Re: GR frequency shift formula

mluttgens@xxxxxxxxxx wrote:
mluttgens@xxxxxxxxxx wrote:
vanep@xxxxxxx wrote:
vanep@xxxxxxx wrote:
The only inadequacies that will be acknowledged, in this newsgroup, are
exhibited by cranks such as yourself.

Marcel Luttgen said:

"Here is my approximate solution of the following problem.
Do GR experts agree?

Two non-rotating spherical celestial objects 1 and 2 are orbiting along

their common center of gravity.
The masses and radii of the objects are respectively M1, R1 and M2,R2.
The distance between centers is d.
Object 1 emits a light of frequency Nu1. What is the frequency Nu2
of the light received by an observer situated on object 2 at a distance

d - (R1+R2) from the emitter?


Example: M1 = M* ( M* = 1 solar mass = 1.989E+33 g),
R1 = 5 Km,
M2 = 5 M*,
R2 = 20 km,
d = 50 km.


M1 = 1.989E+33 'g
M2 = 1.989E+33 * 5 'g
R1 = 500000! 'cm
R2 = 2000000! 'cm
d = 5000000! 'cm


G = 6.673E-08 'Universal gravitational constant (CGS)
c = 2.998E+10 'Speed of light in cm/s


M1 = G * M1 / c ^ 2
M2 = G * M2 / c ^ 2


Nu2/Nu1 = 1 - M1*(1/R-1/(d-R2)-1/(2 * d)) + M2*(1/R2-1/(d-R1)-1/(2*d))
= 0.9


In this case, one has a red shift!"
End of Marcel said.


It's been pointed out to Marcel that his formula will not work for
strong gravitational fields. The reason being his formula doesn't
account for relativistic effects.

It accounts for relativistic effects, but part of them are ignored in
my
approximate formula.

It's interesting to note the
relationship between Marcel's M2 and M1 are similar to the relationship
between the Earth and Moon. ie the center of system mass [barycenter]
is inside the radius of M2 just as the barycenter of the Earth-Moon
system is inside the radius of the Earth.

Yes, but I purposely chose the short distance of 50000 km to get clear
relativistic effects.

Also note that M2 and M1 are
fictitious objects with exceedingly small probability of existing in
this universe. But if they could....

Sure, but this is the fun of thought experiments.


In geometric units:

r_barycenter = r_total [ M1/(M2 + M1) ], r_total = d.

= 50,000m [ 1477m / (7385m + 1477m) ] = 8333.333m from the center of
M2.

Yes.


As M1 orbits the barycenter we can favorably compare the orbit of M2,
around barycenter, to the orbit of Earth around the Earth-Moon
barycenter. To find a reasonable approximation for predicting delta
period wrt a pulse of light emitted on the surface of M1 and received
on the surface of M2 the key parameters become the positions in the
gravitational field where the pulse of light is emitted and received.

Since both M1 and M2 are spherically symmetric and non-rotating we can
make the approximation using the Schwarzschild metric.

Otherwise, the GR solution would have been much more difficult, if not
impossible, to derive.


M2 velocity as it orbits the barycenter is exceedingly small compared
to the velocity of M1 as it orbits the barycenter.

It is not "extremely" small. You should take it into account.


Setting M2 = M**, M1 = M*, r2 = r**, r1 = r*, distance center of M1 to
barycenter = r***

v shell_M* = [ M** / ( r*** -2M**) ] ^1/2 = [ 7385m / (41667m -
14770m) ] ^1/2 = .524c

Are you sure of your formula? I found .35 c.
Would you use the same formula to get the orbital velocity of the Earth
around
the Sun?


For M** we can use the first component of the Schwarzschild metric:

dtshell_M** = ( 1 - 2M**/r** ) ^1/2 dt, where dt = the GR Schwarzschild
bookkeeper coordinate time measured in flat spacetime at infinity
[boundary condition].

For M* we can use a formula derived from the Schwarzschild metric and
the effective potential of GR's equation of motion:

dtshell_M* = ( 1 - 3M**/r*** ) ^1/2 dt

____________________________________________________

Derivation:

Starting with the 2 dimensional form of the Schwarzschild metric and
setting dr = 0:

dTau^2 = (1 - 2M/r)dt^2 - 0 - r^2 dphi^2

Determining a useful expression for dphi:

L/m = r^2 (dphi/dTau)

dphi = [(L/m)/r^2]dTau

Substituting dphi^2 the metric becomes:

dTau^2 = ( 1 - 2M/r)dt^2 - [(L/m)^2/r^2]dTau^2

Dividing through by dt^2 will give the ratio (dTau/dt)^2:

(dTau/dt)^2 = (1 - 2M/r) - [(L/m)/r]^2(dTau/dt)^2

Simplified:

(dTau/dt)^2 = (1 - 2M/r) / [1 + (L/m)^2/r^2]

Using the equation of motion:

(dr/dTau)^2 = (E/m)^2 - (1 - 2M/r)[1 + (L/m)^2/r^2]

and taking the derivative of the effective potential wrt r, dividing
through by r^4 to express dV/dr as a quadratic, solving for 0, then r^2
can be expressed:

Correction: ".........., multiply through by r^4 to express dV/dr as a
quadratic, ........:


r^2 = [(L/m)^2]r - 3(L/m)^2

Substituting [(L/m)^2]r - 3(L/m)^2 for r^2 into (dTau/dt)^2 = [..]
results in:

(dTau/dt)^2=(1 - 2M/r)/[1 + (L/m)^2)/[(L/m)^2)r - 3(L/m)^2]

Simplified:

dTau/dt = (1 - 3M/r)^1/2

_______________________________________________________________

Bact to Marcel's folly:

Here you are a little nasty! But thank you nevertheless.


