −1 0 0)
>> > (0 0 −1 0)
>> > (0 0 0 −1)
>>
>> That is the signature of Minkowski space(time), not its metric.
>
> No, its the full expression for the Minkowski metric tensor, crank.

No, I do not think so.

> Since you don’t understand general relativity, you really shouldn’t be
> making assertions about it.

I am not discussing general relativity. What I posted is (supposed to be)
the metric for Minkowski space(time), so I am obviously discussing special
relativity.

I wonder who’s the crank here.

> In fact I am convinced by that statement and by your
> defining the Minkowski metric tensor with different units

AISB, it is not required for all components of a metric to have the same
unit.

However, in natural units where c = 1, my metric is identical to your metric
as then c² = 1. So I am at a loss to understand the fuss you are making
about this.


PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

[David Waite]

I already told you, defining x0=ct it becomes
> > (1 0 0 0)

[David Waite]

On Sunday, February 28, 2016 at 10:34:03 AM UTC-7, Thomas 'PointedEars' Lahn wrote:
> David Waite wrote:
>
> > On Sunday, February 28, 2016 at 10:19:12 AM UTC-7, Thomas 'PointedEars'
> > Lahn wrote:
> >> David Waite wrote:
> >> > Granted use of spherical coordinates for example would necessitate
> >> > different units for different elements, but for the rectilinear case
> >> > your posing there and therefor as a standard for simplicity you should
> >> > just use x0=ct so your example should be
> >> >
> >> > (1 0 0 0)
> >> > g_µν = (0 −1 0 0)
> >> > (0 0 −1 0)
> >> > (0 0 0 −1)
> >>
> >> That is the signature of Minkowski space(time), not its metric.
> >
> > No, its the full expression for the Minkowski metric tensor, crank.
>
> No, I do not think so.

Yeah, but you're a crank that apparently doesn't know what a rank2 covariant metric tensor is, but calls the Minkowski metric tensor a signature. So it doesn't matter what you think. Anyway the signature is (+,-,-,-).

[Thomas 'PointedEars' Lahn]

David Waite wrote:

> On Sunday, February 28, 2016 at 10:26:31 AM UTC-7, Thomas 'PointedEars'
> Lahn wrote:
>> Thomas 'PointedEars' Lahn wrote:
>> > I think I am following convention here. To recapture, I made the
>> > following substitutions to arrive and the summantion notation (now
>> > superscript not following paranthesis meaning the index in a
>> > contravariant tensor, not an exponent):
>> >
>> > ds² = (c)² (dt)² − (dx)² − (dy)² − (dz)²
>> > = (c)² (dx⁰)² − (dx¹)² − (dx²)² − (dx³)²
>> > = g₀₀ dx⁰ dx⁰ − g₁₁ dx¹ dx¹ − g₂₂ dx² dx² − g₃₃ dx³ dx³
>>
>> Sorry, I meant of course:
>>
>> = g₀₀ dx⁰ dx⁰ + g₁₁ dx¹ dx¹ + g₂₂ dx² dx² + g₃₃ dx³ dx³
>>
>> > How is “g” going to look like if I “define x0 by ct rather than t”?
>> >
>> > There is no “t” there; there is “dt”.

>> […]

>
> I already told you, defining x0=ct it becomes
>> > (1 0 0 0)
>> > g_µν = (0 −1 0 0)
>> > (0 0 −1 0)
>> > (0 0 0 −1)

Unsursprisingly, you have not answered the question that I asked.

[David Waite]

On Sunday, February 28, 2016 at 10:50:29 AM UTC-7, Thomas 'PointedEars' Lahn wrote:
> David Waite wrote:
>
> > On Sunday, February 28, 2016 at 10:26:31 AM UTC-7, Thomas 'PointedEars'
> > Lahn wrote:
> >> Thomas 'PointedEars' Lahn wrote:
> >> > I think I am following convention here. To recapture, I made the
> >> > following substitutions to arrive and the summantion notation (now
> >> > superscript not following paranthesis meaning the index in a
> >> > contravariant tensor, not an exponent):
> >> >
> >> > ds² = (c)² (dt)² − (dx)² − (dy)² − (dz)²
> >> > = (c)² (dx⁰)² − (dx¹)² − (dx²)² − (dx³)²
> >> > = g₀₀ dx⁰ dx⁰ − g₁₁ dx¹ dx¹ − g₂₂ dx² dx² − g₃₃ dx³ dx³
> >>
> >> Sorry, I meant of course:
> >>
> >> = g₀₀ dx⁰ dx⁰ + g₁₁ dx¹ dx¹ + g₂₂ dx² dx² + g₃₃ dx³ dx³
> >>
> >> > How is “g” going to look like if I “define x0 by ct rather than t”?
> >> >
> >> > There is no “t” there; there is “dt”.
> >> […]
> >
> > I already told you, defining x0=ct it becomes
> >> > (1 0 0 0)
> >> > g_µν = (0 −1 0 0)
> >> > (0 0 −1 0)
> >> > (0 0 0 −1)
>
> Unsursprisingly, you have not answered the question that I asked.

