GR THEORY IS NOT EVEN FALSE!
mlut...@wanadoo.fr
Excerpts from:
GTR Tests - The Pound-Rebka-Snider Experiment
Dr. Adrian Sfarti
"1. Abstract
Einstein predicted a change in the energy of photons
in the proximity of a gravitational field, the change
being directly proportional with the distance from
the gravitational source. In the early 60's Pound and
Rebka have set to verify Einstein's prediction."
"The experiment was set up in the Harvard tower.
The Harvard tower is just 22.6 meters, so the fractional
gravitational red shift between the light frequency f
at bottom of the tower and the frequency f0 at the top
predicted by GRT given by the formula
(f-f0)/f0 = gd/c^2 = 2.45*10^-15 (1)
where g =~ 9.81 m/s^2, d is the tower
height and c is the speed of light in void.
In reality Pound and Rebka measured the energy
difference DeltaE = h(f-f0) (2)
h being the Plank constant.
From (1) and (2) we obtain
DeltaE/E = (f-f0)/f0 = gd/c^2
Comparing the energy shifts on the upward and downward
paths gives a predicted difference
(DeltaE/E)down - (DeltaE/E)up = 2 gd/c^2 = 4.9*10^-15
The cleverness of the experiment lies in the fact
that it sidesteps the measurement of the frequencies
at the top and of the bottom of the tower and it
replaces such measurement with a very high precision
energy measurement and here is where the Mossbauer effect
comes into play. The measured difference was 5.1x10^-15."
GR THEORY IS FALSE!
___________________
According to the approximate formula (f-f0)/f0 = gd/c^2
above, photons gain energy during their downward trip.
The corresponding blue shift is (DeltaE/E)down = gd/c^2.
Photons moving upward *must* lose an *identical*
quantity of energy, hence the corresponding red shift =
(DeltaE/E)up = -gd/c^2.
If some formula gave a blue shift whose absolute value
were different from the absolute value of the red shift,
it would be physically false.
Let A = sqrt(1-2GM/(R*c^2))
Let B = sqrt(1-2GM/((R+d)*c^2))),
where G is the gravitational constant, M the mass
of the Earth, R the radius of the Earth and d the
height of the tower.
According to the full GR formula,
f/f0 = A/B, to which corresponds a blue shift
f/f0 - 1 = A/B - 1 = (A-B) / B
As f0/f = B/A, the corresponding red shift =
f0/f - 1 = B/A - 1 = -(A-B) / A
The fact that (A-B)/B is different from (A-B)/A
implies that the energy gain of photons moving
downward is different from the energy loss of
photons moving upward, which is physically wrong.
So stupidely wrong that one could say that
GR THEORY IS NOT EVEN FALSE!
Marcel Luttgens
Dono
>
> f/f0 = A/B, to which corresponds a blue shift
> f/f0 - 1 = A/B - 1 = (A-B) / B
>
> As f0/f = B/A, the corresponding red shift =
> f0/f - 1 = B/A - 1 = -(A-B) / A
>
> The fact that (A-B)/B is different from (A-B)/A
Marcel,
You are such a patented imbecile.
You would expect that |f-f0|=|f0-f|
i.e. the delta of energies moving up and moving down to be the same.
Why on Erath would you expect that the delta of energies normalized to
DIFFERENT energies (E and E0 respectively) to be the same. I.E., why
would an imbecile like you expect that :
|f/f0-1|=|f0/f-1|?
I.e., why an imbecile like you would expect that:
|(f-f0)/f0|=|(f-f0)/f|
I.e. , why would an persistent denier like you expect that:
|(E-E0)/E0|=|(E-E0)/E|
Don't you see that the SAME delta_Energy (E-E0) is normalized to
different energies (E0 and E, respectively)?
Dirk Van de moortel
040c9fbc-0b91-49b9...@a1g2000hsb.googlegroups.com
Only if you conveniently "forget" *again* that
(f-f0)/f0 = gd/c^2
is a first order Taylor approximation for small d.
>
> Let A = sqrt(1-2GM/(R*c^2))
> Let B = sqrt(1-2GM/((R+d)*c^2))),
>
> where G is the gravitational constant, M the mass
> of the Earth, R the radius of the Earth and d the
> height of the tower.
>
> According to the full GR formula,
>
> f/f0 = A/B, to which corresponds a blue shift
> f/f0 - 1 = A/B - 1 = (A-B) / B
>
> As f0/f = B/A, the corresponding red shift =
> f0/f - 1 = B/A - 1 = -(A-B) / A
>
> The fact that (A-B)/B is different from (A-B)/A
> implies that the energy gain of photons moving
> downward is different from the energy loss of
> photons moving upward, which is physically wrong.
Try using
A' = sqrt(1-2GM/((R-d)*c^2))
B' = sqrt(1-2GM/(R*c^2))
and get the first order Taylor.
And then, to continue the exercise, try the first order Taylor on
A = sqrt(1-2GM/(R*c^2))
B = sqrt(1-2GM/((R+d)*c^2)),
>
> So stupidely wrong that one could say that
> GR THEORY IS NOT EVEN FALSE!
>
> Marcel Luttgens
No no... *something* is very stupid allright.
Dirk Vdm
Dono
SperM.hotmail.com> wrote:
> mluttg...@wanadoo.fr <mluttg...@wanadoo.fr> wrote in message
>
> 040c9fbc-0b91-49b9-bb9e-b9c7be97a...@a1g2000hsb.googlegroups.com
You missed his biggest error :-)
Dirk Van de moortel
7fb3f4da-08d2-411e...@e10g2000prf.googlegroups.com
[snip]
> You missed his biggest error :-)
Don't think so.
Check the definition of gravitational redshift as calculated
with A and B.
Dirk Vdm
Dono
>
> > On Mar 17, 7:33 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> > SperM.hotmail.com> wrote:
>
> [snip]
>
> > You missed his biggest error :-)
>
> Don't think so.
> Check the definition of gravitational redshift as calculated
> with A and B.
>
> Dirk Vdm
That is minor compared to this:
http://groups.google.com/group/sci.physics.relativity/msg/43d4a0faf3395929?dmode=print
:-)
Dirk Van de moortel
Marcel's reasoning with A/B-1 and B/A-1 are as good as correct.
See sample problem 3 of
http://www.eftaylor.com/pub/chapter2.pdf
Proof that the first order approximations of |A/B-1| and |B/A-1| are
equal:
A = sqrt( 1 - 2 G M / (R c^2) )
= sqrt( 1 - 2 g R/c^2 )
~ 1 - g R/c^2
B = sqrt( 1 - 2 G M / ((R+d) c^2) )
= sqrt( 1 - 2 g R / ((1+d/R) c^2) )
~ 1 - g R/c^2 / (1+d/R)
~ 1 - g R/c^2 + g/c^2 d
giving
A/B - 1 ~ - g/c^2 /(1-gR/c^2) d + order(d^2)
and
B/A - 1 ~ + g/c^2 /(1-gR/c^2) d + order(d^2)
On the other hand:
A' = sqrt( 1 - 2 G M / ((R-d) c^2) )
= sqrt( 1 - 2 g R / ((1-d/R) c^2) )
~ 1 - g R/c^2 / (1-d/R)
~ 1 - g R/c^2 - g/c^2 d
B' = sqrt( 1 - 2 G M / (R c^2) )
= sqrt( 1 - 2 g R/c^2 )
~ 1 - g R/c^2
giving
A'/B' - 1 ~ - g/c^2 /(1-gR/c^2) d + order(d^2)
and
B'/A' - 1 ~ + g/c^2 /(1-gR/c^2) d + order(d^2)
So we have 4 identical first order approximations which can,
if we further ignore the factor 1/(1-gR/c^2), write as
|A/B-1| = |B/A-1| = |A'/B'-1| = |B'/A'-1| ~ g/c^2 d
Dirk Vdm
Dono
>
> > That is minor compared to this:
>
>
> > :-)
>
> Marcel's reasoning with A/B-1 and B/A-1 are as good as correct.
wrote.
