Re: Odd derivation of the GR gravitational shift formula

On Jun 8, 2:13 am, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On Jun 7, 3:35 pm, mluttg...@xxxxxxxxxx wrote:





On Jun 6, 3:04 am, xxein <xxe...@xxxxxxxxxxxxx> wrote:

On Jun 5, 7:37 pm, mluttg...@xxxxxxxxxx wrote:

Odd derivation of the GR gravitational shift formula

The derivation is ultimately an odd mixture
of special relativity with a Newtonian formula!

Fromhttp://en.wikipedia.org/wiki/Escape_velocity:

"The simplest way of deriving the formula
for escape velocity is to use conservation
of energy. Imagine that a spaceship of
mass m is at a distance r from the center
of mass of the planet, whose mass is M.
Its initial speed is equal to its escape
velocity, ve. At its final state, it will
be an infinite distance away from the planet,
and its speed will be negligibly small
and assumed to be 0. Kinetic energy K and
gravitational potential energy Ug are the
only types of energy that we will deal with,
so by the conservation of energy, <....>
1/2 mv^2 = GMm/r
ve= sqrt(2GM/r)."

xxein1: Correct.  But it is a velocity, not a speed - despite Wiki.
Imagine shooting off a projectile tangent to the horizon.  It would
spend more time in a greater gravitation before it could escape.  That
is why the velocity is perpendicular.

Note that the kinetic energy formula
K = 1/2 mv^2 is *not* the relativistic
formula K = mc^2(1/(1-sqrt(v^2/c^2)) - 1),
which *should* be used in the derivation
of the GR formula giving the gravitational
shift. However, the approximate formula
v = sqrt(2GM/r) is used in the GR shift
formula:
shift = sqrt(1-2GM/c^d2)/sqrt(1-2GM/c^2d1)-1,
which can be written
shift = sqrt(1-v2^2/c^2)/sqrt(1-v1^2/c^2)-1,
where v2=sqrt(2GM/d2) and v1=sqrt(2GM/d1)
are velocities at distances d2 and
d1 from the center of mass of an object
of mass M, attained by an object of mass m
falling from infinity, and which corresponds to
the escape velocities.
Conclusively, the GR shift formula, using
K = 1/2 mv^2
instead of K = mc^2(1/(1-sqrt(v^2/c^2)) - 1),
can only be approximate!

Let's also note that
sqrt(1-v2^2/c^2)/sqrt(1-v1^2/c^2)
is simply a SR ratio of clock rates. Indeed,
t2 = tf  * sqrt(1-v2^2/c^2)
t1 = tf  * sqrt(1-v1^2/c^2)
t2/t1 = sqrt(1-v2^2/c^2)/sqrt(1-v1^2/c^2),
see below:

http://en.wikipedia.org/wiki/Gravitational_shift

"The constant is chosen to make the clock
rate at infinity equal to 1.

Good.

Einstein's result was based on gravitational
time dilation; the Gravitational redshift
and blueshift frequency ratios are the inverse
of each other, suggesting that the "seen"
frequency-change corresponds to the actual
difference in underlying clockrate."

Good.

http://en.wikipedia.org/wiki/Gravitational_time_dilation

"Gravitational time dilation is manifested
in accelerated frames of reference or, by
virtue of the equivalence principle, in the
gravitational field of massive objects
t0 = tf  * sqrt(1-2GM/rc^2)
t0 is the proper time between events A
and B for a slow-ticking observer within
the gravitational field,
tf is the proper time between events A
and B for a fast-ticking observer distant
from the massive object (and therefore
outside of the gravitational field),
G is the gravitational constant,
M is the mass of the object creating the
gravitational field,
r is the radial coordinate of the observer
(which is analogous to the classical distance
from the center of the object, but is actually
(sic) a Schwarzschild coordinate),
c is the speed of light"

Marcel Luttgens

So what you might want to know is that a clock (of 0 velocity) falling
in gravity, from a very great (infinite) distance, will have a
clockrate of 1.  It will maintain that clockrate until it meets a
resistance.  In this case the clock is going where the space which
contains it goes.  In this case it is the "energy space".  It is
simply following the energy equilibrium which changes.  The clock
flows within this pocket of energy flow.  The speed of light is local
within this pocket.  We have many evidences for this, but a lot are
misconstrued to conform to silly theory.

In the opposite scenario, however, launching an object at ve, it must
go against the energy flow.  It's clockrate will be determined by a
velocity through the inflow that amounts to 2*ve wrt d (or r).  This
will of course go to zero and a clockrate of 1 when it reaches an
infinite distance.

But there is another hitch.  The universe is expanding.  We cannot say
'infinite distance' without this consideration.  But we can generally
say that galaxies are separate enough to achieve this notion of
infinite distance wrt gravity even though we still know that some
galaxies are (or have been) colliding.

It becomes a degree of consideration of any forces that might exist
(without becoming specific in the nature of the how's and why's they
exist).

Would you like to take a crack at the how's and why's using Mr. Wizard
as a base of knowledge?  Who has this knowledge?  Does anyone yet?

I think we are just floating in a sea of uncertainty with the gall to
claim a position within it.

The moron Dirk Van de moortel wrote:

Thank you for your comments.

Don't.
They suck big time.
As do yours.

Dirk Vdm

Thus implying that the relativistic kinetic energy formula
K = mc^2(1/(1-sqrt(v^2/c^2)) - 1) should not be used in
the derivation of the shift.
He is clearly not afraid of contradictions.

Marcel Luttgens

SR cannot describe gravitation, persistent idiot

Then explain why the obtained shifts are so
close for "normal" objects. If you can't, the
'idiot' is you.

When the relativistic kinetic energy formula
K = mc^2(1/(1-sqrt(v^2/c^2)) - 1) is
used instead of K = 1/2 mv^2 in the
derivation of the GR gravitational shift
formula, the shift observed at d2 of light
emitted at a distance d1, where d2 and d1
are distances from the center of mass of
an object of mass M, is given by the formula
Shift (new GR formula) =
(GM/(c^2*d2)+1) / (GM/(c^2*d1)+1) - 1, against
Shift (classical GR formula) =
sqrt(1-2GM/c^2d2) / sqrt(1-2GM/c^2d1) - 1.
For instance,

1) shift observed on Earth of light emitted
by the Sun (d1 = Sun's radius, d2 =
distance Sun-Earth)

Shift (new GR formula) = -2.11237 * 10^-6
Shift (classical GR formula) =
-2.11236 * 10^-6

2) shift observed on Earth of light emitted
by a neutron star of solar mass and
Earth radius (= d1), whose distance from
Earth d2 = 1000000 * distance Sun-Earth:

Shift (new GR formula) = -2.31787 * 10^-4
Shift (classical GR formula) =
-2.31867 * 10^-4

Clearly, the 2 formulas give almost
identical shifts for "normal" objects.
However, the calculated shifts will be very
different for hypothetical objects like black
holes. Interestingly, the "new" GR formula
doesn't lead to singularities.

Marcel Luttgens
.

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