I did, twice, fleebrain, but you didn't answer mine. This is the Thomas 'PointedEars' Lahn challenge just for you.

[David Waite]

On Sunday, February 28, 2016 at 10:13:27 AM UTC-7, David Fuller wrote:
> Space is (gravitational time dilation)

You might as well say the color of the sky is flavorful, lunatic.

[Thomas 'PointedEars' Lahn]

David Waite wrote:

> On Sunday, February 28, 2016 at 10:34:03 AM UTC-7, Thomas 'PointedEars'
> Lahn wrote:
>> David Waite wrote:
>> > On Sunday, February 28, 2016 at 10:19:12 AM UTC-7, Thomas 'PointedEars'
>> > Lahn wrote:
>> >> David Waite wrote:
>> >> > Granted use of spherical coordinates for example would necessitate
>> >> > different units for different elements, but for the rectilinear case
>> >> > your posing there and therefor as a standard for simplicity you
>> >> > should just use x0=ct so your example should be
>> >> >
>> >> > (1 0 0 0)
>> >> > g_µν = (0 −1 0 0)
>> >> > (0 0 −1 0)
>> >> > (0 0 0 −1)
>> >>
>> >> That is the signature of Minkowski space(time), not its metric.
>> >
>> > No, its the full expression for the Minkowski metric tensor, crank.
>> No, I do not think so.
>
> Yeah, but you're a crank that apparently doesn't know what a rank2
> covariant metric tensor is,

How did you get that idea?

> but calls the Minkowski metric tensor a signature.

What you have posted is _not_ “the full expression for the Minkowski metric
tensor”, period.

> Anyway the signature is (+,-,-,-).

Or (−,+,+,+), which is another way of saying that

η_µν = diag(+1,−1,−1,−1)

or

η_µν = diag(−1,+1,+1,+1).

Guess what diag(…) expands to…

<http://mathworld.wolfram.com/MetricTensor.html> (once again)

[David Waite]

Being childish isn't a logical response Mr.Spock. Its the rank 2 covariant Minkowski metric tensor, and no matter how much you act childish and scream nah ah and stamp your feet, you're not going to change that fact. You were wrong. Get over it.

[David Waite]

On Sunday, February 28, 2016 at 10:50:29 AM UTC-7, Thomas 'PointedEars' Lahn wrote:
> David Waite wrote:
>
> > On Sunday, February 28, 2016 at 10:26:31 AM UTC-7, Thomas 'PointedEars'
> > Lahn wrote:
> >> Thomas 'PointedEars' Lahn wrote:
> >> > I think I am following convention here.

You're following crapwiki actually. There actually isn't one single set of conventions for general relativity's tensor calculus because I can count on my fingers and toes the number of people who have understood general relativity since Einstein. I'm just telling you what you should be doing.

> >> > There is no “t” there; there is “dt”.

So it just occurs to me that not only am I talking to a crank that makes assertions about tensors he knows nothing about but that he doesn't even have calculus as a prerequisite. Ok, so dt means an infinitesimal displacement in t. As such,
c*dt=dct
c²*dt²=(dct)² or for those who aren't to stupid to get it, dct².
So you define for the Minkowski spacetime
x⁰=ct
x¹=x
x²=y
x³=z
and the Minkowski rank2 covariant metric tensor expressed as a matrix IS THEN,

>> > (1 0 0 0)
>> > g_µν = (0 −1 0 0)
>> > (0 0 −1 0)
>> > (0 0 0 −1)

Nice and elegant, with all the same units for all the elements since so long as you're using rectilinear coordinates, it isn't necessary to do otherwise and its signature is (+,-,-,-).

[David Waite]

On Sunday, February 28, 2016 at 11:12:32 AM UTC-7, Thomas 'PointedEars' Lahn wrote:
>..

Just to spell it out as pedantically as you were being to Jan, the line element ds then is given by
ds²=dct²-dx²-dy²-dz²
or
ds²=(dx⁰)²-(dx¹)²-(dx²)²-(dx³)²
and often in a synonymous way this is referred to as just the metric because all of the elements of the rank 2 covariant expression for the metric tensor can be directly read off of this expression for the line element as its elements are the coefficients of its differential squared terms.

