Equivalent Klein-Gordon form for gravitational red shift

Gene Partlow · science forum beginner Joined: 26 Apr 2005 Posts: 1

This may be of interest or at least amusing to some relativists:

Exploring an assumption in a larger model led me to create
a Klein-Gordon type equation which turns out to be algebrai-
cally identical to the fully relativistic gravitational red-shift re-
lation in GR. [These are respectively (1) and (2) below.] This
K-G approach yields results completely identical to those of
the standard GR calculation for any given setting, including
the relativistic regime near the event horizon of a black hole.
However, the assumptions are <quite> different...

...GR assumes invariant 'rest' mass for any particle emitting
or absorbing a photon regardless of its elevation in a poten-
tial well of mass M. It further assumes that the photon's
wavelength increases or decreases as it travels respectively
up or down through that field.

The K-G approach uses what seems to be an equally valid
point of view: particle 'rest' masses are not invariant but vary
directly with the strength of the local g-field, ie: with increasing
curvature of space-time.

Specifically...

(m1)(C1) = (m2)(C2) ,

where m1 and m2 are the 'rest' mass of, say, an iron-57 nu-
cleus at two different distances r1 and r2 from center-of-mass
of M, and C1 and C2 are the local velocity of light, eg:

C1,2 = Co[1 - 2GM/(r1,2*Co^2)] .

G is the gravitational constant and Co the velocity of light in
field-free space. The K-G relation for r1 is

(E1)^2 = [m1*(C1)^2]^2 + [m1e*v1*C1]^2 ,

where E1 is the total energy of the particle, and

m1e = m1 / [1 - 2GM/(r1*Co^2)]^.5 , a form of the relativistic

mass.

v1 = [2GM/(r1*Co^2)]^.5

(m1e*v1 is the usual relativistic momentum, = p.)

It turns out that this requires that the total energy E of a particle
-at rest- in the g-field of M varies INVERSELY with the magni-
tude of that field.

It is fairly straightforward, though tedious, to show that

(1)
(E1-E2)/E1 (= 1 - E2/E1)

is algebraically identical to the grav'l red shift yielded by stan-
dard GR, where

(2)
(delta nu)/nu = [(1 - Rs/r1)^.5 - (1 - Rs/r2)^.5] /(1-Rs/r1)^.5

= 1 - (1 - Rs/r2)^.5 / (1 - Rs/r1)^.5

The left hand is the fractional change in frequency of photons
moving from r1 to r2. And Rs is 2GM/Co^2 , the Schwarzschild
radius of M; r1 can be restated as 2GM / (v1)^2 . Likewise with
r2.

The above identity may seem blitheringly obvious to a relativist
(and at first sight, perhaps useless) but the interpretation of (1)
is interesting and turns the standard GR point of view on its
head...

...The total energy E of the particle actually decreases stead-
ily with increasing local g-field, until it reaches zero at the event
horizon of a black hole. ALL of its actual energy has now gone
over into the gravitational field of the black hole itself and the
particle-as-particle has ceased to exist. Paradoxically, the
particle's increase in 'rest' mass with increase in local g-field
winds up having less and less importance in the overall energy
budget. What matters gravitationally is the total energy content
per unit mass of each particle...this quantity approaches zero
as the local potential approaches -C^2 at the event horizon.

(One can argue from classical physics that any particle "at-rest"
cannot have a relativistic mass since it cannot have velocity v1
in that frame. Yet the formal identity between (1) and (2) may
mean that quantum uncertainty's non-stop jitteryness + the local
potential may be at play together here, even at rest.)

Notice that another odd aspect of (1)'s point of view is that an
emitted photon traveling either up or down in a g-field is neither
red-shifted nor blue-shifted. Its energy stays constant over the
interval. What matters for example is that when emitted at a
lower elevation by an iron-57 nucleus, the photon already re-
flects a lower energy value for nuclear transition X than for the
same transition in an identical iron-57 nucleus higher up in the
field, and therefore having more intrinsic energy, due to the
weaker g-field there. In other words, in this view, it is not the
photon which is red-shifted but the emitting particle itself whose
total energy is 'red-shifted' relative to that of an identical particle
higher up in the field.

The above may simply be stating the obvious, though the notion
of a photon whose energy is not affected by traveling in or out of
a gravitational field or by changing space-time curvature is un-
derstandably hard to accept. But this odd point of view has nice
consequences in some other areas of physics.

Cheers,
Gene

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