Marcel's contention is that if he uses symbolic computations (NOT
approximations) he doesn't get |A/B-1|=|B/A-1|
And the reason is very simple, since A=E and B=E_0
Dirk Van de moortel
> On Mar 17, 8:45 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
>>
>>> That is minor compared to this:
>>
>>> http://groups.google.com/group/sci.physics.relativity/msg/43d4a0faf33...
>>
>>> :-)
>>
>> Marcel's reasoning with A/B-1 and B/A-1 are as good as correct.
>
> Very nice and correct but not relevant
Highly relevant.
>, you need to re-read what he
> wrote.
When you realize that you are flatly wrong about something,
the best thing to do, is to say "Oops!" and admit it *immediately*.
> Marcel's contention is that if he uses symbolic computations (NOT
> approximations) he doesn't get |A/B-1|=|B/A-1|
>
> And the reason is very simple, since A=E and B=E_0
"And the reason is very simple"- good grief. A and B are
dimensionless gravitational time dilation factors and are in no way
directly related to energy.
Even when you fail to realize that you are flatly wrong about something,
the best thing to do, is to at least think "Oops!".
Are you Androcles in disguise?
Dirk Vdm.
Dono
Dick,
How stupid are you? Really? Sometimes you remind me of Marcel, is this
a country kind of thing?
Marcel is using the fact that f/f0=E/E0, his whole argument is built
around the fact that A=E and B=E0. You are a smart guy who acts
occasionally like the dumbest fuck.
Dono
"The fact that (A-B)/B is different from (A-B)/A
implies that the ENERGY gain of photons moving
downward is different from the ENERGY loss of
photons moving upward, which is physically wrong. "
Once again: for Marcel A=E and B=E0.
Learn how to read, Dick.
Dirk Van de moortel
You act exactly like Androcles, and that is no compliment.
Your Dono-alias is severely compromized.
Dirk Vdm
Lady Chacha
> Dono,
>
> You are such a patented imbecile.
> I.E., why
> would an imbecile like you expect that :
> I.e., why an imbecile like you would expect that:
> I.e. , why would an persistent denier like you expect that:
1,2,3,4
http://www.albinoblacksheep.com/flash/youare
http://www.albinoblacksheep.com/flash/youare
http://www.albinoblacksheep.com/flash/youare
http://www.albinoblacksheep.com/flash/youare
--
Dono is concubine Lady Chacha
Lady Chacha
> You missed biggest error :-)
http://www.secularsobriety.org/mdavey_drunk2.jpg
Lady Chacha
> That is minor compared to this:
http://www.helinium.nl/trolltech.gif
>
> :-)
Lady Chacha
Lady Chacha
> Learn how to read, Dono.
1 result for "How Dono became a usenet troll - part 1 of 400"
http://groups.google.com/group/sci.physics.relativity/search?
group=sci.physics.relativity&q=How+Dono+became+a+usenet+troll+-+part+1+of
+400&qt_g=Search+this+group
Lady Chacha
> How stupid are you Dono? Really?
http://www.albinoblacksheep.com/flash/youare
Lady Chacha
> When you realize that you are flatly wrong about something, the best
> thing to do, is to say "Oops!" and admit it *immediately*.
Good advice!
Will be Dono stupidicies of last weeks listed in your immortal fumbles
page. Or is that too honest for you Dirk :-)
> Even when you fail to realize that you are flatly wrong about something,
> the best thing to do, is to at least think "Oops!". Are you Androcles in
> disguise?
Dono does not think. End.
Dono
Learn how to read:
http://groups.google.com/group/sci.physics.relativity/msg/43d4a0faf3395929?dmode=print
Dono
Eric Gisse
[snip idiocy]
Do you ever get bored misunderstanding an approximation? There are
more interesting things to misunderstand.
Dono
This time he's not using an approximation, his contention is that |f/
f0-1|=|E/E0-1| is not equal to |f0/f-1|=|E0/E-1|
In other words, he is trying to force E=E0 and he wonders what went
wrong. Dirk missed this one too.
Dirk Van de moortel
0ecd2a2f-4210-49e0...@i29g2000prf.googlegroups.com
At least he stopped misunderstanding special relativity
trivialities ;-)
Dirk Vdm
Dono
How did Juan R.Alvarez-Gonzalez start using Lady Caca as a
sockpuppet?
Lady Chacha
Lady Chacha
> How did you start using Dono as a sockpuppet?
Lady Chacha
Lady Chacha
> [snip idiocy]
2,130 results for Eric Gisse idiot
http://groups.google.com/group/sci.physics.relativity/search?
group=sci.physics.relativity&q=Eric+Gisse+idiot&qt_g=Search+this+group
Lady Chacha
Lady Chacha
> You act exactly like Androcles, and that is no compliment. Your
> Dono-alias is severely compromized.
dirk you know is not so "exactly" as you say. Androcles is listed in your
inmortal fumbles page, Dono is not and will not.
And everyone knows why dirk
http://en.wikipedia.org/wiki/Academic_dishonesty
xxein
> Excerpts from:
>
> GTR Tests - The Pound-Rebka-Snider Experiment
> Dr. Adrian Sfarti
>
> "1. Abstract
> Einstein predicted a change in the energy of photons
> in the proximity of a gravitational field, the change
> being directly proportional with the distance from
> the gravitational source. In the early 60's Pound and
> Rebka have set to verify Einstein's prediction."
>
> "The experiment was set up in the Harvard tower.
> The Harvard tower is just 22.6 meters, so the fractional
> gravitational red shift between the light frequency f
> at bottom of the tower and the frequency f0 at the top
> predicted by GRT given by the formula
>
> (f-f0)/f0 = gd/c^2 = 2.45*10^-15 (1)
> where g =~ 9.81 m/s^2, d is the tower
> height and c is the speed of light in void.
>
> In reality Pound and Rebka measured the energy
> difference DeltaE = h(f-f0) (2)
> h being the Plank constant.
>
> From (1) and (2) we obtain
>
> DeltaE/E = (f-f0)/f0 = gd/c^2
>
> Comparing the energy shifts on the upward and downward
> paths gives a predicted difference
>
> (DeltaE/E)down - (DeltaE/E)up = 2 gd/c^2 = 4.9*10^-15
>
> The cleverness of the experiment lies in the fact
> that it sidesteps the measurement of the frequencies
> at the top and of the bottom of the tower and it
> replaces such measurement with a very high precision
> energy measurement and here is where the Mossbauer effect
> comes into play. The measured difference was 5.1x10^-15."
>
> GR THEORY IS FALSE!
> ___________________
>
> According to the approximate formula (f-f0)/f0 = gd/c^2
> above, photons gain energy during their downward trip.
> The corresponding blue shift is (DeltaE/E)down = gd/c^2.
>
> Photons moving upward *must* lose an *identical*
> quantity of energy, hence the corresponding red shift =
> (DeltaE/E)up = -gd/c^2.
>
> If some formula gave a blue shift whose absolute value
> were different from the absolute value of the red shift,
> it would be physically false.
>
> Let A = sqrt(1-2GM/(R*c^2))
> Let B = sqrt(1-2GM/((R+d)*c^2))),
>
> where G is the gravitational constant, M the mass
> of the Earth, R the radius of the Earth and d the
> height of the tower.
>
> According to the full GR formula,
>
> f/f0 = A/B, to which corresponds a blue shift
> f/f0 - 1 = A/B - 1 = (A-B) / B
>
> As f0/f = B/A, the corresponding red shift =
> f0/f - 1 = B/A - 1 = -(A-B) / A
>
> The fact that (A-B)/B is different from (A-B)/A
> downward is different from the energy loss of
> photons moving upward, which is physically wrong.
