Re: Are There Unresolver Foundational Issues With GR

On Apr 8, 11:45 am, "JanPB" <film...@xxxxxxxxx> wrote:
On Apr 6, 5:53 pm, "Koobee Wublee" <koobee.wub...@xxxxxxxxx> wrote:

I have been gone for over a week. Coming back, I am very surprised
that you would pile on your BS and present it as a gourmet dish.

ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dO^2 [2]

If you are describing the same invariant spacetime, ds^2, then your
dr, dt must be different due to your coordinate transformation.

Yes, I have recycled the letters "r" and "t" from the first formula
(the one with coefficients f, g, h, k) into the formulas [1] and [2].
BTW, did you notice the nonzero cross term "dr dt"? _This_ is the
source of the computational complexity, not k(r,t) which is trivially
replaced by r^2 (here I'm recycling "r" as usual).

What happened to the cross terms (dO dt)? You have assumed t0o much
to arrive at a solution that is thoroughly dependent on the assumption
you have made, and yet you don't realize it. <shrug>

ds^2 = A^2 dt'^2 + B^2 dr'^2 + C^2 dO'^2 [3]

I don't agree with (r^2 dO^2).

It's simply switching from r to r', where:

r' = sqrt(k(r,t))

where A(r,t) and B(r,t) are some functions which are never zero (or
else the metric would not be rank-4).

Rank-4?

Brush up on your linear algebra. Here it's synonymous with
"nondegenerate".

NB: this local coordinate change happens to rely on dividing by
certain function, so any solutions for metric coefficients we may
obtain by proceeding from these two general forms may not be defined
over the spacetime events for which the function we divided by is
zero.

This transformation is not necessary a division transformation. Why
do you think it is so?

It's not _just_ a division - I said it only relied on it. The entire
procedure involves cooking up an integrating factor. Here is a quickie
summary. We have:

ds^2 = f(r,t) dt^2 + g(r,t) dr^2 + h(r,t) dt dr + k(r,t) dO^2 (*)

...and we want to get rid of the cross-term involving dt dr. We
consider the 1-form:

f dt + (h/2) dr

More assumptions. You cannot make these assumptions and claiming that
you have not made any assumptions to derive at the Birkhoff's
theorem. That is why in Birkhoff'theorem it lays out the following
assumptions before the proof begins.

** If static, independent of t
** If spherically symmetric, independent of O

[...]


You have reduced the spacetime from 4 variables to 3, remember?

I haven't reduced anything in this sense. There are still 4 variables
t,r,theta,phi. The Ricci rt-component just happens to involve dB/dt,
so it's obviously identically zero if you presume B is a function of r
only. But R_rt is not identically zero in general. Just compute (or
read that PDF doc) and see for yourself.

...with the remaining equations being either identities or equivalent
to those four ones. (For example, R_phi phi = R_theta theta.)

The exact forms of the four left-hand sides are, few pages of
calculations later, respectively (excuse the mess, it's the fourth,
simplest, equation that's interesting to us):

(1/AB)(-d/dt((1/A)*dB/dt) + d/dr((1/B)*dA/dr) + (2/rB)*dA/dr) = 0
(1/AB)( d/dt((1/A)*dB/dt) - d/dr((1/B)*dA/dr)) - (2/rB)*d/dr(1/B) = 0
-+(1/rAB^2)*dA/dr -+ (1/rB)*d/dr(1/B) + (1/r^2)*(1 -+ 1/B^2) = 0
(2/rAB^2)*dB/dt = 0

There is no way you can get ((2/rAB^2)*dB/dt=0) from (3).

That's right. You don't get the fourth equation out of the remaining
three. It's just a fourth independent equation (there are 10 of them
in general, 6 of which are either equivalent to the 4 above or are
identities 0=0).

Actually, with time derivative, these equations are a lot more
complicated than you have shown above.

That's because you use Christoffel symbols for your calculations. I
use Cartan's connection forms which are based on orthonormal moving
frames instead of coordinate frames. This technology is well-worth
learning as it offers significant simplification of the calculations.
At the "bottom" both Christoffel and Cartan-based quantities are the
same of course, but Cartan's approach makes clever use of certain
hidden SKEW-symmetries (rather than Christoffel symmetries in the
lower indices) which means you have less quantities to deal with: for
example, there are only 6 independent entries in a 4x4 skew-symmetric
matrix vs. 10 entries in a 4x4 symmetric matrix. Etc. MTW has a
chapter on Cartan's calculus and books like Wald use this technique
extensively.