[Thomas 'PointedEars' Lahn]

David Waite wrote:

> On Sunday, February 28, 2016 at 10:50:29 AM UTC-7, Thomas 'PointedEars'
> Lahn wrote:
>> Unsursprisingly, you have not answered the question that I asked.
>
> I did, twice,

I think we can agree on that you have answered the question that I asked in
this subthread *in a way* twice as you have posted *a possible, correct*
metric for Minkowski spacetime.

I had previously been unaware that

c dt = d(ct)

holds for constants c and differentials dt [1], so that

ds² = c² dt² − dx² − dy² − dz²
= (c dt)² − dx² − dy² − dz²
= (d(ct))² − dx² − dy² − dz²
= (dx⁰)² − (dx¹)² − (dx²)² − (dx³)² (x⁰ := ct)
= g₀₀ (dx⁰)² + g₁₁ (dx¹)² + g₂₂ (dx²)² + g₃₃ (dx³)²

with

(1 0 0 0)
g_µν = (0 −1 0 0) = diag(+1,−1,−1,−1)

(0 0 −1 0)
(0 0 0 −1)

which is probably what you mean. It then happens that the metric (tensor) g
is identical to the signature of the space when written as diagonal
matrix/tensor η, whereas both are generally related by the identity

g_αβ ≡ ∂ξ^α∕∂x^µ ∂ξ^β∕∂x^ν η_αβ

(see also: <http://mathworld.wolfram.com/MetricTensor.html>).

However, you have _not_ answered my *other* question regarding the form of
ds² in Einstein summation convention then, which is why I had to figure that
out for myself:

ds² = g_µν dx^µ dx^ν = c² dt² − dx² − dy² − dz²

still holds with your metric g and definition of x⁰, as can be seen above.

Had you been more forthcoming and less name-calling, we could have resolved
this earlier.

I maintain, however, that the metric that I posted also is standard notation
and not in any way wrong, and I have to reiterate that with natural units
where c = 1 (which is common in relativistic physics), our metrics are
identical.

See also: <http://mathworld.wolfram.com/MinkowskiMetric.html>

> fleebrain, but you didn't answer mine. […]

And do you think that your insults will effect that I consider your red
herring, or reading further postings from you at all?

So much for “logical responses”.


PointedEars
___________
[1] <https://en.wikipedia.org/wiki/Differential_of_a_function#Properties>
<https://de.wikipedia.org/wiki/Differential_(Mathematik)#Rechenregeln>

[Emigdio Sophronius]

Thomas 'PointedEars' Lahn wrote:

> I had previously been unaware that
>
> c dt = d(ct)

What an idiot. lol. You are strong, man, in tensors.

[David Waite]

On Sunday, February 28, 2016 at 12:40:38 PM UTC-7, Thomas 'PointedEars' Lahn wrote:
> However, you have _not_ answered my *other* question regarding the form of
> ds² in Einstein summation convention then, which is why I had to figure that
> out for myself:
>
> ds² = g_µν dx^µ dx^ν = c² dt² − dx² − dy² − dz²

You didn't have questions you had false assertions which you ignored the answer to. As I said plainly stated,

x⁰=ct
x¹=x
x²=y
x³=z

ds²=dct²-dx²-dy²-dz²

then you get

ds²=(dx⁰)²-(dx¹)²-(dx²)²-(dx³)²

>> > (1 0 0 0)
>> > g_µν = (0 −1 0 0)

>> > (0 0 −1 0)
>> > (0 0 0 −1)

and its signature is (+,-,-,-).

And YOU still haven't answered my question and until you do you should stop being a crank making assertions about things you obviously don't know anything about. Here is an exact solution to Einstein's field equations(one of mine)

ds^2={dct^2/[(1-(A*|z|/c^2))^2]}-[(1-(A*|z|/c^2))^(2n)]dx^2-[(1-(A*|z|/c^2))^(2n/(n-1))]dy^2-[(1-(A*|z|/c^2))^((2*(n^2-2n+2))/(n-1))]dz^2
which is vacuum for |z| greater than 0. Find the stress-energy tensor for the sheet at z=0.

Tell me when you're going to admit you don't know crap about the subject matter and so can't answer it so I can put the answer back up at my site as I temporarily removed it so you couldn't get it from me.

[Emigdio Sophronius]

David Waite wrote:

> On Sunday, February 28, 2016 at 12:44:24 PM UTC-7, Emigdio Sophronius

> Ah so you need to go back to elementary school and learn ppmdas.
> dct=d(ct), moron.

I called him an idiot, but since you insist, David Waste :) Is c not a
constant, idiot!