>
> GR THEORY IS NOT EVEN FALSE!
>
> Marcel Luttgens
xxein: Not even close.
+/-2.45e-15 is the correct measurement (up and down). But it has
another explanation.
Here's how. A single photon may or may not have a different energy
from one source or another. That becomes trivial here. Even Bilge
had to agree that all photons were of the same energy.
The issue is propagation frequency and whose clock position is
measuring it. P-R had two different clockrates to measure the
frequency. One at the top and one at the bottom. That's all they
measured. Period.
Their general conclusion was wrong. Light did not gain or lose
energy, their clocks were of different timerates. They admitted such
(in full agreement to gravity) and still tried to insist that light
gained/lost energy bwo frequency measured.
There are two additional things to consider here. One is that just
moving in a gravitational 'field' doesn't/cannot change the frequency
of the emittance. The timerate of the cause of the emittance,
however, is affected along with the position of the measuree.
The second is that as clocks in a deeper gravity slow down, they will
measure the frequency to be higher and even more energetic. Hence
photons measured 'here' (in higher gravity) have more energy than
photons measured at the top of the tower.
But I can still realise that photons (like anything else) when falling
to gravity, will gain an energy. But the energy is only relative to
those who measure it from a 'stopped' position. Iow, you could simply
accelerate toward a light source, have a time dilation AND more
measured impact energy added to it.
P-R only measured the upstream (not that the downstream would be
significantly different). But they only measured a frequency
difference! It was only frequency that they could use to measure an
energy. They couldn't (and didn't) measure any energy in a direct
fashion. And the clockrates tell the story.
Now light DOES have a finite velocity. It remains c. But in gravity,
it goes with everything everything else. It is affected the same, as
if space were a fluid.
Well? I don't think 'space' is a fluid. I think that the energy that
occupies it IS. Maybe we can call that an energy-space. We can't
define a space without something in it, can we?
So? Where is P-R in this? They want to say that light loses energy
as it climbs against gravity. As far as they were able to measure,
the clockrates account for all the difference. So why is it insisted
that light loses energy?
This is where relativity steps in and gives its description. It
doesn't bother me that it tells me what I measure. It bothers me
because it doesn't have a physic!
I don't know if anyone cares (seems not), but I do.
xxein
>
> - Show quoted text -
xxein: I can't pull up your reply, Dirk.
mlut...@wanadoo.fr
Dear Dono,
You are of course wholly right. I am generally trying not to troll,
but sometimes, the matter is so funny that I can't resist.
Anyhow, if somebody weighting 50 kg (probably a "she") travels
from A to B, and adds to her (his) kit 20 kg during her (his) trip,
she (he)
will weigtht 70 kg at B, so the weight shift , related to the
*original* weight,
will be (70-50)/50 = 0.4.
Otoh, if the same person weighting 70 kg with a 20 kg load goes from
B to A but dops 20 kg during his (her) trip, the weight shift ,
related to
the *original* weight, will be (50-70)/70 = -2/7, which is of course
different from 0.4. In that sense, GR is wholly right.
Btw, the Pound & Rebka experiment doesn't vindicate GR,
as one obtains the same (approximate) result when hypothetizing that
the photon mass = h*freqency/c^2.
Marcel Luttgens
Dono
> On Mar 17, 10:06 pm, Dono <sa...@comcast.net> wrote:
>
> > On Mar 17, 1:43 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > > On Mar 17, 4:13 am, mluttg...@wanadoo.fr wrote:
> > > [snip idiocy]
>
> > > Do you ever get bored misunderstanding an approximation? There are
> > > more interesting things to misunderstand.
>
> > This time he's not using an approximation, his contention is that |f/
> > f0-1|=|E/E0-1| is not equal to |f0/f-1|=|E0/E-1|
>
> > In other words, he is trying to force E=E0 and he wonders what went
> > wrong. Dirk missed this one too.
>
> Dear Dono,
>
> You are of course wholly right. I am generally trying not to troll,
> but sometimes, the matter is so funny that I can't resist.
Good, so stop here. It is good that , for oonce, you understood your
error.
mlut...@wanadoo.fr
Be honest, does the Pound & Rebka experiment
vindicate GR, as one obtains the same (approximate)
result when hypothetizing that the photon mass = h*freqency/c^2?
Marcel
Dono
>
> Be honest, does the Pound & Rebka experiment
> vindicate GR, as one obtains the same (approximate)
> result when hypothetizing that the photon mass = h*freqency/c^2?
>
> Marcel
Back to this fixation , Marcel? I thought that I cured you of it long
ago. Apparently not. Try taking your pills, you sound now like
Y.Parrot.
Dono
> Be honest, does the Pound & Rebka experiment
> vindicate GR, as one obtains the same (approximate)
> result when hypothetizing that the photon mass = h*freqency/c^2?
>
> Marcel
This very good paper shows how only "naive" (a nice term for ignorant)
people attempt to explain the results of the Pound-Rebka experiment by
attributting "mass" hf/c^2 to the photon. Read the paper, you might
finally stop posting garbage:
mlut...@wanadoo.fr
The author wrote:
"A naive (but obviously wrong!) way to derive
the formula for the redshift is to ascribe to the
photon with energy E a mass m = E/c2 and to apply to
the photon a non-relativistic formula
DeltaE = -mDelta(phi) treating it like a stone.
Then the relative shift of photon energy is
DeltaE/E = -Delta(phi)/c2, which coincides
with the correct result. But this coincidence
cannot justify the absolutely thoughtless application
of a nonrelativistic formula to an ultrarelativistic
object."
Note that he recognized that both results are identical.
And indeed, with a tower of 10 km (big enough to
minimize the computation errors), one gets following
shifts from top to bottom
- with my formula: 1.0912*10^-12
- with the GR formula: 1.0912*10^-12
Such coincidence is easily explained:
The GR formula is
f_top/f_bottom =
sqrt((1-2G*M/(R+H)c^2)/(1-2G*M/R*c^2))
Let A = sqrt(1-2G*M/R*c^2)
=~ 1-G*M/R*c^2
Let B = sqrt(1-2G*M/(R+H)c^2)
=~ 1-G*M/(R+H)c^2
B/A =~ (1-G*M/(R+H)c^2) * (1+G*M/R*c^2)
=~ 1 + G*M/R*c^2 - G*M/(R+H)c^2
=~ 1 + GM/c^2 (1/R - 1/(R+H))
=~ 1 + GM/c^2 * H/(R(R+H))
So, the approximate GR shift formula
GM/c^2 * H/(R(R+H)) is identical to the formula
obtained when considering that the photon recovers
an energy amount corresponding to its loss
of potential energy between the top and the bottom
of the tower.
But when one takes into consideration the
fact that the pseudo mass of the photon
continuously increases during the downward trip,
the shift formula becomes
exp(GM/c^2 * H/(R*(R+H)))-1.
This exact formula also gives a shift of
1.0912*10^-12.
Of course, the GR formula and "mine" will
give different results for objects with the
same radius but much greater masses than
the Earth's mass.
Which formula is the right one is impossible
to tell in the absence of experimental results
with very massive objects.
GRists will use GR theory to explain that their
formula is the right one. Moreover, they will
claim that their formula f_top/f_bottom =
sqrt((1-2G*M/(R+H)c^2)/(1-2G*M/R*c^2))
is only valid in the weak field limit, but
they seem unable to show a general formula
valid for all fields.
They should think twice before labelling
sceptics as ignorants.
Marcel Luttgens
andrewg...@gmail.com
>
> So, the approximate GR shift formula
> GM/c^2 * H/(R(R+H)) is identical to the formula
> obtained when considering that the photon recovers
> an energy amount corresponding to its loss
> of potential energy between the top and the bottom
> of the tower.