To find the ratio [ dtshell_M** / dtshell_M* ]

dtshell_M* / ( 1 - 3M**/r*** ) ^1/2 = dt = dtshell_M** / ( 1 -
2M**/r** )^1/2


dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/r*** )^1/2


Could you give your solution in terms of M* and M** ?
Don't forget that the pulse is emitted from the surface of M1. Where is
R1 (your r*)
in your formula?

A simple way to prove that your formula is false:

Let M* = M** = 7383.5 m
r* = r** = 20000 m
d = 200 000 000 m (thus r*** = 100 000 000 m)

As d is so big, your formula reduces to
dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 = .511,
instead of a value very close to 1.
(the relativistic effects are negligible).

You set d = 50km = 50,000m.
The formula you just wrote doesn't make any sense. The reason is you
don't understand what dtshell represents.

dtshell = dTau
dt = Schwarzschild bookkeeper coordinate time measured in flat
spacetime far away [ie at infinity]. The boundary condition. The
formula is the first component of the metric.

dtshell = ( 1 - 2M/r )^1/2 dt

The ratio would be:

dtshell/dt = ( 1 - 2M/r )^1/2

For the ratio [dtshell_M** / dtshell_M*]:

dtshell_M**/( 1 - 2M**/r** )^1/2 = dt = dtshell_M*/( 1 - 2M*/r* )^1/2

So:

[dtshell_M** / dtshell_M*] = ( 1 - 2M**/r** )^1/2 / ( 1 - 2M*/r* )^1/2
= .511371 / .6396874

= .799408

My approximation acknowledged that M* is essentially a satellite of
M**. The satellite shell velocity [measured with a clock on the shell]
is .524c. The derivation resulting in dTau/dt = ( 1 - 3M/r )^1/2
accounts for that. So the final ratio becomes

dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/r*** )^1/2
= .511370707 / .639687422 = .74727606

Even so; I agree with Tom that doing this type of approximation, for
strong field physics, isn't the best way to go. But at the least it
accounts for major strong field relativistic effects.

Essentially this is what you did:

dtshell/dt = ( 1 - 2M/r )^1/2

You approximated:

dtshell/dt = [1 - 2M/2r ] = 1 - M/r

dtshell_M** / dtshell_M* = [ 1 - M**/r** ] / [ 1 - M*/r* ] = .63075 /
..7046 = .8952

So why does this weak field approximation work so well for our solar
system [all weak gravitational fields].

For the Earth the ratio dtshell/dt becomes:

dtshell/dt = [1 - 2(.00444m)/6.371E6 ]^1/2 = .999999999 / 1 =
..999999999

Which is exceptionally close to the Newton prediction of 1.

Aside: The velocity you calculated appears to have been calculated
using the boundary clock with r = 50,000m [even then I get .3843c]. For
the Schwrazschild geometry that is referred to as the bookkeeper
velocity.

The thing I find disturbing, about this conversation, is your
willingness to discount GR as a viable tool for doing physics. It
hasn't failed an empirical test to date and I don't think you've
actually studied the theory. Hopefully this analysis will clarify why
the weak field approximation, and Newtonian gravitational physics,
doesn't work in the strong field.

Best wishes
Bruce

Marcel Luttgens



= .511370707 / .639687422 = .74727606

Throughout this thread Marcel has claimed GR isn't a useful theory for
doing relativistic physics. Since he knows very little about GR it's
difficult to understand how he came to such a conclusion. It is well
understood that weak field approximations are useless for describing
strong field physics. Apparently well understood by everybody but
Marcel Luttgen.

I am skeptical about your elaborations.

Marcel Luttgens

.

Relevant Pages

  • Re: GR frequency shift formula
    ... account for relativistic effects. ... is inside the radius of M2 just as the barycenter of the Earth-Moon ... To find a reasonable approximation for predicting delta ... Marcel Luttgens ...
    (sci.physics.relativity)
  • Re: GR frequency shift formula
    ... is inside the radius of M2 just as the barycenter of the Earth-Moon ... As M1 orbits the barycenter we can favorably compare the orbit of M2, ... To find a reasonable approximation for predicting delta ... Marcel Luttgens. ...
    (sci.physics.relativity)
  • Re: GR frequency shift formula
    ... their common center of gravity. ... account for relativistic effects. ... To find a reasonable approximation for predicting delta ... Marcel Luttgens ...
    (sci.physics.relativity)
  • Re: GR frequency shift formula
    ... their common center of gravity. ... account for relativistic effects. ... To find a reasonable approximation for predicting delta ... Marcel Luttgens ...
    (sci.physics.relativity)
  • Re: GR frequency shift formula
    ... account for relativistic effects. ... is inside the radius of M2 just as the barycenter of the Earth-Moon ... As M1 orbits the barycenter we can favorably compare the orbit of M2, ... Marcel Luttgens ...
    (sci.physics.relativity)