So, if you use some other connection coefficients that are so abstract
in nature compared with Christoffel's which are more tangible you end
up with something exponentially simpler. Yeah, right. Give me a
break. There is nothing as realistic as the Christoffel symbols.
<shrug>

[...]

Conveniently pulling a rabbit out of the hat? At least tell us what
mathematical errors lead to B=constant/A?

We will leave that in a future discussion. I am very tired tonight.

Also, you have conveniently assumed (C^2 = r^2). See (3).

This is a trivial coordinate change.

This is not coordinate change.

I'm beginning to suspect where
you (and Crothers) get this idea of metric being coordinate-dependent.

I don't recall Mr. Crothers is saying the metric is coordinate
dependent. As far as I know, I am the only one so far.

It's the problem of physical units, isn't it?

No.

You both think that
changing units of measurement is the same as changing coordinates,
right?

Not I.

So when I change my unit of length - so you both think - the
numerical values of, say, vectors lengths change also, hence the
values of dot products change, and these values ARE the metric, so the
metric changes too! Am I getting close?

No. That is the problem here. There is no coordinate change. We are
still talking about using the same spherically symmetric coordinate
system.

That is how the Birkhoff's theorem as the author of Wikipedia defined
it.

Assuming (C^2 = r^2) which translates to spherically symmetric. For
every C^2, there is a new set of metric that satisfies the field
equations. Have you finally understood Mr. Crothers?

C^2 makes no difference. Mr. Crothers is a confused soul.

Perhaps, he is. However, you are even more confused than Mr. Crothers
is.

Assuming (static) which translates to all elements of the metric
independent of time.

This whole exchange started with you knocking Birkhoff's theorem and
how it implied the metric had to be static. So don't assume now that
the metric is static.

I still maintain the assumptions of static and spherically
symmetricity must be established before going further with Birkhoff's
theorem. After making these assumptions, the theorem appears to be
correct with the Schwarzschild metric, or Schwarzschild's original
metric, or even the metric Mr. Crothers is championing over the
Schwarzschild metric. However, there are just other metrics that
would invalidate Birkhoff's theorem. I have given some of these
metrics that do so in the past.

Then, you end up with much, much, much, and much simpler equations
described below.

** (r / B^2) (dB/dr) - (1 / B) + 1 = 0
** (r / A / B) (dA/dr) + (1 / B) - 1 = 0

Well, wonderful, but again - we are discussing how Birkhoff says
spherical symmetry in Schwarzschild coordinates implies independence
of t (which you mistook for static). We cannot assume that which we
want to prove!

These are actually not Ricci tensor equations but rather Einstein
tensor equations. With all these simplifications, they are all the
same.

For the spacetime described below,

ds^2 = A dt^2 - B dr^2 - C dO^2

Where (C = r^2)

You will see that you can easily solve for B to be the following.

** B = 1 / (1 + K / r)

Where (K = integration constant)

And A follows.

** A = K' (1 + K / r)

Where (K' = another integration constant)

For (C = (r + K)^2),

You get

** A = K' / (1 + K / r)
** B = 1 + K / r

In this case, you don't get a black hole.

You do - you simply neglected to use the correct domain (you changed
coordinates but kept the domain the same - this is like saying "0
Celsius equals 0 Kelvin, therefore absolute zero doesn't exist").
You've changed the domain of r from r>0 to r>-K.

For (C = ((r^3 + K^3)^(1/3) - K)^2),

You end up with Schwarzschild's original solution which does not
manifest any black holes either.

There are infinite numbers of solutions more.

You've just altered the domain to exclude the black hole. This is not
a different solution, it's just a subset of the good old Schwarzschild
with a restriction of the radial coordinate.

I have not alter any domains. I just presented more independent
solutions where each one is just as valid as the Schwarzschild metric
in which you have worshipped like a God. In the meantime, the
coordinate of choice is still the familiar spherically symmetric polar
coordinate system for each of the solutions I have presented so far
--- no domain change nor change of coordinate system, just independent
solutons to the field equations.

.

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