> Marcel Luttgens
Photons are not small billiard balls , Marcel. Give it a rest (at
least you stopped wondering why A/B-1 is not equal to B/A-1)
mlut...@wanadoo.fr
I am wondering why you referred to that clearly
brainwashed GRist:
"Photons and static gravity
L.B.Okun
ITEP, Moscow, 117218, Russia",
instead of to Steven Weinberg, one of the greatest
scientist of our times, who wrote in "Gravitation
and Cosmology", 1972, J. Wiley & Sons, p. 85:
"This result can be interpreted as saying that
a photon in a gravitational field has "kinetic energy"
hNu and "potential energy" hNuPhi, their sum remaining
constant."
Iow, according to you, Steven Weinberg wrote 'garbage'!
In fact, one can consider, as did Steven Weinberg, that
the gravitational shift can be understood as a consequence
of quantum theory and energy conservation.
But then, both theories should lead to a same shift,
which is not the case.
Let's take the simplest case where the light emitter
is situated at an infinite distance from the receiver
situated on an object of mass M and radius R.
Then the GR shift is given by the relation
S_GR = 1 / SQR(1 - 2 * G * M / (R * c ^ 2)) - 1,
whereas the quantum shift is given by
S_quantum = e ^ (G * M / (R * c ^ 2)) - 1.
Hereafter are a few results, where R is always
a radius identical to the Earth's radius:
1) M = Earth's mass
S_GR = 6.962909 * 10^-10
S_quantum = 6.962909 * 10^-10
2) M = solar mass
S_GR = 2.319212 * 10^-4
S_quantum = 2.318674 * 10^-4
3) M = 10 solar masses
S_GR = 2.326499 * 10^-3
S_quantum = 2.321095 * 10^-3
4) M = 100 solar masses
S_GR = 2.402278 * 10^-2
S_quantum = 2.345489 * 10^-2
Which theory is the right one in this case?
In the absence of experimental results, it is
impossible to decide, but one thing is sure,
quantum supporters are no more ignorant than
GR advocates. Let's remember that, till now,
quantum theory has never been experimentally
demonstrated wrong.
Dono, do you really consider that Steven Weinberg
is a stupid crackpot?
Marcel Luttgens
Dono
No, imbecile
SR produces the same results as GR in the weak field approximation
LIMIT. Do you know what that means?
<rest of your computational masturbation snipped>
mlut...@wanadoo.fr
SR??? You seem rather lost!
And what is the GR formula applicable to strong fields?
Bye, crackpot!
Dono
<idiocies snipped>
Imbecile,
The values you cited are from QED.
mlut...@wanadoo.fr
What is the GR formula applicable to strong fields,
for instance, when M = 100 solar masses?
Marcel Luttgens
mlut...@wanadoo.fr
Recognize that, contrary to quantum theory, GR has
no usable formula allowing to calculate the shift
of light in strong gravitational fields.
Marcel Luttgens
Dono
Marcel,
You are still not taking your pills. What is "the shift
of light in strong gravitational fields" ?
Eric Gisse
Yea, it does. You simply don't know about it because you don't
understand how to derive the formulas.
mlut...@wanadoo.fr
But you seemingly do!
Then you numerically could solve the following little problem:
A light emitter is situated at an infinite distance from
a receiver situated on an object of mass M and radius R.
If R is the Earth's radius and M = 100 solar masses,
what is the frequency shift according to an observer
situated on Earth?
If you don't like an "infinite distance", you could for
instance assume that the light source is situated at
a distance R from the observer.
Without a numerical solution, everybody could conclude
that you are simply trolling.
Marcel Luttgens
Dirk Van de moortel
ff1dca6c-d330-4980...@z38g2000hsc.googlegroups.com
> On Apr 6, 12:40 am, Eric Gisse <jowr...@gmail.com> wrote:
>> On Apr 5, 5:21 am, mluttg...@wanadoo.fr wrote:
>>
>>> On Apr 4, 12:58 am, Dono <sa...@comcast.net> wrote:
>>
>>>> On Apr 3, 2:04 pm, mluttg...@wanadoo.fr wrote:
>>>> <idiocies snipped>
>>
>>>> Imbecile,
>>
>>>> The values you cited are from QED.
>>
>>> Recognize that, contrary to quantum theory, GR has
>>> no usable formula allowing to calculate the shift
>>> of light in strong gravitational fields.
>>
>>> Marcel Luttgens
>>
>> Yea, it does. You simply don't know about it because you don't
>> understand how to derive the formulas.
>
>
> But you seemingly do!
>
> Then you numerically could solve the following little problem:
>
>
> A light emitter is situated at an infinite distance from
> a receiver situated on an object of mass M and radius R.
>
> If R is the Earth's radius and M = 100 solar masses,
> what is the frequency shift according to an observer
> situated on Earth?
frequency blueshift factor = sqrt(1 - 2 G M / (R c^2))
See sample problem 3 of
http://www.eftaylor.com/pub/chapter2.pdf
>
> If you don't like an "infinite distance", you could for
> instance assume that the light source is situated at
> a distance R from the observer.
>
> Without a numerical solution, everybody could conclude
> that you are simply trolling.
Numerical examples are for children and crackpots.
Dirk Vdm
Dono
>
> If R is the Earth's radius and M = 100 solar masses,
> what is the frequency shift according to an observer
> situated on Earth?
>
Old fart,
M=100xslar masses is not weak field by any stretch of imagination,
this is why you will get small discrepancies betwen SR (QED) and GR.
You have already been told twice
mlut...@wanadoo.fr
SperM.hotmail.com> wrote:
> mluttg...@wanadoo.fr <mluttg...@wanadoo.fr> wrote in message
>
> ff1dca6c-d330-4980-9f7e-944332530...@z38g2000hsc.googlegroups.com
>
>
>
>
>
> > On Apr 6, 12:40 am, Eric Gisse <jowr...@gmail.com> wrote:
> >> On Apr 5, 5:21 am, mluttg...@wanadoo.fr wrote:
>
> >>> On Apr 4, 12:58 am, Dono <sa...@comcast.net> wrote:
>
> >>>> On Apr 3, 2:04 pm, mluttg...@wanadoo.fr wrote:
> >>>> <idiocies snipped>
>
> >>>> Imbecile,
>
> >>>> The values you cited are from QED.
>
> >>> Recognize that, contrary to quantum theory, GR has
> >>> no usable formula allowing to calculate the shift
> >>> of light in strong gravitational fields.
>
> >>> Marcel Luttgens
>
> >> Yea, it does. You simply don't know about it because you don't
> >> understand how to derive the formulas.
>
> > But you seemingly do!
>
> > Then you numerically could solve the following little problem:
>
> > A light emitter is situated at an infinite distance from
> > a receiver situated on an object of mass M and radius R.
>
> > If R is the Earth's radius and M = 100 solar masses,
> > what is the frequency shift according to an observer
> > situated on Earth?
>
> frequency blueshift factor = sqrt(1 - 2 G M / (R c^2))
> See sample problem 3 of
> http://www.eftaylor.com/pub/chapter2.pdf
>
Such formula is not valid for strong fields!
>
> > If you don't like an "infinite distance", you could for
> > instance assume that the light source is situated at
> > a distance R from the observer.
>
> > Without a numerical solution, everybody could conclude
> > that you are simply trolling.
>
> Numerical examples are for children and crackpots.
No, they allow to determine if formulas are really
applicable!
Marcel Luttgens
>
> Dirk Vdm
mlut...@wanadoo.fr
Of course, it is *not* weak field, that's the reason
why I chose that problem! What I want from GR theory
is a formula applicable to strong fields, with a numerical
solution, otherwise it would be mere blah!
