Re: charged black hole metric

Your quotes are messed up this time. I have no time to fix those so we
have to go by context (which should be pretty obvious :-) )

Androcles wrote:
> "JanPB" <filmart@xxxxxxxxx> wrote in message
> news:1124348843.533769.7280@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Androcles wrote:

> > What's the speed of light at x' as it reverses direction in
> > reflection?
>
> It's undefined because the reversal is presumed instantaneous.
>
> Very good.
> In SR the velocity of light is undefined.

No. What's undefined is the speed at the instant of reflection. You
have written it above yourself.

> Since
> it's instantaneous it contributes nothing to the elapsed times, hence
> it doesn't enter into the "tau" equation.
>
>
> Very good.
> So we can't use c in 1/sqrt(1-v^2/c^2) because c is undefined.

Nonsense. The speed of light is undefined only at the instant of
reflection. Since this instant takes no time, this little factoid is of
no consequence for the "tau" equation. Everywhere else the speed of
light works just fine.

> > Therefore, both instances of x' are present and nonzero on the right
> > hand side (and on the left as well, obviously).
>
>
> Huh?
> I only see one coordinate and one length on the RHS.

That's what I meant - "both instances".

> Tell you what.
> Let's take another ruler, turn it 180 degrees and lay it beside the
> first,
> so that 0 is beside 12 inches (or 30 cm)
> Does that change the physics?
> It shouldn't, but it does change the coordinates.
> Knock yourself out with my ruler instead of yours.

And I derived exactly the same equation this way.

(Skipping all the Sagnac irrelevancies here.)

> > | The light is sent to the mirror and then it goes back.
> >
> > Right. Total distance 0, total time t.
>
> Light going back and forth is total distance zero? Hellooo?? When I fly
> from LA to Berlin and back - have I just travelled zero distance?
>
>
>
> Coordinate LA to coordinate LA.
> Yep... sure looks like zero distance to me. Hello??? Anyone home?

But to calculate elapsed time the light ray takes to go from the
k-origin to the mirror and then back you do NOT calculate just the
k-origin (or Los Angeles). For the outgoing trip you use k-origin and
the mirror, for the incoming one you use the mirror and the origin. Two
speeds, c+v and c-v.

> > | He simply differentiates with respect to x' at 0. If it bothers you,
> > | the same result exactly can be obtained by pure algebra.
> >
> > You've missed the point.
>
> The point is to calculate what tau is.
>
>
> We already know t1 = x'/(c-v) and t2 = x'/(c+v), why do we need them
> to be equal?
>
> Ahhhh, the whim of Einstein.

Man, you are confused and arrogant!

> we establish by definition that the "time" required by light to travel
> from A to B equals the "time" it requires to travel from B to A.
> [end quote]
> Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
>
> That's a phuckwit definition if ever I saw one.

It doesn't matter what you think. You have already proved to everyone
this much. A person who cannot do simplest mathematics or physics
simply cannot make this sort of statements - common courtesy should
indicate this much to you.

> > | The differentiation is completely inessential here and is used only
> > to
> > | save the trees and the reader's time.
> >
> > Of course, but you have the wrong reason.
> > This is done as a hoax to seem as if he knows what he's doing.
>
> Nonsense. Patent nonsense. First of all, it is way TOO EASY to have
> ever been even tenable as a hoax. Anybody reading it in 1905 would've
> known immediately a mistake was made - assuming one was made at this
> level. (In which case the paper would have never made it through the
> Annalen der Physik peer review process in the first place.)
>
> I can't change factual history. You are spouting nonsense now.

My point was that it WAS NOT any "factual history" - it couldn't have
been. Just like Johann Sebastian Bach could not mistake a trumpet for a
harpsichord. Yes, a mistake of the type you suggest would be THAT
obvious to absolutely everyone in academia in 1905. Zero chance of this
sort of thing squeezing past the publisher.

You have an absurdly inflated notion of what constitutes difficult
mathematics.

> > And he does know. He is not a half-wit.
> > Too late, you've already missed it.
> > Back up aways.
> > What's that half?
> > "we establish by definition that the "time" required by light to
> > travel
> > from A to B equals the "time" it requires to travel from B to A."
> >
> > Two sections earlier, that is.
> > It's a DEFINITION.
> >
> > He already knows the times, x'/(c+v) and x'(c-v)
> > Now he's fucking wi th them, and your head as well.
>
> It's amazing to me how you simply presume - just like that! - that if
> you don't understand something, it means that it's wrong. Incredible.
>
>
> That what you are doing. You don't understand mathematics,
> you want to blame me. A typical phuckwit knee-jerk reaction, that.

Words again. Talk is cheap. I can say that I'm the Queen of England.
See? It works! What counts is your standing to criticize SR. I claim
you have none. Proof: you can't answer simple mathematics and physics
questions. Hence you are not qualified to debate this stuff.

> EXPLANATION: the "DEFINITION" you quote presumes the time required to
> travel from A to B and from B to A is measured by clocks that are
> *stationary* with respect to the observer.
>
> Sure. x'/(c-v) and x'/(c+v) have to be made equal.

No, they do not have to be made equal. In fact they are not equal. The
elapsed times between emissions and reflections are OTOH equal when
measured by the k-system. Different measurements done by different sets
of instruments, no contradiction.

> Nothing to do with physics, just a dumb mathematical GAME of "Let's
> pretend".

Just a dumb game called "let's pretend I know some physics" and "let's
post insulting crap to the Net" (insulting to Einstein - I am pretty
much made of Teflon).

> The clocks used in measuring x', t, x'/(c-v), etc., are *NOT*
> stationary in k.
>
> Who said they were?

Non sequitur. You said that a pair of unequal numbers must be made
equal. I pointed out this was wrong, explaining that different sets of
clocks are being used to get these pairs of numbers - that's why they
are different.

> Had you BOTHERED to read my proof, you would have seen
> Setting the time a = x'/(c-v) and b =x'/(c+v) for clarity
>
> tau(a+b) = 2*tau(a)

Yes, but this proof is wrong because it's based on arbitrarily
resetting one of the x' in the "tau" equation to zero. This naturally
immediately results in nonsense, just like arbitrary resetting "4" to
"5" would result in "2+2=5".

> > He's got you eating out of his hand, buddy. And he's dead.
>
> Words, words, words. And when I ask you simplest math or physics
> questions you cannot do them.
>
> Phuckwit.

Well - you cannot do this stuff. So why should your criticism carry any
weight?

--
Jan Bielawski

.

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