Marcel Luttgens
Dirk Van de moortel
Of course it is, imbecile.
The first order Taylor approximation is not valid for
strong fields. How stupid can you actually be?
Hundreds of times you were given a pointer to this
chapter. Hundreds of times you could have actually
looked at it. But of course, you also should have
studied chapter 1, and we all know that you can't
even grasp that one.
>>
>>> If you don't like an "infinite distance", you could for
>>> instance assume that the light source is situated at
>>> a distance R from the observer.
>>
>>> Without a numerical solution, everybody could conclude
>>> that you are simply trolling.
>>
>> Numerical examples are for children and crackpots.
>
> No, they allow to determine if formulas are really
> applicable!
You don't understand the meanings of the variables
in the formulas, imbecile.
Dirk Vdm
Dirk Van de moortel
frequency blueshift factor = sqrt(1 - 2 G M / (R c^2))
See sample problem 3 of
http://www.eftaylor.com/pub/chapter2.pdf
Numerical examples are for children and crackpots.
Dirk Vdm
Dono
Now you have been told the third time, old fart.
mlut...@wanadoo.fr
You are the *imbecile* and a *poor* mathematician.
If you had numerically applied your formula
frequency blueshift factor = sqrt(1 - 2 G M / (R c^2)) ,
you would immediately have realized that it gives false
results.
But this was not even necessary, you simply had to look
at the formula. As 2 G M / (R c^2) is generally very small,
you systematically get a blueshift of about 1 !!!
For instance, with M = 100 solar masses and R = the Earth's
radius, you get a blueshift of 0.98, whereas the correct result is
about 2.4 * 10^-2.
Marcel Luttgens
>
> Dirk Vdm
Dirk Van de moortel
Read the chapter.
> But this was not even necessary, you simply had to look
> at the formula. As 2 G M / (R c^2) is generally very small,
> you systematically get a blueshift of about 1 !!!
> For instance, with M = 100 solar masses and R = the Earth's
> radius, you get a blueshift of 0.98, whereas the correct result is
> about 2.4 * 10^-2.
1-0.98 = 2*10^(-2)
I told you, numbers are for C&C.
Dirk Vdm
mlut...@wanadoo.fr
You are even more incompetent as I thought!
According to you, the blue shift is given by 1 - 0.98
(from 1 - sqrt(1 - 2 G M / (R c^2)).
But 1 - sqrt(1 - 2 G M / (R c^2)) can never be a shift!
Remember that you tried to calculate the blueshift
of a light emitted at an infinite distance and observed
at the surface of an object of mass M and radius R.
So, according to you,
f_surface / f_infinite = sqrt(1 - 2 G M / (R c^2)),
where sqrt(1 - 2 G M / (R c^2)) is what you called
the frequency *blueshift* factor.
Then the frequency shift
= f_surface / f_infinite - 1
= sqrt(1 - 2 G M / (R c^2)) - 1,
which is negative and thus represents a *red* shift, not
a *blue* one.
You have no idea about the way to calculate a shift.
Marcel Luttgens
Dirk Van de moortel
Dirk Van de moortel
ftflba$fj4$1...@news-pa1.hpl.hp.com
... on (twenty) second thought, don't.
You won't understand.
Dirk Vdm
mlut...@wanadoo.fr
Not only don't you understand -thus be able to correctly
calculate- a frequency shift, but you don't realize that
GR formulas using a "shift factor" like
sqrt((1-2G*M/(R*c^2)/(1-2G*M/(R+H)c^2))
are only valid in the weak field limit!
Eric Gisse who, replying to my comment
"Contrary to quantum theory, GR has
no usable formula allowing to calculate the shift
of light in strong gravitational fields.", claimed
on Apr. 6, "Yea, it does.", was well aware that
the GR shift factor is different for strong fields.
Instead of abusing people and amply demonstrating
your incompetence, you should try to learn a little
more about GR theory.
Marcel Luttgens
Dirk Van de moortel
Marcel, I am not interested in the way you misunderstood
Eric, or perhaps Eric misunderstood you. In the course of
the years you have shown to fail to understand just about
anything anyone ever explained to you, and most people
also failed to understand just about anything you ever tried
to explain to them.
You are like Ken Seto - locked off from the rest of the
world. You /cannot/ /communicate/. Period.
>
> Instead of abusing people and amply demonstrating
> your incompetence, you should try to learn a little
> more about GR theory.
Find another hobby, Marcel. Your self-inflicted stupidity
is embarrassing. I repeat, do *not* read that second
chapter. You will simply fail to understand it, just like
you have failed to understand the first chapter.
Dirk Vdm
carlip...@physics.ucdavis.edu
[...]
> Not only don't you understand -thus be able to correctly
> calculate- a frequency shift, but you don't realize that
> GR formulas using a "shift factor" like
> sqrt((1-2G*M/(R*c^2)/(1-2G*M/(R+H)c^2))
> are only valid in the weak field limit!
That's incorrect. This is the full, strong field expression for the
red shift outside a spherically symmetric, static body. For a
derivation, see, for example,
Wald, _General Relativity_, section 6.3
Carroll, _Spacetime and Geometry_, section 5.5
Hughston and Tod, _An Introduction to General Relativity_, chapter 16
Weinberg, _Gravitation and Cosmology_, section 3.5
Foster and Nightingale, _A Short Course in General Relativity_, section 4.3
Synge, _Relativity: The General Theory_, section 7.9
You're free to argue that GR is wrong about this, but the GR prediction,
for fields of arbitrary strength, is unambiguous.
Steve Carlip
Dirk Van de moortel
ftg540$hcq$1...@skeeter.ucdavis.edu
Nothing helps with Marcel :-)
Not even chapter 2 of
http://www.eftaylor.com/pub/chapter2.pdf
Dirk Vdm
Eric Gisse
Actually that's the exact shift factor that is valid for all r > 2GM.
>
> Eric Gisse who, replying to my comment
> "Contrary to quantum theory, GR has
> no usable formula allowing to calculate the shift
> of light in strong gravitational fields.", claimed
> on Apr. 6, "Yea, it does.", was well aware that
> the GR shift factor is different for strong fields.
No shit it looks different - approximations and the exact formula tend
to be a little dissimilar.
>
> Instead of abusing people and amply demonstrating
> your incompetence, you should try to learn a little
> more about GR theory.
I've learned enough of GR to be satisfied with my internal knowledge
of the subject. I'm going to teach myself some quantum field theory
next.
>
> Marcel Luttgens
mlut...@wanadoo.fr
I wanted to be 100% sure. At least Vdm had one thing right.
Thank you very much for those precisions and references.
The fact that the frequency shift calculated by
assigning a pseudo-mass hf/c^2 and a pseudo-potential
energy to photons is the same as the one obtained from
the approximate GR factor
(1-G*M/(R*c^2)/(1-G*M/(R+H)c^2).
poses a theoretical problem:
How is it possible to physically explain the discrepancy
with the full GR expression? Iow, why do the results
obtained with quantum theory differ from the GR ones.
Marcel Luttgens
Dirk Van de moortel
It *IS* the full GR expression.
Steve told you yesterday:
|<quote>
| This is the full, strong field expression for the
| red shift outside a spherically symmetric, static body. For a
| derivation, see, for example,
|
| Wald, _General Relativity_, section 6.3
| Carroll, _Spacetime and Geometry_, section 5.5
| Hughston and Tod, _An Introduction to General Relativity_, chapter 16
| Weinberg, _Gravitation and Cosmology_, section 3.5
| Foster and Nightingale, _A Short Course in General Relativity_, section 4.3
| Synge, _Relativity: The General Theory_, section 7.9
|</quote>
I told you yesterday:
|<quote>
| You are like Ken Seto - locked off from the rest of the
| world. You cannot communicate. Period.
|</quote>
Dirk Vdm
Dirk Van de moortel
> mlut...@wanadoo.fr <mlut...@wanadoo.fr> wrote in message
> 1bd99152-11de-4349...@24g2000hsh.googlegroups.com
>> On Apr 8, 6:05 pm, carlip-nos...@physics.ucdavis.edu wrote:
>>> mluttg...@wanadoo.fr wrote:
>>>
>>> [...]
>>>
>>>> Not only don't you understand -thus be able to correctly
>>>> calculate- a frequency shift, but you don't realize that
>>>> GR formulas using a "shift factor" like
>>>> sqrt((1-2G*M/(R*c^2)/(1-2G*M/(R+H)c^2))
>>>> are only valid in the weak field limit!
>>>
>>> That's incorrect. This is the full, strong field expression for the
>>> red shift outside a spherically symmetric, static body. For a
>>> derivation, see, for example,
>>>
>>> Wald, _General Relativity_, section 6.3
>>> Carroll, _Spacetime and Geometry_, section 5.5
>>> Hughston and Tod, _An Introduction to General Relativity_, chapter 16
>>> Weinberg, _Gravitation and Cosmology_, section 3.5
>>> Foster and Nightingale, _A Short Course in General Relativity_, section 4.3
>>> Synge, _Relativity: The General Theory_, section 7.9
>>>
>>> You're free to argue that GR is wrong about this, but the GR prediction,
>>> for fields of arbitrary strength, is unambiguous.
>>>
>>> Steve Carlip
>>
>> I wanted to be 100% sure.
You are too stupid to be sure of anything.
>> At least Vdm had one thing right.
>>
>> Thank you very much for those precisions and references.
>> The fact that the frequency shift calculated by
>> assigning a pseudo-mass hf/c^2 and a pseudo-potential
>> energy to photons is the same as the one obtained from
>> the approximate GR factor
>> (1-G*M/(R*c^2)/(1-G*M/(R+H)c^2).
>> poses a theoretical problem:
>> How is it possible to physically explain the discrepancy
>> with the full GR expression?
>
> It *IS* the full GR expression.
I overlooked the fact that Marcel left out the square roots,
so this is not the full GR expression.
Marcel now just throws away the square roots and calls
it an approximation.
And then Marcel asks how it is possible to explain the
discrepancy between an expression and his approximation
of that expression.
Good grief.
Dirk Vdm
mlut...@wanadoo.fr
SperM.hotmail.com> wrote:
> Dirk Van de moortel <dirkvandemoor...@ThankS-NO-SperM.hotmail.com> wrote in message
> WC%Kj.80742$833.9...@newsfe17.ams2> > 1bd99152-11de-4349-8d9b-eac7a3e41...@24g2000hsh.googlegroups.com
Dirk, you are especially dense.
As the quantum theory leads a formula, which is
mathematically identical to the approximate
( = valid for weak fields) GR formula, I am simply
wondering about the physical explanation
of the discrepancy with the results obtained
with the full GR expression.
Marcel Luttgens
Dirk Van de moortel
mlut...@wanadoo.fr
Eric, this remark was not directed to you.
mlut...@wanadoo.fr
If you consider that GR is the only right answer!
Marcel Luttgens
Paul B. Andersen
Hm.
Read Marcel's statement like this:
"The fact that the frequency shift calculated by
assigning a pseudo-mass hf/c^2 and a pseudo-potential
energy to photons is [..](1-G*M/(R*c^2)/(1-G*M/(R+H)c^2)
poses a theoretical problem:
How is it possible to physically explain the discrepancy
with the full GR expression?"
Marcel seems to think that the failure of his pseudo Newtonian
approach to give an exact result is a "theoretical problem" which
should be "physically explained". :-)
--
Paul
Eric Gisse
Since when does quantum theory have G?
>
> Marcel Luttgens
Androcles
--
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/
"Paul B. Andersen" <paul.b....@hiadeletethis.no> wrote in message
news:47FCCF37...@hiadeletethis.no...
That's all very well, but what time is it on the Moon right now?
Err... shithead. :-)
carlip...@physics.ucdavis.edu
[...]
> The fact that the frequency shift calculated by
> assigning a pseudo-mass hf/c^2 and a pseudo-potential
> energy to photons is the same as the one obtained from
> the approximate GR factor
> (1-G*M/(R*c^2)/(1-G*M/(R+H)c^2).
Note that this is already a weak field approximation --
you've left out factors of 2 and square roots.
> poses a theoretical problem:
> How is it possible to physically explain the discrepancy
> with the full GR expression? Iow, why do the results
> obtained with quantum theory differ from the GR ones.
I don't see any problem. Your "quantum theory" method
assumes that gravity can be described by a Newtonian
potential. In a weak field, this is a good approximation;
in a strong field, it isn't.
Steve Carlip
Dono
> I don't see any problem. Your "quantum theory" method
> assumes that gravity can be described by a Newtonian
> potential. In a weak field, this is a good approximation;
> in a strong field, it isn't.
>
> Steve Carlip
Several of us have already told the old fart this, he doesn't accept
it.
mlut...@wanadoo.fr
Of course, the "quantum" formula gives results,
which are different from the GR strong field formula.
The question is why? Could the GR formula not be
applicable to light? Indeed,
The GR frequency shift factor valid in all fields is
sqrt((1-(2GM/d1)/c^2)/(1-(2GM/d2)/c^2), where
d1 and d2 represent distances from the center of
a spherical body of mass M.
As 2GM/d1 and 2GM/d2 are the square of the
escape speeds v1^2 and v2^2 of a body of mass
m << M (for instance a rocket), the GR formula
can be written
sqrt((1-v1^2/c^2)/(1-v2^2/c^2)).
Such formula makes sense for a body of mass m,
but most probably not for light, as its speed is always c.
Marcel Luttgens
Eric Gisse
Why do you keep blabbering about GR while using Newtonian formulae and
Newtonian insights?
>
> Marcel Luttgens
mlut...@wanadoo.fr
"Such formula makes sense for a body of mass m,
but most probably not for light, as its speed is always c."
Do you really think that an escape velocity has a meaning
for light?
Marcel Luttgens
Eric Gisse
Why should I help you learn when you have demonstrated that you are
both unwilling and unable to understand what you are told?
mlut...@wanadoo.fr
I don't need your help.
You are cornered and trying to escape!
Use simple logic, and you will realize that the GR shift
formula cannot be used for light, as it uses escape
velocities, a concept which has no meaning for light.
Marcel Luttgens
Eric Gisse
Really? Show me how the derivation of the "GR shift formula" uses
escape velocities in any way.
Dirk Van de moortel
Luttgens, the GR shift formula
sqrt( (1-(2 G M/d1)/c^2) / (1-(2 G M/d2)/c^2 )
*IS* used for light.
No one cares if the expressions sqrt( 2 G M/d ) also happen to
be escape velocities. It just means that anything faster than
sqrt( 2 G M/d ) can escape the mass - and light is faster, so
it can escape. So what?
Good grief - you are stupid. It almost hurts.
Dirk Vdm
Dono
>
> I don't need your help.
> You are cornered and trying to escape!
> Use simple logic, and you will realize that the GR shift
> formula cannot be used for light, as it uses escape
> velocities, a concept which has no meaning for light.
>
> Marcel Luttgens
See the answer here:
mlut...@wanadoo.fr
can be written
sqrt((1-v1^2/c^2)/(1-v2/c^2)),
where v1 = sqrt(2GM/d1) and v2 = sqrt(2GM/d2),
which are escape velocities.
It is true that the derivation of the "GR shift formula"
doesn't use escape velocities in any way, but the fact
that those velocities appear, coincidentally or not,
in the formula cannot mathematically be neglected.
For light, the escape velocity is always c, hence the
factor reduces to sqrt((1-c^2/c^2)/(1-c^2/c^2)) = 0/0
for any distance d from the center of the massive body,
meaning that the formula doesn't mathematically
determinate a solution for light alone.
However, if a light is for instance emitted by a device
situated at the top of a tower of height H, the
formula would apply, as the device would have an
escape velocity v = sqrt(2GM/(R+H)). Let's note
that the escape velocity of the receiving device situated
at the Earth surface is sqrt(2GM/R).
Marcel Luttgens
Eric Gisse
No, and no.
First off, the formula you write is wrong. I won't bother deriving it
for you /again/, but f'/f = sqrt( [1 - 2GM/r] / [ 1 - 2GM / (r + h) ])
for a radial signal from r to r+h in the Schwarzschild manifold. Note
that Schwarzschild is not "all fields". Note that those are not
velocities in any way.
Next, the units are wrong for velocities. You don't know this since
you don't know what you are doing, but most GR formulae are written in
c = 1 units which suppresses some of the units. In reality, 2GM/r with
c restored is 2GM/rc^2 which if you pay attention - is unit free.
Finally, YOU ARE A MORON. The escape velocities you are trying to
refer to are - once again - Newtonian concepts that do not apply to
GR. This has been explained to you before but in your arrogant
stupidity you both fail to recognize and fail to understand why you
are wrong.
>
> It is true that the derivation of the "GR shift formula"
> doesn't use escape velocities in any way, but the fact
> that those velocities appear, coincidentally or not,
> in the formula cannot mathematically be neglected.
This is physics, not mathematics. Plus what you see is an artifact of
the system of units chosen.
>
> For light, the escape velocity is always c, hence the
> factor reduces to sqrt((1-c^2/c^2)/(1-c^2/c^2)) = 0/0
> for any distance d from the center of the massive body,
> meaning that the formula doesn't mathematically
> determinate a solution for light alone.
The only thing it means is that you are a fucking moron who doesn't
know what he is talking about. They are not escape velocities.
>
> However, if a light is for instance emitted by a device
> situated at the top of a tower of height H, the
> formula would apply, as the device would have an
> escape velocity v = sqrt(2GM/(R+H)). Let's note
> that the escape velocity of the receiving device situated
> at the Earth surface is sqrt(2GM/R).
Let us also note that you don't actually know how to perform the
relevant computation in GR so all you know how to use are Newtonian
formulas that don't apply.
>
> Marcel Luttgens
Dirk Van de moortel
[snip]
> The GR frequency shift factor valid in all fields
> can be written
> sqrt((1-v1^2/c^2)/(1-v2/c^2)),
> where v1 = sqrt(2GM/d1) and v2 = sqrt(2GM/d2),
> which are escape velocities.
>
> It is true that the derivation of the "GR shift formula"
> doesn't use escape velocities in any way, but the fact
> that those velocities appear, coincidentally or not,
> in the formula cannot mathematically be neglected.
Oh help.
>
> For light, the escape velocity is always c, hence the
> factor reduces to sqrt((1-c^2/c^2)/(1-c^2/c^2)) = 0/0
> for any distance d from the center of the massive body,
> meaning that the formula doesn't mathematically
> determinate a solution for light alone.
Oh help!
>
> However, if a light is for instance emitted by a device
> situated at the top of a tower of height H, the
> formula would apply, as the device would have an
> escape velocity v = sqrt(2GM/(R+H)). Let's note
> that the escape velocity of the receiving device situated
> at the Earth surface is sqrt(2GM/R).
OH HELP!
Dirk Vdm
mlut...@wanadoo.fr
Yes, and yes.
> First off, the formula you write is wrong. I won't bother deriving it
> for you /again/, but f'/f = sqrt( [1 - 2GM/r] / [ 1 - 2GM / (r + h) ])
> for a radial signal from r to r+h in the Schwarzschild manifold. Note
> that Schwarzschild is not "all fields". Note that those are not
> velocities in any way.
>
> Next, the units are wrong for velocities. You don't know this since
> you don't know what you are doing, but most GR formulae are written in
> c = 1 units which suppresses some of the units. In reality, 2GM/r with
> c restored is 2GM/rc^2 which if you pay attention - is unit free.
I know this perfectly.
Read what I wrote:
The GR frequency shift factor valid in all fields
can be written
sqrt((1-v1^2/c^2)/(1-v2/c^2)),
where v1 = sqrt(2GM/d1) and v2 = sqrt(2GM/d2),
which are escape velocities.
When d1 = R and d2 = R+H, v1^2 = 2GM/R and
v1^2 = 2GM/(R+H)^2, hence the shift factor is
sqrt((1-2GM/Rc^2)/(1- 2GM/((R+H)c^2)).
This is identical to your GR formula
f'/f = sqrt( [1 - 2GM/r] / [ 1 - 2GM / (r + h) ]), whith
c=1.
>
> Finally, YOU ARE A MORON. The escape velocities you are trying to
> refer to are - once again - Newtonian concepts that do not apply to
> GR. This has been explained to you before but in your arrogant
> stupidity you both fail to recognize and fail to understand why you
> are wrong.
Do you know that Newtonian concepts are incorporated in GR?
> > It is true that the derivation of the "GR shift formula"
> > doesn't use escape velocities in any way, but the fact
> > that those velocities appear, coincidentally or not,
> > in the formula cannot mathematically be neglected.
>
> This is physics, not mathematics. Plus what you see is an artifact of
> the system of units chosen.
When you chose the same unit system, there is no
artifact. And any "physical" formula should be "mathematically"
interpretable.
> > For light, the escape velocity is always c, hence the
> > factor reduces to sqrt((1-c^2/c^2)/(1-c^2/c^2)) = 0/0
> > for any distance d from the center of the massive body,
> > meaning that the formula doesn't mathematically
> > determinate a solution for light alone.
>
> The only thing it means is that you are a fucking moron who doesn't
> know what he is talking about. They are not escape velocities.
Mathematically, they are escape velocities.
Physically, one could claim that they are
"latent" escape velocities.
> > However, if a light is for instance emitted by a device
> > situated at the top of a tower of height H, the
> > formula would apply, as the device would have an
> > escape velocity v = sqrt(2GM/(R+H)). Let's note
> > that the escape velocity of the receiving device situated
> > at the Earth surface is sqrt(2GM/R).
>
> Let us also note that you don't actually know how to perform the
> relevant computation in GR so all you know how to use are Newtonian
> formulas that don't apply.
>
You are already a brainwashed GRist, alas...
Marcel Luttgens
Dirk Van de moortel
[snip]
> You are already a brainwashed GRist, alas...
Give it up, Marcel.
Take a walk.
Take a bag and collect dog turds.
Dirk Vdm
Eric Gisse
[...]
> > Next, the units are wrong for velocities. You don't know this since
> > you don't know what you are doing, but most GR formulae are written in
> > c = 1 units which suppresses some of the units. In reality, 2GM/r with
> > c restored is 2GM/rc^2 which if you pay attention - is unit free.
>
> I know this perfectly.
>
> Read what I wrote:
>
> The GR frequency shift factor valid in all fields
> can be written
> sqrt((1-v1^2/c^2)/(1-v2/c^2)),
> where v1 = sqrt(2GM/d1) and v2 = sqrt(2GM/d2),
> which are escape velocities.
Sure - in NEWTONIAN GRAVITY under different assumptions.
>
> When d1 = R and d2 = R+H, v1^2 = 2GM/R and
> v1^2 = 2GM/(R+H)^2, hence the shift factor is
> sqrt((1-2GM/Rc^2)/(1- 2GM/((R+H)c^2)).
> This is identical to your GR formula
> f'/f = sqrt( [1 - 2GM/r] / [ 1 - 2GM / (r + h) ]), whith
> c=1.
>
>
>
> > Finally, YOU ARE A MORON. The escape velocities you are trying to
> > refer to are - once again - Newtonian concepts that do not apply to
> > GR. This has been explained to you before but in your arrogant
> > stupidity you both fail to recognize and fail to understand why you
> > are wrong.
>
> Do you know that Newtonian concepts are incorporated in GR?
No Newtonian concepts are "incoporated" in GR.
[...]
>
> > The only thing it means is that you are a fucking moron who doesn't
> > know what he is talking about. They are not escape velocities.
>
> Mathematically, they are escape velocities.
Wrong. There is no such concept of escape velocities in math - that's
physics. Furthermore, the best you can claim is that a part of the
expression, when you isolate it from other parts, has units of
velocity. But if you keep all the units, it is unitless. If you
isolate out M, it has units of momentum.
> Physically, one could claim that they are
> "latent" escape velocities.
No.
>
> > > However, if a light is for instance emitted by a device
> > > situated at the top of a tower of height H, the
> > > formula would apply, as the device would have an
> > > escape velocity v = sqrt(2GM/(R+H)). Let's note
> > > that the escape velocity of the receiving device situated
> > > at the Earth surface is sqrt(2GM/R).
>
> > Let us also note that you don't actually know how to perform the
> > relevant computation in GR so all you know how to use are Newtonian
> > formulas that don't apply.
>
> You are already a brainwashed GRist, alas...
Thanks for proving my point.
>
> Marcel Luttgens
Dono
SperM.hotmail.com> wrote:
> > You are already a brainwashed GRist, alas...
>
> Give it up, Marcel.
> Take a walk.
> Take a bag and collect dog turds.
>
> Dirk Vdm
When you are good, you are very good :-)
Dono
>
> You are already a brainwashed GRist, alas...
>
> Marcel Luttgens
Watch it Marcel, you are starting to sound like Ken Shito. Same form
of Alzheimer :-)
mlut...@wanadoo.fr
Read a good textbook on GR theory.
>
> [...]
>
>
>
> > > The only thing it means is that you are a fucking moron who doesn't
> > > know what he is talking about. They are not escape velocities.
>
> > Mathematically, they are escape velocities.
>
> Wrong. There is no such concept of escape velocities in math - that's
> physics. Furthermore, the best you can claim is that a part of the
> expression, when you isolate it from other parts, has units of
> velocity. But if you keep all the units, it is unitless.
The square of any velocity divided by c^2
is of course unitless.
The GR frequency shift factor valid in all fields
corresponds to
sqrt((1-v1^2/c^2)/(1-v2/c^2)),
where v1 = sqrt(2GM/d1) and v2 = sqrt(2GM/d2),
which are "potential" or "latent" escape velocities.
For instance, the light emitter situated at the top
of a tower (cf. the Pound & Rebka experiment) has
a "potential" escape velocity, but it doesn't *move
at all* relatively to the towers's bottom.
Nevertheless, GR uses in its shift factor the gamma
factors of special relativity!
This makes no sense at all, as sqrt(1-v^2/c^2) = 1
when v = 0.
Thus, the GR frequency shift factor reduces to 1!
Conclusively, the acceleration of gravity doesn't
affect the frequency of light.
The blue shift observed by Pound & Rebka is not
"explained" by GR. It is the consequence of a tranfer
of potential energy.
Marcel Luttgens
Dirk Van de moortel
Take that bag and take that walk, pathetic imbecile.
Dirk Vdm
Dono
> The blue shift observed by Pound & Rebka is not
> "explained" by GR. It is the consequence of a tranfer
> of potential energy.
>
> Marcel Luttgens
You are deteriorating rapidly, Marcel. At least, in the past, you
managed to write some math (albeit incorrect). Now, you are down to
writing antirelativistic SLOGANS. Time to check in into that Alzheimer
clinic, Marcel. The time is now.
Eric Gisse
_Gravitation_ by Misner, Thorne, and Wheeler?
_Spacetime and Geometry_ by Carroll?
Since you are so sure the concepts are there, you should be able to
either concisely explain them yourself or produce a literature
reference that does.
>
>
>
> > [...]
>
> > > > The only thing it means is that you are a fucking moron who doesn't
> > > > know what he is talking about. They are not escape velocities.
>
> > > Mathematically, they are escape velocities.
>
> > Wrong. There is no such concept of escape velocities in math - that's
> > physics. Furthermore, the best you can claim is that a part of the
> > expression, when you isolate it from other parts, has units of
> > velocity. But if you keep all the units, it is unitless.
>
> The square of any velocity divided by c^2
> is of course unitless.
So why is it an escape velocity? This is not Newtonian gravitation -
there are no "escape velocities".
> The GR frequency shift factor valid in all fields
> corresponds to
> sqrt((1-v1^2/c^2)/(1-v2/c^2)),
> where v1 = sqrt(2GM/d1) and v2 = sqrt(2GM/d2),
> which are "potential" or "latent" escape velocities.
Why are you repeating something known to be wrong?
>
> For instance, the light emitter situated at the top
> of a tower (cf. the Pound & Rebka experiment) has
> a "potential" escape velocity, but it doesn't *move
> at all* relatively to the towers's bottom.
"Potential escape velocity" makes no sense. Try again - this time,
without making up words to describe things.
>
> Nevertheless, GR uses in its shift factor the gamma
> factors of special relativity!
Wrong in a new and interesting way! There is only one correct answer -
are you trying to work through the solution space by brute force?
> This makes no sense at all, as sqrt(1-v^2/c^2) = 1
> when v = 0.
Of course it makes no sense. You took the proportionality factor for
redshift in the Schwarzschild manifold, then added an asinine
interpretation that isn't supported by either the math or the physics.
Then you really took off with it.
> Thus, the GR frequency shift factor reduces to 1!
Actually in a way that does make sense though the starting assumption
is wrong and stupid. The only way to get GM/r to be zero is for r -->
\infty or M = 0. I think it is cute that you can start with something
really,really stupid and still have a chance to make it say something
useful and still fuck it up.
>
> Conclusively, the acceleration of gravity doesn't
> affect the frequency of light.
There is no force due to gravity in GR.
> The blue shift observed by Pound & Rebka is not
> "explained" by GR. It is the consequence of a tranfer
> of potential energy.
What potential energy? Why are you shoehorning yet more Newtonian
concepts like potential energy, escape velocity, and acceleration due
to gravity into a theory that has none of them?
[...]
mlut...@wanadoo.fr
See for instance "Gravitation and Cosmology", 1972, J. Wiley & Sons:
pp 77 sq, 4. The Newtonian limit
pp 151 sq, 1. Derivation of the Field Equations,
etc...
The GRists should try to get along with each other:
For instance, Steven Weinberg, in "Gravitation
and Cosmology", 1972, J. Wiley & Sons, p. 85,
considers that "... a photon in a gravitational field
has "kinetic energy" hNu and "potential energy" hNuPhi,
their sum remaining constant."
This implies that the photon's frequency is *not constant*
during its trip in a gravitational field.
Other claim that " in a static gravitational field the
frequency of the photon is *a constant*; it is equal to
the frequency at emission." !!!!
They explain the gravitational shift by "the increase
of the energy difference between levels in atoms or
nuclei (in general, by the increase of the rate of clocks)
with the increase of their distance from the gravitating
body. The constant energy of photon appears as redshifted
with respect to the blueshifted energy difference of atomic levels."
(cf. Conclusions of L.B.Okun in
http://arxiv.org/PS_cache/hep-ph/pdf/0010/0010120v2.pdf )
Either the frequency of the photon varies when the
photon moves in a gravitational potential, or the photon's
energy is constant during the trip. If it is not constant,
the explanation of the observed shift by a modification
of the rate of clocks is wrong, and one should conclude
that clocks' rate are not affected by gravitational potentials.
Marcel Luttgens